Laplace equation in polar coordinates.

  • Thread starter yungman
  • Start date
  • #1
5,560
198
[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
 
Last edited:

Answers and Replies

  • #3
5,560
198
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
Science Advisor
Insights Author
Gold Member
17,546
8,539
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]
 
  • Like
Likes 1 person
  • #5
UltrafastPED
Science Advisor
Gold Member
1,912
216
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.
 

Related Threads on Laplace equation in polar coordinates.

Replies
3
Views
1K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
2
Views
3K
Replies
0
Views
951
Replies
2
Views
1K
Replies
2
Views
1K
Replies
1
Views
908
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
5
Views
5K
Top