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 ) \mathrm d x$

where


 * $ \displaystyle P \left ( { x } \right ) > 0 \quad \forall x \in \left [ { a \,. \,. \, b } \right ]$

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


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

iff the interval $ \left [ { a \,. \,. \, b } \right ] $ 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


 * $ \displaystyle 0 = \int _a^b \frac{ \mathrm d}{ \mathrm d x} \left ( { \omega h^2 } \right ) \mathrm d x$

Let $ \omega $ be a solution to the equation $ \displaystyle P \left ( { Q + \omega' } \right ) = \omega^2$

Then

Suppose


 * $ \displaystyle h' +\frac{ \omega h }{ P }=0 $

Then


 * $ h \left ( { a } \right ) = 0 \implies h \left ( { x } \right ) = 0 \quad \forall x \in  \left [ { a \,. \,. \,b } \right] $

due to Existence-Uniqueness Theorem for First-Order Differential Equation.

This implies an infinite number of conjugate points.

Conditions of the theorem do not allow any conjugate points.

Hence


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

and


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

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 \left [ { 0 \,. \,. \, 1 } \right ] $

$ \displaystyle \int_a^b \left ( { Ph^2 + Q h'^2 } \right ) \mathrm d x  > 0 $ by statement of theorem.

Since there are no conjugate points in $ \left [ { a \,. \,. \,b } \right ] $, $ h \left ( { x } \right ) > 0 \quad \forall x \in \left ( { a \,. \,. \,b } \right ) $.

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 \left [ { 0 \,. \,. \, 1 } \right ] $

The corresponding Euler's Equation is


 * $ \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 \left [ { 0 \,. \,. \,1 } \right ] \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 $ \left [ { a \,. \,. \,b } \right ] $.

In other words:


 * $ \exists \tilde a \in \left [ { a \,. \,. \,b } \right ] : 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 condition in the theorem.

Therefore, $ \tilde a \ne b $.

Thus, any other conjugate point may reside only in $ \left [ { a \,. \,. \, b } \right ] $.

Consider the following set of all points $ \left ( { x, t } \right )$:


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

If it terminates inside the rectangle $\left [ { a \,. \,. \, b } \right ] \times \left [ { 0 \,. \,. \, 1 } \right ]$, it contradicts the continuity of $ h \left ( { x, t } \right )$ in the interval $ t \in \left [ { 0 \,. \,. \, 1 } \right ] $.

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

Hence, there are no conjugate points in the interval $ \left [ { a \,. \,. \, b } \right ] $.