Solving first order nonlinear PDE with method of characteristics I wonder how to solve the following PDE:
$$
\frac {\partial u}{\partial x}+\left(x-y^2\right)\frac {\partial u}{\partial y}=0.
$$
I tried the method of characteristics, but it seems like the following Riccati equation
$$
\frac {d y}{dx}=x-y^2
$$
is hard to solve by hand. I've tried it via WoloframAlpha, and the solution involves Bessel function.
I'm thinking of that if $\varphi(x,y)=C$ solves the above Riccati equation, then the general solution will be $u(x,y)=g(\varphi(x,y))$, where $g$ is any differentiable function of one variable.
If I've figured out what $\varphi(x,y)$ is, and have given the condition $u(x,y)\to 1$ as $x,y\to \infty$. Is there any possible additonal condition can help me to decide what $g$ is?
 A: $$
\frac {\partial u}{\partial x}+\left(x-y^2\right)\frac {\partial u}{\partial y}=0.
$$
There is no big difficulty to solve the PDE for the general solution (without the boundary conditions). The difficulty appears with the special conditions that you specify.
The Charpit-Lagrange characteristic OQEs are :
$$\frac{dx}{1}=\frac{dy}{x-y^2}=\frac{du}{0}$$
A first characteristic equation ( implied by $du=0$ ) obviously is :
$$u=c_1$$
A second characteristic equation comes from solving $\frac{dx}{1}=\frac{dy}{x-y^2}$ leadind to :
$$\frac{\text{Ai}'(x)-y\;\text{Ai}(x)}{\text{Bi}'(x)-y\;\text{Bi}(x)}=c_2$$
The solution of the PDE on form of implicit equation  $c_1=F(c_2)$ explicitly leads to :
$$\boxed{u(x,y)=F\left(\frac{\text{Ai}'(x)-y\;\text{Ai}(x)}{\text{Bi}'(x)-y\;\text{Bi}(x)} \right)}$$
$F$ is an arbitray function.
Ai$(x)$ and $Bi$(x) are the Airy functions. https://mathworld.wolfram.com/AiryFunctions.html
$\text{Ai}'(x)=\frac{d}{dx}\text{Ai}(x)\quad$ and $\quad\text{Bi}'(x)=\frac{d}{dx}\text{Bi}(x)$
The difficulty is to find a function $F$ such as the specified conditions be satisfied.
Condition $u(x,y)=1\text{ as }x,y\to +\infty$
$\text{Ai}(x)\to 0\quad;\quad \text{Ai}'(x)\to 0\quad;\quad\text{Bi}(x)\to \infty\quad;\quad\text{Bi}'(x) \to \infty.$
$$\frac{\text{Ai}'(x)-y\;\text{Ai}(x)}{\text{Bi}'(x)-y\;\text{Bi}(x)} \to 0.$$
Fuctions such as $F(0)=1$ are convenient. Thus infinity many functions can be chosen to satisfy the first condition alone.
Condition : $u(x,y)=0\text{ as }y\to -\infty\text{ and }x\text{ bounded above}$
As $y\to -\infty\quad$ all $\quad\text{Ai}(x)$ , $\text{Ai}'(x)$ , $\text{Bi}(x)$ , $\text{Bi}'(x)$ are bounded with $x>0$ bounded.
$$\frac{\text{Ai}'(x)-y\;\text{Ai}(x)}{\text{Bi}'(x)-y\;\text{Bi}(x)} \sim \frac{\text{Ai}(x)}{\text{Bi}(x)}$$
The condition requires $F\left(\frac{\text{Ai}(x)}{\text{Bi}(x)} \right)=0\quad$ any $x$. Thus $F$ should be the constant function equal zero.
The two conditions appear contradictory which seems to lead to the conclusion : "No solution of the problem with the specified conditions". This remains to be checked. A much more thorough study would be necessary.
