Consider the following 3D potential: $$V(r)=\begin{cases}-V_0 \ \ \ \ \ \ & r \leq a\\ 0 &r > a\end{cases}$$ we want to find the eigenfunctions for $\ell=0$, in particular we are interested in the bounded states. Of course we get: $$\left[\frac{p_r}{2m}+V(r)\right]R(r)=ER(r)$$ and so with: $$u(r):=rR(r)$$ $$-\frac{\hbar ^2}{2m}u(r)+V(r)u(r)=Eu(r)$$ so for bounded states we get: $$u(x)=\begin{cases}A\sin{\alpha r}+B\cos{\alpha r} \ \ \ \ \ \ & r \leq a\\Ce^{-\beta r} & r > a\end{cases}$$ I kind of breeze trough this introductory part, I hope it's all clear. Now I have a problem: I would like to impose continuity and continuity of the derivative in $a$, to get more information about the shape and the energy of the solution, in particular I would like to find the number of bounded state in function of the value of $V_0$; but the calculation that derives from this procedure seems pretty hard. But: looking for the resolution for this exercise I found out that $u(r)$ must be $0$ in $r=0$, and this of course implies that in the first region the function becomes a $\sin$. $$u(x)=\begin{cases}A'\sin{\alpha r} \ \ \ \ \ \ & r \leq a\\Ce^{-\beta r} & r > a\end{cases}$$ and this, other than being the correct system of equations, greatly simplify the calculations. However: I don't understand why $u(r)$ must be zero in $r=0$.
So my question is: Why must $u(r)$ be zero in $r=0$?
