- #1
yungman
- 5,755
- 292
1) If [itex]u(r,\theta,\phi)=\frac{1}{r}[/itex], is [itex]\frac{\partial{u}}{\partial {\theta}}=\frac{\partial{u}}{\partial {\phi}}=0[/itex] because [itex]u[/itex] is independent of [itex]\theta[/itex] and[itex] \;\phi[/itex]?
2) If [itex]u(r,\theta,\phi)=\frac{1}{r}[/itex], is:
[tex]\nabla^2u(r,\theta,\phi)=\frac{\partial^2{u}}{\partial {r}^2}+\frac{2}{r}\frac{\partial{u}}{\partial {r}}+\frac{1}{r^2}\frac{\partial^2{u}}{\partial {\theta}^2}+\frac{\cot\theta}{r^2}\frac{\partial{u}}{\partial {\theta}}+\frac{1}{r^2\sin\theta}\frac{\partial^2{u}}{\partial {\phi}^2}=\frac{\partial^2{u}}{\partial {r}^2}+\frac{2}{r}\frac{\partial{u}}{\partial {r}}=0[/tex]
Because [itex]\frac{\partial{u}}{\partial {\theta}}=\frac{\partial{u}}{\partial {\phi}}=0[/itex].
Thanks
2) If [itex]u(r,\theta,\phi)=\frac{1}{r}[/itex], is:
[tex]\nabla^2u(r,\theta,\phi)=\frac{\partial^2{u}}{\partial {r}^2}+\frac{2}{r}\frac{\partial{u}}{\partial {r}}+\frac{1}{r^2}\frac{\partial^2{u}}{\partial {\theta}^2}+\frac{\cot\theta}{r^2}\frac{\partial{u}}{\partial {\theta}}+\frac{1}{r^2\sin\theta}\frac{\partial^2{u}}{\partial {\phi}^2}=\frac{\partial^2{u}}{\partial {r}^2}+\frac{2}{r}\frac{\partial{u}}{\partial {r}}=0[/tex]
Because [itex]\frac{\partial{u}}{\partial {\theta}}=\frac{\partial{u}}{\partial {\phi}}=0[/itex].
Thanks