Polar form of Laplace’s Equation

Jul 31

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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?
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About samjshah

I am a high school math teacher in Brooklyn, New York. I enjoy getting students excited about math by being math’s loudest and most passionate cheerleader.

Posted on July 31, 2008, in Uncategorized. Bookmark the permalink. 10 Comments.

  1. Mr. K | July 31, 2008 at 6:03 am

    I think you’re making it too hard. (If it makes you feel better, I slowed down at part 3)

    You’ve got x = r cos theta. Ignore the y part for now, and differentiate x & r:

    dx = dr cos theta

    will lead to

    dx/dr = cos theta.

    and if you differentiate x & theta, you get

    dx = r d (cos theta) = r (- sin theta) d theta

    which becomes

    dx/d theta = -r sin theta.

    Reply
  2. samjshah | July 31, 2008 at 6:29 am

    Hi Mr. K,

    I do tend to make things tougher than they need be. But I am still confused about your solution, because we need to get

    dr/dx=cos(theta)

    and what you get

    dx/dr=cos(theta).

    When you take x=r cos(theta) and say “differentiate x&r” what do you mean? Take the derivative of both sides with respect to…

    Am I missing something? (Usually I am.)

    Sam.

    Reply
  3. Mr. K | July 31, 2008 at 4:37 pm

    you’re right. i should have had my reading glasses on. plus, my answers didn’t make sense.

    now i’m just as lost as you, if not more.

    Reply
  4. Robert | August 7, 2008 at 7:06 pm

    I’ve been looking at this problem for an hour or so and I can’t think of a simpler or more intuitive approach that what you came up with. Given that this is Anton you’re talking about, a solution involving solving a system of partial derivatives probably is considered “intuitive”.

    Reply
  5. Robert | August 7, 2008 at 7:09 pm

    Correction: I’m also trying to work through your optimization problem, which is from Anton, so I just assumed the problem in this post is Anton as well.

    Carry on.

    Reply
  6. samjshah | August 7, 2008 at 10:29 pm

    Thanks! Yup, it’s definitely from Anton.
    I just figured I was missing something simple, but I’m glad I wasn’t!

    Reply
  7. Kajetan Sikorski | November 2, 2009 at 9:34 am

    r^2 = x^2 + y^2
    2 r dr/dx = 2 x
    2 r dr/dx = 2 r cos(theta)
    dr/dx = cos(theta)

    Reply
  8. Miguel-Angel Manrique | October 11, 2011 at 7:16 pm

    (In case the LaTeX doesn’t show, here’s a link to a web viewer: http://vclab.atp.ruhr-uni-bochum.de/software/HotEqn/HotEqn.html .)

    Using the relations $\theta = \arctan(\frac{y}{x})$, $x = r \cos (\theta)$, and $y = r \sin (\theta)$, you can take the partial of $\theta$ with respect to x to get

    $$ \frac{\partial\theta}{\partial x} = \frac{1}{1 + \frac{y}{x}} \cdot \frac{\partial}{\partial x}\left[ \frac{y}{x} \right] = \dots = – \frac{\sin \theta}{r}$$,

    and do something similar for $ \frac{\partial\theta}{\partial y}$. This is a good problem. It got a lot of old gears turning for me.

    Great blog samjshah. I’m sure it’s useful around the world.

    Best, Miguel

    Reply

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