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Math Help - PDE using D'Alembert's Solution

  1. #1
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    PDE using D'Alembert's Solution

    Use D'Alembert's solution to solve the following

    u_{xx} - \frac{1}{c^2} u_{tt} = 4

    subject to the initial conditions u(x, 0) = \sin (x) and u_t (x, 0) = c \ \cos (x).

    Note that u_{xx} - c^{-2} u_{tt} = 4u_{\xi \eta} with \xi = x + ct and \eta = x - ct
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  2. #2
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    With your change of variables

    4 u_{\xi \eta} = 4 gives u_{\xi \eta} = 1.

    Integrating twice gives

    u = \xi \eta + F(\xi) + G(\eta)

    or

    u = x^2 - c^2 t^2 + F(x+ct) + G(x-ct).

    Imposing the two boundary conditions gives

    x^2 + F(x) + G(x) = \sin x,\;\;\;(1)
    cF'(x)-cG'(x) = c\cos x,\;\;\;(2).

    Differentiating the first and then solving for F' and G' gives

    F'(x) = \cos x - x,\;\;\;G' = - x.

    Thus, F(x) = \sin x - \frac{1}{2}x^2 + c_1,\;\;\;G(x) = - \frac{1}{2} x^2 + c_2.

    Finally, we have

    u = x^2 - c^2 t^2 + \sin(x+ct) - \frac{1}{2}(x+ct)^2 - \frac{1}{2}(x-ct)^2<br />
=-2 c^2t^2 + \sin(x+ct) + c_1+c_2

    but imposing the first BC (1) gives c_1+c_2 = 0, so the final solution is

    u = -2 c^2t^2 + \sin(x+ct).
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    Quote Originally Posted by Danny View Post
    With your change of variables

    4 u_{\xi \eta} = 4 gives u_{\xi \eta} = 1.

    Integrating twice gives

    u = \xi \eta + F(\xi) + G(\eta)

    or

    u = x^2 - c^2 t^2 + F(x+ct) + G(x-ct).

    Imposing the two boundary conditions gives

    x^2 + F(x) + G(x) = \sin x,\;\;\;(1)
    cF'(x)-cG'(x) = c\cos x,\;\;\;(2).

    Differentiating the first and then solving for F' and G' gives

    F'(x) = \cos x - x,\;\;\;G' = - x.

    Thus, F(x) = \sin x - \frac{1}{2}x^2 + c_1,\;\;\;G(x) = - \frac{1}{2} x^2 + c_2.

    Finally, we have

    u = x^2 - c^2 t^2 + \sin(x+ct) - \frac{1}{2}(x+ct)^2 - \frac{1}{2}(x-ct)^2<br />
=-2 c^2t^2 + \sin(x+ct) + c_1+c_2

    but imposing the first BC (1) gives c_1+c_2 = 0, so the final solution is

    u = -2 c^2t^2 + \sin(x+ct).

    Hi Danny,
    Thanks for your reply, but I'm having difficulties understanding this.

    I can't see where you've used d'Alembert's solution i.e.

    u(x,t) = \frac{1}{2} \big ( f (x + ct) + f(x - ct) \big ) + \frac{1}{2c} \int_{x-ct}^{x+ct} g(s) ds

    Could you please go into detail with each step? I really want to understand this.
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  4. #4
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    Quote Originally Posted by JoernE View Post
    Hi Danny,
    Thanks for your reply, but I'm having difficulties understanding this.

    I can't see where you've used d'Alembert's solution i.e.

    u(x,t) = \frac{1}{2} \big ( f (x + ct) + f(x - ct) \big ) + \frac{1}{2c} \int_{x-ct}^{x+ct} g(s) ds

    Could you please go into detail with each step? I really want to understand this.
    I used it after making the change of variables. Let me derive my solution with the D'Alembert solution.

    First the D'Alembert solution applies for

    u_{xx} - \frac{1}{c^2} u_{tt} = 0

    But your PDE has a nonhomogeneous term.

    u_{xx} - \frac{1}{c^2} u_{tt} = 4

    So we'll find a particlar solution - u_p = -2c^2 t^2 works. So let u = v + u_p where now v satisfies

    v_{xx} - \frac{1}{c^2} v_{tt} = 0,

    v(x,0) = \sin x,\;\;\; v_t(x,0) = c\cos x

    D'Almbert solution

    v(x,t)  = \frac{1}{2} \left( f (x + ct) + f(x - ct) \right) + \frac{1}{2c} \displaystyle \int_{x-ct}^{x+ct} g(s) ds

    and your IC's gives f(x) = \sin x and g(x) = c\cos x

    v(x,t)  = \frac{1}{2} \left( \sin (x + ct) + \sin(x - ct) \right) + \frac{1}{2c} \displaystyle \int_{x-ct}^{x+ct} c\cos(s) ds <br />
 = \sin (x + ct)

    so

    u(x,t) = -2c^2t^2 + \sin(x+ct).
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    so we'll find a particular solution u_p = -2c^2 t^2 works
    Thanks again. I'm just curious how you arrive at this particular solution. I now understand the other parts.
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  6. #6
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    Quote Originally Posted by JoernE View Post
    Thanks again. I'm just curious how you arrive at this particular solution. I now understand the other parts.
    Assume that u = F(t) only and substitute into your PDE. You'll get an simple ODE for F(t) from wich my particular solution will emerge. (Note - you can supress the constants of integration, it only makes things more complicated).
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