Mostre que a equação de Laplace em coordenadas cilíndricas é
u_{\rho \rho}+\frac{2}{\rho} u_{\rho}+\frac{1}{r^{2}} u_{\theta \theta}+\frac{1}{\rho^{2} \operatorname{sen}^{2} \varphi} u_{\theta \theta}+\frac{\cot \varphi}{r^{2}} u_{\varphi}=0 .
Sugestão: use x=\rho \operatorname{sen} \varphi \cos \theta, y=\rho \operatorname{sen} \varphi \operatorname{sen} \theta, z=\rho \cos \varphi e a regra da cadeia.

Passo 1

O enunciado pede para mostrar que, ao escrever a equação de Laplace
\nabla^{2}u = u_{xx} + u_{yy} + u_{zz} = 0
em coordenadas esféricas (\rho,\varphi,\theta) definidas por
x = \rho\sin\varphi\cos\theta,\quad y = \rho\sin\varphi\sin\theta,\quad z = \rho\cos\varphi,
obtemos
u_{\rho\rho}+\frac{2}{\rho}u_{\rho} +\frac{1}{\rho^{2}}\,u_{\varphi\varphi} +\frac{\cot\varphi}{\rho^{2}}\,u_{\varphi} +\frac{1}{\rho^{2}\sin^{2}\varphi}\,u_{\theta\theta} =0.
Para isso, vamos usar a regra da cadeia ou, de forma mais sistemática, os fatores de escala em coordenadas esféricas.

Passo 2

Em coordenadas esféricas ortogonais (q_{1},q_{2},q_{3})=(\rho,\varphi,\theta) o elemento de linha é
ds^{2} = d\rho^{2}+(\rho\,d\varphi)^{2}+(\rho\sin\varphi\,d\theta)^{2},
logo os fatores de escala são
h_{\rho}=1,\quad h_{\varphi}=\rho,\quad h_{\theta}=\rho\sin\varphi.
Nessas coordenadas, a fórmula geral para o Laplaciano de uma função u é
\nabla^{2}u =\frac{1}{h_{\rho}h_{\varphi}h_{\theta}}\Bigl[ \frac{\partial}{\partial \rho}\bigl(h_{\varphi}h_{\theta}\,\partial_{\rho}u\bigr) +\frac{\partial}{\partial \varphi}\bigl(h_{\theta}h_{\rho}\,\partial_{\varphi}u\bigr) +\frac{\partial}{\partial \theta}\bigl(h_{\rho}h_{\varphi}\,\partial_{\theta}u\bigr) \Bigr].

Passo 3

Substituindo h_{\rho}=1, h_{\varphi}=\rho, h_{\theta}=\rho\sin\varphi na expressão acima:
\nabla^{2}u =\frac{1}{\rho^{2}\sin\varphi}\Bigl[ \frac{\partial}{\partial \rho}\bigl(\rho^{2}\sin\varphi\,u_{\rho}\bigr) +\frac{\partial}{\partial \varphi}\bigl(\rho\sin\varphi\,u_{\varphi}\bigr) +\frac{\partial}{\partial \theta}\bigl(\rho\,u_{\theta}\bigr) \Bigr].
Note que
\frac{\partial}{\partial \theta}\bigl(\rho\,u_{\theta}\bigr) =\rho\,u_{\theta\theta},
pois \rho independe de \theta. Então
\nabla^{2}u =\frac{1}{\rho^{2}\sin\varphi}\Bigl[ \sin\varphi\,\frac{\partial}{\partial \rho}\bigl(\rho^{2}u_{\rho}\bigr) +\rho\,\frac{\partial}{\partial \varphi}\bigl(\sin\varphi\,u_{\varphi}\bigr) +\rho\sin\varphi\,u_{\theta\theta} \Bigr].

Passo 4

Agora expandimos as derivadas:
\frac{\partial}{\partial \rho}\bigl(\rho^{2}u_{\rho}\bigr) =2\rho\,u_{\rho}+\rho^{2}u_{\rho\rho}, \quad \frac{\partial}{\partial \varphi}\bigl(\sin\varphi\,u_{\varphi}\bigr) =\cos\varphi\,u_{\varphi}+\sin\varphi\,u_{\varphi\varphi}.
Assim
\nabla^{2}u =\frac{1}{\rho^{2}\sin\varphi}\Bigl[ \sin\varphi\,(2\rho\,u_{\rho}+\rho^{2}u_{\rho\rho}) +\rho\bigl(\cos\varphi\,u_{\varphi}+\sin\varphi\,u_{\varphi\varphi}\bigr) +\rho\sin\varphi\,u_{\theta\theta} \Bigr].
Separando cada termo e simplificando:
\nabla^{2}u =\underbrace{u_{\rho\rho}+\frac{2}{\rho}u_{\rho}}_{\text{parte radial}} +\underbrace{\frac{1}{\rho^{2}}\;u_{\varphi\varphi} +\frac{\cot\varphi}{\rho^{2}}\;u_{\varphi}}_{\text{parte polar}} +\underbrace{\frac{1}{\rho^{2}\sin^{2}\varphi}\;u_{\theta\theta}}_{\text{parte azimutal}}.
Portanto, impondo \nabla^{2}u=0, obtemos exatamente
u_{\rho\rho}+\frac{2}{\rho}u_{\rho} +\frac{1}{\rho^{2}}\,u_{\varphi\varphi} +\frac{\cot\varphi}{\rho^{2}}\,u_{\varphi} +\frac{1}{\rho^{2}\sin^{2}\varphi}\,u_{\theta\theta} =0.

Resposta

u_{\rho\rho}+\frac{2}{\rho}u_{\rho} +\frac{1}{\rho^{2}}\,u_{\varphi\varphi} +\frac{\cot\varphi}{\rho^{2}}\,u_{\varphi} +\frac{1}{\rho^{2}\sin^{2}\varphi}\,u_{\theta\theta} =0.

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