Laplacian in polar coordinates (idea)

Question

As we know using chain rule, we can compute the Laplacian in polar coordinates: for $u=u(r,\theta)$ it holds $$\Delta u=u_{rr}+\frac{1}{r}u_r+\frac{1}{r^2}u_{\theta \theta}$$ There is something behind this formula that I don't get. If we forget about Cartesian coordinate system and transformation formulas, considering $u=u(r,\theta)$ "independently" then why shouldn't we have $\Delta u=u_{rr}+u_{\theta \theta}$. Isn't this how the Laplacian is defined? What fails here? Thank you.

Answer 1

The Laplacian is initially defined in Cartesian coordinates. If you switch to polar coordinates, then you are still dealing with the same operator, so you need to use the chain rule to get a form which only involves derivatives with respect to $r$ and $\theta$. Thus the operator $u \mapsto u_{rr} + u_{\theta \theta}$ is completely different from the Laplacian. It's also a pretty physically useless operator, because usually $r$ has units of length, so usually $u_{rr}$ and $u_{\theta \theta}$ have different units, and therefore cannot be added.

It is a bit confusing that we use the same name for the function $u$ in Cartesian and polar coordinates. But the two are not actually the same mathematical function (even though they represent the same physical quantity in different coordinate systems). Using the same name is really an abuse of notation. If we were being really precise, we would introduce $w(r,\theta)=u(r \cos(\theta),r \sin(\theta))$. Here $w$ is "the polar coordinate representation of $u$". Then the chain rule is really telling us that

$$\Delta u(r \cos(\theta),r\sin(\theta)) = w_{rr}(r,\theta) + \frac{1}{r} w_r(r,\theta) + \frac{1}{r^2} w_{\theta \theta}(r,\theta).$$