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Thread: wave equation

  1. #1
    Junior Member
    Joined
    Apr 2009
    Posts
    58

    wave equation

    The question is Solve the one dimensional wave equation

    $\displaystyle \frac{1}{c^2}$

    on $\displaystyle 0<=x<=3$, subject to the conditions

    $\displaystyle u(x,0) = \frac{1}{2}x$, $\displaystyle u_t(x,0)=x(3-x), u(0,t)=u(3,t)=0$

    well i started it, but im not sure if this is right coz i don't really understand wave or heat equations SIGH...



    $\displaystyle
    u_tt(x,t) = X''(x)T(t) $ and $\displaystyle u_tt(x,t) = X(x)T''(t)
    $
    $\displaystyle \frac{1}{c^2} $

    Case 1 let $\displaystyle v = w^2 > 0 $
    $\displaystyle
    \frac{X''(x)}{X(x)} = w^2 $
    $\displaystyle
    X''(x) = w^2 X(x) = 0 $

    Using Auxilliary Equations:
    $\displaystyle
    m^2 - w^2 = 0$
    So $\displaystyle m = +- w $
    And $\displaystyle X(x) = C_1 e^{wx} + C_2 e^{-wx}$

    As $\displaystyle u(0,t) = u(3,t) = 0, X(0) = V(3) = 0$
    Therefore, $\displaystyle C_1 + C_2 = 0$ and $\displaystyle C_1 = - C_2$
    Therefore, no solutions...

    Case 2 $\displaystyle v = w^2=0$
    $\displaystyle X''(x) - 0X(x) = 0$
    Auxilliary Equations $\displaystyle m^2 - 0 = 0$, $\displaystyle m = +- 0 $
    Therefore, $\displaystyle X(x) = C_3 e^0 + C_4 e^{-0} $
    As $\displaystyle X(0) = X(2) = 0, X(0) = C_3, X(3) = 2C_3 + C_4 $
    So $\displaystyle 2C_3 + C_4 = 0 $
    Therefore, no solution as $\displaystyle 2C_3 = -C_4$

    Case 3 $\displaystyle v = -w^2$ as $\displaystyle v<0$
    $\displaystyle
    \frac{X''(x)}{X(x)} = -w^2 $
    $\displaystyle X''(x) + w^2X(x) = 0 $

    Auxilliary Equation: $\displaystyle m^2 + w^2 = 0, m = +-iw $\
    So $\displaystyle C_5coswx +C_6sinwx = 0$
    $\displaystyle C_5 = 0, $ So $\displaystyle
    C_5cos3w + C_6sin3w = 0$
    $\displaystyle sin3w = 0$
    Therefore, $\displaystyle 3w = \frac{n\Pi}{3}$ and $\displaystyle V = \frac{n^2\Pi^2}{9}$
    Thus, $\displaystyle X(x) = B sin $ $\displaystyle \frac{xn\Pi}{3}$

    Now Im lost, am i doing it right so far, if so, then how do i go about completing this, coz i have no idea what to do now.. Thank you
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  2. #2
    Member
    Joined
    May 2009
    Posts
    127
    Case 1: correct.

    Case 2: you have come to the right result but via the wrong method. For case 2 you have:

    $\displaystyle \frac{d^2 X}{dx^2} = 0$

    so integrating twice gives

    $\displaystyle X = A x +B$

    and to satisfy BCs $\displaystyle A = B = 0$ hence no solution as you concluded (for the wrong reasons).

    Case 3: you have correctly found that:

    $\displaystyle X_n (x) = B_n \sin \left( \frac{n \pi}{3} x \right)$

    Consequently

    $\displaystyle \frac{d^2 T}{dt^2} + c^2 \left(\frac{n \pi}{3}\right)^2 T = 0$

    so then, after finding the auxiliary equation, we find the solution for $\displaystyle T$ is

    $\displaystyle T_n = C \sin \left(\omega_n c t \right) + D \cos \left(\omega_n c t \right)$

    where $\displaystyle \omega_n = \frac{n \pi}{3}$ so the general solution is given by:

    $\displaystyle u(x,t) = \sum_{n=0}^{n=\infty} a_n(\sin \left(\omega_n c t \right) + b_n \cos \left(\omega_n c t \right)) \sin \left( \omega_n x \right)$

    Applying the initial conditions on u(x,0) and u'(x,0) we have

    $\displaystyle \sum_{n=1}^{n=\infty} c_n \sin \left( \omega_n x \right) = \frac{1}{2} x$

    where $\displaystyle c_n = a_n b_n$ and

    $\displaystyle \sum_{n=1}^{n=\infty} a_n \omega_n c \sin \left( \omega_n x \right) = x(3-x)$.

    In the first of these summations you can find $\displaystyle c_k$ by multiplying by $\displaystyle \sin( \omega_k x)$ on both sides and then integrating wrt x from 0 to 3. Doing this should give you that:

    $\displaystyle c_n = (-1)^{n+1} \frac{3}{n \pi}$ .

    I may have made arithmetic errors here so double check the result in case.

    For the second summation it's exactly the same procedure to give you $\displaystyle a_n$. At the end substitute back into your general solution the results for $\displaystyle a_n$ and $\displaystyle c_n = a_n b_n$ and you're done!

    An alternative, and easier way would have been to transform the problem to Fourier space and then solve.

    Hope that helped!
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  3. #3
    Junior Member
    Joined
    Apr 2009
    Posts
    58
    that's awesome!! Thank you SO much
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