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 in $\closedint a b$.

Let:

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


Then $\map \beta x$ is differentiable.

Furthermore:

$ \map {\beta'} x=\map \alpha x\quad\forall x\in\closedint a b$.


Proof

Using integration by parts allows us to factor out $\map h x$:

\(\displaystyle \int_a^b \sqbrk{\map \alpha x \map h x+\map\beta x\map {h'} x }\d x\) \(=\) \(\displaystyle \int_a^b \map \alpha x\map h x\d x+\int_a^b \map \beta x\d \map h x\) $\quad$ where $\d \map h x=\map {h'} x\d x$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \displaystyle \int_a^b \map \alpha x \map h x \d x + \map \beta x \map h x \bigg \rvert_a^b - \int_a^b \map h x \d \map \beta x\) $\quad$ integration by parts $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \displaystyle \int_a^b \sqbrk{\map \alpha x-\map {\beta'} x}\map h x \d x\) $\quad$ $\map h a=0$, $\map h b=0$ $\quad$

Hence, the problem has been reduced to

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

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