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Homework Help: Partial differential equation

  1. Jun 21, 2011 #1
    1. The problem statement, all variables and given/known data

    solve uxx = utt

    3. The attempt at a solution

    using the method of separation of variables I get

    X''(x) - λ*X(x) = 0
    T''(t) - λ*Τ(t) = 0

    For λ = 0 I get X(x) = Ax + B where A and B are constants
    T(t) = Dt + E where D and E are constants

    hence u(x,t) = (Ax + B)*(Dt + E)

    for λ>0 I get u(x,t) = [PLAIN]http://img690.imageshack.us/img690/8068/asdqu.gif [Broken] [Broken][/URL] where c1,c2,C1,C2 constants

    for λ = -b^2<0 I get (E*cos(bx) + F*sin(bx))*(G*cos(bt) + H*sin(bt)) where E,F,G,H constants

    hence the solution

    is from the superposition principle

    u(x,t) = (Ax + B)*(Dt + E) + [PLAIN]http://img690.imageshack.us/img690/8068/asdqu.gif [Broken] [Broken][/URL] + (E*cos(bx) + F*sin(bx))*(G*cos(bt) + H*sin(bt))


    please is this correct? i think its wrong, because im not sure about the last part

    also our professor has only the [PLAIN]http://img690.imageshack.us/img690/8068/asdqu.gif [Broken] [Broken][/URL] as a reply, but i think its wrong, because we have to check for all λ to find all the possible solutions
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Jun 21, 2011 #2
    λ is fixed. You either have λ > 0 or λ < 0, so you have either the solution
    with exp[√(λ) x] or the one with the sines and cosines. Apart from that, you are correct but for one (I assume) typo: it's not exp[√(λx)] but exp[√(λ) x].

    (PS: Of course you could also have λ = 0, but that appears seldom in reality.)
     
  4. Jun 21, 2011 #3

    phyzguy

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    Note that given any function f, f(x-t) or f(x+t) satisfies this equation.
     
  5. Jun 21, 2011 #4

    hunt_mat

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    Are you given any boundary conditions?
     
  6. Jun 22, 2011 #5
    thanks for your answers, no there are no boundary conditions

    there is another exercise with boundary conditions and I get a simple result..
     
  7. Jun 22, 2011 #6

    HallsofIvy

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    Without boundary or initial conditions, the most general possible solution is Af(x-t)+ Bf(x+t) where f is any twice differentiable function of a single variable, A and B constants, as phyzguy said.

    That can be written in the form Jncik gives initially by summing over all possible values of [itex]\lambda[/itex].
     
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