Quasi-Linear PDE

Question

I'm trying to solve the following PDE:

$$\begin{align}xu(x, y)\partial_xu(x, y) + yu(x, y)\partial_yu(x, y) &= x^2 + y^2 + (u(x, y))^2\\ u(1, y) &= y^2\end{align}$$

I wrote it in the form $$xv\partial_xv + yv\partial_yv + x^2+y^2+v^2 = 0$$ then I wrote $$\begin{align}x'(t) &= x(t)v(t)\\ y'(t) &= y(t)v(t)\\ v'(t) &= x(t)^2 + y(t)^2+ v(t)^2\end{align}$$

I need two first integrals; one was easy by taking $x'y - y'x = 0$ and I got $$\Phi_1(x,y) = -x(t)y(t)$$ but I'm not succeeding finding the second, can anyone help please? thank you.

Answer 1

$$x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=\frac{x^2+y^2+u^2}{u}$$ The Charpit-Lagrange system of characteristic ODEs is : $$\frac{dx}{x}=\frac{dy}{y}=\frac{u}{x^2+y^2+u^2}du$$ A first characteristic equation comes from $\frac{dx}{x}=\frac{dy}{y}$ : $$\frac{y}{x}=c_1$$ A second characteristic equation comes from $\frac{dx}{x}=\frac{u}{x^2+y^2+u^2}du=\frac{u}{x^2+(c_1x)^2+u^2}du$

Solve this ODE for $u(x)$. Hint : Let $X=x^2$ and $U=u^2$ . The result is : $$\frac{u^2}{x^2}-2(1+c_1^2)\ln|x|=c_2$$ $$\frac{u^2}{x^2}-2\left(1+\left(\frac{y}{x}\right)^2\right)\ln|x|=c_2$$ The general solution of the PDE expressed on implicit form $c_2=F(c_1)$ is : $$\frac{u^2}{x^2}-2\left(1+\left(\frac{y}{x}\right)^2\right)\ln|x|=F\left(\frac{y}{x}\right)$$ $F$ is an arbitrary function in the general case. ($F$ must be determined in order to satisfy the boundary condition). $$u^2=2\left(x^2+y^2\right)\ln|x|+x^2 F\left(\frac{y}{x}\right)$$ Condition : $u(1,y)=y^2$

$u(1,y)^2=y^4=2\left(1+y^2\right)\ln|1|+1^2 F\left(\frac{y}{1}\right)\quad\implies\quad F(y)=y^4$ for any variable $y$. So the function $F$ is determined : $$F(\chi)=\chi^4$$ We put it into the above general solution where $\chi=\frac{y}{x}$ :

$$u^2=2\left(x^2+y^2\right)\ln|x|+x^2 \left(\frac{y}{x}\right)^4$$ $$u(x,y)=\sqrt{2\left(x^2+y^2\right)\ln|x|+\frac{y^4}{x^2}}$$

Answer 2

If we write the equation in polar coordinates using

$$x\partial_x + y\partial_y = r\partial_r$$

we get the following expression

$$ru\partial_r u = r^2 + u^2$$

which effectively makes this an ODE. Now use the substitution $u = rv$

$$r^2 v (v + r\partial_r v) = r^2+r^2v^2 \implies v\partial_r v = \frac{1}{r}$$

Integrating both sides gives

$$v^2 = \log( r^2 )+ f(\theta) \implies u^2 = r^2\log (r^2f(\theta))$$

Now we can plug in the initial conditions to find $f(\theta)$. Plugging in via brute force in Cartesian coordinates will not get us anywhere. Instead we'll have to be clever in polar coordinates by noticing the following relationships:

$$u = r^2\sin^2\theta \hspace{20 pt} r\cos\theta = 1$$

Using both of them gets us

$$r^2\sin^4\theta = \log \left(r^2f(\theta)\frac{\cos^2\theta}{\cos^2\theta}\right) = \log\left(\frac{f(\theta)}{\cos^2\theta}\right)$$

in order to get rid of the $r$, we need equal powers of $r$ on both sides to cancel. To do this we'll utilize the second relationship again

$$r^2\sin^4\theta = 1^2 \log\left(\frac{f(\theta)}{\cos^2\theta}\right) = r^2\cos^2\theta \log\left(\frac{f(\theta)}{\cos^2\theta}\right)$$

$$ \implies f(\theta) = \cos^2\theta \exp\left(\frac{\sin^4\theta}{\cos^2\theta}\right)$$

Plugging in, we get our answer in polar coordinates to be

$$u^2 = r^2\log(r^2\cos^2\theta) + r^2\sin^2\theta\frac{\sin^2\theta}{\cos^2\theta}$$

which in Cartesian is equal to

$$u^2 = (x^2+y^2)\log(x^2) + \frac{y^4}{x^2}$$

Answer 3

Calling $v = u^2$ we have

$$ xv_x+y v_y = 2(v+x^2+y^2),\ \ \ v(1,y) = y^4 $$

now changing variables according

$$ \cases{ r^2= x^2+y^2\\ \theta=\frac yx } $$

we have $v(x,y) = w(r,\theta)$

and

$$ r w_r - 2w=2r^2 $$

this equation is solved as a ODE and has as solution

$$ w(r,\theta) = 2r^2\ln r+r^2\phi(\theta) $$

now considering the condition

$$ v(1,y) = w\left(\sqrt{1+y^2}, y\right)=y^4\Rightarrow \phi\left(\frac yx\right) = \frac{y^4}{x^2 \left(x^2+y^2\right)}-\ln \left(1+\frac{y^2}{x^2}\right) $$

and finally

$$ v(x,y) = u^2(x,y) = \frac{y^4}{x^2}+\left(x^2+y^2\right) \ln \left(x^2+y^2\right)-\left(x^2+y^2\right) \ln \left(1+\frac{y^2}{x^2}\right) = \frac{y^4}{x^2}+(x^2+y^2)\ln(x^2) $$