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Homework Help: Heat equation/energy

  1. Dec 29, 2009 #1
    Suppose that u satisfies the heat equation [tex]u_t = u_x_x[/tex] on the unit interval [0,1] subject
    to Neumann conditions [tex]u_x(0,t) = u_x(1,t) = 0[/tex]. Show that the energy
    [tex]E(t) =\int_{0}^{1}u(x,t)^2dx[/tex] is decreasing and convex. Show also that u(x,t) approaches a constant as
    [tex]t\rightarrow\infty[/tex]
    .

    attempt at solution:


    Separation of variables (!?): X=Acoskx+Bsinkx.
    Sub in Neumann conditions: X=ACosnπx
    ...T=exp(-n^2π^2t^2)
    Combine linearly: [tex]u=E_0+ \sum_{n=1}^\infty[/tex] E_n e^(-n^2π^2t^2) Cosnπx
    Therefore E(t)=[tex]\int_{0}^{1}(E_0+ \sum_{n=1}^\infty E_n exp^(-n^2\pi^2t^2) Cosn\pi x)^2dx[/tex]
    Therefore E(t)=[tex]\int_{0}^{1}(E_0)^2+(\sum_{n=1}^\infty E_n exp^(-n^2\pi^2t^2) Cosn\pi x)^2dx[/tex]
    Therefore [tex]E(t)=F_0+\frac{1}{2}\sum_{n=1}^\infty E_n exp^(-2n^2\pi^2t^2) [/tex]
    ...
    Apparently this is wrong. Using separation of variables and Fourier series is not OK for all initial data. Can you come up with an argument that does not involve solving the problem explicitly? I was told as a hint, to start with the formula for energy and differentiate it directly with respect to t and then try to show that the derivative is negative (integration by parts should help here)....What!!!? Perhaps EXPLAIN this suggestion. Thanks
     
    Last edited: Dec 29, 2009
  2. jcsd
  3. Dec 29, 2009 #2

    Dick

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    It means what it says. Find dE/dt by differentiating inside the integral. Now use the heat equation to swap the time derivative of u for a double x derivative of u. Integrate by parts... Just try it.
     
  4. Dec 29, 2009 #3
    Thats very helpful. Thank you. However, I am left with

    [tex]\int_{0}^{1}u(x,y)\frac{\delta^2u}{\delta x^2}dx[/tex]

    How on earth do you solve this !!?
     
    Last edited: Dec 29, 2009
  5. Dec 29, 2009 #4

    Dick

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    d(u_x)=u_xx*dx. Pick your 'parts' to be u_x(x,t) and u(x,t).
     
  6. Dec 29, 2009 #5
    Hi Dick. I'm really appreciative of your time. You have been a great help so far.

    I was under the impression that integration by parts was only used when a direct substituion for the parts (u, dv) was used ... taken directly from the given equation.

    Is what you have just suggested now not int by substitution?
     
  7. Dec 29, 2009 #6

    Dick

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    Well, no, it's parts. u*dv=d(u*v)-v*du with u=u(x,t) and v=u_x(x,t). You are going to get a boundary term u*u_x evaluated between 0 and 1.
     
  8. Dec 29, 2009 #7
    Slow and steady wins the race! Again, very informative.

    I am left with [tex] [u\frac{\delta u}{\delta x} ]_{0}^{1} - \int_{0}^{1} \frac{\delta u}{\delta x}du [/tex].

    As you rightly stated I am left with a boundary term u*u_x evaluated between 0 and 1. Can anything be done with the [tex] \int_{0}^{1} \frac{\delta u}{\delta x}du [/tex] !?

    or have i done it wqrong and last term should be [tex] \int_{0}^{1} [\frac{\delta u}{\delta x}]^2dx [/tex] ?

    Either way what is the name of this section of maths we are dealing with so i can look it up
     
    Last edited: Dec 29, 2009
  9. Dec 29, 2009 #8

    Dick

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    They are both right. Like I said it's integration by parts. The final integral is u_x*du=u_x*u_x*dx. And there's not much to do with the integral of the square except notice that it's nonnegative.
     
    Last edited: Dec 29, 2009
  10. Dec 31, 2009 #9
    So [tex] E(t)= [u\frac{\delta u}{\delta x} ]_{0}^{1} - \int_{0}^{1} \frac{\delta u}{\delta x}du [/tex] is the final answer? Does this prove energy convex and decreasing? How show u approaches const. as [tex]t \rightarrow \infty [/tex] ?
     
  11. Dec 31, 2009 #10

    Dick

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    i) That's not E(t), it's dE(t)/dt. ii) Evaluate the boundary term. iii) Tell me why the integral is positive. Now tell me something about E(t).
     
  12. Dec 31, 2009 #11
    Ok, so when the boundary conditions are applied [tex] dE(t)/dt= - \int \int_{0}^{1} \frac{\delta u}{\delta x}du dt[/tex] Now what? Because rate of change of energy is minus this implies energy decreasing? What of u approaching const as [tex]t \rightarrow \infty[/tex] ?
     
    Last edited: Dec 31, 2009
  13. Dec 31, 2009 #12

    Dick

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    There's no dt in the integral. Go back and remember how you got here. You differentiated with respect to time first. That's why it is dE/dt. And yes, that's negative because the integral can be written as the integral of a square. What square?
     
  14. Dec 31, 2009 #13
    So [tex] E(t)= -2 \int \int_{0}^{1} [\frac{\delta u}{\delta x}]^2dx dt[/tex] Now what!?
     
  15. Dec 31, 2009 #14

    Dick

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    Write it as dE/dt equals the negative of the dx integral, you dropped the minus sign. Forget the dt integral. Doesn't that tell you E is decreasing? Is E bounded below?
     
  16. Dec 31, 2009 #15
    Ok so [tex] dE(t)/dt= -2 \int_{0}^{1} [\frac{\delta u}{\delta x}]^2 dx [/tex] proves energy convex and decreasing. How show u approaches const. as [tex]t \rightarrow \infty [/tex] ?

    (Thanks a million by the way)
     
  17. Dec 31, 2009 #16

    Dick

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    Well, it does prove E(t) is decreasing. It doesn't prove it's convex. To show it's convex you need E''(t)>=0. You might guess you'd do it in a similar way that you showed E'(t)<=0. But just knowing that E(t) is decreasing and bounded below (why is it bounded below?), tells you E(t) approaches a constant as t->infinity, right? What does that tell you about E'(t) as t->infinity?
     
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