Cauchy problem, method of characteristics

**math2011** · September 3rd 2011, 06:12 AM

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?

**Danny** · September 4th 2011, 09:30 AM

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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