PDES: SEPARATION OF VARIABLES AND OTHER METHODS
yzxV r^0r 1+
−
Figure 21.12 The arrangement of images forsolving Laplace’s equation in
the half-spacez>0.We may evaluate this normal derivative by writing the Green’s function (21.94) explicitly
in terms ofx,yandz(andx 0 ,y 0 andz 0 ) and calculating the partial derivative with respect
tozdirectly. It is usually quicker, however, to use the fact that§
∇|r−r 0 |=r−r 0
|r−r 0 |; (21.96)
thus
∇G(r,r 0 )=r−r 0
4 π|r−r 0 |^3−
r−r 1
4 π|r−r 1 |^3.
Sincer 0 =(x 0 ,y 0 ,z 0 )andr 1 =(x 0 ,y 0 ,−z 0 ) the normal derivative is given by
−∂G(r,r 0 )
∂z=−k·∇G(r,r 0 )=−
z−z 0
4 π|r−r 0 |^3+
z+z 0
4 π|r−r 1 |^3.
Therefore on the surfacez= 0, and writing out the dependence onx,yandzexplicitly,
we have
−∂G(r,r 0 )
∂z∣
∣∣
∣
z=0=
2 z 0
4 π[(x−x 0 )^2 +(y−y 0 )^2 +z^20 ]^3 /^2.
Inserting this expression into (21.95) we obtain the solutionu(x 0 ,y 0 ,z 0 )=z 0
2 π∫∞
−∞∫∞
−∞f(x, y)
[(x−x 0 )^2 +(y−y 0 )^2 +z^20 ]^3 /^2dx dy.An analogous procedure may be applied in two-dimensional problems. For§Since|r−r 0 | (^2) =(r−r 0 )·(r−r 0 ) we have∇|r−r 0 | (^2) =2(r−r 0 ),from which we obtain
∇(|r−r 0 |^2 )^1 /^2 =^1
2
2(r−r 0 )
(|r−r 0 |^2 )^1 /^2
=r−r^0
|r−r 0 |
.
Note that this result holds in twoandthree dimensions.