EDP no lineal de primer orden

Question

Estoy tratando de resolver una EDP

$x (u_x)^2 + u_y = y $

$u(x,0)=x$

utilizando el método de las características (es decir $p=u_x$ , $q=u_y$ y la PDE anterior es $xp^2+q-y=0$ . Lo he intentado muchas veces pero no lo consigo. Se agradece la ayuda.

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} x\pars{\mrm{u}_{x}}^{2} + \,\mrm{u}_{y} = y \implies \bbx{\ds{x\pars{\varphi_{x}}^{2} + \,\varphi_{y} = 0}}\,,\quad \varphi\pars{x,y} \equiv \,\mrm{u}\pars{x,y} - {1 \over 2}\,y^{2}\,,\quad \varphi\pars{x,0} = x \end{align}

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$$ 0 = x\,\mrm{f}'^{2}\pars{x}\,\mrm{g}^{2}\pars{y} + \,\mrm{f}\pars{x}\,\mrm{g}'\pars{y} \implies x\,{\mrm{f}'^{2}\pars{x} \over \mrm{f}\pars{x}} = -\,{\mrm{g}'\pars{y} \over \,\mrm{g}^{2}\pars{y}} = \mu $$

$$ {\mrm{f}'\pars{x} \over \mrm{f}^{1/2}\pars{x}} = {\root{\mu} \over x^{1/2}}\,, \quad {\mrm{g}'\pars{y} \over \mrm{g}^{2}\pars{y}} = -\mu $$

$$ 2\,\mrm{f}^{1/2}\pars{x} = 2\root{\mu}x^{1/2} + 2a \quad -\,{1 \over \mrm{g}\pars{y}} = -\mu y - b $$

\begin{align} \varphi\pars{x,y} & = {\pars{\root{\mu}x^{1/2} + a}^{2} \over \mu y + b} = {\mu x + 2a\root{\mu}x^{1/2} + a^{2} \over \mu y + b}\implies x = \varphi\pars{x,0} = {\mu x + 2a\root{\mu}x^{1/2} + a^{2} \over b} \\[5mm] \implies & b = \mu\,,\quad a = 0 \implies \varphi\pars{x} = {x \over y + 1}\implies \bbx{\ds{\mrm{u}\pars{x,y} = {x \over y + 1} + {1 \over 2}\,y^{2}}} \end{align}