Necessary and Sufficient Condition for Quadratic Functional to be Positive Definite

Theorem
The quadratic functional:


 * $\displaystyle\int_a^b\paren{Ph'^2+Qh^2}\rd x$

where:


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

is positive definite for all $\map h x$:


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

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

Necessary Condition
Let there be $\map \omega x$ :


 * $\displaystyle\map \omega x \in C^1\closedint a b$.

Then:

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


 * $\map P {Q + \omega'} = \omega^2$

Then:

In other words:

Suppose


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

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


 * $\map h a=0\implies \map h x=0\quad\forall x\in\closedint a b$

This implies an infinite number of conjugate points.

By assumption, there are no conjugate points.

Hence


 * $\map h x\ne 0\quad\forall x\in\openint a b$

and


 * $\displaystyle P\paren{h'+\frac{\omega h} P}^2>0$

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

Sufficient Condition
Consider the functional:


 * $\displaystyle \int_a^b \sqbrk{t\paren{Ph^2+Qh'^2}+\paren{1-t}h'^2}\rd x\quad\forall t\in\closedint 0 1$

By assumption:


 * $\displaystyle\int_a^b\paren{Ph'^2+Qh^2}\rd x>0$

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


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

Hence


 * $\displaystyle\int_a^b\sqbrk{t\paren{Ph'^2+Qh^2}+\paren{1-t}h'^2}\rd x>0\quad\forall t\in\closedint 0 1$

The corresponding Euler's Equation is

which is equivalent to


 * $\displaystyle -\frac \d {\d x} \lbrace\sqbrk{tP+\paren{1-t} }h'\rbrace+tQh=0$

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


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

Suppose there exists a conjugate point $\tilde a$ to $a$ in $\closedint a b$.

In other words:


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

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

Suppose $\tilde a=b$.

Then by lemma,


 * $\displaystyle\int_a^b\paren{Ph'^2+Qh^2}\rd x=0$

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 $paren{x,t}$:


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

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

By implicit function theorem, $\map x t$ is continuous.

By hypothesis, $\paren{\tilde a,1}$ 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 $\map h {x,t}$ in the interval $t\in\closedint 0 1$.

If it intersects the line segment $x=b,0\le t\le 1$, then by lemma 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 $\exists t_0:\paren{\map h {x,t_0}=0}\land\paren{ \map {h_x} {x,t_0}=0}$.

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 $\left({\tilde a, 1}\right)$ does not exist, since it belongs to this curve.

Hence, there are no conjugate points in the interval $\closedint a b$.