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Separation of variables:

  1. Apr 5, 2012 #1
    1. The problem statement, all variables and given/known data
    Given the partial differential equation:

    2u/∂x2 = ∂2u/∂t2 , where x[0;L]

    Use separation of variables to find the solution that satisfies the boundary conditions:
    ∂u/∂x (x=0) = ∂u/∂x (x=L) = 0

    2. Relevant equations
    The separation of variables method.


    3. The attempt at a solution
    I think I have found a way to do the problem. There are just minor things that I want to clear up.
    So let's jump into it:

    Assuming a solution of the form u(x,y) = X(x)T(t)

    gives the equations:

    X'' = -k2X
    T'' = -k2T

    with the solutions:

    X = Acos(kx)+Bsin(kx)
    T=Ccos(kx)+Dsin(kx)

    and with the boundary conditions we must have that;

    X'(0)=X'(L)=0

    which gives:

    -Aksin(0)+Bkcos(0) = 0
    =>
    Bk=0
    which must imply that B=0
    From that we get:
    -Aksin(kL)=0
    =>
    kL = (n+½)∏
    =>
    kn = (n+½)∏/L

    So the general solution is:

    ƩAcos(knx)T

    where sum is from -∞ to ∞.

    Is this correct? Now my teacher has uploaded a paper with solutions and in his expression there is no A. Is this just because the A has been absorbed into the constants of T, or shouldn't it be there at all?

    Also, I find it kind of weird to be choosing the constant -k2. I do so because I've been told that, but why do you that? Also my teacher notes, that choosing k2 would instead yield a trivial solution? Can anyone explain to me why this is and why you don't just choose a trivial constant c?
     
  2. jcsd
  3. Apr 5, 2012 #2
    Looks good to me. Indeed, you can just rescale your Tk function to absorb Ak.

    Have you actually done the separation of variables step? If you do, you get into a situation where you have

    (something depending on only t) = (something depending on only x).

    As this should hold for any values of x and t, both sides of the equation need to be constant. This constant may be positive or negative, but in this case, you wouldn't be able to satisfy the boundary conditions with a positive constant.
     
  4. Apr 5, 2012 #3
    Oh right. So if you chose k^2 then you'd get the characteristic equation:
    a^2 - k^2 = 0
    =>
    a=+-k
    which would not be a solution involving sines. Is that just what he means?
     
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