Necessary Condition for Integral Functional to have Extremum for given function/Lemma
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Theorem
Let $\map \alpha x, \map \beta x$ be real functions.
Let $\map \alpha x, \map \beta x$ be continuous on $\closedint a b$.
Let:
- $\forall \map h x \in C^1: \ds \int_a^b \paren {\map \alpha x \map h x + \map \beta x \map {h'} x} \rd x = 0$
subject to the boundary conditions:
- $\map h a = \map h b = 0$
Then $\map \beta x$ is differentiable.
Furthermore:
- $\forall x \in \closedint a b: \map {\beta'} x = \map \alpha x$
Proof
Using Integration by Parts allows us to factor out $\map h x$:
\(\ds \int_a^b \paren {\map \alpha x \map h x + \map \beta x \map {h'} x} \rd x\) | \(=\) | \(\ds \int_a^b \map \alpha x \map h x \rd x + \int_a^b \map \beta x \rd \map h x\) | where $\d \map h x = \map {h'} x \rd x$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \int_a^b \map \alpha x \map h x \rd x + \bigintlimits {\map \beta x \map h x} a b - \int_a^b \map h x \rd \map \beta x\) | Integration by Parts | |||||||||||
\(\ds \) | \(=\) | \(\ds \int_a^b \paren {\map \alpha x - \map {\beta'} x} \map h x \rd x\) | as $\map h a = 0$, $\map h b = 0$ |
Hence the problem has been reduced to:
- $\ds \int_a^b \paren {\map \alpha x - \map {\beta'} x} \map h x \rd x = 0$
Since If Definite Integral of $\map a x \map h x$ vanishes for any $C^0 \map h x$ then $C^0 \map a x$ vanishes, the conclusion is that in the considered interval $\closedint a b$ it holds that:
- $\map \alpha x = \map {\beta'} x$
$\blacksquare$
Sources
- 1963: I.M. Gelfand and S.V. Fomin: Calculus of Variations ... (previous) ... (next): $\S 1.3$: The Variation of a Functional. A Necessary Condition for an Extremum