# 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?

• If you let $y(x) = g'(x)/g(x)$, you get Airy's equation. Commented Jul 14, 2022 at 0:14

$$\frac {\partial u}{\partial x}+\left(x-y^2\right)\frac {\partial u}{\partial y}=0.$$ 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.

• You are right. I corrected my answer. This doesn't change the conclusion. Commented Jul 21, 2022 at 7:39
• @Analyst_311419 . Of course in the general case. But presently the question isn't clear enough about the boundary conditions. Moreover it seems that the wording of the question was modified after my answer making it partly obsolete. Commented Oct 17, 2022 at 9:34