Day: July 31, 2008

Polar form of Laplace’s Equation

As you might know, this summer I’m prepping for the multivariable calculus course I’m teaching next year. When going through the textbook, I’m attempting some of the “challenging” problems near the end of the section.

Today, when refreshing myself with the chain rule, I came across a problem which tested my intuition. And I’m afraid I’ve lost what little intuitive mojo I had years ago. I’m going to reproduce the problem below.

Question:

Suppose that the equation z=f(x,y) is expressed in the polar form z=g(r,\theta) by making the substitution x=r \cos \theta and y=r\sin \theta.

  1. View r and \theta as functions of x and y and use implicit differentiation to show that:
    \frac{\partial r}{\partial x}=\cos\theta
    \frac{\partial \theta}{\partial x}=-\frac{\sin\theta}{r}
  2. View r and \theta as functions of x and y and use implicit differentiation to show that:
    \frac{\partial r}{\partial y}=\sin\theta
    \frac{\partial \theta}{\partial y}=\frac{\cos\theta}{r}
  3. Use the results in parts (1) and (2) to show that:
    \frac{\partial z}{\partial x}=\frac{\partial z}{\partial r}\cos \theta-\frac{1}{r}\frac{\partial z}{\partial \theta}\sin\theta
    \frac{\partial z}{\partial y}=\frac{\partial z}{\partial r}\cos \theta+\frac{1}{r}\frac{\partial z}{\partial \theta}\cos\theta
  4. Use the results in part (3) to show that:
    (\frac{\partial z}{\partial x})^2+(\frac{\partial z}{\partial y})^2=(\frac{\partial z}{\partial r})^2+\frac{1}{r^2}(\frac{\partial z}{\partial \theta})^2
  5.  Use the result in part (4) to show that if z=f(x,y) satisfies Laplace’s equation
    \frac{\partial^2 z}{\partial x^2}+\frac{\partial^2 z}{\partial y^2}=0
    then  z=g(r,\theta) satisfies the equation
    \frac{\partial^2 z}{\partial r^2}+\frac{1}{r^2}\frac{\partial^2 z}{\partial \theta^2}+\frac{1}{r}\frac{\partial z}{\partial r}=0
    and conversley. The latter equation is called the polar form of Laplace’s equation.
Here’s my issue. I can do parts (3)-(5) fine. But I got stumped for a good 15 minutes on the parts (1) and (2) and I’m sad about it.
So my lament is: wow, I suck.
And my question is: is there an easier way to solve this?
My solution is the following:
We know y=r\sin\theta, so differentiating with respect to x we get
(1) 0=\frac{\partial r}{\partial x}\sin\theta+\frac{\partial \theta}{\partial x}r\cos\theta
We know x=r\cos\theta, so differentiation with respect to x we get
(2) 1=\frac{\partial r}{\partial x}\cos \theta-\frac{\partial \theta}{\partial x}r\sin\theta
We now have a system of two equations, which we can solve for \frac{\partial r}{\partial x} and \frac{\partial \theta}{\partial x}. Going through the motion yields the right answer.
So yeah, looking back, its all makes sense. And I feel sheepish for posting this, because this is pretty easy to do. But it doesn’t seem simple. There is a part which calls for intuition: to know that we have to get two equations and then solve them together. I’m guessing there’s an easier way to do this, that doesn’t involve solving a system of equations. Is there?