Difficult engineering second order DE, any pointers? I have the following engineering DE:
$$rR''+R'+\alpha r(R^2_0-r^2)\lambda^2R=0$$
Where $R(r)$ is Real, $r \geq 0$, $\alpha >0$.
Boundary conditions $R(R_0)=0$ and $\Big(\frac{dR}{dr}\Big)_{r=0}=0$.
It looks a bit like a Sturm–Liouville equation but seems to be above my pay grade.
Any help would be welcome.

Background:
I'm solving the Fourier heat equation for laminar flow though a tube with constant wall temperature and radius $R_0$. Here $T(r,x)$ is the temperature, minus that wall temperature.
The PDE is (with boundaries as above):
$$\frac1r \frac{d}{dr}\Big(r\frac{\partial T}{\partial r}\Big)=\frac{v(r)}{\kappa}\frac{\partial T}{\partial x}$$
$$\frac{1}{rv(r)} \frac{d}{dr}\Big(r\frac{\partial T}{\partial r}\Big)=\frac{1}{\kappa}\frac{\partial T}{\partial x}$$
Ansatz:
$$T(r,x)=R(r)X(x)$$
$$\frac{d}{dr}\Big(r\frac{\partial T}{\partial r}\Big)=\frac{d}{dr}\Big(rXR'\Big)=XR'+rXR''$$
$$\frac{1}{rv(r)}(XR'+rXR'')=\frac{1}{\kappa}RX'$$
$$\frac{1}{rv(r)}\Big(\frac{R'}{R}+r\frac{R''}{R}\Big)=\frac{1}{\kappa}\frac{X'}{X}=-\lambda^2$$
Velocity distribution (laminar flow):
$$v(r)=2\bar{v}\big(1-\frac{r^2}{R^2_0}\big)=\alpha (R^2_0-r^2\big)$$
$$\implies rR''+R'+\alpha r(R^2_0-r^2)\lambda^2R=0$$
 A: $$rR''+R'+\alpha r(R^2_0-r^2)\lambda^2R=0$$
Where $R(r)$ is Real, $r \geq 0$, $\alpha >0$.
Boundary conditions $R(R_0)=0$ and $\Big(\frac{dR}{dr}\Big)_{r=0}=0$.
Obviously the solution $R(r)=0$ is convenient : it satisfies the ODE and the boundary conditions.
But this trivial solution isn't certainly what is expected.
This supposes that another solution could exist for the ODE and the boundary conditions. Hence, the problem would have not one, but at least two solutions.
This draw to think that there is something fishy in the wording of the problem (something missing or not quite exact ? One cannot say). If it comes from the modeling of a physical problem, it might be judicious to re-examine the modeling and may be correct a bit the form of the ODE or the boundary conditions.
This suspicion is strengthened by the analytical solving (below). The function $R(r)$ can be expressed in terms of confluent hypergeometric function (or related functions). The boundary conditions leads to an unique solution $R(r)=0$.



A: Let $u = r^2$, then $$R_r=2rR_u$$$$R_{rr}=4uR_{uu}+2R_{u}$$
So, $$r^2R_{rr}+rR_r+\alpha\lambda^2r^2(R_0^2-r^2)R = 0$$ is transformed into
$$4u^2R_{uu}+4uR_u+\alpha\lambda^2(R_0^2-u)uR=0$$
$$uR_{uu}+R_u+\frac{\alpha\lambda^2}{4}(R_0^2-u)R=0$$
Let $R=e^{-ku}W$, where $4k^2 = \alpha\lambda^2$,
$$R_u=(-kW+W_u)e^{-ku}$$
$$R_{uu}=(k^2W-2kW_u+W_{uu})e^{-ku}$$
Then,
$$uW_{uu}+(1-2ku)W_u+k^2R_0^2W-kW=0$$
Finally, let $v = 2ku$.
$$2kvW_{vv}+2k(1-v)W_v+k^2R_0^2W-kW=0$$
$$vW_{vv}+(1-v)W_v+\frac{kR_0^2-1}{2}W=0$$
This equation fits the form as described here: http://mathworld.wolfram.com/LaguerreDifferentialEquation.html
EDIT:
Here is another solution:
http://eqworld.ipmnet.ru/en/solutions/ode/ode0211.pdf
Boundary conditions:
$r=R_0$ corresponds to $v=2kR_0^2=4\rho+2$ where $\rho=\frac{kR_0^2-1}{2}=\frac{\sqrt{\alpha}\lambda R_0^2-2}{4}$. So, the goal is to find a linear combination of $L_\rho(4\rho+2)$, where $L_\rho$ is the Laguerre polynomial, that satisfies both spatial boundary conditions. I don't know how to find $\rho$ such that $L_\rho(4\rho+2)=0$. If you take any two polynomials, though, you can solve $$c_1 L_{\rho_1}(4\rho_1+2)+c_2 L_{\rho_2}(4\rho_2+2) = 0$$ and
$$c_1 L'_{\rho_1}(0)+c_2 L'_{\rho_2}(0) = 0$$
Generally, there is one more initial condition $T(r,0)$ that you can use to match the linear combination of solutions. In other words, sticking with integer $\rho$, $$\sum_{m=0}^\infty a_m L_m(r)=T(r, 0), \space r \le R_0,$$ from which you can find the parameters $a_m$ by using the orthogonality of $L_m$ with respect to the weight $e^{-r}$. Or $$\int dr \space e^{-r}L_n(r)\sum_{m=0}^\infty a_m L_m(r) = a_n (n!)^2 = \int dr \space e^{-r}L_n(r) T(r, 0)$$ Since you don't have $T(r, 0)$, I suppose you can pick any arbitrary set of $L_{\rho}$ and choose weights that satisfy the spatial boundary conditions. 
