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I was reading in Wikipedia about Rotational invariance and noticed that the two-dimensional Laplacian operator $\nabla^2 = \frac{\partial^2 }{\partial x^2} + \frac{\partial^2 }{\partial y^2}$ is thought to be invariant under rotations. I was trying to prove this for a given function $f\in \mathbb{R}^2$ but I couldn't find a way.

Let's assume we have a 2D given function $f(x,y)$ in Cartesian coordinates. I am trying to show that the Laplacian operator is rotational invariant, which means that:

$$\nabla^{2}_{xy} f = f_{xx}+f_{yy}=f_{x^\prime x^\prime}+f_{y^\prime y^\prime }=\nabla^{2}_{x^\prime y^\prime} f$$

Which is the right way to approach this?

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  • $\begingroup$ I wrote an answer to a similar question a while back math.stackexchange.com/questions/3258431/… Bu this treats the general $n$-dimensional case directly $\endgroup$
    – peek-a-boo
    Apr 19, 2020 at 9:03
  • $\begingroup$ @peek-a-boo Hmm this post seems a bit more complicated. I am still confused. Would it be easy for you to write down an answer for my case? Thanks in advance! $\endgroup$
    – Rog Fed
    Apr 19, 2020 at 9:53

2 Answers 2

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In this 2-dimensional case, everything is much simpler, I agree. In fact you can even write down explicitly what a general rotation looks like. So, suppose you have two sets of coordinates; $(x,y)$ and $(u,v)$, where one is obtained from another by a rotation, say of angle $\phi$: \begin{align} \begin{cases} u &= x\cos \phi - y \sin \phi \\ v &= x \sin \phi + y \cos \phi \end{cases} \end{align} Now, using the chain rule, we find that \begin{align} \dfrac{\partial}{\partial x} &= \dfrac{\partial u}{\partial x} \dfrac{\partial }{\partial u} + \dfrac{\partial v}{\partial x} \dfrac{\partial}{\partial v} \\ &= \cos \phi \dfrac{\partial}{\partial u} + \sin \phi \dfrac{\partial}{\partial v} \end{align} and similarly, \begin{align} \dfrac{\partial}{\partial y} &= -\sin \phi \dfrac{\partial}{\partial u} + \cos \phi \dfrac{\partial}{\partial v} \end{align} Now, try to calculate $\dfrac{\partial^2}{\partial x^2}$ and $\dfrac{\partial^2}{\partial y^2}$ similarly, and then add them up. You should find in a few lines of algebra (after using $\sin^2 + \cos ^2 = 1$ a couple of times) that \begin{align} \dfrac{\partial^2}{\partial x^2} + \dfrac{\partial^2}{\partial y^2} = \dfrac{\partial^2}{\partial u^2} + \dfrac{\partial^2}{\partial v^2} \end{align}


Edit: Answering Question in comments

We have \begin{align} \dfrac{\partial^2 f}{\partial x^2} &=\dfrac{\partial}{\partial x} \left(\dfrac{\partial f}{\partial x} \right) \end{align} Now, temporarily define $g$ as \begin{align} g:= \dfrac{\partial f}{\partial x} = \dfrac{\partial u}{\partial x} \dfrac{\partial f }{\partial u} + \dfrac{\partial v}{\partial x} \dfrac{\partial f}{\partial v} = \cos \phi \dfrac{\partial f}{\partial u} + \sin \phi \dfrac{\partial f}{\partial v} \end{align} So, \begin{align} \dfrac{\partial ^2 f}{\partial x^2} &= \dfrac{\partial g}{\partial x} \\ &= \dfrac{\partial u}{\partial x} \cdot \dfrac{\partial g}{\partial u} + \dfrac{\partial v}{\partial x} \cdot \dfrac{\partial g}{\partial v} \\ &= \cos \phi \dfrac{\partial g}{\partial u} + \sin \phi \dfrac{\partial g}{\partial v} \\ &= \cos \phi \dfrac{\partial }{\partial u} \left( \cos \phi \dfrac{\partial f}{\partial u} + \sin \phi \dfrac{\partial f}{\partial v}\right) + \sin \phi \dfrac{\partial }{\partial v} \left( \cos \phi \dfrac{\partial f}{\partial u} + \sin \phi \dfrac{\partial f}{\partial v} \right) \\ &= \cos^2 \phi \dfrac{\partial ^2 f}{\partial u^2} + 2\cos \phi \sin \phi \dfrac{\partial ^2 f}{\partial u \partial v} + \sin^2 \phi \dfrac{\partial ^2 f}{\partial v^2} \end{align} where in the last line, I expanded everything, and used equality of mixed partials. If you do a similar thing with $y$, you'll get a $-2 \sin \phi \cos \phi$ term instead.

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  • $\begingroup$ Thanks for the answer!Nice and easy explanation. Now It makes a lot more sense to me. Could you please give me a link about the certain chain rule you mention (with just the partial derivatives)? Because I am not familiar with it. In particular, I have seen alternative forms of the chain rule and would like to make it more clear. $\endgroup$
    – Rog Fed
    Apr 19, 2020 at 18:37
  • $\begingroup$ @RogFed I'm sure you've seen the following version of the chain rule (or something similar): if $f = f(u,v)$ and $u = u(x,y)$ and $v = v(x,y)$ then $\dfrac{\partial f}{\partial x} = \dfrac{\partial f}{\partial u} \cdot \dfrac{\partial u}{\partial x} + \dfrac{\partial f}{\partial v} \cdot \dfrac{\partial v}{\partial x}$. Now, simply "remove" $f$ from everything, so that we consider this as an "equality of differential operators". This is how I wrote the first equation. $\endgroup$
    – peek-a-boo
    Apr 19, 2020 at 19:19
  • $\begingroup$ I got that. I am just finding difficulties in calculating the second partial derivatives: $\dfrac{\partial^2}{\partial x^2}$ and $\dfrac{\partial^2}{\partial y^2}$ . I must be doing something wrong with the chain rule,I guess. Could you include one of these calculations to your answer? It would be really helpful. I'm gonna accept the answer as well. $\endgroup$
    – Rog Fed
    Apr 19, 2020 at 19:27
  • $\begingroup$ @RogFed how about you show your calculations, then I can tell you where you've gone wrong/ how to correct it. I suggest putting the $f$ back everywhere to make everything obvious. $\endgroup$
    – peek-a-boo
    Apr 19, 2020 at 19:28
  • $\begingroup$ I am not havving difficulties with the $f$ notation. I just don't get how to apply the partial derivative on the first chain rule. I don't actually know how to use the distributive property with the derivatives, so I haven't yet managed to end up with something to show. $\endgroup$
    – Rog Fed
    Apr 19, 2020 at 19:34
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A rotation from one system of Cartesian coordinates $x_i$ to another with coordinates $y_J$ satisfies $x_i=R_{iJ}y_J$, and hence a chain rule of the form $dx_i=R_{iJ}dy_J$, where we sum over repeated indices and the orthogonal matrix $R$ satisfies $RR^T=I$, or in terms of the Kronecker delta$R_{iJ}R_{kJ}=\delta_{ik}$. First derivatives obey$$R_{iJ}\partial_ifdy^J=\partial_i fdx^i=df=\partial_Jfdy^J\implies R_{ij}\partial_i=\partial_J.$$So$$\partial_J\partial_L=R_{iJ}R_{kL}\partial_i\partial_k\implies\nabla^{\prime2}=\partial_J\partial_J=R_{iJ}R_{kJ}\partial_i\partial_k=\partial_i\partial_i=\nabla^2.$$

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  • $\begingroup$ Thanks for the answer! $\endgroup$
    – Rog Fed
    Apr 19, 2020 at 18:51

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