# Thread: Cauchy problem, method of characteristics

1. ## 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?

2. ## 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.