# 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): $$v(r)=2\bar{v}\big(1-\frac{r^2}{R^2_0}\big)=\alpha (R^2_0-r^2\big)$$ $$\implies rR''+R'+\alpha r(R^2_0-r^2)\lambda^2R=0$$

• Tried $R= a+ b r^n ?$ – Narasimham Jul 24 '16 at 19:50
• Sorry but that doesn't fit. – Gert Jul 24 '16 at 20:18

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.

• Fantastic. Well done. Full marks! – Gert Jul 25 '16 at 0:45
• Note that the main difficulty isn't to find the general solution of the ODE, The real problem is the calculus of a particular solution which satisfies the boundary conditions. – JJacquelin Jul 25 '16 at 7:47
• I suspected there was a problem with the boundary conditions too, which was the reason that I did not proceed with the analysis beyond the general form of the solution. R'(R0) = 0, for instance, could provide a non-trivial solution. I have seen these problems occasionally arise in domains bounded by concentric circles... I'd be curious to know what the application is. – player100 Jul 25 '16 at 10:05
• @player: the ODE is obtained after separation of Fourrier's heat equation (PDE) for a tube with laminar flow and constant wall temperature: $\frac1r \frac{d}{dr}\Big(r\frac{\partial T}{\partial r}\Big)=\frac{v(r)}{\kappa}\frac{dT}{dx}$, where $v(r)$ is the parabolic velociiy distribution of the fluid. Separate with $T(r,x)=R(r)X(x)$. – Gert Jul 25 '16 at 12:31
• I found that only for $R(R_0)=0$ only works for $\frac{kR_0^2}{2}=1$, as only the Laguerre polynomial $n=1$ has a root for $r=R_0$. Can you see that too? Sorry to be a PITA. – Gert Jul 25 '16 at 19:29

$$rR''+R'+\alpha r(R^2_0-r^2)\lambda^2R=0$$

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 $R(r)=0$.

• Based on the previous answer I worked out that $\lambda_n=\frac{4n}{R^2_0\sqrt{\alpha}}$ with $n=0,1,2,3,...$ Am I wrong? The $R(R_0)=0$ is the Constant surface temperature (CST) boundary condition. It's very common in heat transfer problems (see my comment above). Certainly $R(0) < +\infty$ and $R(0) > 0$. Thank you! – Gert Jul 25 '16 at 12:46
• Using $\lambda_n=\frac{4n}{R^2_0\sqrt{\alpha}}$ with 'player100s' solution I get manageable Laguerre polynomials. And the $\lambda_n$ eigenvalues also work with the $X(x)$ ODE, so I get $T(r,x)=\Sigma_{n=1}^{\infty}R_n(r)X_n(x)$\$– Gert Jul 25 '16 at 12:54 • @gert, I see now. I think you are right... you can find modes that correspond to zero crossings of the hypergeometric function. – player100 Jul 25 '16 at 13:24 • @player: I'll include the background into the question. Again, thanks so much for your help! – Gert Jul 25 '16 at 13:29 • @Gert : Why didn't you specify in the initial wording of your question that$\lambda$is not any real number but is related to$R_0$and$\alpha$? This is the missing piece of information. I was suspecting something like this and I wrote at end of my answer "except in case of particular values of$\lambda,\alpha,R_0$". Why don't you say at first that$\lambda$isn't a unique and given parameter like$R_0$and$\alpha\$, but in fact is a set of computable eigenvalues? – JJacquelin Jul 25 '16 at 13:38