Laplace in Polar Coordinates: Solving the Polar Coordinate Equation

In summary, the conversation was about solving a polar coordinate equation using variation of parameters and homogeneous boundaries with \vartheta. The equation had the form U(r,\vartheta)=Uo(r,\vartheta)+\sumcnRn(r)\varthetan(\vartheta) and the steps taken involved finding the values for \varthetan(\vartheta) and Rn(r). The conversation ended with the question of how to solve for Cn in the equation Cnr4n.
  • #1
dp182
22
0

Homework Statement


solve the polar coord equation


Homework Equations


Urr+(1/r)Ur+(1/r2)U[itex]\vartheta[/itex][itex]\vartheta[/itex]=0; 1[itex]\leq[/itex]r[itex]\leq[/itex]2, 0[itex]\leq[/itex][itex]\vartheta[/itex][itex]\leq[/itex][itex]\pi[/itex]/4
U(1,[itex]\vartheta[/itex])=0 U(2,[itex]\vartheta[/itex])=[itex]\vartheta[/itex]
U[itex]\vartheta[/itex](r,0)=0 U([itex]\vartheta[/itex],[itex]\pi[/itex]/4)=0
and has the form
U(r,[itex]\vartheta[/itex])=Uo(r,[itex]\vartheta[/itex])+[itex]\sum[/itex]cnRn(r)[itex]\vartheta[/itex]n([itex]\vartheta[/itex])

The Attempt at a Solution


so I used variation of parameters and the homogeneous boundaries with [itex]\vartheta[/itex] to get [itex]\vartheta[/itex]n([itex]\vartheta[/itex])=cos(4n[itex]\vartheta[/itex]) then taking r2R''+rR'-cR=0 and using the Euler equation to get r(r-1)+r-c which is r=[itex]\pm[/itex]c, c=16n2
so that will give me Rn(r)=Cnr4n+dnr-4n and knowing condition U(1,[itex]\vartheta[/itex])=0 bn=0 so what's left is Cnr4n I got stuck here how do I solve for Cn do I try subbing in conditions to solve or is there some equation I am missing
and then using the
 
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  • #2
form U(r,\vartheta)=Uo(r,\vartheta)+\sumcnRn(r)\varthetan(\vartheta) I'm not sure where to go from here
 

1. What is Laplace in polar coordinates?

Laplace in polar coordinates is a mathematical technique used to solve differential equations in two dimensions. It involves converting the Cartesian coordinates (x, y) to polar coordinates (r, θ) and using the Laplace operator (∇²) in the resulting equation.

2. Why is Laplace in polar coordinates useful?

Laplace in polar coordinates is useful because it simplifies the solution of certain types of differential equations, particularly those with circular or symmetric boundaries. It also allows for the use of separation of variables, which can make the problem easier to solve.

3. How do you convert Laplace in Cartesian coordinates to polar coordinates?

To convert Laplace in Cartesian coordinates to polar coordinates, you can use the following formula:

∇² = ∂²/∂x² + ∂²/∂y² = (1/r) ∂/∂r (r ∂/∂r) + (1/r²) ∂²/∂θ²

4. What are the boundary conditions for Laplace in polar coordinates?

The boundary conditions for Laplace in polar coordinates depend on the specific problem being solved. However, they typically involve specifying the value of the function at certain points or along certain curves within the polar coordinate system.

5. Can Laplace in polar coordinates be extended to three dimensions?

Yes, Laplace in polar coordinates can be extended to three dimensions by using spherical coordinates (r, θ, φ) instead of just polar coordinates. The Laplace operator in spherical coordinates is written as:

∇² = (1/r²) ∂/∂r (r² ∂/∂r) + (1/r²sinθ) ∂/∂θ (sinθ ∂/∂θ) + (1/r²sin²θ) ∂²/∂φ²

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