Results 1 to 2 of 2

Thread: Cauchy problem, method of characteristics

  1. #1
    Junior Member
    Joined
    Mar 2011
    Posts
    72

    Cauchy problem, method of characteristics

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

    Attempt:

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

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

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

    Now solve the characteristic equations.

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

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

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

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

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


    Is the above attempt correct?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,470
    Thanks
    83

    Re: Cauchy problem, method of characteristics

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

Similar Math Help Forum Discussions

  1. Method of Characteristics
    Posted in the Calculus Forum
    Replies: 0
    Last Post: Nov 11th 2011, 07:16 AM
  2. [SOLVED] Method of Characteristics: Prove the u and v is a solution
    Posted in the Differential Equations Forum
    Replies: 17
    Last Post: Oct 25th 2010, 10:04 AM
  3. [SOLVED] Method of Characteristics
    Posted in the Differential Equations Forum
    Replies: 4
    Last Post: Oct 16th 2010, 07:57 AM
  4. 1st order PDE through Method of Characteristics
    Posted in the Differential Equations Forum
    Replies: 0
    Last Post: Jun 30th 2010, 05:51 PM
  5. Method of characteristics
    Posted in the Differential Equations Forum
    Replies: 0
    Last Post: Feb 7th 2010, 12:06 AM

Search Tags


/mathhelpforum @mathhelpforum