Bernstein's Theorem on Unique Global Solution to y''=F(x,y,y')/Lemma 2

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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:

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

for all $x \in \closedint a c$, where:

$\map y a = a_1$
$\map y c = c_1$


Then the following bound holds:

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


Proof

As a consequence of $y = \map F {x, y, y'}$ we have:

\(\ds y\) \(=\) \(\ds \map F {x, y, y'}\)
\(\ds \) \(=\) \(\ds \map F {x, y, y'} - \map F {x, \dfrac {a_1 \paren {c - x} + c_1 \paren {x - a} } {c - a}, y'} + \map F {x, \dfrac {a_1 \paren {c - x} + c_1 \paren {x - a} } {c - a}, y'}\)
\(\text {(2)}: \quad\) \(\ds \) \(=\) \(\ds \paren {\map y x - \dfrac {a_1 \paren {c - x} + c_1 \paren {x - a} } {c - a} } \map {F_y} {x, \psi, y'} + \map F {x, \dfrac {a_1 \paren {c - x} + c_1 \paren {x - a} } {c - a},y'}\) Mean Value Theorem with respect to $y$

where:

$\psi = \dfrac {a_1 \paren {c - x} + c_1 \paren {x - a} } {c - a} + \theta \paren {\map y x - \dfrac {a_1 \paren {c - x} + c_1 \paren {x - a} } {c - a} }$

and:

$0 < \theta < 1$


Note that the term $\chi$, defined as:

$\chi = \map y x - \dfrac {a_1 \paren {c - a} + c_1 \paren {x - a} } {c - a}$

vanishes at $x = a$ and $x = c$.

Unless $\chi$ is identically zero, there exists a point $\xi \in \openint a b$ such one of the following holds:

$\chi$ attains a positive maximum at $\xi$
$\chi$ attains a negative minimum at $\xi$.

In the first case:

$\map {y} \xi \le 0$,
$\map {y'} \xi = \dfrac {c_1 - a} {c - a}$

which implies:

\(\ds 0\) \(\ge\) \(\ds \map {y} \xi\)
\(\ds \) \(=\) \(\ds \map F {\xi, \dfrac {a_1 \paren {c - \xi} + c_1 \paren {\xi - a} } {c - a}, \dfrac {c_1 - a_1} {c - a} } + \map {F_y} {x, \map \psi \xi, \map {y'} x} \paren {\map y \xi - \dfrac {a_1 \paren {c - \xi} + c_1 \paren {\xi - a} } {c - a} }\) equation $(2)$
\(\ds \) \(\ge\) \(\ds \map F {\xi, \dfrac {a_1 \paren {c - \xi} + c_1 \paren {\xi - a} } {c - a}, \dfrac {c_1 - a_1} {c - a} } + k \paren {\map y \xi - \dfrac {a_1 \paren {c - \xi} + c_1 \paren {\xi - a} } {c - a} }\) Assumption of Theorem


Hence:

$\map y \xi - \dfrac {a_1 \paren {c - \xi} + c_1 \paren {\xi - a} } {c - a} \le -\dfrac 1 k \map F {\xi, \dfrac {a_1 \paren {c - \xi} + c_1 \paren {\xi - a} } {c - a}, \dfrac {c_1 - a_1} {c - a} }$

The negative minimum part is proven analogously.

Hence, the following holds:

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

$\blacksquare$


Sources


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