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Math Help - I don't know how to start this one

  1. #1
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    I don't know how to start this one

    Find the solutions to \bigtriangledown^2 u = 0 in two and three dimensions subject to the following boundary conditions

    (a) u(x) = u_0 for ||x|| = a and u = u_1 for ||x|| = b

    (b) u(x) = u_0 for ||x|| = a and \bigtriangledown u \cdot n = k for ||x|| = b

    Note that the boundary conditions do not depend on \theta or \phi (in 3D)
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  2. #2
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    Since the boundary conditions are independent of \theta or \theta and \phi, I might suggest switching the PDEs into polar coordinates and looking for solution in terms of the radius only.
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  3. #3
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    Quote Originally Posted by Jimmy_W View Post
    Find the solutions to \bigtriangledown^2 u = 0 in two and three dimensions subject to the following boundary conditions

    (a) u(x) = u_0 for ||x|| = a and u = u_1 for ||x|| = b

    (b) u(x) = u_0 for ||x|| = a and \bigtriangledown u \cdot n = k for ||x|| = b

    Note that the boundary conditions do not depend on \theta or \phi (in 3D)
    I'm having trouble understanding exactly what the initial conditions are saying here.

    If I had to solve this PDE with conditions (another question I have):

    u(0,y) = 0

    u(1,y) = 0

    u(x,0) = 1

    u(x,1) = 1

    I would do it as follows:

    Assume u(x,y) = X(x)Y(x):

    \frac{\partial^2}{\partial x^2} (X(x)Y(y)) + \frac{\partial^2}{\partial y^2}(X(x)Y(y)) = 0

    X''(x)Y(y) + X(x) Y''(y) = 0

    - \frac{X''(x)}{X(x)} = \frac{Y''(y)}{Y(y)} = k^2

    So

    X''(x) = -k^2 X(x) and Y''(y) = k^2 Y(y)

    Solving these,

    X(x) = C_1 \cos(kx) + C_2 \sin(kx) and Y(y) = C_3 e^{ky} + C_4 e^{-ky}

    and u(x,y) = X(x) Y(y) so

    u(x,y) = C_1 \cos(kx) + C_2 \sin(kx)) (C_3 e^{ky} + C_4 e^{-ky}) .........(Eqn 1)

    Then applying all conditions and skipping a few steps the solution is

    \displaystyle u(x,y) = \sum_{n=1}^{\infty} 2 \frac{1 - (-1)^n}{n \pi (e^{n \pi} - e^{-n \pi})} \sin (n \pi x) (e^{n \pi y} - e^{-n \pi y})


    But I'm getting confused with the initial conditions here, and how to go from Eqn (1) onwards with the intitial conditions onwards i.e.

    (a) u(x) = u_0 for ||x|| = a and u = u_1 for ||x|| = b

    (b) u(x) = u_0 for ||x|| = a and \bigtriangledown u \cdot n = k for ||x|| = b


    For the 3d case, I have

    \frac{\partial u}{\partial x^2} + \frac{\partial u}{\partial y^2} + \frac{\partial u}{\partial z^2} = 0

    Assume u(x,y,z) = X(x)Y(y)Z(z)

    Then

    \frac{d^2 X(x)}{dx^2} Y(y)Z(z) + X(x) \frac{d^2 Y(y)}{dy^2} Z(z) + X(x)Y(x) \frac{d^2 Z(z)}{dz^2} = 0

    Dividing through

    \displaystyle \frac{1}{X(x)} \frac{d^2 X(x)}{dx^2} + \frac{1}{Y(y)} \frac{d^2 Y(y)}{dy^2} + \frac{1}{Z(z)} + \frac{d^2 Z(z)}{dz^2}

    Which results in 3 ODE's

    \frac{d^2 X(x)}{dx^2} - aX(x) = 0

    \frac{d^2 Y(x)}{dy^2} - bY(y) = 0

    \frac{d^2 Z(z)}{dz^2} + (a+b)Z(z) = 0

    which have solutions

    X(x) = C_1 e^{\pm \sqrt{a}x}

    Y(x) = C_2 e^{\pm \sqrt{a}y}

    Z(z) = C_3 e^{\pm i \sqrt{a+b}z}

    So, due to superposition,

    u(x,y,z) = C e^{\pm \sqrt{a}x} e^{\pm \sqrt{a}y} e^{\pm i \sqrt{a+b}z}

    which is all I have...I don't understand the form of the initial conditions.
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  4. #4
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    Quote Originally Posted by Danny View Post
    Since the boundary conditions are independent of \theta or \theta and \phi, I might suggest switching the PDEs into polar coordinates and looking for solution in terms of the radius only.
    Here is my attempt for general solution in polar co-ordinates in 2d..

    \frac{\partial^2 u}{\partial r^2} + \frac{1}{r} \frac{\partial u}{\partial r} + \frac{1}{r^2} \frac{\partial^2 u}{\partial \theta^2} = 0

    Assume u(r, \theta) = R(r)\Theta (\theta)

    Skipping some steps, I get

    R(r) = C_1 r^k + C_2 r^{-k}

    and

    \Theta(\theta) = C_3 \cos(k \theta) + C_4 \sin (k \theta)

    so

    u(r, \theta) = (C_1 r^k + C_2 r^{-k})(C_3 \cos(k \theta) + C_4 \sin (k \theta))

    I may be able to continue, but again, I'm having trouble interpreting the initial conditions (a) and (b).
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