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PDE arbitrary function F(x+t)

  1. Oct 4, 2006 #1
    Solve [tex]u_{xx}-3u_{xt}-4u_{tt}=0[/tex] with initial conditions [tex]u(x,0)=x^2, u_t(x,0)=e^x[/tex].

    I got that u is an arbitrary function F(x+t), which makes no sense. I factored the operator into [tex](\partial/\partial x+\partial/\partial t)(\partial/\partial x-4\partial/\partial t)u=0[/tex], but I can't get anywhere.
     
  2. jcsd
  3. Oct 4, 2006 #2
    Say that

    [tex]
    (\partial/\partial x-4\partial/\partial t)u=g
    [/tex]

    Then your equation becomes,

    [tex]
    (\partial/\partial x+\partial/\partial t)g=0
    [/tex]

    Solve for g, then solve your first equation.
     
  4. Oct 4, 2006 #3
    Let [tex]x'=x-4t, t'=-4x-t[/tex]. At some point I have [tex]u_{x'}=f(\frac{3x'+5t'}{-17})/17[/tex]. Am I on the right track?
     
  5. Oct 4, 2006 #4

    HallsofIvy

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    Why does that make no sense? You have a pde that is of second order in both x and t but you give only initial conditions when t= 0. Without boundary conditions on x, you will not have a specific solution.
     
  6. Oct 4, 2006 #5
    Well I'm copying down exactly what's written in the textbook, Partial Differential Equations: An Introduction by W. A. Strauss, S2.2 Problem 9.
     
  7. Oct 4, 2006 #6

    arildno

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    And why should your textbook make no sense?
     
  8. Oct 4, 2006 #7
    Ok this second level questioning confuses me.
     
  9. Oct 4, 2006 #8
    Because there exists a unique solution at the back of the book:

    [tex]\frac{4}{5}(e^{x+t/4}-e^{x-t})+x^2+\frac{t^2}{4}[/tex].
     
    Last edited: Oct 4, 2006
  10. Oct 4, 2006 #9
    [tex]
    \left(\partial_{x}-a^{-1}\partial_{t}\right)\left(\partial_{x}+\partial_{t}\right)u\left(x,t\right)=0
    [/tex]

    Hence the solution is the sum of two arbitrary functions with arguments [itex]x-t[/itex] and [itex]x+at[/itex],

    [tex]
    u\left(x,t\right)=F\left(x-t\right)+G\left(x+at\right)
    [/tex]

    We have the conditions

    [tex]
    \begin{align*}
    u\left(x,0\right)& =x^{2}\\
    u_{t}\left(x,0\right)& =e^{x}
    \end{align*}
    [/tex]

    Using [itex]u\left(x,0\right)[/itex]

    [tex]
    \begin{align*}
    x^{2}&=F\left(x\right)+G\left(x\right) \\
    G\left(x\right)&=x^{2}-F\left(x\right)
    \end{align*}
    [/tex]

    Therefore,

    [tex]
    u\left(x,t\right)=F\left(x-t\right)+\left(x+at\right)^{2}-F\left(x+at\right)
    [/tex]

    Next, using [itex]u_{t}\left(x,0\right)[/itex]

    [tex]
    \begin{align*}
    e^{x}& =-F^{\prime}\left(x\right)+2ax-aF^{\prime}\left(x\right) \\
    F^{\prime}\left(x\right)& =\frac{2ax-e^{x}}{\left(1+a\right) }\\
    F\left(x\right)& =\frac{ax^{2}-e^{x}}{\left(1+a\right)} + c
    \end{align*}
    [/tex]

    So

    [tex]
    \begin{align*}
    u\left(x,t\right)& =\frac{a\left(x-t\right)^{2}-e^{\left(x-t\right)}}{\left(1+a\right)}+c+\left(x+at\right)^{2} -\frac{a\left(x+at\right)^{2}-e^{\left(x+at\right)}}{\left(1+a\right)} - c\\
    &=\frac{e^{\left(x+at\right)}-e^{\left(x-t\right)}}{\left(1+a\right)}+\frac{\left(x+at\right)^{2}\left(1+a\right)+a\left(x-t\right)^{2}-a\left(x+at\right)^{2}}{\left(1+a\right)}\\
    &=\frac{e^{\left(x+at\right)}-e^{\left(x-t\right)}}{\left(1+a\right)}+\frac{x^{2}\left(1+a\right)+\left(1+a\right)at^{2}}{\left(1+a\right)}\\
    &=\frac{e^{\left(x+at\right)}-e^{\left(x-t\right)}}{\left(1+a\right)}+x^{2}+at^{2}
    \end{align*}
    [/tex]
     
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