Mathematical Methods for Physics and Engineering - Matematica.NET

PDES: SEPARATION OF VARIABLES AND OTHER METHODSzV+r 0yx−r 1Figure 21.12 The arrangement of images for solving Laplace’s equation inthe half-space z>0.We may evaluate this normal derivative by writing the Green’s function (21.94) explicitlyin terms of x, y and z (and x 0 , y 0 and z 0 ) and calculating the partial derivative with respectto z directly. 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 04π|r − r 0 | − r − r 13 4π|r − r 1 | . 3Since r 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)= −k · ∇G(r, r 0 )∂z= − z − z 04π|r − r 0 | + z + z 03 4π|r − r 1 | . 3Therefore on the surface z = 0, and writing out the dependence on x, y and z explicitly,we have− ∂G(r, r 0)2z 0∂z ∣ =z=04π[(x − x 0 ) 2 +(y − y 0 ) 2 + z .0 2]3/2 Inserting this expression into (21.95) we obtain the solutionu(x 0 ,y 0 ,z 0 )= z ∫ ∞ ∫ ∞0f(x, y)dx dy. ◭2π −∞ −∞ [(x − x 0 ) 2 +(y − y 0 ) 2 + z0 2 ]3/2An 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(r − r 0 )2 (|r − r 0 | 2 ) 1/2 = r − r 0|r − r 0 | .Note that this result holds in two and three dimensions.760