# Bernstein's Theorem on Unique Extremal

## 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 $a \ne b$.

## Proof

### Lemma 1 (Uniqueness)

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:

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

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:

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

$\Box$

### 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

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

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

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

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:

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

$\Box$

### 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

 $(3):\quad$ $\displaystyle \map y x-\dfrac{a_1\paren{c-x}+c_1\paren{x-a} }{c-a}+m$ $=$ $\displaystyle \dfrac{\Ln u}{2\cal A}$ $\quad$ $\quad$ $(4):\quad$ $\displaystyle -\map y x+\dfrac{a_1\paren{c-x}+c_1\paren{x-a} }{c-a}+m$ $=$ $\displaystyle \dfrac{\Ln v}{2\cal A}$ $\quad$ $\quad$

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 $(3)$ & $(4)$ with respect to $x$:

 $(5):\quad$ $\displaystyle \map {y'} x-\dfrac{c_1-a_1}{c-a}$ $=$ $\displaystyle \dfrac{u'}{2\cal Au}$ $\quad$ $\quad$ $(6):\quad$ $\displaystyle -\map {y'} x+\dfrac{c_1-a_1}{c-a}$ $=$ $\displaystyle \dfrac{v'}{2\cal Av}$ $\quad$ $\quad$

Differentiate again:

 $(7):\quad$ $\displaystyle \map {y''} x$ $=$ $\displaystyle \dfrac{u''}{2\cal Au}-\dfrac{u'^2}{2\cal Au^2}$ $\quad$ $\quad$ $(8):\quad$ $\displaystyle -\map {y''} x$ $=$ $\displaystyle \dfrac{v''}{2\cal Av}-\dfrac{v'^2}{2\cal Av^2}$ $\quad$ $\quad$

By assumption:

 $\displaystyle \size{\map F {x,y,y'} }$ $=$ $\displaystyle \size{\map {y''} x}$ $\quad$ $\quad$ $\displaystyle$ $\le$ $\displaystyle \map \alpha {x,y} \map {y'^2} x+\map \beta {x,y}$ $\quad$ $\quad$ $\displaystyle$ $\le$ $\displaystyle \cal A\map {y'^2} x+\cal B$ $\quad$ $\quad$ $\displaystyle$ $\le$ $\displaystyle 2\cal A\map {y'^2} x+\cal B$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle 2\cal A {\paren{\map{y'} x-\dfrac{c_1-a_1}{c-a} }^2}+2\cal A\paren{\dfrac{c_1-a_1}{c-a} }^2-2\map{y'} x\dfrac{c_1-a_1}{c-a}+\cal B$ $\quad$ $\quad$ $(9):\quad$ $\displaystyle$ $\le$ $\displaystyle 2\cal A\paren{\map {y'} x-\dfrac{c_1-a_1}{c-a} }^2+\cal B_1$ $\quad$ $\quad$

where

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

Then:

 $\displaystyle y''$ $=$ $\displaystyle \dfrac{u''}{2\cal Au}-\dfrac{u'^2}{2\cal Au^2}$ $\quad$ Equation $(7)$ $\quad$ $\displaystyle$ $\ge$ $\displaystyle -2\cal A\paren{y'\paren x-\dfrac{c_1-a_1}{c-a} }^2-\cal B_1$ $\quad$ Inequality $(9)$ $\quad$ $\displaystyle$ $=$ $\displaystyle -2\cal A\paren{\dfrac{u'}{2\cal Au} }^2-\cal B_1$ $\quad$ Equation $(5)$ $\quad$

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

 $\displaystyle u''$ $\ge$ $\displaystyle -2\cal AB_1u$ $\quad$ $\quad$

Similarly:

 $\displaystyle y''$ $=$ $\displaystyle -\dfrac{v''}{2\cal Av}+\dfrac{v'^2}{2\cal Av^2}$ $\quad$ Equation $(8)$ $\quad$ $\displaystyle$ $\le$ $\displaystyle 2\cal A\paren{\map {y'} x-\dfrac{c_1-a_1}{c-a} }^2+\cal B_1$ $\quad$ Inequality $(9)$ $\quad$ $\displaystyle$ $=$ $\displaystyle 2\cal A\paren{\dfrac{v'}{2\cal Av} }^2+\cal B_1$ $\quad$ Equation $(6)$ $\quad$

Multiply the inequality by $-2\cal Av$ and simplify:

 $\displaystyle v''$ $\ge$ $\displaystyle - 2\cal AB_1v$ $\quad$ $\quad$

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:

 $\displaystyle \map {u'^2} x$ $\le$ $\displaystyle 2\cal AB_1\paren{\map {u^2} \xi-\map {u^2} x}$ $\quad$ $\quad$ $\displaystyle$ $\le$ $\displaystyle 2\cal AB_1\map {u^2} \xi$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle 2\cal AB_1\exp\sqbrk{4\cal A\paren{m+\map y x-\dfrac{a_1\paren{c-x}+c_1\paren{x-a} }{c-a} } }$ $\quad$ Equation $(3)$ $\quad$ $\displaystyle$ $\le$ $\displaystyle 2\cal AB_1e^{8\cal Am}$ $\quad$ Lemma 2 $\quad$ $\displaystyle \leadsto \ \$ $\displaystyle \dfrac{\map {u'^2} x}{\map {u^2} x}$ $\le$ $\displaystyle 2\cal AB_1 e^{8\cal Am}$ $\quad$ $u \ge 1$ $\quad$ $\displaystyle \leadsto \ \$ $\displaystyle 4{\cal A}^2 \paren{\map {y'} x-\dfrac{c_1-a_1}{c-a} }^2$ $\le$ $\displaystyle 2\cal AB_1 e^{8\cal Am}$ $\quad$ Equation $(5)$ $\quad$

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.

$\Box$

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

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 $AP$, and $\alpha$ can be found from

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

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 $QB$.

$\blacksquare$

## Source of Name

This entry was named for Sergei Natanovich Bernstein.