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


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

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:

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:


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

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

Due to 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.

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


 * BernsteinCurve.png

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 $QB$.