Re: Integro-differential analog of Love's equation via power series

*To*: mathgroup at smc.vnet.net*Subject*: [mg79843] Re: [mg79823] Integro-differential analog of Love's equation via power series*From*: DrMajorBob <drmajorbob at bigfoot.com>*Date*: Mon, 6 Aug 2007 03:54:03 -0400 (EDT)*References*: <25633792.1186373747971.JavaMail.root@m35>*Reply-to*: drmajorbob at bigfoot.com

The code below is incomplete or otherwise doesn't work (at this machine, anyway). f is undefined when Solve is executed, so f appears in "sol". Then, when f[x_] is defined, we get infinite recursion. Bobby On Sun, 05 Aug 2007 04:00:01 -0500, chuck009 <dmilioto at comcast.com> wrote: > Hello guys, > > I modified Paul Abbott's code (you mind Paul?) to solve the > integro-differential analog of Love's equation: > > y'[x]=1+1/pi Integrate[y[t]/((x-t)^2+1),{t,-1,1}] > > using a simple power series: > > 1. I have to specify an initial condition y0=1/2=c0 > > 2. The left side now contains the derivative of the power series. > > 3. The right side is similar to the integral equation except that now > I'm solving for just > c1 through cn although I keep c0 in the matrix for now unless someone > can suggest a better way. > > 4. When I use Solve, I replace c0 with 1/2 and also specify that now > I'm only solving for {c1, c2,...cn}. I use Drop to remove c0 from the > list. > > 5. For the check I explicityly calculate the derivative of the > resulting power series and then plot the difference lhs-rhs of the IDE . > The result with 40 equations and 40 unknowns is less than 10^{-6} execpt > for spikes at the end-points and also receive a "row-reduce of > badly-conditioned matrix". > > So, anyone wants to look at it and suggest more concise ways of coding > it or explain why I would obtain a badly-conditioned matrix? > > a = -1; > b = 1; > y0 = 1/2; > n = 40; > > xs = N[Table[x, {x, -1, 1, (b - a)/(n - 1)}], 6]; > cs = Table[Subscript[c, k], {k, 0, n}]; > > lhs = cs . Table[(i - 1)*xs^(i - 2), {i, 1, n + 1}]; > > rhs = f[xs] + (1/Pi)*cs . Table[NIntegrate[ > Evaluate[t^i/((xs - t)^2 + 1)], {t, -1, 1}], {i, 0, n}]; > > sol = Solve[lhs == rhs /. Subscript[c, 0] -> y0, Drop[cs, 1]] > > f[x_] = Sum[Subscript[c, i]*x^i, {i, 0, n}] /. First[sol] /. > Subscript[c, 0] -> y0 > > Plot[f[x], {x, -1, 1}] > fd[x_] = D[f[x], x] > Plot[fd[x], {x, -1, 1}] > > Plot[1 + (1/Pi)*NIntegrate[f[t]/((x - t)^2 + 1), {t, -1, 1}] - > fd[x], {x, -1, 1}, PlotRange -> All]; > >