15.2 LINEAR EQUATIONS WITH VARIABLE COEFFICIENTS
Use Green’s functions to solve
d^2 y
dx^2+y=f(x), (15.69)subject to the one-point boundary conditionsy(0) =y′(0) = 0.We again require (15.67) to hold and so again we assume a Green’s function of the form
G(x, z)={
A(z)sinx+B(z)cosx forx<z,
C(z)sinx+D(z)cosx forx>z.However, we now requireG(x, z) to obey the boundary conditionsG(0,z)=G′(0,z)=0,
which implyA(z)=B(z) = 0. Therefore we have
G(x, z)={
0forx<z,
C(z)sinx+D(z)cosx forx>z.Applying the continuity conditions onG(x, z) as before now gives
C(z)sinz+D(z)cosz=0,
C(z)cosz−D(z)sinz=1,which are solved to give
C(z)=cosz, D(z)=−sinz.So finally the Green’s function is given by
G(x, z)={
0forx<z,
sin(x−z)forx>z,and the general solution to (15.69) that obeys the boundary conditionsy(0) =y′(0) = 0 is
y(x)=∫∞
0G(x, z)f(z)dz=
∫x0sin(x−z)f(z)dz.Finally, we consider how to deal with inhomogeneous boundary conditionssuch asy(a)=α,y(b)=βory(0) =y′(0) =γ,whereα, β, γare non-zero. The
simplest method of solution in this case is to make a change of variable such that
the boundary conditions in the new variable,usay, are homogeneous, i.e.u(a)=
u(b)=0oru(0) =u′(0) = 0 etc. Fornth-order equations we generally require
nboundary conditions to fix the solution, but thesenboundary conditions can
be of various types: we could have then-point boundary conditionsy(xm)=ym
form=1ton, or the one-point boundary conditionsy(x 0 )=y′(x 0 )=···=
y(n−1)(x 0 )=y 0 , or something in between. In all cases a suitable change of variable
is
u=y−h(x),whereh(x)isan(n−1)th-order polynomial that obeys the boundary conditions.