Necessary and Sufficient Condition for Quadratic Functional to be Positive Definite

From ProofWiki
Jump to navigation Jump to search



$\map P x : \closedint a b \to \R$
$\map h x : \closedint a b \to \R$.

Let $\map h x$ be continuously differentiable $\forall x \in \closedint a b$.


$\forall x \in \closedint a b: \map P x > 0$


$\ds \forall \map h x : \map h a = \map h b = 0 : \int_a^b \paren {P h'^2 + Q h^2} \rd x > 0$

if and only if the interval $\closedint a b$ contains no points conjugate to $a$.


Necessary Condition

Let $\map \omega x : \closedint a b \to \R$ be a continuously differentiable mapping.


\(\ds 0\) \(=\) \(\ds \bigintlimits {\omega h^2} a b\) Boundary Conditions for $h$
\(\ds \) \(=\) \(\ds \int_a^b \map {\frac \d {\d x} } {\omega h^2} \rd x\) Fundamental theorem of calculus

Let $\omega$ be a solution to the following equation:

$\map P {Q + \omega'} = \omega^2$


\(\ds P h'^2 + Q h^2 + \map {\frac \d {\d x} } {\omega h^2}\) \(=\) \(\ds P h'^2 + 2 h h' \omega + \paren {Q + \omega'} h^2\)
\(\ds \) \(=\) \(\ds P h'^2 + 2 h h' \omega + \frac {\omega^2} P h^2\)
\(\ds \) \(=\) \(\ds P \paren {h' + \frac \omega P h}^2\)
\(\ds \) \(\ge\) \(\ds 0\)

In other words:

\(\ds \int_a^b \paren {P h'^2 + Q h^2} \rd x\) \(=\) \(\ds \int_a^b \paren {P h'^2 + Q h^2 + \map {\frac \d {\d x} } {\omega h^2} } \rd x\)
\(\ds \) \(=\) \(\ds \int_a^b P \paren {h' + \frac \omega P h}^2 \rd x\)
\(\ds \) \(\ge\) \(\ds 0\)

Now we will show, that this expression never vanishes.

Aiming for a contradiction, suppose that it does vanish.


$h' + \dfrac \omega P h = 0$

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.


$\forall x \in \openint a b: \map h x \ne 0$


$P \paren {h' + \dfrac {\omega h} P}^2 > 0$

Thus, a functional formed by a definite integral of positive definite function is positive definite.


Sufficient Condition

Consider the functional:

$\forall t \in \closedint 0 1: \displaystyle \int_a^b \sqbrk {t \paren {P h^2 + Q h'^2} + \paren {1 - t} h'^2} \rd x$

By assumption:

$\displaystyle \int_a^b \paren { Ph'^2 + Q h^2} \rd x > 0$

Since there are no conjugate points in $\closedint a b$:

$\forall x \in \openint a b: \map h x > 0$


$\forall t \in \closedint 0 1: \displaystyle \int_a^b \sqbrk {t \paren {P h'^2 + Q h^2} + \paren {1 - t} h'^2} \rd x > 0$

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, $a \ne \tilde a$.

Aiming for a contradiction, suppose $\tilde a = b$.

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 $\closedint a b$.



1963: I.M. Gelfand and S.V. Fomin: Calculus of Variations ... (previous) ... (next): $\S 5.26$: Analysis of the Quadratic Functional $\int_a^b \paren {P h'^2 + Q h^2} \rd x$