Necessary and Sufficient Condition for Quadratic Functional to be Positive Definite

== Theorem ==
Let
 
:$\map P x : \closedint a b \to \R$ 
:$\map h x : \closedint a b \to \R$.
Let $\map h x$ be continuously differentiable $\forall x \in \closedint a b$.
Suppose:
:$\forall x \in \closedint a b: \map P x > 0$
Then:
:$\ds \forall \map h x : \map h a = \map h b = 0 : \int_a^b \paren {P h'^2 + Q h^2} \rd x > 0$
 the interval $\closedint a b$ contains no points conjugate to $a$.
== Proof ==
=== Necessary Condition ===
Let $\map \omega x : \closedint a b \to \R$ be a continuously differentiable mapping.
Then:
 
Let $\omega$ be a solution to the following equation:
:$\map P {Q + \omega'} = \omega^2$
Then:
In other words:
Now we will show, that this expression never vanishes.
 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.
=== Sufficient Condition ===
Consider the functional:
:$\forall t \in \closedint 0 1: \ds \int_a^b \sqbrk {t \paren {P h^2 + Q h'^2} + \paren {1 - t} h'^2} \rd x$
By assumption:
:$\ds \int_a^b \paren {Ph'^2 + Q h^2} \rd x > 0$ 
Since there are no conjugate points in $\closedint a b$:
:$\forall x \in \openint a b: \map h x > 0$
Hence:
:$\forall t \in \closedint 0 1: \ds \int_a^b \sqbrk {t \paren {P h'^2 + Q h^2} + \paren {1 - t} h'^2} \rd x > 0$
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, $a \ne \tilde a$.
 $\tilde a = b$.
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.
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 $\exists t_0:\map {h_x} {a, t_0} = 0$
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 $\closedint a b$.
== Sources ==
: $\S 5.26$: Analysis of the Quadratic Functional $\int_a^b \paren {P h'^2 + Q h^2} \rd x$
Category:Calculus of Variations