Talk:Power series solution of differential equations

Latest comment: 5 years ago by JimCropcho

In the section Example usage, for the first and second derivatives of f, shouldn't the summation start with index 1 and 2 respectively? Notice how this author starts the summation of the derivatives on higher indices: http://tutorial.math.lamar.edu/Classes/DE/SeriesSolutions.aspx 78.82.194.195 (talk) 21:53, 7 March 2011 (UTC)Reply

The last paragraph of the Nonlinear Equations section reads:

"A limitation of the power series solution shows itself in this example. A numeric solution of the problem shows that the function is smooth and always decreasing to the left of [eta=1], and zero to the right. At [eta=1], a slope discontinuity exists, a feature which the power series is incapable of rendering, for this reason the series solution continues decreasing to the right of instead of suddenly becoming zero."

In the following, I use the operator D to mean "derivative", in order to avoid typographical issues with the usual apostrophe character ...

In fact there are two separate solutions in different regions. When eta<1, the power series as shown gives a correct solution. When eta>1, the separate (trivial) solution F=0 is another correct solution. Now at eta=1, the equation: F D.D.F + 2 (D.F)^2 + (eta-1) D.F + D.F = 0 reduces to an identity, since the factor (eta-1) zeroes the third term, the constraint F(1) = 0 zeroes the first term, and the remaining terms factor as D.F (2 D.F+1) = 0, which is an identity in view of the constraint D.F(1) = -1/2. So at eta=1, the value of D.D.F is unconstrained, and this allows two different solutions to be "stitched together" across the slope discontinuity. This feature also makes continuation of a numeric solution very unreliable when going across the boundary at eta=1. 86.4.253.180 (talk) 02:07, 1 June 2013 (UTC) 86.4.253.180 (talk) 02:15, 1 June 2013 (UTC)Reply

"A simpler way using Taylor series": is this even real? No sources are cited, I cannot find any other reference this method on the internet, and some (but not all) back-of-the-envelope trials I have done, have failed to reproduce known correct answers. — Preceding unsigned comment added by JimCropcho (talkcontribs) 23:22, 15 December 2018 (UTC)Reply