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Solving boundary value problem (Wave Equation)

  1. Oct 24, 2012 #1
    1. The problem statement, all variables and given/known data

    Show that the boundary-value problem $$u_{tt}=u_{xx}\qquad u(x,0)=2f(x)\qquad u_t(x,0)=2g(x)$$ has the solution $$u(x,t)=f(x+t)+f(x-t)+G(x+t)-G(x-t)$$ where ##G## is an antiderivative/indefinite integral of ##g##. Here, we assume that ##-\infty<x<\infty## and ##t\geq 0##

    2. Relevant equations

    I know that the solution for a wave equation of the form ##u_{tt}=a^2u_{xx}## is ##u(x,t)=f(x+at)+g(x-at)##

    3. The attempt at a solution

    Using the "knowledge" above, I tried to take the first partial derivative of ##u(x,t)=f(x+t)+g(x-t)## with respect to ##t## since ##a=1##. I got ##u_t(x,t)=f'(x+t)-g'(x-t)## Then using the conditions ##u(x,0)=2f(x)## and ##u_t(x,0)=2g(x)##, I substituted 0 for ##t## and set them equal to ##2f(x)## and ##2g(x)##.

    I came up with ##f(x)=g(x)=\frac{1}{2}(f'(x)-g'(x))##, which I thought could also be ##\frac{1}{2}(f'(x)-G(x))##. But then I got stuck because I have no idea how to get to $$u(x,t)=f(x+t)+f(x-t)+G(x+t)-G(x-t)$$

    Part of my confusion is that I only have initial conditions and no boundary conditions, and all the examples I can find have boundary conditions as well, not just initial.

    Thanks so much if you can help :smile: I might just be missing something because I'm in freak out mode about my test tomorrow on this stuff! This is a numerical methods for differential equations class by the way.
     
  2. jcsd
  3. Oct 24, 2012 #2

    LCKurtz

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    Let's say ##u(x,t) = F(x+t) + G(x-t)## since we don't want to confuse the general solution with the ##f## and ##g## given in the problem. And we have ##a=1##.

    Good so far, but let's use the capital letters.

    ##u(x,t)=F(x+t)+G(x-t)##
    ##u_t(x,t)=F'(x+t)-G'(x-t)##
    This should give you

    ##F(x)+G(x) = 2f(x)##
    ##F'(x)-G'(x) = 2g(x)##

    Part of your trouble is confusing the capital F and G with the little f and g. What you want to do next is integrate the equation ##F'(x)-G'(x) = 2g(x)## giving$$
    F(x) - G(x) = 2\int g(x)\, dx =2 H(x)$$
    where I am just using ##H## for the antiderivative of ##g##.

    Now work with these two equations:

    ##F(x)+G(x) = 2f(x)##
    ##F(x) - G(x) = 2H(x)##

    to solve for ##F## and ##G##.
     
    Last edited: Oct 24, 2012
  4. Oct 24, 2012 #3
    Wow thank you so so much! :smile:

    From the 2 equations I came up with $$F(x)=f(x)+H(x)$$ and $$G(x)=f(x)-H(x)$$ Then adding them gives me essentially exactly what I need but in the problem the solution ##u(x,t)=f(x+t)+f(x−t)+G(x+t)−G(x−t)## uses lower case f and capital G, those aren't like typos or something. You were exactly right that I was thinking that the ##f## and ##g## from the general solution and the conditions were the same thing, when apparently they aren't :redface: So could I just use completely different letters/symbols for the general solution to keep the variables separate?
     
  5. Oct 24, 2012 #4

    LCKurtz

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    You're welcome
    Yes, that is what you should do. When I wrote up my original reply I hadn't noticed you had already used G or I wouldn't have used it either.
     
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