Prove or disprove that Weierstrass $\wp$ function is holonomic

Question

Recall that a holonomic function $f$ (say over $\mathbb C$) is one that is a solution to a differential equation of the form:

$$p_0(z)f(z)+p_1(z)f'(z)+p_2(z)f''(z)+\dots+p_k(z)f^{(k)}(z) = 0$$

where the $p_i$ are polynomials in $\mathbb C[z]$ for $i=1,\dots,k$

The Weierstrass $\wp$ function (defined on a given lattice $L$) has this identity $\wp'^2 = g_L(\wp)$ where $g_L$ is some cubic polynomial with non-zero discriminant. My question is whether it is known and whether it is easy to show that this (i.e. the given $\wp$ function) is or is not a holonomic funtion. I have seen how one shows how an algebraic function is a holonomic function and I thought one can use this method with some variation to show that the Weierstrass function is holonomic. This naive thought does not seem to help me though.

From what I have read, the inverse of the Weierstrass function (the elliptic integral) is holonomic. I am not an expert in this area either so I do not know how this was proven and to be honest I doubt if this help to show if Weierstrass $\wp$ is holonomic.

Answer 1

The Weierstraß $\wp$-function has a pole of order $2$ at all lattice points. Hence $\wp^{(m)}$ has a pole of order $m+2$ at all lattice points. If $p_0,p_1,\dotsc, p_{k-1}$ are polynomials, the function

$$h(z) = \sum_{m = 0}^{k-1} p_m(z)\wp^{(m)}(z)$$

has at most a pole of order $k+1$ at the lattice points, but if $p_k$ is a nonzero polynomial, then

$$p_k(z)\wp^{(k)}(z)$$

has a pole of order $k+2$ at all but (possibly) finitely many lattice points.

Therefore the entire meromorphic function

$$\sum_{m = 0}^k p_m(z) \wp^{(m)}(z)$$

has infinitely many poles of exact order $k+2$, in particular, it is not identically $0$. Thus the Weierstraß $\wp$-function is not holonomic.