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Math Help - Solution of the Wave Equation

  1. #1
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    Solution of the Wave Equation

    Hi,

    I'm having a bit of trouble understanding D'Alembert's solution to the wave equation.

    The Problem

    \displaystyle \frac{\partial^2 u}{\partial t^2} = c^2 \frac{\partial^2 u}{\partial x^2}

    Or, in short notation:

    \displaystyle u_{tt} = c^2 u_{xx}

    Find  \displaystyle u(x,t) subject to the initial conditions:

     \displaystyle u(x,0) = f(x)

    \displaystyle u_t(x,0) = g(x)

    The Solution (the part I understand)

    The first stage is to express the equation in characteristic coordinates, which, once performed, gives:

    \displaystyle  u_{\xi \eta} = 0

    Here,  \displaystyle \xi = x + ct and  \displaystyle \eta = x-ct .

    Integrating the equation w.r.t both variables gives us the simple general solution:

    \displaystyle  u(\xi, \eta) = F_1(\xi)+F_2(\eta)

    Or, back in cartesian coordinates:

    \displaystyle  u(x,t) = F_1(x+ct) + F_2(x-ct) (1)

    Now, differentiating this, we get:

    \displaystyle  u_t(x,t) = cF_1'(x+ct) - cF_2'(x-ct) (2)

    Applying the initial conditions to equations (1) and (2), we get:

    \displaystyle  u(x,0) = F_1(x) +F_2(x) = f(x) (3)

     \displaystyle u_t(x,0) = cF_1'(x) - cF_2'(x) = g(x) (4)

    The Solution (part I don't understand)

    Now, the book I am using then makes the following statement:

    If we integrate equation (4), we get:

     \displaystyle \int_a^x g(\tau)d\tau =  cF_1(x) - cF_2(x)

    To be honest, this is the only part I don't understand. Where do the other terms which results from the integration disappear to? i.e., by my working, I would get:

    \displaystyle  \int_a^x g(\tau)d\tau =  \int_a^x (cF_1'(\tau) - cF_2'(\tau)) d\tau

     \displaystyle = \biggg(cF_1(\tau)-cF_2(\tau)\biggg)^x_a

     \displaystyle = cF_1(x) - cF_1(a) -cF_2(x) + cF_2(a)

    How do they manage to conclude that  \displaystyle cF_2(a) - cF_1(a) = 0 ??

    Last edited by Phugoid; April 28th 2011 at 06:13 AM.
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  2. #2
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by Phugoid View Post
    If we integrate equation (4), we get:

     \displaystyle \int_a^x g(\tau)d\tau =  cF_1(x) - cF_2(x)
    I believe what the book has done from equation 4 is integrated the middle terms and the RHS

    cF_1(x) - cF_2(x)=\int_a^x g( \tau) d \tau + F_1(0)-F_2(0)

    where the ast two terms are constants of integration which are a function of t and additionally we know u_t (x,0) hence the constants go to zero.

    Can you continue on?
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  3. #3
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    Quote Originally Posted by bugatti79 View Post
    I believe what the book has done from equation 4 is integrated the middle terms and the RHS

    cF_1(x) - cF_2(x)=\int_a^x g( \tau) d \tau + F_1(0)-F_2(0)

    where the ast two terms are constants of integration which are a function of t and additionally we know u_t (x,0) hence the constants go to zero.

    Can you continue on?
    Sorry, I don't fully understand what you mean.

    Could you elaborate?
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  4. #4
    Senior Member bugatti79's Avatar
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    From \displaystyle u(x,t) = F_1(x+ct) + F_2(x-ct) (1) you can see that F1 and F2 are both functions of x and t. Therefore when integrating equation 4 wrt to x the resulting constants of integration must be a function of t.

    I have attached my version using your initial conditions. This should help you find the general solutio u(x,t)
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  5. #5
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    Quote Originally Posted by bugatti79 View Post
    From \displaystyle u(x,t) = F_1(x+ct) + F_2(x-ct) (1) you can see that F1 and F2 are both functions of x and t. Therefore when integrating equation 4 wrt to x the resulting constants of integration must be a function of t.

    I have attached my version using your initial conditions. This should help you find the general solutio u(x,t)
    Ah, I see... so the constants don't necessarily disappear as might be suggested by the book, but eventually cancel out at a later stage in the derivation.

    Thanks for that, cleared a lot up.
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