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Weierstraß Elliptic Function

\begin{figure}\begin{center}\BoxedEPSF{Weierstrass.epsf scaled 850}\end{center}\end{figure}

\begin{figure}\begin{center}\BoxedEPSF{WeierstrassReIm.epsf scaled 700}\end{center}\end{figure}

\begin{figure}\begin{center}\BoxedEPSF{WeierstrassPrimeReIm.epsf scaled 770}\end{center}\end{figure}

The Weierstraß elliptic functions are elliptic functions which, unlike the Jacobi Elliptic Functions, have a second-order Pole at $z=0$. The above plots show the Weierstraß elliptic function $\wp(z)$ and its derivative $\wp'(z)$ for invariants (defined below) of $g_2=0$ and $g_3=0$. Weierstraß elliptic functions are denoted $\wp(z)$ and can be defined by

\wp(z) = {1\over z^2} + \setbox0=\hbox{$\scriptstyle{m, n=-\...
...a_1-2n\omega_2)^2}-{1\over (2m\omega_1+2n\omega_2)^2}}\right].
\end{displaymath} (1)

Write $\Omega_{mn}\equiv 2m\omega_1+2n\omega_2$. Then this can be written
\wp(z)=z^{-2}+\setbox0=\hbox{$\scriptstyle{m, n}$}\setbox2=\...
...}'}_{\kern-\wd4 m, n} [(z-\Omega_{mn})^{-2}-\Omega_{mn}^{-2}].
\end{displaymath} (2)

An equivalent definition which converges more rapidly is

\wp(z)=\left({\pi\over 2\omega_1}\right)^2\left[{-{1\over 3}...
...}^\infty \csc^2\left({{n\omega_2\over n_1}\pi}\right)}\right].
\end{displaymath} (3)

$\wp(z)$ is an Even Function since $\wp(-z)$ gives the same terms in a different order. To specify $\wp$ completely, its periods or invariants, written $\wp(z\vert\omega_1,\omega_2)$ and $\wp(z;g_2,g_3)$, respectively, must also be specified.

The differential equation from which Weierstraß elliptic functions arise can be found by expanding about the origin the function $f(z)\equiv \wp(z)-z^{-2}$.

\wp(z)-z^{-2}=f(0)+f'(0)z+{\textstyle{1\over 2!}} f''(0)z^2+...
...ver 3!}}f'''(0)z^3+{\textstyle{1\over 4}}f^{(4)}(0)z^4+\ldots.
\end{displaymath} (4)

But $f(0)=0$ and the function is even, so $f'(0)=f'''(0)=0$ and
f(z)=\wp(z)-z^{-2}={\textstyle{1\over 2!}} f''(0)z^2+{\textstyle{1\over 4}}f^{(4)}(0)z^4+\ldots.
\end{displaymath} (5)

Taking the derivatives
$\displaystyle f'$ $\textstyle =$ $\displaystyle -2\Sigma'(z-\Omega_{mn})^{-3}$ (6)
$\displaystyle f''$ $\textstyle =$ $\displaystyle 6\Sigma'(z-\Omega_{mn})^{-4}$ (7)
$\displaystyle f'''$ $\textstyle =$ $\displaystyle -24\Sigma'(z-\Omega_{mn})^{-5}$ (8)
$\displaystyle f^{(4)}$ $\textstyle =$ $\displaystyle 120\Sigma'(z-\Omega_{mn})^{-6}.$ (9)

$\displaystyle f''(0)$ $\textstyle =$ $\displaystyle 6\Sigma'\Omega_{mn}^{-4}$ (10)
$\displaystyle f^{(4)}(0)$ $\textstyle =$ $\displaystyle 120\Sigma'\Omega_{mn}^{-6}.$ (11)

Plugging in,
\wp(z)-z^{-2}=3\Sigma'\Omega_{mn}^{-4} z^2+5\Sigma'\Omega_{mn}^{-6}z^4+{\mathcal O}(z^6).
\end{displaymath} (12)

Define the Invariants
$\displaystyle g_2$ $\textstyle \equiv$ $\displaystyle 60\Sigma'\Omega_{mn}^{-4}$ (13)
$\displaystyle g_3$ $\textstyle \equiv$ $\displaystyle 140\Sigma'\Omega_{mn}^{-6},$ (14)

\wp(z)=z^{-2}+{\textstyle{1\over 20}} g_2z^2+{\textstyle{1\over 28}}g_3z^4+{\mathcal O}(z^6)
\end{displaymath} (15)

\wp'(z)=-2z^{-3}+{\textstyle{1\over 10}} g_2z+{\textstyle{1\over 7}} g_3z^3+{\mathcal O}(z^5).
\end{displaymath} (16)

Now cube (15) and square (16)
\wp^3(z) = z^{-6}+{\textstyle{3\over 20}}g_2z^{-2}+{\textstyle{3\over 28}}g_3+{\mathcal O}(z^2)
\end{displaymath} (17)

{\wp'}^2(z) = 4z^{-6}-{\textstyle{2\over 5}} g_2z^{-2}-{\textstyle{4\over 7}}g_3+{\mathcal O}(z^2).
\end{displaymath} (18)

Taking (18) minus $4\times$ (17) cancels out the $z^{-6}$ term, giving

$\displaystyle {\wp'}^2(z)-4\wp^3(z)$ $\textstyle =$ $\displaystyle \left({-{\textstyle{2\over 5}}-{\textstyle{3\over 5}}}\right)g_2z...
...t({-{\textstyle{4\over 7}}-{\textstyle{3\over 7}}}\right)g_3 +{\mathcal O}(z^2)$  
  $\textstyle =$ $\displaystyle -g_2z^{-2}-g_3+{\mathcal O}(z^2)$ (19)

{\wp'}^2(z)-4\wp^3(z)+g_2z^{-2}+g_3={\mathcal O}(z^2).
\end{displaymath} (20)

But, from (5)
\wp(z)=z^{-2}+{\textstyle{1\over 2!}} f''(0)z^2+{\textstyle{1\over 4}}f^{(4)}(0)z^4+\ldots,
\end{displaymath} (21)

so $\wp(z)=z^{-2}+{\mathcal O}(z^2)$ and (20) can be written
{\wp'}^2(z)-4\wp^3(z)+g_2\wp(z)+g_3={\mathcal O}(z^2).
\end{displaymath} (22)

The Weierstraß elliptic function is analytic at the origin and therefore at all points congruent to the origin. There are no other places where a singularity can occur, so this function is an Elliptic Function with no Singularities. By Liouville's Elliptic Function Theorem, it is therefore a constant. But as $z\to 0$, ${\mathcal O}(z^2)\to 0$, so

\end{displaymath} (23)

The solution to the differential equation
\end{displaymath} (24)

is therefore given by $y=\wp(z+\alpha)$, providing that numbers $\omega_1$ and $\omega_2$ exist which satisfy the equations defining the Invariants. Writing the differential equation in terms of its roots $e_1$, $e_2$, and $e_3$,
\end{displaymath} (25)

2\ln(y') = \ln 4+\sum_{r=1}^3 \ln(y-e_r)
\end{displaymath} (26)

{2y''\over y'} = y'\sum_{r=1}^3 (y-e_r)^{-1}
\end{displaymath} (27)

{2y''\over {y'}^2} = \sum_{r=1}^3 (y-e_r)^{-1}
\end{displaymath} (28)

2{{y'}^2y'''-y''(2y'y'')\over {y'}^4} = -y'\sum_{r=1}^3 (y-e_r)^{-2}
\end{displaymath} (29)

{2y'''\over {y'}^3}-{4{y''}^2\over {y'}^4} = -\sum_{r=1}^3 (y-e_r)^{-2}.
\end{displaymath} (30)

Now take (30) divided by 4 plus [(30) divided by 4] quantity squared,

\left({{y'''\over 2{y'}^3}-{{y''}^2\over {y'}^4}}\right)+\le...
...{-2} +{1\over 16} \left[{\,\sum_{r=1}^3 (y-e_r)^{-1}}\right]^2
\end{displaymath} (31)

{3{y''}^2\over 4{y'}^4}-{y'''\over 2{y'}^3} = {\textstyle{3\...
...-e_r)^{-2}-{\textstyle{3\over 8}} y\prod_{r=1}^3 (y-e_r)^{-1}.
\end{displaymath} (32)

The term on the right is half the Schwarzian Derivative.

The Derivative of the Weierstraß elliptic function is given by

$\displaystyle \wp'(z)$ $\textstyle =$ $\displaystyle {d\over dz} \wp(z) = -2\sum_{m,n} {1\over (z-\Omega_{mn})^3}$  
  $\textstyle =$ $\displaystyle -2z^{-3}-2 \setbox0=\hbox{$\scriptstyle{m,n}$}\setbox2=\hbox{$\di...
\mathop{{\sum}'}_{\kern-\wd4 m,n} (z-\Omega_{mn})^{-3}.$ (33)

This is an Odd Function which is itself an elliptic function with pole of order 3 at $z=0$. The Integral is given by
z=\int_{\wp(z)}^\infty (4t^3-g_2t-g_3)^{-1/2}\,dt.
\end{displaymath} (34)

A duplication formula is obtained as follows.

$\displaystyle \wp(2z)$ $\textstyle =$ $\displaystyle \lim_{y\to z} \wp(y+z) ={1\over 4} \lim_{y\to z} \left[{\wp'(z)-\wp'(y)\over \wp(z)-\wp(y)}\right]^2-\wp(z)-\lim_{y\to z} \wp(y)$  
  $\textstyle =$ $\displaystyle {1\over 4}\lim_{h\to 0} \left[{\wp(z)-\wp'(z+h)\over \wp(z)-\wp(z+h)}\right]^2-2\wp(z)$  
  $\textstyle =$ $\displaystyle {1\over 4}\left\{{\left[{\lim_{h\to 0} {\wp'(z)-\wp'(z+h)\over h}}\right]\left[{\lim_{h\to 0} {h\over \wp(z)-\wp(z+h)}}\right]}\right\}^2-2\wp(z)$  
  $\textstyle =$ $\displaystyle {1\over 4}\left[{\wp''(z)\over\wp'(z)}\right]^2-2\wp(z)$ (35)

A general addition theorem is obtained as follows. Given

\end{displaymath} (36)

\end{displaymath} (37)

with zero $y$ and $z$ where $z\not\equiv \pm y \,({\rm mod\ }2\omega_1,2\omega_2)$, find the third zero $\zeta$. Consider $\wp'(\zeta)-A\wp(\zeta)-B$. This has a pole of order three at $\zeta=0$, but the sum of zeros ($=0$) equals the sum of poles for an Elliptic Function, so $z+y+\zeta=0$ and $\zeta=-z-y$.
\end{displaymath} (38)

\end{displaymath} (39)

Combining (36), (37), and (39) gives
\left[{\matrix{\wp(z) & \wp'(z) & 1\cr \wp(y) & \wp'(y) & 1\...
... -1\cr B\cr}}\right] = \left[{\matrix{0\cr 0\cr 0\cr}}\right],
\end{displaymath} (40)

\left\vert\matrix{\wp(z) & \wp'(z) & 1\cr \wp(y) & \wp'(y) & 1\cr \wp(z+y) & -\wp(z+y) & 1\cr}\right\vert = 0.
\end{displaymath} (41)

Defining $u+v+w=0$ where $u\equiv z$ and $v\equiv y$ gives the symmetric form
\left\vert\matrix{\wp(u) & \wp'(u) & 1\cr \wp(v) & \wp'(v) & 1\cr \wp(w) & \wp(w) & 1\cr}\right\vert = 0.
\end{displaymath} (42)

To get the expression explicitly, start again with
\end{displaymath} (43)

where $\zeta=z, y, -z-y$.
\end{displaymath} (44)

But ${\wp}^2(\zeta)=4\wp^4(\zeta)-g_2\wp(\zeta)-g_3$, so
\end{displaymath} (45)

The solutions $\wp(\zeta)\equiv z$ are given by
\end{displaymath} (46)

But the sum of roots equals the Coefficient of the squared term, so
\wp(z)+\wp(y)+\wp(z+y)={\textstyle{1\over 4}}A^2
\end{displaymath} (47)

\end{displaymath} (48)

A={\wp'(z)-\wp'(y)\over \wp(z)-\wp(y)}
\end{displaymath} (49)

\wp(z+y)={1\over 4}\left[{\wp(z)-\wp'(y)\over \wp(z)-\wp(y)}\right]^2-\wp(z)-\wp(y).
\end{displaymath} (50)

Half-period identities include

$\displaystyle x$ $\textstyle \equiv$ $\displaystyle \wp({\textstyle{1\over 2}}\omega_1) = \wp(-h \omega_1+\omega_1)
= e_1+{(e_1-e_2)(e_1-e_3)\over \wp(-{\textstyle{1\over 2}}\omega_1)-e_1}$  
  $\textstyle =$ $\displaystyle e_1+{(e_1-e_2)(e_1-e_3)\over x-e_1}.$ (51)

Multiplying through,
\end{displaymath} (52)

\end{displaymath} (53)

which gives
$\displaystyle \wp({\textstyle{1\over 2}}\omega_1)$ $\textstyle =$ $\displaystyle {\textstyle{1\over 2}}\left\{{2e_1\pm\sqrt{4{e_1}^2-4[{e_1}^2-(e_1-e_2)(e_1-e_3)]}}\right\}$  
  $\textstyle =$ $\displaystyle e_1\pm\sqrt{(e_1-e_2)(e_1-e_3)}.$ (54)

From Whittaker and Watson (1990, p. 445),

\wp'({\textstyle{1\over 2}}\omega_1) = -2\sqrt{(e_1-e_2)(e_1-e_3)}(\sqrt{e_1-e_2}+\sqrt{e_1-e_3}\,).
\end{displaymath} (55)

The function is Homogeneous,

\wp(\lambda z\vert\lambda \omega_1,\lambda\omega_2)=\lambda^{-2}\wp(z\vert\omega_1,\omega_2)
\end{displaymath} (56)

\wp(\lambda z; \lambda^{-4}g_2,\lambda^{-6}g_3) = \lambda^{-2}\wp(z; g_2,g_3).
\end{displaymath} (57)

To invert the function, find $2\omega_1$ and $2\omega_2$ of $\wp(z\vert\omega_1,\omega_2)$ when given $\wp(z;g_2,g_3)$. Let $e_1$, $e_2$, and $e_3$ be the roots such that $(e_1-e_2)/(e_1-e_3)$ is not a Real Number $>1$ or $<0$. Determine the Parameter $\tau$ from

{e_1-e_2\over e_1-e_3} = {{\vartheta_4}^4(0\vert\tau) \over {\vartheta_3}^4(0\vert\tau)}.
\end{displaymath} (58)

Now pick
A\equiv {\sqrt{e_1-e_2}\over {\vartheta_4}^2(0\vert\tau)}.
\end{displaymath} (59)

As long as ${g_2}^3\not=27g_3$, the periods are then
2\omega_1=\pi A
\end{displaymath} (60)

2\omega_2={\pi\tau\over A}.
\end{displaymath} (61)

Weierstraß elliptic functions can be expressed in terms of Jacobi Elliptic Functions by

\wp(u;g_2,g_3)=e_3+(e_1-e_3) {\rm ns}^2\left({u\sqrt{e_1-e_3}\,, \sqrt{{e_2-e_3}\over {e_1-e_3}}\,}\right),
\end{displaymath} (62)

$\displaystyle \wp(\omega_1)$ $\textstyle =$ $\displaystyle e_1$ (63)
$\displaystyle \wp(\omega_2)$ $\textstyle =$ $\displaystyle e_2$ (64)
$\displaystyle \wp(\omega_3)$ $\textstyle =$ $\displaystyle -\wp(-\omega_1-\omega_2)=e_3,$ (65)

and the Invariants are
$\displaystyle g_2$ $\textstyle \equiv$ $\displaystyle 60\Sigma'\Omega_{mn}^{-4}$ (66)
$\displaystyle g_3$ $\textstyle \equiv$ $\displaystyle 140\Sigma' \Omega_{mn}^{-6}.$ (67)

Here, $\Omega_{mn}\equiv 2m\omega_1-2n\omega_2$.

An addition formula for the Weierstraß elliptic function can be derived as follows.

\wp(z+\omega_1)+\wp(z)+\wp(\omega_1) = {1\over 4} \left[{\wp...
...a_1)}\right]^2 = {1\over 4} {{\wp'}^2(z)\over [\wp(z)-e_1]^2}.
\end{displaymath} (68)

\wp'(z)=4\prod_{r=1}^3 [\wp(z)-e_r],
\end{displaymath} (69)

$\displaystyle \wp(z+\omega_1)$ $\textstyle =$ $\displaystyle -\wp(z)-e_1+{1\over 4} {4\prod_{r=1}^3 [\wp(z)-e_r]\over [\wp(z)-e_1]^2}$  
  $\textstyle =$ $\displaystyle -\wp(z)-e_1+{[\wp(z)-e_2][\wp(z)-e_3]\over \wp(z)-e_1}.$  

Use $\sum_{r=1}^3 e_r=0$,

$\displaystyle \wp(z+\omega_1)$ $\textstyle =$ $\displaystyle e_1+{[-2e_1-\wp(z)][\wp(z)-e_1]\over \wp(z)-e_1}+ {\wp^2(z)-\wp(z)(e_2+e_3)+e_2e_3\over \wp(z)-e_1}$  
  $\textstyle =$ $\displaystyle e_1+{-\wp(z)(e_1+e_2+e_3)+e_2e_3+2{e_1}^2\over \wp(z)-e_1}.$ (71)

But $\sum_{r=1}^3 e_r=0$ and
\end{displaymath} (72)

\wp(z+\omega_1) = e_1 +{(e_1-e_2)(e_1-e_3)\over \wp(z)-e_1}.
\end{displaymath} (73)

The periods of the Weierstraß elliptic function are given as follows. When $g_2$ and $g_3$ are Real and ${g_2}^3-27{g_3}^2>0$, then $e_1$, $e_2$, and $e_3$ are Real and defined such that $e_1>e_2>e_3$.

$\displaystyle \omega_1$ $\textstyle =$ $\displaystyle \int_{e_1}^\infty (4t^3-g_2t-g_3)^{-1/2}\,dt$ (74)
$\displaystyle \omega_3$ $\textstyle =$ $\displaystyle -i\int_{-\infty}^{e_3} (g_3+g_2t-4t^3)^{-1/2}\,dt$ (75)
$\displaystyle \omega_2$ $\textstyle =$ $\displaystyle -\omega_1-\omega_3.$ (76)

The roots of the Weierstraß elliptic function satisfy

\end{displaymath} (77)

\end{displaymath} (78)

\end{displaymath} (79)

where $\omega_3\equiv -\omega_1-\omega_2$. The $e_i$s are Roots of $4t^3-g_2t-g_3$ and are unequal so that $e_1\not=
e_2\not= e_3$. They can be found from the relationships
\end{displaymath} (80)

e_2e_3+e_3e_1+e_1e_2=a_1=-{\textstyle{1\over 4}}g_2
\end{displaymath} (81)

e_1e_2e_3=-a_0={\textstyle{1\over 4}}g_3.
\end{displaymath} (82)

See also Equianharmonic Case, Lemniscate Case, Pseudolemniscate Case


Abramowitz, M. and Stegun, C. A. (Eds.). ``Weierstrass Elliptic and Related Functions.'' Ch. 18 in Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, 9th printing. New York: Dover, pp. 627-671, 1972.

Fischer, G. (Ed.). Plates 129-131 in Mathematische Modelle/Mathematical Models, Bildband/Photograph Volume. Braunschweig, Germany: Vieweg, pp. 126-128, 1986.

Whittaker, E. T. and Watson, G. N. A Course in Modern Analysis, 4th ed. Cambridge, England: Cambridge University Press, 1990.

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© 1996-9 Eric W. Weisstein