The Laplace Equation in Polar Coordinates

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The discussion revolves around converting the Laplace equation from Cartesian to polar coordinates. The equation \(\frac{\partial^2f}{\partial x^2}+\frac{\partial ^2f}{\partial y^2}= 0\) is to be shown equivalent to \(\frac{\partial^2f}{\partial r^2}+\frac{1}{r^2} \frac{\partial ^2f}{\partial \theta^2} + \frac{1}{ r} \frac{\partial f}{\partial r}= 0\). Participants discuss using the chain rule for partial derivatives, expressing \(f\) in terms of \(r\) and \(\theta\), and the challenges of applying the chain rule for second-order derivatives. There is confusion regarding the correct application of derivatives and the transformation process, particularly in handling the terms correctly. The conversation highlights the complexities of transitioning between coordinate systems in differential equations.
thejinx0r
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Homework Statement



\frac{\partial^2f}{\partial x^2}+\frac{\partial ^2f}{\partial y^2}= 0


Homework Equations



Show that the equation above is equal to:
\frac{\partial^2f}{\partial r^2}+\frac{1}{r^2} \frac{\partial ^2f}{\partial \theta^2} + \frac{1}{ r} \frac{\partial f}{\partial r}= 0

The Attempt at a Solution



So, let f=(r,\theta)
r = \sqrt{x^2+y^2}
\theta = tan^{-1}(y/x)
then by the chain rule, partial f partial r is:

http://www.texify.com/img/%5CLARGE%5C%21%5Cfrac%7B%5Cpartial%20f%7D%7B%5Cpartial%20r%7D%20%5Cfrac%7Bx%7D%7B%20%5Csqrt%7Bx%5E2%20%2B%20y%5E2%7D%20%7D%20%2B%20%5Cfrac%7B%5Cpartial%20f%7D%7B%5Cpartial%20r%7D%20%5Cfrac%7By%7D%7B%20%5Csqrt%7Bx%5E2%20%2B%20y%5E2%7D%20%7D.gif

And I have something similar for partial f partial theta,
but I'm not sure if what I'm doing is right...

Because, I applied the chain rule to get the top part, (which I do not know how to write a matrix in latex :S).

But I've never applied the chain rule twice to get a second order differential.
So I'm stuck there. I tried to replace df/dr with d^f/dr^2 and similarly with the theta and multiplying it by the gaussian matrix (dr/dx dr/dy ; dt/dx dt/dy) [dt = d theta].

But then I realized that I would not get the 3rd term because of the way I just changed df/dr with d^f/dr^2 which is probably wrong on my part ...
 
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One of your LaTeX images contained an error and didn't load properly. Was it supposed to be:

\frac{\partial f}{\partial r}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial r}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial r}

If so, then just use this again:

\frac{\partial ^2 f}{\partial r^2}= \frac{\partial }{\partial r} \left( \frac{\partial f}{\partial r} \right)=\frac{\partial}{\partial x} \left( \frac{\partial f}{\partial r} \right) \cdot \frac{\partial x}{\partial r}+\frac{\partial }{\partial y} \left( \frac{\partial f}{\partial r} \right) \cdot \frac{\partial y}{\partial r}

Substitute \frac{\partial f}{\partial r} into this, and compute \frac{\partial x}{\partial r} and \frac{\partial y}{\partial r}.
 
So I fixed up the first post.

But it does help me do the second derivative and shows that I sort of did write it out wrong

:)
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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