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Math Help - Laplaces Equation using polar coordinates

  1. #1
    LHS
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    Laplaces Equation using polar coordinates

    I would be very grateful if someone could explain how you go out finding the form of the solution to this problem?

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    Quote Originally Posted by LHS View Post
    I would be very grateful if someone could explain how you go out finding the form of the solution to this problem?

    Have your tried expanding the solution is a Fourier series?

    \displaystyle \Delta T(x,y)=\left(\frac{\partial T^2}{\partial r^2} +\frac{1}{r}\frac{\partial T}{\partial r}+\frac{1}{r^2}\frac{\partial T^2}{\partial \theta^2}\right)=0
    in polar coordinates
    Now use separation of variables with T(r,\theta)=R(r)\Theta(\theta)
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    LHS
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    So by expressing cos^2(theta) as 1/2+1/2 cos(2*theta) I can equate with the general solution to laplace in polars? which is

    T(x,y)=A[0]+B[0]log(r)+ (the infinite summation from n=1 of) (A[n]r^n+B[n]r^-n)(C[n]cos(n theta)+D[n]sin(n theta))?
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    Quote Originally Posted by LHS View Post
    So by expressing cos^2(theta) as 1/2+1/2 cos(2*theta) I can equate with the general solution to laplace in polars? which is

    T(x,y)=A[0]+B[0]log(r)+ (the infinite summation from n=1 of) (A[n]r^n+B[n]r^-n)(C[n]cos(n theta)+D[n]sin(n theta))?
    I think we are on the same page, but just to be sure

    if you separate the equations you get

    r^2R''+rR'-\lambda^2R=0 and

    \Theta''+\lambda^2\Theta=0

    So

    R(r)=c_1r^\lambda+c_2r^{-\lambda}

    Since the solution must be bounded as r goes to infinity c_1=0

    This gives

    R(r)=c_2r^{-\lambda}

    Solving the theta equation gives

    \Theta(\theta)=d_1\sin(\lambda \theta)+d_2\cos(\lambda \theta)

    So

    T_\lambda(r,\theta)=r^{-\lambda}[d_1\sin(\lambda \theta)+d_2\cos(\lambda \theta)]

    Also note that \displaystyle \cos^2(\theta)=\frac{1}{2}+\frac{1}{2}\cos(2\theta  )

    Use this and orthogonality to get the solution.
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  5. #5
    LHS
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    Ah thanks, yes that's much clearer now!
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