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Laplace Eq. Cylinder and 3D Heat Equation

  1. Oct 20, 2008 #1
    Sorry, it does not seem that Latex is not compiling my code right so I will try my best to be clear.

    1. The problem statement, all variables and given/known data

    The curved surface of a cylinder of radius a is grounded while the end caps at z = ± L/2 are
    maintained at opposite potentials ψ (r,θ, ± L/2)= ± V(r,θ).

    Suppose that a simple solid brick with uniform initial temperature
    is immersed at time t = 0 in a heat bath. The temperature ψ (r,t) within the material
    satisfies

    ∂ψ = ∇^2 ψ
    k∂t

    with ψ(r,0)=T1 and ψ(at surface,t)=T2

    I really just want to clear up some interpretations.

    2. Relevant equations

    Laplaican for Cylindrical and Rectangular Coordinates.

    3. The attempt at a solution

    I want to assume that grounded means the potential at the curved surface is zero, but I cannot find it in my notes or book or even online. After that it's just putting it in the equation, I have to solve both exterior and interior cases, nothing new.

    The second question is really confusing me in choosing the appropriate separation constants. Now we solved the 3-D Laplace equation in rectangular coordinates. The separation constants I choose were

    a^2 – b^2 – c^2 = 0 this implies that a^2 = b^2 + c^2

    this gives you a sin/cos solutions in two directions and a sinh/cosh in one solution one direction. I choose the sinh/cosh solution for the direction w/ the non-homogeneous boundary, which you repeat six times to get a complete solution. Now for the question given, I want to choose

    – a^2 – b^2 – c^2 – d= 0 this implies that

    d= – a^2 – b^2 – c^2

    so, all my solutions are in the sin/cos form except T??

    We have T' + k (d) T = 0, since a,b,c are > 0 (other cases are trivial solutions), then d < 0 and k > 0, so I can have an exp(), solution.

    I was really wanting a way to work in a sinh/cosh solution, so that I could use what I got from the previous homework, but I guess it is just a little modification. I guess that makes since b/c it is some form of a wave equation. It seems non-intuitive to me.
     
    Last edited: Oct 21, 2008
  2. jcsd
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