Necessary and Sufficient Condition for Quadratic Functional to be Positive Definite

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Theorem

The quadratic functional:

$\displaystyle\int_a^b\paren{Ph'^2+Qh^2}\rd x$

where:

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

is positive definite for all $\map h x$:

$\map h a=\map h b=0$

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


Proof

Necessary Condition

Let there be $\map \omega x$ :

$\displaystyle\map \omega x \in C^1\closedint a b$.

Then

\(\displaystyle 0\) \(=\) \(\displaystyle \omega h^2\big\vert_a^b\) boundary conditions for $h$
\(\displaystyle \) \(=\) \(\displaystyle \int_a^b \frac \d {\d x} \paren{\omega h^2}\rd x\)

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

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

Then:

\(\displaystyle Ph'^2+Qh^2+\frac \d {\d x} \paren{\omega h^2}\) \(=\) \(\displaystyle Ph'^2+2hh'\omega+\paren{Q+\omega'}h^2\)
\(\displaystyle \) \(=\) \(\displaystyle Ph'^2+2hh'\omega+\frac {\omega^2}Ph^2\)
\(\displaystyle \) \(=\) \(\displaystyle P\paren{h'+\frac \omega P h}^2\)
\(\displaystyle \) \(\ge\) \(\displaystyle 0\)

In other words:

\(\displaystyle \int_a^b\paren{Ph'^2+Qh^2}\rd x\) \(=\) \(\displaystyle \int_a^b\paren {Ph'^2+Qh^2+\frac \d {\d x }\paren{\omega h^2} }\rd x\)
\(\displaystyle \) \(=\) \(\displaystyle \int_a^b P\paren{h'+\frac \omega P h}^2\rd x\)
\(\displaystyle \) \(\ge\) \(\displaystyle 0\)

Suppose

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

By Existence-Uniqueness Theorem for First-Order Differential Equation.

$\map h a=0\implies \map h x=0\quad\forall x\in\closedint a b$

This implies an infinite number of conjugate points.

By assumption, there are no conjugate points.

Hence

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

and

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

Thus, a definite integral of positive definite function is positive definite.

$\Box$


Sufficient Condition

Consider the functional:

$\displaystyle \int_a^b \sqbrk{t\paren{Ph^2+Qh'^2}+\paren{1-t}h'^2}\rd x\quad\forall t\in\closedint 0 1$

By assumption:

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

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

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

Hence

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

The corresponding Euler's Equation is

\(\displaystyle 2Qht-\frac \d {\d x}\sqbrk{2tPh'+2h'\paren{1-t} }\) \(=\) \(\displaystyle 0\)

which is equivalent to

$\displaystyle -\frac \d {\d x} \lbrace\sqbrk{tP+\paren{1-t} }h'\rbrace+tQh=0$

Let $\map h {x,t}$ be a solution to this such that

$\forall t\in\closedint 0 1\quad \map h {a,t}=0,\map {h_x} {a,t}=1$

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, $a\ne\tilde a$.

Suppose $\tilde a=b$.

Then by lemma,

$\displaystyle\int_a^b\paren{Ph'^2+Qh^2}\rd x=0$

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 $paren{x,t}$:

$ \lbrace \paren{x,t}:\paren{\forall x\in\closedint a b} \paren{\forall t\in\closedint 0 1} \sqbrk{\map h {x,t}=0}\rbrace$

If it is non-empty, it represents a curve in $x-t$ plane, such that $\map {h_x} {x, t}\ne 0$.

By implicit function theorem, $\map x t$ is continuous.

By hypothesis, $\paren{\tilde a,1}$ lies on this curve.

Suppose, the curve starts at this point.

The curve can terminate either inside the rectangle or its boundary.

If it terminates inside the rectangle $\closedint a b\times\closedint 0 1$, it implies that there is a discontinuous jump in the value of $ h $.

Therefore, it contradicts the continuity of $\map h {x,t}$ in the interval $t\in\closedint 0 1$.

If it intersects the line segment $x=b,0\le t\le 1$, then by lemma it vanishes.

This contradicts positive-definiteness of the functional for all $t$.

If it 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 it 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 it 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 $\left({\tilde a, 1}\right)$ does not exist, since it belongs to this curve.

Hence, there are no conjugate points in the interval $\closedint a b$.

$\blacksquare$


Sources

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