Bernstein's Theorem on Unique Global Solution to y''=F(x,y,y')/Lemma 3

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:


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

Suppose that for $x \in I$:
 * $\map {y''} x = \map F {x, y, y'}$

where:
 * $\map y a = a_1$
 * $\map y c = c_1$

Then:
 * $\forall x \in I: \exists M \in \R: \size {\map {y'} x - \dfrac {c_1 - a_1} {c - a} } \le M$

Proof
Let $\AA$ and $\BB$ be the least upper bounds of $\map \alpha {x, y}$ and $\map \beta {x, y}$ respectively in the rectangle $a \le x \le c$, $\size y \le m + max \set {\size a_1, \size c_1}$

where:


 * $m = \dfrac 1 k \max \limits_{a \mathop \le x \mathop \le b} \size {\map F {x, \dfrac {a_1 \paren {c - x} + c_1 \paren {x - a} } {c - a}, \dfrac {c_1 - a_1} {c - a} } }$

Suppose that $\AA \ge 1$.

Let $u, v$ be real functions such that:

From Lemma 2, for $x \in I$ the s of $(3)$ and $(4)$ are not negative.

Thus:


 * $\forall x \in I: u, v \ge 1$

Differentiate equations $(3)$ and $(4)$ $x$:

Differentiate again:

By assumption:

where:


 * $\BB_1 = \BB + 2 \AA \paren {\dfrac {c_1 - a_1} {c - a} }^2$

Then:

Multiply the inequality by $2 \AA u$ and simplify:

Similarly:

Multiply the inequality by $-2 \AA v$ and simplify:

Note that $\map u a = \map u c$ and $\map v a = \map v c$.

From Intermediate Value Theorem it follows that


 * $\exists K \subset I: \forall x_0 \in K: \map {y'} {x_0} - \dfrac {c_1 - a_1} {c - a} = 0$

Points $x_0$ divide $I$ into subintervals.

Due to $(5)$ and $(6)$ both $\map {u'} x$ and $\map {v'} x$ maintain sign in the subintervals and vanish at one or both endpoints of each subinterval.

Let $J$ be one of the subintervals.

Let functions $\map {u'} x$, $\map {v'} x$ be zero at $\xi$, the right endpoint.

The quantity:


 * $\map {y'} x - \dfrac {c_1 - a_1} {c - a}$

has to be either positive or negative.

Suppose it is positive in $J$.

From $(5)$, $u'$ is not negative.

Multiply both sides of $(10)$ by $u'$:


 * $u'' u' \ge - 2 \AA \BB_1 u u'$

Integrating this from $x \in J$ to $\xi$, together with $\map {u'} \xi = 0$, yields:


 * $-\map {u'^2} x \ge - 2 \AA \BB_1 \paren {\map {u^2} \xi - \map {u^2} x}$

Then:

Hence:


 * $\forall x \in J: \size {\map {y'} x - \dfrac {c_1 - a_1} {c - a} } \le \sqrt {\dfrac {\BB_1} {2 \AA} } e^{4 \AA m}$

Similar arguments for aforementioned quantity being negative.