3
$\begingroup$

If you have a differential equation with polynomial coefficients then quite remarkably Mathematica can find a solution. For example for the second order differential equation

$ y''(t)+b y'(t)+\left(\text{c0}+\text{c1} t+\text{c2} t^2+\text{c3} t^3\right) y(t) $

where the coefficient of y(t) is a polynomial in t we can get a solution from DSolve

eqn = y''[t] + b y'[t] + (c0 + c1 t + c2 t^2 + c3 t^3) y[t] == 0;

sol = y[t] /. First@DSolve[eqn, y[t], t]

enter image description here

By clicking on the icon we can get some more information

enter image description here

If we put in values we can plot

vals = {c0 -> 10, c1 -> 0.5, c2 -> 0.2, c3 -> 0.1, b -> 0.1, 
   C[1] -> 0, C[2] -> 1};
Plot[Evaluate[sol /. vals], {t, 0, 10}]

enter image description here

The solution may be used to do algebraic calculation. Here I take the derivative and plot on the phase plane.

dsol = D[sol, t];
ParametricPlot[Evaluate[{sol, dsol} /. vals], {t, 0, 10}, 
 AspectRatio -> 1]

enter image description here

However, I would really like to know what the function looks like. FunctionExpand and DifferentialRootReduce do nothing.

sol // FunctionExpand
DifferentialRootReduce[sol, t]

I know that the function behind DifferentialRoot is probably so complicated and long that it is difficult to understand and do anything useful with. However, the general form may be useful. I am curious and would like to see the function behind the icon. Can this be done?

Improve this question
$\endgroup$
7
  • 3
    $\begingroup$ You are right, the analytic solution of such an ODE [homogeneous second order differential equation with non-constant coefficients] is too complicated. However, it exists and the algorithm is known due to Kovacic. For higher orders it is known that linear ODEs do not have a general solution Picard–Vessiot theory. $\endgroup$
    – yarchik
    Commented Oct 9, 2021 at 21:10
  • $\begingroup$ If in your equation c3=0, the solution can be expressed in terms of linear combination of BesselJ and BesselY of order 1/3. $\endgroup$
    – yarchik
    Commented Oct 9, 2021 at 21:16
  • $\begingroup$ Thanks for all the comments. We can get a glimpse of the solution by working out the series around some point, This shows some structure but is not quite what I was hoping for. $\endgroup$
    – Hugh
    Commented Oct 11, 2021 at 8:39
  • $\begingroup$ Your may also find relevant Nikiforov-Uvarov formalism arxiv.org/pdf/quant-ph/0604021.pdf . $\endgroup$
    – Acus
    Commented Nov 9, 2021 at 6:25
  • $\begingroup$ @Acus, thank you. If you know something about this area a chat would be good. $\endgroup$
    – Hugh
    Commented Nov 9, 2021 at 16:46

1 Answer 1

Reset to default
1
$\begingroup$

The manual/help for DifferentialRoot says " FunctionExpand will attempt to expand DifferentialRoot objects in terms of ordinary special and elementary functions."

Test it for various combinations with Manipulate. E.g. for all c4 == 0 or for (c1 and c2 and c3 and b) == 0 you get explicit functions.

eqn = y''[t] + b y'[t] + (c0 + c1 t + c2 t^2 + c3 t^3) y[t] == 0;

ysol[d1_, d2_, b_, c0_, c1_, c2_, c3_] = 
  y /. First@DSolve[eqn && y[0] == d1 && y'[0] == d2, y, t]

Manipulate[{Plot[
  Evaluate[ysol[d1, d2, b, c1, c2, c3, c4][t]], {t, 0, 7}, 
  PlotRange -> All], 
   ysol[d1, d2, b, c1, c2, c3, c4][t] // FunctionExpand[#, Assumptions -> t > 0] &}, {{d1, 1}, 0, 
   4, 1/100, Appearance -> "Labeled"}, {{d2, 2}, 0, 3, 1/100, 
 Appearance -> "Labeled"}, {{b, 1}, 0, 3, 1/100, 
 Appearance -> "Labeled"}, {{c1, 1}, 0, 4, 1/100, 
 Appearance -> "Labeled"}, {{c2, 1}, 0, 4, 1/100, 
 Appearance -> "Labeled"}, {{c3, 1}, 0, 4, 1/100, 
 Appearance -> "Labeled"}, {{c4, 1}, 0, 4, 1/100, 
 Appearance -> "Labeled"}, ContinuousAction -> False]
Improve this answer
$\endgroup$
1
  • $\begingroup$ Thanks. I had looked at this. It is not clear how the simpler forms progress to the one I want to see. Any ideas? $\endgroup$
    – Hugh
    Commented Oct 11, 2021 at 8:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.