# confusion about the laplacian in polar coordinates

I have a confusion about the following. If we have a function $$f:\mathbb{R}^2 \to \mathbb{R}$$ of class $$C^2$$ and we want to calculate the laplacian of this function polar coordinates.

I don't understand what it means. Does it mean we need to calculate :

$$\frac{\partial}{\partial x} (\frac{\partial f(r \cos \theta, r \sin \theta)}{\partial x}) + \frac{\partial}{\partial y} (\frac{\partial f(r \cos \theta, r \sin \theta)}{\partial y})$$

Or does it mean we need to calculate :

$$\frac{\partial ^2 f}{\partial x^2}(r \cos \theta, r \sin \theta) + \frac{\partial ^2 f}{\partial y^2}(r \cos \theta, r \sin \theta)$$

Moreover, I am really confused since I don't understand the fondamental difference between these two expressions.

Thank you !

• A function $f(r\cos(\theta),r\sin(\theta))$ does not have a partial derivative with respect to $x$. Jan 27, 2019 at 9:24

The basic interpretation is the second one. The Laplacian $$\Delta f(x,y)$$ is a function of $$x$$ and $$y$$, which is defined as $$\frac{\partial^2 f}{\partial x^2}(x,y) + \frac{\partial^2 f}{\partial y^2}(x,y)$$, and expressing a function in polar coordinates just means substituting $$x=r \cos \theta$$ and $$y = r \sin \theta$$. So first compute the second partials with respect to $$x$$ and $$y$$, then substitute.
However, the chain rule gives you another way of obtaining the same result, by performing a different sequence of operations. This involves first expressing $$f$$ in polar coordinates, and then taking partial derivatives of that expression with respect to $$r$$ and $$\theta$$. (Not with respect to $$x$$ and $$y$$ – that's sort of pointless, because if you wanted to do that, you should have done it directly on $$f(x,y)$$ instead of changing to polar coordinates to begin with.) How this works is explained in several other answers on this site, for example here or here.
• Thank you ! So it means that we have : $\Delta f(x,y) = \frac{\partial ^2 f}{\partial x}(x,y) + \frac{\partial^2f}{\partial y}(x,y)$, and this can be expressed in polar coordinates by substituing so we have : $\Delta f(r \cos \theta, r \sin \theta) = \frac{\partial ^2 f}{\partial x}(r \cos \theta,r \sin \theta) + \frac{\partial^2f}{\partial y}(r \cos \theta,r \sin \theta)$ which is the laplacian in polar coordinates. But an other way to expressed this laplacian in polar coordinates is : Jan 27, 2019 at 12:13
• $\frac{\partial}{\partial r} (\frac{\partial f(r \cos \theta, r \sin \theta)}{\partial r}) + \frac{\partial}{\partial \theta} (\frac{\partial f(r \cos \theta, r \sin \theta)}{\partial \theta})$ which can be calculated using the chain rule. Jan 27, 2019 at 12:13