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Suppose $(x, y)$ and $(p, q)$ are coordinates in the plane related by rotation around a fixed point $(a, b)$, as follows: $$\begin{bmatrix} p\\ q\end{bmatrix} = \begin{bmatrix} \cos(t) & -\sin(t) \\ \sin(t) & \cos(t) \end{bmatrix} \begin{bmatrix} x-a \\ y-b \end{bmatrix}$$ where $t$ is the rotation angle. Applying the chain rule show that $u(p(x,y), q(x,y)) $ satisfies $u_{xx}+u_{yy}=0$ iff $u_{pp}+u_{qq}= 0$. Where would I use chain rule in this problem? I am kind of confused because $u$, $p$, and $q$ are all functions of two variables.

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marked as duplicate by user99914, jkabrg, Namaste, Sahiba Arora, John Doe Feb 4 '18 at 19:58

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1) Since $u(x,y) = u(p(x,y), q(x,y))$, you have (by the chain rule) $$ \frac{\partial u}{\partial x} = \frac{\partial u}{\partial p} \frac{\partial p}{\partial x} + \frac{\partial u}{\partial q} \frac{\partial q}{\partial x}. $$ The second derivative: $$ \frac{\partial^2 u}{\partial x^2} = \frac{\partial}{\partial x} \left(\frac{\partial u}{\partial x} \right) = \frac{\partial^2 u}{\partial p^2} \left(\frac{\partial p}{\partial x} \right)^2+ \frac{\partial u}{\partial p} \frac{\partial^2 p}{\partial x^2}+ 2 \frac{\partial^2 u}{\partial p \partial q} \frac{\partial p}{\partial x} \frac{\partial q}{\partial x} + \frac{\partial^2 u}{\partial q^2} \left(\frac{\partial q}{\partial x} \right)^2+ \frac{\partial u}{\partial q} \frac{\partial^2 q}{\partial x^2}. $$ The similar equalities you can obtain for $u_y$ and $u_{yy}$.

2) Note that $$ p = (x-a) \, \cos t - (y-b) \, \sin t,\\ q = (x-a) \, \sin t + (y-b) \, \cos t $$ Therefore, $$ \frac{\partial p}{\partial x} = \cos t, \quad \frac{\partial^2 p}{\partial x^2} = 0,\\ \frac{\partial q}{\partial x} = \sin t, \quad \frac{\partial^2 q}{\partial x^2} = 0. $$ Substituting these equalities (and the corresponding for $p_y$, $q_y$) to the expression for $u_{xx}$ (and $u_{yy}$), and summing them, you will finally get what you want.

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