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Heat conductor problem

  1. Apr 9, 2012 #1
    1. The problem statement, all variables and given/known data

    2012-04-09010508am.png

    2. Relevant equations

    All explained in my solution attempt

    3. The attempt at a solution

    Let me know what's going wrong here. I don't feel like I should be getting something as small as t = .05 seconds for an answer.

    2012-04-09011324am.png
     
  2. jcsd
  3. Apr 9, 2012 #2
    Or maybe that's not weird, given the massive amplitude of 100 in the initial condition?
     
  4. Apr 9, 2012 #3
    What do you geniuses think?
     
  5. Apr 10, 2012 #4

    Dick

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    I think you made that more complicated than it needs to be. Your x dependence is already in the form of a sin. There's no need for the fourier series part. Just do the separation of variables. And they gave you α^2=1.158, not α=1.158. You don't want to square it again.
     
  6. Apr 10, 2012 #5
    My bad on the σ2 thing.

    So, from the fact that that G(t)=G(0)e2π2k2/L2sin(kπ/L) where k is any integer, can I finish this problem? I don't see how I can since I don't have a value for G(0) and since there are infinitely many solutions. Our course notes make the G(0) absorb into a constant in an infinite series (see below) in one example.

    Untitled.gif








    Thoughts?
     
  7. Apr 10, 2012 #6
    Additional problem:

    (My answers are in bold.)

    Consider the heat equation utt=uxx, 0 < x < 1 with boundary conditions u(0,t)=u(1,t)=1 and initial condition u(x,0)=f(x).

    (a) Using physical intuitions only, write down the steady state temperature profile uss(x). Explain your answer.

    The "steady state temperature profile" is the one in which temperature does not depend on time. Thus it remains constant at each point in the conductor for all t. But since the ends remain a constant 1, it only makes physical sense that u(x,t)=1 for all t≥0 and x ε (0, 1). Thus uss(x)=1.


    (b) Using the transformation v=u-uss, show that v satisfies the same heat equation with a different initial temperature distribution.

    We have v(x,t)=u(x,t)-1; hence v(0,t)=v(1,t)=0, v(x,0)=0 and obviously vt=vxx.

    (c) Solve for v(x,t) using separation of variables. What is the physical meaning of v?

    Not sure how to do this one. Any ideas?
     
  8. Apr 10, 2012 #7

    Ray Vickson

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    What is stopping you from setting t = 0 in the formula for u(x,t) and finding G(0) by knowing that u(x,0) = 100*sin(πx)?

    RGV
     
  9. Apr 10, 2012 #8
    I see. My general solution will still be an infinite sum, though. Right?
     
  10. Apr 10, 2012 #9

    Ray Vickson

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    What is the Fourier series of the function f(x) = sin(πx), 0 < x < 1?

    RGV
     
  11. Apr 10, 2012 #10


    Let me know if my revised answer is correct.

    screen-capture-6-5.png
     
  12. Apr 10, 2012 #11

    Dick

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    Separation of variables is an 'ansatz'. Look it up, it's a handy term. You don't have to care one bit about the general solution if you can guess the form of the solution and verify it.
     
  13. Apr 10, 2012 #12
    So my u(x,t)=100e22tsin(∏x) checks out?

     
  14. Apr 10, 2012 #13

    Dick

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    You can check that it satisfies the PDE and the boundary conditions, can't you? Just plug it in. That's the "verified later" part.
     
    Last edited: Apr 11, 2012
  15. Apr 11, 2012 #14
    Sweet. What about the additional question I posted in this thread? Any thoughts or concerns?
     
  16. Apr 11, 2012 #15

    Dick

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    For c), now I think they want to do the Fourier series of f(x) and write down a general solution, like your were doing for the original problem.
     
  17. Apr 11, 2012 #16
    Got it. Am I on the right track with parts a) and b)?
     
  18. Apr 11, 2012 #17

    Dick

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    Seems fine to me.
     
  19. Apr 11, 2012 #18
    Bottom line: I get the trivial solution u(x,t)=0 for all 0≤x≤1 and t≥0. Right?
     
  20. Apr 11, 2012 #19

    Dick

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    Well, no. You will only get the trivial solution if f(x)=1. Otherwise, you can expand f(x)-1 in a sine series and treat each term like you did with the first problem.
     
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