=== Lemma 1 (Uniqueness) === {{AimForCont}} 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: {{begin-eqn}} {{eqn | l = \map {\delta''} x        | r = \map {\phi_2''} x - \map {\phi_1''} x        | c = Definition of $\delta$ }} {{eqn | r = \map F {x, \phi_2, \phi_2'} - \map F {x, \phi_1, \phi_1'}       | c = as $y'' = \map F {x, y', y''}$ }} {{eqn | r = \map F {x, \phi_1 + \delta, \phi_1' + \delta'} - \map F {x, \phi_1, \phi_1'}       | c = Definition of $\delta$ }} {{eqn | n = 1       | r = \map \delta x F_y^* + \map {\delta'} x F_{y'}^*       | c = Multivariate Mean Value Theorem {{WRT}} $y, y'$ }} {{end-eqn}} 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: :$\map {\phi_1} x = \map {\phi_2} x$ {{qed|lemma}} === 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: {{begin-eqn}} {{eqn | l = y''       | r = \map F {x, y, y'} }} {{eqn | r = \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'} }} {{eqn | n = 2       | r = \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'}       | c = Mean Value Theorem {{WRT}} $y$ }} {{end-eqn}} 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: :$0 < \theta < 1$ {{explain|the meaning of the square brackets above}} 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: {{begin-eqn}} {{eqn | l = 0       | o = \ge       | r = \map {y''} \xi }} {{eqn | r = \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} }       | c = equation $(2)$ }} {{eqn | o = \ge       | r = \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} }       | c = Assumption of Theorem }} {{end-eqn}} {{explain|the meaning of the square brackets above}} 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:  :$\size {\map y x - \dfrac {a_1 \paren {c - x} + c_1 \paren {x - a} } {c - a} } \le \dfrac 1 k \max \limits_{a \mathop \le x \mathop \le b} \size {\map F {x, \dfrac {a_1 \paren {c - x} + c_1 \paren {x - a} } {c - a}, \dfrac {c_1 - a_1} {c - a} } }$ {{qed|lemma}} === 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  {{begin-eqn}} {{eqn | n = 3       | l = \map y x - \dfrac {a_1 \paren {c - x} + c_1 \paren {x - a} } {c - a} + m       | r = \dfrac {\Ln u} {2 \AA} }} {{eqn | n = 4       | l = -\map y x + \dfrac {a_1 \paren {c - x} + c_1 \paren {x - a} } {c - a} + m       | r = \dfrac {\Ln v} {2 \AA} }} {{end-eqn}} {{explain|The precise meaning of $\Ln$ in this context}} Due to Lemma 2, for $x \in I$ the {{LHS}}s of $(3)$ and $(4)$ are not negative.  Thus: :$\forall x \in I: u, v \ge 1$ Differentiate equations $(3)$ and $(4)$ {{WRT|Differentiation}} $x$: {{begin-eqn}} {{eqn| n = 5      | l = \map {y'} x - \dfrac {c_1 - a_1} {c - a}      | r = \dfrac {u'} {2 \AA u}  }} {{eqn| n = 6      | l = -\map {y'} x + \dfrac {c_1 - a_1} {c - a}      | r = \dfrac {v'} {2 \AA v} }} {{end-eqn}} Differentiate again:  {{begin-eqn}} {{eqn| n = 7      | l = \map {y''} x      | r = \dfrac {u''} {2 \AA u} - \dfrac {u'^2} {2 \AA u^2} }}  {{eqn| n = 8      | l = -\map {y''} x      | r = \dfrac {v''} {2 \AA v} - \dfrac {v'^2} {2 \AA v^2} }} {{end-eqn}} By assumption: {{begin-eqn}} {{eqn | l = \size {\map F {x, y, y'} }       | r = \size {\map {y''} x} }} {{eqn | o = \le       | r = \map \alpha {x, y} \map {y'^2} x + \map \beta {x, y} }} {{eqn | o = \le       | r = \AA \map {y'^2} x + \BB }} {{eqn | o = \le       | r = 2 \AA \map {y'^2} x + \BB }} {{eqn | r = 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 }} {{eqn | n = 9       | o = \le       | r = 2 \AA \paren {\map {y'} x - \dfrac {c_1 - a_1} {c - a} }^2 + \BB_1 }} {{end-eqn}}  where: :$\BB_1 = \BB + 2 \AA \paren {\dfrac {c_1 - a_1} {c - a} }^2$ Then: {{begin-eqn}} {{eqn | l = y''       | r = \dfrac {u''} {2 \AA u} - \dfrac {u'^2} {2 \AA u^2}       | c = Equation $(7)$ }} {{eqn | o = \ge       | r = -2 \AA \paren {\map {y'} x - \dfrac {c_1 - a_1} {c - a} }^2 - \BB_1       | c = Inequality $(9)$ }} {{eqn | r = -2 \AA \paren {\dfrac {u'} {2 \AA u} }^2 - \BB_1       | c = Equation $(5)$ }} {{end-eqn}} Multiply the inequality by $2 \AA u$ and simplify: {{begin-eqn}} {{eqn | l = u''       | o = \ge       | r = -2 \AA \BB_1 u }} {{end-eqn}}   Similarly:  {{begin-eqn}} {{eqn | l = y''       | r = - \dfrac {v''} {2 \AA v} + \dfrac {v'^2} {2 \AA v^2}       | c = Equation $(8)$ }} {{eqn | o = \le       | r = 2 \AA \paren {\map {y'} x - \dfrac {c_1 - a_1} {c - a} }^2 + \BB_1       | c = Inequality $(9)$ }} {{eqn | r = 2 \AA \paren {\dfrac {v'} {2 \AA v} }^2 + \BB_1       | c = Equation $(6)$ }} {{end-eqn}} Multiply the inequality by $-2 \AA v$ and simplify: {{begin-eqn}} {{eqn | l = v''       | o = \ge       | r = -2 \AA \BB_1 v }} {{end-eqn}} 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: {{begin-eqn}} {{eqn | l = \map {u'^2} x       | o = \le       | r = 2 \AA \BB_1 \paren {\map {u^2} \xi - \map {u^2} x} }} {{eqn | o = \le       | r = 2 \AA \BB_1 \map {u^2} \xi }} {{eqn | r = 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} } }       | c = Equation $(3)$ }} {{eqn | o = \le       | r = 2 \AA \BB_1 e^{8 \AA m}       | c = Lemma 2 }} {{eqn | ll= \leadsto       | l = \dfrac {\map {u'^2} x} {\map {u^2} x}       | o = \le       | r = 2 \AA \BB_1 e^{8 \AA m}       | c = $u \ge 1$ }} {{eqn | ll= \leadsto       | l = 4 \AA^2 \paren {\map {y'} x - \dfrac {c_1 - a_1} {c - a} }^2       | o = \le       | r = 2 \AA \BB_1 e^{8 \AA m}       | c = Equation $(5)$ }} {{end-eqn}} {{explain|square brackets in the above}} 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. {{qed|lemma}} Consider a plane with axes denoted by $x$ and $y$: :500px 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 $QB$. {{qed}} {{Namedfor|Sergei Natanovich Bernstein|cat = Bernstein, Sergei}}