Lamé function

Remarks

Example: for $\nu = 1$ and $h = 1 + k^2$, the function $w = \operatorname{sn} z$ is a Lamé polynomial. Check: from $\operatorname{sn}' = \operatorname{cn} \operatorname{dn}$, $\operatorname{cn}^2 = 1 - \operatorname{sn}^2$ and $\operatorname{dn}^2 = 1 - k^2 \operatorname{sn}^2$, one gets $\operatorname{sn}'' = -(1 + k^2) \operatorname{sn} + 2k^2 \operatorname{sn}^3$, so $w'' + (1 + k^2 - 2k^2 \operatorname{sn}^2) w = 0$. In the same way $\operatorname{cn} z$ is a solution for $h = 1$ and $\operatorname{dn} z$ for $h = k^2$. Non-example: for $\nu = 1$ and $h = 0$, the function $\operatorname{sn} z$ is not a solution, since the left side equals $-(1 + k^2) \operatorname{sn} z$, which is not identically $0$. For each nonnegative integer $n$ there are exactly $2n + 1$ values of $h$ that admit a Lamé polynomial of degree $n$. This matches the dimension $2n + 1$ of the space of spherical harmonics of degree $n$. An equivalent form of the equation uses the Weierstrass function $\wp$ in place of $k^2 \operatorname{sn}^2$, and in the variable $\operatorname{sn}^2 z$ it becomes a special case of Heun's equation.