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I am trying to express Laplace's equation in terms of polar coordinates. That is, $$ \frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2}=0,\\ x=r\cos\theta,\\ y=r\sin\theta. $$ My book immediately concludes that it is $$ \frac1r\frac{\partial}{\partial r}\left(r\frac{\partial u}{\partial r}\right)+\frac1{r^2}\frac{\partial^2u}{\partial\theta^2}=0, $$ but leaves us with no insight as to how that was obtained.

Any hint would be greatly appreciated!

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  • $\begingroup$ it's the multivariate chain rule. $\endgroup$ – James S. Cook Sep 27 '12 at 16:48
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    $\begingroup$ In this answer I show how to obtain $\partial^2 f/\partial x\partial y$, and next apply the result to $f=\theta$. Following the same reasoning you can obtain $\partial^2 f/\partial x^2$ and $\partial^2 f/\partial y^2$. $\endgroup$ – enzotib Sep 27 '12 at 16:57
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My favorite method for this is to use Gauss's theorem in reverse, so to speak. For brevity, let me write $\Delta u$ for the Laplacian and $u_r$, $u_\theta$ etc for the partial derivatives. Note that $\Delta u$ is the divergence of $\nabla u$, so Gauss's theorem says $$ \iint_\Omega\Delta u\,dx\,dy=\int_{\partial\Omega}\mathbf{n}\cdot\nabla u\,ds .$$ Apply this to the domain $\Omega$ given by $r_1<r<r_2$ and $\theta_1<\theta<\theta_2$, and note that

  • On the boundary $r=r_2$, $\mathbf{n}\cdot\nabla u=u_r$ and $ds=r_2\,d\theta$
  • On the boundary $r=r_1$, $\mathbf{n}\cdot\nabla u=-u_r$ and $ds=r_1\,d\theta$
  • On the boundary $\theta=\theta_2$, $\mathbf{n}\cdot\nabla u=u_\theta/r$ and $ds=dr$
  • On the boundary $\theta=\theta_1$, $\mathbf{n}\cdot\nabla u=-u_\theta/r$ and $ds=dr$

so Gauss becomes $$\begin{aligned}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}\Delta u\cdot r\,dr\,d\theta &=\int_{\theta_1}^{\theta_2}\bigl(r_2u_r(r_2,\theta)-r_1u_r(r_1,\theta)\bigr)\,d\theta\\ &\quad+\int_{r_1}^{r_2}\frac{u_\theta(r,\theta_2)-u_\theta(r,\theta_1)}{r}\,dr\\ &=\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(ru_r)_r\,dr\,d\theta +\int_{r_1}^{r_2}\int_{\theta_1}^{\theta_2}\frac{u_{\theta\theta}}{r}\,d\theta\,dr\\ &=\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}\Bigl((ru_r)_r+\frac{u_{\theta\theta}}{r}\Bigr)\,dr\,d\theta \end{aligned}$$ Since this holds for all choices of the limits, the integrands must be the same, so $$\Delta u\cdot r=(ru_r)_r+\frac{u_{\theta\theta}}{r}.$$ Now divide by $r$.

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  • $\begingroup$ Sorry, how does one find the normal vector $\textbf{n}$? $\endgroup$ – user135520 Dec 6 '16 at 22:43
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    $\begingroup$ @user135520 The boundary of $\Omega$ consists of four curves, where two are circular arcs and two are parts of rays from the origin. So the calculation of $\mathbf{n}$ is straightforward: On $r=r_2$, we have $\mathbf{n}=(x,y)/\sqrt{x^2+y^2}$, while on $r=r_1$, we have $\mathbf{n}=(-x,-y)/\sqrt{x^2+y^2}$. Similarly, on $\theta=\theta_2$, it is $\mathbf{n}=(-y,x)/\sqrt{x^2+y^2}$, and on $\theta=\theta_2$, it is $\mathbf{n}=(y,-x)/\sqrt{x^2+y^2}$. But really, a picture explains it better than formulas. $\endgroup$ – Harald Hanche-Olsen Dec 7 '16 at 14:30
  • $\begingroup$ I'm a little confused now as to why $u_{r} = \textbf{n} \cdot \nabla{u}$? I'm thinking that since $u_{r} = u_{x}\cos(\theta)$, we can think of $\textbf{n}=(\cos(\theta), \sin(\theta))$ as an element of the unit circle and maybe then that product works out? $\endgroup$ – user135520 Feb 27 '18 at 18:08
  • $\begingroup$ @user135520 No, since (with some standard abuse of notation) $u(r,\theta)=u(r\cos\theta,r\sin\theta)$, we get $u_r=u_x\cos\theta+u_y\sin\theta=(\cos\theta,\sin\theta)\cdot(u_x,u_y)$. $\endgroup$ – Harald Hanche-Olsen Feb 27 '18 at 18:31
  • $\begingroup$ see people.whitman.edu/~hundledr/courses/M367/LaplaceInPolar.pdf $\endgroup$ – Aleksas Domarkas Jun 22 '18 at 5:12

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