AN ABSTRACT OF THE THESIS OF
JACKSON HENRY BELLOfor the MASTER OF SCIENCE (Name) (Degree)
in MATHEMATICS presented on September 29, 1972 (Major) (Date)
Title:Solution by the Method of G. C. Evans of the Volterra Integral Equation Corresponding to the Initial Value Problem for a Non-Homogeneous Linear Differential Equation with Constant CoefficientsRedacted for Privacy Abstract approved. A. T. Lonseth
In the first chapter of this thesis, several methods are used to solve ann-thorder linear ordinary differential equation with con-
stant coefficients together withnknown initial values.The first method is the standard elementary method where the general solution
of the differential system is found as a sum of two solutions u and
vwhereu is the solution of the homogeneous part of the ordinary
differential equation andv is any particular solution of the non- homogeneous differential equation.The method is not strong enough to find a particular solution for every function that might be given as the non-homogeneous term of the ordinary differential equation and so we
try a more powerful approach for finding v; hence the Lagrange's method of variation of parameters.Following this, the method of Laplace transforms is employed to solve the differential system. In the second chapter then-thorder linear ordinary differ- ential equation is converted into a Volterra integral equation of second kind and in the next chapter, the idea of the resolvent kernel of an integral equation is introduced with some proofs of the existence and convergence of the resolvent kernel of the integral equation.The method of solving the Volterra integral equation by iteration is briefly discussed. The fourth chapter is devoted to solving the Volterra integral equation with convolution type kernel by the method of E. T. Whittaker, but the method is found to be very involved, and as a result, a method suggested by G. C. Evans (1911) is employed in calculating the resol- vent kernels for kernels made up of sums of two exponential functions (the method of iteration was applied to the same problem but it was tedious--it took about 20 pages of writing) and finally the method pro- vides an easier way for calculating the resolvent kernel of the Volterra integral equation corresponding to ann-thorder linear ordinary differential equation with constant coefficients. Solution by the Method of G. C. Evans of the Volterra Integral Equation Corresponding to the Initial Value Problemfor a Non-Homogeneous Linear Differential Equation with Constant Coefficients
by Jackson Henry Bello
A THESIS submitted to Oregon State University
in partial fulfillment of the requirements for the degree of Master of Science
June 1973 APPROVED: Redacted for Privacy
Professor of Mathematics in charge of major Redacted for Privacy
Acting Chairman oFtepartment of Mathematics
Redacted for Privacy
Dean of Graduate School
Date thesis is presented September 29, 1972 Typed by Clover Redfern for Jackson Henry Bello ACKNOWLEDGMENTS
I am expressing special thanks to Professor A. T. Lonseth for the very sound lectures he delivered on integral equations, for provid- ing this thesis problem and for helpful corrections and suggestions in the course of the development of this thesis. For the helpful comments and advice from professor R.B. Saunders, I am expressing my appreciation. I also want to express my gratitude to professor Iya Abubakar of Ahmadu Bello University in Nigeria, for nominating me for an AFGRAD fellowship, and to the officials of the African American Institute for the financial support.
Jackson H. Bello TABLE OF CONTENTS
Chapter Page
I.SOLUTION OF THE n-th ORDER NON-HOMOGENEOUS LINEAR DIFFERENTIAL EQUATION WITH CONSTANT COEFFICIENTS 1 1.1. Solution by the Standard Elementary Method 1 1.2. Lagrange's Solution by Variation of Parameters 10 1.3. Solution by the Method of Laplace Transforms 16
II.CONVERSION OF n-th ORDER NON-HOMOGENEOUS DIFFERENTIAL EQUATION WITH CONSTANT COEFFICIENTS INTO LINEAR INTEGRAL EQUATION OF VOLTERRA TYPE 23 2.1. Classification of Linear Integral Equation 23 2.2. Transformation of n-th Order Non- Homogeneous Linear Differential Equation with Constant Coefficients into Integral Equation of Volterra Type 24
III.SOLUTION OF THE VOLTERRA INTEGRALEQUATION BY THE METHOD OF ITERATION 30 3. I. The Resolvent Kernel 39 3.2. The Resolvent Kernel R(x, t;).) is a Solution of the Volterra Integral Equation 40 IV. SOLUTION OF INTEGRAL EQUATION BY WHITTAKER'S METHOD 43 V. CALCULATION OF THE RESOLVENT KERNELBY THE METHOD OF G. C. EVANS 51 5. I. Solution of the Volterra Integral Equation Corresponding to the Initial Value Problem for a Non-Homogeneous Linear Ordinary Differential Equation 59
SUMMARY 67
BIBLIOGRAPHY 69 SOLUTION BY THE METHOD OF G. C. EVANS OF THE VOLTERRA INTEGRAL EQUATION CORRESPONDING TO THE INITIAL VALUE PROBLEM FOR A NON- HOMOGENEOUS LINEAR DIFFERENTIAL EQUATION WITH CONSTANT COEFFICIENTS
I. SOLUTION OF THE n-th ORDER NON- HOMOGENEOUS LINEAR DIFFERENTIAL EQUATION WITH CONSTANT COEFFICIENTS
1. 1. Solution by the Standard. Elementary Method
The solution- -if there is one- -to the initial value problem of the form:
(Dy)(x) a any(n)(x)+ an-1y(n-1)(x) +... +ay(x) = f(x). 0
(1) 1 y(0) = yo,y (0) yo, ,y(11-1)(0)= wherey(i)(x),i = 1, 2, ...,n,denote thei-thderivative ofy(x) with respect toxand thea.' s,j = 0, 1, 2, ... , nare real func- J tions ofx, together with the boundary conditions expressed in (2), can be obtained in several ways.In this section, we shall discuss its solution by the standard elementary method. For this method, one finds the "general solution" of the linear differential equation as the sum of two functionsu(x)and v(x) whereu(x)is the "general solution" (meaning that it contains assignable parameters) of the homogeneous equation 2
+an_ly(n-1) (x)+ ...+ ay(1)(x)+ a y(x) = 0 . (3) ay(n)(x) 1 0 whilev(x)is any particular solution of the non-homogeneous equa- tion (1). We see that expressing the "general solution" of the linear differential equation as the sum of two functionsu(x)andv(x),i.e.,
y(x) = u(x) + v(x) makes sense since, if one has the problem(Dy)(x) = f(x)together with initial values, it follows from the expression forthe general solution that (Dy)(x) = (D(u+v))(x)
= (Du)(x) + (Dv)(x)
= 0 + f(x).
We shall now assume that the coefficientsa0, al, ..., anare constants and that the homogeneous equation in(3) has a solution of the form: u(x) = exs (4)
If we differentiate equation (4) ntimes we obtain 3 (1) sx u (x) = se
u(2)(x)=s2esx (5)
u(n)(x)=snesx
Substituting the set of functions in (4) and (5) into equation (3) yields: sx sx an + al se + ae = 0 snesx + an-lsn-1 esx + 0 from which esxra L n+anan- 1+... +a Is+a 0] =0. (5. 1) - 1
The exponentialesxnever vanishes and so the polynomial P(s): P(s) = an sn + an-lsn-1+ ...+ als + a0 must be zero, so that we have the equationfor
P(s) = 0 (6)
Equation (6) hasnsolutions in the complex field, which may be real or non-real.If non-real, they occur in complex conjugate pairs.Thensolutions may all be distinct or some of them may be repeated.The methods for solving equation (6) will not be discussed here since they may be found in many books dealing with solutions of algebraic equations.For our purpose in this section, we shall 4 assume first that the roots are real and distinct and shall denote them by mi,m2, ,mn. The general solution of the homogeneous equa- tion (3) can therefore be written thus:
mx m2x m x 1 u(x) = Ale A2. e + + A (7) ne
theA.' s,swhere i = 1, 2, ...,n,arenassignable parameters. If some of the real roots of equation (6) are repeated, then the function (7) will no longer be the general solution of the homogeneous equation (3).
We recall that if s is a repeated root of the second order linear differential equation
(2) (1)(x)+ ay(x) = 0, a2Y (x) + aly 0 then two linearly independent solutions areesxandxesx.Simi- larly if the roots1, say,is repeatedktimes for then-th order differential equation, then
s x 1 lx 2six1 k e ,xe ,x-lse 1Xe and all linearly independent solutions of the homogeneous equation (3). More generally, suppose that equation (6) has m distinct real roots sl, s2, ..., s, m < n,wheres.has multiplicityn.,so that 5
n 1 =< n. <= n, = n, and j=1
P(s) =ii(s-s.) , j =1 then
n. ms.x I 1 i -1 u(x) = e . ixj i=1 j=1
Also if a non-real complex rootX + iµoccurs with multi- plicityk, its complex conjugateX - iµalso occursktimes.
There will therefore be 2kcomplex-valued roots of (6) and corres- ponding to these, we can find 2k real-valued solutions of (3) by noting that the real and imaginary parts of
xk -le (X±ill)x e(X.±ip.),xe(X±ip.)x ... are all linearly independent solutions and that
eip.x= cos p.x + i sin p.x
Hence the real valued solutions are
eXxcos µx, eXxsin Xx xe cos µx, xe sin p.x,
k 1 k-leX.x x eXxcos µx, x sin µx . 6
Thus one has foru(x)a "general solution" which may contain real exponentials, real exponentials times polynomials (in the repeated case), and real sines and cosines, possibly also multiplied by polynomials. The real trouble comes in finding av(x).This can be done in only a few cases by the method of "undetermined coefficients." Such cases include those wheref(x) is a polynomial, or a linear combi- nation of exponentials or a linear combination of sines and cosines, possibly multiplied by powers ofxor by exponential functions, etc.
This limitation on the functionf(x) is a serious shortcoming of the "method of undetermined coefficients," and other methods have been devised which will handle more general problems. One such will be illustrated now by an example.It presents the solutiony(x)in terms of an integral involvingf(x).
Example.Consider the special differential problem of order 2 in whichf(x)is not specified:
y"(x) + 5y'(x) + 6y(x) = f(x). (i)
y(0) y0, yi(0) = yo
For the homogeneous equation
y"(x) + 5y'(x) + 6y(x) = 0, let us try the solution Xx y = e .
Then
X.2+ 5X + 6 = 0,
(X +3)(X +2) = 0, and so
X.1 = -3 andX2 = -2.
Therefore the general solution of the non-homogeneous equation is
y(x) = u(x) + v(x)
= Ae-3x+ A 2e-2x + v(x). 1
Rewriting the non-homogeneous equation (i) using the operator nota-
tiony' = Dy andy" =D2ywe have
(D2+5D +6)y(x) = f(x) or (D+3)(D+2)y(x) = f(x)
and if we let
(D+2)y(x) = g(x) we have
(D+3)g(x) = f(x) or 8
g'(x) + 3g(x) = f(x), whence (g(x)e3x)'= f(x),
and so x 3x g(x)e = f(s)ds + A. 0
From this x g(x) = e-1x f(s)ds +Ae-3x `j0
where Ais the constant of integration.Substituting this back in (iii) we obtain -3x S (D+2)y(x) = e f(s)ds +Ae-3x 0 and so -3x y(x)e2x= {e-3tS +A73 + B 0 0
where B is the second constant of integration.Thus
-2x Ae-5x+ Be y(x) =sxe-Zx xe-3tdtf(s)ds 3 0 s
x -3x--3s -5x -2x e f(s)ds -Ae + Be e-2x(- 3 3
Thus
x-2x-3se-5x Ae-5x -2x y(x) (e )f(s)ds - Be (iv) J 3 3 0 3+ 9 Substituting the initial values from (ii) we have forx = 0;
y(0) = + B = yo (v)
Differentiating equation (4) with respect toxwe obtain
-2x-3s -5x -2x-3x -5x x e e e )f(s)]dsds + [( e )f(x)] Yi(x) f( 3 3 3 3 0 -5x -2x +5Ae - 2Be 3
Putting the valuex = 0yields:
00) =A5 - 2B = (vi) 3 0
Multiplying equation (v) by 2 we have
2A+ 2B = 2y0 (vii) 0 and adding (vi) and (vii) yields
A = 2y0 + y'o, and putting this in (vii) we have
2B = 2y+ 2(2y+yi ) 0 3 0 0 from which
5 1 , B =70 +70
Equation (iv) therefore becomes 10
x -2x-3 s -5x e y(x) = ( e )f(s)ds 3 3 0
-5x 5 4.1 , -2x 'TY° -3-YOIe
This is the solution of the differential problem in terms of an integral involving the unspecified functionf(x). Still more general is J. L. Lagrange's method of variation of parameters, which will be described in the next section.It applies also to the case of variable coefficients, provided one can find a set of fundamental solutions of the homogeneous equation.If the coeffi- cients {a.(x)}11can be expanded in power series, a whole new 0 approach is possible (method of Frobenius), but it will not be dis- cussed in this thesis.
1. 2. Lagrange's Solution by Variation of Parameters
The method discussed in Section 1. 1 can be used to determine the general solution of any non-homogeneous equation with constant coefficients provided the non-homogeneous term is of suitable form. In this section we shall discuss a more powerful approach for finding the particular solutionv(x)since, as we pointed out in the last section, the method of undetermined coefficients can be applied in only a few cases.The method is known as Lagrange's method for finding the particular solution of the non-homogeneous system of the 11 form:
+ a_iy(n-1)(x) + ...+ a oy(x) = f(x) (*) any(n)(x) n with (n-1) -1 y(0) = yo,y(1)(0)= yO, . , y (0) yno
For this method it is essential thatnlinearly independent solutions of the homogeneous differential equation
any(n)(x)+ a y(n-1)(x)+ ...+ ay(x) = 0 (1) n-1 0 discussed in Section 1.1 be known.Let them beul (x)' u 2(x) '...'un(x) so that the general solution of equation(1) is
u(x) = Au 1(x) + Au 2(x) + ... + Au(x) (2) 1 2 nn
The method of variation of parameters involves replacing the constants
Al' A2, ., Anby functionsP (x) P2(x) P (x)ofxrespec- tively.Thus we shall assume that the solution of equation (1) is of the form:
u(x) = P(x)u P(x)u 2(x) + ... +Pn(x)un(x) 1 1(x) + 2
or n
u(x) = P.(x)u.(x) . (3) i=1
The problem then reduces to that of findingnsuitable 12 functionsP1 (x)(x), P(x) P (x)such that the functionu(x) 2 " defined by (3) satisfies equation (1).To do this we need a set of independent conditions, and(n-1)of them could be selected rather arbitrarily so as to simplify the calculations.The last condition arises from the fact that equation (3) must satisfy equation (1). Differentiating equation (3) we obtain:
n n (1) (1) u(1)(x)= P.(x)u.(x) + P.(x)u.(x) . 1 i=1 i=1
Since we can select(n-1)conditions arbitrarily, suppose we let the first be n
(x)u.( )= 0 , l)( i=1 thus n (1) u(1)(x) = P.(x)u.(x) . (4) 1 1 i=1
Differentiating equation (4) we have:
n n (1) (1) (2) u(2)(x)= (xP.)11. (x) + P.(x)u.(x), 1 1 1 1 i=1 i =1 and again suppose we set n (1) (1) P.(x)u.(x) = 0, 1 i=1 13 to obtain n (2) u(2)(x)= P.(x)u.(x) . (5) 1 1 i=1
We could continue the process of differentiation and each time if we set the term involvingP(.1)'s(i = 1, 2, ,n-1)to zero, after ndifferentiations we would have
n n (n) (1) (n-1) u(n)(x)= P.(x)u.(x) + P.(x)u. (x) (6) 1 1 1 1 i=1 i=1
We shall now substitute all the expressions for the nderiva- tives ofu(x)into equation (*) to obtain:
n (1) (n-1) (n)(x) + a P.(x)u. (x) P.(x)u. n i=1 i =1
n (n-2) + a P.(x)u.(n1)(x) + a P.(x)u. (x) n- n- 1 1 i= 1 i=1
n n 1)(x) + + a P. (x)Li. + a P.(x)u.1 (x) = f(x). 1 i=1 i=1
out each of theP.(xFactorizing )'swe shall obtain: 14 (1) Pl(x)ianur (n)(x)+a u(n- 1)(x)+. .. +a u(x)+a u (x)} 1 n-1 1 1 1 0 1
+ P (x){au(n)(x)+a u(n-1)(x)+... +au(1)(x)+a u (x)} 2 n 2 n-1 2 2 2 0 2 (n-1) (1) + ... + P(x){a u(n)(x)+a u (x)÷... +a u(x)+au(x)} n n n n-1 n 1 1 0 n
1) n-1) + an P(. (x)u( (x) = f(x) (7) 1 1 i= 1
We see that the expressions in the braces now vanishbecause
u.(xeach) is a solution of equation (1) .
Thus, n (1) (n-1) a P.(x)u. (x) = f(x) (8) n/ i=1 is ourn-thcondition.The first(n-1)conditions can be sum- marized as: n ( (k) P.1)(x)u.(x) = 0, k = 0, 1, 2, ..., (n-2), 1 i=1
can be and from these conditions, the functionsP1(x),...,P (x) determined as
x Ain(t)f(t) 131 dt, .(x) = an 0
whereA. (t)is the cofactor of the element an V 0,i = 1, 2, ..., n; in in thenthrow andithcolumn ofW(t)andW(t) V 0.is the 15
Wronskian of the independent functionsu.(x)whose linear combi- i nation isu(x)in (2).W(t) is defined by the determinant
u u un 1 2 (1) (1) u(1) u2 .. un 1 (2) (2) W(t) =ul 11(2) un (9) 2
(n-1) (n-1) (n-1 u u un 1 2
The particular solutionv(x)is thus given by
u.(x)A.(t) X i=1 v(x) = f(t)dt; anW(t) 0
an V 0, W(t) 0, i = 1, 2, ...,n.
The general solution of the non-homogeneous equation (*) is therefore given by y(x) = u(x) + v(x)
u.(x)A.(t) x i=1 A u (x) +...+ Au(x) + f(t)dt, 1 1 nn a W(t) t)0 n
an 0,W(t) 0is defined by (9) andA.(t) is the cofactor of the in element in thenthrow andithcolumn ofW(t).The independent 16 functions u.(x),i = 1, 2, , n, are the independent solutions of the homogeneous equation (1), and theAi's,i = 1, 2, ,naren assignable parameters.
1.3. Solution by the Method of Laplace Transforms
Definition:The Laplace transform of a piecewise continuous functionf(t)for allt > 0will be denoted byL{f(t)}and is defined by the following expression:
co L{f(t)} = (Lf)(s) e-stf(t)dt (1) 0
It is evident from (1) thatL{f(t)}is a function of a parameters. This definition is valid whenever the improper integral converges and for our discussion in this section we shall assume thats is real. To establish that the Laplace transform off(t)as defined by equation (1) does in fact exist,the followingsufficient conditions have to be imposed on the functionf(t).We shall suppose that
(a)fis piecewise continuous on the interval0 < t <
for allTE > 0, (*) (b) I f(t) I
The problem therefore reduces to that of showing that the inte- gral in equation (1) converges for all values of the parameters > p. This we can show by simply splitting the integral in two parts, thus:
oo L{f(t)}=5e-stf(t)dt 0 and so oo-stf(t)dt L{f(t)} e-stf(t)dt +.S1e (2) 0 for some fixed real constant N > 0. Condition (a) in (*) imposed onf(t) shows clearly that the N-st first integral e f(t)dtexists and so our problem is narrowed 0 c°-st down to that of showing that e f(t)dtconverges. Observing the .cN functione-stf(t)and applying condition (b) imposed onf(t)we
-s ., have tf(t) 1 < Ce-stePtfrom which we see that e-stf, We know that e(p-s)tdtconverges for
Assume that the functionf, and its derivatives (1)(2) (n -1) f ,f , ,f are continuous andf(n)is piecewise continu- ous on any interval0 < t < X'and thatf(1), f(2), ...,f(n-1)all satisfy the conditions (a) and (b) of (*), namely that
f(i)(t) I
sf( n-2)(0) _f(n-1)(0). L{f(n)(t)}=snL{f(t)} sn-1 f(0) (4) The proof for this could be found in many books on differential equa- tions [8, p. 228]. We shall now apply the method of Laplace transform to solve our 19 nthorder linear differential equation with constantcoefficients, i.e.,
any(n)(x)+ an-1y(n-1)(x) + ...+ a0y(x) = f(x) (5) 1 } y(n)(0) = yy(n-1)(0)=yno-, ..., y(0) = y0.
Knowing that ifkis a constant theL{ky(x)} = kL {y(x)}and that ifP(x)andQ(x) are two functions suchthatP(x) = Q(x), then by the uniqueness of Laplace transformsL{P(x)} = L{Q(x)}, we shall now take the Laplacetransform of each term of equation(5), applying formula (4) for each derivative,knowing that for two func-
tionsy andy2,L{y+y2} = L{y} + L {y2 }, we obtain 1 1 1
(n-1) L{y(x)} = L{f(x)} , a nL{y(n)(x)} +an-1L{y (x)} + ...+a0 and so
an[snL{y(x)}-sn-ly(0)-sn-2y(1)(0)- -sy(n-2) (0)-y(n-1)(0)]
+ an -sn-2y(0)-s11-3y(1)(0)-. -sy (0) -y (0)] (n-3)(0)] + an-2[sn-2 L{y(x)}-sn-3y(0)-sn-4y(1)(0) -sy(n-4) (0)-y
a2[s2L{y(x)}-sy(0)-y(1)(0)]+ a[sL{y(x)}-y(0)] + ...+ 1
a0[L{y(x)}1 = L{f(x)}.
We shall rewrite this, factorizing outtheL{y(x)}to obtain: 20
1+. sn .. +als+a 0]L{y(x)} [ansn+an-1
r n-1 2+. Lans .. +a]y(0) +an- lsn- 1
(1)(0) [ansn-2+an-lsn-3+... +a ]y
2)(0) (n-1)(0) .. -[ans+an_}y(n- -[an]y = L{f(x)} ,
and solving forL{y(x)}we have:
(1) (n-1) [Pl(s)]y(0)+P (s)y(0)+... +any (0)+L{f(x)} L{y(x)} (6) 1+. rasn+ansn- .. +a ls+a 0] L n - 1 where
r n-1 P(s) = Lans +an-lsn-2÷...+al] 1
sn-3+...+a2] P2(s) =[a nsn-2+an-1
etc. Recalling that the Laplace transform off(x)is given by -sx L{f(x)} =Se f(x)dx,we see thatL{f(x)}is a function ofs, 0 a.Is and we shall denote it byQ(s)say. We also note that the i and the contributions of the derivatives ofy(x)evaluated at zero,
are all constants.The expression forL{y(x)}is therefore strictly
a function of s as we expect from the definition of Laplace trans - form. We already pointed out the linearity of Laplace operator and its 21 inverse, and so the rest of the problem isthat of rearranging the right hand side of the equation into sums offunctions ofs, such that the inverse transform of each term isrecognizable from tables of Laplace transforms. A method often employed tosimplify the right hand side of (6) is the method of partial fractions,and if this is done, the inverse transform of each fraction couldbe found from tables of Laplace transforms [9, 10].As an example let us suppose that after simplifying the right hand side of (6) we obtained
1 a L{y(x)} = - s -a 2 2 2 2 s+a s-a
From tables of Laplace transforms, the inversetransform which we shall denote byL-1of the right hand side are respectively
at, e s >a
1[[ ] = sin(at), s > 0 L 2+a 2 s
L1[[ ] = cosh(at),s > al s-a2
and so, again by the uniqueness ofLaplace transforms
-1la s2] L1{y(x)}= L [ L-1[ 2]+ L-1[2 s s +a s -a
from which 22 eat y(x} sin(at) +cosh(at).
The solution of our non-homogeneousordinary differential equation with constant coefficientswould therefore be
n-1 [13 (s)Jy+LP(s)jy1+...+anyo+Q(s) 1 0 2 0 y(x) L-1 ras +a sn-1+...+as+a ] L n-1 1 0
It should be noted that thedenominator in the solution is exactlythe auxiliary equation obtained in equation(6) of Section 1.1.This will always be the case, and there shallalways be the need to find the roots of this auxiliary equationregardless of which method is employed to solve the non-homogeneouslinear differential equation with constant coefficients. 23
II.CONVERSION OF n-th ORDER NON-HOMOGENEOUS DIFFERENTIAL EQUATION WITH CONSTANTCOEFFICIENTS INTO LINEAR INTEGRAL EQUATION OF VOLTERRATYPE
2. 1. Classification of Linear Integral Equation
The phrase "linear integral equation" is a broad termthat encompasses large families ofintegral equations.To narrow this down, we need to specify what "TYPE" of integralequation, and fur- thermore, any type of integral equation could be any oneof many "KINDS," and so we talk of an integral equation being of any oneof 1st, 2nd or 3rd kind. An integral equation whose limits of integration arenot constant but variable such as:
x a(x)y(x) = b(x)+.54k(x, s)y(s)ds 0
is known as an integral equation of VOLTERRATYPE.If the limits of integration are fixed such as:
1 a(x)y(x) = b(x)+.14k(x, s)y(s)ds, 0
it is known as the FREDHOLM TYPE ofintegral equation. The nature of the coefficienta(x)determines what "KIND"
the equation is.Ifa(x) = 0then we have an equation of the 1st kind. 24
Ifa(x)is always positive or always negative we have an integral equation of the 2nd kind, and a 3rd kind is obtained ifa(x)vanishes at one or more isolated points.
If,in the integral equation,b(x) = 0,we say that the equation is homogeneous.This chapter will be devoted to transformation of then-thorder linear non-homogeneous differential equation with constant coefficients into an integral equation of VOLTERRA TYPE.
2. 2. Transformation of n-th Order Non-Homogeneous Linear Differential Equation with Constant Coefficients into Integral Equation of Volterra Type
The n-th order non-homogeneous linear differential equations with constant coefficients
ay(n)(x)+ an-ly(n-1)(x) + ...+ a y(1)(x)+ a y(x) = F(x) n 1 0
together with ( 1 ) (n-1)(0) (n-2)(0) y =yn-1,y =yn-2 ,y(0) = yo '...
can be transformed into a Volterra integralequation of the form
u(x) = f(x) + k(x, t)u(t)dt (2) 0
by the following procedure.Suppose we let
y(n)(x) = u(x) (3) 25
We shall further define
-1 -2 -1 -1 (Df)(x) =.51 f(s)ds, (Df)(x) = (D(Df))(x), 0 etc.Performing successive integrationsofu(x)we thereforehave
-1 D u(x) = u(t)dt, 0
-1 D-zu(x) (D1(1)-1u))(x)= D u(t)dt =S x(x-t)u(t)dt (4) 0 0
-1 -(n-1) (x-t)n-1 D-nu(x) (D (D u))(x)=Sx u(t)dt 0
Furthermore integrating both sides of equation(3) we obtain
(n_1) -1 n-1 y (X) = D ti(x) Yo
n- 1 where is the constant of integration obtained from YO n-1 (n-1)(0)= y in equation (1).Integrating successively and each 0 time employing the boundary conditions in(1) we obtain:
-2 2 y(n-2)(x) =y(n-1)x+ yn0 + D u(x), 0
n-3 -3 (n-3) n-1x2 n-2x + yo + Du(x), (x) YO 7.2 + YO n-1xn-1 n-2xn-2 n y(x) = y0 + .. .+ y + D 26
Substituting the values of eachD-iu(x),i = 1, 2, ...,nfrom (4) into the last set of equations yields:
n- 1 u(t)dt y(n-1)(x) y0 0
y(n n-1 n2 -2)(x)= yox + yo (x-t)u(t)dt 0
n-1xn-1 n-2xn-2 Y(x) YO (n-1) ! YO (n-2) ! ...+YOx YO
x (x-11-10 u(t)dt, (n-1) ! 0 from which we further see that
(n-1) -1 a y (x) = an-1y° +an-lSu(t)dt n-1 0
x n-1 n-2 an-2y(n-2)(x) = any x + a,y +an-2.c(x-t)u(t)dt -2 0 n-G 0 0
n-1xn-1 n-2xn-2 a0y(x) = ay0 + ay0 + ...+ aoyx'o 0 (n-1) ! 0 (n-2) !
+ a0y0 + a0 (n-U!u(t)dt 'Co
Adding both sides of these equations we have 27 (n-1)( (n-2) (1)( x) + an-2y (x) + + ay x)+ a0y(x) an-ly 1
n-1 n-1 n- Z 1 xn-1 =an_ly0 + a (y x+ + a (yn- +.-+y;x+yo) 1.1-2 0 y0 )4 00 (n-1) ! n-1 (x-t)i + u(t)dt. an-i-1(i)! 0 i=0
We can simplify this as
an-1y(n-1)(x) + an-2y(n-2)(x) + ...+ aiy(1)(x)+ a0y(x)
= f(x) k(x, t)u(t)dt (5) 0 where n-1 n-1 n-1 n-2 n-1 x f(x) = a y0 +... +y0) n-1 + an-2(y0 x+YO ) + + a0(Y0 (n--7TT! (6) and
k(x, t) = (7) n-i-1 (i)! i=1
However, we already sety(n)(x)= u(x)and so
(n) any )= an u(x) (8)
Adding equations (5) and (8) we have 28
any(n)(x) + a n-ly(n-1) (x) + + ay(1)(x)+ ay(x) 1 0
= anu(x) + f(x)+.1k(x, t)u(t)dt 0
= F(x) from equation (1).Thus
anu(x) + f(x) k(x, t)u(t)dt = F(x) , 0
and since thea s.' are constant coefficients, suppose we divide through by-anto obtain
-f(x)+F(x) x u(t)dt . u(x) ( an ) + -an 0
This can be further written as
u(x) = W(x) K(x, t)u(t)dt (9) 0
IF(x) = 1 where (F(x)-f(x)).an Equation (9) is a Volterra integral equation with kernel
n-1 K(x, t) = (10) 29 We have therefore converted the original non-homogeneous lin- ear differential equation with constant coefficient into a Volterra integral equation expressed in (9) with its kernelK(x, t)given by equation (10).If the integral equation has a solution,it will furnish one for the differential problem. 30
III.SOLUTION OF THE VOLTERRA INTEGRAL EQUATION BY THE METHOD OF ITERATION
In Chapter II, the n-th order non-homogeneous linear differ- ential equation
(n-1) (1)(x) an-ly (x) + ...+ ay + a0y(x) = F(x) a ny(n)(x) 1 together with (1) (n-1) n-1 (n-2) n-2 (0) yo , ,y(0) = yo (3) YO was converted into a Volterra integral equation of the form
y(x) = f(x) + X.5. x-k(x,t)y(t)dt (2) 0
0 < x < cr,wheref(x)is a function ofxaryl k(x, t)is a function ofxandt, and as a matter of convenience, we are introducing the parameter X. which we shall assume real.In this chapter we shall look into the method of solving equation (2) (and hence solving the differential equation in (1)) by the method of itera- tion. Substitution fory(t)in equation (2) gives 31
y(x) = f(x) + X k(x, t)f(t)+YX. k(t, s)y(s)d sdt, 0 0
x = f(x) + k(x, t)f(t)dt + k(x, t)5-k(t, s)y(s)ds dt, 0 0 0
= f(x) + k(x, t)f(t)dt +X.2 (k(x, t)k(t, s))dty(s)ds, 0 0 s
(2)(x, = f(x) + Xsk(x, t)f(t)dt + X k s)y(s)ds, 0 0
(2)(x, where we shall definek s)by
k(2)(x, s) =fk(x, t)k(t, s)dt .
The equation therefore becomes, on substitutings = t,ds = dt,
2 9 x(2)(x, y(x) = f(x) + X k (x, t)f(t)dt + k t)y(t)dt (3) 0 0
The process of repeated substitution of expression fory(t)as done in the last step is known as "ITERATION." We shall iterate again substituting value fory(t)in equation (3) to obtain: 32
xk (2)(x, y(x) = f(x) + X.51 k(x, t)f(t)dt +X2 t)f(t)+X k(t,$)y(s)dsdt 0 0 0
x S (2)(x, = f(x) + ,..c )(x, t)f(t)dt + X2 k t)f(t)dt 0 0
+X3 S k(2)(x,t) Sk(t,s)y(s)ds/dt 0 0
xk (2)(x, = f(x) + X Jk(x, t)f(t)dt +X2 t)f(t)dt 0 0
x x 3 + X k (x, t)k(t, s)dty(s)ds 0
= f(x) + X Jk(x, t)f(t)dt +X2 k(2)(x,t)f(t)dt 0 0
+3 k(3)(x,s)y(s)ds 0 where x(2)(x, k(3)(x,s) =Sk t)k(t, s)dt (3. 1)
If we lets = t,ds = dt,we have
2y x(2) y(x) = f(x) +Jk(x, t)f(t)dt + k (x, t)f(t)dt 0 0
X3 SX(3) + (x, t)y(t)dt (4) 0 33
This process of iteration could be carried on, and afternsubstitu- tions we would have
x y(x) = f(x) + X k(x, t)f(t) + X k (x, t)f(t)dt 0 0 (5) n+1 r (n+1)(x, + ...+ Xnxk(n)(x, t)f(t)dt + X s)y(s)ds 0 0 where
k(n+1)(x,s) = Jxk(n)(x,t)k(t, s)dt and (6)
k(1)(x, t) = k(x, t)
Simplifying equation (5) we have:
(n+1) y(x) = f(x) + xXik(i)(x,t)f(t)dt + X.n+1S xk (x, t)y(t)dt(7) 0 0 i= 1
We see that equation (7) satisfies our original Volterra equation (2) since the former has been deduced from the latter.However, observation of equation (7) suggests that we may be generating an
infinite series fory(x)of the form
00 x y(x) = f(x) + X lk (x, t)f(t)dt (8) i=1 34 This is the case, but to establish this it is necessary to show that the solution of this equation converges to some useable function and satisfies equation (2).In addition we ought to show that this solu- tion is unique and that the iterated kernels as defined by equation (6) actually do exist. We saw from equation (3.1) above that
k(3)(x,s) = k(2)(x,t)k(t, s)dt .
We shall assume that the integrand is uniformly continuous in 0 < t < XforX < oo, so that the expression may be Riemann- integrable for allx > 0.It will also be a continuous function ofx on0 < x < Xfor eacht.This shows thatk(3)(x,s)will exist and will have the same continuity properties.The argument is also true fork(n)(x,s)forn > 2.We therefore see that iff(x)is a continuous function on0 < x < X,then all the terms of the series expressed in equation (8) exist and are continuous functions ofx. Under our continuity assumption fork(x, t)andf(x)in the above paragraph, each of the sequence of partial sum{yn(x)} given by n x(i)k(i)(x, {yn(x)} = f(x) + t)f(t)dt (9) 0 i= n = 1, 2, ... ,is continuous on0 < x < 00.The problem thus reduces 35 to that of showing that{yn(x)}is uniformly convergent on the closed interval0 < x < X for some X < 00,since this being so implies that the limit function must also be continuous. We shall employ
Cauchy's theorem stating that for{yn(x)}to converge uniformly on
[0, X]it is necessary and sufficient that, given anyE > 0, there exists an integer N(E) such that if n > N(E)then
yn+p(x)-yn(x) I< E for every positive integerpand everyxin
[0, X]. However we see that from equation (9),
n+p yn+p(x) - yn(x) = Xlk(1)(x,t)f(t)dt (10) 0 i=n+1
Sincef(t)was assumed continuous, this means thatIVO' has some maximum value which we shall denote byF(X).Further, let us assume thatk(x, t)is bounded on the interval0 < t < x < X and that I k(x, t) I< M(X), whereM(X) < for0 < X <00 . Substituting these in (10) leads to
n+p (x)I < F(X) olk(i)(.,t)Icit I Yn+p(x)-yn = 0 i=n+1 at each pointxin the interval[0, X]. We note that 36
I k(x, t) I dt < M(X) Jdt = M(X) x , 0 0 and x x 2(x, I k t) I dt = I k(x, s)k(s , t)ds I dt S`0 0 t
2(X) M 2 < M2(X) S (x-t)dt 2! 0
Using equation (6),it could be shown by the process of induction that
xn , n = 1,2, .... k(n)(x,t) dt < Mn(X)n! 0
We shall put all these values in (11) to obtain
n+p . k M(X)xi (12) n+p(x)-y n(x) I< F(X) (i)! i=n+1 and it is valid for allxin[0, X].However we note that the sum on the right hand side of (12) is a segment of the Maclaurin series for [M(X) the exponentiale which we know from Calculus converges uniformly on each bounded interval.Therefore given anyE > 0,any
N(E) can be found so that forn > N(E)andp > 0,an integer,
I yn+p(x)-yn(x) I< E for allxin[0, X].
This shows uniform convergence for the partial sum{yn(x)}on the 37 interval[0, X], to a continuous limit function. The uniqueness of the solution of the Volterra equation can be shown simply by assuming that there exist two distinct continuous solutions and that the difference between them is denoted byw(x). We know that this difference ought to satisfy the homogeneous Volterra equation
w(x) = X Jk(x, t)w(t)dt 0 since if the two solutions are denoted byyi(x)andy2(x)we have
x y(x) = f(x) + X k(x, t)y(t)dt 1 1 0 and
y2(x) = f(x) + X k(x, t)y2(t)dt 0
Subtracting one equation from the other we have:
yi(x) - y2(x) = w(x) = X k(x, t)[yi(t)-y2(t)ldt 0 from which
w(x) = X k(x, t)w(t)dt. 0
Iterating this n-times yields 38
x w(x) skni(x, t)w(t)dt,n = 2, 3, ..., 0 and using the same notation for bounds onk(n)(x,t)as done in showing convergence above, we have
Mn( 1 w(x) 1< xn max I w(t)11 n! 0 < t < x
We notice that the right hand side is just a term of the
Maclaurin series for exponential (M(X) x X)multiplied bymaxi w(t)I. We also know that each term of a convergent series approaches zero, mn(x)xn xn and sinceexp(M(X) xX) converges, the single term n! must approach zero, hence
Mn(X)xn XnImax w(t) I -- 0 n. as n~ 00.
This means thatw(x) = 0on[0, X], and we thus show that there is only one solution to the Volterra integral equation defined by equa- tion (2). Having thus shown that the iterated kernels defined by equation (6) do exist, and are uniformly convergent on the interval[0, X],and that the solution of the Volterra integral equation is unique, we shall from now on accept that 39
y(x) = f(x) + ik(i)(x,t)f(t)dt (13) i=1 wherek(1)(x,t) is the iterated kernel of equation (2).
3.1. The Resolvent Kernel
It was shown preceding this section that the solution of the Volterra integral equation, which was deduced from ourn-th order non-homogeneous linear differential equation with constant coeffi- cients, was
oo x ik(i)(x, y(x) = f(x) + t)f(t)dt (1) 0 i= and, of course since we have shown that the infinite series expressed in equation (5) of the preceding section converges uniformly, we can from now interchange the summation and the integration signs.Thus equation (1) becomes:
co y(x) = f(x) ik(i)(x, t))f(t)dt (2) o and we shall further simplify it as 40
y(x) = f(x) + A R(x, t; X.)f(t)dt (3) 0 where oo i-1(i) R(x, t;X) = k(x, t) (4) i=1
R(x, t; X) is called "The Resolvent Kernel" for the kernelk(x, t). As we shall see in Chapter V, the resolvent kernel plays an important role in finding solutions to some Volterra integral equations that would have been otherwise involved or too complicated to solve.First we need to establish that this resolvent kernelR(x, t;)%.) actually satisfies the Volterra integral equation as defined in Chapter II.
3. 2. The Resolvent Kernel R(x, t;)q is a Solution of the Volterra Integral Equation
R(x, t;).) = k(x, t) + k(x, t;X)dT, (5) 0
Proof: Let us consider an integral of the form
rck(x,)R(; t;X)cl = k(x, ) t t
The uniform convergence of the series t) enables us i=1 to interchange the summation and the integral signs so that 41
00 xk(x,)Rg, td.)cg = Xi-1 k (x,Qk(i)g,t)d i=1
00 i-lk(i+1)(x, t),
applying equation (6) of the first section of Chapter III, i. e.,
(n+1) (1) (n) k (x, t) k (x, t)cl
Thus
oo i-lk(i+1)(x, k(x, ,)R(r,, = t) t i=1
oo xi-2k(i)(x,
i=2 and so 00 x k(x, gR(,t,;X )(% = X1-1k(i)(x,t), t i=2
oo i-1(i) (1)(x, (1) k(x, t)+k t) - k(x,t) i=2 (1) = R(x, td.) - k (x, t), since 00 00 i-1(i) i-1(i) R(x, t;X) = k(x, t) + k(1)(x,t) k(x, t), i=2 i=1 42 andk(1)(x,t) = k(x, t).Rearranging, the integral finally becomes
R(x, t;X) = k(x, t) + A Jk(x, t;X),:g . (6) t
Equation (6) is of the same form as our original Volterra equation if we replacey(x)byR(x, t;X)andf(x)byk(x, t).We shall further impose thatk(x, t) 0 forx < tto take care of the lower limit in equation (6).From now on, and especially in Chapter V, we shall use this version of the Volterra integral equation (6) involving the resolvent kernelR(x, t;X)since it is evident thatR(x, t;X)is a solution of the original Volterra integral equation and hence a solu- tiOn of our original initial value problem from which the Volterra integral equation was deduced. 43
IV. SOLUTION OF INTEGRAL EQUATION BY WHITTAKER'S METHOD
In this chapter we shall briefly discuss E. T. Whittaker's numerical method of solving integral equations of the form:
y(x) y(s)k(x-s)ds = f(x) (1) 0 where the kernelk(x, s)is of special form known as a "convolution- type" kernel. We shall represent the kernelk(x)as a sum ofn exponential functions ofx, thus
qx q qnx 1 2x k(x) = Qe + Q2e + ...+ Qne (2) 1
where the coefficientsQ1, Q2, ,Qnof the exponentials and the coefficientsqv q2, ...,qnofxin the exponentials are constants. Using this expression for the kernel, we shall show that a solution of the integral equation (1) is of the form:
y(x) = f(x) K(x-s)f(s)ds (3) 0 where p x p2x pn 1 K(x) =Pe +Pe + ...+ Pne (4) 1 2 another sum ofnexponential functions whereP1, ...,Pnand 44
P1' 'Pnare to be determined in equation (3).We note that the integral equation given by (3) has a unique solution since it is of Volterra type.Furthermore it was shown in Section 3. 2 that the resolvent kernel K(x)satisfies the Volterra integral equation
K(x) K(x)k(x-s)ds = k(x) (*) 0
If we put back values fork(x)andK(x) as defined by equations (2) and (4) respectively we obtain:
p x p pnx 1 2x P + P + + Pne le 2e p s 1 p q1(x-s) qn(x-s 2 Pte e Qle +... +Qne ds 0 +...+Pne
x q x ql 2 q x = Q + Q + + Qnen le 2e
Expanding this by multiplying out the product under the integral sign we have:
px pnx 1 Ple + P2e + + Pne x+pls-ci lq x+Pls-qls 2 2s +S xf [PQ e + +PIQnecinx P1 I 1 1 2 0 q lx+p2s -qls q2x+P2x-q2s iP2Q1e +P2Q2e +...+P2 Q necinx+P2x-cins 45
qix+pns -qi s qx+ps -q q x+p s-q s Qle +P Q e2 n 2s+...+P Q en n n n n 2 n n Ida q x q x qnx Q lei + Q 2ez + ..+ Qne .
Performing the integrations we obtain:
pnx + Pep2x+ + Pne P lepl 2
kPQ1 qx+ (p-q 1)5 s=x PQ2 q 2x+ (p S = X 1 1 1 1 1-q 2)s
l-c11 s=0 (p1 -q2) x= 0
P1 Qn qnx+(p S = X 1-q n)s +. + e x=0
P2Q1 cilx+(P2-cil)ss=x P2Q2 ci2x+(P2-c12)s + + )e (132 -c11) s=0 `F2 si2 ==x0
P2 Qn qnx+(p-q n)sS = X +.. + )e 2 1.'2 `in s=0
PnQIqix+(pn-qi)s pnQnqnx+(pn-pn)sS =X + + e +...+ e Pn 1 x=0 wn
ql x q x qnx 2 = Qe + Q + + Qne . 1 2e
Evaluating the definite integrals we have: 46
plx Pte + P 2eP2x + + PnenP x
P1 Q1 pQn pxqnx Plxqlx P1Q2 Plxq2x 1 1 (e -e )+ (e -e ) +...+ -e (1)1-q1) (131 -q2) (Prqn)
PQn P2Q1 P2x_ qlx)+ P2Q2 (eP2x_eq2x)+...4. 2 qnx + (e p2x-e ) (132-q1) (132-(12) (P2-qn) PQpx qxP Q px q x P Q px qx n 2 n 2 nn n n +...+ (e -e )+ (e -e )+...+ 1(e -e ) (Pn-q1) (Pn-q2) 43n-qn1
qx q qnx 1 2x = Qe + Q + + Q e . ( 9) 1 2e
P1 x Equating coefficients ofe on both sides of the equation yields:
PQ1 PQ2 PQn 1 1 1 P + + + = 0 , 1 (pl-q1) (131-q2) (131 -qn) and so Q2 Qn 1 p(1+ + ...+ )= 0. (P 1 (131-c11) (131-q2) 1qn)
We do not wantP1to vanish otherwise we shall be contradicting our assumption that the functionK(x) defined by equation (4) consists of nexponential functions.Therefore
Q1 Q2 Qn 1+ + + = 0 (10) (P1 -cid (131-q2) plqn) 47
By similar reasoning, after equating coefficients ofep2Xin equation (9) we would obtain:
Q2 Qn 1 1 + + + = 0 (p2 -q1) (13 (P2-q2) 2-cln)
The process could be carried out n-times and each time equating pix coefficients of e , i = 1,2, ...,n,in equation (9) and we would obtain:
Q2 Qn Q1 1+ + + = 0 (12) (Pi-ql)(pi q2) (Piqn) i = 1,2,3, ... , n.Thus, we see thatply p2, ...,pnare the roots of the algebraic equation in
Q2 Qn Q1 + + = 0 (13) 1+ (x-qn) (x-ql)(x-2)c1 and from this, we can determine the values ofP1' P2' ' Pn. lq x Returning to equation (9) and equating coefficients of e on both sides of the equation we obtain:
PQ1 PQ1 PnQI 1 2 -( + + )= Q1 (13l qi) (132-`11) (Pnq1) from which P1 P2 Pn Q1(1+ ) = 0. (prqd(p2 q1) (13nI) 48 By similar reasoning done in the forgoing paragraphs we do not want
Q1 to vanish, hence
P P2 Pn 1 + + + ... + = 0 (14) pl-c111 132-c111 (Pncl 1)
q 2x Equating coefficients of e on both sides of (9) also yields
p P2 Pn 1 1 + + + ... + 0 (15) (131-q2) (132-c12) (pn-c12)
q.x and equating the coefficients of a general exponential eJ we have:
P Pn 1 Pp2-c1j)2 1 + + + ... + = 0 (16) (p1-qj) ( (pri-cj) j = 1,2, ...,n.Again we see thatql,q2, ...,qnare the roots of the algebraic equation inz:
P P2 Pn 1 1+ + + ... + =0 (17) (p1-z) (p2-z) (Pn-z) and this will enable us to determine their values.
Ifpi, p2, , pn , andqi, q2, 1 qn are known they could be used in the set of 2nequations given by equation (12) and (16) to determine our coefficientsPP... Pnof the functionK(x) l' 2 that we are constructing. Hence ifp1, p2, ..., pnand 49
P1,P2,...,Pnare determined, they could be put back in our expres- sion forK(x)and we thus see that the resulting expression for
K(x) given by equation (4) will satisfy the integral equation (*). The problem therefore reduces to that of finding the value for
K(x)and this can be done by eliminatingP1, P2, ...,Pndetermi- nantly from the n-equations defined by (16) together with (4).From these equations we have:
1 1 1 (Pi -qi)(p2 -q1) (Pn-q1)
1 1 1 (p2 -q2) K(x)(P1-q2) (pn-q2)
1 1 1 (1°1 -qn)(132-qn) (Pn-qn)
1 1 1 1 (Pn-q1) (p1 -q1)
1 1 1 pix 1 p2X = -e (P2 -q2) (Pn-q2) e (P1-q2)
1 1 1 1 1 1 ... (1°2-qn) (Pn-qn) (Prqn) `Priqn)
and factorizing these determinants [5, p. 380] we have 50
(1)1-q1"1-c12) (131-qn) (132-c11"2-c12) (p2-cin) K(x) = ePlx eP2x (P1-132"1-133) (pl-Pn) (P2-131)(P2-133) (132 Pn)
(pn-c11)(Pn-c12). (pn-qn) - ePnx (18) (Pn-P1)(Pn-P2). (Pn-Pn-1)
With this value ofK(x), equation (*) and hence equation (3) is a solution of our original integral equation defined by (1). The method, as a whole is very much involved and so in the next chapter we shall discuss a method proposed by G. C. Evans in 1911 [2] for finding the resolvent kernels (and hence solving the integral equation containing the kernel) for kernels that satisfy differential equations together with known initial conditions.The method will therefore be applicable in treating a kernel of Whittaker's exponential- sum type since it will always satisfy a linear ordinarydifferential equation with constant coefficients. 51
V. CALCULATION OF THE RESOLVENT KERNEL BY THE METHOD OF G. C. EVANS
The resolvent kernelR(x, t;X)of the Volterra integral equation
y(x) = f(x) + XSk(x, t)y(t)dt (i) 0 corresponding to the initial value problem
any(n)(x)+ a y(n-1)(x) + ...+ ay(1)(x)+ ay(x) = f(x) n-1 1 0
(n -1) n -1(n-2)-1 n-2 (0) = y ( 0 )= yo , y(0) = yo y ,y ' was found to be
oo i-lk(i)(x, R(x, t;)) = t) i=1 wherek(x, t)is the kernel of the integral equation (i).We pointed out that different kernelsk(x, t)will produce different resolvent kernels according to equation (ii).The method of iteration could be employed to solve for the resolvent kernels for some simple kernels of the form k(x, t) = 1 or k(x, t) = k(x-t) =em(x-t) 52 However the work involved in calculating resolvent kernels for ker- nels of the form m (x-t) m(x-t) 2 k(x, t) = k(x-t) = Ale 1 + A2e is formidable, and certainly for more involved kernels, finding the resolvent kernels by the method of iteration does not help. We hope that there is a way of finding the resolvent kernel by some other method and so for this chapter we shall discuss the method suggested by G.C. Evans in 1911. Some kernelsk(x, t)do in fact satisfy linear differential equations with known initial conditions, and it is to this family of kernels that we shall apply the method of G.C. Evans. Before we employ the method to solve ourn-thorder non- homogeneous linear differential equation with constant coefficients, we shall apply it to solve for the resolventkernel for the kernel
m(x-t) m(x-t) 1 + A 2 k(x, t) = k(x-t) = Ale 2e where it was almost impossible to solve for it by the method of iteration. We shall assume, without loss of generality that m>m2' 1 andAl' A2' ml' m2 are real constants.This kernelk(x, t)is in fact a solution of the linear differential system: 53
k"(x-t) (m1 +m2)k'(x -t) + mmk(x-t) = 0, 1 2 k(0) = Al + A2 ' (1) le(0) = mA1 + m2A2 1
The general integral equation satisfied by the resolvent kernel
R(x, t;X)which may be represented byR(x, t;X) = r(x-t;X)as we saw in Chapter III is,
r(x-t;X) = k(x-t) + XSIk(x-u)r(u-t;X)du (2)
t
Differentiating twice with respect toxyields:
(x-t;X) = k (x-t) + X kx(x-u)r(u-t;X)du + Xk(0)r(x-t;X) (2a) t and
(x-t;X) = k(x-t) + X k(x-u)r(u-t;X)du rxx xx xx
+ Xkx(0)r(x-t;X) + Xk(0)rx(x-t;X) (3)
Substituting (1) in (3) gives
r(x-t;X.) = {(m1 +m2)k'(x- t)- m1m2k(x -t)} xx
+XS1{(mi+m2)k'(x-u)-m1m2k(x-u)}r(u-t;X)du
+ X.(m1A1+m2A2)r(x-t;X) + X.(A1+A2)rx(x-t;X) (4) 54
For simplicity we shall leave the(x -t;))out of the resolvent kernels and(x-t)out of the kernelsk(x-t). We therefore have:
x + X {(ml +m2)kx-m1m2k}rdu rxx= kxx t
+ X {m1A1 +m2A2 }r + k(Al+A2)rx (5)
Multiplying (2) by(mim2)yields;
(m m 2)r = (mm 2)k + k(mm2).5.krdu (6) I 1
Adding (5) and (6) leads to;
x r + (m1m2)r = kxx + X (ml +m2)kxrdu + k(m1A+m2A xx 1 2)r t
+ X.(A1+A2)rx + (m1m2)k . (7)
From (2a), substituting (1) gives;
r= k+ X J k (x-u)r(u-t; k)du + k (A+A (8) x x 1 2)r t and multiplying (8) by(mi+m2)gives;
(ml +m2)rx = (ml +m2)kx + k(m+m ) J kxrdu+k(m+m )(A+A 1 1 1 2)r (9) 55 Subtracting (7)-(9) leads to;
rxx(rillm2)r(rnl+m2)rx
= k + {(m+m)kxr}du + X(mA+mA xx Sx 1 2 1 1 22)r t x + X(A1+A2)rx + (m1m2)k (n1 +rn2)kxX(mi+m2 kxrdu
"rnl+rn2)(Al+A2)r
= {kxx-(m1+m2)kx+(m1m2)k} + X(A1+A2)rx
+ X.{(m1Al+m2A2)-(mi+m2)(Ai+A2)}r
= k(A1 +A2)rx + X{(m1A1 +m2A2)-(mi+m2)(A1 +A2)}r
Since{kxx-(m1 +m2)kx+mim2k} = 0from (1), leading to the differ- ential equation inr:
rxx - {(m+m)+X(A+A)}r 1 2 1 2 x
+ {(mirr2)-X.(m1A1 +m2A2)+X(mi+m2)(A1 +A2)}r = 0,
and simplifying gives:
r -{(m+m)+X(A ) }rx + {(m1m2) +X(m1A2 +m2A1) }r = 0 xx 1 2 1+AZ (10) which for simplicity may be written as:
rxx- arx + br = 0 56 where the constants
a = {(m1tinz)+X(A1 +A2)} (12) b = {(m1m2)+X(m1A2 +m2A1)} .
In solving equation (11) assume that the solution is of the form
r =es(x-t) and differentiating with respect tox, we obtain:
r =ses(x-t) and r 2 s(x-t) xx= s e
Substituting these in the differential equation (11) yields:
2 s - as + b = 0, with solution 1 /2] [a±(a2-4b) s 2
2 Now to find the value of (a-4b)from (12), we have that:
a2 4b = {(m1 +m2) +X(A1 +A2)}2-4{(mm)+X(mA+m A1)/ I 2 1 2
2 =(ml2+m2+2m1 m2) +X2(A1 +A2)2+ 2(m1 +m2)k(A1 +A2) - 57
- 4 (m m )- 4 X. (m A +m 2A i)
2 2 2 = [m+m-2(mm2)] + X(A+A2)2+ 2X.mA1 + 2XmA2 1 2 1 1 1 1
+ 2kmAl + 2kmA2 - 4kmA2 - 4 kmA1 2 2 1 2
2(A = (m-m)2+ X +A)2+ 2XmA1 - 2Xm1A2 1 2 1 2 1
- 2X.mA1 + 2Xm2A2 2
2 2 2 = (m1 -m2)+ X(A+A2)+ 2XA 1(m-m2) - 2X.A -m2) 1 1 2(ml
= (m-m 2)2 +X2(A1 +A2)2+ 2(m-m)X(A-A2) . 1 1 2 1
Unfortunately, it does not come to a nice perfect square as hoped, but anyway, the two roots of equation (11) are:
1 s = [ {(m+m)+X(A+A 2)} 1 2 1 2 1
2 2 2 +{(m-m ) +X.(A+A ) +2X(ml -m)(A1 -A2)}1 /2] 1 2 1 2 2 and
1 s = [ {(m )+X(A 2 2 l+m2 1+A 2)1
-{(m )2+X2(A+A)2+2X(m )(A1 -A)}1 /2} 1-m 2 1 2 1-m 2 2
So the general solution of (11), which is the resolvent kernel, is:
s(x-t) s(x-t) 1 2 + C r(x-t; X) Cle 2e 58 whereC1, C2are real constants.Therefore,
R(x, t; X)
22(A1+A2) 2 [{(ml+m2)+X(A1+A2)}+{(m1-m2)+X +2X(m1-m2)(ArA2)/2Rx-t) = Cle 1 22(A1+A2) z+2X(m1-m2)(A1-A2)} 1/2 2[{(mi+m2)+X(A1 +A2)}-{(m1-m2)+X ix-t) +C 2e
To eliminate the constantsC1, C2,we see from (2) that if
X = 0 then; r(x-t; 0) = k(x-t)
m(x-t) m(x-t) = Ale 1 + A2e 2 (14)
Also from (13) putting X. = 0 gives:
m(x t) m(x-t) 1 2 R(x-t; 0) = C + C (15) le 2e
Comparing (14) and (15) shows that:
Al = C1, and A2= C2.
Thus the resolvent kernel is given by the expression 59
R(x, t; X) 22(Ai+A2) 2 +2X(m1-ma)(Ai-A2/21 [{(mi+m2)+X(Ai+A2)}+{(m1-m2)+X (x-t) ) = Ale 2 2 ii-m2)+X (A tA2)} -{(m1 -m2)+X2(A1 +A2)+2X (m 1-m2)(AfA2)1121(x-t) +A 2e
As mentioned before, calculating this resolvent kernel by the method of iteration was almost impossible because of the amount of work involved.The method of G. C. Evans has provided a way out.
5.1. Solution of the Volterra Integral Equation Corresponding to the Initial Value Problem for a Non-Homogeneous Linear Ordinary Differential Equation
In Chapter II, we converted the initial value problem for a non- homogeneous linear differential equation with constant coefficients
any(n)(x)+an-ly(n-1)(x) + ... + a0y(x) = f(x) (1) (n-1)(0) n-1 y yo y(0) = y0 into a Volterra integral equation of the form:
x y(x) = f(x) + X Jk(x, t)y(t)dt (2) 0 where we found the kernel of equation (2) to be 60
n-1 an-1-1. (x-t)i k(x, t) = - (3) an (i)! i=0
In Chapter III, we introduced the idea of the resolvent kernel
R(x, t;X)for kernelk(x, t)and we showed that the resolvent kernel actually satisfies the Volterra integral equation
R(x, t;X) = k(x, t) + S. k(x,)RR, t;X)d. t
In other words, in solving fory(x),the solution of our initial value problem, it is sufficient to solve for R(x, t;X-) since equation (2) on page 30 is solved by equation (3) on page 40. In solving (1), our problem therefore reduces to that of finding the resolvent kernelR(x, t;X.), and, as mentioned in the preceding section, the method of iteration will do us no practical good. We shall find it by the method of G. C. Evans. First, we need to find out what differential equation together with suitable boundary conditions is satisfied by the kernel given in equation
(3).To do this, let us write out the terms of the kernelk(x, t)and then differentiatentimes. From equation (3), 61
n-1an-i-1 k(x,t) = (x-t)1 -an (i)! i=0
2 1 (x-t) (x-t) (x-t)3+...+a(x-t)n-1 = - a +a +a +a a n -1 n -2 1! 2! n-4 3! 0(n-1)!
Differentiating with respect toxwe have:
2 (1) 1 (x-011-2 k (x-t) = - a +a (x-t)+an-4 + .+ a an n-2n-3 2! 0 (n-2) !j and n-3 (2) 1 (x-t)2 (x-t) k (x-t) = - an-3+an-4(x-t)+an-5 + + a a 2! 0(n-3)! n
(n-1) a k (x-t) = , 0an from which
k(n)(x-t)= 0.
Our differential system, satisfied byk(x-t)is therefore:
k(n)(x-t)= 0
k(0) = -an-1 /an
k(1)(0)(0) = -a /an n-2 (4) k(2)(0) = -an-3/an
(n-1) k (0) = -a /a 0 n
Having obtained these, we are now in a position to apply the 62 method of G. C. Evans to solve for the resolvent kernelR(x, t;X) for the kernelk(x-t).As we pointed out before, the resolvent kernel
R(x, t;X)is a solution of the Volterra integral equation:
R(x, t;X) = k(x, t) + X 1k(x, u)R(u, t;X)du (5) 0 and sincek(x, t)is of the formk(x-t)in equation (3) we may assume that the resolvent kernelR(x, t;X)is of the formr(x-t;X) so that rewriting equation (5) we have:
r(x-t;).) = k(x-t) + X.sk(x-u)r(u-t;))du (6) 0
To simplify our writing suppose we just writerforr(x-t;X)and
r(u-t;X)andkfork(x-t)andk(x-u),thus obtaining
x r k + X krdu (7) 0
Differentiating this with respect tox, we obtain
x (1) (1) (1) r = k + k rdu + Xk(0)r , 0 and differentiating again leads to 63
(2) (2) (2) (1)(0)r (1) r = k + k rdu + Xk + Xk(0)r . 0
A third differentiation yields:
(3) (3) (2) (1) (1) r =k(3)+ rdu + Xk(0)r + Xk (0)r 0 (2) + Xk(0)r .
We can see a general pattern building up, and after the i-th differen- tiatiori we would obtain:
j) (i-j-1) r(i)=k(i)+XS + X (0)r 0 j=0
In particular, after then-thdifferentiation we have
n-1 (n) (n) (n-i-1) r(n)=k k rdu+X (0)r (8) 0 i=0
Equation (8) can be further simplified since we already showed that k(n)= 0and so
(n) S (n) k + X k rdu = 0. 0
Equation (8) therefore becomes 64
n-1 (n-i-1) r(n)- k(0)r = 0 (9) i=0
Writing out a few terms we have
(n-1) (1) (n 2) (n-1) r(n) Xk(0)r Xic (0)r Xk (0)r = 0 (10)
However, from our expressions for the derivatives ofk(x-t) we see that
k(0) = a /a n-1n (1)(0) k = an-2n/a (11) k(2)(0)(0) = - an-3/an
(n-1)(0) k = - a /a 0 n Equation (10) therefore becomes:
an-1 r(n) an -1(n-1) (n-2) a0 +X. r (x-t;X) + X r (x-t;X)+...+ X r = 0. an an (12)
This is the differential equation satisfied by the resolvent kernel. We can solve this equation provided we havenboundary conditions, since the differential equation is of ordern.To find the conditions we substituteX = 0in the general equation for the derivatives ofr: i.e., in 65
i-1 (i-j-1) r(i)=k(i)+ X k(i)rdu+ (0)r `j0 to obtain
r(i)(x-t,0) =k(i)(x-t),i = 0,1,2, ... , n-1.
We shall further substitutex-t = 0to obtain the set ofnboundary conditions: r(0,0) = k(0) = -an-1 /anfrom (11), and (1)(0,0) r = k(1)(0) = -an-2 /anfrom (11),
r(n-1)(0,0) =k(n-1)(0) = -a/a from (11). 0 n
Our problem therefore reduces to that of solving the homogene- ous linear differential equation with constantcoefficients, together with n boundary conditions:
(n)(x-t;X) + Xa r(n-1)(x-t;X) + = 0 anr n-1 -+ xaor r(0, 0) = -a /a n -1n r(1)(0,0) = -a /a n-2n (13) r(2)(0,0)= -a /a n-3n
r(n-1)(0,0) = -a /a 0 n
Several methods for solving such equations were discussed in 66 detail in Chapter I and so any of them could be readily applied. We already pointed out that the resolvent kernelR(x-t;X)is in fact a solution of the Volterra integral equation deduced from our original initial value problem; therefore once the resolvent kernel R(x-t;X) is determined, we essentially have the solution of our Volterra inte- gral equation corresponding to our original initial value problem for a non-homogeneous linear differential equation with constant coefficients. 67
SUMMARY
Different methods have been employed to solve the initial value problem for a non-homogeneous linear differential equation with constant coefficients
y(n-1)(x) + + ay(x) = f(x) any(n)(x) + an-1 0
n-1 y(0) = yo,y(1)(0)= . ,y(n-1)(0)= y0
We saw that the method of "undetermined coefficients" has its limitations since it could be applied in only a few cases, and so we sought a more powerful method--the Lagrange's method of variationof parameters--for finding a particular solutionv(x)of the non- homogeneous differential equation. The method of Laplace transform was also employed in solving the differential system.It is a nice method but we would almost always have to have a table of Laplace transforms handy every time we want to solve such differential systems. We further converted our differential system into a Volterra integral equation and found that solving by the method of iteration was not the best thing to do because of the work involved.We tried Whittaker 's numerical method but here again we saw that the method was very much involved. 68 The limitations encountered with these various methods led us into developing the idea of the resolvent kernel of the Volterra integral equation and calculating it by the method suggested by G. C. Evans in 1911, having shown that the resolvent kernel actually is a solution of the Volterra integral equation corresponding to the initial value prob- lem for a non-homogeneous linear differential equation with constant coefficients. 69
BIBLIOGRAPHY
1. Boyce, W.E., and. Diprima, R. C.; Elementary Differential Equa- tions and Boundary Value Problems. John Wiley and Sons, Inc., New York.1969. 2. Doetsch, G.; Guide to Applications of Laplace Transforms. Translation editor: W. McA. Fairbairn. New York, Van Nostrand. 1961. 3. Evans, G. C.;Sul calcolo del nucleo dell'equazione risolvente per una data equazione integrale.R. Acc. dei Lincei., Vol. XX, 1911. 4. Ince, E. L.; Ordinary Differential Equations.Longmans, Green and Co. LTD. London, E. C. 4. N. Y.1927. 5. Lonseth, A. T. ,and Rall, L.B.; Linear Integral Equations (yet to be published). 6. Nixon, F.E.; Handbook of Laplace Transforms and Tables. Englewood Cliffs, N. J Prentice-Hall, 1960. 7. Tricomi, F. G. ;Differential Equations.Translated by Elizabeth A. McHarg, London, Blackie; 1961. 8. Volterra, V.; Theory of Functionals and of Integral and Integro- Differential Equations.Dover Publications, Inc. New York, 1958. 9. Whittaker, E. T. ,and Watson, G. N. ;Modern Analysis.4th edition, Cambridge Univ. Press.1935.