Necessary Condition for Integral Functional to have Extremum/Two Variables/Lemma

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Theorem

Let $D \subset \R^2$.

Let $\Gamma$ be the boundary of $D$.

Let $\alpha : D \to \R$ be a continuous mapping.

Let $h : D \to \R$ be a twice differentiable mapping such that $\map h \Gamma = 0$.

Suppose for every $h$ we have that:

$\ds \iint_D \map \alpha {x, y} \map h {x,y} \rd x \rd y = 0$.


Then:

$\ds \forall x, y \in D : \map \alpha {x, y} = 0$


Proof

Aiming for a contradiction, suppose that:

$\ds \exists x_0,y_0 \in D : \map \alpha {x_0,y_0} > 0$

$\alpha$ is continuous in $D$.

Therefore, there exists a closed ball $B^-_{\epsilon}$ defined by:

$\map {B^-_{\epsilon}} {x_0, y_0} := \set {\tuple{x,y} \in D : \paren {x - x_0}^2 + \paren {y - y_0}^2 \le \epsilon^2}$

such that:

$\forall x,y \in B^-_{\epsilon} : \map \alpha {x,y} > 0$

Choose $\map h {x, y}$ in the following way:

$\map h {x,y} = \begin {cases} 0, \forall x, y \notin B^-_{\epsilon} \\ \sqbrk {\epsilon^2 - \paren {x - x_0}^2 - \paren {y - y_0}^2}^3, \forall x,y \in B^-_{\epsilon} \end{cases}$

Such a choice for $\map h {x,y}$ satisfies the conditions of the lemma.

But then both $\alpha$ and $h$ are positive inside this ball.

Hence, the integral is positive.

This contradicts assumptions of the lemma.

Hence:

$\ds \forall x, y \in D : \map \alpha {x,y} = 0$

$\blacksquare$


Mistake


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The author uses a slightly different form of the function $h$:

$\map h {x,y} = \begin{cases} 0, \forall x, y \notin B^-_{\epsilon} \\ \sqbrk {\paren {x - x_0}^2 + \paren {y - y_0}^2 - \epsilon^2}^3, \forall x,y \in B^-_{\epsilon} \end{cases}$

Then he concludes that for points inside the circle the integrand is positive.

$\alpha$ is positive by assumption.

However, $h$ is not positive inside since $\paren {x - x_0}^2 + \paren {y - y_0}^2 < \epsilon^2$.

The result is negative, and stays negative after being raised to the $3$rd power.


Sources

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