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So here I have an assignment about higher order derivatives and the chain rule, and a relation to be proved: Show that for a rotation in the plane$$\begin{bmatrix}u\\v \end{bmatrix} =\begin{bmatrix}\cos\theta & -\sin\theta\\\sin\theta & \cos\theta\end{bmatrix} \begin{bmatrix}x\\y \end{bmatrix}$$

and any twice differentiable function $f,$ there holds $$\frac{\partial^2f}{\partial x^2} + \frac{\partial^2f}{\partial y^2} = \frac{\partial^2f}{\partial u^2}+\frac{\partial^2f}{\partial v^2}.$$ What I got so far is that $ f(u,v)=f(x\cos \theta-y\sin\theta, x\sin\theta + y\cos\theta) $ from the matrix multiplication. But I don't really understand how to get $\frac{\partial f}{\partial u} $and $\frac{\partial f}{\partial v}$.

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    $\begingroup$ I think you meant $(u,v)=(x\cos\theta-y\sin\theta,x\sin\theta+y\cos\theta)$, not $f(u,v)$, and $\dfrac{\partial f}{\partial u}=\dfrac{\partial f}{\partial x}\dfrac{\partial x}{\partial u}+\dfrac{\partial f}{\partial y}\dfrac{\partial y}{\partial u}$ $\endgroup$ – J. W. Tanner May 24 '20 at 21:55
  • $\begingroup$ @J.W.Tanner Yes, I meant (u,v).. But what is $ \frac{\partial x}{\partial u} $ and $ \frac{\partial y}{\partial u} $ etc. since x and y do not consist u and v? Or did I misunderstand? $\endgroup$ – Angelia Servais May 24 '20 at 22:31
  • $\begingroup$ you could invert the matrix to get $(x,y)=(u\cos\theta+v\sin\theta,-u\sin\theta+v\cos\theta)$ $\endgroup$ – J. W. Tanner May 24 '20 at 22:37
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The relation between $(u,v)$ and $(x,y)$ is written down by you, and then by the chain rule we could get

$$\frac {\partial f}{\partial x} = \frac {\partial f}{\partial u} \frac {\partial u}{\partial x}+ \frac {\partial f}{\partial v} \frac {\partial v}{\partial x} = \frac {\partial f}{\partial u}\cos\theta + \frac {\partial f}{\partial v}\sin\theta, $$ and likewise you can get $$\frac {\partial f}{\partial y}= -\frac {\partial f}{\partial u}\sin\theta+ \frac {\partial f}{\partial v}\cos\theta. $$ Following this and we can do chain rule again to obtain $$\frac {\partial^2f}{\partial x^2} = \frac {\partial^2f}{\partial u^2}\cos^2\theta+2 \frac {\partial^2 f}{\partial u\partial v}\cos\theta\sin\theta+ \frac {\partial^2f}{\partial v^2}\sin^2\theta $$ and $$ \frac {\partial^2 f}{\partial y^2}= \frac {\partial^2f}{\partial u^2}\sin^2\theta-2 \frac {\partial^2 f}{\partial u\partial v}\cos\theta\sin\theta+ \frac {\partial^2f}{\partial v^2}\cos^2\theta, $$ by which the conclusion follows. Note that this proves the Laplacian is invariant under the rotation.

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