The Laplace Equation in Polar Coordinates

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SUMMARY

The discussion focuses on transforming the Laplace equation from Cartesian coordinates to polar coordinates, specifically demonstrating that the equation \(\frac{\partial^2f}{\partial x^2}+\frac{\partial ^2f}{\partial y^2}= 0\) is 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 the application of the chain rule for partial derivatives and the challenges of deriving second-order derivatives in polar coordinates. The conversation highlights the importance of correctly substituting derivatives and understanding the transformation process.

PREREQUISITES
  • Understanding of partial derivatives
  • Familiarity with the Laplace equation
  • Knowledge of polar coordinates and their relationship to Cartesian coordinates
  • Proficiency in using the chain rule for differentiation
NEXT STEPS
  • Study the derivation of the Laplace equation in polar coordinates
  • Learn about the chain rule in the context of multivariable calculus
  • Explore applications of the Laplace equation in physics and engineering
  • Practice converting between Cartesian and polar coordinates in various mathematical contexts
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Students and professionals in mathematics, physics, and engineering who are working with differential equations, particularly those focusing on the Laplace equation and coordinate transformations.

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 ...
 
Last edited by a moderator:
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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

:)
 

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