Necessary and Sufficient Condition for Quadratic Functional to be Positive Definite/Dependent on N Functions

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Theorem

Let $K$ be a functional, such that:

$\displaystyle K \sqbrk {\mathbf h} = \int_a^b \paren {\mathbf h' \mathbf P \mathbf h' + \mathbf h \mathbf Q \mathbf h} \rd x$

where:

$\mathbf h$ is an $N$-dimensional vector
$\mathbf Q$ is a $N \times N$ matrix
$\mathbf P$ is a $N\times N$ symmetric positive definite matrix.


Let $\closedint a b$ be such that it does not contain a point conjugate to $a$.

Then:

$\forall \mathbf h: \map {\mathbf h} a = \map {\mathbf h} b = 0: K \sqbrk {\mathbf h} > 0$

if and only if the above holds.



Proof

Necessary Condition

Let $\mathbf W$ be an arbitrary differentiable symmetric matrix.

Then

\(\displaystyle 0\) \(=\) \(\displaystyle \int_a^b \map {\frac \d {\d x} } {\mathbf h \mathbf W \mathbf h} \rd x\)
\(\displaystyle \) \(=\) \(\displaystyle \int_a^b \mathbf h \mathbf W' \mathbf h \rd x + 2 \int_a^b \mathbf h' \mathbf W \mathbf h \rd x\)

Suppose, $\mathbf W$ is such that:

$\mathbf Q + \mathbf W' = \mathbf W \mathbf P^{-1} \mathbf W$

Then:

\(\displaystyle \int_a^b \paren {\mathbf h' \mathbf P \mathbf h' + \mathbf h \mathbf Q \mathbf h} \rd x\) \(=\) \(\displaystyle \int_a^b \paren {\mathbf h' \mathbf P \mathbf h' + 2 \mathbf h' \mathbf W \mathbf h + \mathbf h \mathbf Q \mathbf h + \mathbf h \mathbf W' \mathbf h} \rd x\)
\(\displaystyle \) \(=\) \(\displaystyle \int_a^b \paren {\mathbf h' \mathbf P \mathbf h' + 2 \mathbf h' \mathbf W \mathbf h + \mathbf h \mathbf W \mathbf P^{-1} \mathbf W \mathbf h} \rd x\)
\(\displaystyle \) \(=\) \(\displaystyle \int_a^b \paren {\mathbf P^{1/2} \mathbf h' + \mathbf P^{1/2} \mathbf h' \mathbf P^{-1/2} \mathbf W \mathbf h + \mathbf P^{-1/2} \mathbf W \mathbf h}^2 \rd x\)



Note that:

$\mathbf P^{1/2} \mathbf h' + \mathbf P^{-1/2} \mathbf W \mathbf h \ne 0$

unless:

$\forall x\in\closedint a b: \map {\mathbf h} x = 0$

However, this contradicts the absence of conjugate points.

Hence:

$K > 0$

$\Box$


Sufficient Condition

Consider the following functional:

$\displaystyle \int_a^b \paren {K t + \mathbf h' \paren {1 - t} \mathbf h'} \rd x$

The corresponding Euler's equations are:

$-\map {\dfrac \d {\d x} } {t \mathbf P \mathbf h' + \paren {1 - t} \mathbf h'} + t \mathbf Q \mathbf h = 0$

Suppose the interval $\closedint a b$ contains a point $\tilde a$ conjugate to $a$.

Then the determinant $\size {h_{ij} }$ vanishes.

Therefore there exists a linear combination of $h_i$ not identically equal to zero such that $\map {\mathbf h} {\tilde a} = 0$.

Furthermore, since the Euler's equations are continuous with respect to $t$, so is the solution of this equation.


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

By lemma, $K$ vanishes.

This contradicts the positive definiteness of $K$.

Therefore, $\tilde a \ne b$.

Thus, for $t = 1$ the conjugate point may only reside in $\openint a b$.



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