Laplace Operator in Polar Coordinates: Steps & Solutions

In summary, the Laplace operator in polar coordinates takes the form shown in the given equation. To find this result, one must use the chain rule for multivariable differentiation, substituting in the values for x and y in terms of r and θ. This will result in the correct partial derivatives for both x and y, which can then be used to find the second derivatives and the final form of the Laplace operator in polar coordinates.
  • #1
tsamocki
20
0

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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  • #2
Can you show your work where you differentiate the first time?
 
  • #3
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:

[itex]\partial[/itex] / [itex]\partial[/itex]x= -rsinθ+cosθ

For the y partial derivative:

[itex]\partial[/itex] / [itex]\partial[/itex]y= 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?
 
  • #4
tsamocki said:
For the x partial derivative:

[itex]\partial[/itex] / [itex]\partial[/itex]x= -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:
[tex]{\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}[/tex]

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

FAQ: Laplace Operator in Polar Coordinates: Steps & Solutions

What is the Laplace Operator in polar coordinates?

The Laplace Operator, also known as the Laplacian, is a mathematical operator used in vector calculus to describe the curvature and gradient of a function in a given coordinate system. In polar coordinates, the Laplacian is expressed as a combination of partial derivatives with respect to the radial and angular coordinates.

How do you solve for the Laplace Operator in polar coordinates?

To solve for the Laplace Operator in polar coordinates, you must first express the function in terms of the radial and angular coordinates. Then, take the partial derivatives of the function with respect to each coordinate and plug them into the Laplace Operator formula. Finally, simplify the resulting equation to solve for the Laplacian.

What are the steps for solving the Laplace Operator in polar coordinates?

The steps for solving the Laplace Operator in polar coordinates are:

  1. Express the function in terms of the radial and angular coordinates.
  2. Take the partial derivatives of the function with respect to each coordinate.
  3. Plug the derivatives into the Laplace Operator formula.
  4. Simplify the resulting equation to solve for the Laplacian.

What are the applications of the Laplace Operator in polar coordinates?

The Laplace Operator in polar coordinates has various applications in physics, engineering, and mathematics. It is used to describe the motion of fluids, heat transfer, and electromagnetic fields. It also plays a crucial role in solving boundary value problems and partial differential equations.

What are some common mistakes when solving for the Laplace Operator in polar coordinates?

Some common mistakes when solving for the Laplace Operator in polar coordinates include:

  • Forgetting to express the function in terms of the correct coordinate system.
  • Making errors in taking the partial derivatives of the function.
  • Not simplifying the resulting equation properly.
  • Mistakes in applying the Laplace Operator formula.

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