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Solve Uxx-3Uxt-4Utt=0 (hyperbolic)

  1. Oct 4, 2004 #1
    solve Uxx-3Uxt-4Utt=0 (hyperbolic) help!!

    solve Uxx-3Uxt-4Utt=0 with u(x,0)=x^2 and Ut(x,0)=e^x

    I know that this is hyperbolic since D=(-1.5)^2+4 >0 so I have to transform the variables x and t linearly to obtain the wave equation of the form
    (Utt-c^2Uxx=0). The above equation is equivalent to:

    (d/dx - 1.5 d/dt)*(d/dx - 1.5 d/dt)u - 6.25 d^2u/dt^2 = 0

    let x=b
    let t=-1.5b + 2.5a
    Thus,
    Ub=Ux - (1.5) Ut
    Ua=2.5 Ut

    thus Ubb-Uaa=0. This is where I am stuck..

    I know the general solution is u(a,b)=f(a+b)+g(a-b)
    also the explicit solution is u(a,b)=(1/2)*[φ(a+b)+φ(a-b)]*(1/2c)*(integral
    ψ(s)ds from a-b to a+b).
    where u(a,0)=φ(a) and Ub(a,0)=ψ(a).

    The solution is (4/5)*[e^(x+t/4)-e^(x-t)]+x^2+(1/4)*t^2
    but how to obtain it?
     
    Last edited: Oct 4, 2004
  2. jcsd
  3. Oct 4, 2004 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    "I know the general solution is u(a,b)=f(a+b)+g(a-b)
    also the explicit solution is u(a,b)=(1/2)*[φ(a+b)+φ(a-b)]*(1/2c)*(integral
    ψ(s)ds from a-b to a+b).
    where u(a,0)=φ(a) and Ub(a,0)=ψ(a)."

    Excuse me? That the first time φ has appeared. What is φ(x)??
     
  4. Oct 4, 2004 #3
    u(a,0)=φ(a) and Ub(a,0)=ψ(a)

    These are the initial conditions that would satisfy the explicit solution, in terms of a and b. φ and ψ ar functions.

    Now what functions they are, that is where I need help, if I need them at all that is.
     
  5. Oct 5, 2004 #4
    Problem solved, thanks for taking the time to look at it
     
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