Switching to polar coordinates in differential equation Hi guys i need to help with this excercises. :)

If 
  $$\begin{cases}
x=r\cos \theta\\
y=r \sin\theta
\end{cases}$$
  prove that the equation:
  $$x^2\frac{\partial^2z}{\partial x^2}+2xy\frac{\partial^2z}{\partial x\partial y}+y^2\frac{\partial^2z}{\partial y^2}=0$$
  is equivalent to 
  $$r^2\frac{\partial^2z}{\partial \theta^2}=0.$$
OK I´VE TRIED

this could be helpful ? then, what do i make?
I tried of isolate  $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$
$$\frac{\partial z}{\partial x}=\frac{\partial z}{\partial r}cos\theta-\frac{\partial z}{\partial \theta}(\frac{1}{r})sin\theta$$and
$$\frac{\partial z}{\partial y}=\frac{\partial z}{\partial \theta}(\frac{1}{r})cos\theta+\frac{\partial z}{\partial r}sin\theta$$
mm and we know that
$$\frac{\partial z}{\partial r}=\frac{\partial z}{\partial x}cos\theta+\frac{\partial z}{\partial y}sin\theta$$
$$\frac{\partial z}{\partial \theta}=-\frac{\partial z}{\partial x}rsin\theta+\frac{\partial z}{\partial y}rcos\theta$$
but I do not get anything.
How can I do to get the result??
help.
 A: You can convert the first differential equation from Cartesian coordinates to polar coordinates, or you can convert the second differential equation from polar coordinates to Cartesian coordinates.
I think the second way is slightly easier.
First, observe that by the product rule,
$$r \frac{\partial}{\partial r} \left(r \frac{\partial u}{\partial r}\right)
 = r \left(\frac{\partial r}{\partial r} \frac{\partial u}{\partial r}
           + r \frac{\partial}{\partial r}
                \left(\frac{\partial u}{\partial r}\right)\right)
 = r \frac{\partial u}{\partial r} + r^2 \frac{\partial^2 u}{\partial r^2},$$
that is,
$$ r^2 \frac{\partial^2 u}{\partial r^2}
 = r \frac{\partial}{\partial r} \left(r \frac{\partial u}{\partial r}\right)
   - r \frac{\partial u}{\partial r}. \tag 1$$
We can convert the right-hand side of this equation to Cartesian coordinates
as follows.
Let $x = r\cos\theta$ and $y = r\sin\theta$. Then for any $u$,
\begin{align}
r\frac{\partial u}{\partial r}
&= r\left(\frac{\partial u}{\partial x} \frac{\partial x}{\partial r}
         + \frac{\partial u}{\partial y} \frac{\partial y}{\partial r}\right) \\
&= r\left(\frac{\partial u}{\partial x} \cos\theta
         + \frac{\partial u}{\partial y} \sin\theta\right) \\
&= x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y}.
\end{align}
Apply this formula,
$r\frac{\partial u}{\partial r}
 = x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y}$,
with $r\frac{\partial u}{\partial r}$ in place of $u$:
\begin{align}
r\frac{\partial}{\partial r}\left(r\frac{\partial}{\partial r}u\right)
&= x \frac{\partial}{\partial x} \left(r\frac{\partial}{\partial r}u\right)
   + y \frac{\partial}{\partial y} \left(r\frac{\partial}{\partial r}u\right) \\
&= x \frac{\partial}{\partial x} 
 \left(x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y}\right)
   + y \frac{\partial}{\partial y}
 \left(x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y}\right)
\end{align}
Working out the various terms,
\begin{align}
x \frac{\partial}{\partial x} \left(x \frac{\partial u}{\partial x}\right)
 &= x \frac{\partial u}{\partial x} + x^2 \frac{\partial^2 u}{\partial x^2}, \\
x \frac{\partial}{\partial x} \left(y \frac{\partial u}{\partial y}\right)
 &= xy \frac{\partial^2 u}{\partial x \partial y}, \\
y \frac{\partial}{\partial y} \left(x \frac{\partial u}{\partial x}\right)
 &= xy \frac{\partial^2 u}{\partial x \partial y}, \text{ and}\\
y \frac{\partial}{\partial y} \left(y \frac{\partial u}{\partial y}\right)
 &= y \frac{\partial u}{\partial y} + y^2 \frac{\partial^2 u}{\partial y^2}.
\end{align}
Therefore
\begin{align}
r\frac{\partial}{\partial r}\left(r\frac{\partial}{\partial r}u\right)
&= x^2 \frac{\partial^2 u}{\partial x^2}
  + 2 xy \frac{\partial^2 u}{\partial x \partial y}
  + y^2 \frac{\partial^2 u}{\partial y^2}
  + x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} \\
&= x^2 \frac{\partial^2 u}{\partial x^2}
  + 2 xy \frac{\partial^2 u}{\partial x \partial y}
  + y^2 \frac{\partial^2 u}{\partial y^2}
  + r \frac{\partial}{\partial r} u,
\end{align}
that is,
$$ r\frac{\partial}{\partial r}\left(r\frac{\partial}{\partial r}u\right)
    -  r \frac{\partial u}{\partial r}
  = x^2 \frac{\partial^2 u}{\partial x^2}
    + 2 xy \frac{\partial^2 u}{\partial x \partial y}
    + y^2 \frac{\partial^2 u}{\partial y^2}. \tag 2$$
Taking Equations $(1)$ and $(2)$ together, we find that
$$ r^2 \frac{\partial^2 u}{\partial r^2}
  = x^2 \frac{\partial^2 u}{\partial x^2}
    + 2 xy \frac{\partial^2 u}{\partial x \partial y}
    + y^2 \frac{\partial^2 u}{\partial y^2}.$$
This is a little stronger than the fact that was to be proved, but it
certainly does imply that the equations
$x^2 \frac{\partial^2 u}{\partial x^2}
    + 2 xy \frac{\partial^2 u}{\partial x \partial y}
    + y^2 \frac{\partial^2 u}{\partial y^2} = 0$
and $r^2 \frac{\partial^2 u}{\partial r^2} = 0$ are equivalent.
A: In the formula for $\partial z\over\partial\theta$, replace the function $z$ with $\partial z\over\partial x$, then with $\partial z\over\partial y$, and finally with $\partial z\over\partial\theta$. The resulting three equations can be combined to get the result you want.
A: It is just a booring exercise to compute the partial derivatives in polar coordinates (below, the steamroller technique). May be one can find a smarter method ? 

