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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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    In [this answer](http://math.stackexchange.com/questions/192445/simple-partial-differentiation-x-r-cos-theta-and-y-r-sin-theta/192712#192712) 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 $\partial^2 f/\partial y^2$.2012-09-27

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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 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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    see http://people.whitman.edu/~hundledr/courses/M367/LaplaceInPolar.pdf2018-06-22