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:

$\forall \map h x \in C^1: \displaystyle \int_a^b \paren {\map \alpha x \map h x + \map \beta x \map {h'} x} \d x = 0$

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 $\map h x$:

\(\displaystyle \int_a^b \paren {\map \alpha x \map h x + \map \beta x \map {h'} x} \rd x\) \(=\) \(\displaystyle \int_a^b \map \alpha x \map h x \rd x + \int_a^b \map \beta x \rd \map h x\) where $\d \map h x = \map {h'} x \rd x$
\(\displaystyle \) \(=\) \(\displaystyle \int_a^b \map \alpha x \map h x \rd x + \bigintlimits {\map \beta x \map h x} a b - \int_a^b \map h x \rd \map \beta x\) Integration by Parts
\(\displaystyle \) \(=\) \(\displaystyle \int_a^b \paren {\map \alpha x - \map {\beta'} x} \map h x \rd x\) as $\map h a = 0$, $\map h b = 0$

Hence the problem has been reduced to:

$\displaystyle \int_a^b \paren {\map \alpha x - \map {\beta'} x} \map h x \rd x = 0$

Since If Definite Integral of a(x)h(x) vanishes for any C^0 h(x) then C^0 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