PDE: What is the most general solution of $F_xF_y=1$ for a real function, $F(x,y)$?

Question

WolframAlpha gives the simple solution, $F(x,y)=cx+\dfrac{y}{c}+c'$ with two constants $c$ and $c'$ .

Is this the most general solution?

Answer 1

It is true that the only $C^2$-smooth global (defined on $\mathbb R^2$) solution is linear. But there are many more solutions defined on proper subdomains, such as halfplanes.

Since the derivatives $F_x$, $F_y$ do not vanish, they have constant sign. Let's assume they are positive and introduce a function $u$ such that $F_x=e^u$ and $F_y=e^{-u}$. The only constraint on $u$ is that the field $e^u\vec \imath +e^{-u}\vec\jmath$ is conservative. In differential form (this is where I need $F\in C^2$) this yields $u_ye^u=-u_x e^{-u}$, or $$e^{-u} u_x+e^{u}u_y =0 \tag1$$ This is another first order PDE, but it's better than the one we started with: it is quasilinear and as such, can be treated with the method of characteristics.

Indeed, consider a level curve $\Gamma = \{u=c\}$. According to (1), the vector $e^{-c} \vec\imath+e^{c}\vec\jmath$ is orthogonal to $\nabla u$ on $\Gamma$, and therefore is parallel to $\Gamma$. That is to say, $\Gamma$ is a line with slope $e^{2c}$. Since level curves cannot intersect, we see that any global solution of (1) must be constant, which corresponds to $F$ being linear.

On the other hand, for any smooth decreasing function $g:\mathbb R\to\mathbb R $ we have a solution of (1) in the upper half-plane such that $u(x,0)=g(x)$, because the characteristic lines $y=e^{2g(t)}(x-t)$ do not intersect in the upper half-plane. From $u$ we recover $F$ by integration, as in multivariable calculus. Since $F_x(x,0)= e^{g(x)}$, we can arrange for $F$ to be any $C^2$ concave increasing function on the $x$-axis.

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I think that the same conclusion about global solutions (i.e., they are all linear) is true under mere $C^1$ assumption (which is the natural degree of regularity here). Maybe something along the lines of this question about $F_xF_y=0$ can work.

Answer 2

With reference to http://eqworld.ipmnet.ru/en/solutions/fpde/fpde3301.pdf, I have another nice approach:

$F_xF_y=1$

$F_y-\dfrac{1}{F_x}=0$

$F_{xy}+\dfrac{F_{xx}}{F_x^2}=0$

Let $u=F_x$ ,

Then $u_y+\dfrac{u_x}{u^2}=0$

$u_x+u^2u_y=0$

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{dx}{dt}=1$ , letting $x(0)=0$ , we have $x=t$

$\dfrac{du}{dt}=0$ , letting $u(0)=u_0$ , we have $u=u_0$

$\dfrac{dy}{dt}=u^2=u_0^2$ , letting $y(0)=g(u_0)$ , we have $y=u_0^2t+g(u_0)=u^2x+g(u)$ , i.e. $u=G(u^2x-y)$

But $u(x,y)$ is difficult to express explicitly, so does for $F(x,y)$ .