PDE in cylindrical coordinates with semi-infinite boundary conditions in r direction I've been toiling over this fluid mechanics problem but I found myself hopeless.

An infinitely long rod of radius R is suddenly rotated about its axis in an infinite pool of Newtonian liquid, with constant angular velocity $\Omega$. Calulate the velocity $u_{\theta}(r,t)$.

Now, obviously the governing equation would be like this:
$$\frac{\partial u_{\theta}}{\partial t}=\nu \left(\frac{\partial ^2 u_{\theta}}{\partial ^2 r}+\frac{1}{r}\frac{\partial  u_{\theta}}{\partial  r}-\frac{u_{\theta}}{r^2}\right)$$
with boundary and initial conditions:
$$u_{\theta}(R,t)=R\Omega$$
$$u_{\theta}(\infty,t)=0$$
$$u_{\theta}(r,0)=0$$
I introduced $u'_{\theta}=\frac{R^2\Omega}{r}-u_{\theta}$ to make the boundary conditions homogeneous.       
Thus,
$$\frac{\partial u'_{\theta}}{\partial t}=\nu \left(\frac{\partial ^2 u'_{\theta}}{\partial ^2 r}+\frac{1}{r}\frac{\partial  u'_{\theta}}{\partial  r}-\frac{u'_{\theta}}{r^2}\right)$$
$$u'_{\theta}(R,0)=
0$$$$u'_{\theta}(\infty,0)=0$$
$$u'_{\theta}(r,0)=\frac{R^2\Omega}{r}$$ 
Next, I separated the variables;
$$u'_{\theta}=X(r)T(t)$$ 
By substituting, I obtained the general solution for $X(r)$;$$X(r)=c_1 J_1\left(\frac{\alpha r}{R}\right) + c_2 Y_1\left(\frac{\alpha r}{R}\right)$$
With boundary condition on $r=R$, it is clear that $c_1 J_1(\alpha) + c_2 Y_1(\alpha)=0$.
However, I can't make use of the second boundary condition, since $J_1(\infty) = Y_1(\infty)=0$ which is trivial!
How can I solve this semi-infinite boundary condition problem in cylindrical coordinates?
 A: It seems that your difficulty comes when you apply the boundary conditions to a particular solution of the PDE instead of to the general solution. There is little chance that only one particular solution could satisfy all conditions.
The following isn't a complete answer, but only the calculus of the general solution.
$$\frac{\partial u'_{\theta}}{\partial t}=\nu (\frac{\partial ^2 u'_{\theta}}{\partial ^2 r}+\frac{1}{r}\frac{\partial  u'_{\theta}}{\partial  r}-\frac{u'_{\theta}}{r^2})$$
Search for particular solutions on the form :
 $$U=X(r)T(t)$$
We use a different symbol $U$ instead of $u'_\theta$ in order to not confuse a particular solution with the general solution.
$$XT'=\nu (X''T+\frac{1}{r}X'T-\frac{1}{r^2}XT)$$
$$\frac{T'}{T}=\nu \left(\frac{X''}{X}+\frac{1}{r}\frac{X'}{X}-\frac{1}{r^2}\right)=\text{constant}=-\lambda\nu$$
$\frac{T'}{T}=\lambda\quad$ leads to $\quad T(t)=e^{-\lambda\nu t}$
$r^2X''+rX'+(\lambda r^2-1)X=0\quad$ leads to $\quad X(r)=c_1 J_1(\sqrt{\lambda}\:r)+c_2 Y_1(\sqrt{\lambda}\:r)$
$$U(\lambda,r,t)=e^{-\lambda\nu t}\left(C_1 J_1(\sqrt{\lambda}\:r)+C_2 Y_1(\sqrt{\lambda}\:r) \right)$$
The particular case $\lambda=0$ corresponds to the equation $r^2X''+rX'-X=0$ which solution is $X=c_1r+c_2\frac{1}{r}$ 
$$U(0,r,t)=c_1r+c_2\frac{1}{r}$$
The general solution is the linear combination of all those solutions :
$$u'_\theta=c_1r+c_2\frac{1}{r}+\sum_{\forall\lambda\neq 0} e^{-\lambda\nu t}\left(C_1(\lambda) J_1(\sqrt{\lambda}\:r)+C_2(\lambda) Y_1(\sqrt{\lambda}\:r) \right)$$
For each value of $\lambda$ the coefficients $C_1$ and $C_2$ can be different. So they are arbitrary functions of $\lambda$.
Of course, often there is no need to take the infinite number of values of $\lambda$. Keeping only a few convenient ones is often sufficient to satisfy the conditions. The others are eliminated in setting the corresponding coefficients $C(\lambda)$ equal to $0$.
