My textbook states the following:

Consider a vibrating circulate membrane. The displacement $u(r, \theta, t)$ from equilibrium satisfies the wave equation

$$\nabla^2 u = \dfrac{1}{c^2} \dfrac{\partial^2{u}}{\partial{t}^2},$$

$0 \le r \le a$,

with boundary condition $u(a, \theta, t) = 0$ for some initial conditions.

In polar coordinates, the wave equation becomes

$$\dfrac{1}{r} \dfrac{\partial}{\partial{r}} \left( r \dfrac{\partial{u}}{\partial{r}} \right) + \dfrac{1}{r^2} \dfrac{\partial^2{u}}{\partial{\theta}^2} = \dfrac{1}{c^2} \dfrac{\partial^2{u}}{\partial{t}^2}$$

My question is, how does

$$\nabla^2 u = \dfrac{1}{c^2} \dfrac{\partial^2{u}}{\partial{t}^2}$$


$$\dfrac{1}{r} \dfrac{\partial}{\partial{r}} \left( r \dfrac{\partial{u}}{\partial{r}} \right) + \dfrac{1}{r^2} \dfrac{\partial^2{u}}{\partial{\theta}^2} = \dfrac{1}{c^2} \dfrac{\partial^2{u}}{\partial{t}^2}$$? This wasn't shown.

I would appreciate it if someone could please show this.

  • $\begingroup$ Look up the definition of the Laplacian in Polar coordinates. There are many ways to prove it. $\endgroup$
    – Dylan
    Aug 25, 2018 at 15:38
  • $\begingroup$ See here:math.stackexchange.com/a/2871849/135643 $\endgroup$
    – Dylan
    Aug 28, 2018 at 8:40

1 Answer 1


This may not be the easiest (maybe by far not the easiest) way to arrive at the right result, but to me this always seemed to be the most intuitive way of deriving any equation in any given coordinate system, so I thought this may help you as much as it helped me.

First, some clarifications:

$$ \vec{x} = x(r, \phi) \vec{e}_x + y(r, \phi)\vec{e}_y = r(x,y)\vec{e}_r+\phi(x,y)\vec{e}_{\phi}, $$ where $$ \begin{align} x(r, \phi) &= r\cos\phi\\ y(r, \phi) &= r\sin\phi\\ r(x,y) &=\sqrt{x^2+y^2}\\ \phi(x,y) &=\arctan\left( \frac{y}{x} \right). \end{align} $$ First we want to find the basis vectors $\vec{e}_{\phi}$ and $\vec{e}_{r}$ as functions of $r$, $\phi$, $\vec{e}_x$ and $\vec{e}_y$. We find them by:

$$ \begin{align} \vec{e}_r&:=\frac{ \frac{\partial \vec{x}}{\partial r} }{ \left| \left| \frac{\partial \vec{x}}{ \partial r} \right| \right| } = \cos\phi\vec{e}_x + \sin\phi\vec{e}_y \\ \vec{e}_{\phi}&:=\frac{ \frac{\partial \vec{x}}{\partial \phi} }{ \left| \left| \frac{\partial \vec{x}}{ \partial \phi} \right| \right| } = -\sin\phi\vec{e}_x + \cos\phi\vec{e}_y. \end{align} $$

With this we can find the basis vectors $\vec{e}_x$ and $\vec{e}_y$ as functions of $r$, $\phi$, $\vec{e}_r$ and $\vec{e}_{\phi}$. (For example by inverting the coefficient matrix of the above system of linear equations - which is very easy since it consists only of a matrix of rotation.) This gives us: $$ \begin{align} \vec{e}_x &= \cos\phi\vec{e}_r - \sin\phi\vec{e}_{\phi} \\ \vec{e}_y &= \sin\phi\vec{e}_r + \cos\phi\vec{e}_{\phi}. \end{align} $$

Next step is to write the partial derivatives $\frac{\partial}{\partial i} =: \partial_i$ in terms of the new coordinates. This is achieved by: $$ \begin{align} \partial_x &= \frac{\partial r}{ \partial x} \partial_r + \frac{\partial \phi}{\partial x}\partial_{\phi} = \cos\phi\partial_r - \frac{\sin\phi}{r}\partial_{\phi} \\ \partial_y &= \frac{\partial r}{ \partial y} \partial_r + \frac{\partial \phi}{\partial y} \partial_{\phi} = \sin\phi\partial_r + \frac{\cos\phi}{r}\partial_{\phi}. \end{align} $$

Now it should be easy to calculate the gradient to be: $$ \vec{\nabla} = \vec{e}_x\partial_x + \vec{e}_y\partial_y = \vec{e}_r\partial_r + \vec{e}_{\phi}\frac{1}{r}\partial_{\phi}. $$

But since your question includes the laplacian $\Delta= \mathrm{div}(\mathrm{grad}) $ we also must know what it means to take the divergence of a vector field $\vec{A}=A_r(r,\phi)\vec{e}_r+A_{\phi}(r,\phi)\vec{e}_{\phi} = A_x(x,y)\vec{e}_x + A_{y}(x,y)\vec{e}_{y}$ given in polar coordinates. Since $\mathrm{div}(\vec{A})=\partial_xA_x+\partial_yA_y$, we only need to find $A_x$ and $A_y$ as functions of $r$, $\phi$, $A_r$ and $A_{\phi}$: $$ A_x(r, \phi)=\langle \vec{e}_x, \vec{A} \rangle = \cos\phi A_r - \sin\phi A_{\phi} \\ A_y(r, \phi)=\langle \vec{e}_y, \vec{A} \rangle = \sin\phi A_r + \cos\phi A_{\phi}. $$

Now we can calculate $$ \mathrm{div}(\vec{A})=\partial_xA_x+\partial_yA_y=\frac{1}{r}\partial_{\phi}A_{\phi} + \frac{1}{r}A_r + \partial_r A_r. $$

If we now apply this result to the gradient, which we already calculated, we arrive at: $$ \begin{align} \Delta=\mathrm{div}(\mathrm{grad})&=\partial_x \vec{\nabla}_x + \partial_y \vec{\nabla}_y\\ &=\frac{1}{r}\partial_{\phi}\vec{\nabla}_{\phi} + \frac{1}{r}\vec{\nabla}_r + \partial_r \vec{\nabla}_r\\ &=\frac{1}{r}\partial_{\phi}\left( \frac{1}{r} \partial_{\phi} \right) + \frac{1}{r}\partial_r+\partial_r(\partial_r)\\ &=\frac{1}{r^2}\partial_{\phi}^2+\frac{1}{r}\partial_r+\partial_r^2. \end{align} $$

Now we can see that the wave equation $\Delta u(x,y,t)=\frac{1}{c^2}\partial_t^2u(x,y,t)$ indeed becomes $$ \begin{align} \Delta u(r,\phi,t)&=\left[\frac{1}{r^2}\partial_{\phi}^2+\frac{1}{r}\partial_r+\partial_r^2\right] u(r,\phi,t)\\ &=\frac{1}{r^2}\partial_{\phi}^2u+\frac{1}{r}\partial_ru+\partial_r^2u\\ &=\frac{1}{r^2}\partial_{\phi}^2u+\frac{1}{r}\partial_r\left(r \partial_ru\right) = \frac{1}{c^2} \partial_t^2 u. \end{align} $$


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