Laplace Operator in Polar Coordinates: Steps & Solutions

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Homework Statement



The Laplace operator Δ is defined by: Δ=
c65dd028c1c9fb80a8288ca893e949da.png


Show in polar coordinates r and Θ, that the Laplace operator takes the following form:

http://upload.wikimedia.org/wikipedia/en/math/0/7/a/07a878276cffd0c680f3f827204aba24.png

Homework Equations



x=rcos(Θ), y=rsin(Θ), r ≥ 0, Θ ∈ [0,2∏]

The Attempt at a Solution



It seems simple; convert the x and y in terms of polar coordinates, and differentiate twice. However, when i attempt to differentiate twice, in respect to r and theta, my r disappears and the trig function simply reverts.
 
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Can you show your work where you differentiate the first time?
 
I like Serena said:
Can you show your work where you differentiate the first time?

Sorry for the delayed response, been studying for other finals.

Here is where i am stuck.

For the x partial derivative:

\partial / \partialx= -rsinθ+cosθ

For the y partial derivative:

\partial / \partialy= sinθ+rcosθ

I know this is incorrect because i was informed by my professor that there is a nested chain ruled inside a product rule in the second differentiation (or something similar). d/dr d/dx +d/dθ d/dx is the format for the x partial derivative, and just replacing x with y gives you the format for the y partial derivative.

Where am i going wrong?
 
tsamocki said:
For the x partial derivative:

\partial / \partialx= -rsinθ+cosθ

What did you differentiate? That seems to be missing.
And how did you get this result?


tsamocki said:
I know this is incorrect because i was informed by my professor that there is a nested chain ruled inside a product rule in the second differentiation (or something similar). d/dr d/dx +d/dθ d/dx is the format for the x partial derivative, and just replacing x with y gives you the format for the y partial derivative.

Where am i going wrong?

Yes, the chain rule for multivariable differentiation is:
{\partial \over \partial x}f(u(x,y), v(x,y))={\partial f \over \partial u}{\partial u\over \partial x}+{\partial f \over \partial v}{\partial v \over \partial x}

You should do something similar to f(x(r,θ), y(r,θ))...
 
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