48 Chapter 0 Ordinary Differential Equations
has one and only one solution, unless there is a nontrivial solution of
d^2 u
dx^2+k(x)du
dx+p(x)u= 0 , l<x<r,that satisfies(i)and(ii).
When a unique solution exists, it is given by Eqs.(17)and(18).
Example.
The boundary value problem
d^2 u
dx^2 +u=−^1 ,^0 <x<π,
u( 0 )= 0 , u(π )= 0 ,does not have a unique solution, according to the theorem, becauseu(x)=
sin(x)is a nontrivial solution of the problem
d^2 u
dx^2+u= 0 , 0 <x<π,
u( 0 )= 0 , u(π )= 0.Indeed, if we try to follow through the construction, we find thatu 1 (x)=
sin(x)and alsou 2 (x)=sin(x)(or a multiple thereof ), and so all three quanti-
ties in Eq. (19) are 0.
On the other hand, suppose we try to obtain a solution by the usual method.
The general solution of the differential equation is
u(x)=− 1 +c 1 cos(x)+c 2 sin(x).However, application of the boundary conditions leads to the contradictory
requirements
− 1 +c 1 =0and− 1 −c 1 = 0.Thus, in this case, there simply is no solution to the problem stated.
If the differential equation (1) has a singular point atx=lorx=r(or
both), a Green’s function may still be constructed. The boundary condition (2)
or (3) would be replaced by a boundedness condition, which would also apply
tou 1 oru 2 as the case may be.
Example.
Construct Green’s function for the problem
1
xd
dx(
xdudx)
=f(x), 0 <x< 1 ,u( 0 ) bounded, u( 1 )= 0.