Converting differential equation $U_{xx} - U_{yy} = 0$ to polar coordinates Let $U \in C^2(\mathbb {R}), U_{xx} - U_{yy} = 0$. Convert it to polar coordinates $x = r\cos(\phi), y = r\sin(\phi)$.
I tried to solve it by this way:
$\frac{\partial U}{\partial r} = \frac{\partial U}{\partial x}\frac{\partial x}{\partial r}+ \frac{\partial U}{\partial y}\frac{\partial y}{\partial r}$
$U_r = U_x\cos(\phi) + U_y\sin(\phi)$
similarly:
$U_\phi = -U_xr\sin(\phi)+U_y\cos(\phi)$
$U_{rr} = \cos(\phi)(U_{xx}\cos(\phi) + U_{xy}\sin(\phi))+\sin(\phi)(U_{yx}\cos(\phi) + U_{yy}\sin(\phi))$
$U_{r\phi} = \cos(\phi)(-U_{xx}r\sin(\phi)+U_{xy}r\cos(\phi)) - \sin(\phi)U_x + sin(\phi)(-U_{xy}r\sin(\phi)+U_{yy}r\cos(\phi))+\cos(\phi)U_y$
$U_{\phi\phi} = -r\sin(\phi)(-U_{xx}r\sin(\phi) + U_{xy}r\cos(\phi)) - U_xr\cos(\phi) + r\cos(\phi)(-U_{xy}r\sin(\phi) + U_{yy}r\cos(\phi)) - U_yr\sin(\phi)$
 A: The differential operators can be written as:
$
\left\{ \begin{matrix} \frac{\partial}{\partial r} \\ \frac{\partial}{\partial \phi} \end{matrix} \right\} = 
\left[ \begin{matrix}  \frac{\partial x}{\partial r} & \frac{\partial y}{\partial r} \\ \frac{\partial x}{\partial \phi} & \frac{\partial y}{\partial \phi} \end{matrix}  \right]
\left\{ \begin{matrix} \frac{\partial}{\partial x} \\ \frac{\partial}{\partial y} \end{matrix} \right\}
$
So you can obtain $\frac{\partial}{\partial x}$ and $\frac{\partial}{\partial y}$ as a function of $r$ and $\phi$. To do that, evaluate the terms in the square matrix and invert it:
$
\left\{ \begin{matrix} \frac{\partial}{\partial x} \\ \frac{\partial}{\partial y} \end{matrix} \right\}
 = 
\left[ \begin{matrix}  \frac{\partial x}{\partial r} & \frac{\partial y}{\partial r} \\ \frac{\partial x}{\partial \phi} & \frac{\partial y}{\partial \phi} \end{matrix}  \right]^{-1}
\left\{ \begin{matrix} \frac{\partial}{\partial r} \\ \frac{\partial}{\partial \phi} \end{matrix} \right\}
$
You end up with:
$
\frac{\partial}{\partial x} = cos\phi \frac{\partial}{\partial r} - \frac{\sin\phi}{r}\frac{\partial}{\partial\phi}\\
\frac{\partial}{\partial y} = \sin\phi \frac{\partial}{\partial r} + \frac{cos\phi}{r}\frac{\partial}{\partial\phi}\\
$
Now, use these relations to rewrite your differential equation. Observe that
$
\frac{\partial^2 U}{\partial x^2}=\frac{\partial}{\partial x}\frac{\partial U}{\partial x}
$
so, write $\partial^2/\partial x^2$ as:
$
\left( cos\phi \frac{\partial}{\partial r} - \frac{\sin\phi}{r}\frac{\partial}{\partial\phi} \right) \left( cos\phi \frac{\partial}{\partial r} - \frac{\sin\phi}{r}\frac{\partial}{\partial\phi} \right)U
$
$
\left( cos^2\phi \frac{\partial^2}{\partial r^2} - \frac{\sin\phi cos\phi}{r}\frac{\partial}{\partial r}\frac{\partial}{\partial \phi} - \frac{\sin\phi cos\phi}{r}\frac{\partial}{\partial \phi}\frac{\partial}{\partial r} + \frac{\sin^2\phi}{r}\frac{\partial^2}{\partial\phi^2} \right)U
$
Notice the cross derivatives $\partial^2/\partial r\partial\phi$ and $\partial^2/\partial \phi\partial r$ are equal, so you can write:
$
\left( cos^2\phi \frac{\partial^2}{\partial r^2} - 2\frac{\sin\phi cos\phi}{r}\frac{\partial^2}{\partial r\partial \phi} + \frac{\sin^2\phi}{r}\frac{\partial^2}{\partial\phi^2} \right)U
$
So,
$
U_{xx}=
\left( cos^2\phi U_{rr} - 2\frac{\sin\phi cos\phi}{r}U_{r\phi} + \frac{\sin^2\phi}{r}U_{\phi\phi} \right)
$
Similarly for y, 
$
U_{yy}=
\left( \sin^2\phi U_{rr} + 2\frac{\sin\phi cos\phi}{r}U_{r\phi} + \frac{cos^2\phi}{r}U_{\phi\phi} \right)
$
and your differential equation can be written in polar coordinates as:
$
U_{xx}-U_{yy}=\left( cos^2\phi U_{rr} - 2\frac{\sin\phi cos\phi}{r}U_{r\phi} + \frac{\sin^2\phi}{r}U_{\phi\phi} \right) - \left( \sin^2\phi U_{rr} + 2\frac{\sin\phi cos\phi}{r}U_{r\phi} + \frac{cos^2\phi}{r}U_{\phi\phi} \right)=0
$
Rearranging terms, you get:
$
U_{xx}-U_{yy}=(cos^2\phi-\sin^2 \phi) U_{rr} - \frac{4\sin\phi cos\phi}{r}U_{r\phi} + \frac{(\sin^2\phi-cos^2\phi)}{r}U_{\phi\phi}=0
$
A: Use
$$
r = \sqrt{x^2 + y^2} ~~~ \phi = \tan^{-1}\frac{y}{x}
$$
So that
\begin{eqnarray}
\frac{\partial u}{\partial x} &=& \frac{\partial u}{\partial r}\frac{\partial r}{\partial x} + \frac{\partial u}{\partial \phi}\frac{\partial \phi}{\partial x}  \\
&=& \frac{x}{r} \frac{\partial u}{\partial r} -\frac{y}{r^2} \frac{\partial u}{\partial \phi} \\
&=& \cos\phi \frac{\partial u}{\partial r} - \frac{\sin \phi}{r}\frac{\partial u}{\partial \phi}
\end{eqnarray}
Can you take it from here?
A: From the equations for $U_r$ and $U_\phi,$ solve for $U_x$ and $U_y.$ Differentiate the latter, take their difference and set it to zero. This is neater, and faster. Also, note that $U_{xy}=U_{yx}$ since $U$ is twice continuously differentiable.
