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!

  • $\begingroup$ it's the multivariate chain rule. $\endgroup$ Sep 27, 2012 at 16:48
  • 2
    $\begingroup$ 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, 2012 at 16:57

1 Answer 1


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 $r$.

  • $\begingroup$ Sorry, how does one find the normal vector $\textbf{n}$? $\endgroup$
    – user135520
    Dec 6, 2016 at 22:43
  • 1
    $\begingroup$ @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. $\endgroup$ Dec 7, 2016 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, 2018 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 $u_r=u_x\cos\theta+u_y\sin\theta=(\cos\theta,\sin\theta)\cdot(u_x,u_y)$. $\endgroup$ Feb 27, 2018 at 18:31
  • $\begingroup$ see people.whitman.edu/~hundledr/courses/M367/LaplaceInPolar.pdf $\endgroup$ Jun 22, 2018 at 5:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.