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Partial DIfferential Equations

  1. Oct 29, 2005 #1
    Solve using separation of variables and find particular solution
    [tex] \frac{\partial^2 u}{\partial t^2} - \frac{\partial^2 u}{\partial x^2} - u = 0 \ for\ 0 <x<1, t >0 [/tex]
    [tex] \frac{\partial u}{\partial t} (x,0) = 0 [/tex]
    [tex] 0,t) = u(1,t) = 0 [/tex]
    to assume u(x,t) = X(x) T(t)
    then [tex] \frac{X''}{X} = \frac{T'' - T}{T} = \lambda [/tex]
    solving for X(x) yields [tex] X(x) = C_{1} \cos{\sqrt{\lambda} x} + C_{2} \sin{\sqrt{\lambda} x} [/tex]
    and i get C1 = 0 and C2 assumed to be 1 and [itex] \lambda = \sqrt{n \pi} [/itex]
    now for T(t),
    [tex] T'' - (1 + n^2 pi^2) T = 0 [/tex]
    now here is where i am stuck...
    is the answer just the
    [tex] C_{1} \cos{\sqrt{1 + n^2 \pi^2} t}+ C_{2} \sin{\sqrt{1 + n^2 \pi^2} t} [/tex]
    i'm not quite sure what to do after this
    the answer in my book is [tex] \cos{\sqrt{n^2 \pi^2 -1}} t} \sin (n \pi x) [/tex]
    i got the x part right... but what about the t part
     
    Last edited: Oct 29, 2005
  2. jcsd
  3. Oct 29, 2005 #2

    Physics Monkey

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    Hi stunner5000pt,

    Sorry, I got all confused there, and I read your differential equation backwards. Your problem is simply that [tex] \lambda < 0 [/tex] in order to obtain sines and cosines and meet the boundary condition for the x part. In reality, you should have [tex] - \lambda = n^2 \pi^2 [/tex] and not [tex] \lambda = n^2 \pi^2 [/tex].

    Once you have solved the t part, apply the boundary condition you haven't used yet.
     
    Last edited: Oct 29, 2005
  4. Oct 29, 2005 #3
    the DE is
    [tex] \frac{\partial^2 u}{\partial t^2} - \frac{\partial^2 u}{\partial x^2} - u = 0 [/tex]

    2 negative signs

    what is wrong with the equation?
     
  5. Oct 29, 2005 #4

    Physics Monkey

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    Yeah, my fault, I read it as
    [tex]
    \frac{\partial^2 u}{\partial x^2} - \frac{\partial^2 u}{\partial t^2} - u = 0,
    [/tex]
    and then realized my mistake as I was thinking about the solution.
     
  6. Oct 29, 2005 #5
    uh.. the t part was the part at which i was stuck.

    the condition given is [tex] \frac{\partial u}{\partial t} (x,0) = 0 [/tex]
    im not quite sure howto relate this to what i have already...
     
  7. Oct 29, 2005 #6

    Physics Monkey

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    First, as I said, your general solution to the t part in your first post is incorrect, it should be
    [tex]
    T(t) = C_1 \sin{(\sqrt{n^2 \pi^2 -1} \,t )} + C_2 \cos{(\sqrt{n^2 \pi^2 - 1}\, t)}
    [/tex]
    Now, regarding the boundary condition, what do you mean you don't know how to relate it to what you have? Just take the derivative of [tex] u(x,t) = X(x)T(t) [/tex] with respect to t and set it equal to zero for t=0.
     
  8. Oct 29, 2005 #7
    why should the generla solution for T be that?
     
  9. Oct 29, 2005 #8

    Physics Monkey

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    Read my post 2 above, and try to figure it out.
     
  10. Oct 30, 2005 #9
    so why should [itex] - \lambda = n^2 \pi^2 [/itex]
    i m not sure about the 'conditions on the x part' to which you are referring. And wouldnt the solution become imaginary if that was the case?
     
  11. Oct 30, 2005 #10

    Physics Monkey

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    What is the solution to
    [tex]
    \frac{1}{X}\frac{d^2 X}{dx^2} = \lambda,
    [/tex]
    isn't it
    [tex]
    X = C_1 e^{\sqrt{\lambda} x} + C_2 e^{-\sqrt{\lambda} x}.
    [/tex]
    you get real exponentials and not real sines and cosines if [tex] \lambda > 0 [/tex]. If you try applying the boundary conditions with real exponentials then you find [tex] C_1 = C_2 = 0 [/tex]. Take it from here.
     
  12. Oct 30, 2005 #11
    i get it now lambda mustb e imaginary iof you are going to use Eler's identity and expand E into cos and sin. ANd thus carries onto the T(t) part where C1-0 and C2 =1 and the solution is what appears in the book

    thank you for your help! I appreciate it!
     
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