1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Science Advisor

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

    hunt_mat

    User Avatar
    Homework Helper

    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

    User Avatar
    Staff Emeritus
    Science Advisor

    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].
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Partial differential equation
Loading...