Sufficient Conditions for Weak Extremum

Theorem
Let $J$ be a functional such that:


 * $\ds J \sqbrk y = \int_a^b \map F {x, y, y'} \rd x$
 * $\map y a = A$
 * $\map y b = B$

Let $y = \map y x$ be an extremum.

Let the strengthened Legendre's Condition hold.

Let the strengthened Jacobi's Necessary Condition hold.

Then the functional $J$ has a weak minimum for $y = \map y x$.

Proof
By the continuity of function $\map P x$ and the solution of Jacobi's equation:


 * $\exists \epsilon > 0: \paren {\forall x \in \closedint a {b + \epsilon}:\map P x > 0} \land \paren {\tilde a \notin \closedint a {b + \epsilon} }$

Consider the quadratic functional:


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

together with Euler's equation:


 * $-\dfrac \rd {\rd x} \paren{\paren {P - \alpha^2} h'} + Q h = 0$

The Euler's equation is continuous $\alpha$.

Thus the solution of the Euler's equation is continuous $\alpha $.

Since:


 * $\forall x \in \closedint a {b + \epsilon}: \map P x > 0$

$\map P x$ has a positive lower bound in $\closedint a {b + \epsilon}$.

Consider the solution with $\map h a = 0$, $\map {h'} 0 = 1$.

Then


 * $\exists \alpha \in \R: \forall x \in \closedint a b: \map P x - \alpha^2 > 0$

Also:


 * $\forall x \in \hointl a b: \map h x \ne 0$

By Necessary and Sufficient Condition for Quadratic Functional to be Positive Definite:


 * $\ds \int_a^b \paren {\paren {P - \alpha^2} h'^2 + Q h^2} \rd x > 0$

In other words, if $c = \alpha^2$, then:


 * $(1): \quad \exists c > 0: \displaystyle \int_a^b \paren {P h'^2 + Q h^2} \rd x > c \int_a^b h'^2 \rd x$

Let $y = \map y x$ be an extremal.

Let $y = \map y x + \map h x$ be a curve, sufficiently close to $y = \map y x$.

By expansion of $\Delta J \sqbrk {y; h}$ from lemma $1$ of Legendre's Condition:


 * $\ds J \sqbrk {y + h} - J \sqbrk y = \int_a^b \paren {P h'^2 + Q h^2} \rd x + \int_a^b \paren {\xi h'^2 + \eta h^2} \rd x$

where:


 * $\ds \forall x \in \closedint a b: \lim_{\size h_1 \mathop \to 0} \set {\xi,\eta} = \set {0, 0}$

and the limit is uniform.

By Schwarz inequality:

Notice that the integral on the right does not depend on $x$.

Integrate the inequality $x$:

Let $\epsilon \in \R_{>0}$ be a constant such that:


 * $\size \xi \le \epsilon$, $\size \eta \le \epsilon$

Then:

Thus, by $(1)$:


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

while by $(2)$:


 * $\ds \int_a^b \paren {\xi h'^2 + \eta h^2} \rd x$

can be made arbitrarily small.

Thus, for all sufficiently small $\size h_1$, which implies sufficiently small $\size \xi$ and $\size \eta$, and, consequently, sufficiently small $\epsilon$:


 * $J \sqbrk {y + h} - J \sqbrk y = \int_a^b \paren {P h'^2 + Q h^2} \rd x + \int_a^b \paren {\xi h'^2 + \eta h^2} \rd x > 0$

Therefore, in some small neighbourhood $y = \map y x$ there exists a weak minimum of the functional.