Spherical harmonics : \begin{equation} Y_{m}^{n}(\theta,\phi) = N_{}\mathcal{P}_{m}^{n}(\cos \phi) e^{in\theta}, \end{equation} are the solution of the two angular equations (Legendre associated eq and let's say second order diff eq): \begin{equation} \begin{cases} \frac{d}{dx}\left[(1-x^2)\frac{d\tilde{\mathcal{P}}_l^n(x)}{dx}\right]+\left[n(n+1)-\frac{m^2}{1-x^2}\right]\tilde{\mathcal{P}}_m^n(x)=0\text{,}\\ \frac{\partial^2 \Theta}{\partial \theta^2}+m^2\Theta=0.\\ \end{cases} \end{equation}
Or on a sphere domain, the order $m$ and degree $n$ are integers and I want to find why using boundary condition (BC). I did it for the degree $n$, which have on a a sphere a continuity BC : \begin{equation} \Theta(0) = \Theta(2\pi). \end{equation} Which result that for integers $n$ , periodicals $\Theta$ functions verifying the BC.
I can't find the BC which prove that $m$ is integer on a sphere for Associated Legendre equation which should be something like : \begin{equation} \begin{cases} \mathcal{P}_{m}^{n}(\cos 0) = ??\\ \mathcal{P}_{m}^{n}(\cos \pi) = ??\\ \end{cases} \end{equation}
Any thoughts ?
It looks as if you got your $n's$ and $m's$ a little mixed up.
It's quite simple. The solution to the differential equation $$y(\theta)'' = -m^2y(\theta)$$ is $$y(\theta)=e^{im\theta}$$ In order to have sensible solutions, the value for $y(0)$ and $y(2\pi)$ need to be the same, as $\theta = 0$ and $\theta = 2\pi$ (or, for that matter, any multiple of $2\pi$), will give the same point in space for fixed $r, \phi$. Hence $1=e^{2im\pi}$, or $2im\pi$ is a multiple of $2\pi i$, hence $m$ is an integer.
$n$ being integer follows from the fact that otherwise the (associated) Legendre equation has no solution at all.
Plugging in integers $m$ and $n$ into the Legendre equation, then leads to the spherical harmonics $Y^m_n (\theta,\phi)$.