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Separation of Variables / Boundary Conditions

  1. Apr 3, 2007 #1
    1. The problem statement, all variables and given/known data

    The edges of a square sheet of thermally conducting material are at x=0, x=L, y= -L/2 and y=L/2

    The temperature of these edges are controlled to be:
    T = T0 at x = 0 and x = L
    T = T0 + T1sin(pi*x/L) at y = -L/2 and y = L/2

    where T0 and T1 are constants. The temperature obeys Laplace's equation grad² T (x,y) = 0

    (a) Find the general solution to the equation d²X(x) / dx² = 0
    (b) Use the method of separation of variables to find the solution to Laplace's equation that objeys the boundary conditions. You'll need to consider a superposition of two solutions: one with a separation constant equal to 0 and a second for which the separation constant is nonzero.

    2. Relevant equations



    3. The attempt at a solution

    Expanding the LaPlace's equation, I get my two ODEs:

    d²X / dx² - k²X = 0
    d²Y / dy² + k²Y = 0

    Because one of the boundary conditions has a sin(pi*x/L) term in, I've written the general solution for d²X / dx² - k²X = 0 as:
    X(x) = Asin(kx) + Bcos(kx) (think that's right...)

    However I'm having major problems with part (b)... my natural assumption would be that the entire general solution reads as:
    T(x,y) = [Asin(kx) + Bcos(kx)][Ce^ky + De^ky]

    However using that I can't zero enough terms given my boundary conditions to get anywhere. HELP!!!!
     
    Last edited: Apr 4, 2007
  2. jcsd
  3. Apr 4, 2007 #2

    HallsofIvy

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    The fact that the general solution to your equation for X is Acos(kx)+ Bsin(kx) is because of the differential equation, not the boundary condition!

    In order to satisfy the boundary conditions, k cannot be just any number. This is basically an "eigenvalue" problem with k as eigenvalue. For one thing, X(0)= T0 and X(L)= T0. What values of k can give a function that satisfies both of those?
    Once you know that, use those same values of k for the Y equation. Finally, the general solution is a SUM of X(x)Y(y) for all k.
     
  4. Apr 4, 2007 #3
    Hmm, i'm not sure I understand. If I ignore the boundary condition when specifying a general solution for X, I would tend to write it as Ae^kx + Be^kx as that's the general solution for an ODE of d²X / dx² - k²X = 0 where k>0. Likewise, the gen. sol. for Y would be what I currently have for X, Ccos(ky)+ Dsin(ky) (again, where k>0).

    The only reason I wrote it as X=Acos(kx)+ Bsin(kx) is because I have some notes with an example where it states the form of the solution is altered because boundary conditions inolve sin(2pi*x/a) so the solution for X should also included sines/cosines...
     
  5. Apr 4, 2007 #4
    Ever so sorry, I've missed something off in the transcription of the question: part (a) should read

    Find the general solution to the equation d²X(x) / dx² = 0

    Now i'm even more confused!!
     
  6. Apr 4, 2007 #5

    HallsofIvy

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    Ah, I misread your problem and reversed X and Y.

    If your equation is X"- k2X= 0, then the general solution is of the form X= C1ekx+ C2e-kx (for real k) no matter what the boundary condition is! Similarly, if the equation is Y"+ k2Y= 0 then the general solution is of the form Y(y)= D1 cos(ky)+ D2 sin(ky) (again for real k) no matter what the boundary condition is.

    However, in separating the X and Y in the partial differential equation, you get X"/X+ Y"/Y= 0 and, at that point, you have a choice as to which to make equal to k2 and which -k2. That can be guided by looking at the boundary conditions.

    (But that choice only simplifies the problem- doing it the other way will be harder but give an equivalent solution. If there are no sine, cosine, or exponentials in the boundary conditions, you can take your choice and get two very different sums that give exactly the same values!)

    But I have no idea what is meant by "You'll need to consider a superposition of two solutions: one with a separation constant equal to 0 and a second for which the separation constant is nonzero."

    If you let T(x,y)= X(x)Y(y), then, just as you did, you can reduce that to (1/X)X"+ (1/Y)Y"= 0. If one of those has "separation constant" k, if, for example (1/X)X"= k2, then the other must have separation constant -k2 just as you said. You CAN'T have one zero and the other non-zero!
     
  7. Apr 4, 2007 #6
    ^ Exactly, that's what I'm thinking. I've consulted a textbook and it too has the statement "...the only essential requirement being that k has the same value in both parts of the solution, ie, the part depending on x and the part depending on y". So that would indeed indicate that one can't be zero and the other non-zero...?

    However, what confuses me more right now is that part (a) states...
    "Find the general solution to the equation d²X(x) / dx² = 0 "

    There's no mention of a constant in there (implying k is zero I guess), so surely the general solution in that case is just Ax + B. Yikes, this question's killing me!
     
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