# Conditions for Strong Minimum of Functional

## 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:

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

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:

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

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

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

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

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

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 $\rd\mathbf y=\boldsymbol\psi\rd x$ $\displaystyle$ $=$ $\displaystyle \int_{\gamma^*}\map E {x,\mathbf y,\boldsymbol\psi,\mathbf y'}\rd x$ 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 value in the interval of integration.

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

Therefore:

$\Delta J\ge 0$

$\blacksquare$