Re: solve the following problem

*To*: mathgroup at smc.vnet.net*Subject*: [mg69376] Re: solve the following problem*From*: p-valko at tamu.edu*Date*: Sat, 9 Sep 2006 03:26:58 -0400 (EDT)*References*: <edqr42$t4p$1@smc.vnet.net>

Alexandr Alexandr wrote: > How would I solve the following problem with > Mathematica? > x'[to] + x[to-1] == 0. I used "to" (original time) instead of time beacuse I will need a new time. The new time is t -> to-1 I will denote the solution by y[t] (the solution starts at t = 0, thet is at original time to=1) We will also need an initial segment of the solution denoted by x0[t] for -1 <= t <= 0 The Laplace transform of the solution is obtained as In: Solve[s Y[s] - x0[1] == - LaplaceTransform[x0[t](1 - UnitStep[t - 1]), t, s] - Y[s]Exp[-s], Y[s]] // First we get: Y[s] -> (E^s*(-LaplaceTransform[x0[t], t, s] + LaplaceTransform[UnitStep[-1 + t]*x0[t], t, s] + x0[1]))/(1 + E^s*s) This is still not very useful, we have to use the initial segment. For instance I assume that the original segment is x0[t_]: Sin[ Pi t ]. Then F[s_] = Y[s] /. (Solve[s Y[s] - x0[0] == - LaplaceTransform[x0[t](1 - UnitStep[t - 1]), t, s] - Y[s]Exp[-s], Y[s]] // First) /. x0[t_] -> Sin[p t] ; In: F[s]//InputForm Out: (E^s*(-1 - s^(-1) + 1/(E^s*s)))/(1 + E^s*s) is the Laplace transform of the unknown solution. Unfortunately, we are not able to invert it symbolically. But we can invert it numerically. I will use here the GWR rutine (available from MathSource) to calculate the solution y[t] at discrete time points t =1, 2 and 3 : In[97]:= GWR[F,1] GWR[F,2] GWR[F,3] The result is Out[97]= -0.63660861559252271027 Out[98]= -0.3183693616578363197 Out[99]= 0.2236507922295097426 Best Regards Peter