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

Theorem
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:


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

Then one and only one integral curve of the equation $y'' = \map F {x, y, y'}$ passes through any two points $\tuple {a, A}$ and $\tuple {b, B}$ such that $a \ne b$.

Lemma 3
Consider a plane with axes denoted by $x$ and $y$:


 * BernsteinCurve.png

Put the point $A \tuple {a, a_1}$.

Through this point draw an arc of the integral curve such that $\map {y'} a = 0$.

On this arc put another point $D \tuple {d, d_1}$.

For $x \ge d$ draw the straight line $y = d_1$.

Put the point $B \tuple {b, b_1}$.

For $y \ge d_1$ draw the straight line $x = b_1$.

Denote the intersection of these two straight lines by $Q$.

Then the broken curve $DQB$ connects points $D$ and $B$.

Choose any point of $DQB$ and denote it by $P \tuple {\xi, \xi_1}$.

Consider a family of integral curves $y = \map \phi {x, \alpha}$, passing through the point $A$, where $\alpha = \map {y'} a$.

For $\alpha = 0$ the integral curve concides with $AD$.

Suppose point $P$ is sufficiently close to the point $D$.

By Lemma 1, there exists a unique curve $AP$.

Then, $\alpha$ can be found uniquely from:


 * $d_1 = \map \phi {\xi, \alpha}$.

Due to uniqueness and continuity, it follows that $\xi$ is a monotonic function of $\alpha$.

Hence, $\alpha$ is a monotonic function of $\xi$.

Put the point $R$ in between of $D$ and $Q$.

Suppose, that, except for $R$, any point of $DR$ can be reached by the aforementioned procedure.

When $\xi$ approaches the abscissa $r$ of $R$, $\alpha$ monotonically approaches a limit.

If it is different from $\pm \dfrac \pi 2$, point $R$ is attained.

By assumption, $R$ is not attained.

Thus:


 * $\ds \lim_{\xi \mathop \to r} \alpha = \pm \dfrac \pi 2$

In other words, as $P$ approaches $R$, the derivative of $\map y x$ joining $A$ to $P$ will not be bounded at $x = a$.

This contradicts the bounds from Lemma 2 and 3, and the fact that the difference of abscissas of $A$ and $P$ does not approach $0$.

Therefore, $R$ can be reached.

Similar argument can be repeated for the line segment $QB$.