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BELL'S MATHEMATICAL SERIES
ADVANCED SECTION ( £(■-*
General Editor-. WILLIAM P. MILNE, M.A., D.Sc^ Professor of Mathematics, Leeds University
AN ELEMENTARY TREATISE
ON DIFFERENTIAL EQUATIONS AND
THEIR APPLICATIONS
y
G. BELL AND SONS, LTD.
LONDON : PORTUGAL ST., KINGSWAY CAMBRIDGE : DEIGHTON, BELL & CO. NEW YORK : THE MACMILLAN COM- PANY BOMBAY : A. H. WHEELER & CO.
AN ELEMENTARY TREATISE ON
DIFFERENTIAL EQUATIONS
AND THEIR APPLICATIONS
BY
H. T. H. PIAGGIO, M.A., D.Sc.
PROFESSOR OF MATHEMATICS, UNIVERSITY COLLEGE, NOTTINGHAM FORMERLY SENIOR SCHOLAR OF ST. JOHN'S COLLEGE, CAMBRIDGE
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LONDON
G. BELL AND SONS, LTD.
1920
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Glasgow: printed at the university press by robert maclehose and co. ltd.
PREFACE
" The Theory of Differential Equations," said Sophus Lie, " is the most important branch of modern mathematics." The subject may be considered to occupy a central position from which different lines of development extend in many directions. If we travel along the purely analytical path, we are soon led to discuss Infinite Series, Existence Theorems and the Theory of Functions. Another leads us to the Differential Geometry of Curves and Surfaces. Between the two lies the path first discovered by Lie, leading to continuous groups of transformation and their geometrical interpretation. Diverging in another direction, we are led to the study of mechanical and electrical vibrations of all kinds and the important phenomenon of resonance. Certain partial differential equations form the start- ing point for the study of the conduction of heat, the transmission of electric waves, and many other branches of physics. Physical Chemistry, with its law of mass-action, is largely concerned with certain differential equations.
The object of this book is to give an account of the central parts of the subject in as simple a form as possible, suitable for those with no previous knowledge of it, and yet at the same time to point out the different directions in which it may be developed. The greater part of the text and the examples in the body of it will be found very easy. The only previous knowledge assumed is that of the elements of the differential and integral calculus and a little coordinate geometry. The miscellaneous examples at the end of the various chapters are slightly harder. They contain several theorems of minor importance, with hints that should be sufficient to enable the student to solve them. They also contain geometrical and physical applications, but great care has been taken to state the questions in such a way that no knowledge of physics is required. For instance, one question asks for a solution of a certain partial
vi PREFACE
differential equation in terms of certain constants and variables. This may be regarded as a piece of pure mathematics, but it is immediately followed by a note pointing out that the work refers to a well-known experiment in heat, and giving the physical meaning of the constants and variables concerned. Finally, at the end of the book are given a set of 115 examples of much greater difficulty, most of which are taken from university examination papers. [I have to thank the Universities of London, Sheffield and Wales, and the Syndics of the Cambridge University Press for their kind per- mission in allowing me to use these.] The book covers the course in differential equations required for the London B.Sc. Honours or Schedule A of the Cambridge Mathematical Tripos, Part II., and also includes some of the work required for the London M.Sc. or Schedule B of the Mathematical Tripos. An appendix gives sugges- tions for further reading. The number of examples, both worked and un worked, is very large, and the answers to the un worked ones are given at the end of the book.
A few special points may be mentioned. The graphical method in Chapter I. (based on the MS. kindly lent me by Dr. Brodetsky of a paper he read before the Mathematical Association, and on a somewhat similar paper by Prof. Takeo Wada) has not appeared before in any text-book. The chapter dealing with numerical integration deals with the subject rather more fully than usual. It is chiefly devoted to the methods of Runge and Picard, but it also gives an account of a new method due to the present writer.
The chapter on linear differential equations with constant co- efficients avoids the unsatisfactory proofs involving " infinite con- stants." It also points out that the use of the operator D in finding particular integrals requires more justification than is usually given. The method here adopted is at first to use the operator boldly and obtain a result, and then to verify this result by direct differentiation.
This chapter is followed immediately by one on Simple Partial Differential Equations (based on Riemann's " Partielle Differential - gleichungen "). The methods given are an obvious extension of those in the previous chapter, and they are of such great physical importance that it seems a pity to defer them until the later portions of the book, which is chiefly devoted to much more difficult subjects.
In the sections dealing with Lagrange's linear partial differential equations, two examples have been taken from M. J. M. Hill's recent paper to illustrate his methods of obtaining special integrals.
PREFACE vii
In dealing with solution in series, great prominence has been given to the method of Frobenius. One chapter is devoted to the use of the method in working actual examples. This is followed by a much harder chapter, justifying the assumptions made and dealing with the difficult questions of convergence involved. An effort has been made to state very clearly and definitely where the difficulty lies, and what are the general ideas of the somewhat complicated proofs. It is a common experience that many students when first faced by a long " epsilon-proof " are so bewildered by the details that they have very little idea of the general trend. I have to thank Mr. S. Pollard, B.A., of Trinity College, Cambridge, for his valuable help with this chapter. This is the most advanced portion of the book, and, unlike the rest of it, requires a little know- ledge of infinite series. However, references to standard text-books have been given for every such theorem used.
I have to thank Prof. W. P. Milne, the general editor of Bell's Mathematical Series, for his continual encouragement and criticism, and my colleagues Mr. J. Marshall, M.A., B.Sc, and Miss H. M. Browning, M.Sc, for their work in verifying the examples and drawing the diagrams.
I shall be very grateful for any corrections or suggestions from those who use the book.
H. T. H. PIAGGIO.
University College, Nottingham, February, 1920.
CONTENTS
Historical Introduction
PAOK
XV
CHAPTER I
INTRODUCTION AND DEFINITIONS. ELIMINATION. GRAPHICAL REPRESENTATION
ART.
1-3. Introduction and definitions 1
4-6. Formation of differential equations by elimination - - 2 7-8. Complete Primitives, Particular Integrals, and Singular
Solutions 4
9. Brodetsky and Wada's method of graphical representation - 5
10. Ordinary and Singular points 7
Miscellaneous Examples on Chapter I - - - - 10
CHAPTER II
EQUATIONS OF THE FIRST ORDER AND FIRST DEGREE
11. Types to be considered 12
12. Exact equations 12
13. Integrating factors 13
14. Variables separate ........ 13
15-17. Homogeneous equations of the first order and degree - - 14
18-21. Linear equations of the first order and degree - - - 16
22. Geometrical problems. Orthogonal trajectories - - - 19
Miscellaneous Examples on Chapter II - - - 22
CHAPTER III LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS
23. Type to be considered -
24. Equations of the first order
25 25
CONTENTS
ART.
25. Equations of the second order
26. Modification when the auxiliary equation has imaginary or
complex roots
27. The case of equal roots -••---..
28. Extension to higher orders -
29. The Complementary Function and the Particular Integral -
30-33. Properties of the operator D
34. Complementary Function when the auxiliary equation has repeated roots
35-38. Symbolical methods of finding the Particular Integral. Ten- tative methods and the verification of the results they give
39. The homogeneous linear equation
40. Simultaneous linear equations ------
Miscellaneous Examples on Chapter III. (with notes on mechanical and electrical interpretations, free and forced vibrations and the phenomenon of resonance)
CHAPTER IV SIMPLE PARTIAL DIFFERENTIAL EQUATIONS
41. Physical origin of equations to be considered
42-43. Elimination of arbitrary functions and constants -
44. Special difficulties of partial differential equations -
45-46. Particular solutions. Initial and boundary conditions -
47-48. Fourier's Half-Range Series
49-50. Application of Fourier's Series in forming solutions satisfying given boundary conditions ------
Miscellaneous Examples on Chapter IV. (with notes on the conduction of heat, the transmission of electric waves and the diffusion of dissolved salts)
CHAPTER V
EQUATIONS OF THE FIRST ORDER, BUT NOT OF THE FIRST DECREE
51. Types to be considered -----...
52. Equations solvable for p
53. Equations solvable for y
54. Equations solvable for x
CONTENTS XI
CHAPTER VI SINGULAR SOLUTIONS
ART. PAGE
55. The envelope gives a singular solution ----- 6"> 56-58. The c-discriroinant contains the envelope (once), the node- locus (twice), and the cusp-locus (three times) 60 59-64. The ^-discriminant contains the envelope (once), the tac-locus
(twice), and the cusp-locus (once) - - - - - 71
65. Examples of the identification of loci, using both discriminants 75
66-67. Clairaut's form : - - 76
Miscellaneous Examples on Chapter VI - - - - 78
CHAPTER VII
MISCELLANEOUS METHODS FOR EQUATIONS OF THE SECOND AND HIGHER ORDERS
68. Types to be considered 81
69-70. y or x absent 82
71-73. Homogeneous equations 83
74. An equation occurring in Dynamics 85
75. Factorisation of the operator - - - - - - 86
76-77. One integral belonging to the complementary function known 87
78-80. Variation of Parameters 88
81. Comparison of the different methods ----- 90
Miscellaneous Examples on Chapter VII. (introducing the Normal form, the Invariant of an equation, and the Schwarzian Derivative) 91
CHAPTER VIII
NUMERICAL APPROXIMATIONS TO THE SOLUTION OF DIFFERENTIAL EQUATIONS
82. Methods to be considered ------- 94
83-84. Picard's method of integrating successive approximations - 94
85. Numerical approximation direct from the differential equa- tion. Simple methods suggested by geometry 97 86-87. Runge's method 99
88. Extension to simultaneous equations 103
89. Methods of Heun and Kutta 104
90-93. Method of the present writer, with limits for the error - - 105
xii CONTENTS
CHAPTER IX
SOLUTION IN SERIES. METHOD OF FROBENIUS
ART. PAOK
94. Frobenius' form of trial solution. The indicia] equation - 109
95. Case I. Roots of indicial equation unequal and differing by a
quantity not an integer - - 110
96. Connection between the region of convergence of the series
and the singularities of the coefficients in the differential equation 112
97. Case II. Roots of indicial equation equal - - - - 112
98. Case III. Roots of indicial equation differing by an integer,
making a coefficient infinite - - - -. - - 114
99. Case IV. Roots of indicial equation differing by an integer,
making a coefficient indeterminate 116
100. Some cases where the method fails. No regular integrals - 117
Miscellaneous Examples on Chapter IX. (with notes on the hypergeometric series and its twenty-four solu- tions) 119
CHAPTER X
EXISTENCE THEOREMS OF PICARD, CAUCHY, AND FROBENIUS
101. Nature of the problem 121
102. Picard's method of successive approximation - - - 122
103-105. Cauchy's method 124
106-110. Frobenius' method. Differentiation of an infinite series with
respect to a parameter - - - - - - -127
CHAPTER XI
ORDINARY DIFFERENTIAL EQUATIONS WITH THREE VARIABLES AND THE CORRESPONDING CURVES AND SURFACES
111. The equations of this chapter express properties of curves and
surfaces 133
112. The simultaneous equations dx/P=dylQ=dz/R • - - 133
113. Use of multipliers 135
114. A second integral found by the help of the first - - - 136
115. General and special integrals 137
CONTENTS xiii
ART. FAOB
116. Geometrical interpretation of the equation
Pdx+Qdy + Rdz=0 - - - - 137
117. Method of integration of this equation when it is integrable - 138
118-119. Necessary and sufficient condition that such an equation
should be integrable 139
120. Geometrical significance of the non-integrable equation - 142
Miscellaneous Examples on Chapter XI - - - 143
CHAPTER XII •
PARTIAL DIFFERENTIAL EQUATIONS OF THE FIRST ORDER. PARTICULAR METHODS
121-122. Equations of this chapter of geometrical interest - - - 146
123. Lagrange's linear equation and its geometrical interpretation 147
124. Analytical verification of the general integral - - - 149
125. Special integrals. Examples of M. J. M. Hill's methods of
obtaining them 150
126-127. The linear equation with n independent variables - - - 16*1
128-129. Non-linear equations. Standard I. Only p and q present - 153
130. Standard II. Only p, q, and z present 153
131. Standard III. f(x, p)=F(y, q) 154
132. Standard IV. Partial differential equations analogous to
Clairaut's form 154
133-135. Singular and General integrals and their geometrical signifi- cance. Characteristics 155
136. Peculiarities of the linear equation 158
Miscellaneous Examples on Chapter XII. (with a note on
the Principle of Duality) 160
CHAPTER XIII
PARTIAL DIFFERENTIAL EQUATIONS OF THE FIRST ORDER. GENERAL METHODS
137. Methods to be discussed 162
138-139. Charpit's method 162
140-141. Three or more independent variables. Jacobi's method - 165
142. Simultaneous partial differential equations - 168
Miscellaneous Examples on Chapter XIII - - - 170
xiv CONTENTS
CHAPTER XIV
PARTIAL DIFFERENTIAL EQUATIONS OF THE SECOND AND HIGHER ORDERS
ART. PAQ
143. Types to be considered 171]
144. Equations that can be integrated by inspection. Determina-
tion of arbitrary functions by geometrical conditions - 17!
145-151. Linear partial differential equations with constant coefficients 171.
152-153. Examples in elimination, introductory to Monge's methods - 17!>
154. Monge's method of integrating Rr + Ss + Tt = V - - - 181
155. Monge's method of integrating Rr + Ss + Tt + U(rt -s-) = V - 18.'! 156-157. Formation of Intermediate Integrals 183
158. Further integration of Intermediate Integrals - - - 186
Miscellaneous Examples on Chapter XIV. (with notes on the vibrations of strings, bars, and membranes, and on potential) 188
APPENDIX A
Necessary and sufficient condition that the equation Mdx+Ndy=0 should be exact - - 191
APPENDIX B
An equation with no special integrals ----- 192
APPENDIX C
The equation found by Jacobi's method .of Art. 140 is
always integrable -----... 193
APPENDIX D
Suggestions for further reading -----.. 194
Miscellaneous Examples <>n the Whole Book (with notes on solution by definite intregals, asymptotic series, the Wronskian, Jacobi's last multiplier, finite difference equations, Hamilton's dynamical equations, Foueault's pendulum, and the perihelion of Mercury) - - . 195
Answers t<> the Examples j
,S|JKX xxiii
HISTORICAL INTRODUCTION
The study of Differential Equations began very soon after the invention of the Differential and Integral Calculus, to which it forms a natural sequel. Newton in 1676 solved a differential equation by the use of an infinite series, only eleven years after his discovery of the fluxional form of the differential calculus in 1665. But these results were not published until 1693, the same year in which a differential equation occurred for the first time in the work of Leibniz * (whose account of the differential calculus was published in 1684).
In the next few years progress was rapid. In 1694-97 John Bernoulli f explained the method of " Separating the Variables," and he showed how to reduce a homogeneous differential equation of the first order to one in which the variables were separable. He applied these methods to problems on orthogonal trajectories. He and his brother Jacob ft (after whom " Bernoulli's Equation " is named) succeeded in reducing a large number of differential equa- tions to forms they could solve. Integrating Factors were probably discovered by Euler (1734) and (independently of him) by Fontaine and Clairaut, though some attribute them to Leibniz. Singular Solutions, noticed by Leibniz (1694) and Brook Taylor (1715), are generally associated with the name of Clairaut (1734). The geo- metrical interpretation was given by Lagrange in 1774, but the theory in its present form was not given until much later by Cayley (1872) and M. J. M. Hill (1888).
The first methods of solving differential equations of the second or higher orders with constant coefficients were due to Euler. D'Alembert dealt with the case when the auxiliary equation had equal roots. Some of the symbolical methods of finding the par- ticular integral were not given until about a hundred years later by Lobatto (1837) and Boole (1859).
The first partial differential equation to be noticed was that giving the form of a vibrating string. This equation, which is of the second order, was discussed by Euler and D'Alembert in 1747. Lagrange completed the solution of this equation, and he also
* Also spelt Leibnitz. f Also spelt Bcrnouilli. ft Also known as James.
xv
XVI HISTORICAL INTRODUCTION
dealt, in a series of memoirs from 1772 to 1785, with partial dif- ferential equations of the first order. He gave the general integral of the linear equation, and classified the different kinds of integrals possible when the equation is not linear.
These theories still remain in an unfinished state ; contributions have been made recently by Chrystal (1892) and Hill (1917). Other methods for dealing with partial differential equations of the first order were given by Charpit (1784) and Jacobi (1836). For higher orders the most important investigations are those of Laplace (1773), Monge (1784), Ampere (1814), and Darboux (1870).
By about 1800 the subject of differential equations in its original aspect, namely the solution in a form involving only a finite number of known functions (or their integrals), was in much the same state as it is to-day. At first mathematicians had hoped to solve every differential equation in this way, but their efforts proved as fruitless as those of mathematicians of an earlier date to solve the general algebraic equation of the fifth or higher degree. The subject now became transformed, becoming closely allied t<5 the Theory of Functions. Cauchy in 1823 proved that the infinite series obtained from a differential equation was convergent, and so really did define a function satisfying the equation. Questions of convergency (for which Cauchy was the first to give tests) are very prominent in all the investigations of this second period of the study of dif- ferential equations. Unfortunately this makes the subject very abstract and difficult for the student to grasp. In the first period the equations were not only simpler in themselves, but were studied in close connection with mechanics and physics, which indeed were often the starting point of the work.
Cauchy's investigations were continued by Briot and Bouquet (1856), and a new method, that of " Successive Approximations," was introduced by Picard (1890). Fuchs (1866) and Frobenius (1873) have studied linear equations of the second and higher orders with variable coefficients. Lie's Theory of Continuous Groups (from 1884) has revealed a unity underlying apparently disconnected methods. Schwarz, Klein, and Goursat have made their work easier to grasp by the introduction of graphical con- siderations, and a recent paper by Wada (1917) has given a graphical representation of the results of Picard and Poincarr. Runge (1895) and others have dealt with numerical approximations.
Further historical notes will be found in appropriate places throughout the book. For more detailed biographies, see Rouse Ball's Short History of Mathematics.
CHAPTER I
INTRODUCTION AND DEFINITIONS. ELIMINATION. GRAPHICAL REPRESENTATION
^=-P2y, (i)
1. Equations such as
dx2
Miyf-g <3>
dv= t u)
dx y^l+x1)'
dt2 dx2' {0)
involving differential coefficients, are called Differential Equations.
2. Differential Equations arise from many problems in Algebra, Geometry, Mechanics, Physics, and Chemistry. In various places in this book we shall give examples of these, including applications to elimination, tangency, curvature, envelopes, oscillations of mechanical systems and of electric currents, bending of beams, conduction of heat, diffusion of solvents, velocity of chemical reactions, etc.
3. Definitions. Differential equations which involve only one independent variable,* like (1), (2), (3), and (4), are called ordinary.
Those which involve two or more independent variables and partial differential coefficients with respect to them, such as (5), are called partial.
* In equations (1), (2), (3), (4) x is the independent and y the dependent variable. In (5) x and ( are the two independent variables and y the dependent.
p.d.e. a S
2 DIFFERENTIAL EQUATIONS
An equation like (1), which involves a second differential co- efficient, but none of higher orders, is said to be of the second order (4) is of the first order, (3) and (5) of the second; and (2) of the third.
The degree of an equation is the degree of the highest differential coefficient when the equation has been made rational and integral as far as the differential coefficients are concerned. Thus (1), (2), (4) and (5) are of the first degree.
(3) must be squared to rationalise it. We then see that it is of
d2v the second degree, as j~ occurs squared.
Notice that this definition of degree does not require x or y to occur rationally or integrally.
Other definitions will be introduced when they are required.
4. Formation of differential equations by elimination. The
problem of elimination will now be considered, chiefly because it gives us an idea as to what kind of solution a differential equation may have.
We shall give some examples of the elimination of arbitrary constants by the formation of ordinary differential equations. Later (Chap. IV.) we shall see that partial differential equations may be formed by the elimination of either arbitrary constants or arbitrary functions.
5. Examples.
(i) Consider x = A cos (pt- a), the equation of simple harmonic motion. Let us eliminate the arbitrary constants A and a.
Differentiating, -=- = -pA sin (pt - a)
d2x and -572 = - p2A cos (pt -a)= - p2x.
d2x Thus -j-g = -p2x is the result required, an equation of the second
order, whose interpretation is that the acceleration varies as the distance from the origin.
(ii) Eliminate p from the last result.
d x dx
Differentiating again, -5-3 = - p2 -57 •
d?x \ dx d x \
Hence -373 -57 = - -p' J = ■■-,■■{ • x, (from the last result).
3,'X Q.X d X
Multiplying up, x . -=-3 » -j- • j-§, an equation of the third order.
ELIMINATION 3
(iii) Form the differential equation of all parabolas whose axis is the axis of x.
Such a parabola must have an equation of the form
y2 = ia(x-h).
Differentiating twice, we get
i.e. yd£=2a,
(L 11 i '(Lti\
and Vj\+ \y) = ^> wn^cn w °f the second order.
Examples for solution.
Eliminate the arbitrary constants from the following equations : (1) y = Ae2x + Be~2x. (2) y = A cos 3x + B sin 3x.
(3) y = AeBx. (4) y = Ax + A*.
(5) If x2 + y2 = a2,^ prove that j-= --, and interpret the result geometrically. &
(6) Prove that for any straight line through the origin -»^, and interpret this. * dx
d2u
(7) Prove that for any straight line whatever -r\ = 0. Interpret
this. dx
6. To eliminate n arbitrary constants requires (in general) a differ- ential equation of the n^ order. The reader will probably have arrived at this conclusion already, from the examples of Art. 5. If we differentiate n times an equation containing n arbitrary con- stants, we shall obtain (n + 1) equations altogether, from which the n constants can be eliminated. As the result contains an nth differ- ential coefficient, it is of the nth order.*
* The ?'-^ument in the text is that usually given, but the advanced student will notice some weak points in it. The statement that from any (n + 1) equations n quantities can be eliminated, whatever the nature of those equations, is too sweeping. An exact statement of the necessary and sufficient conditions would be extremely complicated.
Sometimes less than (n + 1) equations are required. An obvious case is y = (A + B)x, where the two arbitrary constants occur in such a way as to be really equivalent to one.
A less obvious case is y2=2Axy + Bx2. This represents two straight lines through the origin, say y = mlx and y=m2x, from each of which we easily get
-= j-, of the first instead of the second order. The student should also obtain x dx
this result by differentiating the original equation and eliminating B. This will
give . , .
[y-x£)(y-Ax) = 0.
4 DIFFERENTIAL EQUATIONS
7. The most general solution of an ordinary differential equation of the nth order contains n arbitrary constants. This will probably seem obvious from the converse theorem that in general n arbitrary con- stants can be eliminated by a differential equation of the nth order. But a rigorous proof offers much difficulty.
If, however, we assume * that a differential equation has a solution expansible in a convergent series of ascending integral powers of x, we can easily see why the arbitrary constants are n in number.
Consider, for example, ^-| = ^, of order three.
Assume that y = a0 + a1x+a2^ + ... + aM— f + ... to infinity. Then, substituting in the differential equation, we get
X2 Xn~3... x2 xn~x
so a3=av
ai = a2,
tt5=a3=0!l>
/ X3 X5 \ (x2 X* T6 \
Hence y-*+^+S+fl + "0+<2!+II + fl+"0
= aQ+ai sinh x + a2 (cosh x - 1), containing three arbitrary constants, a0, ax and a2. Similar reasoning applies to the equation
^y=f(r „ <k ^i dn~ly\ dxn j\x> y> dx> dxv ••> dxn~ij-
In Dynamics the differential equations are usually of the second
order, e.g. -j-f +p2y=0, the equation of simple harmonic motion.
To get a solution without arbitrary constants we need Iwo con- ditions, such as the value of y and dyjdt when t = 0, giving the initial displacement and velocity.
8. Complete Primitive, Particular Integral, Singular Solution. The solution of a differential equation containing the full number of arbitrary constants is called the Complete Primitive.
Any solution derived from the Complete Primitive by giving particular values to these constants is called a Particular Integral.
* The student will see in later chapters that this assumption is not always justifiable.
GRAPHICAL REPRESENTATION
Thus the Complete Primitive of -t4=-j r dx3 dx
is y = aQ+a1 sinh x+a2 (cosh x - 1),
or ?/ = c + «j sinh x+a2 cosh a;, where c = aQ- a2,
or y=c+aea!+&e~a;, where a = l(a1+a2) and 6 = |(a2-«i)-
This illustrates the fact that the Complete Primitive may often be written in several different (but really equivalent) ways. The following are Particular Integrals : ,
y=4, taking c = 4, a1=a2=0;
y = 5smhx, taking 0^=5, c = a2=0; y = 6 cosh z - 4, taking a2 = 6, ax =0, c = - 4 ; y = 2+ex-3e~x, taking c = 2, a = l, 6= -3. In most equations every solution can be derived from the Com- plete Primitive by giving suitable values to the arbitrary constants. Bowever, in some exceptional cases we shall find a solution, called a Singular Solution, that cannot be derived in this way. These will be discussed in Chap. VI.
Examples for solution.
Solve by the method of Art. 7 :
« ■£* •
« 3--*
(3) Show that the method fails for ■£-— -.
x ' ax x
[log as cannot be expanded in a Maclaurin series.]
(4) Verify by elimination of c that y = ca; + - is the Complete Primitive
of v = x -T- + 1 / -^ . Verify also that y2 = ix is a solution of the differential y dx I dx J *
equation not derivable from the Complete Primitive {i.e. a Singular
Solution). Show that the Singular Solution is the envelope of the
family of lines represented by the Complete Primitive. Illustrate by
a graph.
9. Graphical representation. We shall now give some examples of a method * of sketching rapidly the general form of the family of curves representing the Complete Primitive of
* Duo to Dr. S. Brodetsky and Prof. Takeo Wada.
6
DIFFERENTIAL EQUATIONS
where f{x, y) is a function of x and y having a perfectly definite finite value * for every pair of finite values of x and y.
The curves of the family are called the characteristics of the equation. ,
Ex. (i)
Here
Now a curve has its concavity upwards when the second differential coefficient is positive. Hence the characteristics will be concave up above y = \, and concave down below this line. The maximum or minimum points lie on x=0, since dy/dx = 0 there. The characteristics near y = l, which is a member of the family, are flatter than those further from it.
These considerations show us that the family ie of the general form shown in Fig. 1.
y
Fig. 1
Ex. (ii)
Here
dy
d2y dy
-~=^r + ex = y + 2ea
dx2 dx
We start by tracing the curve of maxima and minima y + e* = 0, and the curve of inflexions y + 2ex — 0. Consider the characteristic through the origin. At this point both differential coefficients are positive, so as x increases y increases also, and the curve is concave upwards. This gives us the right-hand portion of the characteristic marked 3 in Fig. 2. If we move to the left along this we get to the
•Thus excluding a function hke yjx, which is indeterminate when a;=0 and
GRAPHICAL REPRESENTATION
curve of minima. At the point of intersection the tangent is parallel to Ox. After this we ascend again, so meeting the curve of inflexions. After crossing this the characteristic becomes convex upwards. It still ascends. Now the figure shows that if it cut the curve of minima again
y
Fig. 2.
the tangent could not be parallel to Ox, so it cannot cut it at all, but becomes asymptotic to it.
The other characteristics are of similar nature.
|
Examples for solution. |
||
|
Sketch the characteristics of |
: |
|
|
(1) |
dx |
y{\-x). |
|
(2) |
dx |
x2y. |
|
(3) |
dy = dx |
y+x2. |
10. Singular points. In all examples like those in the last article, we get one characteristic, and only one, through every point
dv d2v
of the plane. By tracing the two curves -g- =0 and j\ =0 we can
easily sketch the system.
If, however, f(x, y) becomes indeterminate for one or more
points (called singular points), it is often very difficult to sketch the
8 DIFFERENTIAL EQUATIONS
system in the neighbourhood of these points. But the following examples can be treated geometrically. In general, a complicated analytical treatment is required.*
Ex. (i).
■4-=-- Here the origin is a singular point. The geo-
CLOO X
metrical meaning of the equation is that the radius vector and the tangent have the same gradient, which can only be the case for straight
Fig. 3.
lines through the origin. As the number of these is infinite, in this case an infinite number of characteristics pass through the singular point.
Ex.(ii). *--?, i.e. *.*--l.
ax y x Gfe
This means that the radius vector and the tangent have gradients
whose product is -1, i.e. that they are perpendicular. The char- acteristics are therefore circles of any radius with the origin as centre.
* See a paper, " Graphical Solution," by Prof. Takeo Wada, Memoirs of the College of Science, Kyoto Imperial University, Vol. II. No. 3, July 1917.
GRAPHICAL REPRESENTATION 9
In this case the singular point may be regarded as a circle of zero radius, the limiting form of the characteristics near it, but no characteristic of finite size passes through it.
Bx.(iH). P'*^-
v ax x + ky
Writing dy/dx=>ta,n\fs, y/x = tan 6, we get
, tan 0-k tan^ = l+fctan0'
i.e. tan \f, + k tan \\r tan 0 = tan 6-k, tan 0 - tan \[s
i.e.
k,
1 + tan 6 tan \/r
i.e. tan (d-\fs) = k, a constant.
The characteristics are therefore equiangular spirals, of which the singular point (the origin) is the focus.
Fig. 5.
These three simple examples illustrate three typical cases.
Sometimes a finite number of characteristics pass through a singular
point, but an example of this would be too complicated to give
here.*
* See Wada's paper.
10 DIFFERENTIAL EQUATIONS
MISCELLANEOUS EXAMPLES ON CHAPTER I.
Eliminate the arbitrary constants from the following :
(1) y = Aex + Berx + C.
(2) y = Aex + Be2x + C<?x.
[To eliminate A, B,C from the four equations obtained by successive differentiation a determinant may be used.]
(3) y = ex (A cos x + B sin x),
(4) y = c cosh -, (the catenary).
c
Find the differential equation of
(5) All parabolas whose axes are parallel to the axis of y.
(6) All circles of radius a.
(7) All circles that pass through the origin.
(8) All circles (whatever their radii or positions in the plane xOy). [The result of Ex. 6 may be used.]
(9) Show that the results of eliminating a from
2y=xd£+ax> (1)
) dy
and b from y = x-j--bx2, (2)
d y dy are in each case x2^A,-2x^- + 2y = 0 (3)
dx2 dx J v '
[The complete primitive of equation (1) must satisfy equation (3), since (3) is derivable from (1). This primitive will contain a and also an arbitrary constant. Thus it is a solution of (3) containing two constants, both of which are arbitrary as far as (3) is concerned, as a does not occur in that equation. In fact, it must be the complete primitive of (3). Similarly the complete primitives of (2) and (3) are the same. Thus (1) and (2) have a common complete primitive.]
(10) Apply the method of the last example to prove that
y+y = 2aex dx
■and y-^~ = 2be-x
* dx
have a common complete primitive.
(11) Assuming that the first two equations of Ex. 9 have a common
dlJ
complete primitive, find it by equating the two values of ~ in terms
of x, y, and the constants. Verify that it satisfies equation (3) of Ex. 9.
(12) Similarly obtain the common complete primitive of the two equations of Ex. 10.
MISCELLANEOUS EXAMPLES 11
(13) Prove that all curves satisfying the differential equation
ax \dx/ ax*
cut the axis of y at 45°.
(14) Find the inclination to the axis of x at the point (1, 2) of the two curves which pass through that point and satisfy
(|)V-2z + *s.
(15) Prove that the radius of curvature of either of the curves of Ex. 14 at the point (1, 2) is 4.
(16) Prove that in general two curves satisfying the differential equation
•0EN2+i-«
pass through any point, but that these coincide for any point on a certain parabola, which is the envelope of the curves of the system.
(17) Find the locus of a point such that the two curves through it satisfying the differential equation of Ex. (16) cut (i) orthogonally ; (ii) at 45°.
(18) Sketch (by Brodetsky and Wada's method) the characteristics of
ax
CHAPTER II EQUATIONS OF THE FIRST ORDER AND FIRST DEGREE
11. In this chapter we shall consider equations of the form
M+N^=0, ax
where M and N are functions of both x and y.
This equation is often written,* more symmetrically, as Mdx+Ndy=0.
Unfortunately it is not possible to solve the general equation of this form in terms of a finite number of known functions, but we shall discuss some special types in which this can be done.
It is usual to classify these types as
(a) Exact equations ; •
(b) Equations solvable by separation of the variables ; -
(c) Homogeneous equations ; •
(d) Linear equations of the first order. •
The methods of this chapter are chiefly due to John Bernouilli of Bale (1667-1748), the most inspiring teacher of his time, and to his pupil, Leonhard Euler, also of Bale (1707-1783). Euler made great contributions to algebra, trigonometry, calculus, rigid dynamics, hydrodynamics, astronomy and other subjects.
12. Exact equations, f
Ex. (i). The expression ydx + xdy is an exact differential.
Thus the equation ydx + xdy = 0,
giving d{yx)=0,
i.e. yx = c, is called an exact equation.
* For a rigorous justification of the use of the differentials dx and dy see Hardy's Pure Mathematics, Art. 136.
t For the necessary and sufficient condition that Mdx + Ndy = 0 should be exact see Appendix A.
12
EQUATIONS OF FIRST ORDER AND FIRST DEGREE 13
Ex. (ii). Consider the equation tan y . tfo + tan x . dy = 0.
This is not exact as it stands, but if we multiply by cos x cos y it becomes sin y COs x dx + sin x cos y dy = 0,
which is exact.
The solution is sin y sin x = c. y
13. Integrating factors. In the last example cos x cos y is called an integrating factor, because when the equation is multiplied by it we get an exact equation which can be at once integrated.
There are several rules which are usually given for determining integrating factors in particular classes of equations. These will be found in the miscellaneous examples at the end of the chapter. The proof of these rules forms an interesting exercise, but it is generally easier to solve examples without them.
14. Variables separate.
dx Ex. (i). In the equation — =tan y . dy, the left-hand side involves
x only and the right-hand side y only, so the variables are separate. Integrating, we get log x = - log cos y + c, i.e. log (x cos y) = c, x cos y = ec = a, say.
Ex. (ii). | = 2x*/.
The variables are not separate at present, but they can easily be made so. Multiply by dx and divide by y. We get
— =2xdx.
V Integrating, log y = x2 + c.
As c is arbitrary, we may put it equal to log a, where a is another arbitrary constant.
Thus, finally, y = aex2.
Examples for solution.
(1) (12x + 5y-9)dx + (5x + 2y-4)dy = 0.
(2) {cos x tan y + cos (x + y)} dx.+ {sin x sec2 y + cos (x + y)} dy = 0.
(3) (sec x tan x tan y - ex) dx + sec x sec2 y dy -■= 0. * (4) (x + y) (dx - dy) =dx + dy. . (5) ydx-xdy + Sx^e^dx = 0.
(6) y dx - x dy = 0. • " (7) (sin x + cos x) dy + (cos x - sin x) dx = 0.
•<8) g=*y.
,{9) y dx-x dy = xy dx. (10) tan x dy = cot y dx.
14 DIFFERENTIAL EQUATIONS
15. Homogeneous equations. A homogeneous equation of the first order and degree is one which can be written in the form
dy=f(y\
dx J \x/ To test whether a function of x and y can be written in the form of the right-hand side, it is convenient to put
y
-=v or y=vx.
00
If the result is of the formf(v), i.e. if the x's all cancel, the test is satisfiM.
-, ... dy x2 + y2 , dy \+v2 m, . -. . ,
Ex. (i). j^= g becomes -f-= . This equation is homo-
geneous. dx Zx dx l
diJ w da
Ex. (ii). ^=^2 becomes -^- = xvz. This is not homogeneous.
CLOO 00 (LOO —
16. Method of solution. Since a homogeneous equation can be
dy reduced to ^f-=f(v) by putting y=vx on the right-hand side, it is
natural to try the effect of this substitution on the left-hand side also. As a matter of fact, it will be found that the equation can always be solved * by this substitution (see Ex. 10 of the miscel- laneous set at the end of this chapter).
Ex. (i). f^ = ^.
w dx 2x2
Put y=vx,
i.e. -jr**v+z-z-> i for if y is a function of x, so is v). dx dx v 9 '
_,, . dv 1 + v2
The equation becomes v + x-z- = — ~ — ,
Separating the variables,
i.e. 2x dv = (1 + v2 - 2v) dx. 2dv dx
(v-l)2 x
— 2
Integrating, — - = logx + c.
y -2 _ -2 - 2x _ 2x
u' v~x' ° vzi~y_1~y-x~x-y'
X
Multiplying by x-y, 2x = (x - y) (log x + c) .
* By " solved " we mean reduced to an ordinary integration. Of course, this integral may not be expressible in terms of ordinary elementary functions.
EQUATIONS OF FIRST ORDER AND FIRST DEGREE 15
Ex. (ii). (x + y) dy + (x - y) dx = 0.
m,. . dy y-x
This gives ~ = - — •
dx y + x
Putting y = vx, and proceeding as before, we get
dv v-1
dx v + l
dv v-l v2 + l
i.e. x^-= — --« = -.
dx v + l v + l
(v + l)dv dx
Separating the variables, -
i.e.
v2 + l x v dv dv dx
v2 + l v2 + l X Integrating, - £ log (v2 + 1) - tan-1v - log x + c,
i.e. 2 log x + log (v2 + 1) + 2 tan-1?; + 2c = 0, logx2(,y2 + l)+2tan~M+a = 0, putting 2c = a.
Substituting for v, log {y2 + x2) + 2 tan-1 - + a = 0.
•2/
17. Equations reducible to the homogeneous form.
-n /•* mi ^ dy y-x + \,
Ex. (i). The equation ■#=- =
u dx y + x + b
is not homogeneous.
This example is similar to Ex. (ii) of the last article, except that
y-x . , -, i y-x + \
is replaced by -.
y+x r J y+x+5
Now y-x=0 and y + x = 0 represent two straight lines through the origin.
The intersection of y-x+l=0 and y + x + 5 = 0 is easily found to be (-2, -3).
Put x = X - 2 ; y=Y -3. This amounts to taking new axes parallel to the old with ( - 2, - 3) as the new origin.
Then y-x + l = Y-X and y + x + 5=Y + X.
Also dx = dX and dy = dY.
The equation becomes -t^> = -^ — =>• * dX Y + X
As in the last article, the solution is
log(Y2 + X2)+2tan-1^ + a = 0, A
i.e. log[(2/ + 3)2 + (z + 2)2] + 2tan-1^ + a = 0.
x + A
16 DIFFERENTIAL EQUATIONS
Ex. (u). /=^ -.
dx y-x + o
This equation cannot be treated as the last example, because the lines y-x+\=0 and y-x + 5 = 0 are parallel.
As the right-hand side is a function oiy-x, try putting y-x = z,
dy _dz dx dx'
The equation becomes 1 + ^- = - — -,
' m z + 5
dz -4 i.e. -j- = — =. dx z + 5
Separating the variables, (z + 5) dz = - 4 dx.
Integrating, |z2 + 5z = - 4sc + c,
i.e. z2 + 10z + 8x = 2c.
Substituting for z, (y - x) 2 + 10 (y - x) + 8x = 2c,
i.e. (y-x)2 + \0y -2x = a, putting 2c = a.
Examples for solution.
•(1) {2x-y)dy = (2y-x)dx. [Wales.]
(2) {x*-y2)^- = xy. [Sheffield.]
^-(3)2tH+& [MatL Tripos-]
.(4) xfx = y + V(x* + y2)-
dy_2x + 9y-20 1 ' dx 6x + 2y-10* (6) (12» + 21y-9)flte + (47a? + 40y + 7)dy-0.
> dy_3x-iy-2 [ ' dx 3x-4y-3"
(8) (jB + 2y)(ia5-dy)=*B + dy. 18. Linear equations. The equation ~£+Py = ®'
where P and Q are functions of x (but not of y), is said to be linear
of the first order.
', . dy 1 „
A simple example is -^ + - . y =x2.
{
EQUATIONS OF FIRST ORDER AND FIRST DEGREE 17
If we multiply each side of this by x, it becomes
xix+y=x>
ie' dx^=X^
Hence, integrating, xy = \x* + c.
We have solved this example by the use of the obvious integrating factor x.
19. Let us try to find an integrating factor in the general case. If R is such a factor, then the left-hand side of
Rfx + RPy = RQ
is the differential coefficient of some product, and the first term
R -j- shows that the product must be Ry.
Put, therefore, R^+RPy=-^(Ry) =#^ + y~.
This gives RPy = y-^,
i.e. Pdx = -n, li
i.e.
ipdx=logR,
[Pdx
This gives the rule : To solve -j- +Py = Q, multiply each side by
\pdx
e , which will be an integrating factor.
20. Examples.
(i) Take the example considered in Art. 18.
ty . 1 2
-T-+- .y=<c2.
ax x. HereP = -, so \Pdx = logx, and elo«r=x.. Thus the rule gives the same integrating factor that we used before.
(ii) g + 2a*/ = 2<r*\
Here P = 2x, \Pdx = x2, and the integrating factor is e**.
P.D.E. B
:-
18 DIFFERENTIAL EQUATIONS
Multiplying by this, e*2 ~- + 2xex%y = 2,
Integrating, yex* = 2x + c,
y = (2x + c)e-°fi.
(is, t+%-*2*
Here the integrating factor is e3x.
Multiplying by this, e3* ^ + ^xy ~ &x>
a i.e. j-(yeZx)-e5x.
Integrating, ye3 x = -te5 x + c,
y = -Lg2* + ce-3*.
21. Equations reducible to the linear form.
Ex. (i). xy-^-yh-*.
Divide by yz, so as to free the right-hand side from y.
ttT 1 1 dy ^
We get .tjS. jg.^,
1 1 <Z /l
$-*•
Putting ^ = 2» 2a» + -=- = 2e~-
i/2 2 cfoVy5 1 <fc
This is linear and, in fact, is similar to Ex. (ii) of the last article with 2 instead of y.
Hence the solution is z = (2x + c) e~x\
i.e. — = {2x + c)e~x\
y2
ei*2
'J{2x+o)
This example is a particular case of " Bernoulli's Equation "
where P and Q are functions of x. Jacob Bernouilli or Bernoulli ol Bale (1654-1705) studied it inl695.
EQUATIONS OF FIRST ORDER AND FIRST DEGREE 19
Ex. (ii).
(2*-i(y) d£+y=o.
fix
This is not linear as it stands, but if we multiply by -=-, we get
2z-i(y+</|=o, •
dx 2x ,. „ i.e. -T-+— = 10w2. dy y
This is linear, considering y as the independent variable.
Proceeding as before, we find the integrating factor to be y2, and
the solution 2 o *, ,
y2x = 2y5 + c, ,
i.e. a: = 2?/3 + c?/"-2. Examples for solution.
/(I) (* + a)j|-8y-(a> + a)«. [Wales.]
%(2) a;cosa;^ + «/(a;sincc + cosaj) = l. [Sheffield.]
•(3) xloga?^ + «/ = 21ogcc. (4) x2y - x* j- = y* cos x.
-7(5) y + 2fx = f(x-l).
.(6) (* + 2jfl|[-y.
• (7) dx + xdy = e~v sec2y dy.
22. Geometrical Problems. Orthogonal Trajectories. We shall now consider some geometrical problems leading to differential equations.
y
Ex. (i). Find the curve whose subtangent is constant. The sttbtangent TN = PN cot yj, = y ~ .
20 DIFFERENTIAL EQUATIONS
Hence y -y- - k,
*dy
dx = k—,
y
x + c = k log y,
putting the arbitrary constant c equal to k log a.
Ex. (ii). Find the curve such that its length between any two points PQ is proportional to the ratio of the distances of Q and P from a fixed point 0.
If we keep P fixed, the arc QP will vary as OQ.
Use polar co-ordinates, taking 0 as pole and OP as initial line. Then, if Q be (r, 6), we have s — ^r>
But, as shown in treatises on the Calculus, (ds)* = (rdd)z + (dr)2. Hence, in our problem,
k* (dr) 2 = (rd6)2 + (dr)*,
i.e. <Z0=±V(&2-1)-
ldr
= , say,
a r
giving r = cea0, the equiangular spiral.
Ex. (iii). Find the Orthogonal Trajectories of the family of semi- cubical parabolas ay2 = x3, where a is a variable parameter.
Two families of curves are said to be orthogonal trajectories when every member of one family cuts every member of the other at right angles.
We first obtain the differential equation of the given family by eliminating a.
Differentiating ay2 = x3,
we get 2ay -^- = 3x2,
whence, by division, ~:r=- (1)
' J y dx x
Now ^=tan \ls, where \lr is the inclination of the tangent to the dx T
axis of x. The value of \\r for the trajectory, say rfs', is given by
\fr = \Js' ± \ir, i.e. tan \js= -cot yj/',
i e. — for the given family is to be replaced by - -j- for the trajectory. dx ay
EQUATIONS OF FIRST ORDER AND FIRST DEGREE 21
Making this change in (1), we get
_2dx 3
ydy = x
2xdx + 3y dy = 0,
2x2 + 3y2 = c,
a family of similar and similarly situated ellipses.
Ex. (iv). Find the family of curves that cut the family of spirals r = a9 at a constant angle a.
As before, we start by eliminating a.
This gives -j- = 6.
Now — j- =tan <f>, where 0 is the angle between the tangent and the
radius vector. If (/>' is the corresponding angle for the second family,
0' = 0±a,
, tan <j> ± tan a 6 + k
tan <A =- — ~- — ■ = - — r7:,
r l+tan0tana 1 - kQ
putting in the value found for tan <f> and writing k instead of ±tan a. Thus, for the second family,
rdd^ 6 + k dr ~l-kO'
The solution of this will be left as an exercise for the student. The result will be found to be
r = c(6 + k)kl+1e-k9.
Examples for solution.
(1) Find the curve whose subnormal is constant.
(2) The tangent at any point P of a curve meets the axis of x in T. Find the curve for which OP — PT, 0 being the origin.
-(3) Find the curve for which the angle between the tangent and radius vector at any point is twice the vectorial angle.
(4) Find the curve for which the projection of the ordinate on the normal is constant.
Find the orthogonal trajectories of the following families of curves :
(5) x2-y2 = a2. • .(6) x$ + y* = a$. (7) px2 + qy2 = a2, (p and q constant).
a6
.(8) rd = a. (9) r =
l+<9*
(10) Find the family of curves that cut a family of concentric circles at a constant angle a.
22 DIFFERENTIAL EQUATIONS
MISCELLANEOUS EXAMPLES ON CHAPTER II.
(I) (3y*-x)d£ = y. (2) *d£ = y + 2V(y2-x*).
(3) tan x cos y dy + sin y dx + e8'11 x dx =0.
(4) x*ijt + Zy* = xyK [Sheffield.]
.(5) tffx=yz+yW(y2-x*)>
(6) Show that 4- -+*+{ x ' da; hx + by+f
represents a family of conies.
(7) Show that ydx-2xdy = 0
represents a system of parabolas with a common axis and tangent at the vertex.
y (8)Showthat (4x + 3«/ + l) dx + {3x + 2y + l) dy=0 represents a family of hyperbolas having as asymptotes the lines x + y = 0 and 2x + y + l=0.
(9) If J + 2«/tanz = sina:
and y = 0 when x = \ir, show that the maximum value of y is ^.
[Math. Tripos.]
(10) Show that the solution of the general homogeneous equation of the first order and degree £ =f ( - ) is
. f dv
log x= \-TT-\ +C»
6 Jf(v)-V
where v = y/x.
(II) Prove that xhyk is an integrating factor of
py dx + qxdy + xmyn (ry dx + sxdy)=0
h + 1 Jc + l , h + m + 1 k + n + l
if = — — and = •
p q r s
Use this method to solve
3y dx -2xdy + x*y-l(\0y dx - 6x dy) = 0.
(12) By differentiating the equation
Cf(xy) + F(xy) d(xy) +1 *, if(xy)-F(xy) xy *y
Verifythat xy{f(xy)-F(xy)}
MISCELLANEOUS EXAMPLES 23
is an integrating factor of
f(xy) ydx + F (xy) xdy = 0. Hence solve (x2y2 + xy + l)ydx-(x2y2-xy + l)xdy=0.
(13) Prove that if the equation M dx + N <fo/ = 0 is exact,
dN = dM dx By ' [For a proof of the converse see Appendix A.]
(14) Verify that the condition for an exact equation is satisfied by
(Pdx + Qdy)e$Ax)dx = 0
Hence show that an integrating factor can always be found for Pdx + Qdy = 0
if if^.m
Qldy dx] is a function of x only. Solve by this method
(xz + xy4) dx + 2ysdy = 0.
(15) Find the curve (i) whose polar subtangent is constant ;
(ii) whose polar subnormal is constant.
(16) Find the curve which passes through the origin and is such that the area included between the curve, the ordinate, and the axis of x is k times the cube of that ordinate.
(17) The normal PG to a curve meets the axis of x in 0. If the distance of 0 from the origin is twice the abscissa of P, prove that the curve is a rectangular hyperbola.
(18) Find the curve which is such that the portion of the axis of x cut off between the origin and the tangent at any point is proportional to the ordinate of that point.
(19) Find the orthogonal trajectories of the following families of curves: (i) (x-l)2 + y2 + 2ax = 0,
•(ii) r = a0, (iii) r = a + cos n$, and interpret the first result geometrically.
(20) Obtain the differential equation of the system of confocal conies x2 i y2 _
a2 + \ b2 + \ and hence show that the system is its own orthogonal trajectory.
(21) Find the family of curves cutting the family of parabolas y2 = iax at 45°.
24 DIFFERENTIAL EQUATIONS
(22) If u + iv =f(x + iy), where u, v, x and y are all real, prove that the families u = constant, v = constant are orthogonal trajectories.
., ., • d2u d2u x d2v d2v
Also prove that 3—0+2-2 = 0 = 5-^ + aTi-
r ox1 ayi ox1 oy*
[This theorem is of great use in obtaining lines of force and lines of constant potential in Electrostatics or stream lines in Hydrodynamics. u and v are called Conjugate Functions.]
(23) The rate of cteca'y~oi radium Is proportional to the amount remaining. Prove that the amount at any time t is given by
A=AQe~kt.
(24) If j =g(} -p) and v = 0 if * = 0, prove that
v = &tanh %-• k
[This gives the velocity of a falling body in air. taking the resistance of the air as proportional to v2. As t increases, v approaches the limiting value k. A similar equation gives the ionisation of a gas after being subjected to an ionising influence for time t. ]
(25) Two liquids are boiling in a vessel. It is found that the ratio of the quantities of each passing off as vapour at any instant is pro- portional to the ratio of the quantities still in the liquid state. Prove that these quantities (say x and y) are connected by a relation of the form y = cxk.
[From Partington's Higher Mathematics for Students of Chemistry, p. 220.]
CHAPTER III LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS
23. The equations to be discussed in this chapter are of the form
dny dn~xy dy ., . ...
where f(x) is a function of x, but the p's are all constant.
These equations are most important in the study of vibrations of all kinds, mechanical, acoustical, and electrical. This will be illustrated by the miscellaneous examples at the end of the chapter. The methods to be given below are chiefly due to Euler and D'Alembert.*
We shall also discuss systems of simultaneous equations of this form, and equations reducible to this form by a simple transformation.
24. The simplest case ; equations of the first order. If we take n = l and /(#)=(), equation (1) becomes
jPo|+Ay=o, (2)
i.e. p0^+Pidx=0,
or p0 log y + pxx = constant,
so log y = ~Pix/pQ + constant
- -p^/po+logA, say, giving y = Ae~ PlX/Po.
25. Equations of the second order. If we take n = 2 and f(x) = 0, equation (1) becomes
ftS+Ai+fty"° (3)
* Jean-le-Rond D'Alembert of Paris (1717-1783) is best known by " D'Alem- bert's Principle " in Dynamics. The application of this principle to the motion of fluids led him to partial differential equations.
25
26 DIFFERENTIAL EQUATIONS
The solution of equation (2) suggests that y = Aemx, where m is some constant, may satisfy (3).
With this value of y, equation (3) reduces to
Aemx(pQm2 +pjm +p2) =0. Thus, if m is a root of
p0m2+p1m+p2=0, (4)
■y = Aemx is a solution of equation (3), whatever the value of A.
Let the roots of equation (4) be a and {3. Then, if a and /3 are unequal, we have two solutions of equation (3), namely y=AeoX and y=Bepx. Now, if we substitute y =AeaZ + Bepx in equation (3), we shall get AeaX(p0a2 +pia +p2) +Be>3x(p0/32 +p±p +p2) =0, which is obviously true as a and (3 are the roots of equation (4).
Thus the sum of two solutions gives a third solution (this might have been seen at once from the fact that equation (3) was linear). As this third solution contains two arbitrary constants, equal in number to the order of the equation, we shall regard it as the general solution.
Equation (4) is known as the " auxiliary equation."
Example.
To solve 2 -r\ + 5 —■ + 2y =0 put y = Aemx as a trial solution. This
leads to Aemx(2m2 + 5m + 2)--=0,
which is satisfied by m = - 2 or - f . The general solution is therefore
y = Ae-2x + Be~ix. 26. Modification when the auxiliary equation has imaginary or complex roots. When the auxiliary equation (4) has roots of the form p + iq, p-iq, where i2 = - 1, it is best to modify the solution
y=Ae{p+iq)z+Be(p-i'i)x) (5)
so as to present it without imaginary quantities.
To do this we use the theorems (given in any book on Analytical Trigonometry) enx = cos qx + 1 sm qX}
e - lix = cos qx - % sin qx. Equation (5) becomes
y - epx {A (cos qx + i sin qx) +B(cosqx-i sin qx) } = epx{E cos qx+F sin qx\ writing E for A +B and F for i(A -B). E and F are arbitrary
LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS 27
constants, just as A and B are. It looks at first sight as if F must be imaginary, but this is not necessarily so. Thus, if A = l+2i and B = \ -2i, E = 2 and F = -4.
Example. ^_6^+1%m0
leads to the auxiliary equation
m2- 6m + 13 = 0, whose roots are m = 3 ± 2i.
The solution may be written as
• y = Ae^+^x + Be^~2i'>x, or in the preferable form
y = eSx{E cos 2x + F sin 2x), or again as ?/ = CeZx cos (2# - a),
' where C cos a = E and C sin a = F,
so that C = J(E2 + F2) and tan a = F/E.
27. Peculiarity of the case of equal roots. When the auxiliary equation has equal roots a=/3, the solution
y = AeaX+Befix reduces to y = {A + B) eaX.
Now A + B, the sum of two arbitrary constants, is really only a single arbitrary constant. Thus the solution cannot be regarded as the most general one.
We shall prove later (Art. 34) that the general solution is y = (A+Bx)eaX.
28. Extension to orders higher than the second. The methods of Arts. 25 and 26 apply to equation (1) whatever the value of n, as long as/(x)=0.
The auxiliary equation is
m3-6m2 + llm-6 = 0, giving m = l, 2, or 3.
Thus y = Aex + Be2x + Ce3x.
Ex.(ii). U'8y = °'
The auxiliary equation is m3 - 8 = 0,
i.e. (m-2)(m2 + 2m + 4)=0, giving m = 2 or -l±i\/S.
Thus y = Ae2x + e~x(E cos x^/3 + F sin x\/3),
or y = Ae2x + Ce~x cos (x\/3- a).
28 DIFFERENTIAL EQUATIONS
Examples for solution.
Solve
y,., d2s .ds J d2s . ds _
(7) «^ + 2&i & 2y_a
(8) What does the solution to the last example become if the initial conditions are fly
y = l, -p = 0 when x = 0,
and if y is to remain ^finite when x= + co ? Solve
<»»3-»3+"»-*-v t .
•(H) ^ + 8y=0. \| .(12)g-64<,=0.
,72/3 J/3
* (13) Z-T-g +#0=0, -given that 0 = a and tt=0 when t=0.
[The approximate equation for small oscillations of a simple pen- dulum of length I, starting from rest in a position inclined at a to the vertical.]
(14) Find the condition that trigonometrical terms should appear in the solution of ^2S fe
mdT*+kdi+cs=0-
[The equation of motion of a particle of mass m, attracted to a fixed point in its line of motion by a force of c times its di -ance from that point, and damped by a frictional resistance of k times its velocity. The condition required expresses that the motion should be oscillatory. e.g. a tuning fork vibrating in air where the elastic force tending to restore it to the equilibrium position is proportional to the displacement and the resistance of the air is proportional to the velocity.]
(15) Prove that if k is so small that k2/mc is negligible, the solution of the equation of Ex. (14) is approximately e~kt/2m times what it would be if k were zero.
[This shows that slight damping leaves the frequency practically unaltered, but causes the amplitude of successive vibrations to diminish in a geometric progression. ]
LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS 29
(16) Solve L^ + R^ + Q-O, given that Q=QQ and ^ = 0 when
* = 0, and that CR2<4:L.
[Q is the charge at time t on one of the coatings of a Leyden jar of capacity C, whose coatings are connected when t = 0 by a wire of resist- ance R and coefficient of self-induction L. ]
29. The Complementary Function and the Particular Integral. So far we have dealt only with examples where the f(x) of equation (1) has been equal to zero. We shall now show the relation between the solution of the equation when f(x) is not zero and the solution of the simpler equation derived from it by replacing f(x) by zero. To start with a simple example, consider the equation
It is obvious that y=x is one solution. Such a solution, con- taining no arbitrary constants, is called a Particular Integral. Now if we write y=x+v, the differential equation becomes
0 2g+R(l.+£)+«t,+.>-5+*
n d%* K dv _ n
' giving v = Ae~2x + Be~ix,
T so that y=x+Ae-2x+Be-*-x.
The terms containing the arbitrary constants are called the Complementary Function.
This can easily be generalised. If y = u is a particular integral of
dny {?n_1v dy £. » ia.
*lf+AjSA + ~+**£+**-fW (6)
so that Po^a+Pi^=i + -+Pn-i-^.+PnU=f(x), (7)
put y=u + v in equation (6) and subtract equation (7). This gives dnv dn~xv dv _ /ox
V°fan+Pl^+---+Pn-llx+PnV=0 (8)
If the solution of (8) be v = F(x), containing n arbitrary con- stants, the general solution of (6) is
y = u+F(x), and F(x) is called the Complementary Function.
30 DIFFERENTIAL EQUATIONS
Thus the general solution of a linear differential equation with constant coefficients is the sum of a Particular Integral and the Com- plementary Function, the latter being the solution of the equation obtained by substituting zero for the function of x occurring.
Examples for solution.
Verify that the given functions are particular integrals of the follow- ing equations, and find the general solutions :
j I *•> p-4+*>-<"- <2> 3 ■• - g-i3!+i»
• (3)2sin3a;; j\ + iy = - 10 sin Sx.
For what values of the constants are the given functions particular integrals of the following equations ? ^
(4)ae»*; g + 13^ + 42y»ll2e» ^,>V'X
d2s \j / d2v
- (5) aeU > ij2+9s= QOe~t- &v (6) a sin px ' ri + y = 12 sin 2cc*
V-^7) a sin px + b cos px ; -=-| + 4 j^ + 3y = 8 cos x - 6 sin x.
<«> •' §+5l+6^12-
Obtain, by trial, particular integrals of the following :
.(11) 0 + 9y = 4Osin5*. <12> H"8l + 9!' = 40sill5a;-
» <13) § + 8! + 2»
30. The operator D and the fundamental laws of algebra. When a particular integral is not obvious by inspection, it is convenient to employ certain methods involving the operator D, which stands
for -j-. This operator is also useful in establishing the form of the
complementary function when the auxiliary equation has equal
roots.
d2 d3
D2 will be used for j-2, D3 for -7-3, and so on.
LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS 31
The expression 2 -y-| + 5 -j- + 2y may then be written
2D*y + 5Dy+2y, or (2D*+bD + 2)y.
We shall even write this in the factorised form (2D + l)(D+2)y, factorising the expression in D as if it were an ordinary algebraic quantity. Is this justifiable ?
The operations performed in ordinary algebra are based upon three laws :
I. The Distributive Law
m(a + b)=ma + mb ; II. The Commutative Law
ab = ba ; III. The Index Law am . an=am+n. Now D satisfies the first and third of these laws, for D(u+v)=Du+Dv, and Dm . Dnu=Dm+n . u
(m and n positive integers). As for the second law, D (cu) = c (Du) is true if c is a constant, but not if c is a variable.
Also Dm (Dnu) = Dn (Dmu)
(m and n positive integers). Thus D satisfies the fundamental laws of algebra except in that it is not commutative with variables. In what follows we shall write F(D) s p0D» +PlD^ + ... +pn_1D +pn,
where the p's are constants and n is a positive integer. We are justified in factorising this or performing any other operations depending on the fundamental laws of algebra. For an example of how the commutative law for operators ceases to hold when negative powers of D occur, see Ex. (iii) of Art. 37.
31. F(D)eax=eaxF(a). Since
Deax = aeax,
D2eax=a2eax, '
and so on, ,^
F(D) eax = (p0Dn +p1Dn~1 + ... +pn-xI) +pn) eax
^(p^+p^-1 + ... +pn.1a +pn) eax
=ea*F(a).\
32 DIFFERENTIAL EQUATIONS
32. F(D){eaxV} =eaxF(D +a) V, where V is any function of x. By Leibniz's theorem for the nth differential coefficient of a product, D»{eP*V} = (Dneax) V + w(D"-1eoa') {DV)
+ ln(n-l)(Dn-*eax)(D*V) + ... + eax(DnV) =ane°*Y +nan-1eazDV +\n{n - l)a»-V*Da7 + ... +eaxDnV =eax(an+nanr1I>k + hn(n - l)an~2D2 + ...+Dn)V = ea*(Z)+a)nF. Similarly Dn~1{eaxV}=eax(D+a)n-1V, and so on. Therefore F(D){eaxV} = (p0D» +PlD»~l + ... +pn.1D +pn){eaxV}
= eax{p0(D+a)n +p1(D+a)n~1 + ... +pn-x{D +a) +pn}V = e*xF(D + a)V.
33. F(D2) cos ax =F( - a2) cos ax. Since
D2 cos ax = -a2 cos ax,
Dioosax = (- a2)2 cos ax, and so on,
F(D2) cos ax = (p0D2n +p1D2n~2 + ... +pn-1D2 +pn) cos ax
= {Po(~ a2)n +Pi(-a2)n~1 + ... +pn-i( -«2) +Pn} cos «# — F( -a2) cos ax. Similarly F (D2) sinax=F(- a2) sin ax.
34. Complementary Function when the auxiliary equation has equal roots. When the auxiliary equation has equal roots a and a, it may be written m2 _ 2ma + a2 = 0.
The original differential equation will then be
i.e. (D2-2aZ>+a2)*/=0,
(D-a)2y=0 :...(9)
We have already found that y=AeaX is one solution. To find a more general one put y=e*?V, where V is a function of x. By Art. 32,
(D -a)*{e?*V} =eaX(D -a +a)2V = (T?D2V. Thus equation (9) becomes
D2V=0, i.e. V = A+Bx, so that y = eax(A+Bx).
m
LINEAR EQL\. X)NSTANT COEFFICIENTS 33
Similarly the equation (D - a)py = 0 reduces to DPV =0,
giving V = {Ax + A& + A &? + . . . + Ap^'1),
and y = eaX (Ax + A^c + A zx2 + . . . + Ap^xP^x).
When there are several repeated roots, as in
(D-anD-(3)*(D-yyy=0, (10)
we note that as the operators are commutative we may rewrite the equation in the form
iD-PnD-yY{{D-aVy}=0, which is therefore satisfied by any solution of the simpler equation
(D-ayy=o '. (ii)
Similarly equation (10) is satisfied by any solution of
(D-P)*y=0, (12)
or of (D-y)ry=0 (13)
The general solution of (10) is the sum of the general solutions of (11), (12), and (13), containing together (p+q+r) arbitrary constants.
Ex. (i). Solve (D*-8D2 + 16)y = 0, (p.
^
i.e. (Z)2-4)2?/ = 0. The auxiliary equation is (m2-4)2 = 0,
m = 2 (twice) or -2 (twice). Thus by the rule the solution is
y = (A + Bx) e2x + (E + Fx) e~2x.
Ex. (ii). Solve (D2 + l)2*/ = 0.
The auxiliary equation is (m2 + l)2 = 0,
m — i (twice) or -i (twice). Thus y = {A + Bx)eix + {E+Fx)e~ix,
or better y = (P+ Qx) cos x + ( R + Sx) sin x.
Examples for solution.
\h) {D* + 2D* + D2)y = 0. >< $) (-D6 + 3Z>* + 32)a + l)y = 0.
^) (Di-2D3 + 2D2-2D + l)y = 0. $4) (4Z>5-3Z>3- D2) t/ = 0. s
(5) Show that
F (D2){P cosh ax + Q sinh ax) = F(a2) (P cosh ax + Q sinh ax) .
(6) Show that (D - a)4n(eax sin px) =p*neax sin px.
35. Symbolical methods of finding the Particular Integral when f(x) =eax. The following methods are a development of the idea of treating the operator D as if it were an ordinary algebraic quan-
p.d.e. c
34 DIFE
tity. We shall proceed tentatively, at first performing any opera- tions that seem plausible, and then, when a result has been obtained in this manner, verifying it by direct differentiation. We shall use
the notation Y7Jy\ /(*) *° denote a particular integral of the equati „ F(D)y-f{z).
(i) If f(x) =eax, the result of Art. 31,
F(D) e™ = eaxF (a)
1
"
suggests that, as long as F(a)=j=0, ^y-r eax may be a value of
F(u)v — v_—"» F(Dy
This suggestion is easily verified, for
W^}-9lSrVyArt.3i.
>F(a) J F{a) = eax. (ii) If F (a) =0, {D-a) must be a factor of F(D). Suppose that F(D)=(D -a)p<f>(D), where 0(a)=/=O. Then the result of Art. 32,
F (D) {eax V } = eax F (D + a) V, suggests that the following may be true, if 7 is 1,
1-^^ 1 _, 1 l(*x.l\ e™ 1 ,
/jw*~--. pax j y _
F(D) (D-a)P<f>(D) (D-a)p\</>(a){ <j>(a) Dp
e?x xv
adopting the very natural suggestion that jz is the operator inverse to D, that is the operator that integrates with respect to x, while y- integrates p times. Again the result obtained in this tentative manner is easily verified, for
■♦^K^S' byArt-32'
= e?x, byArt. 31.
LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS 35
In working numerical examples it will not be necessary to repeat the verification of our tentative methods.
Ex. (i). (D + 3)2y = 50e*x.
The particular integral ia
(Z> + 3)2 (2+3)
Adding the complementary function, we get y = 2e2x + {A + Bx)e~3x. Ex. (ii). (Z>-2)2i/=.50e2*.
If we substitute 2 for D in jj: — ^ 50e2x, we get infinity. But using the other method,
(D]_2)2 • 50e2* = 50e2ie -^-2 . 1 = 5Qe2* , \x2 - 25x2e2*. \
Adding the complementary function, we get y = 25x2e2x + (A + Bx)e2x.
Examples for solution. Solve v?
m)(D2 + 6D + 25)y=^lQi:^x. rfft) (D2 + 2pD + p2 + q2)y = eax.
:($) (D2-9)?/ = 54^^ m (D3-D)y = ex + e-x.
(5) (D2-p2)y = a'2o&px. (6) (D* + iD2 + lD) y = 8e~2x.
36. Particular Integral when f (x) =cos ax. From Art. 33,
(j> (D2) cos ax = <f> ( - a2) cos ax. This suggests that we may obtain the particular integral by writing - a2 for D2 wherever it occurs.
Ex. (i). (D2 + SD + 2) y = cos 2x.
1 1 1
. COS 2x = ; t-=: - . COS 2x = 7rF: rr . cos 2x.
D2 + 3D + 2 -4 + 3Z) + 2 3D-2
To get D2 in the denominator, try the effect of writing
1 _3Z> + 2 3ZT^2~9Z>2-4'
suggested by the usual method of dealing with surds. This gives
— ^r — i cos 2x = - xV(3-D cos 2x + 2 cos 2x) - oo - 4
= -tL0.(-6sin2a; + 2cos2a;)
^^(Ssu^aj-cc^a;).
36 DIFFERENTIAL EQUATIONS
Ex. (ii). (Di + 6D2 + nD + 6)y = 2ein3x.
2 sin 3x = 2 — ^= — — : — r-r^ — - sin 3x
D3 + 6D2 + llD + 6 -9D-54- LD + 6
1
Z)-24 D + 24
Z>2-576
_ i
sin 3z sin 3x
5 s
-(3 cos 3a; + 24 sin Sx)
= - T-^T(cos 3x + 8 sin 3x). We may now show, by direct differentiation, that the results obtained are correct.
If this method is applied to
[<f> (D2) + D\J, (D2) ] y = P cos ax + Q sin ax, where P, Q and a are constants, we obtain
<j> ( - a2) . (P cos ax + Q sin ax)+a\p- {-a2) . (P sin ax -Q cos ax) {^(-a2)}2 + a2{yjr(-a2)}2.
It is quite easy to show that this is really a particular integral, provided that the denominator does not vanish. This exceptional case is treated later (Art. 38).
Examples for solution.
Solve /
.$& {D + l)y = 10sm2x. fa (Z)2-5Z) + 6) </ = 100sin ix.
^(3) (Z)2 + 8D + 25)i/ = 48 cos a: -16 sin a. v<4) (D2 + 2D + 401) y = sin 20z + 40 cos 20x.
(5) Prove that the particular integral of
d2s _7 ds „
-tt + 2* -t; + V s = a cos at
at2 at
may be written in the form b cos (qt - e),
where b = a/{(p2-q2)2 + ik2q2}h and tan € = 2kq/(p2-q2).
Hence prove that if q is a variable and &, p and a constants, 6 is greatest when q = y/{p2 - 2k2) = p approx. if k is very small, and then e = 7r/2 approx. and b = a/2kp approx.
[This differential equation refers to a vibrating system damped by a force proportional to the velocity and acted upon by an external periodic force. The particular integral gives the forced vibrations and the complementary function the free vibrations, which are soon damped out (see Ex. 15 following Art. 28). The forced vibrations have the greatest amplitude if the period 2-n-fq of the external force is very nearly equal to that of the free vibrations (which is
LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS 37
2ir/y/{p2-k2)=2ir/p approx.), and then e the difference in phase between the external force and the response is approx. tt/2. This is the important phenomenon of Resonance, which has important applications to Acoustics, Engineering and Wireless Telegraphy.]
37. Particular integral when f(x) =xm , where m is a positive integer.
In this case the tentative method is to expand j^ in a series of ascending powers of D.
Ex.(i). -^^ = 1.(1+^)-^
= i(*2-i). Hence, adding the complementary function, the solution suggested
for {D*+4)y = z*
is y = \(x2-^)+A cos 2x4-5 sin 2x.
Ex. (ii).
D2-iD + S ^ = £ \i^d ~ 3TI)) **> b? Partial fractions,
f / D D2 D3 Z)4 M
= i{(l + Z) + Z)2 + Z)3 + Z).4 + ...)-i(l+- + T+-+-+...)}^
Adding the complementary function, the solution suggested for (D2-iD + 3)y = x3 is y^^xS + ^ + zg-x + ^ + Aet + Be?*.
s* m dhb^T)96x2=w ■ Uwrtx°}
-W.-jTj^-g). from Ex. (i),
-■i(S-t)
= 2x4-6x2. Hence the solution of D2(Z)2 + 4) y = 96x2 should be
y = 2x*-6x2 + A cos 2x + Z?sin 2x + E + Fx. Alternative method.
= (24:D-2-6 + %D2-...)x2 = 2x*-6x2 + 3.
38 DIFFERENTIAL EQUATIONS
This gives an extra term 3, which is, however, included in the complementary function.
* The method adopted in Exs. (i) and (ii), where F(D) does not contain D as a factor, may be justified as follows. Suppose the expan- sions have been obtained by ordinary long division. This is always possible, although the use of partial fractions may be more convenient in practice. If the division is continued until the quotient contains Dm, the remainder will have Dm+1 as a factor. Call it <p(D) . Dm+1. Then
^=CQ + c1D + c2D* + ...+cmn™ + *iD]}-]^+1 (1)
This is an algebraical identity, leading to
l = F(D){c0 + c1D + ctD* + ...+cmDm} + <f>{D) . D"*1 (2)
Now equation (2), which is true when D is an algebraical quantity, is of the simple form depending only on the elementary laws of algebra, which have beffc shown to apply to the operator D, and it does not involve the difficulties which arise when division by functions of D is concerned. Therefore equation (2) is also true when each side of the equation is regarded as an operator. Operating on xm we get, since Dm+1xm = 0,
xm=F(D){{c(t + c1D + c2D2 + ...+cmDm)xm}, (3)
which proves that the expansion obtained in (1), disregarding the remainder, supplies a particular integral of F(D)y=xm.
It is interesting to note that this method holds good even if the expansion would be divergent for algebraical values of D.
To verify the first method in cases like Ex. (iii), we have to prove that 1
i.e. (c0D-r + c1D~r+1 + c2D-r+2 + ... +cmD~r+m) xm, is a particular integral of {F(D) . Dr}y = xm,
i.e. that {F (D) . Dr} {(cQD-r + c^*1 + c2D~r+2
+ ...+cmD-r+m)xm}=xm (4)
Now {F( D) . D'} u m F(D) . {Dru},
also Dr{(c8D ~ '■+*) xm} = (cgD*) xm ;
hence the expression on the left-hand side of (4) becomes
F(D){(c0 + c1D + c2D* + ...+cmD™)xm} = x'», by (3),
which is what was to be proved.
In the alternative method we get r extra terms in the particular integral, say (Cffi+1Z)^^+...+W2>»)*«
These give terms involving the (r-l)th and lower powers of x. But these all occur in the complementary function. Hence the first method is preferable.
* The rest of this article should be omitted on a first reading.
LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS 39
Note that if D~hi denotes the simplest form of the integral of u, without any arbitrary constant,
D-1(DA) = D-1. 0=0,
while i>{2H.l)-X>.*-l,
so that D(D-1.l)^D-1.(Z).l).
Similarly Dm (D-m . xn)=^D-m(Dm . xn), if m is greater than n.
So when negative powers of D are concerned, the laws of algebra are not always obeyed. This explains why the two different methods adopted in Ex. (iii) give different results.
Examples for solution.
Solve -j(l) (D + \)y = %*. /(2) (D2 + 2D)y = 2ix.
^(3) (D2-6D + 9)y = 5±x + l8. ' (4) (D4-6D3 + 952)?/ = 54a: + 18. ,
v/(5) (D2-D-2)y = U-76x-±8x2.
JG) (IP-D2-2D)y = U-76x-±8x2. U
38. Particular integrals in other simple cases. We shall now give some typical examples of the evaluation of particular integrals in simple cases which have not been dealt with in the preceding articles. The work is tentative, as before. For the sake of brevity, the verification is omitted, as it is very similar to the verifications already given.
Ex. (i). (D2 + 4)y=sm2x.
We cannot evaluate -^ — j sin 2x by writing - 22 for D2, as in
Art. 36, for this gives zero in the denominator. But i sin 2x is the imaginary part of e2ix, and
eax = e2ix t if as m Art. 35,
Z)2 + 4 (Z> + 2*')2 + 4
— ■ x _ JL
L_.i ±-S*> -x.i
1 I. Dv*
ca".
-'"OT-{(»-B+™--)-U P)
"e 4*D 4*
«■ - |«x(cos 2a; + 1 sin 2x) ;
40 DIFFERENTIAL EQUATIONS
hence, picking out the imaginary part,
™ — r sin 2x = - \x cos 2x.
Adding the complementary function, we get
y = A cos 2x + B sin 2x - \x cos 2x. Ex. (ii). (D2 - 52) + 6) y = e2xx*.
(D2-5D + 6) \2-D 3-D/ ^
=e2x(4-r^)^ *
= e2* ( - ^ - 1 - D - D2 - D3 - D* - ...)
a?
= e2*( - \x* - x3 - 3x2 - 6x - 6). Adding the complementary function, we get
y = A<?x - e**{\x* + x3 + 3x* + 6x- B), including the term - 6e2x in Be2x.
Ex. (iii). (Z)2 - 6D + 13) y = 8<?x sin 2x.
. 8<?x sin 2x = 8eSx,,„ ..,„ * -^ — r^ . sin 2x
(D2-6D + 13)" {(D + 3)2-6(Z) + 3) + 13}
= 8e3a:-K5 — 7sin2z
= 8e3a:( - \x cos 2x) (see Ex. (i) ) = - 2xe?x cos 2x. Adding the complementary function, we get
y = e?x(A cos 2x + B sin 2x - 2x cos 2x).
These methods are sufficient to evaluate nearly all the particular integrals that the student is likely to meet. All other cases may be dealt with on the lines indicated in (33) and (34) of the miscel- laneous examples at the end of this chapter.
Examples for solution.
Solve "(1) (Z)2 + l)</ = 4cosa;. (2) (D-l) y = (x + 3) e*x.
y\Z) (D!i-3LD-2)y = 5i0x3e-x. y(i) (DM- 2D + 2) y » 2e-*«n x. f:
(5) {D^+}J^ji=2ixcosx. (6) (W^~D)y = \2ex + 8 sin x -2x.
(7) (D2-6D + 25)y = 2e3*cos4:r + 8e3a:(l-2:r)sin4x. .
39. The Homogeneous Linear Equation. This is the name given to the form {p(fcnBn +p1xn~1Dn-1 + . . . + pn) y =f (x).
It reduces to the type considered before if we put x = e'.
LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS 41
Ex. (x3D3 + 3x2D2 + xD)y = 2ix2.
Put x = el,
dx_ t_ dt~6=X>
t. _. d dt d 1 d
so that D = dZ="d-xjr-Jt'
\xdt) x*dt + x dt x2\ dt^dt2).' x2\ dt + dt2) a*\ dt + dt2J+x2 \ dt + dt2/ x*\ dt + dt2)+x3\ dt2 + dt?) x*\ dt dt2 dtV'
d u
thus the given differential equation reduces to -^ = 2ie2\
giving y = A + Bt + Ct2 + 3eit
= A + B\ogx + C(logx)2 + 3x2.
Another method is indicated in (28)-(30) of the miscellaneous examples at the end of this chapter. The equation
p0(a+bx)nDny +p1(a + bx)n-1Dn~1y + ... +pny =f(x) can be reduced to the homogeneous linear form by putting z = a +bx, giving ^ =dy^dydz=bdy
y dx dz dx dz '
Examples for solution. /(I) x2^-2xd£ + 2y = M. / (2) ^g + 9x| + 25^50.
(3) x3^ + 3:K2§ + a;| + 8^ = 65c0s(l0^)-
W ^dx* + Z^dx* + X dx2 Xdx + y~[°gX-
(5) (1+2^2-6(1 +2x)g + 16^r8 (l+2a:)«.
(6) (l + x)2<^ + (l+x)d£ + y==ico8log(l+x).
42 DIFFERENTIAL EQUATIONS
40. Simultaneous linear equations with constant coefficients. The method will be illustrated by an example. We have two de- pendent variables, y and z, and one independent variable x.
D stands for -»-, as before. ax
Consider (52)+4) y -(2D + l)z = e~x, (1)
(D + S)y- 3z =be~x (2)
Eliminate z, as in simultaneous linear equations of elementary algebra. To do this we multiply equation (1) by 3 and operate on -equation (2) by (22) + 1).
Subtracting the results, we get
{3 (52) +4) - (22) + 1)(D + 8)} y = Se~x - (2D + 1) 5e~*, i.e. (-2D2-2D + ±)y = Ser*t or (D2jD-2)y = -4<r*
Solving this in the usual way, we get
y = 2e-x+Aex+Be~2x. The easiest way to get z in this particular example is to use •equation (2), which does not involve any differential coefficients of z. Substituting for y in (2), we get
Ue~x + 9Aex + QBe~2x - 3z = 5e~*, so that z = 36-* + 3Aex + 2Be~2x.
However, when the equations do not permit of such a simple method of finding z, we may eliminate y. In our case this gives { - (2) +8) (2D + 1) + 3(52) + 4)} y = {D + S)e~x - (52) + 4)5e"*, i.e. (-22)2-22) + 4)z = 12e-*, -giving z = 3e~x + Eex + Fe~2x.
To find the relation between the four constants A, B, E, and F, substitute in either of the original equations, say (2). This gives (2) + 8) (2e~x + Aex + Be~2x) - 3 {3e~x + Eex + Fe~2x) = 5e~x, i.e. (9 A - ZE) ex + (6# - 32?) e~2x = 0, whence E = ?>A and F = 2B,
so z = Se~x + Eex + Fe~2x = 3e~x + 3^e* + 2Be~2x, as before.
Examples for solution.
(1) Dy-z = 0, (2) (Z>- 17) t/ + (22)- 8)2 = 0,
(D-l)y-(D + \)z = 0. , (132)- 53)?/- 22 = 0.
<3) (22)2-2) + 9)y-(Z)2 + 2) + 3)2 = 0) (22)2 + D + 7) y -{D2-D + 5)2=0.
LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS 43
<4) (D + \)y = z + ex, (5) {D2 + b)y-d = -36cos 7x,
{D + l)z = y + ex. p>n)2z = 99co8 7x.
(6) (2.D + l)«/ + (Z> + 32)z = 91e-* + 147sin2a; + 135cos2a;, y-(D-8)z = 29e~x + 47 sin 2x + 23 cos 2x.
MISCELLANEOUS EXAMPLES ON CHAPTER III.
Solve /O) (Z)-l)32/ = 16e3*. (2) (4D2 + 12Z) + 9)i/ = 144a*fH
u <3) (D* + 6IP + nD2 + 6D) y = 20e~2x sin x.
<4) (D3-D2 + 4:D-i)y = 68exam2x.
(5) (D*-6D2-8D-S)y = 256(x + l)(?x.
(6) (ZH-8Z)2-9)i/ = 50sinh2a;. (7) (Z>»-2Z)2 + 1) */ = 40cosh x. (8) (Z)-2)22/ = 8(x2 + e2* + sin2x). (9) (D-2)2y = 8x2e2x Bin 2x.
<10) (7)2 + l)?/ = 3cos2a; + 2 8in3x.
<11) (D* + lOD2 + 9)y = 96am2xcosx.
(12) (D-a)ay = ax, where a is a positive integer.
ax2 x ax x2 dx2 x dx
<15> % = f ^ (, + l)2g + (, + l)gH2x + 3)(2x + 4).
§ + 4§ + 4, = 25* + W. «**-* £-*;>* <l«)£ + I-0; <§+*-0.
(21) Show that the solution of (D2n+1-l)y = 0 consists of Je* and n pairs of terms of the form
efx (Br cos sx + Cr sin sx),
. 2ttt . . 2xr
where c = cos ^ and s = sin - ,
2w + 1 2n + 1
r taking the values 1, 2, 3 ... n successively.
(22) If (D-a)u = 0,
(D-a) v = u, and (D- a)y = v,
find successively w, w, and y, and hence solve (D-a)3</ = 0
/
44 DIFFERENTIAL EQUATIONS
(23) Show that the solution of
(D-a)(D-a-h)(D-a-2h)y=0
can be written Aeax + B^x- — r— -+Ceax- ^ '-.
h h*
Hence deduce the solution of (D- a)zy = 0.
[This method is due to D'Alembert. The advanced student will notice that it is not quite satisfactory without further discussion. It is obvious that the second differential equation is the limit of the first, but it is not obvious that the solution of the second is the limit of the solution of the first.]
oz o z
(24) If {D-a)3emx is denoted by z, prove that z, =— , and =— ^ all
vanish when m = a.
Hence prove that eax, xe?x, and x2eax are all solutions of (D - a)3y =0.
[Note that the operators {D-af and -~— are commutative.]
.-„, .., , , cos ax - cos (a + h) x
(25 Show that ; ,.2 v .
x (a + h)2-a2
is a solution of (D2 + a2) y = cos (a + h) x.
Hence deduce the Particular Integral of (D2 + a2)y = cos ax.
[This is open to the same objection as Example 23.]
(26) Prove that if V is a function of x and F(D) has its usual meaning,
(i) Dn[xV] =xDnV + nDn-W\ (ii) F(D)[xV] = xF{D)V + F'(D)V;
(iii)^-rsri-s — 7- F{D) v-
{m} F(D)[ - J F(D) V [F(D)]2 '
(iv) <p(D)[xnV] = xntp{D)V + nxn-1</>'(D)V + --- +nC,.xn-r<f>r(D)V where <p{D) stands for •
(27) Obtain the Particular Integrals of (i) (D- l)y = xe2x,
(ii) (D + l)y = x2 coax, by using the results (iii) and (iv) of the last example.
(28) Prove, by induction or otherwise, that if 6 stands for x 3-,
x»^=e(d-l)(0-2)...(0-n + l)y.
(29) Prove that
(i) F(6)xm = xmF{m); xm
(u) jrgf ~r$sj- provided F(m)±°'-
(iii» m[xmV]=^-FW^)v-
where V is a function of x.
LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS 45
(30) By using the results of the last question, prove that the solu- tion of d2 d ..-,-. „ „
x2j\-ix~+%y = x5 is $x5 + Axa + Bxb,
-where a and b are the roots of m{m - 1) - 4m + 6 = 0,
i.e. 2 and 3.
(31) Given that (D-l)y = e2x,
prove that (D-l)(D-2)y = 0.
By writing down the general solution of the second differential equation (involving two unknown constants) and substituting in the first, obtain the value of one of these constants, hence obtaining the solution of the first equation.
d2y
(32) Solve j^2+p2y = sin ax by the method of the last question.
(33) If ux denotes eax I ue~ax dx,
u2 denotes ebx I u^er** dx,
etc., prove the solution of F(D)y = u, where F(D) is the product of n factors.
(D-a)(D-b)...
may be written y = un-
This is true even if the factors of F(D) are not all different. Hence solve (D-a)(D-b)y = eaxlog x.
(34) By putting „ ~ into partial fractions, prove the solution of F(D)y = u may be expressed in the form
2-^rrreaa;l ue~axdx, F'{a) J
provided the factors of F(D) are all different.
[If the factors of F(D) are not all different, we get repeated inte- grations.]
Theoretically the methods of this example and the last enable us to solve any linear equation with constant coefficients. Unfortunately, unless u is one of the simple functions (products of exponentials, sines and cosines, and polynomials) discussed in the text, we are generally left with an indefinite integration which cannot be performed.
If u =f(x), we can rewrite eax I ue~ax dx
in the form f(t)ea^-^dt,
where the lower limit Jc is an arbitrary constant.
46 DIFFERENTIAL EQUATIONS
(35) (i) Verify that
1 f * y = ~\ f(t)ainp(x-t)dt
is a Particular Integral of
[Remember that if a and b are functions of x,
(ii) Obtain this Particular Integral by using the result of the last example.
(iii) Hence solve (Z)2 + l)i/ = cosec x.
(iv) Show that this method will also give the solution of
(in a form free from signs of integration), if f(x) is any one of the func- tions tan x, cot x, sec x).
(36) Show that the Particular Integral of j~ + p2y = k cos pt repre- sents an oscillation with an indefinitely increasing amplitude.
[This is the phenomenon of Resonance, which we have mentioned before (see Ex. 5 following Art. 36). Of course the physical equatic of this type are only approximate, so it must not be assumed thatjpie oscillation really becomes infinite. Still it may become too^pge for safety. It is for this reason that soldiers break step on c^£hg a bridge, in case their steps might be in tune with the natural osOTlation of the structure. ]
(37) Show that the Particular Integral of
-^ + 2h-^ + (h2 + p2)y = Jce-htcospi
k represents an oscillation with a variable amplitude —te~ht.
Find the maximum value of this amplitude, and show that it is very large if h is very small. What is the value of the amplitude after an infinite time ?
[This represents the forced vibration of a system which is in reson- ance with the forcing agency, when both are damped by friction. The result shows that if this friction is small the forced vibrations soon become large, though not infinite as in the last example. This is an advantage in some cases. If the receiving instruments of wireless telegraphy were not in resonance with the Hertzian waves, the effects would be too faint to be detected.]
LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS 47
(38) Solve ^~n*y=0-
[This equation gives the lateral displacement y of any portion of a thin vertical shaft in rapid rotation, x being the vertical height of the portion considered. ]
(39) If, in the last example,
-i = y=0 when x=0 and x = l,
prove that y = E(coa nx - cosh nx) + F(ain nx - sinh nx) and cos nl cosh nl = 1 .
[This means that the shaft is supported at two points, one a height I above the other, and is compelled to be vertical at these points. The last equation gives n when I is known.]
(40) Prove that the Complementary Function of
becomes negligible when t increases sufficiently, while that of
dzy d2y _
oscillates with indefinitely increasing amplitude.
[An equation of this type holds approximately for the angular velocity of the governor of a steam turbine. The first equation corre- sponds to a stable motion of revolution, the second to unstable motion or " hdyting." See the Appendix to Perry's Steam Engine. ]
(41) fSgove that the general solution of the simultaneous equations :
md£=Ve-HedJ,
dt2 dt
m dt*~Me dt' where m, V, H, and e are constants, is
x = A + B cos {at - a),
V
y = ■= t + C + B sin ( cot - a),
He
where w = — and A, B, C, a are arbitrary constants.
Given that — = -^ = x = y=0 when t = 0, show that these reduce to
x=— -(1 -COS tot), ton.
V .
y= —jj (cot - sin tot), the equations of a cycloid.
48 DIFFERENTIAL EQUATIONS
[These equations give the path of a corpuscle of mass m and charge e repelled from a negatively-charged sheet of zinc illuminated with ultra-violet light, under a magnetic field H parallel to the surface. V is the electric intensity due to the charged surface. By finding ex- perimentally the greatest value of x, Sir J. J. Thomson determined
IV m
— =, from which the important ratio — is calculated when V and H are
<*>H e
known. See Phil. Mag. Vol. 48, p. 547, 1899.] (42) Given the simultaneous equations,
- (1*1 1 ..dH* Ix „
. (1*1 ' ^(Z2/, I2
where Lv L2, M, cv c2, E and p are constants, prove that Ix is of the
form ax cos pt + Ax cos (mt -a) + B1 cos (nt - /3),
and 1 2 of the form
a2 cos pt + A2 cos (mt -a) + B2 cos (nt - /3),
E
where ai = TFa(l -P2c2L2)>
EM 3
<X2— , p C]C2,
k denoting the expression
(LXL2 - M2) cxc2^ - (LlCl + L2c2)p2 + 1 ;
m and n are certain definite constants ; A1} Bv a and /3 are arbitrary constants ; and A2 is expressible in terms of Ax and B2 in terms of A2.
Prove further that m and n are real if Lv L2, M, cv and c2 are real and positive.
[These equations give the primary and secondary currents Ix and 72 in a transformer when the circuits contain condensers of capacities Cj and c2. Lx and L2 are the coefficients of self-induction and M that of mutual induction. The resistances (which are usually very small) have been neglected. E sin pt is the impressed E.M.F. of the primary.]
CHAPTER IV SIMPLE PARTIAL DIFFERENTIAL EQUATIONS
41. In this chapter we shall consider some of the ways in which partial differential equations arise, the construction of simple par- ticular solutions, and the formation of more complex solutions from infinite series of the particular solutions. We shall also explain the application of Fourier's Series, by which we can make these complex solutions satisfy given conditions.
The equations considered include those that occur in problems on the conduction of heat, the vibrations of strings, electrostatics and gravitation, telephones, electro-magnetic waves, and the diffusion of solvents.
The methods of this chapter are chiefly due to Euler, D'Alembert, and Lagrange.*
42. Elimination of arbitrary functions. In Chapter I. we showed how to form ordinary differential equations by the elimination of arbitrary constants. Partial differential equations can often be formed by the elimination of arbitrary functions.
Ex. (i). Eliminate the arbitrary functions /and F from
y=f(x-«l) + F(x + at) (!)
We get ^ =f\x - at) + F'(x + at) •
and <^=f"(x-al) + F"(x + at) (2)
Similarly -~ — - af'(x - at) + aF'(x + at)
and yi = a2f"{x-at)+a2F"(x + at) (3)
* Joseph Louis Lagrange of Turin (1730-1813), the greatest mathematician of the eighteenth century, contributed largely to every branch of Mathematics. He created the Calculus of Variations and much of" the subject of Partial Differential Equations, and he greatly developed Theoretical Mechanics and Inlinitesimal Calculus.
p.d.b. 49 u
C*o
50 DIFFERENTIAL EQUATIONS
From (2) and (3), g = I g : (4)
a partial differential equation of the second order.* Ex. (ii). Eliminate the arbitrary function / from
■-/©■
dz dz n so sc^+w^ =0.
dx * By
Examples for solution.
/ Eliminate the arbitrary functions from the following equations : y (1) z=f(x + ay). /(2) z=f(x + iy) + F(x-iy),wheTei*=-l.
v/ (3) z =f (x cos a + y sin a - at) + F(x cos a + y sin a + at). /(4) z=f(x2-y2). v (5) z = eax+bvf(ax-by).
- (6) -*/©•
43. Elimination of arbitrary constants. We have seen in Chapter I. how to eliminate arbitrary constants by ordinary differential equations. This can also be effected by partials.
Ex. (i). Eliminate A and p from z = Aept sin px.
d2z
— = - p2Aept sin px,
d2z 2 = p2Aept sin px ;
d!
dx2+dt2
Eliminate a, b, and c from
z = a(x + y) + b(x-y) +abt + c.
dz We get =-=a + b,
dz 7 — =a-b, ay
dz ,
=-=00.
dt
* Tin's equation holds for the transverse vibrations of a stretched string. The most general solution of it is equation (1), which represents two waves travelling with spied a, one to the right and the other to the left.
PARTIAL DIFFERENTIAL EQUATIONS 51
But (a + o)2-(a-&)2 = 4a&.
(!)'-(!)'-£•
Examples for solution.
Eliminate the arbitrary constants from the following equations : <i(l) z= Ae-P2t cos px. */(2) z = Ae-?1 cos qx sin ry, where p2 = q2 + rl. y/{Z) z = ax + (\-a)y + b. (4) z = ax + by + a2 + b2.
'(5) 2 = (x-a)2 + (t/-6)2. (6) az + b = a2x + y.
44. Special difficulties of partial differential equations. As we have already stated in Chapter I., every ordinary differential equation of the nth order may be regarded as derived from a solution con- taining n arbitrary constants* It might be supposed that every partial differential equation of the nth order was similarly derivable from a solution containing n arbitrary functions. However, this is not true. In general it is impossible to express the eliminant of n arbitrary functions as a partial differential equation of order n. An equation of a higher order is required, and the result is not unique."!*
In this chapter we shall content ourselves with finding particular solutions. By means of these we can solve such problems as most commonly arise from physical considerations. J We may console ourselves for our inability to find the most general solutions by the reflection that in those cases when they have been found it is often extremely difficult to apply them to any particular problem. §
♦It will be shown later (Chap. VI.) that in certain exceptional cases an ordinary differential equation admits of Singular Solutions in addition to the solution with arbitrary constants. These Singular Solutions are not derivable from the ordinary solution by giving the constants particular values, but are of quite a different form.
tSee Edwards' Differential Calculus, Arts. 512 and 513, or Williamson's Differential Calculus, Art. 317.
X The physicist will take it as obvious that every such problem has a solution, and moreover that this solution is unique. From the point of view of pure mathematics, it is a matter of great difficulty to prove the first of these facts : this proof has only been given quite recently by the aid of the Theory of Integral Equations (see Heywood and Frechet's L' Equation de Fredholm et ses application.? a la Physique Math6matique). The second fact is easily proved by the aid of Green's Theorem (see Carslaw's Fourier's Series and Integrals, p. 206).
§For example, Whittaker has proved that the most general solution of Laplace's equation twit t&v 7VV
3a;2 dy2 3z2 ~
V= I f(x cos t + y sin t + iz, t) dt,
but if we wish to find a solution satisfying certain given conditions on a given surface, we generally use a solution in the form of an infinite series.
52 DIFFERENTIAL EQUATIONS
45. Simple particular solutions.
d2z 1 dz Ex. (i). Consider the equation ,-2 = -2r (which gives the con- duction" of heat in one dimension). This equation is linear. Now, in the treatment of ordinary linear equations we found exponentials very useful. This suggests z = e"u:+nt as a trial solution. Substituting in the differential equation, we get
a2 *
which is true if n = m2a2.
ThUs e""'+"l2a2< is a solution. Changing the sign of m, e'mx+nfiaH is also a solution.
Ex. (ii). Find a solution of the same equation that vanishes when t= +oo .
In the previous solutions t occurs in ew2a2t. This increases with t, since m2a2 is positive if m and a are real. To make it decrease, put m = ip, so that m2a2 = - p2a2.
This gives eipx~p2aH as a solution.
Similarly e-^-^"2' is a solution.
Hence, as the differential equation is linear, e~p2a2t(Aeipx + Be~ipx) is a solution, which we replace, as usual, by
e~p'kl\E cos px + F sin px).
d2z d2z Ex. (iii). Find a solution of ^-^ + ^~^=^ which shall vanish when
y = + oo , and also when x = 0. ^
Putting z = emx+ny, we get (m2 + n2)emx+nv = 0, so m2 + n2=0.
The condition when y = + oo demands that n should be real and negative, say n= -p.
Then m = ± ip.
Hence e~vy{Aeipx + Be~ipx) is a solution,
i.e. e~vi'(E cos px.-\- F sm px) is a solution.
But z = 0 if x = 0, so E = 0.
The solution required is therefore Fe~P!/sin px.
Examples for solution.
O^U u it
(1) ~-| = =-£-, given that y = 0 when x = + oo and also when t = + oo .
d2z 1 92z
(2) — - =-x =-j, given that z is never infinite (for any real values of v ox2 a2 oy2
x or y), and that 2=0 when x = 0 or y = 0.
(3) \-a „-=0, given that z is never infinite, and that =- =0 when
K ' dx dy s ox
■x=*y=>(X
PARTIAL DIFFERENTIAL EQUATIONS 53
d2V d2V d2V
(4) 5 ,2 +2_2 + "^T=:^' giyen that V=0 when z=+oo, when
y = - oo , and also when 2 = 0.
(5) p-g = - - , given that V is never infinite, and that V = C and
dV dV dV . ,
x— •»-=-= -5-= 0 when x = y = z=0.
ox ay oz
d2V d2V dV
(6) -2^" + "2~i= ar> given that V =0 when £ = + oo , when x=0 or
I, and when ?/ = 0 or I.
46. More complicated initial and boundary conditions.* In Ex. (iii) of Art. 45, we found Fe~py sin fx as a solution of
dx2+dy2~ ' satisfying the conditions that z=0 if y= + oo or if £=0.
Suppose that we impose two extra conditions,"}- say 2=0 if x = l and 2 = Ix - x2 if «/ =0 for all values of x between 0 and I. The first condition gives sin pi =0,
i.e. pi = nw, where n is any integer. For simplicity we will at first take I = w, giving p = n, any integer. The second condition gives F sin px = irx- x2 for all values of x between 0 and ir. This is impossible.
However, instead of the solution consisting of a single term, we may take
Fxe~y sin x + F2e~2y sin 2x + F3e-3y sin 3x + . . . ,
since the equation is linear (if this is not clear, cf. Chap. TIL Art. 25), giving p the values 1, 2, 3. ... and adding the results.
By putting y = 0 and equating to irx - x2 we get FA sin x + F2 sin 2x + F3 sin 3x + . . . = ttx-x2 for all values of & between 0 and tr.
The student will possibly think this equation as impossible to satisfy as the other, but it is a remarkable fact that we can choose values of the F's that make this true.
This is a particular case of a more general theorem, which we now enunciate.
* As t usually denotes time and x and y rectangular coordinates, a condition such as z = 0 when t = 0 is called an initial condition, while one such as z = 0 if a; = 0, or if x = l, or if y = x, is called a boundary condition.
fThis is the problem of finding the steady distribution of temperature in a semi-infinite rectangular strip of metal of breadth I, when the infinite sides are kept at 0° and the base at (Ix - x2)°.
54 DIFFERENTIAL EQUATIONS
47. Fourier's Half-Range Series. Every function of x which satisfies certain conditions can be expanded in a convergent series of the form
/ (x) = a1 sin x + a2 sin 2x + a3 sin 3x + . . . to inf.
for all values of x between 0 and -k (but not necessarily for the extreme values x=0 and x = ir).
This is called Fourier's * half-range sine series.
The conditions alluded to are satisfied in practically every physical problem."]"
Similarly, under the same conditions f(x) may be expanded in a half-range cosine series
l0 + Zx cos x + 12 cos 2x + 13 cos 3x + . . . to inf.
These are called half-range series as against the series valid between 0 and 2ir, which contains both sine and cosine terms.
The proofs of these theorems are very long and difficult. J How- ever, if it be assumed that these expansions are possible, it is easy to find the values of the coefficients.
Multiply the sine series by sin nx, and integrate term by term, § giving
pir PIT rn
I f(x)smnxdx = a1\ sin x sin nx dx + a2 1 sin 2x sin nxdx + ... .
Jo Jo Jo
The term with an as a factor is sin2 nx dx
an I si] Jo
a r 1 • ~\n
(1 -cos 2nx)dx = ~ x - „ sin 2 nx\ 2 L 2/i Jo
2 Jo
= ^an7r.
* Jean Baptiste Joseph Fourier of Auxerre (1768-1830) is best known as the author of La Th4orie analytique de la chaleur. His series arose in the solution of problems on the conduction of heat.
f It is sufficient forf(x) to be single-valued, finite, and continuous, and have only a limited number of maxima and minima between sc = 0 and x = w. However, these conditions are not necessary. The necessary and sufficient set of conditions has not yet been discovered.
X For a full discussion of Fourier's Series, see Carslaw's Fourier's Series and Integrals and Hobson's Theory of Functions.
§ The assumption that this is legitimate is another point that requires justification.
PARTIAL DIFFERENTIAL EQUATIONS 65
The term involving any other coefficient, say ar, is ar sin rx sin nx dx
= -~\ {cos (n - r)x- cos (n+r)x}dx
^arrsia(n-r)x sin (n+r)x~y 2l n-r n + r J0= '
So all the terms on the right vanish except one.
Thus I f(x) sin nx dx = \anir,
2 f /., 01 an = - \ f(%) sin nx dx.
Similarly, it is easy to prove that if
f(x) =b0 + b1 cos x + 62 cos 2o3 + . . . for values of a; between 0 and tt, then
1 r
ttJo and bn = /(«) cos wee (Zx
for values of n other than 0.
48. Examples of Fourier's Series.
(i) Expand ttx-x2 in a half-range sine series, valid between x = 0 and x = 7r.
It is better not to quote the formula established in the last article. Let 7rx - x2 = a1 sin x + a2 sin 2x + a3 sin 3x + . . . .
Multiply by sin nx and integrate from 0 to tt, giving
I (7rx - x2) sin nx dx = an\ sin2 nxdx = ~ an, as before.
Now, integrating by parts,
I (irx-x2) sin nxdx = \ — {ttx - x2) cos nx\ +- (tt - 2x) cos nx dx Jo L n J0 wj0v
= 0 + — = (x - 2x) sin wz + — ~\ sin wa; tZir U2 Jo ™2Jo
2 r t 4 .
= 0 — 5 cos nx = — if n is odd or 0 if « is even.
w L Jo n
o
Thus an= — g if w is odd or 0 if w is even, giving finally
ttx - x2 = - (sin cc + -gV sin 3x + T I T sin 5x + . . . ).
7T
56 DIFFERENTIAL EQUATIONS
(ii) Expand /(as) in a half-range series valid from x =0 to x = ir, where f{x) — mx between x — 0 and x = —
Z
and f(x) = m{7r-x) between x = — and sc = 7r.
In this case f(x) is given by different analytical expressions in different parts of the range.* The only novelty lies in the evaluation of the integrals.
In this case
1 f(x) sin nx dx= I f(x) sin nx dx+\ f(x) sin wx cZx Jo Jo J|
7T
= I mx sin nx dx + I m(7r - x) sin wx dx. Jo J|
We leave the rest of the work to the student. The result is
d-fyy
— (sin z-^sin Sx + ^r sin So;-^8*11 7as + ...).
The student should draw the graph of the given function, and compare it with the graph of the first term and of the sum of the first two terms of this expansion. f
Examples for solution.
Expand the following functions in half -range sine series, valid between x = 0 and x = tt :
(1) 1. (2) x. (3) Xs. (4) cos a. (5) ex.
(6) f(x)=0 from x=0 to x =—, and from x=— to tt,
f(x) = (4:X-7r)(37r-4:x) from x = — to x=-r-.
(7) Which of these expansions hold good (a) for x — 0 ?
(6) for x — 7t ?
49. Application of Fourier's series to satisfy boundary conditions.
We can now complete the solution of the problem of Art. 46. We found in Art. 46 that
Fxe-v sin x+F2e~2" sin 2x +_F3e~3" sin 3a; + ... satisfied all the conditions, if
Fx sin x + F2 sin 2x + F3 sin Sx + . . . = irx - x2 for all values of x between 0 and 7r.
* Fourier's theorem applies even if f(x) is given by a graph with no analytical expression at all, if the conditions given in the footnote to Art. 47 are satisfied.
For a function given graphically, these integrals are determined by arith- metical approximation or by an instrument known as a Harmonic Analyser.
t Several of the graphs will be found in Carslaw's Fonrur's Series mid Inter/rah, Art. 59. More elaborate ones are given in the Phil. May., Vol. 45 (1898).
PARTIAL DIFFERENTIAL EQUATIONS 57
In Ex. (i) of Art. 48 we found that, between 0 and -k,
o
r- (sin x + -jV sm 3a; + y^T sin 5a; + ...) = ttx - a;2.
7T
Thus the solution required is
- {e~'J sin a; + ^Te~Z!l sin 3a; + 1-J--5 e~5j/ sin 5a; + . . .).
7T
50. In the case when the boundary condition involved I instead of ir, we found Fe~vy sin jpx as a solution of the differential equation, and the conditions showed that p, instead of being a positive integer n, must be of the form mr/l.
Thus F^e-^' sin ttx/1 + F2e~in^1 sin 2 irx/l + . . .
satisfies all the conditions if
Ft sin 7rx{l + F2 sin Sirx/l + ...=lx-x2 for all values of x between 0 and I.
I2 I2
Put ttx/1 = z. Then Ix -x2 = — -A-kz-z2). The F's are thus -=
7T 7T
times as much as before. The solution is therefore
8l2
" -3 (e_Tr?//' sin 7ra;/^ + irre~'inv" sm STrai/Z + xi^6"5^' sin 57r:r/^ + •••)•
MISCELLANEOUS EXAMPLES ON CHAPTER IV.
1 -Jt
(1) Verify that V = — re 4^< is a solution of
32V_1 dV Bx2 ~ K dt '
(2) Eliminate A and p from V =Ae~px sin (2p2Kt-px).
dV d2V
(3) Transform -^- = K ^ - hV
3W „d2W t0 . -U=K~W
by putting V = e-htW.
[The first equation gives the temperature of a conducting rod whose surface is allowed to radiate heat into air at temperature zero. The given transformation reduces the problem to one without radiation.]
(4) Transform
dV_Klf2dV\ dW d2W
dt~r2dr\ dr) dt dr2
by putting W = rV.
[The first equation gives the temperature of a sphere, when heat flows radially.]
58 DIFFERENTIAL EQUATIONS
(5) Eliminate the arbitrary functions from
vJ-[f(r-at) + F(r + at)].
(6) (i) Show that if emx+int is a solution of
dV d2V
where n and h are real, then m must be complex.
(ii) Hence, putting m=-g-if, show that V0e-ffx sin (nt -fx) is a solution that reduces to F0sin nt for x = 0, provided K(g2-f2)=h and n = 2Kfg.
(iii) If V=0 when x= +oo , show that if K and n are positive so are g and /.
[In Angstrom's method of measuring K (the " diffusivity "), one end of a very long bar is subjected to a periodic change of temperature V0 sin nt. This causes heat waves to travel along the bar. By measur- ing their velocity and rate of decay n/f and g are found. K is then calculated from K = n/2fg.]
dV d2V
(7) Find a solution of -^- = K^~y reducing to V0 sin nt for x=0
and to zero for x = + oo . dt dx
[This is the problem of the last question when no radiation takes place. The bar may be replaced by a semi-infinite solid bounded by a plane face, if the flow is always perpendicular to that face. Kelvin found K for the earth by this method.]
(8) Prove that the simultaneous equations
are satisfied by V = V0e-^+iJ^+int,
if g*-p = RK-n*LC,
2fg = n(RC + LK),
and V( R + iLn) = V02(K + iCn).
[These are Heaviside's equations for a telephone cable with resist- ance R, capacity C, inductance L, and leakance K, all measured per unit length. / is the current and V the electromotive force.]
(9) Show that in the last question g is independent of n if RC = KL. [The attenuation of the wave depends upon g, which in general
depends upon n. Thus, if a sound is composed of harmonic waves of different frequencies, these waves are transmitted with different degrees of attenuation. The sound received at the other end is therefore
MISCELLANEOUS EXAMPLES 59
distorted. Heaviside's device of increasing L and K to make RC = KL prevents this distortion.]
(10) In question (8), if L = K=0, show that both V and / are propagated with velocity <\/(2n/RC).
[The velocity is given by n/f.]
(11) Show that the simultaneous equations
|
kdPdy 5/3. |
fidadR dQ. |
|
|
c dt dy dz ' |
c dt dy dz ' |
|
|
k dQ da dy |
lxd($_dP dR, |
|
|
* |
c dt dz dx ' |
c dt dz dx |
|
kdR d(3 da c dt dx dy ' |
lxdy_dQ dP, c dt dx dy ' |
|
|
are satisfied by |
P=0; |
a=0; |
|
£=0; |
/3 = /30sin y(x-vt) ; |
|
|
R=R0 sin p (x - |
vt); y = 0; |
|
|
provided that v |
= c/Vk/uL and /50 = |
= -VWfx)K |
[These are Maxwell's electromagnetic equations for a dielectric of specific inductive capacity k and permeability /x. P, Q, R are the components of the electric intensity and a, /3, y those of the magnetic intensity, c is the ratio of the electromagnetic to the electrostatic units (which is equal to the velocity of light in free ether). The solution shows that plane electromagnetic waves travel with the velocity c/^/k/u, and that the electric and magnetic intensities are perpendicular to the direction of propagation and to each other.]
dV d2V
(12) Find a solution of -^- — K ^-j such that
F=/=oo if t= +oo ;
F=0 if x=0 or ir, for all values of t ;
V = ttx-x2 if t = 0, for values of x between 0 and ir. [N.B. Before attempting this question read again Arts. 46 and 49. V is the temperature of a non-radiating rod of length ir whose ends are kept at 0°, the temperature of the rod being initially (ttx-x2)° at a distance x from an end.]
(13) What does the solution of the last question become if the length of the rod is I instead of 7r ?
[N.B. Proceed as in Art. 50.]
(14) Solve question (12) if the condition 7 = 0 for x = 0 or ir is
dV replaced by •=— = 0 for x = 0 or ir.
[Instead of the ends being at a constant temperature, they are here treated so that no heat can pass through them.]
(15) Solve question (12) if the expression ttx-x2 is replaced by 100.
60 DIFFERENTIAL EQUATIONS
dV d2V
(16) Find a solution of -=-=K =-— such that
at ox*
T^=oo if t= +00 ;
F = 100 if x=0 or ir for all values of t ; F = 0 if /=0 for all values of x between 0 and x. [Here the initially ice-cold rod has its ends in boiling water.]
(17) Solve question (15) if the length is I instead of ir. If I increases indefinitely, show that the infinite series becomes the integral
200 r 1
7r J0 a
e KaH sin ax da.
[N.B. This is called a Fourier's Integral. To obtain this residt put (2r + l)Tr/l = a and 2x// = ^a.
Kelvin used an integral in his celebrated estimate of the age of the earth from the observed rate of increase of temperature underground. (See example (107) of the miscellaneous set at the end of the book.) Strutt's recent discovery that heat is continually generated within the earth by radio-active processes shows that Kelvin's estimate was too small.]
dV d2V
(18) Find a solution of -=-==K^~y such that
V is finite when t = + co ;
dV ■]
-=-=0 when x = 0, 1 ox Y for all values of t ;
F=0 when x = lj
V=V0 when t = 0, for all values of x between 0 and I.
[If a small test-tube containing a solution of salt is completely
submerged in a very large vessel full of water, the salt diffuses up out
of the test-tube into the water of the large vessel. If VQ is the initial
concentration of the salt and I the length of test-tube it fills, V gives
the concentration at any time at a height x above the bottom of the
dV test-tube. The condition ^— = 0 when x = 0 means that no diffusion
ox
takes place at the closed end. V = 0 when x = l means that at the top
of the test-tube we have nearly pure water.]
(19) Find a solution of ^y = v2^~ such that
y involves x trigonomctrically ;
?/=0 when x — 0 or it, for all values of t ;
dy
~=0 when 1 = 0, for all values of x ;
ot
y = mx between # = 0 and — ,
y = ?n{TT-x) between £=o and 7r,
for nil values of I.
MISCELLANEOUS EXAMPLES 61
[N.B. See the second worked example of Art. 48.
y is the transverse displacement of a string stretched between two points a distance ir apart. The string is plucked aside a distance nnr/2 at its middle point and then released.]
UtU
* (20) Writing the solution of j~ = D2y, where D is a constant, in
the form
d2y d2y deduce the solution of ^ri=-^4 in the form ox2 ol2
y = exDA+e-xDB,
=tHt in the form at2
y=f(t + x) + F(t-x)
by substituting =- for D, f(t) and F(t) for A and B respectively, and
using Taylor's theorem in its symbolical form
f(t + x) = e*Df{t).
[The results obtained by these symbolical methods should be regarded merely as probably correct. Unless they can be verified by other means, a very careful examination of the argument is necessary to see if it can be taken backwards from the result to the differential equation.
Heaviside has used symbolical methods to solve some otherwise insoluble problems. See his Electromagnetic Theory. J
Q/1J
* (21) From the solution of -— = Dhi, where D is a constant, deduce
. . ay o£y . . . that of jr- — %fir in the form
d2f x2 dif
[This is not a solution unless the series is convergent.] Use this form to obtain a solution which is rational, integral, and algebraic of the second degree in t.
d2y d2y *(22) Transform the equation ^j — ^^i by changing the inde- pendent variables x and i to Z and T, where X = x-at; T = x + at. Hence solve the original equation.
*To be omitted on a first reading.
CHAPTER V
EQUATIONS OF THE FIEST ORDER BUT NOT OF THE FIRST DEGREE
51. In this chapter we shall deal with some special typee of equations of the first order and of degree higher than the first for which the solution can sometimes be obtained without the use of infinite series.
These special types are :
(a) Those solvable for p.
(b) Those solvable for y.
(c) Those solvable for x.
52. Equations solvable for p. If we can solve for p, the equation of the nth degree is reduced to n equations of the first degree, to which we apply the methods of Chap. II.
Ex. (i). The equation p2 + px+py + xy = 0 gives p= -x or p= -y ; from which 2y = - x2 + c1 or x = - log y + c2 ;
or, expressed as one equation,
(2y + x2-Cl)(x + \ogy-c2)=0 (1)
At this point we meet with a difficulty ; the complete primitive apparently contains two arbitrary constants, whereas we expect only one, as the equation is of the first order.
But consider the solution
(2y + x2 - c)(x + log y - c) = 0 (2)
If we are considering only one value of each of the constants c, clt and c2, these equations each represent a pair of curves, and of course not the same pair (unless c = c1=c2). But if we consider the infinite set of pairs of curves obtained by giving the constants all possible values from - oo to + go , we shall get the same infinite set when taken altogether, though possibly in a different order. Thus (2) can be taken as the complete primitive.
62
EQUATIONS OF THE FIRST ORDER 63
Ex. (ii). p2+p -2=0.
Here p = 1 or p = - 2,
giving y=x + c1 or y=-2x + c2.
As before, we take the complete primitive as (y-x-c)(y + 2x-c)=0, not (y-x-c1)(y + 2x-c2)=0.
Each of these equations represents all lines parallel either to y=x or to y = -2x.
Examples for solution. ♦/(I) p2 + p-6=0. ^(2) p2 + 2xp = 3x2. v/(3) p2 = a.B
(4) x + yp2=p(l+xy). (5) p3-p(x2 + xy + y2)+xy(x + y)=0.
v-<6) y2- 2p cosh x + 1=0. 53. Equations solvable for y. If the equation is solvable for y, we differentiate the solved form with respect to x. Ex. (i). p2-py + x=>0.
Solving for y, V=P + ~-
„.„ . . dp 1 x dp
Differentiating, P^ir ^ 5 j~ »
° r dx p pl ax
l\dx x
p) dp p2
This is a linear equation of the first order, considering p as the
independent variable. Proceeding as in Art. 19, the student will obtain
x=p(c+cosh~1p)(p2-l) .
00 —It
Hence, as y=p + -, y =* p + (c + cosh^p) (p2 - 1) .
These two equations for x and y in terms of p give the parametric equations of the solution of the differential equation. For any given value of c, to each value of p correspond one definite value of x and one of y, defining a point. As p varies, the point moves, tracing out a curve. In this example we can eliminate p and get the equation con- necting x and y, but for tracing the curve the parametric forms are as good, if not better.
Ex. (ii). 3p5-py + l=0.
Solving for y, y = 3p* + p~l.
Differentiating, p = I2ps-j- - p~2 .~,
i.e. dx = (12p2-p'3)dp.
Integrating, x = 4^3 + \p~2 + c, ~\
and from above, y = 3pi + p~1. J
The student should trace the graph of this for some particular value of c, say c = 0.
64 DIFFERENTIAL EQUATIONS
54. Equations solvable for x. If the equation is solvable for x,
we differentiate the solved form with respect to y, and rewrite -y-
1 in the form - . V
Ex. p2-py + x=0. This was solved in the last article by solving for y.
Solving for x, x=py- p%.
Differentiating with respect to y,
1 dp dp
which is a linear equation of the first order, considering p as the inde- pendent and y as the dependent variable. This may be solved as in Art. 19. The student will obtain the result found in the last article.
Examples for solution.
(1) x = 4p + 4:p3. (2) p2-2xp + l=0.
\ (3) y=p2x + p. (4) y=x+pz.
(5) p3 + p = ev. (6) 2y+p2 + 2p = 2x(p + l).
(7) p3-p (y + 3)+x = 0. (8) y = pBm p + cosp. -
(9) y=p tan 2> + log cos p. (10) ep~y=p2-l.
(12) Prove that all curves of the family given by the solution of Ex. 1 cut the axis of y at right angles. Find the value of c for that curve of the family that goes through the point (0, 1).
Trace this curve on squared paper.
(13) Trace the curve given by the solution of Ex. 9 with c=0. Draw the tangents at the points given by p = 0, p=l, p = 2 and p = S, and verify, by measurement, that the gradients of these tangents are respectively 0, 1, 2 and 3.
CHAPTER VI SINGULAR SOLUTIONS*
55. We know from coordinate geometry that the straight line
y = mx + — touches the parabola y2 = iax, whatever the value of m.
Consider the point of contact P of any particular tangent. At P the tangent and parabola have the same direction, so they have
a common value of -,-, as well as of x and y.
Fig. 7. But for the tangent m=^-=jp say, so the tangent satisfies the
differential equation y=px+~.
Hence the equation holds also for the parabola at P, where x, y, and p are the same as for the tangent. As P may be any point on the parabola, the equation of the parabola y2 = iax must be a solution of the differential equation, as the student will easily verify.
* The arguments of this chapter will be based upon geometrical intuition. The results therefore cannot be considered to be proved, but merely suggested as probably true in certain cases. The analytical theory presents grave difficulties (see M. J. M. Hill, Proc. Loud. Math. Soc, 1918).
P.U.E. t)5 B
66 DIFFERENTIAL EQUATIONS
In general, if we have any singly infinite system of curves which all touch a fixed curve, which we will call their envelope* and if this family represents the complete primitive of a certain differential equation of the first order, then the envelope represents a solution of the differential equation. For at every point of the envelope x, y, and p have the same value for the envelope and the curve of the family that touches it there.
Such a solution is called a Singular Solution. It does not contain any arbitrary constant, and is not deducible from the Complete Primitive by giving a particular value to the arbitrary constant in it.
Example for solution.
Prove that the straight line y = x is the envelope of the family of parabolas y = x + \{x-c)2. Prove that the point of contact is (c, c), and that p — \ for the parabola and envelope at this point. Obtain the differential equation of the family of parabolas in the form y — x + (p- 1)2, and verify that the equation of the envelope satisfies this.
Trace the envelope and a few parabolas of the family, taking c as 0, 1, 2, etc.
56. We shall now consider how to obtain singular solutions. It has been shown that the envelope of the curves represented by the complete primitive gives a singular solution, so we shall commence by examining the method of finding envelopes.
The general method t is to eliminate the parameter c between f(x, y, c) =0, the equation of the family of curves, and
I-
E.g. if f(x,y,c) = 0 is y-cx--=0, (1)
|=0 is - * + l=0 (2)
giving c = ± l/^x.
*In Lamb's Infinitesimal Calculus, 2nd ed., Art. 155, the envelope of a family is defined as the locus of ultimate intersection of consecutive curves of the family. As thus defined it may include node- or cusp loci in addition to or instead of what wo have called envelopes. (We shall give a geometrical reason for this in Art. 56 ; see Lamb for an analytical proof.)
t See Lamb's Infinitesimal Calculus, 2nd ed., Art. 155. If f(x, y, c) is of the form Lci + Mc + N, the result comes to J\12 = 4LN. Tims, for
1 o
y - ex - - = 0,
9 c '
i. e. c2x - cy + 1 = 0, the result is y2 = 4x.
SINGULAR SOLUTIONS
67
Substituting in (1),
y=±2^/x, or y2 = 4:X.
This method is equivalent to finding the locus of intersection of f(x, y, c)=0, and f(x,y,c + h)=0,
two curves of the family with parameters that differ by a small quantity h, and proceeding to the limit when h approaches zero. The result is called the c-discriminant oif(x, y, c) =0.
57. Now consider the diagrams 8, 9, 10, 11. Fig. 8 shows the case where the curves of the family have no special singularity. The locus of the ultimate intersections
Fig 8.
PQRSTUV is a curve which has two points in common with each of the curves of the family (e.g. Q and R lie on the locus and also on the curve marked 2). In the limit the locus PQRSTUV there- fore touches each curve of the family, and is what we have defined as the envelope.
In Fig. 9 each curve of the family has a node. Two con- secutive curves intersect in three points (e.g. curves 2 and 3 in the points P, Q, and R).
The locus of such points consists of three distinct parts EE', AA', and BB' .
When we proceed to the limit, taking the consecutive curves ever closer and closer, A A' and BB' will move up to coincidence with the node-locus iVAT', while EE' will become an envelope. So
68
DIFFERENTIAL EQUATIONS
in this case we expect the c-discriminant to contain the square of the equation of the node-locus, as well as the equation of the envelope.
E!-^£-»vv«i
Fig. 9.
As Fig. 10 shows, the direction of the node-locus NN' at any point P on it is in general not the same as that of either branch of the curve with the node at P. The node-locus has x and y in common with the curve at P, but not p, so the node-locus is not a solution of the differential equation of the curves of the family.
Fig. 10.
If the node shrinks into a cusp, the loci EE' and NN' of Fig. 10 move up to coincidence, forming the cusp-locus CC of Fig. 11. Now NN' was shown to be the coincidence of the two loci AA' and BB' of Fig. 9, so CC is really the coincidence of three loci, and its equation must be expected to occur cubed in the c-discriminant.
Fig. 11 shows that the cusp-locus, like the node-locus, is not (in general) a solution of the differential equation.
K
FIG. 11.
To sum up, we may expect the c-discriminant to contain (i) the envelope, (ii) the node-locus squared, (iii).« the cusp-locus cubed.
SINGULAR SOLUTIONS
69
The envelope is a singular solution, but the node- and cusp- loci are not (in general *) solutions at all.
58. The following examples will illustrate the preceding results :
Ex. (i). y=p2.
The complete primitive is easily found to be iy = (x-c)2, i.e. c2-2cx + x2-±y = 0.
As this is a quadratic in c, we can write down the discriminant at once as (2z)2 = 4(a;2-4*/),
i.e. y = 0, representing the envelope of the family of equal parabolas given by the complete primitive, and occurring to the first degree only, as an envelope should.
y
Flu. 12.
Ex. (ii).
%y = 2px-2
V
Proceeding as in the last chapter, we get
i.e. px2 - 2p2 = (2x* - ipx)
dp dx'
i.e. a;2-2p = 0 or p = 2x
dp dx'
.(A)
dx dp
— =2 > x p
* We say in general, because it is conceivable that in some special example a node- or cusp-locus may coincide with an envelope or with a curve of the family.
70
DIFFERENTIAL EQUATIONS
log x = 2 log p - log c, cx=p2,
1 8
whence 3y = 2c%* - 2c,
i.e. (3?/ + 2c)2 = 402?, a family of semi-cubical parabolas with their cusps
on the axis of y.
The c-discriminant is (3y - x3)2 = 9y2, i.e. x3(6?/-x3)=0.
The cusp-locus appears cubed, and the other factor represents the envelope.
It is easily verified that 6y = x3 is a solution of the differential equation, while x=0 (giving p = oo) is not.
If we take the first alternative of the equations (a),
i.e. x<
2p=0,
we get by substitution for p in the differential equation
3*/ =4*3, i.e. the envelope.
This illustrates another method of finding singular solutions.
Examples for solution.
Find the complete primitives and singular solutions (if any) of the following differential equations. Trace the graphs for Examples 1-4: (1) ±p2-9x = 0. (2) ip2(x-2) = l.
(3) xp2-2yp + 4x = 0. (5) p2 + 2xp-y = 0. (7) ixp2 + iyp - 1 =0.
(4) ?)2 + ?/2-l=0. (6) xp2-2yp + l=0.
SINGULAR SOLUTIONS 71
59. The p-discriminant. We shall now consider how to obtain the singular solutions of a differential equation directly from the equation itself, without having to find the complete primitive.
Consider the equation x2p2 - yp + 1 =0.
If we give x and y any definite numerical values, we get a quad- ratic for p. For example, if
a = v% y=3, 2^2-3^ + l=0, p=\ or 1.
Thus there are two curves of the family satisfying this equation through every point. These two curves will have the same tangent at all points where the equation has equal roots in p, i.e. where the discriminant y2 - 4x2 =0.
Similar conclusions hold for the quadratic Lp2 + Mp+N=0} where L, M, N are any functions of x and y. There are two curves through every point in the plane, but these curves have the same direction at all points on the locus M2 - 4LN =0.
More generally, the differential equation
f(x, y, p) = L0p" +LlP"-i +L2p-2 + ... +Ln =0, where the L's are functions of x and y, gives n values of p for a given pair of values of x and y, corresponding to n curves through any point. Two of these n curves have the same tangent at all points on the locus given by eliminating p from
J{x,y,p)=o,
l -»■
for this is the condition given in books on theory of equations for the existence of a repeated root.
We are thus led to the ^-discriminant, and we must now in- vestigate the properties of the loci represented by it.
60. The Envelope. The ^-discriminant of the equation
1
y=px + ~
or p2x-py + 1=0 f*-
is y2 = ix.
We have already found that the complete primitive consists of the tangents to the parabola, which is the singular solution. Two of these tangents pass through every point P in the plane, and these tangents coincide for points on the envelope.
72
DIFFERENTIAL EQUATIONS
This is an example of the 59-discriminant representing an envelope. Fig. 15 shows a more general case of this.
FIG. 14.
Consider the curve SQP as moving up to coincidence with the curve PRT, always remaining in contact with the envelope QRU. The point P will move up towards R, and the tangents to the two curves through P will finally coincide with each other and with the tangent at the envelope at R. Thus R is a point for which the p'a of the two curves of the system through the point coincide, and consequently the ^-discriminant vanishes.
U
Fig. 15.
Thus the jo-discriminant may be an envelope of the curves of the system, and if so, as shoAvn in Art. 55, is a singular solution.
61. The tac-locus. The envelope is thus the locus of points where two consecutive curves of the family have the same value of p. But it is quite possible for two non-consecutive curves to touch.
Consider a family of circles, all of equal radius, whose centres lie on a straight line.
SINGULAR SOLUTIONS
73
Fig. 16 shows that the line of centres is the locus of the point of contact of pairs of circles. This is called a,, tac-locus. Fig. 17
fig. 16.
shows circles which do not quite touch, but cut in pairs of neigh- bouring points, lying on two neighbouring loci AA' , BB' . When, we proceed to the limiting case of contact these two loci coincide in the tac-locus TT . Thus the ^-discriminant may be expected to contain the equation of the tac-locus squared.
Fig. 17.
It is obvious that at the point P in Fig. 16 the direction of the tac-locus is not the direction of the two circles. Thus the relation between x, y, and p satisfied by the circles will not be satisfied by the tac-locus, which has the same x and y but a different p at P. In general, the tac-locus does not furnish a solution of tlie differential equation.
62. The circles of the last article are represented by (x + c)2+y2=r2, if the line of centres is Ox.
This gives x+c = Vr2 - y2,
or 1= -yp/Vr2-y2,
i.e.
y2p2 + y2
r2=0.
The ^-discriminant of this is y2(y2 -r2) = 0.
The line y=0 (occurring squared, as we expected) is the tac- locus, y=±r are the envelopes EE' and FF' of Fig. 16; y = ±r, giving p=0, are singular solutions of the differential equation, but y = 0 does not satisfy it.
63. The cusp-locus. The contact that gives rise to the equal roots in p may be between two branches of the same curve instead
74
DIFFERENTIAL EQUATIONS
of between two different curves, i.e. the ^-discriminant vanishes at a cusp.
As shown in Fig. 18, the direction of the cusp-locus at any point P on it is in general not the same as that of the tangent to the cusp, so the cusp-locus is not a solution of the differential equation.
C'
Fig. 18.
It is natural to enquire if the equation of the cusp-locus will appear cubed in the p- discriminant, as in the c-discriminant. To decide this, consider the locus of points for which the two p's are nearly but not quite equal, when the curves have very flat nodes. This will be the locus NN' of Fig. 19. In the limit, when the nodes
Fig. 19.
contract into cusps, we get the cusp-locus, and as in this case there is no question of two or more loci coinciding, we expect the p- discriminant to contain the equation of the cusp-locus to the first power only.
64. Summary of results. The ^-discriminant therefore may be
expected to contain
(i) the envelope,
(ii) the tac-locus squared,
4 (iii) the cusp-locus,
and the c-discriminant to contain
(i) the envelope,
(ii) the node-locus squared,
(iii) the cusp-locus cubed.
SINGULAR SOLUTIONS
75
Of these only the envelope is a solution of the differential equation.
65. Examples.
Ex.(i). ?2(2-3<V)2 = 4(l-t/).
Writing this in the form
fa_ 2-3y
dy-±2V(i-y)'
we easily find the complete primitive in the form
(a>-c)8=ya(l-y). The c-discriminant and ^-discriminant are respectively y2(l-y)=0 and (2-3y)2(l -y)=0.
1 - y=0, which occurs in both to the first degree, gives an envelope ; y=0, which occurs squared in the c-discriminant and not at all in the p-discriminant, gives a node-locus ; 2 - Sy = 0, which occurs squared in the ^-discriminant and not at all in the c-discriminant, gives a tac -locus.
It is easily verified that of these three loci only the equation of the envelope satisfies the differential equation.
no. 20.
Ex. (ii). Consider the family of circles
x2 + y2 + 2cx + 2c2-l=0. By eliminating c (by the methods of Chap. I.), we obtain the differ- ential equation
2y2p2 + 2xyp + x2 + y2 - 1 =0.
76
DIFFERENTIAL EQUATIONS
The c- and ^-discriminants are respectively
x2-2(x2 + y2-l)=0 and x2y2-2y2{x2 + y2-l)=0, i.e. x2 + 2y2-2=0 and y2(x2 + 2y2-2)=0.
x2 + 2y2-2=0 gives an envelope as it occurs to the first degree in both discriminants, while y = 0 gives a tac-locus, as it occurs squared in the p-discriminant and not at all in the c-discriminant.
FIG. 2t.
Examples for solution.
In the following examples find the complete primitive if the differ- ential equation is given or the differential equation if the complete primitive is given. Find the singular solutions (if any). Trace the graphs.
(1) ix(x-l){x-2)p2-{3x2-6x + 2)2 = 0. (2) 4*;>2-(3z-l)2=0.
(3) yp2-2xp + y = 0. (4) 3xp2-6yp + x + 2y=0.
(5) p2 + 2px3-ix2y = 0. '6) p3-±xyp + 8y2=0.
(7) x2 + ?/2-2c£ + c2cos2a = 0. (8) c2 + 2cy -x2 + \ = 0.
(9) c2 + (x + y)c + l-xy = 0. (10) x2 + y2 + 2cxy + c2 - 1 =0.
66. Clairaut's Form.* We commenced this chapter by con- sidering the equation
y = px +
V
♦Alexis Claude Clairaut, (if Paris (1713-176,")), although best known in con- nection with differential equations, wrote chiefly on astronomy.
SINGULAR SOLUTIONS
77
This is a particular case of Clairaut's Form
y=px+f(p) (1)
To solve, differentiate with respect to y,-*.
p=p + {x+f'(p)}-£;
therefore
dp
= 0,
(2)
dx~v> P = C or 0=x+f'(p) (3)
Using (1) and (2) we get the complete primitive, the family of straight lines, y = cx+f(c) (4)
If we eliminate p from (1) and (3) we shall simply get the jo-dis- criminant.
To find the c-discriminant we eliminate e from (4) and the result of differentiating (4) partially with respect to c, i.e.
0=x+f(c) (5)
Equations (4) and (5) differ from (1) and (3) only in having c instead of p. The eliminants are therefore the same. Thus both disoriminants must represent the envelope.
Of course it is obvious that a family of straight lines cannot have node-, cusp-, or tac-loci.
Equation (4) gives the important result that the complete primi- tive of a differential equation of Clairaut's Form may be written down immediately by simply writing c in place of p.
67. Example.
Find the curve such that OT varies as tan \fs, where T is the point in which the tangent at any point cuts the axis of x, \Js is its inclination to this axis, and 0 is the origin.
y
O T
I'm. 22.
therefore x-- = kp, •<
78 DIFFERENTIAL EQUATIONS
From the figure, OT = ON-TN
= x-y cot \fr
V V since tani//-=p;
y
V
i.e. y=px-kp2. This is of Clairaut's Form, so the complete primitive is y = cx- kc2, and the singular solution is the discriminant of this, i.e. x2 = iky.
The curve required is the parabola represented by this singular solution. The complete primitive represents the family of straight lines tangent to this parabola.
Examples for solution.
Find the complete primitive and singular solutions of the following differential equations. Trace the graphs for Examples (1), (2), (4), (7), (8) and (9).
7(1) y=px+p2. y(2) y=px + p3.
-/(3) y=px + cosp. (4) y = px + ^{a2p2 + b2).
^(5) p=log(px-y). (6) sinpxeos y = coapxsiny+p.
(7) Find the differential equation of the curve such that the tangent makes with the co-ordinate axes a triangle of constant area k2, and hence find the equation of the curve in integral form.
(8) Find the curve such that the tangent cuts off intercepts from the axes whose sum is constant.
(9) Find the curve such that the part of the tangent intercepted between the axes is of constant length.
MISCELLANEOUS EXAMPLES ON CHAPTER VI.
Illustrate the solutions by a graph whenever possible.
(1) Examine for singular solutions p2 + 2xp = 3x2.
(2) Reduce xyp2-(x2 + y2-l)p + xy = 0
to Clairaut's form by the substitution X — x2 ; Y = y2.
Hence show that the equation represents a family of conies touching the four sides of a square.
MISCELLANEOUS EXAMPLES 79
(3) Show that xyp2 + (x2-y2-h2)p-xy = 0
represents a family of confocal conies, with the foci at (± h, 0), touching the four imaginary lines joining the foci to the circular points at infinity.
(4) Show by geometrical reasoning or otherwise that the sub- stitution x = aX+bY, y = a'X+b'Y,
converts any differential equation of Clairaut's form to another equation of Clairaut's form.
(5) Show that the complete primitive of 8p3x = y(12p2-9) is (x + c)3 = 3y2c, the p-discriminant y2(9x2-iy2)=0, and the c-dis- criminant y*(9x2 - 4ty2) = 0. Interpret these discriminants.
(6) Reduce the differential equation
x2p2 + yp(2x + y)+y2=0, where p = -r-
to Clairaut's form by the substitution £=y, rj = xy.
Hence, or otherwise, solve the equation.
Prove that y + ix = 0 is a singular solution ; and that y=0 is both part of the envelope and part of an ordinary solution. [London.]
(7) Solve y^iy-Xjj^^ij) » which can be transformed to Clairaut's form by suitable substitutions. [London.]
(8) Integrate the differential equations :
(i) 3{p + x)2 = {p-xf. (ii) y2{\ +ip2)- 2pxy -1=0. In (ii) find the singular solution and explain the significance of any factors that occur. [London.]
(9) Show that the curves of the family
y2-2cx2y + c2(xi-x3)=0
all have a cusp at the origin, touching the axis of x.
By eliminating c obtain the differential equation of the family in the form
ip2x2 (x - 1 ) - ipxy (ix - 3y) + ( 1 6a; - 9) y2 = 0.
Show that both discriminants take the form x3y2=0, but that x=0 is not a solution, while y =0 is a particular integral as well as an envelope.
[This example shows that our theory does not apply without modi- fication to families of curves with a cusp at a fixed point.]
(10) Show that the complete primitive of
represents the family of equal lemniscates of Bernoulli
r2 = a2 cos 2(0- a), inscribed in the circle r = a, which is the singular solution, with the point r = 0 as a node-locus.
80 DIFFERENTIAL EQUATIONS
(11) Obtain and interpret the complete primitive and singular solution of /dr\2
&) "
\d6J
(12) Show that r = cd-c2 is the complete primitive and 4r = #2 the singular solution of dr /gr\2
r'eTe-\Te)-
Verify that the singular solution touches the complete primitive at the point (c2, 2c), the common tangent there making an angle tan-1c vith the radius vector.
CHAPTER VII
MISCELLANEOUS METHODS FOR EQUATIONS OF THE SECOND AND HIGHER ORDERS
68. In this chapter we shall be concerned chiefly with the reduction of equations of the second order to those of the first order. We shall show that the order can always be so reduced if the equation
(i) does not contain y explicitly ;
or (ii) does not contain x explicitly ;
or (iii) is homogeneous.
A special form of equation, of some importance in Dynamics, may be reduced by using an integrating factor.
The remainder of the chapter will be devoted to the linear equation, excluding the simple case, already fully discussed in Chapter III., where the coefficients are merely constants. It will be found that the linear equation of the second order can be reduced to one of the first order if
(i) the operator can be factorised, or (ii) any one integral belonging to the complementary function is known.
If the complete complementary function is known, the equation may be solved by the method of Variation of Parameters. This elegant method (due to Lagrange) is applicable to linear equations of any order.
Further information on linear equations, such as the condition for exact equations, the normal form, the invariantive condition of equivalence, and the Schwarzian derivative, will be found in the form of problems among the miscellaneous examples at the end of the chapter, with hints sufficient to enable the student to work them out for himself.
P.D.E. 81 K
82 DIFFERENTIAL EQUATIONS
We shall use suffixes to denote differentiations with respect to
dhi x, e.g. y2 for tt|, but when the independent variable is any other
than x the differential coefficients will be written in full.
69. y absent. If y does not occur explicitly in an equation of
the second order, write p for yx and J- for y2.
We obtain an equation containing only J-, p, and x, and so of the first order. *
Consider, for example, xy2+y1 = ix.
This transforms into Xj-+p=4:X,
which can be integrated at once
xp = 2x2 + a,
i.e. p = 2x + -. x
By integrating, y = x2 + a log x + b,
where a and 6 are arbitrary constants.
This method may be used to reduce an equation of the nth order not containing y explicitly to one of the (n - l)th.
70. x absent. If x is the absent letter, we may still write p for
. , , dp . dp dy dp dp ^
yv but for y2 we now write V-fy, since V dy=TxWy =^ -y«- The
procedure reduces an equation of the second order without x to one of the first order in the variables p and y. For example, W^Vx
transforms into «/])-r= p2,
from which the student will easily obtain
p=by and y = aehx.
Examples for solution.
(1) 2/2cos2x = l. (2) yyz+y^-yv I (3) yy2 + l=yi*.
(4) Reduce to the previous example, and hence solve
(5) xy3 + y 2 = 12x. (6) 2/n - 23/»-i =gX- (7) Integrate and interpret geometrically
(i+yi2)*_7g
EQUATIONS OF SECOND AND HIGHER ORDERS 83
(8) The radius of curvature of a certain curve is equal to the length of the normal between the curve and the axis of x. Prove that the curve is a catenary or a circle, according as it is convex or concave to the axis of x.
(9) Find and solve the differential equation of the curve the length of whose arc, measured from a fixed point A to a variable point P, is proportional to the tangent of the angle between the tangent at P and the axis of x.
*71. Homogeneous equations. If x and y are regarded as of dimension 1,
yx is of dimension 0, y2 is of dimension - 1,
yz is of dimension - 2, and so on.
We define a homogeneous equation as one in which all the terms are of the same dimensions. We have already in Chap. II. dealt with homogeneous equations of the first order and degree, and in Chap. III. with the homogeneous linear equation
xnyn + Axn-iyn_x + Bxn~2yn_2 + ...+ Exyx + Ky=0 (where A, B, ... B, K are merely constants), for which we used the substitution x = (? or t =log x.
Let us make the same substitution in the homogeneous equation xijy2+xy12=3yy1 (1)
Now v Ji%LjL*y
Ul dx dt xdt'
^ dx x2 dt x dx dt
_ 1 dy 1 dt d2y x2 dt x dx dt2
l^dy \_d?y
x2 dt x2 dt2 ' Substituting in (1) and multiplying by x, we get
y\dt2 dt)+\dt) ~6ydt'
d2y fdii\2 . du
This is an equation, with t absent, similar to those in the last article with x absent.
Arts. 71-73 may be omitted on a first reading.
84 DIFFERENTIAL EQUATIONS
By putting -tt = ?> the student will easily obtain yq=2(y2+b), giving t+c = l\og(y2+b).
Hence y2+b=e^t+c^
= ax4, replacing e4c by another arbitrary constant a.
72. The example of Art. 71 came out easily because it had no superfluous a;'s left after associating x2 with y2 and x with yx. In fact, it could have been written
y(*ty«) +(«&)" -3y(a^i).
But (^+^2)(2/-^i)+a;V«/2=0 (2)
cannot be so written. To reduce this to a form similar to that of the last example, put y=vx, a substitution used for homogeneous equations in Chap. II.
(2) becomes
(x2 + x2v2) (vx - vxx2 - vx) + xiv2(xv2 + 2uj) =0, i.e. -(1 + v2)v1+v2(xv2+2v1)=0, a.
which may be written v2x2v2 = (l -v2)xv1 (3)
"We now proceed as before and put x = e\ giving
dv
dH dv dfi'dt'
(3) becomes ^--^=(1-^-,
AH dv ,.v
l-e' VcU2=di> (4)
an equation with t absent.
, , dv d2v dq
As before, put j-j, &-q&
(4) becomes v2q^=q,
i.e. tq = I (unless q = 0, giving y = ex), dv vi
and x2v2=—
dv = =1_1
dt V a v
,, av dv / a2\,
dt = = (o+- av,
v - a \ v-a/
t=av + a2 log (v-a) + b, and finally log x = ay/x + a2 log (y - ax) - a2 log x + b.
EQUATIONS OF SECOND AND HIGHER ORDERS 85
73. By proceeding as in the last article, we can reduce any homogeneous equation of the second order.
Any such equation can be brought to the form
ftyfayi>m/*)-o-
For example, the equation of Art. 71 when divided by x becomes while that of Art. 72 divided by x3 becomes
(i*5)G-*)+©»-*
The substitutions y =vx and x =e* transform
/ (y/x> Vx> xVi) = 0 to / (v, xvx + v, x2v2 + 2xvj) = 0,
j ,r /./ dv d2v dv\ n
and then to /^, _ +v, _. + _j=0,
an equation with t absent, and therefore reducible to the first order.
Examples for solution.
(1) x2y2-xy1 + y = 0. (2) x2y2-xy1 + 5y = 0.
(3) 2x2yy2 + y2 = x2y12.
(i) Make homogeneous by the substitution y = z2, and hence solve
2x2yy2 + iy2 = x2yx2 + 2xyyv
74. An equation occurring in Dynamics. The form y2=f{y) occurs frequently in Dynamics, especially in problems on motion under a force directed to a fixed point and of magnitude depending solely on the distance from that fixed point.
Multiply each side of the equation by 2yv "We get 2y1y2 = 2f(y)yv
Integrating, y2 = 2 J / (y) £dx=2]f (y) dy.
This is really the equation of energy.
d2x dt2
o or Applying the method to -, z- = - p2x, (the equation of simple
harmonic motion), we get
zdtdt2~=~Zpxdt
Integrating with respect to t,
'dx ,2
(dx 2
( j- ) = -f2x2 + const. =p2(a2 -x2), say
86 DIFFERENTIAL EQUATIONS
Hence
<ti=l 1_
dx p \/(a2-x2)'
1 . , x t = - sin-1- + const., p a
x = a sin (pt+e).
Examples for solution.
(1) y2 = y3-y, given that yx=0 when y = \.
(2) y2 = e2y, given that y = 0 and yx = \ when x=0.
(3) y2 = sec2 y tan y, given that ?/=0 and yx = \ when x = 0. "
(l or nd (Lt
(4) tt— -^Tj given that x = A and -r=0 when < = 0. <w2 x2 at
[h - x is the distance fallen from rest under gravity varying inversely as the square of the distance x from the centre of the earth, neglecting air resistance, etc.]
... d2u P . .
(5) ffli + u = fiw> m the two cases
(i)P = Hu2; (\\)P = ijm*;
given that 0 = -771 = 0 when u — ~, where /x, h, and c are constants. (lu c
[These give the path described by a particle attracted to a fixed point with a force varying inversely as the square and cube respectively of the distance r. u is the reciprocal of r, 6 has its ordinary meaning in polar co-ordinates, /u. is the acceleration at unit distance, and h is twice the areal velocity. ]
)\ 75. Factorisation of the operator. The linear equation. (x + 2)y2 - (2x+5)y1 + 2y = (x + l)ex may be written as
{(x + 2)D2 - (2x +5)2) + 2}y = (x + l)ex,
where D stands for -p, as in Chapter III.
Now the operator in this particular example can be factorised, giving {(x+2)D-l}(D-2)y = (x + l)ex.
Put (D-2)y=v.
Then {(x+2)D -l}0-(x + l)&.
This is a linear equation of the first order. Solving as in Art. 20, we get v = c{x+2)+ex,
i.e. (D-2)y = c(x+2)+ex, another linear equation, giving finally
y = a(2x + 5) + be2x - ex, replacing - \c by a.
EQUATIONS OF SECOND AND HIGHER ORDERS 87
Of course it is only in special cases that the operator can be factorised. It is important to notice that these factors must be written in the right order, as they are not commutative. Thus, on reversing the order in this example, we get
(D-2){(x + 2)D-l}y = {(x+2)D2-(2x + 4)D + 2}y.
Examples for solution.
(1) (x + l)y2 + (x-l)yi-2y=0. (2) xy2 + (x-\)yi-y = 0.
(3) xy2 + (x-l)y1-y = x2.
(4) xy2 + (x2 + l)yl + 2xy = 2x, given that y = 2 and yx = 0 when x=0.
(5) (x2 - 1) y2 - (4a;2 - 3x - 5) yx + (4z2 - 6x - 5) y = e2x, given that y = 1 and «/! = 2 when £ = 0.
76. One integral belonging to the complementary function * known.
When one integral of the equation
y2 + Py, + Qy=0 (1)
is known, say y = z, then the more general equation of the second
order y2 + Py1 + Qy=R, (2)
where P, Q, R are functions of x, can be reduced to one of the first order by the substitution « _ VZt
Differentiating, y1 = v±z + vz1}
y2 = v2z+2v1z1+vz2. Hence (2) becomes
t>gz + vx (2z1 +Pz)+v(z2+Pz1+Qz)=0,
i.e. z^+v^z.+Pz)^, (3)
since by hypothesis z2 + Pz1 +Qz=0.
(3) is a linear equation of the first order in vv
Similarly a linear equation of the nth order can be reduced to one of the (n - 1 )th if one integral belonging to the complementary function is known.
77. Example.
Consider again the equation
(x + 2)y2-(2x + 5)y1 + 2y = (x + l)ex (4)
*The proof of Art. 29 that the general solution of a linear differential equation is the sum of a Particular Integral and the Complementary Function holds good when the coefficients are functions of x as well as in the case when they arc constants.
88 DIFFERENTIAL EQUATIONS
If we notice that y = e2x makes the left-hand side of the equation zero, we can put y = ve2x
giving y1 = (v1 + 2v)e2x,
and y2 = (v2 + 4v, + 4v) e2x.
Substitution in (4) gives
(x + 2)v2e2x + {Mx + 2)-(2x + 5)}v1e2x
+{i(x + 2)-2(2x + 5)+2}ve2x = (x + l)ex,
dv i.e. (x + 2)j1 + {2x + 3) v1 = {x + l)e~x
Solving this in the usual way (by finding the integrating factor) we obtain Vi = e~x + c(x + 2) e~2x.
Integrating, v= - e~x - \c{2x + 5) e~2x + b,
whence y = ve2x = - ex - ±c(2x + 5) + be2x.
Examples for solution.
(1) Show that y2 + Pyx+Qy=0 is satisfied by y = ex if 1+P + Q=0, and by y = x if P + Qx = 0.
(2) x2y2 + xy1-y = 8x3.
(3) x2y2-(x2 + 2x)y1 + (x + 2)y = xsex.
(4) xy2-2 (x + l)y1 + (x + 2) y = (x-2)e2x.
(5) x2y2 + xy1-9y = 0, given that y = x* is a solution.
(6) xy2(x cos x-2 sin x) +(x2 + 2)yx sin x-2y (x sin z + cos x) =0, given that y = x2 is a solution.
78. Variation of Parameters. We shall now explain an elegant but somewhat artificial method for finding the complete primitive of a linear equation whose complementary function is known.
Let us illustrate the method by applying it to the example already solved in two different ways, namely,
(x+2)y2-(2x+5)y1+2y = (x + l)e*, ...(1)
of which the complementary function is y = a(2x +5) +be2x.
Assume that y = (2x+5)A +e2xB, (2)
where A and B are functions of x.
This assumption is similar to, but more symmetrical than, that of Art. 77, viz. : y^^x
Differentiating (2),
y1=(2x +5)At +e2xBl +2A +2e2xB (3)
Now so far the two functions (or parameters) A and B are only connected by the single equation (1). We can make them satisfy the additional equation
(2x+5)A1+e2xBl=0 (4)
EQUATIONS OF SECOND AND HIGHER ORDERS 89
(3) will then reduce to
y1=2A+2e2xB (5)
Differentiating (5),
y2=±eZxB + 2A1+2e2xBl (6)
Substitute these values of y, yv and y2 from equations (2), (5), and (6) respectively in (1). The co-factors of A and B come to zero, leaving
2{x+2) Al+2(x + 2)e2xB1=(x + l)ex (7)
(4) and (7) are two simultaneous equations which we can solve for Ax and Bv giving
4l Bi (x + l)ex __ (x + l)erx
e2x~ -(2x +b)~2e2x(x +2)(1 -2x-5)~ 4(z+2)2' w . (x + l)ex _ exj 1 1 1 •
Mence ^i- 4(* + 2)«"~Tls+2 (^T2)2/'
€x
and. by integration, A = - -r-. ^ + a, where a is a constant.
J & 4 (a; +2)
Similarly,
R = (^x+5){x + l)e-x er*_ f _ _1_ _1 I
x~ , 4(z+2)2 "4 I x+2 (ic+2)2/'
and B = ~{-\-2\+b.
4 \x+2 J
Substituting in (2),
y=v*H -i(£2)+a} +?{^2-2l +be"
= a(2x+5)+be2x-ex.
79. Applying these processes to the general linear equation of
the second order, y2 + Py1 + Qy = R, (1)
of which the complementary function au+bv is supposed known,- a and b being arbitrary constants and u and v known functions of xr
we assume that y=uA+vB, (2)
giving y1=ulA+v1B, (3)
provided that uA1+vB1=0 (4)
Differentiating (3),
y2 = u2A + v2B + u1A1+vlB1 (5)
Substitute for y2, yl and y in (1).
The terms involving A will be A(u2 + Pu1 + Qu), i.e. zero, as by hypothesis, Wg + p% + qu = 0.
Similarly the terms involving B vanish, and (1) reduces to
uxAx +v1Bl =R (6)
90 DIFFERENTIAL EQUATIONS Solving (4) and (6), — * = A = ^_
V -U VUX-UVX*
We then get A and B by integration, say
A=f(x)+a, B=F(x)+b, where / (x) and F (x) are known functions of x, and a and 6 are arbitrary constants.
Substituting in (2), we get finally
y = uf(x) + vF (x) + au + bv.
* 80. This method can be extended to linear equations of any order. For that of the third order,
y3 + Py2 + Qy1 + Ry=S, (1)
of which the complementary function y = au +bv+cw is supposed known, the student will easily obtain the equations
y = uA+vB+wC, (2)
y1=u1A +v1B+w1C, (3)
provided that 0=uA1+vB1+ivC1; (4)
hence y2 = u2A+v2B +w2C, (5)
provided that 0=u1A1+v1B1+w1C1 ; (6)
then yz - u3A + v3B + w3C
+ u2A1+v2B1+w2C1; (7)
by substitution in(l), S=v2A1+v2B1+w2C1 (8)
Aly Bly and Cx are then found from the three equations (4), (6) and (8).
Examples for solution.
(1) 2/2 + V = cosec x. (2) y2 + iy = 4 tan 2z.
<3) y*-y=rh'
(4) x2y2 + xyx - y = x2ex, given the complementary function ax + bx*1.
(5) y3-6ij2 + nyi-6y = e2*.
81. Comparison of the different methods for solving linear equations. If it is required to solve a linear equation of the second order and no special method is indicated, it is generally best to try to guess a particular integral belonging to the complementary function and proceed as in Art. 76. This method may be used to reduce a linear equation of the nth order to one of the (n - l)th.
* To be omitted on a first reading.
EQUATIONS OF SECOND AND HIGHER ORDERS 91
The method of factorisation of the operator gives a neat solution in a few cases, but these are usually examples specially constructed for this purpose. In general the operator cannot be factorised.
The method of variation of parameters is inferior in practical value to that of Art. 76, as it requires a complete knowledge of the complementary function instead of only one part of it. Moreover, if applied to equations of the third or higher order, it requires too much labour to solve the simultaneous equations for Ax> Bv Cx, etc., and to perform the integrations.
MISCELLANEOUS EXAMPLES ON CHAPTER VII.
(1) 2/^2 ~2/i2 + 2/i = 0. <2) xy2 + xy12-y1 = 0.
(3) y»*r*y*-v (4) y„ + «/„-2 = 8cos3z.
(5) (x2 log x - x2)y2- xyt + y = 0.
(6) (x2 + 2x -l)y2- (3x2 + 8x -l)y1 + (2x2 + 6x)y = 0.
(7) Verify that cos nx and sin nx are integrating factors of
y2 + n2y=f{x). Hence obtain two first integrals of
y2 + n2y = sec nx, and by elimination of yx deduce the complete primitive.
(8) Show that the linear equation
Ay + By1 + Cy2 + ...+Syn = T,
where A, B, C, ... T are functions of x, is exact, i.e. derivable imme- diately by differentiation from an equation of the next lower order, if the successive differential coefficients of A, B, C, ... satisfy the relation
A-B1 + C2-...+(-l)nSn = 0.
[N.B. — By successive integration by parts,
]Syndx = Syn^1-S1yn_2 + S2yn_3 + ...+(-l)n-1Sn_1y + }(-l)nSnydx.]
Verify that this condition is satisfied by the following equation, and hence solve it :
(2x2 + 3x) y2 + (6x + 3)y1 + 2y = (x + l)ex.
(9) Verify that the following non-linear equations are exact, and solve them: (i) yy2 + yi2 = 0.
(ii) xyy2 + xy12 + tjy1 = 0.
(10) Show that the substitution y = ve •> IX transforms y2 + Py1+Qy = R, where P, Q, and R are functions of x, into the Normal Form
v2 + Iv = S,
92 DIFFERENTIAL EQUATIONS
where I=Q-\PX-\P\
and S = Re^Pdx.
Put into its Normal Form, and hence solve
y2-ixy1 + (4:X2-l)y= - 3ex* sin 2x.
(11) Show that if the two equations
and z2 + pz1 + qz = 0
reduce to the same Normal Form, they may be transformed into
each other by the relation
\[pdx \\pdx
i.e. the condition of equivalence is that the Invariant I should be the same.
(12) Show that the equations
x2y2 + 2(xz-x)y1 + (l-2x2)y=0 and x2z2 + 2(x3 + x)zl-{l-2x2)z=0
have the same invariant, and find the relation that transforms one into the other. Verify by actually carrying out this transformation.
(13) If u and su are any two solutions of
v2 + lv = 0, (1)
prove that ^=_2^1, (2)
s1 u
and hence that ^-^(-2Y = 2/ (3)
From (2) show that if s is any solution of (3), s^ and ss^ are solutions of (1).
[The function of the differential coefficients of s on the left-hand side of (3) is called the Schivarzian Derivative (after H. A. Schwarz of Berlin) and written {s, x}. It is of importance in the theory of the Hypergeometric Series.]
(14) Calculate the Invariant / of the equation
x2y2 - (x2 + 2x)yx + (x + 2)y=Q. Taking s as the quotient of the two solutions xex and x, verify that
{s,x} = 21,
and that sx and ssf* are solutions of the Normal Form of the original equation.
(15) If u and v are two solutions of
y2+Pyi+Qy=o,
prove that uv2 - vu2 + P{uv1 - vuj) = 0,
and hence that uv1-vu1 — ae * .
Verify this for the equation of the last example.
MISCELLANEOUS EXAMPLES 93
(16) Show that yyx = const, is a first integral of the equation formed by omitting the last term of
By putting yyx = C, where C is now a function of x (in fact, varying the parameter C), show that if y is a solution of the full equation, then
c[=-y2, and hence C2 = const. - \y*,
giving finally y2 = a sin {x\J1 + b).
[This method applies to any equation of the form
*■+*■/(*) +*(y)-o.]
(17) Solve the following equations by changing the independent variable :
d2ti Q'V
(ii) (l+*«)»g + 2»(l+^+4y-a
(18) Transform the differential equation
j~2 cos x + ~j~ sm x ~ %y cos3 X = ^ c°s5 £
into one having z as independent variable, where
z = sin x, and solve the equation. [London.]
(19) Show that if z satisfies
^ + P-=0, dx2 dx
by changing the independent variable from x to z, we shall transform
into d4 + Sy=T'
Hence solve 0 + (l -^)^ + ix2ye-2x = 4:(x2 + x3)e-Zx.
CHAPTER VIII
NUMERICAL APPROXIMATIONS TO THE SOLUTION OF DIFFERENTIAL EQUATIONS
82. The student will have noticed that the methods given in the preceding chapters for obtaining solutions in finite form only apply to certain special types of differential equations. If an equation does not belong to one of these special types, we have to use approxi- mate methods. The graphical method of Dr. Brodetsky, given in Chapter L, gives a good general idea of the nature of the solutionr but it cannot be relied upon for numerical values.
In this chapter we shall first give Picard's * method for getting successive algebraic approximations. By putting numbers in these, we generally get excellent numerical results. Unfortunately the method can only be applied to a limited class of equations, in which the successive integrations can be easily performed.
The second method, which is entirely numerical and of much more general application, is due to Runge.f With proper pre- cautions it gives good results in most cases, although occasionally it may involve a very large amount of arithmetical calculation. "We shall treat several examples by both methods to enable their merits to be compared.
Variations of Runge's method have been given by Heun, Kutta, and the present writer.
83. Picard's method of integrating successive approximations. The
differential equation dv
* E. Picard, Professor at the University of Paris, is one of the most distinguished mathematicians of to-day. He is well known for his researches on the Theory of Functions, and his Cours (Tanalyse is a standard text-book.
t C. Runge, Professor at the University of Gottingen, is an authority on graphical methods.
94
NUMERICAL APPROXIMATIONS 9fr
where y =6 when x-a, can be written
*/ = & + [ f(x,y)dx.
For a first approximation we replace the y inf(x, y) by b ; for a second we replace it by the first approximation, for a third by the second, and so on.
du Ex. (i). J?- = x + y2, where y=0 when x=0.
Here y = I (x + y2) dx.
Jo
First approximation. Put ?/=0 in x + y2, giving
y=\ xdx = \x2. Jo
Second approximation. Put ?/ = \x2 va.x + y2, giving
r
y = (» + iz4) (far = \x2 + z\x5. Jo
Third approximation. Put i/ = \x2 + -^V^5 in cc + «/2, giving
r
1/ = (a? + \& + -itx1 + Thvx10) dx Jo
— ir2 i 1 r5 i 1 y8 i l yll
— 2X ^W*' + TTRr,c +TTTTD"X >
and so on indefinitely.
where t/ = l and z = £ when cc=0.
Here ?/ = l + l 2^ and z = i+l x3(y + z)dx.
Jo Jo
First approximation.
y=*l + I -^cfa; = l+£x,
Jo"
z = \ + [V(l +|) cfc^ + fz4. Jo
Second approximation.
y = l + [ (% + %xi)dx = l+±x + Ts1Tx5,
■96 DIFFERENTIAL EQUATIONS
Third approximation.
y - 1 + f (£ + f x* + -As5 + frx*) dx Jo
- 1 + \x + ^z5 + isVz6 + Tfox9,
z =4 + r a»(4 + \x + f z4 + ^r5 + -fox8) dx Jo
and so on.
Ex. (iii). j^ = x3(^- + y), where */ = l and -j-=\ when x=0.
By putting j-~z, we reduce this to Ex. (ii).
It may be remarked that Picard's method converts the differential equation into an equation involving integrals, which is called an Integral Equation.
Examples for solution.
Find the third approximation in the following cases. For examples (1) and (2) obtain also the exact solution by the usual methods.
\/ (1) ~ = 2y - 2x2 - 3, where y = 2 when x = 0.
CLOC
(2) -~ = 2--, where v = 2 when x = l. dx x
ft-"*-.
where y = 2 and z=0 when x = 0.
(dy
(4)
dx~Z'
dz a .
-1-=x2z + x*y, dx 9
where y = 5 and 2 = 1 when x=0.
(5) y = xzJ-+tfy where y = 5 and / = 1 when x=0. v ' dx2 dx * * dx
84. Determination of numerical values from these approximations.
Suppose that in Ex. (i) of the last article we desire the value of y, correct to seven places of decimals, when x =0-3.
Substituting x =0-3, we get \ (0-3)2 =0-045 from the first approxi- mation.
The second adds ^(0-3)5 =0-0001215, while the third adds ^(O^)8 +ttVtf(0-3)u =0-00000041 ... .
NUMERICAL APPROXIMATIONS
97
Noticing the rapid way in which these successive increments decrease, we conclude that the next one will not affect the first seven decimal places, so the required value is 0-0451219... .
Of course for larger values of x we should have to take more than three approximations to get the result to the required degree of accuracy.
We shall prove in Chap. X. that under certain conditions the approximations obtained really do tend to a limit, and that this limit gives the solution. This is called an Existence Theorem.
Example for solution.
(i) Show that in Ex. (ii) of Art. 83, x = 0-5 gives y = 1-252... and 2=0-526... , while z=0-2 gives y = 1-100025... and 2 = 0-500632... .
85. Numerical approximation direct from the differential equation.
The method of integrating successive approximations breaks down if, as is often the case, the integrations are impracticable. But there are other methods which can always be applied. Consider the problem geometrically. The differential equation
dy
dx
=/fo y)
determines a family of curves (the " characteristics ") which do not intersect each other and of which one passes through every point
Fig. 23.
in the plane.* Given a point P (a, b), we know that the gradient of the characteristic through P is f (a, b), and we want to determine
* This is on the assumption that f(x, y) has a perfectly definite value for every point in the plane. If, however, f(x, y) becomes indeterminate for one or more points, these points are called singular points of the equation, and the behaviour of the characteristics near such points calls for special investigation. See Art. 10.
98 DIFFERENTIAL EQUATIONS
the y =NQ of any other point on the same characteristic, given that x = ON = a + h, say. A first approximation is given by taking the tangent PR instead of the characteristic PQ, i.e. taking
y =2VX +LR=NL+PL tan /_RPL = b+hf{a, b) =b + hf0, say.
But unless h is very small indeed, the error RQ is far from negligible.
A more reasonable approximation is to take the chord PQ as parallel to the tangent to the characteristic through S, the middle point of PR.
Since <# is (a + \h, b + %hf0), this gives
y =NL +LQ=NL + PL tan /_QPL =b+hf{a + \h, b +ffi0).
This simple formula gives good results in some cases, as will be seen from the following examples :
Ex. (i) -^- = x + y2; given that y=0 when x = 0, required y when a; = 0-3. dx
Here a = 6 = 0, A = 0-3, f(x,y)=x + y*.
Therefore
/o=/(a,6)=0, a + £A = 0-15, 6 + P/0 = 0,
giving b + hf{a + \h, 6 + ^/0)=0 + 0-3 x/(0-15, 0) =0-045.
The value found in Art. 84 was 0-0451219... , so the error is 0-00012... , about J per cent.
Ex. (ii). -j- = 2 - - ; given that y — 2 when x = l, find y when x = 1 -2. Here o-l, 6 = 2, A=0-2, /0=2-f=0.
Therefore b + hf(a + $h, 6 + ^/0) = 2 + 0-2 x/(M, 2)
= 2+0-2 x (2--^-) =2-036.... Now the differential equation is easily integrable, giving y = x + -,
so when a? = 1-2 the value of y is 2-033... . The error is 0-003... , which is rather large compared with the increment of y, namely 0-036... .
Ex. (iii). -g = «=/(&, y, z), say,
-^ = x3(y + z)~g(x, y, 2), say;
given that y = \ and 2 = 0-5 when x = 0, find y and 2 when x = 0-5. Here a=0, 6 = 1, c (the initial value of 2) =0-5, A=0-5. Hence /0=/(0, 1, 0-5) =0-5 ; g0=g<0, 1, 0-5)=0.
NUMERICAL APPROXIMATIONS 99
By an obvious extension of the method for two variables, we take
= b + hf{a + \h, b + \hf0, c + ^0) = l+0-5x/(0-25, 1-125, 0-5) = l-2500,
and z = c + hg(a + $h, b + Wo> c + ihgQ)
=0-5 +0-5x0(0-25, 1-125, 0-5) =0-5127.
The accurate values, found as in Art. 84, are
y = 1-252... and 2=0-526....
Thus we have obtained a fairly good result for y, but a verv bad one for z. J
k TheilunJcer*aiuty ab°ut the degree of accuracy of the result deprives the method of most of its value. However, it forms an introduction to articb°re e meth°d 0f RunSe> t0 be explained in the next
Examples for solution.
.-. dy i
{) dx = {x ~y) ~1; ^ven that y = 4 when x = 2-3, obtain the value y = 4-122 when z = 2-7. [Runge's method gives 4-118.]
in\ dy 1 $
{) di^lO™ -1+lo&(x + y)}'> given that y = 2 when x=-l, obtain ;he value y = 2-194 when a-1. [Runge's method gives 2-192.]
ti\ dy y •
{) ti '< x' glventhat2/ = 2whenz = l, obtain the valuey = 2-076
X ion2:. .^ show that y=lx2+i> so that when -1* V ™
86. Runge's method. Suppose that the function of y defined * by
~dx ~$ ^' $' y=b when x = a>
3 denoted by y=F(x).
If this can be expanded by Taylor's theorem,
F(a + h) =F{a) + hF{a) +^F"(a) +~F'"(a) + ... .
Now F\x)=%=f{x,y)=f,^y.
We shall now take the total differential coefficient with respect > x (that is, taking the y in/ to vary in consequence of the variation i x). ^ Let us denote partial differential coefficients by
p-f?, ,-§?, rJ-f s-^f t-d2f- to' * dy' dx*' S~dxdy' t dtf>
id their values when x = a and y =b by Po, qQ, etc.
io^IinodZ.Tt' ThlCh th« diffe^«al equation and the initial con- ■«♦ «?lu 7i * 1 • f functl0n are discussed in Chap. X. The graphical treat nt of the last article assumes that these conditions are satisfied graptU°al treat"
100 DIFFERENTIAL EQUATIONS
Similarly, *»<(*) = (|+| J) (,+,,)
= r+^+/s+/(s+?2+/|F). Thus JP(a+A)-.F(a)
= fc/0 + JA8(p0 +/o?o) +#"(*• +2/>o +/o% +M> +/o?o2) + ••• • (1) The first term represents the first approximation mentioned and rejected in Art. 85.
The second approximation of Art. 85, i.e.
y-b=hf{a+\h,b + \hfQ) = k1} say, may now be expanded and compared with (1).
Now, by Taylor's theorem for two independent variables, f(a+lh,b+±hf0) x
=/o + 1% + Wtfo + 2! (^2/"o + hh2Uo + i Wo) + • • • >
giving A;1=A/0+^2(^o+/o?o)+^3K+2/oSo4o) + (2)
It is obvious that Jcx is at fault in the coefficient of h3.
Our next step is suggested by the usual methods * for the
numerical integration of the simpler differential equation
!=/(*>■ ■
Our second approximation in this case reduces to the Trapezoidal Rule y-b=hf{a+\h).
Now the next approximation discussed is generally Simpson's Rule, which may be written
y-&-i*{/(a)+4/(a+#)+/(a + *)}. If we expand the corresponding formula in two variables, namely \Hh + tfiP+\K b+lhf^+fia+h, b + hf0)}, we easily obtain
¥o + ¥i2(Po+foqo)+hW(ro + 2foSo + t0) + ---, (3)
which is a better approximation than kv but even now has not the coefficient of h3 quite in agreement with (1).
To obtain the extra terms in h3, Runge f replaces hf(a+h,b+hf0)
* See the text- books on Calculus by Gibson or Lamb, f Mathematische Annalen, Vol. XL VI. pp. 167-178.
NUMERICAL APPROXIMATIONS 101
by V" = hf(a +h,b+ V), where k" = hf(a + h,b+ hf0). The modified formula maybe briefly written ^{k' +ikl+k'"}, where k' =hf0, or f kx + ^k2 = kx + 1 (k2 - kj), where k2 = \(k' +k'").
vThe student will easily verify that the expansion of Runge's formula agrees with the right-hand side of (1) as far as the terms in h, h2, and h3 are concerned.
Of course this method will give bad results if the series (1) con- verges slowly.
If yo > 1 numerically, we rewrite our equation
i=KhrF(x'y)-m'
and now F0< 1 numerically, and we take y as the independent variable.
87. Method of solving examples by Runge's rule. To avoid confusion, the calculations should be formed in some definite order, such as the following :
Calculate successively k' =hf0,
k"=hf{a + h,b+k'), k"'=hf(a + h,b+k"), k^hfia + ih, 6+P'), k2 = 2\k +k ), and finally k = k1+^(k2-k1).
Moreover, as kx is itself an approximation to the value required, it is clear that if the difference between k and kv namely I (k2 - kx), is small compared with k± and k, the error in k is likely to be even smaller.
dy Ex. (i). j-=x + y2; given that y=0 when x = 0, find y when x = 0-3.
Here a = 0, 6 = 0, A = 0-3, f(x,y)=x + y2, /0 = 0; tf-Vo-0;
k" = hf{a + h, & + A/)=0-3x/(0-3, 0)=0-3x0-3 =0-0900;
k'" = hf(a + h, b + k") =0-3 x/(0-3, 0-09) =0-3 x (0-3 + 0-0081) =0-0924 ;
kx-hf{a + \h, 6 + P') =0-3 x/(0-15,0) =0-3x0-15 =0-0450;
k2 = \(k' + k'") = \x 0-0924 =0-0462;
and
k = kx +£(&2-fc1) =0-0450 + 0-0004 =0-0454.
As the difference between k = 0-0454 and kx = 0-0450 is fairly small compared with either, it is highly probable that the error in k is less
102 DIFFERENTIAL EQUATIONS
than this difference 0-0004. That is to say, we conclude that the value ( is 0-045, correct to the third place of decimals.
We can test this conclusion by comparing the result obtained in Art. 84, viz. 0-0451219... .
fjbti II — nt*
Ex. (ii). j- ; given that y = \ when sc = 0, find y when x = l.
(too y *t" x
This is an example given in Eunge's original paper. Divide the range into three parts, 0 to 0-2, 0-2 to 0-5, 0-5 to 1. We take a small increment for the first step because / (x, y) is largest at the beginning. First step. a = 0, 6 = 1, /* = 0-2, /0 = 1 ;
k' = hf0 =0-200
k" = hf(a + h, b + k') =0-2 xf (0-2, 1-2) =0-143
k'" = hf{a + h, b + k")=0-2xf(0-2, 1-143) =0-140
kx = hf(a + \h,b + \k')=0-Zxf{0-\, 1-1) =0-167
k2 = ^{k' + k'") = ^x 0-340 =0-170
and k^^ + H^-kj) =0-167 +0-001 =0-168
giving i/ = l-168 when z = 0-2.
Second step.
a = 0-2, 6 = 1-168, 7*=0-3, /0=/(0-2, 1-168) =0-708. Proceeding as before we get i1 = 0-170, &2 = 0-173 and so £ = 0-171, giving y = 1-168 +0-171 = 1-339 when x = 0-5.
Third step. a = 0-5, 6 = 1-339, 7j = 0-5.
We find kt = k2 = k = 0-160, giving i/ = l-499 when x = l. Considering the k and kv the error in this result should be less than 0-001 on each of the first and second steps and negligible (to 3 decimal places) on the third, that is, less than 0-002 altogether.
As a matter of fact, the true value of y is between 1 -498 and 1 -499, so the error is less than 0-001. This value of y is found by integrating the equation, leading to
7T - 2 tan-1 ^ = loge(x2 + y2).
Examples for solution.
Give numerical results to the following examples to as many places of decimals as are likely to be accurate :
(1) y = 77;{y -1 +l°ge(3 + 2/)} ; given that y = 2 when x= -1, find
y when x = l, taking h = 2 (as /is very small). ,
(2) Obtain a closer approximation to the preceding question by taking two steps.
(3) ~ = (x2-y) -1 ; given that y = i when x = 2 -3, find y when x = 2 -7 (a) in one step, (6) in two steps.
NUMERICAL APPROXIMATIONS 103
dtf V
(4) Show that if -^=2-- and y = 2 when as—1, then y = x + -.
dx X X
Hence find the errors in the result given by Runge's method, taking (a) ^=0-4, (b) h=0-2, (c) h=0-l (a single step in each case), and compare these errors with their estimated upper limits.
(5) If E(h) is the error of the result of solving a differential equation of the first order by Runge's method, prove that
Lt EW _ I
h.+o E{nh) n*'
Hence show that the error in a two-step solution should be about ■§■ of that given by one step ; that is to say, we get the answer correct to an extra place of decimals (roughly) by doubling the number of steps.
88. Extension * to simultaneous equations. The method is easily extended to simultaneous equations. As the proof is very similar to the work in Art. 86, though rather lengthy, we shall merely give an example. This example and those given for solution are taken, with slight modifications, from Runge's paper.
Ex. £ = 22-|=/(x, y, z), say,
dz y .
given that y= 0-2027 and 2 = 1-0202 when x = 0-2, find y and z when z = 0-4. Here o = 0-2, 6=0-2027, c = 1-0202, /„=/ (0-2, 0-2027, 1-0202) = 1-027, a0 = 0-2070, A = 0-2; k' = hf 0 = 0-2x 1-027 =0-2054
Z' = ^0 = 0-2x 0-2070 =0-0414
k" = hf(a + h, b + k', c + V) =0-2 x/(0-4, 0-4081, 1-0616) =0-2206 l" = hg(a + h, b + k', c + l') =0-2 xo(0-4, 0-4081, 1-0616) =0-0894 k'" = hf(a + h, b + k", c + O=0-2x/(0-4, 0-4233, 1-1096) =0-2322 l'" = hg(a + h, b + k", c + l") =0-2 xo(0-4, 0-4233, 1-1096) =0-0934 kj_ = hf(a + \h, b + p', c + \V) =0-2 x/(0-3, 0-3054, 1 -0409) =0-2128 l^hfia + lh, b + \lc', c + \l') =0-2xo(0-3, 0-3054, 1-0409) =0-0641 k2 = $(k' + k'") =0-2188
12 = \{V + V") =0-0674
k = k1 + ^(k2-k1)= 0-2128 + 0-0020 =0-2148
l = ^ + l(l2 -Ji) =0-0641 +0-0011 =0-0652
giving y = 0-2027 +0-21 48 = 0-41 75
and 2 = 1 -0202 + 0 -0652 = 1 -0854,
probably correct to the third place of decimals.
*The rest of this chapter may be omitted on a first reading.
104 DIFFERENTIAL EQUATIONS
Examples for solution.
(1) With the equation of Art. 88, show that if y =04175 and 2 = 1-0854 when x =0-4, then y =0-6614 and 2 = 1-2145 (probably correct to the third place of decimals) when x = 0-6.
... dw _ V(l-w2) dr w . ,
(2) — =-2« + -^ '-; j-^—^z rx ; given that w =0-7500
dz r dz v(l ~w )
and r=0-6 when 2 = 1-2145, obtain the values w =0-5163 and r =0-7348
when 2 (which is to be taken as the independent variable) = 1-3745.
Show that the value of r is probably correct to four decimal places, but
that the third place in the value of iv may be in error.
(3) By putting w = cos <f> in the last example and t/ = sin <p, x = r in the example of Art. 88, obtain in each case the equations
dz _ sin d> d<h
^- = tanrf>; 22 = - + cos<b-r-,
dr r r T dr
which give the form of a drop of water resting on a horizontal plane.
89. Methods* of Heun and Kutta. These methods are very similar to those of Runge, so we shall state them very briefly. The
problem is : given that -~ =f(x, y) and y=b when x = a, to find
the increment h of y when the increment of x is h. Heun calculates successively
k'=hf(a,b), k"=hf(a+ih,b+lk'), Jc'"=hf(a + %h,b+ik"), and then takes l(k' +SJc'") as the approximate value of h. Kutta calculates successively, k'=hf(a,b), Jc"=hf{a+ih,b+ik'), k'"=hf(a+lh,b+k" -\k'), k""=hf(a+h,b+k"'-k"+k'), and then takes |(&' +3k" +3k'" +k"") as the approximate value oik.
The approximations can be verified by expansion in a Taylor's series, as in Runge's case.
Example for solution.
n/'U <u nr
Given that -f-=- and y — \ when x=0, find the value of y (to 8
significant figures) when x = 1-2 by the methods of Runge, Heun, and Kutta, and compare them with the accurate value 1-1678417. [From Kutta 's paper. ]
* Zeitschrift fur Mathematik und Physik, Vols. 45 and 46.
NUMERICAL APPROXIMATIONS
105
90. Another method, with limits for the error. The present writer has found * four formulae which give four numbers, between the greatest and least of which the required increment of y must lie. A new approximate formula can be derived from these. When applied to Runge's example, this new formula gives more accurate results than any previous method.
The method is an extension of the following well-known results concerning definite integrals.
91. Limits between which the value of a definite integral lies. Let
F(x) be a function which, together with its first and second differential coefficients, is continuous (and therefore finite) between x=a and x=a + h. Let F"{x) be of constant sign in the interval. In the figure this sign is taken as positive, making the curve concave upwards. LP, MQ, NR are parallel to the axis of y, M is the middle point of LN, and SQT is the tangent at Q. OL=a, LN = h.
|
3 |
|||
|
R |
|||
|
s |
~-^^Q |
||
|
T |
M
FIG. 24.
Then the area PLNR lies between that of the trapezium SLNT and the sum of the areas of the trapezia PLMQ, QMNR.
That
Ca+k
is, F(x)di
J a
ix lies between
hF{a + \h)=A, say, and lh{F(a)+2F(a + lh)+F(a+h)}=B, say.
In the figure F"(x) is positive and A is the lower limit, B the upper. If F"(x) were negative, A would be the upper limit and B the lower.
* Phil. Mag., June 1919. Most of this paper is reproduced here.
106 DIFFERENTIAL EQUATIONS
As an approximation to the value of the integral it is best to take, not the arithmetic mean of A and B, but %B+^A, which is exact when PQR is an arc of a parabola with its axis parallel to the axis of x. It is also exact for the more general case when
F (x) = a + bx + ex2 + ex3, as is proved in most treatises on the Calculus in their discussion of Simpson's Rule.
92. Extension of preceding results to functions denned by differential equations. Consider the function denned by
^ =»/(*» y)> y=b wnen * = a ;
where f(x, y) is subject to the following limitations in the range of values a to a + h for x and 6 - h to b + h for y. It will be seen from what follows below that the increment of y is numerically less than h, so that all values of y will fall in the above range. The limitations are :
(1) f(x, y) is finite and continuous, as are also its first and second partial differential coefficients.
(2) It never numerically exceeds unity. If this condition is not satisfied, we can generally get a new equation in which it is satisfied by taking y instead of x as the independent variable.
(3) Neither cPy/dx3 nor df/dy changes sign.
" Let m and M be any two numbers, such that
Then if the values of y when x is a +\h and a + h are denoted by b +j and b + h respectively,*
-^h^lmh<j<^Mh^ih, (1)
and -&= mh<k<Mh^h (2)
We shall now apply the formulae of the last article, taking y to be the same function as that defined by
Ca+x
y = b + \ F(x)dx,
J a
Ca+h
so that &=| F(x)dx.
We have to express the formulae in terms of / instead of F. Now, F(a) =the value of dy/dx when x = a, so that F(a)=f(a, b).
*The following inequalities hold only if h is positive. If h is negative, they must be modified, but the final result stated at the end of this article is still true.
NUMERICAL APPROXIMATIONS 107
Similarly, F(a + \h) =f(a + \h, b +j),
and F(a + h)=f(a+h, b + k).
Now, if df/dy is positive, so that/ increases with y, the inequalities <1) and (2) lead to
f{a+\h, b+\mh)<f(a + \h, b+j)<f(a+\h, b+\Mh), (3)
and f(a+h,b+mh)<f(a + h,b+k)<f(a + h,b+Mh); (4)
while if df/dy is negative,
/(a + \h, b + \mh) >f(a + %h, b +j)>f(a + \h, b + \Mh), ... (5)
and f(a+h,b+mh)>f(a+h,b+k)>f(a + h,b+Mh) (6)
Thus if F" (x) — dPy/dx3 is positive and df/dy is also positive, the result of Art. 91,
A<k<B,
may be replaced by p<k<Q, (7)
where p = hf(a + \h, b + \mh)
and Q = \h{f{a, b) + 2f(a + ±h,b+ \Mh) +f{a + h,b+Mh)};
while if F" (x) is positive, and df/dy is negative,
P<k<q, (8)
where P = hf(a + \h, b + \Mh)
and q = \h{f{a, b) + 2f(a + hh, b + ^mh) +f(a + h, b+ mh)}. Similarly, if F" (x) and df/dy are both negative,
p>k>Q, (9)
while if F" {x) is negative and df/dy positive,
P>k>q (10)
These results may be summed up by saying that in every case
{subject to the limitations on/ stated at the beginning of this article)
k lies between the greatest and least of the four numbers p, P, q, and Q.
As an approximate formula we use k = %B+^A, replacing B by
Q or q, and A by p or P.
93. Application to a numerical example. Consider the example selected by Runge and Kutta to illustrate their methods,
-t~=- ; w = lwhena;=0.
ax y + x a
It is required to find the increment k of y when x increases by 0-2. Here f(x, y)=(y -x)/(y +x). This function satisfies the con- ditions laid down in the last article.*
WetakeM = l, m = (l-0-2)/(l-2+0-2)=4/7.
* As f (x, y) is positive, y lies between 1 and 1-2. When finding M and m we always take the smallest range for y that we can find
108 DIFFERENTIAL EQUATIONS
Then p =0-1654321,
P =0-1666667, 9=0-1674987,
£=0-1690476. Thus h lies between p and Q. Errors.
|Q + *p =0-1678424, 00000007
Kutta's value 0-1678449, 0-0000032
Kunge's value 0-1678487, 0-0000070
Heun's value 0-1680250, 0-0001833
The second, third, and fourth of these were calculated by Kutta. Now this particular example admits of integration in finite terms, giving
log {x2 +y2)-2 tan-1 (x/y) =0.
Hence we may find the accurate value of k. Accurate value =0-1678417. Thus in this example our result is the nearest to the accurate value, the errors being as stated above.
We may also test the method by taking a larger interval h = \. Of course a more accurate way of obtaining the result would be to take several steps, say ^=0-2, 0-3, and finally 0-5, as Runge does.
Still, it is interesting to see how far wrong the results come for the larger interval.
We take M = 1, m = (l -l)/(2 +1) =0.
Then iQ+iP =0-50000.
True value =0-49828, Errors.
Kutta's value =0-49914, 0-00086
Our value =0-50000, 0-00172
Heun's value =0-51613, 0-01785
Runge's value =0-52381, 0-02553
This time Kutta's value is the nearest, and ours is second.
CHAPTER IX
SOLUTION IN SERIES. METHOD OF FROBENIUS
94. In Chapter VII. we obtained the solution of several equations of the form d2y ndy n A
where P and Q were functions of x.
In every case the solution was of the form y=af(x)+bF(x), where a and b were arbitrary constants.
The functions f(x) and F(x) were generally made up of integral or fractional powers of x, sines and cosines, exponentials, and logarithms, such as
(1 +2x)ex, sin x + x cos x, x*+x , x+\ogx, e':.
The first and second of these functions can be expanded by Maclaurin's theorem in ascending integral powers of x ; the others cannot, though the last can be expanded in terms of 1/x.
In the present chapter, following F. G. Frobenius,* of Berlin, we shall assume as a trial solution
y = x° (a0 + axx + ag? + ... to inf.), where the a's are constants. f
The index c will be determined by a quadratic equation called the Indicial Equation. The roots of this equation may be equal, different and differing by an integer, or different and differing by a quantity not an integer. These cases will have to be discussed separately.
The special merit of the form of trial solution used by Frobenius is that it leads at once to another form of solution, involving log x, when the differential equation has this second form of solution.
* Crelle, Vol. LXXVL, 1873, pp. 214-224.
t In this chapter suffixes will not be used to denote differentiation.
109
110 DIFFERENTIAL EQUATIONS
1
As such a function as ex cannot be expanded in ascending powers of x, we must expect the method to fail for differential equations having solutions of this nature. A method will be pointed out by which can be determined at once which equations have solutions of Frobenius' forms (regular integrals) and for what range of values of x these solutions will be convergent.
The object of the present chapter is to indicate how to deal with examples. The formal proofs of the theorems suggested will be given in the next chapter.
Among the examples will be found the important equations of Bess*l,* Legendre, and Riccati. A sketch is also given of the Hyper- geometric or Gaussian equation and its twenty-four solutions.
95. Case I. Roots of Indicial Equation unequal and differing by a quantity not an integer. Consider the equation
v*+^&-d£-6*y=° w
Put z = Xs (a0 + axx + ag? + ...), where a0=l=0, giving f
dz
j- =a0cxc-x + ax(c + l)x° + a2(c + 2)x°+1 + ... ,
d2z and j^ = a0c(c -\)x°-* -i-a^c + l)cxc~1 +a2(c +2)(c + l)z? + ... .
Substitute in (1), and equate the coefficients of the successive powers of x to zero.
The lowest power of x is a?-1. Its coefficient equated to zero gives
a0{2c(c-l)-c}=0,
i.e. c(2c-3)=0, : (2)
as a0 =f= 0.
* Friedrich Wilhelm Bessel, of Minden (1784-1846), was director of the obser-