# Sufficient Conditions for Weak Extremum

## Theorem

Let $J$ be a functional such that:

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

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:

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

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

together with Euler's equation

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

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

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

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

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

Then

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

Also

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

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

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

$\exists c>0:\displaystyle\int_a^b\paren{Ph'^2+Qh^2}\rd x>c\int_a^b h'^2\rd x\quad\paren {\star}$

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 of Legendre's Condition:

$\displaystyle J\sqbrk{y+h}-J\sqbrk y=\int_a^b\paren{Ph'^2+Qh^2}\rd x+\int_a^b\paren{\xi h'^2+\eta h^2}\rd x$

where

$\displaystyle\forall x\in\closedint a b\quad\lim_{\size {h}_1\to 0}\lbrace{\xi,\eta}\rbrace=\lbrace{0,0}\rbrace$

and the limit is uniform.

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

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

Integrate the inequality with respect to $x$:

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

Let $\epsilon>0$ be a constant such that:

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

Then:

 $\displaystyle \size {\int_a^b \paren {\xi h^2+\eta h'^2} \rd x}$ $\le$ $\displaystyle \int_a^b \size{\xi}h^2 \rd x+\int_a^b \size \eta h'^2 \rd x$ $\quad$ Absolute Value of Definite Integral, Absolute Value Function is Completely Multiplicative $\quad$ $\displaystyle$ $\le$ $\displaystyle \epsilon \int_a^b h'^2 \rd x+\epsilon \frac{\paren {a-b}^2} 2 \int_a^b h'^2 \rd x$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle \epsilon \paren {1+\frac{\paren {b-a}^2} 2} \int_a^b h'^2 \rd x \quad \paren \ast$ $\quad$ $\quad$

Thus, by $\paren \star$

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

while by $\paren \ast$

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

Thus, for all sufficiently small $\size {h}_1$, which implies sufficiently small $\size \xi$ and $\size \eta$, and, consequently, sufficiently small $\epsilon$:
 $\displaystyle J\sqbrk {y+h}-J\sqbrk y$ $=$ $\displaystyle \int_a^b \paren {Ph'^2+Q h^2} \rd x+\int_a^b \paren{\xi h'^2+\eta h^2} \rd x>0$ $\quad$ $\quad$
Therefore, in some small neighbourhood $y=\map y x$ is 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