Cauchy problem, method of characteristics

[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?

[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.

[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?

[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}}

[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.