I've already asked on a forum but can't find a global solution on $\mathbb{R}$.
Let $f \in C^{1}(R^{2}, R)$. We suppose the existence of $(x_{0}, y_{0}) \in R^{2}$ with $$f(x_{0}, y_{0}) = 0(1)$$ and we also suppose $|\frac{\partial f}{\partial y}| > |\frac{\partial f}{\partial x}|$. In particular, $\frac{\partial f}{\partial y}$ is not zero in $(x_{0}, y_{0})$, so by the implicit function theorem we find an open neighbourhood $U$ of $x_{0}$ and an open neighbourhood $V$ of $y_{0}$ and $\phi \in C^{1}(U, V)$ with $\forall x \in U, \phi'(x) = \frac{\frac{\partial f}{\partial x}(x, \phi(x))}{\frac{\partial f}{\partial y}(x, \phi(x))} = F(x, \phi(x))$, where $F = \frac{\partial_{x}f}{\partial_{y}f} \in C^{0}(R^{2}, R)$. This is equivalent to saying that $\phi(x_{0}) = y_{0}$ and $\forall x \in U, f(x, \phi(x)) = 0$ (because of (1)).
I'm looking for a $\phi$ which is a $C^{1}$ solution, but defined on all $\mathbb{R}$ (not just on U). Have you got an idea please?
There are theorems saying that a maximum solution is necessarily defined on all $\mathbb{R}$ (otherwise it diverges near the frontier), but $F$ is not a local Lipschitzian function.
Thank you in advance and have a nice afternoon.
First, the correct right hand side of the differential equation should be $$F(x,y) = - \, \frac{\frac{\partial f}{\partial x}(x,y)}{\frac{\partial f}{\partial y}(x,y)},$$ where due to the inequality $$\left| \frac{\partial f}{\partial y}(x,y) \right| > \left| \frac{\partial f}{\partial x}(x,y) \right| \geq 0$$ the partial derivative $\frac{\partial f}{\partial y}(x,y)$ is never zero, so it is always positive or always negative.
The existence and uniqueness of the solution of the initial value problem for the differential equation $$\frac{d\phi}{dx} = F(x,\phi)$$ $$\phi(x_0)=y_0$$ for any $(x_0,y_0) \in \mathbb{R}^2$ comes from the existence and uniqueness of the function $\phi(x)$ defined as $$f(x,\phi(x)) = f(x_0,y_0)$$ by the Implicit Function Theorem. That is why despite the fact that $F(x,y)$ is just a continuous function, the differential equation $\phi'=F(x,ϕ)$ still has a unique local solution for any initial value. For that reason we have existence and uniqueness without any local Lipschitz property for F. Therefore, the theorem of extension of solutions in a compact set can be applied to this case.
More specifically, the latter theorem Implies that if $C$ is a compact set of $\mathbb{R}^2$ and $(x_0,y_0) \in C$, then the unique solution $\phi(x)$ satisfying the initial value problem $$\frac{d \phi}{dx}(x) = F(x,\phi(x))$$ $$\phi(x_0) = y_0$$ is defined on an interval $(a-\varepsilon,b+\varepsilon) \ni x_0$ so that $(x,\phi(x)) \in C$ for all $x \in [a,b]$ and $(x,\phi(x)) \in \mathbb{R}^2\setminus C$ for all $x \in (a-\varepsilon, a) \cup (b, b+\varepsilon)$.
In your case, the inequality between the absolute values of the partial derivatives of $f$ implies that $|F(x,y)| < 1$ for all points $(x,y) \in \mathbb{R}^2$, or in other words $-1 < F(x,y) < 1$. Therefore we can construct the compact set $$C_T = \{(x,y) \in \mathbb{R}^2 \, | \, -T \leq x \leq T \,\, \text{ and } \,\, y_0 - (x-x_0) \leq y \leq y_0 + (x - x_0) \}$$
Since $(x_0,y_0) \in C_T$, the solution $\phi(x)$ of the initial value problem defined above has to be defined on some interval $(a-\varepsilon, b+\varepsilon)$ so that for $x \in [a,b]$ it stays in the compact $C_T$ until it reaches its boundary and then for $x \in (a-\varepsilon, a) \cup (b, b+\varepsilon)$ it leaves $C_T$. Clearly the points $(a,\phi(a))$ and $(b,\phi(b))$ are at the boundary of $C_T$. Now, if we assume that say $(b,\phi(b))$ lies on the part of the boundary given by the equation $y = y_0 + (x-x_0)$, that is $\phi(b) = y_0 + (b-x_0)$, then $\phi'(b) \geq 1$ which means $\phi'(b) = F(b,\phi(b)) \geq 1$ which is a contradiction. Absolutely analogously we can show that the $(b,\phi(b))$ cannot be on the boundary defined by the equation $y = y_0 - (x-x_0)$. Therefore $(b,\phi(b))$ should be on the boundary $x = T$, which means that $b=T$. Absolutely analogously we can show that $a=-T$. Thus, the solution $\phi(x)$ is defined on the interval $[-T,T]$ for any real number $T>0$. Therefore, $\phi(x)$ is defined on the whole real line, i.e. for $x \in (-\infty,\infty) = \mathbb{R}$.