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Solve the following Cauchy problem

[tex]\displaystyle \frac{1}{2x}u_x + xu u_y + u^2 = 0[/tex],

subject to

[tex]\displaystyle u(x,x) = \frac{1}{x^2}, x > 0[/tex].

Attempt:

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

The initial conditions are [tex]x(0,s) = s, y(0,s) = s, u(0,s) = \frac{1}{s^2}[/tex].

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

Now solve the characteristic equations.

[tex]\displaystyle \frac{dx}{dt} = \frac{1}{2x}[/tex]

[tex]2x dx = dt[/tex]

[tex]x^2 = t + f_1(s)[/tex].

Apply initial condition to get [tex]f_1(s) = s^2[/tex] and hence [tex]x = \sqrt{t + s^2}[/tex].

[tex]\displaystyle \frac{du}{dt} = -u^2[/tex]

[tex]\frac{1}{u^2} du = - dt[/tex]

[tex]u^{-1} = t + f_2(s)[/tex]

[tex]u = \frac{1}{t + f_2(s)}[/tex].

Apply initial condition to get [tex]f_2(s) = s^2[/tex] and hence [tex]\displaystyle u = \frac{1}{t + s^2} = \frac{1}{x^2}[/tex]. (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 [tex]y_t = xu[/tex] and we get [tex]\displaystyle y = \frac{1}{\sqrt{t + s^2}}[/tex]. Integrate over t and we get [tex]y = 2\sqrt{t + s^2} + f_3(s)[/tex]. Apply initial condition we get [tex]f_3(s) = -s[/tex] and [tex]y = 2 \sqrt{t + s^2} - s[/tex].

From expressions of x and y obtained above we get

[tex]t = x^2 - s^2[/tex]

[tex]\displaystyle t = \frac{1}{4}(y + s)^2 - s^2[/tex].

Therefore the characteristics is [tex](y + s)^2 = 4 x^2[/tex]. (Do I need this characteristics at all? What should I do with it?)

Is the above attempt correct?

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# Cauchy problem, method of characteristics

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