## 275 -The Laplacian in polar coordinates

In many situations in which symmetries are involved, it is useful to write the equations describing the physical objects under study with respect to frames related to said symmetries. For situations involving rotational symmetry, polar (or cylindrical) coordinates are particularly useful.

For this reason, it is a good exercise to see how to transform something like Laplace equation into polar coordinates.

Remember that the Laplacian of $f(x,y)$ is $\displaystyle \nabla\cdot\nabla f=\frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial y^2}$. If we write $f$ in polar coordinates as $g(r,\theta)=f(r\cos\theta,r\sin\theta)$, we then need to write the Laplacian directly in terms of derivatives with respect to $r$ and $\theta$.

Following a common abuse of notation, I will write $f_r$, etc, for what one should more precisely write $g_r$, etc, with $g$ as above.

Let’s begin by finding $\displaystyle \frac{\partial r}{\partial x}$, $\displaystyle \frac{\partial \theta}{\partial y}$, etc.

Since $r=\sqrt{x^2+y^2}$, then $\displaystyle \frac{\partial r}{\partial x}=\frac12\frac1{\sqrt{x^2+y^2}}\cdot 2x=\frac x{\sqrt{x^2+y^2}}=\frac{r\cos\theta}r=\cos\theta.$

We have $\displaystyle\frac{\partial r}{\partial x}=\cos\theta$. Similarly, $\displaystyle \frac{\partial r}{\partial y}=\sin\theta$.

Since $\displaystyle\tan\theta=\frac yx$, then $\displaystyle\sec^2\theta,\frac{\partial \theta}{\partial x}=-\frac y{x^2}=-\frac{\sin\theta}{r\cos^2\theta}$, so $\displaystyle \frac{\partial \theta}{\partial x}=-\frac{\sin\theta}r$.

Similarly, $\displaystyle\sec^2\theta,\frac{\partial\theta}{\partial y}=\frac1x=\frac1{r\cos\theta}$, so $\displaystyle\frac{\partial\theta}{\partial y}=\frac{\cos\theta}r$.

Before we continue, a word of caution is in order: Notice that $x=r\cos\theta$ so, for example, $\displaystyle \frac{\partial x}{\partial r}=\cos\theta$. This means that $\displaystyle \frac{\partial x}{\partial r}$ is not $\displaystyle \frac{1}{\frac{\partial r}{\partial x}}$, as it is the case with functions of one variable.

Now we use the chain rule: Suppose $h(x,y)$ is a differentiable function of $x,y$. Then we obtain $\displaystyle h_x=h_r\frac{\partial r}{\partial x}+h_\theta\frac{\partial\theta}{\partial x}=h_r\cos\theta-h_\theta\frac{\sin\theta}r$.

Assume now that $f$ is sufficiently differentiable so all the derivatives to follow exist and are continuous, mixed derivatives commute, etc. Apply the above twice, first with $h=f$ and then with $h=f_x$ to obtain $\displaystyle f_{xx}=(f_r\cos\theta-f_\theta\frac{\sin\theta}r)_r\cos\theta-(f_r\cos\theta-f_\theta\frac{\sin\theta}r)_\theta\frac{\sin\theta}r$ or $f_{xx}=(f_{rr}\cos\theta-f_{r\theta}\frac{\sin\theta}r+f_\theta\frac{\sin\theta}{r^2})\cos\theta$ $-(f_{r\theta}\cos\theta-f_r\sin\theta-f_{\theta\theta}\frac{\sin\theta}r-f_\theta\frac{\cos\theta}r)\frac{\sin\theta}r.$

Similarly, $\displaystyle f_{yy}=(f_r\sin\theta+f_\theta\frac{\cos\theta}r)_r\sin\theta+ (f_r\sin\theta+f_\theta\frac{\cos\theta}r)_\theta\frac{\cos\theta}r$ or $f_{yy}=(f_{rr}\sin\theta+f_{\theta r}\frac{\cos\theta}r-f_\theta\frac{\cos\theta}{r^2})\sin\theta$ $+(f_{r\theta}\sin\theta+f_r\cos\theta+f_{\theta\theta}\frac{\cos\theta}r-f_\theta\frac{\sin\theta}r)\frac{\cos\theta}r.$

Thus, $\nabla\cdot\nabla f=\frac{\partial^2 f}{\partial x^2}+frac{\partial^2 f}{\partial y^2}=f_{rr}(\cos^2\theta+\sin^2\theta)$ $+f_{\theta r}(-\frac{\sin\theta\cos\theta}r-\frac{\sin\theta\cos\theta}r+\frac{\cos\theta\sin\theta}r+\frac{\sin\theta\cos\theta}r)$ $+f_\theta(\frac{\sin\theta\cos\theta}{r^2}+\frac{\sin\theta\cos\theta}{r^2}-\frac{\cos\theta\sin\theta}{r^2}-\frac{\sin\theta\cos\theta}{r^2})$ $+f_r(\frac{\sin^2\theta}r+\frac{\cos^2\theta}r)+f_{\theta\theta}(\frac{\sin^2\theta}{r^2}+\frac{\cos^2\theta}{r^2})$ $=f_{rr}+\frac1r f_r+\frac 1{r^2} f_{\theta\theta}.$

We have obtained the Laplacian in polar coordinates as $\displaystyle\nabla\cdot\nabla f=f_{rr}+\frac1r f_r+\frac1{r^2} f_{\theta\theta}.$

Example: Let $f(x,y)=x^2-y^2$. We know from lecture that $f$ is harmonic. Writing $f$ in polar coordinates, we obtain $f=r^2(\cos^2\theta-\sin^2\theta)=r^2\cos(2\theta)$. Then $f_r=2r\cos2\theta$, $f_{rr}=2\cos2\theta$, $f_\theta=-2r^2\sin2\theta$, and $f_{\theta\theta}=-4r^2\cos2\theta$, so $\displaystyle \nabla\cdot\nabla f=f_{rr}+\frac1r f_r+\frac1{r^2} f_{\theta\theta}$ $\displaystyle =2\cos2\theta+\frac1r 2r\cos2\theta+\frac1{r^2}(-4r^2\cos2\theta)=0,$ as expected.

Exercise: To test your understanding of the algebra involved in the operations above, you may want to find an expression for the wave equation $\displaystyle\frac{\partial^2 w}{\partial y^2}=c^2\frac{\partial^2 w}{\partial x^2}$ in polar coordinates.

Cylindrical coordinates: Finally, if $f$ is a function of three variables $x,$ $y,$ and $z,$ so $\displaystyle \nabla\cdot\nabla f=\frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial y^2}+\frac{\partial^2 f}{\partial z^2}$, then we see immediately from the work above that we can write the Laplacian in cylindrical coordinates as $\displaystyle \nabla\cdot\nabla f=f_{rr}+\frac1r f_r+\frac1{r^2} f_{\theta\theta}+f_{zz}$.

### 5 Responses to 275 -The Laplacian in polar coordinates

1. […] proof that the method of partial fractions decomposition works. For 275, you may want to review the polar expression of the Laplacian and how to derive it. For both courses, a safe assumption is that if something was […]

2. AC says:

Thanks for that post!
I was making some mistakes in these computations,
and this has helped!

3. mobi says:

thanks i am really impressed how you explained all this i am undergraduate student i am having problem with spherical polar coordinate conversion how we find derivative of Theta w.r.t x, y,z kindly help me

4. Matt says:

I’m still just a little confused in the second application of the chain rule, when we differentiate to get the second derivative of f in respect to x. Can anyone help me here?

5. […] proof that the method of partial fractions decomposition works. For 275, you may want to review the polar expression of the Laplacian and how to derive it. For both courses, a safe assumption is that if something was […]