Conditions for Strong Minimum of Functional

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Theorem

Let $\mathbf y$ be an $n$-dimensional vector such that $\map {\mathbf y} a = A$ and $\map {\mathbf y} b = B$

Let $J$ be a functional such that:

$\ds J \sqbrk {\mathbf y} = \int_a^b \map F {x, \mathbf y, \mathbf y'} \rd x$

Let $\gamma$ be an extremal curve of $J$.

Let the following be the field of the functional $J$:

$\mathbf y' = \map {\boldsymbol \psi} {x, \mathbf y}$

Let $R$ be an open region containing $\gamma$ and have the field $\boldsymbol \psi$ defined as $\forall \paren {x, \mathbf y} \in R$.

Let $\mathbf w$ be a finite vector.

Suppose that:

$\forall \paren {x, \mathbf y} \in R: \map E {x, \mathbf y, \boldsymbol \psi, \mathbf w}\ge 0$

where $E$ is Weierstrass E-Function.


Then $J$ has a strong minimum for $\gamma$.



Proof

By definition, the increment of $J$ is:

$\ds \Delta J = \int_{\gamma^*} \map F {x, \mathbf y, \mathbf y'} \rd x - \int_\gamma \map F {x, \mathbf y, \mathbf y'} \rd x$

where $\gamma$ and $\gamma^*$ are curves described by $\paren {x, \map {\mathbf y} x}$ and $\paren {x, \paren {\mathbf y^*} x}$ respectively, such that $\map {\mathbf y^*} x - \map {\mathbf y} x = \map {\mathbf h} x$.

Consider Hilbert's invariant integral:

\(\displaystyle \map g {x, \mathbf y}\) \(=\) \(\displaystyle \int_\Gamma \paren {-H \rd x + \mathbf p \rd \mathbf y}\)
\(\displaystyle \) \(=\) \(\displaystyle \int_\Gamma \paren {-\paren {-F + \mathbf y' F_{\mathbf y'} } \rd x + F_{\mathbf y'} \rd \mathbf y}\) Definition of Hamiltonian and Momentum
\(\displaystyle \) \(=\) \(\displaystyle \int_\Gamma \paren {F \rd x - \mathbf y' F_{\mathbf y'} \rd x + F_{\mathbf y'} \rd \mathbf y}\)
\(\displaystyle \) \(=\) \(\displaystyle \int_{\Gamma} \paren {F \rd x - \boldsymbol \psi F_{\mathbf y'} \rd x + F_{\mathbf y'} \rd \mathbf y}\) $\mathbf y' = \map {\boldsymbol \psi} {x, \mathbf y}$

Since the integrand is full differential, the integral does not depend on the shape of $ \Gamma $, but only on its endpoints.

Therefore, for $\Gamma = \gamma$ and $\Gamma = \gamma^*$ the value of the integral is the same.

Since $\map {\mathbf y'} x = \map {\boldsymbol \psi} {x, \mathbf y}$ determines boundary conditions for $\gamma$, $\Gamma = \gamma$ is one of the trajectories of the field $\map {\mathbf y'} x = \map {\boldsymbol \psi} {x, \mathbf y}$.

Hence, $\d \mathbf y$ is constrained by $\d \mathbf y = \boldsymbol \psi \rd x$:

$\ds \map g {x, \mathbf y} = \int_\gamma \map F {x, \mathbf y, \mathbf y'} \rd x$

Thus, $\map g {x, \mathbf y'}$ can be written in two different ways.

$\ds \int_\gamma \map F {x, \mathbf y, \mathbf y'} \rd x = \int_{\gamma^*} \paren {\paren {\map F {x, \mathbf y, \boldsymbol \psi} - \boldsymbol \psi F_{\mathbf y'} \paren {x, \mathbf y, \boldsymbol \psi} } \rd x + F_{\mathbf y'} \paren {x, \mathbf y, \boldsymbol \psi} \rd \mathbf y}$

Substitute this into the expression for $\Delta J$:

\(\displaystyle \Delta J\) \(=\) \(\displaystyle \int_{\gamma^*} \map F {x, \mathbf y, \mathbf y'} \rd x - \int_{\gamma^*} \paren {\paren {\map F {x, \mathbf y, \boldsymbol \psi} - \boldsymbol \psi \map {F_{\mathbf y'} } {x, \mathbf y, \boldsymbol \psi} }\rd x + \map {F_{\mathbf y'} } {x, \mathbf y, \boldsymbol \psi} \rd \mathbf y}\)
\(\displaystyle \) \(=\) \(\displaystyle \int_{\gamma^*} \paren {\map F {x, \mathbf y, \mathbf y'} - \map F {x, \mathbf y, \boldsymbol \psi} - \paren {\mathbf y'-\boldsymbol \psi} \map {F_{\mathbf y'} } {x, \mathbf y, \boldsymbol \psi} }\rd x\) as $\d \mathbf y = \boldsymbol \psi \rd x$
\(\displaystyle \) \(=\) \(\displaystyle \int_{\gamma^*} \map E {x, \mathbf y, \boldsymbol\psi, \mathbf y'} \rd x\) Definition of Weierstrass E-Function

By assumption:

$\map E {x, \mathbf y, \boldsymbol \psi, \mathbf y'} \ge 0$

Hence the integrand is bounded below by $0$ and above by its maximum in the interval of integration.

Hence, the integral is bounded below by $0$ and above by some positive number.

Therefore:

$\Delta J \ge 0$

$\blacksquare$


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