Normalmente, as fórmulas são dadas, sem qualquer dedução. Não porque a dedução em si seja difícil, mas porque os cálculos em si podem ser longos e não trazem nada de novo. O problema deste ponto de vista é que há quem nunca tenha visto nem feito uma dedução!
É um mero exercício de cálculo e de aplicação de principalmente da regra da derivação do produto e da regra da cadeia, para quem quiser fazer.
...E que proponho que se faça! Clicando nos botões podem ver a minha solução e a minha resolução. Sugiro que tente fazê-la primeiro!
O Laplaciano em coordenadas polares
Seja $f:D \subseteq \R^2 \to \R$ uma função real de variável vectorial, ou, como vai ser designação comum neste blog, um campo escalar.Suponhamos que a função é de classe $C^2$ em $D$.
O Laplaciano de $f$ é a divergência do gradiente de $f$. \[ \nabla ^2 f = \nabla \cdot \nabla f = \dfrc{{\partial ^2 f}}{{\partial x{}^2}} + \dfrc{{\partial ^2 f}}{{\partial y{}^2}} \] Considere-se a nova função que se obtém fazendo a mudança de variáveis \[ \left\{ {\begin{array}{l} {x = r\cos \theta } \\ {y = r\sen \theta } \end{array}} \right. \] para $r>0$ e $\theta\in [0,2\pi[ $ por forma a que, a nova definição faça sentido.
Por abuso de linguagem, continuemos a designar a nova função por $f$.
Determine uma nova fórmula para $\nabla ^2 f$ em função das novas variáveis $r$ e $\theta$.
\[\nabla ^2 f=\dfrc{{\partial ^2 f}}{{\partial r^2 }} + \dfrc{1}{r}\dfrc{\partial f}{\partial r} + \dfrc{1}{{r^2 }}\dfrc{{\partial ^2 f}}{{\partial \theta ^2 }}\]
(Carlos Paulo A. de Freitas; 19/07/2019)
(Carlos Paulo A. de Freitas; 19/07/2019)
\[
\left\{ {\begin{array}{l}
{x = r\cos \theta } \\
{y = r\sen \theta }
\end{array}} \right. \Rightarrow
\left\{ {\begin{array}{l}
{r^2 = x^2 + y^2 } \\
\cos \theta = \dfrc{x}{r} \\
\sen \theta = \dfrc{y}{r} \\
\end{array}} \right. \Rightarrow
\left\{ {\begin{array}{l}
{2r\dfrc{\partial r}{\partial x} = 2x} \\
2r\dfrc{\partial r}{\partial y} = 2y \\
- \sen \theta \dfrc{{\partial \theta }}{\partial x} = \dfrc{{r - x\dfrc{\partial r}{\partial x}}}{{r^2 }} \\
\cos \theta \dfrc{{\partial \theta }}{\partial y} = \dfrc{{r - y\dfrc{\partial r}{\partial y}}}{{r^2 }}
\end{array}} \right. \Rightarrow
\left\{ {\begin{array}{l}
{\dfrc{\partial r}{\partial x} = \dfrc{x}{r} = \cos \theta } \\
\dfrc{\partial r}{\partial y} = \dfrc{y}{r} = \sen \theta \\
\dfrc{\partial \theta }{\partial x} = \dfrc{{r - r\cos ^2 \theta }}{{ - r^2 \sen \theta }} = \dfrc{{1 - \cos ^2 \theta }}{{ - r\sen \theta }} = - \dfrc{\sen \theta }{r} \\
\dfrc{\partial \theta }{\partial y} = \dfrc{r - r\sen ^2 \theta }{r^2 \cos \theta } = \dfrc{\cos \theta }{r}
\end{array}} \right.
\]
Portanto:
${\color{red}\dfrc{\partial r}{\partial x}=\cos\theta}$ ; ${\color{green}\dfrc{\partial r}{\partial y}=\sen\theta}$ ; $\color{blue} \dfrc{\partial \theta }{\partial x}=- \dfrc{\sen \theta }{r}$ e $\color{brown} \dfrc{\partial \theta }{\partial y} =\dfrc{\cos \theta }{r}$
Estes resultados agora usam-se neste desenvolvimento: \begin{eqnarray*} { \nabla ^2 f }&{=}&{\dfrc{\partial ^2 f}{\partial x^2} + \dfrc{\partial ^2 f}{\partial y^2}}\\ {}&{=}&{\dfrc{\partial }{\partial x}\left( \dfrc{\partial f}{\partial x} \right) + \dfrc{\partial }{\partial y}\left( {\dfrc{\partial f}{\partial y}} \right)}\\ {}&{=}&{\dfrc{\partial }{\partial x}\underbrace{\left(\dfrc{\partial f}{\partial r}{\color{red}\dfrc{\partial r}{\partial x}} + \dfrc{\partial f}{\partial \theta}{\color{blue} \dfrc{\partial \theta }{\partial x}}\right)}_{\text{regra da cadeia}} + \dfrc{\partial }{\partial y}\underbrace{\left(\dfrc{\partial f}{\partial r}{\color{green}\dfrc{\partial r}{\partial y}} + \dfrc{\partial f}{\partial \theta }\color{brown} \dfrc{\partial \theta }{\partial y}\right)}_{\text{regra da cadeia}} }\\ {}&{=}&{\dfrc{\partial }{\partial x}\left( \dfrc{\partial f}{\partial r}\cos \theta - \dfrc{\partial f}{\partial \theta }\dfrc{\sen \theta }{r} \right) + \dfrc{\partial }{\partial y}\left( \dfrc{\partial f}{\partial r}\sen \theta + \dfrc{\partial f}{\partial \theta}\dfrc{\cos \theta }{r} \right)}\\ {}&{=}&{\dfrc{\partial }{\partial r}\left( {\dfrc{\partial f}{\partial r}\cos \theta - \dfrc{\partial f}{\partial \theta }\dfrc{\sen \theta }{r}} \right){\color{red}\dfrc{\partial r}{\partial x}} + \dfrc{\partial }{\partial \theta }\left( {\dfrc{\partial f}{\partial r}\cos \theta - \dfrc{\partial f}{\partial \theta }\dfrc{\sen \theta }{r}} \right){\color{blue} \dfrc{\partial \theta }{\partial x}}}\\ {}&{ }&{+ \dfrc{\partial }{\partial r}\left(\dfrc{\partial f}{\partial r}\sin \theta + \dfrc{\partial f}{\partial \theta }\dfrc{\cos \theta }{r} \right){\color{green}\dfrc{\partial r}{\partial y}} + \dfrc{\partial }{\partial \theta }\left( {\dfrc{\partial f}{\partial r}\sin \theta + \dfrc{\partial f}{\partial \theta }}\dfrc{\cos \theta }{r} \right){\color{brown} \dfrc{\partial \theta }{\partial y}} }\\ {}&{}&{\text{(regra da cadeia)}}\\ {}&{=}&{\left( \dfrc{\partial ^2 f}{\partial r^2 }\cos \theta + \dfrc{\sen \theta}{r^2 }\dfrc{\partial f}{\partial \theta} - \dfrc{\sen \theta}{r}\dfrc{\partial ^2 f}{\partial r\partial \theta } \right){\color{red}\dfrc{\partial r}{\partial x}} + \left( {\dfrc{\partial ^2 f}{\partial \theta \partial r}\cos \theta - \dfrc{\partial f}{\partial r}\sin \theta - \dfrc{\cos \theta}{r}\dfrc{\partial f}{\partial \theta} - \dfrc{\sen \theta}{r}\dfrc{\partial ^2 f}{\partial \theta ^2 }} \right){\color{blue} \dfrc{\partial \theta }{\partial x}} }\\ {}&{}&{+ \left( \dfrc{\partial ^2 f}{\partial r^2 }\sin \theta - \dfrc{\cos \theta}{r^2 }\dfrc{\partial f}{\partial \theta} + \dfrc{\cos \theta}{r}\dfrc{\partial ^2 f}{\partial r\partial \theta } \right){\color{green}\dfrc{\partial r}{\partial y}} + \left( \dfrc{\partial ^2 f}{\partial \theta \partial r}\sin \theta + \dfrc{\partial f}{\partial r}\cos \theta - \dfrc{\sen \theta}{r}\dfrc{\partial f}{\partial \theta} + \dfrc{\cos \theta}{r}\dfrc{\partial ^2 f}{\partial \theta ^2 } \right){\color{brown}\dfrc{\partial \theta}{\partial y}} }\\ {}&{=}&{\dfrc{\partial ^2 f}{\partial r^2 }\cos ^2 \theta + \dfrc{\sin \theta \cos \theta }{r^2 }\dfrc{\partial f}{\partial \theta} - \dfrc{\sen \theta \cos \theta }{r}\dfrc{\partial ^2 f}{\partial r\partial \theta } - \dfrc{\sen \theta \cos \theta }{r}\dfrc{\partial ^2 f}{\partial \theta \partial r} + \dfrc{\partial f}{\partial r}\dfrc{\sin ^2 \theta }{r} + \dfrc{\sin \theta \cos \theta }{r^2 }\dfrc{\partial f}{\partial \theta} + \dfrc{\sin ^2 \theta }{r^2 }\dfrc{\partial ^2 f}{\partial \theta ^2 } }\\ {}&{ }&{+ \dfrc{\partial ^2 f}{\partial r^2 }\sin ^2 \theta - \dfrc{\sin \theta \cos \theta }{r^2 }\dfrc{\partial f}{\partial \theta} + \dfrc{\sen \theta \cos \theta }{r}\dfrc{\partial ^2 f}{\partial r\partial \theta } + \dfrc{\sin \theta \cos \theta }{r}\dfrc{\partial ^2 f}{\partial \theta \partial r} + \dfrc{\partial f}{\partial r}\dfrc{\cos ^2 \theta }{r} - \dfrc{\sen \theta \cos \theta }{r^2 }\dfrc{\partial f}{\partial \theta} + \dfrc{\cos ^2 \theta }{r^2 }\dfrc{\partial ^2 f}{\partial \theta ^2 } }\\ {}&{=}&{\dfrc{\partial ^2 f}{\partial r^2 } + \dfrc{1}{r}\dfrc{\partial f}{\partial r} + \dfrc{1}{r^2 }\dfrc{\partial ^2 f}{\partial \theta ^2 }} \end{eqnarray*}
${\color{red}\dfrc{\partial r}{\partial x}=\cos\theta}$ ; ${\color{green}\dfrc{\partial r}{\partial y}=\sen\theta}$ ; $\color{blue} \dfrc{\partial \theta }{\partial x}=- \dfrc{\sen \theta }{r}$ e $\color{brown} \dfrc{\partial \theta }{\partial y} =\dfrc{\cos \theta }{r}$
Estes resultados agora usam-se neste desenvolvimento: \begin{eqnarray*} { \nabla ^2 f }&{=}&{\dfrc{\partial ^2 f}{\partial x^2} + \dfrc{\partial ^2 f}{\partial y^2}}\\ {}&{=}&{\dfrc{\partial }{\partial x}\left( \dfrc{\partial f}{\partial x} \right) + \dfrc{\partial }{\partial y}\left( {\dfrc{\partial f}{\partial y}} \right)}\\ {}&{=}&{\dfrc{\partial }{\partial x}\underbrace{\left(\dfrc{\partial f}{\partial r}{\color{red}\dfrc{\partial r}{\partial x}} + \dfrc{\partial f}{\partial \theta}{\color{blue} \dfrc{\partial \theta }{\partial x}}\right)}_{\text{regra da cadeia}} + \dfrc{\partial }{\partial y}\underbrace{\left(\dfrc{\partial f}{\partial r}{\color{green}\dfrc{\partial r}{\partial y}} + \dfrc{\partial f}{\partial \theta }\color{brown} \dfrc{\partial \theta }{\partial y}\right)}_{\text{regra da cadeia}} }\\ {}&{=}&{\dfrc{\partial }{\partial x}\left( \dfrc{\partial f}{\partial r}\cos \theta - \dfrc{\partial f}{\partial \theta }\dfrc{\sen \theta }{r} \right) + \dfrc{\partial }{\partial y}\left( \dfrc{\partial f}{\partial r}\sen \theta + \dfrc{\partial f}{\partial \theta}\dfrc{\cos \theta }{r} \right)}\\ {}&{=}&{\dfrc{\partial }{\partial r}\left( {\dfrc{\partial f}{\partial r}\cos \theta - \dfrc{\partial f}{\partial \theta }\dfrc{\sen \theta }{r}} \right){\color{red}\dfrc{\partial r}{\partial x}} + \dfrc{\partial }{\partial \theta }\left( {\dfrc{\partial f}{\partial r}\cos \theta - \dfrc{\partial f}{\partial \theta }\dfrc{\sen \theta }{r}} \right){\color{blue} \dfrc{\partial \theta }{\partial x}}}\\ {}&{ }&{+ \dfrc{\partial }{\partial r}\left(\dfrc{\partial f}{\partial r}\sin \theta + \dfrc{\partial f}{\partial \theta }\dfrc{\cos \theta }{r} \right){\color{green}\dfrc{\partial r}{\partial y}} + \dfrc{\partial }{\partial \theta }\left( {\dfrc{\partial f}{\partial r}\sin \theta + \dfrc{\partial f}{\partial \theta }}\dfrc{\cos \theta }{r} \right){\color{brown} \dfrc{\partial \theta }{\partial y}} }\\ {}&{}&{\text{(regra da cadeia)}}\\ {}&{=}&{\left( \dfrc{\partial ^2 f}{\partial r^2 }\cos \theta + \dfrc{\sen \theta}{r^2 }\dfrc{\partial f}{\partial \theta} - \dfrc{\sen \theta}{r}\dfrc{\partial ^2 f}{\partial r\partial \theta } \right){\color{red}\dfrc{\partial r}{\partial x}} + \left( {\dfrc{\partial ^2 f}{\partial \theta \partial r}\cos \theta - \dfrc{\partial f}{\partial r}\sin \theta - \dfrc{\cos \theta}{r}\dfrc{\partial f}{\partial \theta} - \dfrc{\sen \theta}{r}\dfrc{\partial ^2 f}{\partial \theta ^2 }} \right){\color{blue} \dfrc{\partial \theta }{\partial x}} }\\ {}&{}&{+ \left( \dfrc{\partial ^2 f}{\partial r^2 }\sin \theta - \dfrc{\cos \theta}{r^2 }\dfrc{\partial f}{\partial \theta} + \dfrc{\cos \theta}{r}\dfrc{\partial ^2 f}{\partial r\partial \theta } \right){\color{green}\dfrc{\partial r}{\partial y}} + \left( \dfrc{\partial ^2 f}{\partial \theta \partial r}\sin \theta + \dfrc{\partial f}{\partial r}\cos \theta - \dfrc{\sen \theta}{r}\dfrc{\partial f}{\partial \theta} + \dfrc{\cos \theta}{r}\dfrc{\partial ^2 f}{\partial \theta ^2 } \right){\color{brown}\dfrc{\partial \theta}{\partial y}} }\\ {}&{=}&{\dfrc{\partial ^2 f}{\partial r^2 }\cos ^2 \theta + \dfrc{\sin \theta \cos \theta }{r^2 }\dfrc{\partial f}{\partial \theta} - \dfrc{\sen \theta \cos \theta }{r}\dfrc{\partial ^2 f}{\partial r\partial \theta } - \dfrc{\sen \theta \cos \theta }{r}\dfrc{\partial ^2 f}{\partial \theta \partial r} + \dfrc{\partial f}{\partial r}\dfrc{\sin ^2 \theta }{r} + \dfrc{\sin \theta \cos \theta }{r^2 }\dfrc{\partial f}{\partial \theta} + \dfrc{\sin ^2 \theta }{r^2 }\dfrc{\partial ^2 f}{\partial \theta ^2 } }\\ {}&{ }&{+ \dfrc{\partial ^2 f}{\partial r^2 }\sin ^2 \theta - \dfrc{\sin \theta \cos \theta }{r^2 }\dfrc{\partial f}{\partial \theta} + \dfrc{\sen \theta \cos \theta }{r}\dfrc{\partial ^2 f}{\partial r\partial \theta } + \dfrc{\sin \theta \cos \theta }{r}\dfrc{\partial ^2 f}{\partial \theta \partial r} + \dfrc{\partial f}{\partial r}\dfrc{\cos ^2 \theta }{r} - \dfrc{\sen \theta \cos \theta }{r^2 }\dfrc{\partial f}{\partial \theta} + \dfrc{\cos ^2 \theta }{r^2 }\dfrc{\partial ^2 f}{\partial \theta ^2 } }\\ {}&{=}&{\dfrc{\partial ^2 f}{\partial r^2 } + \dfrc{1}{r}\dfrc{\partial f}{\partial r} + \dfrc{1}{r^2 }\dfrc{\partial ^2 f}{\partial \theta ^2 }} \end{eqnarray*}
O Laplaciano em coordenadas esféricas
Seja $f:D \subseteq \R^3 \to \R$ um campo escalar.Suponhamos que a função é de classe $C^2$ em $D$.
O Laplaciano de $f$ é a divergência do gradiente de $f$. \[ \nabla ^2 f = \nabla \cdot \nabla f = \dfrc{\partial ^2 f}{\partial x^2} + \dfrc{\partial ^2 f}{\partial y^2}+ \dfrc{\partial ^2 f}{\partial z^2} \] Considere-se a nova função que se obtém fazendo a mudança de variáveis \[ \left\{ {\begin{array}{l} {x = r\sen \theta \cos \phi} \\ {y = r\sen \theta \sen \phi} \\ {z = r\cos \theta} \end{array}} \right. \] para $r>0$, $\theta\in [0,\pi[ $ e $\phi\in [0,2\pi[ $ por forma a que, a nova definição faça sentido.
Por abuso de linguagem, continuemos a designar a nova função por $f$.
Mostre que \[ \nabla ^2 f = \frac{1}{r^2}\left[\frac{\partial}{\partial r} \left( {r^2}\dfrc{\partial f}{\partial r} \right) + \frac{1}{ \sen \theta }\frac{\partial }{{\partial \theta }}\left( {\sen \theta \frac{{\partial f}}{{\partial \theta }}} \right) + \frac{1}{{ \sen ^2 \theta }}\frac{{\partial ^2 f}}{{\partial \phi ^2 }}\right] \]
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