Switching to polar coordinates in differential equation

Question

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.

Answer 1

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.

Answer 2

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 ?

Answer 3

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.