0.5 Green’s Functions 47
u(x)=∫x0sinh(z)sinh( 1 −x)
sinh( 1 ) dz+∫ 1
xsinh(x)sinh( 1 −z)
sinh( 1 ) dz=sinhsinh(^1 (− 1 )x)cosh(z)∣∣
∣∣
x
0+sinhsinh((x 1 ))(
−cosh( 1 −z))∣∣∣
∣
1
x
=sinh(^1 −x)
sinh( 1 )(
cosh(x)− 1)
+sinh(x)
sinh( 1 )(
cosh( 1 −x)− 1)
=sinh(^1 −x)cosh(x)+sinh(x)cosh(^1 −x)
sinh( 1 )−sinh(^1 −x)+sinh(x)
sinh( 1 )= 1 −sinh( 1 −x)+sinh(x)
sinh( 1 ).This, finally, is easily seen to be the correct solution. In this instance, there
are much quicker ways to arrive at the same result. The advantage of the
Green’s function is that it shows how the solution of the problem depends
on the inhomogeneityf(x). It is an efficient way to obtain the solution in
some cases.
Now let us look back over the calculations and see if there is some place they
might fail. Aside from the possibility that the coefficientsk(x)orp(x)in the
differential equation might not be continuous, it seems that division by 0 is
the only possibility of failure. Quantities canceled or divided by were
W(x)=∣∣
∣∣
∣
u 1 (x) u 2 (x)
u′ 1 (x) u′ 2 (x)∣∣
∣∣
∣,
αu 2 (l)−α′u′ 2 (l),
βu 1 (r)+β′u′ 1 (r)in Eqs. (7), (10), and (12), respectively. It can be shown that all three of these
are 0 if any one of them is 0, and, in that case,u 1 (x)andu 2 (x)are proportional.
We summarize in a theorem.
Theorem.Let k(x),p(x),andf(x)be continuous, l≤x≤r. The boundary value
problem
d^2 u
dx^2+k(x)du
dx+p(x)u=f(x), l<x<r,αu(l)−α′u′(l)= 0 , (i)
βu(r)+β′u′(r)= 0 , (ii)