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Math Help - wave equation

  1. #1
    Junior Member
    Joined
    Apr 2009
    Posts
    58

    wave equation

    The question is Solve the one dimensional wave equation

    \frac{1}{c^2}

    on 0<=x<=3, subject to the conditions

     u(x,0) = \frac{1}{2}x, 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...



    <br />
u_tt(x,t) = X''(x)T(t) and u_tt(x,t) = X(x)T''(t)<br />
    \frac{1}{c^2}

    Case 1 let v =  w^2 > 0
    <br />
\frac{X''(x)}{X(x)} = w^2
    <br />
X''(x) = w^2 X(x) = 0

    Using Auxilliary Equations:
    <br />
m^2 - w^2 = 0
    So  m = +- w
    And  X(x) = C_1 e^{wx} + C_2 e^{-wx}

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

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

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

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

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

    so integrating twice gives

    X = A x +B

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

    Case 3: you have correctly found that:

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

    Consequently

    \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 T is

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

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

    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

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

    where c_n = a_n b_n and

    \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 c_k by multiplying by \sin( \omega_k x) on both sides and then integrating wrt x from 0 to 3. Doing this should give you that:

    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 a_n. At the end substitute back into your general solution the results for a_n and 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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