# Cauchy problem, method of characteristics

1. Sep 4, 2011

### math2011

This question is also posted at (with better formatting): http://www.mathhelpforum.com/math-help/f59/cauchy-problem-method-characteristics-187192.html.

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 $$x(0,s) = s, y(0,s) = s, u(0,s) = \frac{1}{s^2}$$.

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. Sep 5, 2011

### hunt_mat

You have for y:
$$\frac{dy}{dt}=\frac{1}{\sqrt{s^{2}+t}}\Rightarrow y(t)-s=\int_{0}^{t}\frac{du}{\sqrt{s^{2}+u}}$$
This will give a different y than you have and it's where you have made a mistake I think. Once you have x and y in terms of s and t ten you should be able to get u in terms of x and y.

3. Sep 5, 2011

### math2011

Why does $$\frac{dy}{dt} = \frac{1}{\sqrt{s^2 + t}}$$ imply $$y(t) - s = \int^t_0 \frac{1}{\sqrt{s^2 + u}} du$$?

How do you get the -s on the LHS of the second equation?

Why is the integration on the RHS only from 0 to t? (I just realized that s > 0, is this related to the integral?)

What is the reason for substituting t with u and integrating from 0 to t?

4. Sep 5, 2011

### hunt_mat

Integrate both sides and use the initial conditions:
$$\int_{0}^{t}\frac{dy}{dt}dt=\int_{0}^{t}\frac{dr}{\sqrt{s^{2}+r}}=\left[ y\right]_{0}^{t}=y(t)-y(0)=y(t)-s$$
Use a change of variables $v=r+s^{2}$ and the integral becomes:
$$\int_{0}^{t}\frac{dr}{\sqrt{s^{2}+r}}=\int_{s^{2}}^{t+s^{2}}\frac{dv}{\sqrt{v}}$$

5. Sep 6, 2011

### math2011

I tried this and it turns out to be the same as what I got before.
$$\int^t_0 \frac{1}{\sqrt{s^2 + r}} dr = \int^{s^2+t}_{s^2} \frac{1}{\sqrt{v}} dv = \left[2 v^{\frac{1}{2}} \right]^{s^2+t}_{s^2} = 2\sqrt{s^2 + t} - 2s$$
$$y(t) = 2\sqrt{s^2 + t} - s$$
Then
$$s = 2x - y$$
$$t = x^2 - (2x - y)^2$$
and
$$u = \frac{1}{t + s^2} = \frac{1}{x^2 - (2x - y)^2 + (2x - y)^2} = \frac{1}{x^2}$$
Maybe this is the correct solution.