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Separation of variables

  1. Nov 13, 2013 #1
    1. The problem statement, all variables and given/known data

    utt = uxx -(25/4)cos((5/2)x)
    ux(0,t) =1
    u(pi,t)= pi
    u(x,0)=x
    ut(x,0)=0

    2. Relevant equations

    u(x,t)=v(x) + w(x,t)

    3. The attempt at a solution

    This is what I did so far:

    u(x,t)=v(x) + w(x,t)
    u(x,0) = v(x) +w(x,0)

    when t is large:
    vxx - (25/4)cos((5/2)x) = 0
    vx = (5/2)sin((5/2)x)
    v(x) = -cos((5/2)x) +x

    when t is not large
    wtt = wxx
    w(x,t)= x-v(x) = cos((5/2)x)

    and I'm not sure what to do after this

    Maybe: X''+ λX = 0
    X'(0) = 1 and X(pi)= pi
     
  2. jcsd
  3. Nov 13, 2013 #2

    LCKurtz

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    What does ##t## being large have to do with anything?

    No. The point of your original substitution is to let ##v(x)## take care of the non-homogeneous terms. I will call ##f(x) = \frac {25}4\cos(\frac 5 2 x)## to save typing. What you want to do is substitute ##u(x,t) = v(x) + w(x,t)## into the equation ##u_{tt}=u_{xx}-f(x)## and its boundary conditions:

    ##u_{tt}=u_{xx}-f(x)## becomes ##w_{tt} = w_{xx}+v''(x) -f(x)## That will be satisfied if we take ##v''(x) = f(x)## and ##w_{tt} = w_{xx}##

    Now ##u_x(0,t) = 1## becomes ##w_x(0,t) + v'(0) = 1##. That will be satisfied if we take ##v'(0)=1## and ##w_x(0,t)=0##.

    ##u(\pi,t)=\pi## becomes ##w(\pi,t) +v(\pi)=1##. That will be satisfied if we take ##v(\pi)=\pi## and ##w(\pi,t) = 0##.

    Now with those conditions on ##v(x)## you should be able to solve for ##v(x)## and you have a homogeneous equation and boundary conditions in ##w##. Try to take it from there.
     
  4. Nov 14, 2013 #3
    Thank you for your help, I m almost done solving this problem and got:

    u(x,0)= Ʃ (cos((2n+1)/2)x)*Dn = x

    sigma from 0 to infinity and Dn is a constant.

    How do I find Dn?
     
  5. Nov 14, 2013 #4

    LCKurtz

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    That isn't correct. You need to show your work so I can see where you fell off the tracks.

    Remember, when you substitute in the initial condition you get$$
    x=u(x,0) = v(x) + w(x,0)$$so your initial condition for ##w(x,t)## is ##w(x,0)=x - v(x)##. What did you get for ##v(x)##?

    Also, did you remember to check whether the eigenvalue zero works?

    [Edit: Added] Your focus should be on solving the ##w(x,t)## system completely. That's where I need to see your work. Once you have figured that out, all you have to do to solve the ##u## system is write ##u(x,t) = v(x) + w(x,t)## because you know both ##v## and ##w##.
     
    Last edited: Nov 14, 2013
  6. Nov 14, 2013 #5
    Sorry that was supposed to be w(x,t) not u(x,t). So:

    Xn=C*cos((2n+1)/2)x
    Tn=D*cos((2n+1)/2)t)+(B*sin((2n+1)/2)t)

    w(x,t) = Ʃ Xn * Tn

    w(x,t) = Ʃcos((2n+1)/2)x)*((D*cos((2n+1)/2)t)+(B*sin((2n+1)/2)t))

    so

    w(x,0)=Ʃcos((2n+1)/2)x)*D

    and

    w(x,0) = cos(5/2)x
     
    Last edited: Nov 14, 2013
  7. Nov 14, 2013 #6

    LCKurtz

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    You need subscripts on the constants, as ##D_n## and ##B_n##. You didn't answer some of my questions in post #4. What did you get for ##v(x)##? Did you check for zero eigenvalues? (This matters as to whether there is a constant term). What does the initial condition for ##w(x,0)## become?

    [Edit: added] Also you need to show what your T(t) eigenvalue problem becomes, with boundary condition, and how you got ##T_n(t)##, because what you have isn't correct.
     
    Last edited: Nov 14, 2013
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