# Solve first order partial diferential equation

Consider the Cauchy problem: $$\left\{\begin{array}{lll} x^2\partial_x u+y^2\partial_yu=u^2\\ u(x,2x)=1 \end{array}\right.$$ It is easy to show that the characteristic equations are given by: $$\frac{dx}{x^2}=\frac{dy}{y^2}=\frac{dz}{px^2+qy^2}=\frac{dp}{2pu-2xp}=\frac{dq}{2qu-2yq}=dt$$ By the Lagrange-Charpit's method, we need a first integral $\Psi$ different to $\Phi=x^2p+y^2q-u^2$.

How I can find $\Psi$? Maybe we should get directly the characteristic curves of the previous system?

Many thanks!

$$\left\{\begin{array}{lll} x^2\partial_x u+y^2\partial_yu=u^2\\ u(x,2x)=1 \end{array}\right.$$ FIRST METHOD :

The characteristics equations are : $$\frac{dx}{x^2}=\frac{dy}{y^2}=\frac{du}{u^2}$$

First characteristic curve, from $\frac{dx}{x^2}=\frac{dy}{y^2} \quad\to\quad \frac{1}{x}-\frac{1}{y}=c_1$

Second characteristic curve, from $\frac{dy}{y^2}=\frac{du}{u^2} \quad\to\quad \frac{1}{y}-\frac{1}{u}=c_2$

General solution expressed on implicit form : $$\Phi\left(c_1\:,\:c_2\right)= \Phi\left(\frac{1}{x}-\frac{1}{y}\:,\:\frac{1}{y}-\frac{1}{u}\right)$$ where $\Phi$ is any differentiable function of two variables.

Solving the implicit equation for its second variable leads to : $$\frac{1}{y}-\frac{1}{u}=F\left(\frac{1}{x}-\frac{1}{y} \right)$$ where $F$ is any differentiable function.

$$u(x,y)=\frac{1}{\frac{1}{y}-F\left(\frac{1}{x}-\frac{1}{y} \right)}$$

With condition $u(x,2x)=1 \quad\to\quad \frac{1}{\frac{1}{2x}-F\left(\frac{1}{x}-\frac{1}{2x} \right)}=1 \quad\to\quad F\left(\frac{1}{2x} \right)= \frac{1}{2x}-1$

This determines the function $F(Z)=Z-1$ and so, $F\left(\frac{1}{x}-\frac{1}{y} \right)= (\frac{1}{x}-\frac{1}{y})-1$

$$u(x,y)=\frac{1}{\frac{1}{y}-\left(\frac{1}{x}-\frac{1}{y}-1 \right)}= \frac{1}{\frac{2}{y}-\frac{1}{x}+1}=$$

$$u(x,y)=\frac{xy}{2x-y+xy}$$

SECOND METHOD :

Change : $\begin{cases} x=\frac{1}{X}\\ y=\frac{1}{Y}\\ u=\frac{1}{U}\end{cases}$ transforms the PDE : $\frac{x^2}{u^2}\frac{\partial u}{\partial x}+\frac{y^2}{u^2}\frac{\partial u}{\partial y}=1$ to :

$$\quad U_X+U_Y=1$$ Solving this elementary linear PDE leads to : $$U=Y+f(X-Y)$$ any differentiable function $f$.

The condition $u(x,2x)=1$ is transformed to $\frac{1}{U(\frac{1}{x},\frac{1}{2x})}=\frac{1}{U(X,\frac{X}{2})}=1 \quad\to\quad U(X,\frac{X}{2} )=1$

$1=\frac{X}{2}+f(X-\frac{X}{2}) \quad\to\quad f(\frac{X}{2})=1-\frac{X}{2}$ which determines $f(Z)=1-Z$

$f(X-Y)=1-(X-Y)=1-X+Y$

$U=Y+f(X-Y)=Y+(1-X+Y)=1-X+2Y$

$\frac{1}{u}=1-\frac{1}{x}+2\frac{1}{y} \quad\to\quad u=\frac{xy}{2x-y+xy}$

$\dfrac{dx}{dt}=x^2$ , letting $x(1)=-1$ , we have $-\dfrac{1}{x}=t$

$\dfrac{dy}{dt}=y^2$ , we have $-\dfrac{1}{y}=t+y_0=-\dfrac{1}{x}+y_0$

$\dfrac{du}{dt}=u^2$ , we have $\dfrac{1}{u}=-t+f(y_0)=\dfrac{1}{x}+f\left(\dfrac{1}{x}-\dfrac{1}{y}\right)$ , i.e. $u(x,y)=\dfrac{1}{\dfrac{1}{x}+f\left(\dfrac{1}{x}-\dfrac{1}{y}\right)}$

$u(x,2x)=1$ :

$\dfrac{1}{\dfrac{1}{x}+f\left(\dfrac{1}{x}-\dfrac{1}{2x}\right)}=1$

$\dfrac{1}{x}+f\left(\dfrac{1}{2x}\right)=1$

$f\left(\dfrac{1}{2x}\right)=1-\dfrac{1}{x}$

$f(x)=1-\dfrac{1}{\dfrac{1}{2x}}=1-2x$

$\therefore u(x,y)=\dfrac{1}{\dfrac{1}{x}+1-2\left(\dfrac{1}{x}-\dfrac{1}{y}\right)}=\dfrac{1}{1+\dfrac{2}{y}-\dfrac{1}{x}}=\dfrac{xy}{xy+2x-y}$