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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!
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
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 $r$.
