46 Chapter 0 Ordinary Differential Equations
Then the formula given in Eq. (16) forusimplifies to
u(x)=∫rlG(x,z)f(z)dz. (18)Example.
Solve the problem that follows by constructing the Green’s function.
d^2 u
dx^2−u=− 1 , 0 <x< 1 ,
u( 0 )= 0 , u( 1 )= 0.First, we must find two independent solutions of the homogeneous differential
equationu′′−u=0 that satisfy the boundary conditions as required. The
generalsolutionofthehomogeneous differential equation is
u(x)=c 1 cosh(x)+c 2 sinh(x).Asu 1 (x)is required to satisfy the condition at the left,u 1 ( 0 )=0, we take
c 1 =0,c 2 =1 and concludeu 1 (x)=sinh(x). The second solution is to sat-
isfyu 2 ( 1 )=0. We may take
u 2 (x)=sinh( 1 )cosh(x)−cosh( 1 )sinh(x)=sinh( 1 −x).The Wronskian of the two solutions is
W(x)=∣∣
∣∣sinh(x) sinh(^1 −x)
cosh(x) −cosh( 1 −x)∣∣
∣∣=−sinh( 1 ).Now, by Eq. (17), the Green’s function for this problem is
G(x,z)=
sinh(z)sinh( 1 −x)
−sinh( 1 ), 0 <z≤x,
sinh(x)sinh( 1 −z)
−sinh( 1 ) , x≤z<1.Furthermore, sincef(x)=−1, the solution, by Eq. (18), is the integral
u(x)=∫ 1
0−G(x,z)dz.To actually carry out the integration, we must break the interval of integration
atx,thusrevertingineffecttoEq.(16).Theresult: