Is there a C^1 function that satisfies certain conditions?

[Calabi]

Homework Statement

Let be $f \in C^{1}(\mathbb{R}^{2}, \mathbb{R})$ and $(x_{0} , y_{0}) \in \mathbb{R}^{2} / f(x_{0}, y_{0}) = 0$. I suppose $|\frac{\partial f}{\partial y}| \ge |\frac{\partial f}{\partial x}|(2)$.
I'm looking for $\phi \in C^{1}(\mathbb{R}, \mathbb{R}) / \forall x \in \mathbb{R} f(x, \phi(x)) = 0$ and $\phi(x_{0}) = y_{0}$.

Homework Equations

$(x_{0} , y_{0}) \in \mathbb{R}^{2} / f(x_{0}, y_{0}) = 0$
$|\frac{\partial f}{\partial y}| \ge |\frac{\partial f}{\partial x}|$

The Attempt at a Solution

If a such solution exists we necssarly have $\forall x \in \mathbb{R}, \phi'(x) \frac{\partial f}{\partial y}(x, \phi(x)) + \frac{\partial f}{\partial x}(x, \phi(x)) = 0 (1)$.
We can stated $x \rightarrow (x, \phi(x))$ is injective so if we look its image we perhaps make a variable changement in (1). But I don't think it's usefull.
And I don't see how the majoration (2) is usefull.

 

[Ray Vickson]

Homework Statement

Let be $f \in C^{1}(\mathbb{R}^{2}, \mathbb{R})$ and $(x_{0} , y_{0}) \in \mathbb{R}^{2} / f(x_{0}, y_{0}) = 0$. I suppose $|\frac{\partial f}{\partial y}| \ge |\frac{\partial f}{\partial x}|(2)$.
I'm looking for $\phi \in C^{1}(\mathbb{R}, \mathbb{R}) / \forall x \in \mathbb{R} f(x, \phi(x)) = 0$ and $\phi(x_{0}) = y_{0}$.

Homework Equations

$(x_{0} , y_{0}) \in \mathbb{R}^{2} / f(x_{0}, y_{0}) = 0$
$|\frac{\partial f}{\partial y}| \ge |\frac{\partial f}{\partial x}|$

The Attempt at a Solution

If a such solution exists we necssarly have $\forall x \in \mathbb{R}, \phi'(x) \frac{\partial f}{\partial y}(x, \phi(x)) + \frac{\partial f}{\partial x}(x, \phi(x)) = 0 (1)$.
We can stated $x \rightarrow (x, \phi(x))$ is injective so if we look its image we perhaps make a variable changement in (1). But I don't think it's usefull.
And I don't see how the majoration (2) is usefull.

It might not be "usefull", but it could be "useful". Anyway, if the inequality $|f_y| >|f_x| \; \forall (x,y)$ were strict (which your given one is not), then you would be allowed to divide by $f_y$, and so get the differential equation
\phi'(x) = -F(x,\phi(x)), \; F(x,y) \equiv \frac{f_x(x,y)}{f_y(x,y)}
If you know some theorems about the existence and smoothness of solutions to differential equations, you could use those in this problem.

You would still need to worry about what happens if the inequality is not strict and there are points where $f_x(x,y) = f_y(x,y) = 0$.

If you do not have access to theorems about differential equations, you would need another method.

 

[Calabi]

Hello : indeed the inequality is trict my beg. So the equation is true.
Anyway I was thinking of https://en.wikipedia.org/wiki/Peano_existence_theorem .
With the condition $\phi(x_{0}) = y_{0}$. But the theorem give a local solution.
Not on $\mathbb{R}^{2}$. And I don't use $f(x_{0}, y_{0}) = 0$.

 

[Ray Vickson]

Hello : indeed the inequality is trict my beg. So the equation is true.
Anyway I was thinking of https://en.wikipedia.org/wiki/Peano_existence_theorem .
With the condition $\phi(x_{0}) = y_{0}$. But the theorem give a local solution.
Not on $\mathbb{R}^{2}$. And I don't use $f(x_{0}, y_{0}) = 0$.

Yes, you do use $f(x_0,y_0) = 0$; do you see where?

 

[Calabi]

In the Azela Peano theorem you just look for condition on $\phi$. So I don't see.
And I still have a local solution.

 

[Calabi]

So is there a general way to solve $\phi'(x) = F(x, \phi(x))$ on all $\mathbb{R}^{2}$ please?

 

[Calabi]

If $g : x \rightarrow f(x, \phi(x))$ is nul with $\phi(x_{0}) = y_{0}$ then we get the equation you wroght.
If we got the equation you wroght then g is constant and as $\phi(x_{0}) = y_{0}$ and $f(x_{0}, y_{0}) = 0$ then g is constant.
But I still have a local solution.