Mathematical Methods for Physics and Engineering - Matematica.NET

23.4 CLOSED-FORM SOLUTIONS23.4.1 Separable kernelsThe most straightforward integral equations to solve are Fredholm equationswith separable (or degenerate) kernels. A kernel is separable if it has the formn∑K(x, z) = φ i (x)ψ i (z), (23.8)i=1where φ i (x) areψ i (z) are respectively functions of x only and of z only and thenumber of terms in the sum, n, is finite.Let us consider the solution of the (inhomogeneous) Fredholm equation of thesecond kind,y(x) =f(x)+λ∫ baK(x, z)y(z) dz, (23.9)which has a separable kernel of the form (23.8). Writing the kernel in its separatedform, the functions φ i (x) may be taken outside the integral over z to obtainn∑∫ by(x) =f(x)+λ φ i (x) ψ i (z)y(z) dz.i=1Since the integration limits a and b are constant for a Fredholm equation, theintegral over z in each term of the sum is just a constant. Denoting these constantsbyc i =the solution to (23.9) is found to be∫ bay(x) =f(x)+λaψ i (z)y(z) dz, (23.10)n∑c i φ i (x), (23.11)where the constants c i can be evalutated by substituting (23.11) into (23.10).◮Solve the integral equationy(x) =x + λ∫ 10i=1(xz + z 2 )y(z) dz. (23.12)The kernel for this equation is K(x, z) =xz + z 2 , which is clearly separable, and using thenotation in (23.8) we have φ 1 (x) =x, φ 2 (x) =1,ψ 1 (z) =z and ψ 2 (z) =z 2 . From (23.11)the solution to (23.12) has the formy(x) =x + λ(c 1 x + c 2 ),where the constants c 1 and c 2 are given by (23.10) asc 1 =c 2 =∫ 10∫ 10z[z + λ(c 1 z + c 2 )] dz = 1 3 + 1 3 λc 1 + 1 2 λc 2,z 2 [z + λ(c 1 z + c 2 )] dz = 1 4 + 1 4 λc 1 + 1 3 λc 2.807