# 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

- 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