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Math Help - Cauchy problem, method of characteristics

  1. #1
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    Cauchy problem, method of characteristics

    Solve the following Cauchy problem
    \displaystyle \frac{1}{2x}u_x + xu u_y + u^2 = 0,
    subject to
    \displaystyle u(x,x) = \frac{1}{x^2}, x > 0.

    Attempt:

    The characteristic equations are \displaystyle x_t = \frac{1}{2x}, y_t = xu, u_t = -u^2.

    The initial conditions are x(0,s) = s, y(0,s) = s and \displaystyle u(0,s) = \frac{1}{s^2}.

    The Jacobian is J = \begin{vmatrix}\frac{1}{2s} & \frac{1}{s} \\1 & 1\end{vmatrix} = - \frac{1}{2s} and hence we expect a unique solution when s \ne \pm \infty and s \ne 0. (Is this correct?)

    Now solve the characteristic equations.

    \displaystyle \frac{dx}{dt} =& \frac{1}{2x} \\ 2x dx =& dt \\ x^2 = t + f_1(s).
    Apply initial condition to get f_1(s) = s^2 and hence x = \sqrt{t + s^2}.

    \displaystyle \frac{du}{dt} = -u^2 \\ \frac{1}{u^2} du = - dt \\ u^{-1} = t + f_2(s) \\ u = \frac{1}{t + f_2(s)}.
    Apply initial condition to get f_2(s) = s^2 and hence \displaystyle u = \frac{1}{t + s^2} = \frac{1}{x^2}. (Is this it for the question? Why is u independent of y? What have I done wrong?)

    Substitute above x and y into characteristic equation y_t = xu and we get \displaystyle y = \frac{1}{\sqrt{t + s^2}}. Integrate over t and we get y = 2\sqrt{t + s^2} + f_3(s). Apply initial condition we get f_3(s) = -s and y = 2 \sqrt{t + s^2} - s.

    From expressions of x and y obtained above we get
    t = x^2 - s^2
    \displaystyle t = \frac{1}{4}(y + s)^2 - s^2.

    Therefore the characteristics is (y + s)^2 = 4 x^2. (Do I need this characteristics at all? What should I do with it?)


    Is the above attempt correct?
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  2. #2
    MHF Contributor
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    Re: Cauchy problem, method of characteristics

    What you have looks good! Once you got u = \frac{1}{x^2}, it wasn't really necessary to go any further. I agree with you, having u independent of y seems odd but it satisfies both the PDE and IC.
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