Mathematical Methods for Physicists: A concise introduction - Site Map

PARTIAL DIFFERENTIAL EQUATIONS
Substituting this into Eq. (10.68) we obtain
or
Z 1
0
G…0; x 0 † d…0†
dx
G…1; x 0 † d…1† Z 1
dx
0
Z
d2 G
1
dx 2 dx ˆ
d2 G
dx 2 dx ˆ G…1; x 0 † d…1†
dx
G…0; x 0 † d…0† Z 1
dx
0
0
G…x; x 0 †…x†
dx
"
G…x; x 0 †…x†
dx:
"
…10:69†
We must now choose a Green's function which satis®es the following equation
and the boundary conditions:
d 2 G
dx 2 ˆ…x x 0 †; G…0; x 0 †ˆG…1; x 0 †ˆ0: …10:70†
Combining these with Eq. (10.69) we ®nd the solution to be
…x 0 †ˆ
Z 1
0
1
" …x†G…x; x 0 †dx: …10:71†
It remains to ®nd G…x; x 0 †. By integration, we obtain from Eq. (10.70)
Z
dG
dx ˆ …x x 0 †dx ‡ a ˆU…x x 0 †‡a;
where U is the unit step function and a is an integration constant to be determined
later. Integrating once we get
Z
G…x; x 0 †ˆ U…x x 0 †dx ‡ ax ‡ b ˆ…x x 0 †U…x x 0 †‡ax ‡ b:
Imposing the boundary conditions on this general solution yields two equations:
From these we ®nd
and the Green's function is
G…0; x 0 †ˆx 0 U…x 0 †‡a 0 ‡ b ˆ 0 ‡ 0 ‡ b ˆ 0;
G…1; x 0 †ˆ…1 x 0 †U…1 x 0 †‡a ‡ b ˆ 0:
a ˆ…1 x 0 †U…1 x 0 †; b ˆ 0
G…x; x 0 †ˆ…x x 0 †U…x x 0 †‡…1 x 0 †x:
…10:72†
This gives the response at x 0 due to a unit source at x. Interchanging x and x 0 in
Eqs. (10.70) and (10.71) we ®nd the solution of Eq. (10.67) to be
…x† ˆ
Z 1
0
1
" …x 0 †G…x 0 ; x†dx 0 ˆ
Z 1
0
1
" …x 0 †‰…x 0 x†U…x 0 x†‡…1 x†x 0 Šdx 0 :
…10:73†
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