Laplace Operator in Polar Coordinates: Steps & Solutions

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SUMMARY

The discussion focuses on deriving the Laplace operator in polar coordinates, specifically the transformation from Cartesian coordinates (x, y) to polar coordinates (r, Θ). The participants emphasize the importance of applying the chain rule and product rule correctly during differentiation. Key equations include the relationships x = r cos(Θ) and y = r sin(Θ). The correct approach involves differentiating the functions with respect to both r and Θ, ensuring that the nested chain rule is applied properly to obtain accurate results.

PREREQUISITES
  • Understanding of multivariable calculus, specifically differentiation techniques.
  • Familiarity with polar coordinates and their relationship to Cartesian coordinates.
  • Knowledge of the Laplace operator and its mathematical significance.
  • Proficiency in applying the chain rule and product rule in calculus.
NEXT STEPS
  • Study the derivation of the Laplace operator in polar coordinates using detailed examples.
  • Learn about the chain rule for multivariable functions in depth.
  • Explore applications of the Laplace operator in physics and engineering contexts.
  • Practice converting between Cartesian and polar coordinates with various functions.
USEFUL FOR

Students studying advanced calculus, particularly those focusing on differential equations and mathematical physics. This discussion is also beneficial for educators seeking to clarify the application of differentiation in polar coordinates.

tsamocki
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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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