Show Laplace operator is rotationally invariant

Question

I'm trying to show the Laplace operator is rotationally invariant. Essentially this boils down to showing $$\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = \frac{\partial^2 f}{\partial u^2} + \frac{\partial^2 f}{\partial v^2}$$

where $$u = x \cos \theta + y \sin \theta$$ $$v = -x \sin \theta + y \cos \theta$$

I think I'm on the right track by noting that $$\frac{\partial^2 f}{\partial x^2} = \frac{\partial}{\partial x}\left(\frac{\partial f}{\partial x}\right) = \frac{\partial}{\partial u}\left(\frac{\partial f}{\partial x}\right)\frac{\partial u}{\partial x} + \frac{\partial}{\partial v}\left(\frac{\partial f}{\partial x}\right)\frac{\partial v}{\partial x}$$ but I'm having difficulty reaching an end game where I show $$\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = \frac{\partial^2 f}{\partial u^2}({\sin}^2 \theta + {\cos}^2 \theta) + \frac{\partial^2 f}{\partial v^2}({\sin}^2 \theta + {\cos}^2 \theta)$$

Answer 1

Here is an answer along the lines of @TeeJay, but with a few more details and in general $\mathbb R^n$ space. So, we need to show that if $u$ is harmonic ($\Delta_y u(y)=0$) then $v(x)=u(Mx)$ is also harmonic ($\Delta_x v(x)=0$) for an orthogonal $M$ (i.e., $M^\top=M^{-1}$).

First I note that if $D^2_y u(y)$ denotes the Hessian matrix, then $$ \Delta_y u(y)={\rm tr}\, D^2_y u(y). $$ Note also that $D^2=\nabla\nabla^T$, where $\nabla$ is the column vector of partial derivatives.

Now the only thing we need is the fact that $\nabla_x u(Mx)=M^\top \nabla_y u(y)|_{y=Mx}$ and that ${\rm tr}\,(AB)={\rm tr}\,(BA)$.

$$ \Delta_x v(x)={\rm tr}(D^2_x v(x))={\rm tr}(D^2_x u(Mx))=\\ {\rm tr}(\nabla_x\nabla_x^Tu(Mx))={\rm tr}(\nabla_x(\nabla_x u(Mx))^\top)=\\ {\rm tr}(\nabla_x(M^\top \nabla_y u(y))^\top)={\rm tr}(\nabla_x(\nabla_y^\top u(y))M)=\\ {\rm tr}(M^\top\nabla_y\nabla_y^\top u(y)M)={\rm tr}(\nabla_y\nabla_y^\top u(y))\\ ={\rm tr}(D^2_y u(y))=\Delta_y u(y)=0 $$ as required.

Answer 2

$$ \frac{\partial f}{\partial y} = \frac{\partial f}{\partial u} \frac{\partial u}{\partial y} + \frac{\partial f}{\partial v} \frac{\partial v}{\partial y} = \frac{\partial f}{\partial u} \sin \theta+ \frac{\partial f}{\partial v} \cos \theta$$

$$\frac{\partial^2f}{\partial y^2}=\frac{\partial}{\partial y} \left ( \frac{\partial f}{\partial y} \right)=\frac{\partial}{\partial y} \left ( \frac{\partial f}{\partial u} \sin \theta + \frac{\partial f}{\partial v} \cos \theta \right)= \frac{\partial}{\partial y} \frac{\partial f}{\partial u} \sin \theta + \frac{\partial}{\partial y} \frac{\partial f}{\partial v} \cos \theta$$

Now the problem is to compute $\frac{\partial}{\partial y} \frac{\partial f}{\partial u}$ and $\frac{\partial } {\partial y} \frac{\partial f}{\partial v}$. We can think of $\frac{\partial}{\partial y} \frac{\partial f}{\partial u}$ as $\frac{\partial}{\partial u} \frac{\partial f}{\partial y}$ and similarly for the latter.

$$\frac{\partial}{\partial u} \frac{\partial f}{\partial y} = \frac{\partial}{\partial u} \left( \frac{\partial f}{\partial u} \sin \theta + \frac{\partial f}{\partial v} \cos \theta \right) = \frac{\partial^2 f}{\partial u^2} \sin \theta + \frac{\partial f}{\partial u \partial v} \cos \theta$$

$$\frac{\partial}{\partial v} \frac{\partial f}{\partial y} = \frac{\partial}{\partial v} \left( \frac{\partial f}{\partial u} \sin \theta + \frac{\partial f}{\partial v} \cos \theta \right) = \frac{\partial^2 f}{\partial u \partial v} \sin \theta + \frac{\partial^2 f}{\partial v^2} \cos \theta$$

Now plug this back in to $\frac{\partial^2f}{\partial y^2}$ giving,

$$\frac{\partial^2f}{\partial y^2}= \left (\frac{\partial^2 f}{\partial u^2} \sin \theta + \frac{\partial f}{\partial u \partial v} \cos \theta \right) \sin \theta + \left (\frac{\partial^2 f}{\partial u \partial v} \sin \theta + \frac{\partial^2 f}{\partial v^2} \cos \theta \right) \cos \theta $$

Repeat the process for $\frac{\partial^2 f}{\partial x^2}$, add together, and you will arrive at the desired result.

Answer 3

In a paper Discrete spherical means of directional derivatives and Veronese maps (arXiv:1106.3691) get get the identity:

$$ \Delta f = \frac{2}{\pi}\int_0^\pi d\phi \frac{\partial^2 f}{\partial\mathbf{e}_\phi^2} $$

The Laplacian is the average of the second directional derivative in all directions. $\mathbf{e}_\phi = (\cos \phi, \sin \phi)$ and

$$ \frac{\partial}{\partial\mathbf{e}_\phi} = \cos \phi \frac{\partial}{\partial x}+ \sin \phi \frac{\partial}{\partial y} = \nabla \cdot (\cos \phi, \sin \phi)$$

Directional derivatives are ways of taking dervatives in directions other than $x$ and $y$ axes.

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If this seems too much, let's just try 3 directions (then try 5, 7, or more):

$$ \Delta f = \frac{1}{3} \bigg[ \frac{\partial^2 f}{\partial x^2} + \big( \underbrace{\cos \tfrac{2\pi}{3}\cdot \frac{\partial^2 f}{\partial x^2} + \sin \tfrac{2\pi}{3}\cdot \frac{\partial^2 f}{\partial y^2}}_{\phi = 2\pi/3}\big) + \big(\underbrace{\cos \tfrac{4\pi}{3}\cdot \frac{\partial^2 f}{\partial x^2} + \sin \tfrac{4\pi}{3}\cdot \frac{\partial^2 f}{\partial y^2}}_{\phi=4\pi/3}\big) \bigg]$$

Answer 4

Off the top of my head, I think an easier way to do this would be to write it all in vector/matrix notation. Under a rotation, $\vec\nabla f\rightarrow R\cdot \vec\nabla f$ with $R$ being your rotation matrix. Then the Laplacian would transform like $\nabla^{2}f=\vec\nabla\cdot \vec\nabla f\rightarrow \vec\nabla\cdot R^{T}R\cdot \vec\nabla f$. Rotation matrices satisfy $R^{T}R=1$ so that should do it. There is a bit more rigor to this but this should be a good starting point.

Answer 5

Consider a function $f: \mathbb{R}^2 \to \mathbb{R}$, say $(x,y) \mapsto f(x,y)$

Rotate $(x,y)$ by $\theta$ to get $(u,v)$ :

$$ \begin{bmatrix} u \\ v \end{bmatrix} = \begin{bmatrix} \cos \theta & -\sin\theta \\ \sin \theta & \cos\theta \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} $$ In reverse, we get: $$ \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} u\cos \theta +v\sin\theta \\ -u\sin \theta + v\cos\theta \end{bmatrix} $$

We can define a new function $\tilde{f}:\mathbb{R}^2\to \mathbb{R}$ that

$$ \tilde{f}(u,v):=f(x(u,v),y(u,v)) $$

If Laplace’s Operator is rotationally invariant, then $\nabla^2 \tilde{f}(u,v)= \nabla^2f(x,y)$. This can be proved by the following:

$$\begin{aligned} &\nabla^2 \tilde{f}(u,v)\\ =&\tilde{f}_{uu}+\tilde{f}_{vv} \\ =&{\partial \over \partial u} \left(f_x {\partial x \over \partial u} + f_y {\partial y \over \partial u} \right) +\tilde{f}_{vv} \\ =&f_{xx} \left({\partial x \over \partial u}\right)^2 + f_{xy} {\partial y \over \partial u}{\partial x \over \partial u}+f_x {\partial^2 x \over \partial u^2} \\ &f_{yy} \left({\partial y \over \partial u}\right)^2 + f_{yx} {\partial x \over \partial u}{\partial y \over \partial u}+f_y {\partial^2 y \over \partial u^2}\\ &+\tilde{f}_{vv}\\ =&f_{xx} \cos^2\theta -2\sin\theta\cos\theta f_{xy} + \sin^2\theta f_{yy} +\tilde{f}_{vv} \\ =&f_{xx} \cos^2\theta -2\sin\theta\cos\theta f_{xy} + \sin^2\theta f_{yy} \\ &f_{xx} \sin^2\theta +2\sin\theta\cos\theta f_{xy} + \cos^2\theta f_{yy} \\ =&f_{xx}+f_{yy} \\ =&\nabla^2f(x,y) \end{aligned}$$

Answer 6

In addition to the existing answers, I would like to point out a simple,elegant proof in coordinates. Let $x^i,y^i:U \to \mathbb{R}^{n}$ be two charts, which are related by a rotation matrix $R \in SO(2)$, or in coordinates $R^{i}_{j}$. We then have \begin{equation} x^i=R^{i}_{j} y^{j} \implies dx^{i}=R^{i}_{j} dy^{j}. \end{equation} More concretely, in your notation we have $x^1=u$,$x^2=v$, $y^1=x$, $y^2=y$, so that my above equations reads as: \begin{equation} \begin{pmatrix} u \\ v \end{pmatrix}=\begin{pmatrix}\cos \theta & \sin \theta \\ - \sin \theta & \cos \theta \end{pmatrix} \begin{pmatrix}x \\ y \end{pmatrix}. \end{equation} As $R$ is a rotation matrix, it satisfies: $R^{i}_{j} R^{j}_{k}=\delta^{i}_{k}.$

Let $f$ be a smooth function. For its differential, we can write: \begin{equation} df=\partial_{i}f dx^{i}=\partial_{j} f dy^{j}. \end{equation} Now, we can use the relation how $x$ and $y$ are related: \begin{equation} \partial_{i}f dx^{i}=\partial_{i} f R^{i}_{j} dy^{j}=\underbrace{R^{i}_{j} \partial_{i}}_{= \partial_{j}, \; \; \text{from the above equality}}f dy^{j}. \end{equation} Therefore, the above implies: \begin{equation} \partial'_{j} \partial'^{j}=R^{i}_{j} R^{j}_{k} \partial_{i} \partial^{k} \implies \nabla ' ^2=\partial'_{j} \partial'^{j}=R^{i}_{j} R^{j}_{k} \partial_{i} \partial^{k} =\delta^{i}_{k} \partial_{i} \partial^{k}=\partial_{i} \partial^{i}=\nabla^2. \end{equation} Recalling the definition of the Laplacian in terms of $\nabla$ concludes the proof: \begin{equation} \Delta':=\nabla'^2, \; \; \Delta:= \nabla^2 \implies \Delta'=\Delta. \end{equation}