In this section we define Weierstrass $h$ functions.
These functions are essentially just the three odd Jacobian (Glaisher) elliptic functions which have a simple pole at the origin.
Handy Preliminary Formulae
Using the DLMF definitions of $\sigma, \omega_i, \eta_i$
we have
\begin{equation*}
\omega_1+\omega_2+\omega_3 = 0\qquad\qquad
\eta_1+\eta_2+\eta_3 = 0\qquad\qquad
\eta_2\omega_1 - \eta_1\omega_2 = \eta_3\omega_2 - \eta_2\omega_3 = \eta_1\omega_3 - \eta_3\omega_1 = \tfrac 1 2 \pi \imath
\end{equation*}
which implies
\begin{equation*}
\eta_1 \omega_1 + \eta_2\omega_2 + \eta_3\omega_3 \space = \space
-2(\eta_1\omega_2 + \eta_2\omega_3 + \eta_3\omega_1) - \tfrac 3 2 \pi \imath \space = \space
-2(\eta_2\omega_1 + \eta_3\omega_2 + \eta_1\omega_3) + \tfrac 3 2 \pi \imath
\end{equation*}
We also have with $i,j,k$ distinct
\begin{equation*}
\sigma(z+2\omega_i) = - e^{2\eta_i (z+\omega_i)} \sigma(z) \qquad\qquad
\sigma(z-\omega_i) = - e^{-2\eta_i z} \sigma(z+\omega_i) \qquad\qquad
\sigma(\omega_j - \omega_i) = e^{-2\eta_i\omega_j}\sigma(\omega_k)
\end{equation*}
and therefore
\begin{equation*}
\wp(z) - \wp(u) \space = \space - \frac {\sigma(z-u)\sigma(z+u)} {\sigma^2(z)\sigma^2(u)} \qquad\qquad
\wp'(z) \space = \space - \frac {\sigma(2z)} {\sigma^4(z)} \qquad\qquad
\wp(z) - e_i \space = \space - \frac {\sigma(z-\omega_i)\sigma(z+\omega_i)} {\sigma^2(z)\sigma^2(\omega_i)} \qquad\qquad
e_j - e_i \space = \space e^{-2\eta_i\omega_j} \frac {\sigma^2(\omega_k)} {\sigma^2(\omega_i) \sigma^2(\omega_j)}
\end{equation*}
and combining eta, sigma and wpwp gives
\begin{equation*}
\kappa \space = \space
\begin{vmatrix}
1 & e_1 & e_1^2 \\
1 & e_2 & e_2^2 \\
1 & e_3 & e_3^2 \\
\end{vmatrix} \space = \space
(e_1-e_2)(e_2-e_3)(e_3-e_1) \space = \space
\frac {e^{\eta_1\omega_1+\eta_2\omega_2+\eta_3\omega_3+\sfrac 3 2 \pi \imath}} {\sigma^2(\omega_1)\sigma^2(\omega_2)\sigma^2(\omega_3)} \space = \space
\frac {\sigma(\omega_1 - \omega_2)\sigma(\omega_2 - \omega_3)\sigma(\omega_3 - \omega_1)} {\sigma^3(\omega_1)\sigma^3(\omega_2)\sigma^3(\omega_3)}
\end{equation*}
We also need these sigma duplication formulae
\begin{equation*}
\sigma(2z) \enspace = \enspace \frac {2\sigma(z)\sigma(z+\omega_1)\sigma(z+\omega_2)\sigma(z+\omega_3)} {\sigma(\omega_1)\sigma(\omega_2)\sigma(\omega_3)} \enspace = \enspace
-\frac {2\sigma(z)\sigma(z-\omega_1)\sigma(z-\omega_2)\sigma(z-\omega_3)} {\sigma(\omega_1)\sigma(\omega_2)\sigma(\omega_3)}
\end{equation*}
Definition Of The $h$ Functions
Define $h_i(z)$ so that it's residue is $+1$ at $z=0$
\begin{equation*}
h_i(z) \enspace = \enspace \sqrt{\wp(z) - e_i} \enspace = \enspace
-e^{\eta_i z} \frac {\sigma(z - \omega_i)} {\sigma(z)\sigma(\omega_i)} \enspace = \enspace
e^{-\eta_i z} \frac {\sigma(z + \omega_i)} {\sigma(z)\sigma(\omega_i)}
\end{equation*}
From hdef we get the basic relations with $i,j,k$ distinct
\begin{equation*}
h_i(z) = \frac 1 z \space + \space \bigO(z) \qquad\qquad
h_i(-z) = -h_i(z) \qquad\qquad
h_i(z + 2\omega_i) = h_i(z) \qquad\qquad
h_i(z + 2\omega_j) = -h_i(z)
\end{equation*}
Taking the product of hdef gives
\begin{equation*}
h_1(z)h_2(z)h_3(z) \enspace = \enspace -\tfrac 1 2 \wp'(z)
\end{equation*}
Differentiating hdef gives
\begin{equation*}
h_i'(z) \enspace = \enspace -h_j(z)h_k(z)
\end{equation*}
Since
\begin{equation*}
h_i^2(z) \space - \space h_j^2(z) \enspace = \enspace e_j \space - \space e_i
\end{equation*}
the differential equation for $h_i(z)$ is
\begin{equation*}
{h_i'}^2(z) \enspace = \enspace h_i^4(z) \enspace + \enspace 3 e_i h_i^2(z) \enspace + \enspace(e_i - e_j)(e_i - e_k)
\end{equation*}
An Addition Formula for $h_i$
The functions $h_1,h_2,h_3$ are equivalent, after scaling, to the Jacobian elliptic functions $\ns,\cs,\ds$ with $\displaystyle k^2 = \frac {e_3-e_1} {e_2-e_1}$.
From this observation, it is easy to determine this addition formulae for them
\begin{equation*}
h_i(u+v) \enspace = \enspace \frac {h_i(u) \thinspace h_i'(v) \space - \space h_i(v) \thinspace h_i'(u)} {h_i^2(u) \space - \space h_i^2(v)}
\enspace = \enspace \frac {h_i^2(\omega_j)h_i^2(\omega_k) \space - \space h_i^2(u)h_i^2(v)} {h_i(u) \thinspace h_i'(v) \space + \space h_i(v) \thinspace h_i'(u)}
\end{equation*}
$K = h_1(\omega_2), \space \sn = K/h_1, \space \cn = h_2/h_1, \space \dn = h_3/h_1$ implies $K^2 = e_2 - e_1$, $\sn^2 + \cn^2 = 1$ and $k^2\sn^2 + \dn^2 = 1$
A well known addition formula for $\sn$ is
\begin{equation*}
\sn(u+v) \enspace = \enspace \frac {\sn^2 u \space - \space \sn^2 v} {\sn u \cn v \dn v \space - \space \sn v \cn u \dn u}
\enspace = \enspace \frac {\sn u \cn v \dn v \space + \space \sn v \cn u \dn u} {1 \space - \space k^2 \sn^2 \sn^2 v}
\end{equation*}
which translates to
\begin{equation*}
\frac K {h_1(u+v)} \enspace = \enspace \frac {K^2/h_1^2(u) - K^2/h_1^2(v)} {K h_2(v) h_3(v) / h_1(u) h_1^2(v) - K h_2(u) h_3(u) / h_1^2(u) h_1(v)} \qquad \implies \qquad
h_1(u+v) \enspace = \enspace \frac {h_1(u)h_2(v)h_3(v) \space - \space h_1(v)h_2(u)h_3(u)} {h_1^2(v) \space - \space h_1^2(u)}
\end{equation*}
and
\begin{equation*}
\frac K {h_1(u+v)} \enspace = \enspace \frac {K h_2(v) h_3(v) / h_1(u) h_1^2(v) + K h_2(u) h_3(u) / h_1^2(u) h_1(v)} {1 - k^2 K^4/h_1^2(u)h_1^2(v)} \qquad \implies \qquad
h_1(u+v) \enspace = \enspace \frac {h_1^2(v) h_1^2(u) \space - \space k^2K^4} {h_1(u)h_2(v)h_3(v) \space + \space h_1(v)h_2(u)h_3(u)}
\end{equation*}
from which follows add_h for $h = h_1$.
By symmetry it is also true for $h_2,h_3$.
References