Derivation of the Polar Laplacian

Claim: Let \(u(x,y)\) be a function in Cartesian coordinates and define \[U(r,\theta) = u(r\cos(\theta), r\sin(\theta)).\] Then, \[U_{rr} + \frac{1}{r}U_r + \frac{1}{r^2}U_{\theta\theta} = u_{xx} + u_{yy}.\] That is, in polar coordinates, the Laplacian operator takes the form \[\Delta u = u_{rr} + \frac{1}{r} u_r + \frac{1}{r^2} u_{\theta\theta}.\]

Computation

We’re going to make extensive use of the multivariable chain rule: \[\frac{d}{dt}f(x(t),y(t)) = \frac{\partial f}{\partial x}\frac{dx}{dt} + \frac{\partial f}{\partial y}\frac{dy}{dt}.\]

When applied to \(U(r,\theta) = u(r\cos(\theta),r\sin(\theta))\) using the more compact subscript notation for partial derivatives, we get:

\[U_r = \textcolor{blue}{u_x \cos(\theta) + u_y \sin(\theta)}.\]

Now, \(u_x\) and \(u_y\) have the same arguments as \(u\). Thus, a similar pattern emerges when we compute \(U_{rr}\), which eases the following computation:

\[\begin{aligned} U_{rr} &= (u_{xx} \cos(\theta) + u_{yx} \sin(\theta)) \cos(\theta) + (u_{yx} \cos(\theta) + u_{yx} \sin(\theta)) \sin(\theta) \\ &= u_{xx}\cos^2(\theta) + u_{yy}\sin^2(\theta) + \textcolor{red}{2u_{xy}\sin(\theta)\cos(\theta)}. \end{aligned}\]

We then do something similar when computing \(U_{\theta}\): \[U_{\theta} = u_{x}(-r\sin(\theta)) + u_{y} (r\cos(\theta)).\]

Our last derivative is \(U_{\theta\theta}\), which is just a little more complicated since we need to use the product rule:

\[\begin{aligned} U_{\theta\theta} &= (u_{xx}(-r\sin(\theta)) + u_{xy} (r\cos(\theta)))(-r\sin(\theta)) + u_x (-r\cos(\theta)) + \\ &\ \cdots (u_{yx}(-r\sin(\theta)) + u_{yy} (r\cos(\theta)))(r\cos(\theta)) + u_y (-r\sin(\theta)) \\ &= u_{xx} r^2 \sin^2(\theta) + u_{yy} r^2 \cos^2(\theta) - \textcolor{red}{2u_{xy}r^2\sin(\theta)\cos(\theta)} - \textcolor{blue}{r(u_x \cos(\theta) + u_y \sin(\theta))}. \end{aligned}\]

Now, when we add up \[U_{rr} + \frac{1}{r} U_r + \frac{1}{r^2} U_{\theta\theta},\] we see that the colored parts cancel one another while the remaining parts add up to exactly the Laplacian!