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 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 1 (Uniqueness)
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:

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:


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

Lemma 2
Suppose


 * $y''\paren x=F\paren{x,y,y'}$

for all $x\in \sqbrk{a,c}$, where


 * $\map y a=a_1,$
 * $\map y c=c_1.$

Then the following bound holds:

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

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

where


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

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

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:


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

Lemma 3
Suppose 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 $\cal A$ and $\cal B$ 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


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

Suppose, that $\cal A\ge 1$.

Let $u,v$ be real functions such that

Due to Lemma 2, for $x\in I$ the LHS's of $(3)$ & $(4)$ are not negative.

Thus,


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

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

Differentiate again:

By assumption:

where


 * $\cal B_1=\cal B+2\cal A\paren{\dfrac{c_1-a_1}{c-a}}^2$

Then:

Multiply the inequality by $2\cal Au$ and simplify:

Similarly:

Multiply the inequality by $-2\cal Av$ 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 $(5)$ & $(6)$ both $\map {u'} x$ & $\map {v'} x$ 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 $u'$:


 * $u'' u'\ge-2\cal AB_1uu'$.

Integrating this from $x\in J$ to $\xi$, together with $u'\paren\xi=0$, yields


 * $-\map {u'^2} x\ge-2\cal AB_1\paren{\map {u^2} \xi-\map {u^2} x}$

Then:

Hence,


 * $\forall x\in J:\size{y'\paren x-\dfrac{c_1-a_1}{c-a}}\le\sqrt{\dfrac{\cal B_1}{2\cal A}}e^{4\cal Am}$

Similar arguments for aforementioned quantity being negative.

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


 * BernsteinCurve.png

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

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

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

For $x\ge d$ draw a line with $y=d_1$.

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

For $y\ge d_1$ draw a line with $x=b_1$.

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 $P\paren{\xi,\xi_1}$.

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:


 * $\displaystyle\lim\limits_{\xi\to r}\alpha=\pm\dfrac{\pi}{2}$.

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

This contradicts the bounds from Lemma 2 & 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 $QB$.