# 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 $a \ne b$.

## 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:

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

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

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

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:

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

where:

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

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:

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

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

$\Box$

### Lemma 3

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

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

Due to Lemma 2, for $x \in I$ the left hand sides of $(3)$ and $(4)$ are not negative.

Thus:

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

Differentiate equations $(3)$ and $(4)$ with respect to $x$:

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

Differentiate again:

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

By assumption:

 $\displaystyle \size {\map F {x, y, y'} }$ $=$ $\displaystyle \size {\map {y''} x}$ $\displaystyle$ $\le$ $\displaystyle \map \alpha {x, y} \map {y'^2} x + \map \beta {x, y}$ $\displaystyle$ $\le$ $\displaystyle \AA \map {y'^2} x + \BB$ $\displaystyle$ $\le$ $\displaystyle 2 \AA \map {y'^2} x + \BB$ $\displaystyle$ $=$ $\displaystyle 2 \AA {\paren {\map {y'} x - \dfrac {c_1 - a_1} {c - a} }^2} + 2 \AA \paren {\dfrac {c_1 - a_1} {c - a} }^2 - 2 \map {y'} x \dfrac {c_1 - a_1} {c - a} + \BB$ $\text {(9)}: \quad$ $\displaystyle$ $\le$ $\displaystyle 2 \AA \paren {\map {y'} x - \dfrac {c_1 - a_1} {c - a} }^2 + \BB_1$

where:

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

Then:

 $\displaystyle y''$ $=$ $\displaystyle \dfrac {u''} {2 \AA u} - \dfrac {u'^2} {2 \AA u^2}$ Equation $(7)$ $\displaystyle$ $\ge$ $\displaystyle -2 \AA \paren {\map {y'} x - \dfrac {c_1 - a_1} {c - a} }^2 - \BB_1$ Inequality $(9)$ $\displaystyle$ $=$ $\displaystyle -2 \AA \paren {\dfrac {u'} {2 \AA u} }^2 - \BB_1$ Equation $(5)$

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

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

Similarly:

 $\displaystyle y''$ $=$ $\displaystyle - \dfrac {v''} {2 \AA v} + \dfrac {v'^2} {2 \AA v^2}$ Equation $(8)$ $\displaystyle$ $\le$ $\displaystyle 2 \AA \paren {\map {y'} x - \dfrac {c_1 - a_1} {c - a} }^2 + \BB_1$ Inequality $(9)$ $\displaystyle$ $=$ $\displaystyle 2 \AA \paren {\dfrac {v'} {2 \AA v} }^2 + \BB_1$ Equation $(6)$

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

 $\displaystyle v''$ $\ge$ $\displaystyle -2 \AA \BB_1 v$

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:

 $\displaystyle \map {u'^2} x$ $\le$ $\displaystyle 2 \AA \BB_1 \paren {\map {u^2} \xi - \map {u^2} x}$ $\displaystyle$ $\le$ $\displaystyle 2 \AA \BB_1 \map {u^2} \xi$ $\displaystyle$ $=$ $\displaystyle 2 \AA \BB_1 \exp \sqbrk {4 \AA \paren {m + \map y x - \dfrac {a_1 \paren {c - x} + c_1 \paren {x - a} } {c - a} } }$ Equation $(3)$ $\displaystyle$ $\le$ $\displaystyle 2 \AA \BB_1 e^{8 \AA m}$ Lemma 2 $\displaystyle \leadsto \ \$ $\displaystyle \dfrac {\map {u'^2} x} {\map {u^2} x}$ $\le$ $\displaystyle 2 \AA \BB_1 e^{8 \AA m}$ $u \ge 1$ $\displaystyle \leadsto \ \$ $\displaystyle 4 \AA^2 \paren {\map {y'} x - \dfrac {c_1 - a_1} {c - a} }^2$ $\le$ $\displaystyle 2 \AA \BB_1 e^{8 \AA m}$ Equation $(5)$

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.

$\Box$

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

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

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 $QB$.

$\blacksquare$

## Source of Name

This entry was named for Sergei Natanovich Bernstein.