example of classical solution of elliptic PDE

Question

I have some problems with the following example of a PDE:

Let $\Omega = \{(\varphi, r)| r \in (0,1), \varphi \in (0, \frac{3}{2}\pi)\}$ which is basically the unitsphere with a quarter piece broken out of it. And $$f(r,\varphi) = 0 \text{ in } \Omega \\ g(r,\varphi) = \sin \left( \frac{2}{3}\varphi \right) \text{ in } \delta\Omega $$ be the given PDE. Now they say: we transform the Laplace-operator with polar coordinates $$x = r \cos(\varphi)\\ y = r \sin \left( \varphi \right)$$ and $u(r,\varphi)=r^ \frac{2}{3} \sin \left( \frac{2}{3} \varphi \right)$ yields a solution of the boundary value problem.

But there is no Laplace-Operator in the PDE. So I must assume, that they also mean $\Delta u =0$ referring to the first of the two equations. But when I check it, I get the following: $$\Delta u = u_{rr} + u_{\varphi\varphi}=-\frac{2}{9}r^{-4/3} \sin(\frac{2}{3} \varphi) -\frac{4}{9} r^{2/3} \sin \left( \frac{2}{3} \varphi \right) $$ Which is not zero. What is the problem here?

They proceed to show $u \notin C^1(\bar{\Omega})$ to point out, that the solution space $C^2(\Omega) \cap C(\bar{\Omega})$ cannot be chosen smaller:

With $u_x$ and $u_y$ bounded, it should be that $u_r$ is bounded, because: $$u_r = u_x x_r + u_y y_r = u_x \cos(\varphi) + u_y \sin(\varphi)$$ But $ u_r = \frac{2}{3} r^{-1/3} \sin \left( \frac{2}{3} \varphi \right) \rightarrow \infty$ as $ r \rightarrow 0$

but what are $u_x$ and $u_y$ in the first place? And why is it possible to split $u_r$ like this? I seem to be all over the place with it.

Answer 1

Note that metric tensor within your parametrisation of the plane is not a unit tensor (hence the name "curvilinear coordinates"). For a generic Riemannian metric $g_{\mu\nu}$ the expression for the Laplace (Laplace-Beltrami) operator is $$\Delta=\sum_{\mu=1}^{d}\sum_{\nu=1}^{d}\frac{1}{\sqrt{|g|}}\frac{\partial}{\partial{x_{\mu}}}\Big(\sqrt{|g|}g^{\mu\nu}\frac{\partial}{\partial{x}_{\nu}}\Big)$$ Where $|g|$ is the determinant of the metric tensor and $d$ is the dimension of the manifold. In the case of Euclidean space $g^{\mu\nu}=\delta^{\mu\nu}$ which indeed yields $$\Delta=\sum_{\mu=1}^{d}\frac{\partial^{2}}{\partial{x^{2}_{\mu}}}$$ In your case $d=2$ and the line element is $$ds^{2}=dr^{2}+r^{2}d\varphi^{2}=\sum_{\mu=1}^{2}\sum_{\nu=1}^{2}g_{\mu\nu}dx^{\mu}dx^{\nu}$$ With $x^{1}=r$ and $x^{2}=\varphi$. So you have $|g|=r^{2}$, $g_{r\varphi}=g_{\varphi{r}}=0$, $g_{rr}=1$ and $g_{\varphi\varphi}=r^{2}$. Hence $g^{rr}=1$ and $g^{\varphi\varphi}=r^{-2}$. Using the above formula we arrive at $$\Delta=\frac{1}{r}\frac{\partial}{\partial{r}}\Big(r\frac{\partial}{\partial{r}}\Big)+\frac{1}{r^{2}}\frac{\partial^{2}}{\partial{\varphi^{2}}}$$ Now $$\Delta\Big[r^{2/3}\sin\Big(\frac{2\varphi}{3}\Big)\Big]=0$$ As required! However, normally, there is no need to compute those Laplacians for various coordinate systems, you can find them on the internet for the popular coordinate systems.