Converting differential equation $U_{xx} - U_{yy} = 0$ to polar coordinates

Question

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:

$U_\phi = -U_xr\sin(\phi)+U_y\cos(\phi)$

$U_{rr} = \cos(\phi)(U_{xx}\cos(\phi) + U_{xy}\sin(\phi))+\sin(\phi)(U_{yx}\cos(\phi) + U_{yy}\sin(\phi))$

$U_{r\phi} = \cos(\phi)(-U_{xx}r\sin(\phi)+U_{xy}r\cos(\phi)) - \sin(\phi)U_x + sin(\phi)(-U_{xy}r\sin(\phi)+U_{yy}r\cos(\phi))+\cos(\phi)U_y$

$U_{\phi\phi} = -r\sin(\phi)(-U_{xx}r\sin(\phi) + U_{xy}r\cos(\phi)) - U_xr\cos(\phi) + r\cos(\phi)(-U_{xy}r\sin(\phi) + U_{yy}r\cos(\phi)) - U_yr\sin(\phi)$

Answer 1

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 $

Answer 2

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?

Answer 3

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.