# Bernstein's Theorem on Unique Global Solution to y''=F(x,y,y')/Lemma 1

## Lemma for Bernstein's Theorem on Unique Global Solution to $y'' = \map F {x, y, y'}$

Let $F$ and its partial derivatives $F_y, F_{y'}$ be real functions, defined on the closed interval $I = \closedint a b$.

Let $F, F_y, F_{y'}$ be continuous at every point $\tuple {x, y}$ for all finite $y'$.

Suppose there exists a constant $k > 0$ such that:

$\map {F_y} {x, y, y'} > k$

Suppose there exist real functions $\alpha = \map \alpha {x, y} \ge 0$, $\beta = \map \beta {x, y}\ge 0$ bounded in every bounded region of the plane such that:

$\size {\map F {x, y, y'} } \le \alpha y'^2 + \beta$

Let $L$ be an integral curve of equation $y'' = \map F {x, y, y'}$ which passes through the two points $\tuple {a, A}$ and $\tuple {b, B}$ such that $a \ne b$.

Then $L$ is unique.

## Proof

Aiming for a contradiction, suppose there are two integral curves $y = \map {\phi_1} x$ and $y = \map {\phi_2} x$ such that:

$\map {y''} x = \map F {x, y, y'}$

Define a mapping $\delta: I \to S \subset \R$:

$\map \delta x = \map {\phi_2} x - \map {\phi_1} x$

From definition it follows that:

$\map \delta a = \map \delta b = 0$

Then the second derivative of $\delta$ yields:

 $\ds \map {\delta''} x$ $=$ $\ds \map {\phi_2''} x - \map {\phi_1''} x$ Definition of $\delta$ $\ds$ $=$ $\ds \map F {x, \phi_2, \phi_2'} - \map F {x, \phi_1, \phi_1'}$ as $y'' = \map F {x, y', y''}$ $\ds$ $=$ $\ds \map F {x, \phi_1 + \delta, \phi_1' + \delta'} - \map F {x, \phi_1, \phi_1'}$ Definition of $\delta$ $\text {(1)}: \quad$ $\ds$ $=$ $\ds \map \delta x F_y^* + \map {\delta'} x F_{y'}^*$ Multivariate Mean Value Theorem with respect to $y, y'$

where:

$F_y^* = \map {F_y} {x, \phi_1 + \theta \delta, \phi_1' + \theta \delta'}$
$F_{y'}^* = \map {F_{y'} } {x, \phi_1 + \theta \delta, \phi_1' + \theta \delta'}$

and $0 < \theta < 1$.

Suppose:

$\forall x \in I: \map {\phi_2} x \ne \map {\phi_1} x$

Then there are two possibilities for $\delta$:

$\map \delta x$ attains a positive maximum within $\tuple {a, b}$
$\map \delta x$ attains a negative minimum within $\tuple {a, b}$.

Denote this point by $\xi$.

Suppose $\xi$ denotes a maximum.

Then:

$\map {\delta''} \xi \le 0$
$\map \delta \xi > 0$
$\map {\delta'} \xi = 0$

These, together with $(1)$, imply that $F_y^* \le 0$.

This is in contradiction with assumption.

For the minimum the inequalities are reversed, but the last equality is the same.

Therefore, it must be the case that:

$\map {\phi_1} x = \map {\phi_2} x$

$\blacksquare$