Results 1 to 6 of 6

Math Help - PDE using D'Alembert's Solution

  1. #1
    Newbie
    Joined
    Aug 2010
    Posts
    10

    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,367
    Thanks
    42
    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).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Aug 2010
    Posts
    10
    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.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,367
    Thanks
    42
    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).
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Aug 2010
    Posts
    10
    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.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,367
    Thanks
    42
    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).
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] D'Alembert PDE
    Posted in the Differential Equations Forum
    Replies: 7
    Last Post: September 1st 2011, 06:06 AM
  2. The wave equation - D'Alembert Solution - Set up Correctly?
    Posted in the Differential Equations Forum
    Replies: 0
    Last Post: May 20th 2011, 12:51 PM
  3. d'alembert solution Help
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: May 16th 2010, 09:37 AM
  4. Verify my understanding of D'Alembert Solution.
    Posted in the Differential Equations Forum
    Replies: 0
    Last Post: February 27th 2010, 10:50 AM
  5. D'Alembert's Solution
    Posted in the Differential Equations Forum
    Replies: 0
    Last Post: November 27th 2009, 05:54 PM

Search Tags


/mathhelpforum @mathhelpforum