Mathematical Methods for Physics and Engineering : A Comprehensive Guide

15.2 LINEAR EQUATIONS WITH VARIABLE COEFFICIENTS

(ii) When considered as a function ofxaloneG(x, z) obeys the specified

(homogeneous) boundary conditions ony(x).

(iii) The derivatives ofG(x, z) with respect toxup to ordern−2 are continuous

atx=z, but the (n−1)th-order derivative has a discontinuity of 1/an(z)

at this point.

Use Green’s functions to solve

d^2 y

dx^2

+y=cosecx, (15.66)

subject to the boundary conditionsy(0) =y(π/2) = 0.

From (15.62) we see that the Green’s functionG(x, z)mustsatisfy

d^2 G(x, z)

dx^2

+G(x, z)=δ(x−z). (15.67)

Now it is clear that forx=zthe RHS of (15.67) is zero, and we are left with the
task of finding the general solution to the homogeneous equation, i.e. the complementary
function. The complementary function of (15.67) consists of a linear superposition of sinx
and cosxandmustconsist of different superpositions on either side ofx=z, since its
(n−1)th derivative (i.e. the first derivative in this case) is required to have a discontinuity
there. Therefore we assume the form of the Green’s function to be

G(x, z)=

{

A(z)sinx+B(z)cosx forx<z,

C(z)sinx+D(z)cosx forx>z.

Note that we have performed a similar (but not identical) operation to that used in the
variation-of-parameters method, i.e. we have replaced the constants in the complementary
function with functions (this time ofz).
We must now impose the relevant restrictions onG(x, z) in order to determine the
functionsA(z),...,D(z). The first of these is thatG(x, z) should itself obey the homogeneous
boundary conditionsG(0,z)=G(π/ 2 ,z) = 0. This leads to the conclusion thatB(z)=
C(z) = 0, so we now have

G(x, z)=

{

A(z)sinx forx<z,

D(z)cosx forx>z.

The second restriction is the continuity conditions given in equations (15.64), (15.65),
namely that, for this second-order equation,G(x, z) is continuous atx=zanddG/dxhas
a discontinuity of 1/a 2 (z) = 1 at this point. Applying these two constraints we have

D(z)cosz−A(z)sinz=0

−D(z)sinz−A(z)cosz=1.

Solving these equations forA(z)andD(z), we find

A(z)=−cosz, D(z)=−sinz.

Thus we have

G(x, z)=

{

−coszsinx forx<z,

−sinzcosx forx>z.

Therefore, from (15.60), the general solution to (15.66) that obeys the boundary conditions