Equivalence of Definitions of Weierstrass E-Function

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Theorem

Let $\mathbf y,\mathbf z,\mathbf w$ be $n$-dimensional vectors.

Let $\mathbf y$ be such that $\map{\mathbf y} a=A$ and $\map{\mathbf y} b=B$.

Let $J$ be a functional such that:

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

The following definitions of the concept of Weierstrass E-Function are equivalent:

Definition 1

The following mapping is known as the Weierstrass E-Function of $J\sqbrk{\mathbf y}$:

$\map E {x,\mathbf y,\mathbf z,\mathbf w}=\map F {x,\mathbf y,\mathbf w}-\map F {x,\mathbf y,\mathbf z}+\paren{\mathbf w-\mathbf z} F_{\mathbf y'}\paren{x,\mathbf y,\mathbf z}$

Definition 2

Let $\theta\in\R:0<\theta<1$.


The following mapping is known as the Weierstrass E-Function of $J\sqbrk{\mathbf y}$:

$\displaystyle \map E {x,\mathbf y,\mathbf z,\mathbf w}=\frac 1 2\sum_{i,k\mathop=1}^n\paren{w_i-z_i}\paren{w_k-z_k}F_{y_i'y_k'} \paren{x,\mathbf y,\mathbf z+\theta\paren{\mathbf w-\mathbf z} }$


Proof

Definition 1 implies Definition 2

By Definition 1:

$\map E {x,\mathbf y,\mathbf z,\mathbf w}=\map F {x,\mathbf y,\mathbf w}-\map F {x,\mathbf y,\mathbf z}+\paren{\mathbf w-\mathbf z}\map {F_{\mathbf y'} } {x,\mathbf y,\mathbf z}$

By Taylor's Theorem, where expansion is done around $\mathbf w=\mathbf z$ and Lagrange form of remainder is used:

$\displaystyle \map F {x,\mathbf y,\mathbf w}=\map F {x,\mathbf y,\mathbf z}+\frac{\partial \map F {x,\mathbf y,\mathbf z} } {\partial\mathbf y'}\paren{\mathbf w-\mathbf z}+\frac 1 2\sum_{i,j\mathop=1}^n\paren{w_i-z_i}\paren{w_j-z_j}\frac{\partial^2 \map F {x,\mathbf y,\mathbf z+\theta\paren{\mathbf z-\mathbf w} } }{\partial y_i'y_j'}$

where $\theta\in\R:0<\theta<1$.

Insertion of this expansion into the definition for Weierstrass E-Function leads to the desired result.


$\Box$

Definition 2 implies Definition 1


Sources