Let $$\Delta=\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}$$ be the Laplace operator on the $(x,y)$-plane. Consider the polar coordinates with $x=r\cos\theta$ and $y=r\sin\theta$. Show that $$\Delta=\frac{\partial^2}{\partial r^2}+\frac1r\frac\partial{\partial r}+\frac1{r^2}\frac{\partial^2}{\partial \theta^2}.$$

I don't know how to change the coordinates for an operator like this one. What is the method that should be used?

  • 1
    $\begingroup$ You need to use chain rule. For example, $\frac{\partial }{\partial r} = \frac{\partial x}{\partial r}\frac{\partial }{\partial x} + \frac{\partial y}{\partial r}\frac{\partial }{\partial y}$. $\endgroup$
    – user99914
    Nov 2, 2013 at 2:07

1 Answer 1



$$\frac{\partial}{\partial x} = \frac{\partial r}{\partial x} \frac{\partial}{\partial r} + \frac{\partial \theta}{\partial x} \frac{\partial}{\partial \theta}$$

where $r=\sqrt{x^2+y^2}$ and $\tan{\theta} = y/x$. Then $\partial r/\partial x = x/r$ and $\partial \theta/\partial x = -y/r^2$, and

$$\frac{\partial}{\partial x} = \frac{x}{r} \frac{\partial}{\partial r} -\frac{y}{r^2} \frac{\partial}{\partial \theta}$$

Similarly, you may show that

$$\frac{\partial}{\partial y} = \frac{y}{r} \frac{\partial}{\partial r} +\frac{x}{r^2} \frac{\partial}{\partial \theta}$$

The Laplacian is then

$$\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} = \left (\frac{x}{r} \frac{\partial}{\partial r} -\frac{y}{r^2} \frac{\partial}{\partial \theta} \right )^2 + \left (\frac{y}{r} \frac{\partial}{\partial r} +\frac{x}{r^2} \frac{\partial}{\partial \theta} \right )^2$$

This should be plenty to go on.

  • $\begingroup$ It will be easier to carry out the computations if you write $\frac{\partial}{\partial x}$ (and $\frac{\partial}{\partial y}$) like here. $\endgroup$
    – Leponzo
    Aug 2, 2020 at 14:05

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