Nonnegative Quadratic Functional implies no Interior Conjugate Points

Theorem
If the quadratic functional


 * $\displaystyle \int_a^b \paren {P h'^2 + Q h^2} \rd x$

where:


 * $\forall x \in \closedint a b: \map P x > 0$

is nonnegative for all $\map h x$:


 * $\map h a = \map h b = 0$

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

In other words, the open interval $\openint a b$ contains no points conjugate to $a$.

Proof
Consider the functional:


 * $\forall t \in \closedint 0 1: \displaystyle \int_a^b \paren {t \paren {P h^2 + Q h'^2} + \paren {1 - t} h'^2} \rd x$

By assumption:


 * $\displaystyle \int_a^b \paren {P h'^2 + Q h^2} \rd x \ge 0$

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


 * $\map {h''} x = 0$

which, along with condition $\map h a = 0$, is solved by:


 * $\map h x = x - a$

for which there are no conjugate points in $\closedint a b$.

In other words:


 * $\forall x \in \openint a b: \map h x > 0$

Hence:


 * $\forall t \in \closedint 0 1: \displaystyle \int_a^b \paren {t \paren {P h'^2 + Q h^2} + \paren {1 - t} h'^2} \rd x \ge 0$

The corresponding Euler's Equation is:


 * $2 Q h t - \map {\dfrac \d {\d x} } {2 t P h' + 2 h' \paren {1 - t} } = 0$

which is equivalent to:


 * $-\map {\dfrac \d {\d x} } {\paren {t P + \paren {1 - t} } h'} + t Q h = 0$

Let $\map h {x, t}$ be a solution to this such that:


 * $\forall t \in \closedint 0 1: \map h {a, t} = 0$
 * $\map {h_x} {a, t} = 1$

Suppose that for $\map h {x, t}$ there exists a conjugate point $\tilde a$ to $a$ in $\closedint a b$.

In other words:


 * $\exists \tilde a \in \closedint a b: \map h {\tilde a, 1} = 0$

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

Suppose $\tilde a = b$.

Then by lemma 1 of Necessary and Sufficient Condition for Quadratic Functional to be Positive Definite:


 * $\displaystyle \int_a^b \paren {P h'^2 + Q h^2} \rd x = 0$

This agrees with the assumption.

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

For $t = 1$, any other conjugate point of $\map h {x, t}$ may reside only in $\openint a b$.

Consider the following set of all points $\tuple {x, t}$:


 * $\set {\tuple {x, t}: \paren {\forall x \in \closedint a b} \paren {\forall t \in \closedint 0 1} \paren {\map h {x, t} = 0} }$

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, $\map x t$ is continuous.

By hypothesis, $\tuple {\tilde a, 1}$ lies on this curve.

Suppose that 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 $\map h {x, t}$ in the interval $t \in \closedint 0 1$.

If it intersects the line segment $x = b, 0 \le t \le 1$, then by lemma 2 of Necessary and Sufficient Condition for Quadratic Functional to be Positive Definite 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 $\exists t_0: \paren {\map h {x, t_0} = 0} \land \paren {\map {h_x} {x, t_0} = 0}$.

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 $\exists t_0: \map {h_x} {a, t_0} = 0$

By Proof by Cases, no such curve exists.

Thus, the point $\tuple {\tilde a, 1}$ does not exist, since it belongs to this curve.

Hence there are no conjugate points of $\map h {x, 1} = \map h x$ in the interval $\openint a b$.