Laplace equation in polar coordinates.

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[tex]\nabla^2 u=\frac {\partial ^2 u}{\partial x^2}+\frac {\partial ^2 u}{\partial y^2}=\frac {\partial ^2 u}{\partial r^2}+\frac{1}{r}\frac {\partial u}{\partial r}+\frac{1}{r^2}\frac {\partial ^2 u}{\partial \theta^2}[/tex]

I want to verify ##u=u(r,\theta)##, not ##u(x,y)##

Because for ##u(x,y)##, it will just be ##\frac {\partial ^2 u}{\partial x^2}+\frac {\partial ^2 u}{\partial y^2}##

Thanks
 
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  • #3
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I know the derivation, I just want to confirm as none of the books or notes specify this.

It should be ##u(r,\theta)## not ##u(x,y)##.
 
  • #4
vanhees71
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Of course for [itex]\vec{r} \neq 0[/itex] you have
[tex]u(\vec{r})=u[\vec{r}(r,\theta)]=u[\vec{r}(x,y)].[/tex]
The vector operators are better defined through covariant integrals than by brute force as in UltrafastPED's source, although it's of course a valid way to find the expression for the covariant differential operators.

A third, more physicist's way, is to use the action principle. Define the action
[tex]A=\int_{\mathbb{R}^2} \mathrm{d}^2 \vec{r} \left [\frac{1}{2}(\vec{\nabla} u)^2+u \Phi \right],[/tex]
where [itex]\Phi[/itex] is an arbitrary external field. Taking the variation you see that the stationary point is
[tex]\frac{\delta A}{\delta u}=0 \; \Rightarrow \; \Delta u=\Phi.[/tex]
To evaluate thus the Laplace operator in polar coordinates, you simply write out the action in these coordinates. You only need the gradient in polar coordinates, which is very easy to derive by using differential forms:
[tex]\mathrm{d} u=\mathrm{d} \vec{x} \cdot \vec{\nabla} u=\mathrm{d} r \vec{e}_r \cdot \vec{\nabla} u + r \mathrm{d} \theta \vec{e}_\vartheta \cdot \vec{\nabla} u=\mathrm{d} r \partial_r u + \mathrm{d} \theta \partial_{\theta} u.[/tex]
Comparison of both sides of the last equation of this line shows that
[tex]\vec{\nabla} u = \vec{e}_r \partial_r u + \vec{e}_{\theta} \frac{1}{r} \partial_{\theta} u.[/tex]
Further you have
[tex]\mathrm{d}^2 \vec{r}=r \mathrm{d} r \mathrm{d} \theta,[/tex]
and thus
[tex]A=\int_0^{\infty} \mathrm{d} r \int_0^{2 \pi} \mathrm{d} \theta r \left [\frac{1}{2}(\partial_r u)^2+\frac{1}{2r^2} (\partial_{\theta}) u + u \Phi \right].[/tex]
Then via the Euler-Lagrange equations you get the stationary point of the action as given by
[tex]\frac{\partial}{\partial r} \left (r \frac{\partial u}{\partial r} \right )+\frac{1}{r} \frac{\partial^2 u}{\partial \theta^2}=r \Phi,[/tex]
and finally dividing by [itex]r[/itex]
[tex]\Phi=\Delta u=\frac{1}{r} \frac{\partial}{\partial r} \left (r \frac{\partial u}{\partial r} \right )+\frac{1}{r^2} \frac{\partial^2 u}{\partial \theta^2}.[/tex]
 
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  • #5
UltrafastPED
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See steps (4) and (5).

If your function is expressed as u(x,y) you should use the cartesian form.
If it is in polar form then you should use the polar form.

Otherwise you have to go through all of the steps (4) and (5), making use of the chain rule.
 

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