# Second order ODE of unknown format problem

I'm looking for a Real function $$R(r)$$ that satisfies:

$$r^2R''+R'+m^2 rR=0$$

where $$m\in\Bbb R$$.

It looks a bit like an Euler DE but it isn't, and a bit like a Bessel DE but isn't either.

Wolfram alpha (link to the ODE) doesn't recognise it and provides no solution unfortunately.

I think I need a substitution like $$r=f(u)$$ to get going.

Any help is appreciated.

A little background -

The equation is the radial equation of two ODEs obtained after separation of variables, of a heat conduction problem (a very thin disk of radius $$R_1$$).

Boundary conditions are:

$$R(R_0)=0$$ $$R'(R_1)=0$$

I upvoted the first answer because it looked like a good idea but it turned out to be incorrect, as I showed in my comment.

In response to commenter 'tomasliam', the Sturm Liouville form of the DE is:

$$\frac{\text{d}}{\text{d}r}\left[e^{-1/r}R'(r)\right]+\frac{e^{-1/r}m^2R(r)}{r}=0$$

On request of @themaker:

A very thin disc of radius $$R_1$$ is at temperature $$T_i$$. It is insulated on both sides, as well as the outer edge.

At $$t=0$$ the area $$[0,R_0]$$ is suddenly heated to $$T_0$$.

What is the temperature evolution $$T(t,r)$$ of the disc (on $$[R_0,R_1]$$)?

Fourier heat equation for the disc, taking symmetry into account:

$$T_t=\frac{\alpha}r\frac{\partial}{\partial r}\Big(r\frac{\partial T}{\partial r}\Big)$$ $$\frac{T_t}{\alpha}=\frac1r(T_{r}+rT_{rr})$$ For homogeneity, we make a substitution:

$$u(t,r)=T(t,r)-T_0$$ $$\frac{u_t}{\alpha}=\frac1r(u_{r}+ru_{rr})$$ Initial: $$u(0,r)=T_i-T_0$$ Boundaries: $$u(t,R_0)=0$$ $$u_r(t,R_1)=0$$ Ansatz: $$u(t,r)=\Theta (t)R(r)$$ Substitute, then divide by $$u$$: $$\frac{\Theta'}{\alpha \Theta}=\frac{1}{r}\frac{R'}R+\frac{R''}R=-m^2$$ $$\frac{1}{r}\frac{R'}R+\frac{R''}R=-m^2$$ $$rR''+R'+m^2 rR=0$$ So it looks an error was made in setting up the original ODE! Mea culpa. The solution of the last equation is: $$R(r)=c_1J_0(mr)+c_2Y_0(mr)$$

• Have you found series solutions or put it into Sturm-Louville form? – tomasliam May 19 at 9:44
• No to the first, yes to the second: wolframalpha.com/input/… – Gert May 19 at 12:21
• For $m=0$, we note that solutions $$R(r) = c_1 \big(r e^{1/r} - \operatorname{Ei}(\tfrac1{r})\big) + c_2$$ where $\operatorname{Ei}$ is the exponential integral are functions of $1/r$. Another hard SL problem is $(xy')' - x\mathrm e^{-x} y =0$, while it even looks simpler than yours.. – EditPiAf May 20 at 16:11
• Yes it does. I'm fairly certain that the problem is strongly singular at $R_0 = 0$. It is definitely not solvable numerically for $R_0 = 0$, but I suspect it might even be infinite analytically – Aleksejs Fomins May 21 at 21:17
• @DanielD.wolframalpha.com/input/… – Gert May 22 at 15:54

HINT.

My attempts to get exact solution have not bring useful results.

At the same time, looks possible to get solution of the Sturm-Liouville equation in the series form.

Let $$E = e^{-{^1/_{\large r}}},\quad F=m^2ER',\quad v=\dfrac1{m^2},\tag1$$ then $$rF' + ER = 0.\tag2$$ Denote $$G_0(r) = \dfrac1rF',\quad G_{n+1} = r^2(G_n\!\!'+vF),\tag3$$ then \begin{align} &G_0 = rF' = - ER,\\ &G_0\!\!'= - E\left(R'+\dfrac1{r^2}R \right) =-vF-\dfrac1{r^2}ER,\quad G_1\!\! = r^2(G_0\!\!'+vF) = -ER,\\ &G_1\!\!'= - E\left(R'+\dfrac1{r^2}R \right) =-vF-\dfrac1{r^2}ER,\quad G_2\!\! = r^2(G_1\!\!'+vF) = -ER,\dots\\ &G_n = -ER,\quad n= 0,1,\dots.\tag4 \end{align}

Assuming $$R(\rho_0) = 0,\quad R'(\rho_0) = q,\quad vF(\rho_0)=qe^{-{^{\large1\!}/{ \rho^\,_0}}},\quad R'(\rho_1)=0,\tag5$$ one can get \begin{align} &G_0\!\!'(\rho_0) = rF'(\rho_0) = 0,\quad F'(\rho_0) = 0,\tag{6.1}\\[4pt] &G_1\!\!' = r^2\big((rF')'+vF\big) = r^3F''+r^2F'+vr^2F,\\[4pt] &G_1\!\!'(\rho_0) = \rho_0^3F''(\rho_0)+\rho_0^2 qe^{-{^{\large1\!}/{ \rho^\,_0}}}= 0,\\[4pt] &F''(\rho_0) = -\dfrac q{\rho_0}\,e^{-{^{\large1\!}/{ \rho^\,_0}}},\tag{6.2}\\[4pt] &G_2\!\!' = r^2\big((r^3F''+r^2F'+vr^2F)'+vF\big)\\[4pt] &= r^5F'''+4r^4F''+r^2(2r+v)F'+r^2(2r+1)vF,\\[4pt] &G_2\!\!'(\rho_0) = \rho_0^5F'''(\rho_0)+\rho_0^2(1-2\rho_0) qe^{-{^{\large1\!}/{ \rho^\,_0}}}= 0,\\[4pt] &F'''(\rho_0) = \dfrac {2\rho_0-1}{\rho_0^3}\,qe^{-{^{\large1\!}/{ \rho^\,_0}}},\dots\tag{6.3}\\[4pt] \end{align}

This recurrent process should obtain Taylor series for $$F(r)$$ and then for $$R(r).$$

Possible problem is applying of the condition to the derivative.

If we take $$R=e^{f(r)}$$ we get: $$e^{f(r)}(r^2f’’(r)+2f’(r)+m^2r)$$ After a little algebraic manipulation and use of $$e^{f(r)}$$ never equaling 0: $$r^2f’’(x)+2f’(r)=-m^2r$$ By integrating and use of integration by parts you might be able to solve it from there.

• Sadly there's an error in your derivations, because: $R''=e^{f(r)}[f''(r)+f'(r)^2]$. So the ODE becomes second degree. – Gert May 16 at 15:41

I have decided to contribute a numerical solution to your equation. I have the derivation of the numerical procedure and python code in jupyter. I can share both if there is interest.

I have performed finite difference discretization of the equation. I have assumed

• $$n = 1000$$ - the number of discretization points
• $$r_0 = 1$$ - the inner edge of the disc
• $$r_n = 5$$ - the outer edge of the disc

Note it is not possible to naively place the inner edge at zero because of singularity.

Using finite differences I have converted the problem into a linear discrete eigenvalue problem for a 3-diagonal matrix. Here is a plot of sorted eigenvalues $$m$$ (not $$m^2$$) by index for progressively improving discretization. It is not unreasonable to hypothesize that the in continuum limit the eigenvalues are a linear function of their index.   These are the first three eigenvectors. As expected, the following eigenvectors continue oscillating more and more.

• Thank you for your work. Are these eigenvectors obtained for $n=1000$? – Gert May 21 at 21:26
• Yes, the eigenvectors are for the case $n=1000$. For $n=500$ the first few look quite simular – Aleksejs Fomins May 21 at 21:29
• So what are the numerical values of the first $3$ eigenvalues? – Gert May 21 at 21:33
• Here are first 10 :) [0.45607236, 1.33735621, 2.23632993, 3.13494659, 4.03305012, 4.930833, 5.82841248, 6.72585358, 7.62319328, 8.52045335] – Aleksejs Fomins May 21 at 21:34
• You seem to show very high confidence levels [$7$ sign. digits and better]. Is that justified? – Gert May 21 at 21:38