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[PDE] Transforming Nonhomogeneous BCs into Homogeneous Ones

  1. Feb 19, 2013 #1
    So there's this problem in my text that's pretty challenging. I can't seem to work out the answer that is given in the back of the book, and then I found a solution manual online that contains yet another solution.

    The problem is a the heat equation as follows:

    PDE: [itex]u_{t} = α^2u_{xx}[/itex]

    BCs: [itex]u(0,t) = 1[/itex]
    [itex]u_x(1,t)+hu(1,t) = 1[/itex]

    IC: [itex]u(x,0) = \displaystyle\sin (πx)+x[/itex]

    Also, [itex]0<x<1[/itex].

    We assume that [itex]u(x,t) = S(x,t)+U(x,t)[/itex] and that [itex]S(x,t) = A(t)[1-x]+B(t)[x][/itex]. By substituting these into the BCs, I get [itex]A(t) = 1[/itex] and [itex]B(t) = \frac{2}{1+h}[/itex]. Now we have [itex]S(x,t) = 1-x+\frac{2x}{1+h} = 1+x\frac{1-h}{1+h}[/itex]. With the steady state solution in place, we construct our homogenous problem as follows:

    PDE: [itex]U_{t} = α^2U_{xx}[/itex]

    BCs: [itex]U(0,t) = 0[/itex]
    [itex]U_x(1,t)+hU(1,t) = 0[/itex]

    IC: [itex]U(x,0) = \displaystyle\sin (πx)+x\frac{2h}{1+h}-1[/itex]

    If I try to solve this one, it turns into an eigenvalue problem which isn't covered until the next section and the IC is a nightmare. Anyways, the book gives me the following answer:

    [itex]x+e^{-(πα)^2t}\displaystyle\sin (πx)[/itex]

    Moreover, the solution manual I found on the top of pg 16 looks as if it's solving an entirely different problem. Any help on this problem would be greatly appreciated.
     
  2. jcsd
  3. Feb 19, 2013 #2

    LCKurtz

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    It looks to me like all your work is correct as far as you went. The book's answer$$
    x+e^{-(πα)^2t}\displaystyle\sin (πx)$$doesn't solve the original problem's boundary conditions, so it isn't the answer to the problem. After all, you would expect the answer to depend on ##h##.
     
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