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MathGroup Archive 2006

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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


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