Invariance of the Laplacian operator under the action of rotations centred at the origin

Question

I need to solve the following problem:

$$\begin{cases} \Delta u(x, y) = 1 \ \ \mathrm{if} \ (x, y) \in D\\ u(x, y) = 0 \ \ \ \ \; \mathrm{if} (x, y) \in D \end{cases}$$

where $D$ is the domain strictly included between the two circles below:

My exercice book simplifies this problem using the polar coordinates:

$$\begin{cases} x = r \cos \theta\\ y = r \sin \theta \end{cases}$$

with $r \in \mathbb{R}^+$ and $\theta \in [0, 2\pi[$. So far, so good.

Now here is the tricky part:

Because of the invariance of the above problem under the action of rotations centred at the origin, we can dedecue that $\partial_\theta u(r, \theta) = 0$.

My book thus claims that the problem can be simplified as:

$$\begin{cases} u''(r) + \frac{1}{r}u'(r) = 1\\ u(a) = u(b) = 0 \end{cases}$$

Could someone explain me please why $\partial_\theta u = 0$ in this context please ? I see that there is a obvious simetry in $D$, but I don't see how to link this with the partial derivative of $u$ with respect to $\theta$.

Thank you in advance.

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Update: considering the comments of @Feynman_00, I decomposed $u(r, \theta)$ in $R(r)\Theta(\theta)$. The equation for $D$ becomes:

$$\frac{\partial^2 u}{\partial r^2} + \frac{1}{r}\frac{\partial u}{\partial r} + \frac{1}{r^2}\frac{\partial^2 u}{\partial \theta^2} = 0$$

$$ \Longleftrightarrow R''(r) \Theta (\theta) + \frac{1}{r} R'(r) \Theta(\theta) + \frac{1}{r^2}R(r) \Theta''(\theta) = 0$$

Now, if $\Theta''(\theta) = 0$, it becomes possible to simplify the equation with only considering $R(r)$. However:

1. I didn't understand how to prove this
2. What about the solutions with $\Theta(\theta)$ ? The exercice asks for a general solution.

Answer 1

Laplacian operator

The Laplacian operator is a scalar operator under rotation, which means that after a rotation$^1$ $$(x',y')=(x,y)\cdot R^T\qquad R\in\mathrm{SO}(3)$$ Consider a $C^2(A)$ function $$f:A\subset\mathbb{R}^2\rightarrow\mathbb{R}$$ Then $$\underbrace{\Delta' f'(x',y')}_{\{\Delta f(x,y)\}'}=\Delta f(x,y)\tag{1}\label{1}$$ where $$f'(x',y'):=f(x,y)(\iff f'(x,y)=f[(x,y)\cdot(R^{-1})^T])\tag{2}\label{2}$$ and $$\Delta'=\frac{\partial^2}{\partial x'^2}+\frac{\partial^2}{\partial y'^2}.$$ You can prove \eqref{1} by using the chain rule. Let me use the $(x_1,x_2)$ notation now $$\frac{\partial}{\partial x'_i}=\sum_j\frac{\partial x_j}{\partial x'_i}\frac{\partial}{\partial x_j}=\sum_jR_{ij}\frac{\partial}{\partial x_j}$$ Thus \begin{align} \Delta' f'(x',y')&\overset{\eqref{2}}{=}\Delta' f(x,y)=\sum_i\frac{\partial^2}{\partial x'^2_i}f(x,y)=\\&=\sum_i\sum_j\sum_k\frac{\partial x_j}{\partial x'_i}\frac{\partial x_k}{\partial x'_i}\frac{\partial^2}{\partial x_j\partial x_k}f(x,y)=\\&=\sum_k\sum_j\underbrace{\sum_iR_{ij} R_{ik}}_{\delta_{kj}}\frac{\partial^2}{\partial x_j\partial x_k}f(x,y)=\\&=\sum_j\frac{\partial^2}{\partial x_j^2}f(x,y)=\Delta f(x,y) \end{align}

Your problem

Up to now we've seen that the laplacian is a scalar operator and that the function $$D\ni(x',y')\mapsto u'(x',y')$$ is a solution to the problem. Having said that, if we rotate the coordinate system, as the domain is rotationally invariant and we have just proved that the equation is too, nothing changes. More technically, the solutions of your equation have to retain the same form. This is not just telling us that $$u'(x',y')=u(x,y)\tag{a}\label{a}$$ which is obviously satisfied as this is a scalar function$^2$. This requires instead that there exists a solution $v$ in the non-rotated frame such that $$u'(x,y)=v(x,y).$$ Moreover, the solution to your problem is unique, so $v\equiv u$ and we have $$u'(x,y)=u(x,y)\tag{b}\label{b}$$ which means that the solution evaluated at the point of coordinates $(x,y)$ in the rotated frame, which is not the same point of the plane as the point of coordinates $(x,y)$ in the non-rotated frame, yields the same value. Now, the points of coordinates $(x,y)$ in the two frames, which again are different points have the same distance from the origin i.e. $$r=\sqrt{x^2+y^2}$$ as we have considered arbitrary rotations, \eqref{b} means that $$u(x,y)=u'(x,y)\overset{\eqref{a}}{=}u[(x,y)\cdot(R^{-1})^T].\tag{c}\label{c}$$ Now, rotation are isometries i.e. distance preserving mappings, so \eqref{c} means that at all points at the same distance $r$ the solution has the same value. Let $f(r,\theta):=u(r\cos\theta, r\sin\theta)$, what I have written above in bold font means that $$\frac{\partial f(r,\theta)}{\partial \theta}=0\implies f(r,\theta)=\tilde{f}(r)$$ for some function $\tilde{f}$.

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$^1$ _{I'm using the transpose notation along with row vectors to write things in a more compact way instead of writing $R\begin{pmatrix}x\\y\end{pmatrix}$.}

$^2$ _{Which is the same condition as in the previous paragraph and means that thesolution in the rotated frame evaluated at the same point (the primed coordinates are the coordinates of the same point in the new coordinate frame) of the domain is the same as in the non-rotated system.}

Answer 2

First, by integration by parts, we can deduce that your PDE has at most one solution. By general theory of the Laplace operator, we can even deduce that it has exactly one solution. Call it $u$. Now let $R$ be a rotation about the origin, so $R^TR = I$. It is well known that $R^* \Delta u = \Delta R^* u$, that is, $\Delta u \circ R = \Delta(u \circ R)$; this can be proven either "geometrically" or by simple computation. Hence if $u$ solves the PDE, then so does $u \circ R$. By uniqueness of solutions to the PDE, we conclude $u = u \circ R$. Since $R$ was arbitrary, we conclude that in polar coordinates, $u(r, \theta) = u(r, \theta + \tau)$ for every $r \in [a, b]$, $\theta, \tau \in [0, 2\pi)$.