IVP $xuu_x+yuu_y=xy$

Question

Solve the following IVP$$ \begin{cases} xuu_x+yuu_y=xy\\ u(x,\sqrt{x})=0\\ \end{cases}$$

By Lagrange we get:

$$\frac{dx}{xu}=\frac{dy}{yu}=\frac{du}{xy}$$

$$\frac{dx}{xu}=\frac{dy}{yu}\to \frac{dx}{x}=\frac{dy}{y}\implies\phi_1=\frac{x}{y}=c_1$$

$$\frac{dx}{xu}=\frac{du}{xy}\to\frac{dx}{u}=\frac{du}{y}\implies\phi_2=2yx-u^2=c_2$$

So the genral sloution is: $$F\left(\frac{x}{y},2yx-u^2\right)=0$$

As $\phi_1$ is not a function of $u$ it can be writing as

$$F\left(\frac{x}{y}\right)=2yx-u^2\implies u^2=2yx-F\left(\frac{x}{y}\right)$$

Using initial condition we get $$ 0=2x\sqrt{x}-F\left(\frac{x}{\sqrt{x}}\right)\implies F\left(\frac{x}{\sqrt{x}}\right)=2x\sqrt{x} $$

So the solution is $$u^2=2yx-2x\sqrt{x}$$

Or I got it wrong somewhere?

Answer 1

As

$$ x \frac{\partial u^2}{\partial x}+y\frac{\partial u^2}{\partial y}=2x y $$

we have

$$ \frac 1y\frac{\partial u^2}{\partial x}+\frac 1x\frac{\partial u^2}{\partial y}=2 $$

now calling $v = u^2$ we have

$$ \frac 1y\frac{\partial v}{\partial x}+\frac 1x\frac{\partial v}{\partial y}=2 $$

with general solution

$$ v = x y+\phi\left(\frac yx\right) $$

but considering the conditions $v(x,\sqrt{x})=0$ we have

$$ \phi\left(\frac yx\right) =-\left(\frac xy\right)^3 $$

and then

$$ v = x y-\frac{x^3}{y^3} $$

hence

$$ u = \sqrt{ x y-\frac{x^3}{y^3}} $$

Answer 2

$$ \begin{cases} xuu_x+yuu_y=xy\\ u(x,\sqrt{x})=0\\ \end{cases}$$

System of characteristic ODEs :

$$\frac{dx}{xu}=\frac{dy}{yu}=\frac{du}{xy}.$$

The equation of a first characteristic curve comes from $\frac{dx}{xu}=\frac{dy}{yu}$ $$\frac{y}{x}=c_1$$

The equation of a second characteristic curve comes from :

$\frac{dx}{xu}=\frac{dy}{yu}=\frac{ydx+xdy}{y(xu)+x(yu)}=\frac{d(xy)}{2xyu}=\frac{du}{xy} \quad\to\quad d(xy)=2udu$ $$xy-u^2=c_2$$ The general solution expressed on the form of implicit equation is : $$\Phi\left(\frac{y}{x}\;,\;xy-u^2\right)=0$$ where $\Phi$ is any differentiable function of two variables.

Solving the implicit equation for the second variable leads to the explicit form of general solution : $$F\left(\frac{y}{x}\right)=xy-u^2\tag{1}$$ where $F$ is any differentiable function. The condition $u(x,\sqrt{x})=0$ implies $$F\left(\frac{\sqrt{x}}{x}\right)=x\sqrt{x} \quad\to\quad F\left(\frac{1}{\sqrt{x}}\right)=x\sqrt{x}=x^{3/2}$$

Hence $$F\left(\frac{y}{x}\right)=\frac{x^3}{y^3}$$

therefore by equation $(1)$

$$\frac{x^3}{y^3}=xy-u^2 \quad\to\quad u^2=xy - \frac{x^3}{y^3}$$

Thus, the solution of the PDE is $$u^2(x,y)=xy - \frac{x^3}{y^3}$$

Answer 3

Your system of characteristic ODEs is correct. And you correctly found a first characteristic equation $\frac{y}{x}=c_1$ .

The mistake is in : $$\frac{dx}{xu}=\frac{du}{xy}\to\frac{dx}{u}=\frac{du}{y}\implies\phi_2=2yx-u^2=c_2$$ because you cannot integrate $2ydx=2udu$

OK $\int 2udu=u^2+$constant.

But $\int 2ydx\neq 2yx$ because $y$ is not constant with respect to $x$.

If you want to use the equation $\frac{dx}{xu}=\frac{du}{xy}$ or equivalently $ydx=udu$ :

with the first characteristic equation $y=c_1x$ then :

$$c_1xdx=udu$$ $$u^2-c_1x^2=c_2$$ $$u^2-\frac{y}{x}x^2=c_2$$ So, a second characteristic equation is : $$u^2-xy=c_2$$ I suppose that you can take it from here.