# 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:

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

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

together with Euler's equation:

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

The Euler's equation is continuous with respect to $\alpha$.

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

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:

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

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

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.

 $\ds \map {h^2} x$ $=$ $\ds \paren {\int_a^x \map {h'} x \rd x}^2$ $\ds$ $=$ $\ds \int_a^x 1^2 d y \int_a^x \map {h'^2} x \rd x$ $\ds$ $\le$ $\ds \paren {x - a} \int_a^x \map {h'^2} x \rd x$ $\ds$ $\le$ $\ds \paren {x - a} \int_a^b \map {h'^2} x \rd x$ $h'^2 \ge 0$

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

Integrate the inequality with respect to $x$:

 $\ds \int_a^b \map{h^2} x \rd x$ $\le$ $\ds \int_a^b \paren {x - a} \rd x \int_a^b \map {h'^2} x \rd x$ $\ds$ $=$ $\ds \frac {\paren {b - a}^2} 2 \int_a^b \map {h'^2} x \rd x$

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

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

Then:

 $\ds \size {\int_a^b \paren {\xi h^2 + \eta h'^2} \rd x}$ $\le$ $\ds \int_a^b \size \xi h^2 \rd x + \int_a^b \size \eta h'^2 \rd x$ Absolute Value of Definite Integral, Absolute Value of Product $\ds$ $\le$ $\ds \epsilon \int_a^b h'^2 \rd x + \epsilon \frac {\paren {a - b}^2} 2 \int_a^b h'^2 \rd x$ $\text {(2)}: \quad$ $\ds$ $=$ $\ds \epsilon \paren {1 + \frac {\paren {b - a}^2} 2} \int_a^b h'^2 \rd x$

Thus, by $(1)$:

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

while by $(2)$:

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

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.
$\blacksquare$
1963: I.M. Gelfand and S.V. Fomin: Calculus of Variations ... (previous) ... (next): $\S 5.27$: Jacobi's Necessary Condition. More on Conjugate Points