Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Need help with a couple of BVP's.

  1. Oct 24, 2006 #1
    I've got a couple of BVP's I'm working on that are giving me some problems. First, I'm asked to solve the 1-D Heat Equation:

    [tex]k \frac{\partial^2 u}{\partial x^2} = \frac{\partial u}{\partial t}[/tex]

    in a thin rod of length L subject to the following:

    BC1) u(0, t) = 0
    BC2) u(L, t) = 0
    IC) 2 different problems. For the first, [itex]u(x,0) = 1 - 2 cos(\frac{9 \pi x}{L})[/itex]. For the second, [itex]u(x,0) = 3 cos(\frac{9 \pi x}{2 L})[/itex].

    I can separate this problem into:

    [tex]\frac{X''}{X} = \frac{T'}{k T} = - \lambda[/tex]

    easily enough and apply the 2 BC's to get:

    [tex]u_{n} (x,t) = A_{n} \sin(\frac{n \pi x}{L}) \exp(\frac{-k n^2 \pi^2 t}{L}) , n=1,2,3,...[/tex]

    My problem is with applying the Initial condition and solving for u(x,t):

    [tex]u(x,t) = \sum \limits_{n=1} ^ {\infty} A_{n} \sin(\frac{n \pi x}{L}) = IC[/tex]

    My book shows me how to do a half-range expansion here, but according to my professor, I shouldn't need to. He even sent us an e-mail changing one of the IC's "so it would be an eigenfunction." How do I get there from here? Is there some easy trig identity to convert those IC's to a sin function I'm just missing?


    My 2nd question involves steady state heat conduction in a 2-D square plate of side L:

    [tex]\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0[/tex]

    Subject To:

    BC1) [itex]\frac{\partial u}{\partial x} = 0[/itex] at x = 0
    BC2) [itex]\frac{\partial u}{\partial x} = 0[/itex] at x = L
    BC3) [itex]\frac{\partial u}{\partial y} = 0[/itex] at y = 0
    BC4) u(x,L) = 5

    I separate the Laplacian into:

    [tex]\frac{X''}{X} = \frac{-Y'}{Y} = - \lambda[/tex]

    and solve using the first 3 BC's, I get [itex]X = c_{1}[/itex] and [itex]Y = c_{4}[/itex] for [itex]\lambda = 0[/itex] and

    [tex]
    X(x) = c_{2} \sin(\frac{n \pi x}{L}) [/tex]

    [tex]Y(y) = c_{4} \cosh(\frac{n \pi y}{L})[/tex]

    for [itex]\lambda > 0[/itex].

    Using superposition, I end up with:

    [tex]u(x,y) = A_{0} + \sum \limits_{n=1} ^ {\infty} A_{n} \sin(\frac{n \pi x}{L}) \cosh(\frac{n \pi y}{L})[/tex]

    When I apply the final BC, u(x,L) = 5, I end up with a solution of u(x,y) = 5. I guess I'm just looking for some sanity checking here. I can certainly see how u = 5 works here, but is it the only solution?

    Thanks in advance,

    Grogs
     
    Last edited: Oct 24, 2006
  2. jcsd
  3. Oct 24, 2006 #2

    HallsofIvy

    User Avatar
    Science Advisor

    (To get the sigma to show use \sum with a small "s".)
    For your first problem, there is a difficulty with your initial conditions: you don't have enough! Since the equation is second order in t, you need two initial conditions. Perhaps you were also told that [itex]u_t(x,0)= 0[/itex]? That is, that the thin rod was deformed to the given initial position and then released "from rest".
    The "complication" is that, since u= 0 at both endpoints, you have written u(x,t) as a sine series in x which is always an odd function by your initial conditions are even functions. That isn't really a problem since your interval is from 0 to L but notice that, according to the first initial condition, u(0, 0)= 1- 2cos(0)= -1 and for the second initial condition, u(0,0)= 3cos(0)= 3, while, according to the boundary conditions, u(0,0)= u(0,t)= 0! You necessarily have a discontinuity at the boundary points.

    Again, you can't expect the solution to be continues at y= L because the initial conditions aren't. Yes u(x,y)= 5 is the only solution.
     
  4. Oct 24, 2006 #3
    Halls, thanks for the reply.

    I'm not sure where you're getting that. This is the heat equation (u = temperature,) not the wave equation. It's 1st order in t -> 1 IC required. The 2 IC's I posted are for 2 different problems where everything is the same, save for the IC's.

    Hmm, interesting. I see where you're coming from with the cosines. There's a mathematical discontinuity between 2 cos(0) and 0 at x = 0. Looking at that leads me to believe there's no way this problem can be solved in the same way as an IC of say, [itex]3 \sin(\frac{5 \pi x}{L}[/itex]. So would using the half-range expression:

    [tex]A_{n} = \frac{2}{L} \int \limits_{0} ^ {L} sin(\frac{n \pi x}{L}) f(x) dx[/tex]

    where f(x) is the IC be the only way to find u(x,y) in this case?
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook