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Homework Help: PDE-Heat Equation with weird boundary conditions help!

  1. Mar 13, 2010 #1
    1. The problem statement, all variables and given/known data
    Consider the Heat Equation: du/dt=k(d2u/dx2), where d is a partial and d2 is the second partial. The B.C.'s are u_x(0,t)=u(0,t) and u_x(L,t)=u(L,t), where u_x is the partial of u with respect to x. The I.C is u(x,0)=f(x)

    Now, consider the Boundary Value Problem X''(x)=-lambda*X(x), note the negative sign, with B.C's X'(0)=X(0) and X'(L)=X(L), where L is the length of a 1D rod(at the very end of the rod).

    Find the eigenvalues and eigenfunctions for lambda>0, lambda=0, and lambda<0

    3. The attempt at a solution

    lambda=0 was an easy one, lambda=0, gave an eigenfunction X=0.
    lambda<0, wasn't too bad either. Since lambda < 0 I get X=Acosh(sqrt(-lambda)*x)+Bsinh(sqrt(-lambda)*x), where A and B are constants. Also, note that lambda is negative, thus the square root will not give any complex numbers. Using the first initial condition X'(0)=X(0), I get that A=sqrt(-lambda)*B and plugging back in again, using the second BC, I get that lambda=-1 and thus the eigenfunction is X=Bcosh(x)+Bsinh(x).

    lambda>0 is the one I'm having trouble with. If do what I did for lambda<0, and use X=Acos(sqrt(lambda)*x)+Bsin(sqrt(lambda)*x), I get that lambda is -1, again, however, lambda is suppose to be greater than 0, and the eigenfunction is complex. Any help would be appreciated. Thanks.
     
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  3. Mar 13, 2010 #2

    LCKurtz

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    For X'' + λX = 0, take the case where λ = μ2> 0, so your equation looks like

    X'' + μ2X = 0

    Instead of using the pair {sin(μx),cos(μx)} try the pair {sin(μx),sin(μ(L-x))}, so

    X(x) = Asin(μx) + Bsin(μ(L-x)).

    When you use your two boundary conditions you should get that A = B and you should find a sequence of values μn giving corresponding eigenvalues λn.
     
  4. Mar 13, 2010 #3
    Thanks for the reply, but should I use the sin(ux), sin(u(L-x))? isn't the solution to this question cos and sin?
     
  5. Mar 13, 2010 #4
    Sorry, I meant why should I
     
  6. Mar 13, 2010 #5

    LCKurtz

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    Because it makes the work easier. sin(ux) and sin(u(L-x)) are linearly independent so they work just as well as sin(ux) and cos(ux). Try it.
     
  7. Mar 13, 2010 #6
    I did try it and I got stuck. If I say X=Asin(ux)+Bsin(u(L-x)) we get X'=Aucos(ux)-Bucos(u(L-x)) and then I get stuck here. I tried both of my BC and couldn't get it to work. If I have X'(0)=X(0), I get Bsin(uL)=Au-Bucos(uL) and I don't know how to simplify that. If I use X'(L)=X(L), I get Asin(uL)=Aucos(uL)-Bu. Again, I can't simplify this.
     
  8. Mar 14, 2010 #7

    LCKurtz

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    Looking at it again this morning, I see I had an arithmetic error. I think your {sine,cosine} pair works:

    X' + u2X = 0
    X(0)=X'(0), X(L) = X'(L)

    X = A cos(ux) + B sin(ux)
    X' = -Au sin(ux) + Bu cos(ux)

    X(0)=X'(0) becomes A = Bu and
    X(L) = X'(L) becomes Acos(uL) + Bsin(ul) = -Au sin(uL) + Bu cos(uL)

    Put A = Bu in the second equation:

    Bu cos(uL) + Bsin(ul) + Bu2sin(uL) - Bu cos(uL) = 0
    B (1 + u2)sin(uL) = 0

    This leads to [itex]u_n = \frac{n\pi}{L}[/itex] with corresponding eigenfunctions.
     
  9. Mar 14, 2010 #8
    I see what you did. Thanks a lot! I just ended up dividing both sides by B sin(uL) without thinking. Thanks. This leads me to a second question, for the lambda < 0, did I do that one right? I followed similar steps for that one. Thanks again. You're a real life saver.
     
  10. Mar 14, 2010 #9

    LCKurtz

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    Your original equation was X'' + λX = 0. The case where λ = u2 > 0 gives the sine-cosine eigenfunctions we have looked at.

    The case λ = - u2 < 0 gives rise to

    X(x) = A cosh(ux) + B sinh(ux)

    If you do the same steps as I did on the sine-cosine one I think you will find that only u = 1 giving λ = -1 works with eigenfunction φ1 = cosh(x) + sinh(x).

    I also think you need to look again at λ = 0; you should find that you can take φ0 = 1.
     
  11. Mar 14, 2010 #10
    How did you get X0=1? If lambda=0, then X=A+Bx, using X'(0)=X(0), you get A=B and so X=A+Ax, using X'(L)=X(L), you get A=A+A(L), since L>0 A=0 and so X0=0? Am I missing something? Thanks.
     
  12. Mar 14, 2010 #11

    LCKurtz

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    No, you aren't missing anything. I had another glitch. What happens is that λ = 0 is not an eigenvalue because it doesn't lead to a non-trivial solution. φ0(x) ≡ 0 is not an eigenfunction
     
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