Mathematical Methods for Physics and Engineering - Matematica.NET

INTEGRAL EQUATIONSsides of (23.51) with y j , we find∑a i 〈y j |y i 〉 = λ ∑ a i〈y j |y i 〉 + λ〈y j |Kf〉. (23.52)λii iSince the eigenfunctions are orthonormal and K is an Hermitian operator,we have that both 〈y j |y i 〉 = δ ij and 〈y j |Kf〉 = 〈Ky j |f〉 = λ −1j 〈y j |f〉. Thus thecoefficients a j are given bya j = λλ−1 j 〈y j |f〉1 − λλ −1 = λ〈y j|f〉jλ j − λ , (23.53)and the solution isy = f + ∑ a i y i = f + λ ∑ 〈y i |f〉λii i − λ y i. (23.54)This also shows, incidentally, that a formal representation for the resolvent kernelisR(x, z; λ) = ∑ iy i (x)y ∗ i (z)λ i − λ . (23.55)If f can be expressed as a linear superposition of the y i ,i.e.f = ∑ i b iy i ,thenb i = 〈y i |f〉 and the solution can be written more briefly asy = ∑ 1 − λλ −1 y i . (23.56)iiWe see from (23.54) that the inhomogeneous equation (23.50) has a uniquesolution provided λ ≠ λ i ,i.e.whenλ is not equal to one of the eigenvalues ofthe corresponding homogeneous equation. However, if λ does equal one of theeigenvalues λ j then, in general, the coefficients a j become singular and no (finite)solution exists.Returning to (23.53), we notice that even if λ = λ j a non-singular solution tothe integral equation is still possible provided that the function f is orthogonalto every eigenfunction corresponding to the eigenvalue λ j ,i.e.〈y j |f〉 =∫ bab iy ∗ j (x)f(x) dx =0.The following worked example illustrates the case in which f can be expressed interms of the y i . One in which it cannot is considered in exercise 23.14.◮Use Schmidt–Hilbert theory to solve the integral equationy(x) = sin(x + α)+λ∫ π0sin(x + z)y(z) dz. (23.57)It is clear that the kernel K(x, z) =sin(x + z) is real and symmetric in x and z and is818