Necessary and Sufficient Condition for Quadratic Functional to be Positive Definite

Theorem
The quadratic functional:


 * $\displaystyle \int_a^b \left({P h'^2 + Q h^2}\right) \rd x$

where:


 * $\forall x \in \closedint a b: P \left({x}\right) > 0$

is positive definite for all $h \left({x}\right)$:


 * $h \left({a}\right) = h \left({b}\right) = 0$

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

Necessary Condition
Let there be $ \omega \left ( { x } \right ) $ :


 * $ \displaystyle \omega \left ( { x } \right ) \in C^1 \left [ { a \,. \,. \,b } \right] $.

Then

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


 * $P \left({Q + \omega'}\right) = \omega^2$

Then:

In other words:

Suppose


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

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


 * $ h \left ( { a } \right ) = 0 \implies h \left ( { x } \right ) = 0 \quad \forall x \in \closedint a b$

This implies an infinite number of conjugate points.

By assumption, there are no conjugate points.

Hence


 * $ h \left ( { x } \right ) \ne 0 \quad \forall x \in \openint a b$

and


 * $ \displaystyle P \left ( {h' +\frac{ \omega h }{ P } } \right )^2 > 0$

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

Sufficient Condition
Consider the functional:


 * $ \displaystyle \int_a^b \left [ { t \left ( { Ph^2 + Q h'^2 } \right ) + \left ( { 1 - t  } \right ) h'^2 } \right ] \mathrm d x \quad \forall t \in \closedint 0 1$

By assumption:


 * $ \displaystyle \int_a^b \left ( { Ph'^2 + Q h^2 } \right ) \mathrm d x  > 0 $

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


 * $ h \left ( { x } \right ) > 0 \quad \forall x \in \openint a b$

Hence


 * $ \displaystyle \int_a^b \left [ { t \left ( { Ph'^2 + Q h^2 } \right ) + \left ( { 1 - t  } \right ) h'^2 } \right ] \mathrm d x > 0 \quad \forall t \in \closedint 0 1$

The corresponding Euler's Equation is

which is equivalent to


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

Let $ h \left ( { x, t } \right ) $ be a solution to this such that


 * $ \forall t \in \closedint 0 1 \quad h \left ( { a, t } \right ) = 0, h_x \left ( { a, t } \right ) = 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: h \left ( { \tilde a, 1 } \right ) = 0$

By definition, $ a \ne \tilde a$.

Suppose $ \tilde a = b$.

Then by lemma,


 * $ \displaystyle \int_a^b \left ( { Ph'^2 + Qh^2 } \right ) \mathrm d 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 $ \left ( { x, t } \right )$:


 * $ \left \{ \left( { x, t } \right) : \left ( { \forall x \in \closedint a b} \right ) \left ( { \forall t \in \closedint 0 1} \right ) \left [ { h \left ( { x, t } \right ) = 0 } \right ] \right \}$

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

By implicit function theorem, $x \left({t}\right)$ is continuous.

By hypothesis, $\left({\tilde a, 1}\right)$ 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 $ h \left ( { x, t } \right )$ 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 : \left ( { h \left ( { x, t_0 } \right )=0 } \right ) \land \left ( { h_x \left ( { x, t_0 } \right )=0  } \right )$.

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 $\exists t_0: h_x \left({a, t_0}\right) = 0$

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 $\closedint a b$.