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 $ iff 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 equation is continuous $ t $, so is the solution of this equation.

Suppose, $ \tilde a = b $.

This contradicts the positive definiteness of $ K $.