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Consider the differential equation \begin{align} u_x^2+xu_y=0.\tag{1} \end{align} Making the so called Legendre transformation $$v=px+qy-u,$$ where $p=u_x,q=u_y$, show that $v$ satisfies the equation $$p^2+qv_p=0.$$ Show that the solution of $(1)$ can be expressed in parametric form as \begin{cases} x=-\frac{p^2}{q} \\ y=\frac{p^3}{3q^2}+f'(q) \\ u=-\frac{p^3}{3q}+qf'(q)-f(q) \end{cases} where $f$ is an arbitrary continuously differentiable function.

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  • $\begingroup$ Have you done the first part? $\endgroup$ – Christopher A. Wong Nov 12 '16 at 9:10
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Note:$$ v_{p}=x \ , \ v_{q}=y $$ So $$p^2 +qv_{p}=0 \Rightarrow v_{p}=\frac{-p^2}{q}=x $$ Now by integrating $v_{p}$ w.r.to p which implies : $$ v=\frac{-p^3}{3q}+f(q) \Rightarrow v_{q}=\frac{p^3}{3q^2}+f'(q)=y$$ Now $$u =px+qy -v = \frac{-p^2}{3q}+qf'(q)-f(q)$$

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Hint:

$u_x^2+xu_y=0$

$\dfrac{u_x^2}{x}+u_y=0$

$\dfrac{2u_xu_{xx}}{x}-\dfrac{u_x^2}{x^2}+u_{xy}=0$

Let $v=u_x$ ,

Then $\dfrac{2v}{x}v_x-\dfrac{v^2}{x^2}+v_y=0$

$\dfrac{2v}{x}v_x+v_y=\dfrac{v^2}{x^2}$

$2xv_x+\dfrac{x^2}{v}v_y=v$

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{dv}{dt}=v$ , letting $v(0)=1$ , we have $v=e^t$

$\dfrac{dx}{dt}=2x$ , letting $x(0)=x_0$ , we have $x=x_0e^{2t}=x_0v^2$

$\dfrac{dy}{dt}=\dfrac{x^2}{v}=x_0^2e^{3t}$ , we have $y=f(x_0)+\dfrac{x_0^2e^{3t}}{3}=f\left(\dfrac{x}{v^2}\right)+\dfrac{x^2}{3v}$

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