Bsp 24

[dj_m.o.h.t.]

Man zeige: u=ln Wurzel (x-a)^2+(y-b)^2 genügt der Differentialgleichung uxx+uyy=0!

[dj_m.o.h.t.]

ux = 1 / Wurzel (x-a)^2+(y-b)^2 * 1/2 * [(x-a)^2 + (y-b)^2)^-1/2 * [2 * (x-a)] = x-a / (x-a)^2 + (y-b)^2

uxx = (x-a)^2+(y-b)^2-(x-a)*2*(x-a) / [(x-a)^2+(y-b)^2]^2 = (y-b)^2 - (x-a)^2 / [(x-a)^2 + (y-b)^2]

uy = y-b / (x-a)^2 + (y-b)^2

uyy = (x-a)^2 - (y-b)^2 / [(x-a)^2 + (y-b)]^2

uxx + uyy = 0

(y-b)^2 - (x-a)^2 + (x-a)^2 - (y-b)^2 / [(x-a)^2 + (y-b)^2]^2 = 0 / [(x-a)^2 + (y-b)^2]^2 = 0 => w. A.