=== Necessary Condition === Let $\map \omega x : \closedint a b \to \R$ be a continuously differentiable mapping. Then: {{begin-eqn}} {{eqn | l = 0       | r = \bigintlimits {\omega h^2} a b       | c = Boundary Conditions for $h$ }} {{eqn | r = \int_a^b \map {\frac \d {\d x} } {\omega h^2} \rd x       | c = Fundamental theorem of calculus }} {{end-eqn}}    Let $\omega$ be a solution to the following equation: :$\map P {Q + \omega'} = \omega^2$ Then: {{begin-eqn}} {{eqn | l = P h'^2 + Q h^2 + \map {\frac \d {\d x} } {\omega h^2}       | r = P h'^2 + 2 h h' \omega + \paren {Q + \omega'} h^2 }} {{eqn | r = P h'^2 + 2 h h' \omega + \frac {\omega^2} P h^2 }} {{eqn | r = P \paren {h' + \frac \omega P h}^2 }} {{eqn | o = \ge       | r = 0 }} {{end-eqn}} In other words: {{begin-eqn}} {{eqn | l = \int_a^b \paren {P h'^2 + Q h^2} \rd x       | r = \int_a^b \paren {P h'^2 + Q h^2 + \map {\frac \d {\d x} } {\omega h^2} } \rd x }} {{eqn | r = \int_a^b P \paren {h' + \frac \omega P h}^2 \rd x }} {{eqn | o = \ge       | r = 0 }} {{end-eqn}} Now we will show, that this expression never vanishes. {{AimForCont}} that it does vanish. Then: :$h' + \dfrac \omega P h = 0$ By Existence-Uniqueness Theorem for First-Order Differential Equation: :$\forall x \in \closedint a b: \map h a = 0 \implies \map h x = 0$ This implies an infinite number of conjugate points. However, the theorem assumes no conjugate points. This is a contradiction. Hence: :$\forall x \in \openint a b: \map h x \ne 0$ and: :$P \paren {h' + \dfrac {\omega h} P}^2 > 0$ Thus, a functional formed by a definite integral of positive definite function is positive definite. {{qed|lemma}} === Sufficient Condition === Consider the functional: :$\forall t \in \closedint 0 1: \displaystyle \int_a^b \sqbrk {t \paren {P h^2 + Q h'^2} + \paren {1 - t} h'^2} \rd x$ By assumption: :$\displaystyle \int_a^b \paren { Ph'^2 + Q h^2} \rd x > 0$  Since there are no conjugate points in $\closedint a b$: :$\forall x \in \openint a b: \map h x > 0$ Hence: :$\forall t \in \closedint 0 1: \displaystyle \int_a^b \sqbrk {t \paren {P h'^2 + Q h^2} + \paren {1 - t} h'^2} \rd x > 0$ The corresponding Euler's Equation is: :$2 Q h t - \map {\dfrac \d {\d x} } {2 t P h'+ 2 h' \paren {1 - t} } = 0$ which is equivalent to: :$-\map {\dfrac \d {\d x} } {\paren {t P + \paren {1 - t} } h'} + t Q h = 0$ Let $\map h {x, t}$ be a solution to this such that: :$\forall t \in \closedint 0 1: \map h {a, t} = 0, \map {h_x} {a, t} = 1$ Suppose there exists a conjugate point $\tilde a$ to $a$ in $\closedint a b$. In other words: :$\exists \tilde a \in \closedint a b: \map h {\tilde a, 1} = 0$ By definition, $a \ne \tilde a$. {{AimForCont}} $\tilde a = b$. Then by the lemma 1: :$\ds \int_a^b \paren {P h'^2 + Q h^2} \rd x = 0$ This contradicts the assumption. Therefore, $\tilde a \ne b$. Thus, for $t = 1$, any other conjugate point may reside only in $\openint a b$. Consider the following set of all points $\tuple {x, t}$: :$\set {\tuple {x, t}: \paren {\forall x \in \closedint a b} \paren {\forall t \in \closedint 0 1} \paren {\map h {x, t} } = 0}$ If it is non-empty, it represents a curve $\paren \star$ in $x - t$ plane, such that $\map {h_x} {x, t} \ne 0$. By the Implicit Function Theorem, $\map x t$ is continuous. By hypothesis, $\tuple {\tilde a, 1}$ lies on this curve. Suppose the curve starts at this point. The curve can terminate either inside the rectangle or its boundary. {{explain|Clarify the below by replacing "it" wherever it occurs with a label to the specific thing that "it" refers to.}} If $\paren \star$ terminates inside the rectangle $\closedint a b \times \closedint 0 1$, it implies that there is a jump discontinuity in the value of $h$. Therefore, it contradicts the continuity of $\map h {x, t}$ in the interval $t \in \closedint 0 1$. If $\paren \star$ intersects the line segment $x = b, 0 \le t \le 1$, then by lemma 2 the considered functional vanishes. :This contradicts positive definiteness of the functional for all $t$. If $\paren \star$ intersects the segment $a \le x \le b, t = 1$, then: :$\exists t_0: \paren {\map h {x, t_0} = 0} \land \paren {\map {h_x} {x, t_0} = 0}$ If $\paren \star$ intersects $a \le x \le b, t = 0$, then Euler's equation reduces to $h'' = 0$ with solution $h = x - a$, which vanishes only for $x = a$. If $\paren \star$ intersects $x = a, 0 \le t \le 1$, then $\exists t_0:\map {h_x} {a, t_0} = 0$ {{Help|explain cases $t {{=}} 1$ and $x {{=}} a$}} By Proof by Cases, no such curve exists. Thus, the point $\tuple {\tilde a, 1}$ does not exist, since it belongs to this curve. Hence, there are no conjugate points in the interval $\closedint a b$. {{qed}} {{improve|1: Transclude the lemmata in the usual {{ProofWiki}} style. <br/> 2: There are a number of proofs by contradiction embedded in this proof. It is important to make it clear exactly which statement is the supposition which leads to the contradiction, and link back to that statement somehow. <br/>3: The proof is in a number of sections but there is no visual indication by (for example) double spacing or use of the qed lemma template where one section ends and the next begins.<br/>4: There are still references to "positive definite", in the title and in the proof without an actual definition of what "positive definite" actually means. It's suboptimal to gloss over this. We really need to add a page defining what a positive definite function actually is in this context -- particularly when the term itself has a far wider interpretation in mathematics in general than "never goes negative".}}