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

Theorem
Let $K$ be a functional, such that:


 * $ \displaystyle K \left [ { \mathbf h } \right ] = \int_a^b \left({\mathbf h' \mathbf P \mathbf h' + \mathbf h \mathbf Q \mathbf h}\right) \mathrm d x $

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

Suppose $\left [ { a \,. \,. \, b } \right ]$ does not contain a point conjugate to $a$.

Then:
 * $\forall \mathbf h : \mathbf h \left({a}\right) = \mathbf h \left({b}\right) = 0 : K \left [ { \mathbf h } \right ] > 0$

the above holds.

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

Then

Suppose, $\mathbf W$ is such that:


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

Then:

Note that:
 * $\displaystyle \mathbf P^{ 1 / 2 } \mathbf h' + \mathbf P^{ -1 / 2 } \mathbf W \mathbf h \ne 0$

unless:
 * $\displaystyle \mathbf h \left({x}\right) = 0 : \forall x \in \left [ { a \,. \,. \, b } \right ] $

However, this contradicts the absence of conjugate points.

Thus, $K > 0$.

Sufficient Condition
Consider the following functional:


 * $\displaystyle \int_a^b \left [ { K t + \mathbf h' \left({1 - t}\right) \mathbf h' } \right ] \mathrm d x$

The corresponding Euler's equations are:


 * $\displaystyle - \frac{ \mathrm d }{ \mathrm d x } \left [ { t \mathbf P \mathbf h' + \left({1 - t}\right) \mathbf h' } \right ] + t \mathbf Q \mathbf h = 0$

Suppose the interval $\left [ { a \,. \,. \, b } \right ]$ contains a point $\tilde a$ conjugate to $a$.

Hence the determinant $\left \vert h_{ i j } \right \vert$ vanishes.

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

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

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 $\left({a \,. \,. \, b}\right)$.