Nonnegative Quadratic Functional implies no Interior Conjugate Points

Theorem
If the quadratic functional


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

where


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

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


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

then the interval $ \left [ { a \,. \,. \, b } \right ] $ contains no inside points conjugate to $ a $.

In other words, the interval $ \left ( { a \,. \,. \, b } \right ) $ contains no points conjugate to $ a $.

Proof
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 ] $

By assumption:


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

For $ t = 1 $, Euler's Equation reads


 * $h'' \left ( { x } \right ) = 0$

which, along with condition $ h \left ( { a } \right ) = 0 $, is solved by


 * $h \left ( { x } \right ) = x - a$

for which there are no conjugate points in $ \left [ { a \,. \,. \,b } \right ] $.

In other words:


 * $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 \ge 0 \quad \forall t \in \left [ { 0 \,. \,. \, 1 } \right ] $

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 \left [ { 0 \,. \,. \,1 } \right ] \quad h \left ( { a, t } \right ) = 0, ~h_x \left ( { a, t } \right ) = 1 $

Suppose that for $ h \left ( { x, t } \right ) $ 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 agrees with the assumption.

Therefore, it is allowed that $ \tilde a = b $.

For $ t = 1$, any other conjugate point of $ h \left ( { x, t } \right ) $ 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 the 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 $\left[{a \,.\,.\, b}\right] \times \left[{0 \,.\,.\, 1}\right]$, 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 \left [ { 0 \,. \,. \, 1 } \right ] $.

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 line 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 of $h \left({x, 1}\right) = h \left({x}\right)$ in the interval $\left({a \,.\,.\, b}\right)$.