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Math Help - Partial Differential Equation -- method of characteristics (ode solve)

  1. #1
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    Partial Differential Equation -- method of characteristics (ode solve)

    I've been asked to solve

    u_t + cos(\pi x) u_x = 0 using the method of characteristics. This amounts to solving the following odes:

    dt/ds = 1, t(0) = 0 which implies t = s and

    dx/ds = cos(\pi x), x(0) = x_0.

    This ode yields (if I'm right):

    |sec(\pi x) + tan(\pi x)| = Ce^{\pi s}.

    However, I cannot solve this explicitly for x, and therefore cannot solve the pde by the method of characteristics? Can anyone shed insight? Have I done something wrong? Is there a missing trig identity somewhere?

    Thanks
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  2. #2
    MHF Contributor chisigma's Avatar
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    Re: Partial Differential Equation -- method of characteristics (ode solve)

    Quote Originally Posted by davismj View Post
    I've been asked to solve

    u_t + cos(\pi x) u_x = 0 using the method of characteristics. This amounts to solving the following odes:

    dt/ds = 1, t(0) = 0 which implies t = s and

    dx/ds = cos(\pi x), x(0) = x_0.

    This ode yields (if I'm right):

    |sec(\pi x) + tan(\pi x)| = Ce^{\pi s}.

    However, I cannot solve this explicitly for x, and therefore cannot solve the pde by the method of characteristics? Can anyone shed insight? Have I done something wrong? Is there a missing trig identity somewhere?

    Thanks
    Your PDE...

    u_{t}+ \cos (\pi x)\ u_{x}=0

    ... is 'nonhomogeneous' and equivalent to then system...

    d t=\frac{d x}{\cos \pi x}

    d u=0 (2)

    ...the solution of which is...

     c_{1}= v(x,t)= t-\frac{1}{\pi}\ \ln\ |\tan (\frac{\pi}{2} x +\frac{\pi}{4})|

    c_{2}= u (3)

    ... so that the solution of (1) is...

    u= \gamma\{v(x,t)\} (4)

    ... where \gamma(*) is any continous function with continous derivative...

    Kind regards

    \chi \sigma
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