1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
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: 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.
     
  2. jcsd
  3. Mar 13, 2010 #2

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook