Sufficient Conditions for Weak Extremum
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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.
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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 with respect to $\alpha$.
Thus the [[Definition::Solution to Differential Equation|solution]] of the Euler's equation is continuous with respect to $\alpha $.
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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:
- $\forall x \in \hointl a b: \map h x \ne 0$
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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:
- $(1): \quad \exists c > 0: \ds \int_a^b \paren {P h'^2 + Q h^2} \rd x > c \int_a^b h'^2 \rd x$
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:
- $\ds \forall x \in \closedint a b: \lim_{\size h_1 \mathop \to 0} \set {\xi,\eta} = \set {0, 0}$
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\(\ds \map {h^2} x\) | \(=\) | \(\ds \paren {\int_a^x \map {h'} x \rd x}^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int_a^x 1^2 d y \int_a^x \map {h'^2} x \rd x\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \paren {x - a} \int_a^x \map {h'^2} x \rd x\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \paren {x - a} \int_a^b \map {h'^2} x \rd x\) | $h'^2 \ge 0$ |
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 Function is Completely Multiplicative | |||||||||||
\(\ds \) | \(\le\) | \(\ds \epsilon \int_a^b h'^2 \rd x + \epsilon \frac {\paren {a - b}^2} 2 \int_a^b h'^2 \rd x\) | ||||||||||||
\(\text {(2)}: \quad\) | \(\ds \) | \(=\) | \(\ds \epsilon \paren {1 + \frac {\paren {b - a}^2} 2} \int_a^b h'^2 \rd x\) |
Thus, by $(1)$:
- $\ds \int_a^b \paren {P h'^2 + Q h^2} \rd x > 0$
while by $(2)$:
- $\ds \int_a^b \paren {\xi h'^2 + \eta h^2} \rd x$
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.
$\blacksquare$
Sources
1963: I.M. Gelfand and S.V. Fomin: Calculus of Variations ... (previous) ... (next): $\S 5.27$: Jacobi's Necessary Condition. More on Conjugate Points