# Solution to differential equation and implicit function

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.

I know the theorem you mention but he need the local Lipschitz condition(as I try to say sorry I'm not English.). which is unfonutunatly not the case in the proplem ennoncee(ther's no more information than the equality.).

Do you think thers an error of ennoncee please? Or a lack of hypohtesis on $F$.

• No, I don't think that there is an error or lack of hypotheses. The existence and uniqueness of solutions of the differential equation comes from the existence and uniqueness of the function defined by the Implicit Function Theorem. That is why despite the fact that $F(x,y)$ is just a continuous function, the differential equation $\phi' = F(x,\phi)$ 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 applies again. Commented Aug 14, 2016 at 18:39
• See the edits in my answer. Commented Aug 14, 2016 at 19:20

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 $x \in (-\infty,\infty) = \mathbb{R}$.

"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 applies again."

Have you got a proof of this please?

Thank you in advance and have a nice morning.