1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

0.5 Green’s Functions 49

The general solution of the corresponding homogeneous equation isu(x)=
c 1 +c 2 ln(x).Thus,wewouldchoose

u 1 (x)= 1 , u 2 (x)=ln(x)

so thatu 1 (x)is bounded atx=0andu 2 (x)is 0 atx=1. The Green’s function
is thus

G(x,z)=

{zln(x), 0 <z≤x,

zln(z), x≤z<1.

A similar procedure is followed if the intervall<x<ris infinite in length.

EXERCISES

In Exercises 1–8, find the Green’s function for the problem stated.

1. d

(^2) u
dx^2 =f(x),0<x<a,
u( 0 )=0, u(a)=0.

2. d

(^2) u
dx^2
=f(x),0<x<a,
u( 0 )=0,
du
dx(a)=0.

3. d

(^2) u
dx^2 −γ
(^2) u=f(x),0<x<a,
du
dx
( 0 )=0, u(a)=0.

4.^1
r

d

dr

(

rdu

dr

)

=f(r),0≤r<c,

u(c)=0, u(r)bounded atr=0.

5.

1

ρ^2

d

dρ

(

ρ^2

du

dρ

)

=f(ρ),0≤ρ<c,

u(c)=0, u(ρ)bounded atρ=0.

6.

d^2 u

dx^2 +

1

x

du

dx−

1

4 x^2 u=f(x),0≤x<a,

u(a)=0, u(x)bounded atx=0.

7. d

(^2) u
dx^2
−γ^2 u=f(x),0<x,
u( 0 )=0, u(x)bounded asx→∞.