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

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4 Answers 4

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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.

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    $\begingroup$ Sorry for the necropost. Say we have f: $R^m \rightarrow R^n $ in $C^2$ and J(f) is the Jacobian of f . Would $JJ^T$ be , in general, the second derivative of f? $\endgroup$
    – gary
    Sep 4, 2018 at 1:45
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    $\begingroup$ Why is it sufficient to only show that harmonic functions are "send" to harmonic functions? $\endgroup$ Sep 21, 2018 at 16:20
  • $\begingroup$ A direct computation is available here people.math.wisc.edu/~feldman/819/sol1.pdf $\endgroup$
    – QA Ngô
    Jun 23, 2021 at 18:54
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$$ \frac{\partial f}{\partial y} = \frac{\partial f}{\partial u} \frac{\partial u}{\partial y} + \frac{\partial f}{\partial v} \frac{\partial v}{\partial y} = \frac{\partial f}{\partial u} \sin \theta+ \frac{\partial f}{\partial v} \cos \theta$$

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

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.

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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.

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  • $\begingroup$ I think this is a potentially elegant answer; please provide more details. $\endgroup$
    – JSycamore
    Nov 3, 2017 at 13:08
  • $\begingroup$ Why you transform the laplacian that way? Can you explain why you put the matrices between the two gradients? Thanks $\endgroup$ Mar 1, 2018 at 17:05
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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.


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

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

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    $\begingroup$ This looks like overkill. I have the feeling the OP will not follow. $\endgroup$ Oct 12, 2014 at 23:55
  • $\begingroup$ I appreciate the answer, but I have to agree with Mark - a bit above my paygrade. $\endgroup$
    – EmutheEmu
    Oct 12, 2014 at 23:57
  • $\begingroup$ @EmutheEmu By chain rule, your equation $u = x \cos \theta + y \sin \theta$ becomes $$\frac{\partial}{\partial u} = \frac{\partial}{\partial x} \cos \theta + \frac{\partial}{\partial y} \sin \theta $$ and a similar formula for second derivatives. Then you can plug in. $\endgroup$
    – cactus314
    Oct 13, 2014 at 0:09
  • $\begingroup$ Thanks, I was able to complete the proof using this. $\endgroup$
    – EmutheEmu
    Oct 13, 2014 at 0:21

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