Solve $y(u_x^2 - u_y^2) + uu_y = 0$ for $u(x,y)$ if $u(2y,y) = 3y$ Consider the PDE
$$\color{blue}{y(u_x^2 - u_y^2) + uu_y = 0}$$
with the initial condition
$$\color{blue}{u(x,y) = 3y \quad \text{on} \quad \Sigma := \{(x,y) \in \Bbb R^2: x - 2y = 0\}}$$

Work: The method of characteristics for a fully non-linear PDE $$F(p,z,x) = 0$$
where $u: \Omega \subset \Bbb R^n \to \Bbb R$, $p = \nabla u$ and $z = u$ gives us the following set of $(2n+1)$ equations:
$$\frac{\mathrm{d}x(s)}{\mathrm{d}s} = \nabla_pF \tag{1}$$
$$\frac{\mathrm{d}p(s)}{\mathrm{d}s} = -p(s) F_z - \nabla_x F \tag{2}$$
$$\frac{\mathrm{d}z}{\mathrm{d}s} = p(s) \cdot \nabla_p F \tag{3}$$
So, in our case, for the given PDE, we have $$F(p,q,z,x,y) := y(p^2 - q^2) + zq$$
where $p = u_x, q = u_y$, and $z = u(x,y)$. The equations are
$$\frac{\mathrm{d}x(s)}{\mathrm{d}s} = 2py \tag{4}$$
$$\frac{\mathrm{d}y(s)}{\mathrm{d}s} = z - 2qy \tag{5}$$
$$\frac{\mathrm{d}p(s)}{\mathrm{d}s} = -pq \tag{6}$$
$$\frac{\mathrm{d}q(s)}{\mathrm{d}s} = -p^2 \tag{7}$$
$$\frac{\mathrm{d}z(s)}{\mathrm{d}s} = 2y(p^2 - q^2) + qz \tag{8}$$
From $(6)$ and $(7)$, we obtain $$\frac{\mathrm{d}p}{\mathrm{d}q} = \frac{q}{p}$$
so that $$p^2 - q^2 = c_1 = \text{constant}$$
The PDE reduces to $$F(p,q,z,x,y) = yc_1 + zq = 0$$
and some quick calculation tells us that $$q = -\frac{yc_1}{z} \quad\text{and}\quad p = \pm \left(c_1 + \frac{y^2 c_1^2}{z^2}\right)^{1/2}$$
Further, $(1)$ and $(3)$ (which is $(8)$ in our case) give
$$\frac{\mathrm{d}z(s)}{\mathrm{d}s} = (p, q)\cdot \left(\frac{\mathrm{d}x(s)}{\mathrm{d}s}, \frac{\mathrm{d}y(s)}{\mathrm{d}s} \right) = \left(\pm \left(c_1 + \frac{y^2 c_1^2}{z^2}\right)^{1/2}, -\frac{yc_1}{z} \right)\cdot \left(\frac{\mathrm{d}x(s)}{\mathrm{d}s}, \frac{\mathrm{d}y(s)}{\mathrm{d}s} \right)$$
$$\implies \frac{\mathrm{d}x(s)}{\mathrm{d}s} = \frac{\pm 1}{(z^2c_1 + y^2c_1^2)^{1/2}} \left(\frac{z\mathrm{d}z(s)}{\mathrm{d}s} + \frac{c_1y\mathrm{d}y}{\mathrm{d}s} \right)$$
which succumbs to a change of variables $w = (z^2c_1 + y^2c_1^2)^{1/2}$ and $2w \mathrm{d}w = 2zc_1 \mathrm{d}z + 2yc_1^2 \mathrm{d}y$, simplifying to
$$\frac{\mathrm{d}x(s)}{\mathrm{d}s} = \frac{\pm 1}{wc_1} \cdot \frac{w\mathrm{d}w(s)}{\mathrm{d}s} = \frac{\pm 1}{c_1} \cdot \frac{\mathrm{d}w(s)}{\mathrm{d}s}$$
After integration, we have
$$z^2 = (u(x,y))^2 = c_1(x + c_2)^2 - y^2 c_1$$
Lastly, using the initial data, $c_1  =3$ and $c_2 = 0$. Therefore, the PDE is solved by
$$u^2(x,y) = 3(x^2 - y^2)$$
which makes sense (since we consider real-valued solutions only) in $$\Omega := \{(x,y) \in \Bbb R^2: |x| > |y|\}.$$

Questions:

*

*Is my solution correct? If yes, is (i) it unique? (ii) $\Omega$ the maximal subset of $\Bbb R^2$ in which the given PDE admits a solution? I would guess so.

*In several places, I have divided by $z, p$, and $q$, which means that I've assumed the said quantities do not vanish. To what extent is this okay to do? Does this mean that I am missing out on some solutions? It is worth noting that $u(x,y)$ as obtained above is indeed nowhere-vanishing on $\Omega$.

*I'd love to know if there are any other (possibly simpler or complicated) ways to go about solving this PDE.

Thank you!
 A: The general equation is
$$
y(u_x^2-u_y^2)+u u_y=0.\tag 1
$$
Write it in the form
$$
y\left(\left(\frac{u_x}{u}\right)^2-\left(\frac{u_y}{u}\right)^2\right)+\frac{u_y}{u}=0
$$
Hence if we set $F=\log u$, then
$$
y(F_x^2-F_y^2)+F_y=0
$$
Next set $F_x=f(x,y)$, then
$$
F_y=\frac{1\pm\sqrt{1-4f(x,y)^2y^2}}{2y}
$$
and the integrability condition is $F_{xy}=F_{yx}$, from which we get
$$
f_y=\pm\frac{2yf\cdot f_x}{\sqrt{1-4y^2f^2}}\tag 2
$$
The solution of this last equation is
$$
g\left(f(x,y),\frac{2xf(x,y)\pm e\sqrt{1-4 y^2 f(x,y)^2}}{4ef(x,y)^2}\right)=c,\tag 3
$$
where $g(x,y)$ is any smooth function. Hence for any $g(x,y)$ if we solve (3) with respect to $f(x,y)$, then always $f(x,y)$ will be a solution of (2). By this way
$$
F_x=f(x,y)\Leftrightarrow F(x,y)=\int f(x,y)dx+g_1(y),
$$
where $g_1(y)$ is any function. Finaly
$$
u(x,y)=\exp{(F(x,y))}=\exp\left(\int^{x}_{c}f(t,y)dt+g_1(y)\right),\tag 4
$$
where $g_1(y)$ is any function and $f(x,y)$ such that
$$
g\left(f(x,y),\frac{2xf(x,y)\pm e\sqrt{1-4 y^2 f(x,y)^2}}{4ef(x,y)^2}\right)=c\tag 5
$$
and $g(x,y)$ is also any function and $e=\pm 1$, $c-$constant (any) .
I think in a particular problem you can set the conditions.
Examples of solutions are with $g(x,y)=x y$, $g(x,y)=x^2 y$, $g(x,y)=x+y$,$\ldots$ etc. You must also carefull which branch cut you are using.
