Mathematica Asked by Eric N on June 6, 2021
I am trying to solve an ODE by subbing in a series form and then looking individually at the coefficients of different powers of the variable. I’m looking at a general form of equation:
$$frac{mathrm{d}^2w}{mathrm{d}z^2} + f_1(z) frac{mathrm{d}w}{mathrm{d}z} + f_0(z)=0$$
I want to fill in the solution form $w(z) = e^{lambda z} z^mu sum_{s=0}^infty a_s z^{-s}$ and then collect coefficients of the various powers of $z$. However when I try to do this the series terms are not being multiplied with any $z$ terms which are multiplying the series from the outside. A simpler example of the problem I’m having is when I input say:
w[z_] = E^(λ z)z^μ Sum[a[s] z^-s, {s, 0, Infinity}]
Coefficient[z^5 w[z], z^μ]
The output of this is to give the coefficient as:
$$e^{lambda z} z^5sum_{s=0}^infty a_s z^{-s}$$
What I want is for the $z^5$ to multiply the $z^mu$ term and also term-by-term with the series terms, and then give me just the coefficient of the resulting $z^mu$ term. So the only corresponding term in the sum is the one of the form $e^{lambda z}z^5z^mu a_5 z^{-5}$, and therefore the coefficient I should be getting is just: $$e^{lambda z} a_{5}$$
Can I not do this type of manipulation or have I just done something wrong here?
Let me show you an example of series solution of ODE with given f0 and f1 for your differential equation.
{f0[z_] = Sin[z], f1[z_] = 1};
deq = Derivative[2][w][z] + f1[z]*Derivative[1][w][z] + f0[z] == 0
For comparison the solution with DSolve at initial conditions.
wdsol = w /. First@DSolve[deq && w[0] == 4 && w'[0] == 3, w, z]
(* Function[{z}, 1/2 E^-z (-5 + 12 E^z + E^z Cos[z] + E^z Sin[z])] *)
Generate a series expression. Don't omitt O[z]^smax+1. LogicalExpand generates equations for each power of z. Solve to get the coefficients a[s]. Get the last coefficients with initial conditions.
w[z_] = Sum[a[s] z^s, {s, 0, 15}] + O[z]^16
le = LogicalExpand[deq]
sol1 = Solve[le, Table[a[s], {s, 2, 15}]]
ww[z_] = Normal[w[z] /. First@sol1]
sol2 = Solve[{ww[0] == 4, ww'[0] == 3}, {a[0], a[1]}]
wle[z_] = ww[z] /. First@sol2
(* 4 + 3 z - (3 z^2)/2 + z^3/3 - z^4/12 + z^5/40 - z^6/240 + z^7/2520 - z^8/20160
+ z^9/120960 - z^10/1209600 + z^11/19958400 - z^12/239500800 +
z^13/2075673600 - z^14/29059430400 + z^15/653837184000 *)
Plot[{wle[z], wdsol[z]}, {z, 0, 7}, PlotStyle -> {Blue, Red}]
Answered by Akku14 on June 6, 2021
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