# Laplace equation in polar coordinates, using matrices

I saw that the method shown below could be used to derive the Laplace equation for polar coordinates using less calculations.

\begin{aligned} &\nabla^{2} u=\frac{\partial^{2} u}{\partial r^{2}}+\frac{1}{r} \frac{\partial u}{\partial r}+\frac{1}{r^{2}} \frac{\partial^{2} u}{\partial \theta^{2}}\\ &\left(\begin{array}{c} \frac{\partial}{\partial r} \\ \frac{\partial}{\partial \theta} \end{array}\right)=\left(\begin{array}{ll} \frac{\partial x}{\partial r} & \frac{\partial y}{\partial r} \\ \frac{\partial x}{\partial \theta} & \frac{\partial x}{\partial \theta} \end{array}\right)\left(\begin{array}{c} \frac{\partial}{\partial x} \\ \frac{\partial}{\partial y} \end{array}\right) \quad \longrightarrow \quad\left(\begin{array}{c} \frac{\partial}{\partial x} \\ \frac{\partial}{\partial y} \end{array}\right)=\left(\begin{array}{ll} \frac{\partial x}{\partial r} & \frac{\partial y}{\partial r} \\ \frac{\partial x}{\partial \theta} & \frac{\partial x}{\partial \theta} \end{array}\right)^{-1}\left(\begin{array}{c} \frac{\partial}{\partial r} \\ \frac{\partial}{\partial \theta} \end{array}\right)\\ &\nabla^{2}=\left(\begin{array}{c} \frac{\partial}{\partial x} \\ \frac{\partial}{\partial y} \end{array}\right) \cdot\left(\begin{array}{c} \frac{\partial}{\partial x} \\ \frac{\partial}{\partial y} \end{array}\right) \end{aligned}

Even after computing the following matrix:

$$\begin{pmatrix}\frac{\partial x}{\partial r} & \frac{\partial y}{\partial r} \\ \frac{\partial x}{\partial \theta} & \frac{\partial x}{\partial \theta} \end{pmatrix}$$ (using $$x=r\cos\theta, y=r\sin\theta$$), I still do not know how the Laplacian $$\nabla^{2} u=\frac{\partial^{2} u}{\partial r^{2}}+\frac{1}{r} \frac{\partial u}{\partial r}+\frac{1}{r^{2}} \frac{\partial^{2} u}{\partial \theta^{2}}$$ is gotten. What especially confuses me is the dot product and how the $$2\times1$$ operator gets applied to a $$2 \times2$$ matrix.

First of all, the idea behind the operator being multiplied on the right by a matrix is that it will still output an operator. For example, operating on a function $$f$$ we would find

\begin{aligned} \left(\begin{array}{c} \frac{\partial}{\partial r} \\ \frac{\partial}{\partial \theta} \end{array}\right)[f]=\left(\begin{array}{ll} \frac{\partial x}{\partial r} & \frac{\partial y}{\partial r} \\ \frac{\partial x}{\partial \theta} & \frac{\partial x}{\partial \theta} \end{array}\right)\left(\begin{array}{c} \frac{\partial}{\partial x} \\ \frac{\partial}{\partial y} \end{array}\right)[f] = \left(\begin{array}{ll} \frac{\partial x}{\partial r} & \frac{\partial y}{\partial r} \\ \frac{\partial x}{\partial \theta} & \frac{\partial x}{\partial \theta} \end{array}\right)\left(\begin{array}{c} \frac{\partial f}{\partial x} \\ \frac{\partial f}{\partial y} \end{array}\right) \end{aligned} \; .

Just note that this vector is really a functional that takes an input.

As for your main question, it looks like you're starting to derive it using the metric tensor $$g_{ij}.$$ It can be thought of as a matrix. You have already written down the Jacobian matrix (or it's transpose depending on who you ask). We will say

$$J^T=\begin{bmatrix} \frac{\partial x}{\partial r} & \frac{\partial y}{\partial r} \\ \frac{\partial x}{\partial \theta} & \frac{\partial x}{\partial \theta} \end{bmatrix}$$

and define $$g_{ij} = J^TJ$$ to be a symmetric, indexable object that gives the components of the resulting matrix from the multiplication $$J^TJ\;.$$ Now let $$g^{ij}$$ denote the matrix inverse of $$g_{ij}$$ and let $$g$$ denote the matrix determinant of $$g_{ij}.$$

From this framework, we can use this "metric" $$g_{ij}$$ to define $$\Delta f = \frac{1}{\sqrt{g}} \sum_i \sum_j \frac{\partial }{\partial x_i}\bigg[ \sqrt{g} \; g^{ij} \frac{\partial f}{\partial x_j} \bigg] \;$$ as the Laplacian of $$f.$$ Do note that you should then take $$(x_1,x_2) = (r,\theta) .$$ Also, $$g_{ij}$$ works out quite nicely for polar coordinates and when all is said and done is

$$g_{ij} = \begin{bmatrix} 1 & 0\\ 0 & r^2 \end{bmatrix} \; .$$

Let me know if this needs any further clarification.

$$\textbf{EDIT:}\;$$ clarification to the OP

The Laplacian is defined as I have written it above for a general coordinate system on any pseudo-Riemannian manifold. This can be worked out for general coordinates as defined above or for a specific coordinate system. As an example (which I will not fully work out) we can use polar coordinates as is relevant to your question.

Consider a scalar function of polar coordinates $$f:(r,\theta)\rightarrow\mathbb{R} \;.$$ We know that in Cartesian coordinates that the Laplacian is defined as

$$\Delta f = \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} \;.$$

We can use the chain and product rules to expand this as

$$\Delta f = \frac{\partial^2 f}{\partial r^2}\bigg[ \frac{\partial r}{\partial x} \bigg]^2 + \frac{\partial f}{\partial r}\frac{\partial^2 r}{\partial x^2} + \frac{\partial^2 f}{\partial \theta^2}\bigg[ \frac{\partial \theta}{\partial x} \bigg]^2 + \frac{\partial f}{\partial \theta}\frac{\partial^2 \theta}{\partial x^2} \\+ \frac{\partial^2 f}{\partial r^2}\bigg[ \frac{\partial r}{\partial y} \bigg]^2 + \frac{\partial f}{\partial r}\frac{\partial^2 r}{\partial y^2} + \frac{\partial^2 f}{\partial \theta^2}\bigg[ \frac{\partial \theta}{\partial y} \bigg]^2 + \frac{\partial f}{\partial \theta}\frac{\partial^2 \theta}{\partial y^2} \; .$$

Using the standard polar coordinates \begin{align} x&=r\cos(\theta)\\ y&=r\sin(\theta) \end{align}

we can evaluate the expression above and it had better come out to be what we expect (and it does). We can do something similar if you prefer the Laplacian definition $$\Delta f = \nabla \cdot \nabla f = \text{div}(\nabla f)$$ but then we would have to work out the polar gradient and polar divergence. If you feel unconvinced, I recommend that you work one of these out and see that it yields the same results as the summation definition I gave above.

• @vbd628 Excellent answer. I think it's worth mentioning that $g$ is the so called metric tensor. – K.defaoite Oct 14 '20 at 9:43
• @K.defaoite I already did :) "As for your main question, it looks like you're starting to derive it using the metric tensor $g_{ij}$". – vb628 Oct 14 '20 at 9:45
• Whoops, I missed that. Nice work anyways. – K.defaoite Oct 14 '20 at 9:46
• This answered some questions but also made some. I still don't get what the dot product does. And what input are we exactly taking in, then? Why is $g$ defined that way: $J^TJ$? And why is the Laplacian defined that way with the square root and double sum? – Hyewon Oct 14 '20 at 9:56
• @Hyewon The "real" definition of $g$ is the matrix containing the pairwise dot products of the basis vectors in a given coordinate system on the manifold (in your case this is just polar coordinates in the plane). It is more of a result that the Jacobian can generate the metric tensor but it seems easier to define it this way given your background. Keep in mind that dot products are a nice way to measure lengths in space and that is what the metric tensor is. It is a rule by which we can measure things in a given coordinate system on a Riemannian manifold. – vb628 Oct 14 '20 at 10:13