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Homework Help: PDE problems

  1. May 1, 2006 #1
    there's something about these PDE:s that I dont understand, cant find out how it really works. Here comes a problem that we can discuss.

    2 equal 0.2m think iron plates got the temperatures 100 and 0 degree C from the beginning. At the time t = 0 are these 2 plates laid next to eachother. Calculate the temperature where the plates touch after 10min (a_iron = 1.5*10^-5 m^2/s)

    the hints in the book says:

    du/dt - a d^2u/dx^2 = 0
    u(0,t) = u(0.4,t) = 0
    u(x,0) = 100 0<x<0.2
    u(x,0) = 0 0.2<x<0.4

    that sounds fair to me, but then they continue with, " Hopfully we reckognice the room-operator and know which eigenfunctions we have. If so we can directly try with a sinus serie and get the solution

    u(x,t) = 200/pi sum( (1-cos(k*pi/2))/k * exp( -ak^2pi^2*t/0.16) * sin(k*pi*x/0.4)"

    how do we get this?

    I mean, with the Stum Liouville operator we write it as A = -d^2/dt^2 so we get

    du/dt + aAu = 0
    and A*fi_k = lamda_k * fi_k
    where we assume that the solution looks like u(x,t)= sum(u_k(t)*fi_k(x))
    and that u_k(t) = (fi_k|s)/(fi_k|fi_k) * exp(-a*lamda_k*t)

    (|) is the scalar product and s is u(x,0).

    when I calculate fi_k I get fi_k = sin(alfa_k*x) where alfa_k = k*pi/0.4
    and lamda_k = alfa_k^2

    but how do I get u_k(t). with the methos I know it depends on u(x,0). And here I got 2 values for that. I probably could write u(x,0) with heaviside, but I have no ide on how to take the scalar product then, so I choose not to :)
     
  2. jcsd
  3. May 1, 2006 #2

    HallsofIvy

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    "Room operator"? I wonder if that isn't a mis-translation of something like "space-operator", i.e the derivative with respect to the space variable x?

    In any case, you could approach this by "separation of variables"- look for a function of the form u(x,t)= A(x)B(t). Then u_xx= A"B and u_t= AB' so the equation becomes AB'- aA"B= 0. Write that as AB'= aA"B and divide both sides by AB: B'/B= aA"/A.
    Now, what can k be? If k= 0, then A"= 0 which has general solution A(x)= Cx+ D. The only way that could be 0 at two different points is if C= 0 and D= 0 which gives A(x)= 0 for all x (the "trivial" solution).
    If k<0, write k= u2 with u a positive number. The equation is aA"= -u2x and the general solution is A(x)= Ceux/a+ De-ux/a. Now A(x)= C+ D= 0 and A(0.4)= Ce0.4u/a+ De-0.4u/a= 0. From C+ D= 0, we get D= -C. Putting that into Ce0.4u/a+ De-0.4u/a= 0 and factoring out the C gives C(e0.4u/a- e-0.4/a)= 0. But e0.4u/a is greater than 1 while e-0.4/a is less than 1 so the quantity in parentheses can't be 0: We must haved C= 0 and then D= 0. Again, we have only the "trivial" solution A(x)= 0.

    If k< 0, write k= -u2 so the equation is aA"= -u2.
    The general solution to that is A(x)= C cos(ux/a)+ D sin(ux/a). A(0)= C= 0 and A(0.4)= D sin(.4u/a)= 0. In order that D NOT be 0 (so we get a non-trivial solution) we must have .4u/a equal to a multiple of [itex]\pi[/itex]: that is, [itex]u= \frac{na\pi}{0.4}[/itex].

    We say that [itex]k= \left(\frac{na\pi}{0.4}\right)^2[/itex] is an eigenvalue of the problem and [itex]sin(\frac{na\pix}{0.4}t[/itex] is an eigenvector or eigenfunction of the problem.

    Once we know that k must be of the form [itex]\left(\frac{na\pi}{0.4}\right)^2[/itex], the equation for B(t) must be [tex]B'(t)= \left(\frac{na\pi}{0.4}\right)^2B[/itex] and has solutions
    [tex]B(t)= e^{\left(\frac{na\pi}{0.4}\right)^2t}[/itex] .
    Putting those together,
    [tex]u(x,t)= Ce^{\left(\frac{na\pi}{0.4}\right)^2t}sin(\frac{na\pix}{0.4}[/tex] is a solution and the general solution is a sum of things like that.

    What they mean when they say "Hopfully we recognise the space-operator and know which eigenfunctions we have. If so we can directly try with a sine series" (I have corrected the English. That's not a criticism of you- you should see my (put the language of your choice here)!) is that when we see y" we immediately think of the eigenvector equation y"= ky and don't have to repeat what I just did above- we know what that the eigenvalues are what I gave and the eigenvector is a sine function.

    No, exponentials are one-to-one functions and cannot be 0 at two different values of x. As I showed above, lambda_k must be negative and the solutions are sine and cosine functions.
     
  4. May 2, 2006 #3
    Hmm, You probably mean k>0 when you write k= u^2. (or maby you mean k>0 at the third case, but I figured this was more likely)

    Where do you get aA"= -u^2x from then? from my point of view I would say it should be aA"/A = u^2, where do you get the x and - from? And why does the /A dissapear? I would say I get a D.E that looks like aA'' - Au^2=0.

    And I get kind of nervous when you dont like the method I tought were correct, with Sturm Liouville and calculating scalar products. Is that totaly wrong? Separation of variables were last weeks method, this week we use SL :) I mean with my method I got sin(k*pi*x/0.4) and that looks kind of close to what you got.
     
  5. May 3, 2006 #4

    HallsofIvy

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    I'm more used to A"+ kA= 0 than A"= kA so, yes, I'm sure I got + and - confused.

    That was a typo. aA"/A= u^2 is the same as aA"= u^2A. (or aA"= -u^2A in the case that k is negative.

    Sturm-Liouville problems include the boundary conditions- you didn't do that here at the point I was criticizing. You were writing the solutions in terms of exponentials and my point was that exponentials can't make the function equal to 0 at the endpoints. Other than the fact that you are using k where I was using n (and the "minor" detail that I left out the x!)
    your solution sin(k*pi*x/0.4) is exactly what I had.
     
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