Creating points while using Mathematica's Table Function - wolfram-mathematica

Question

I'm trying to plot points that I've created in a table in mathematica but for some reason one component of my points seems to have double braces around it while the other only has one as below:
{{x},y},{{x1},y1}....{{xn},yn}
and list plot will not recognize these as points and will not plot them.
Here is my mathematica code:
Remove["Global`*"]
b = .1;
w = 1;
Period = 1;
tstep = 2 Pi/Period;
s = NDSolve[{x''[t] + b x'[t] - x[t] + x[t]^3 - .5 Cos[w t] == 0,
x'[0] == 0, x[0] == 0}, x[t], {t, 0, 1000}, MaxSteps -> Infinity];
x[t_] = x[t] /. s
data = Table[Evaluate[{x'[t], .5}], {t, 0, 1000, tstep}]
ListPlot[data]
I've also tried using the command
ListPlot[Flatten[Table[Evaluate[{x'[t], .5}], {t, 0, 1000, tstep}]]]
to no avail as well as
ListPlot[Table[Evaluate[{Flatten[x'[t]], .5}], {t, 0, 1000, tstep}]]]
How can I remove the {}?

Answer 1

You may try something along these lines:
Clear["Global`*"]
b = .1;
w = 1;
s = NDSolve[{x''[t] + b x'[t] - x[t] + x[t]^3 - .5 Cos[w t] == 0,
x'[0] == 0, x[0] == 0}, x[t], {t, 0, 1000}, MaxSteps -> Infinity];
xr[u_] := ((x[t] /. s[[1]]) /. t -> u)
Plot[(xr'[u]), {u, 0, 30}]
But I am not sure what are you trying to get from the {x'[t], .5} part

Answer 2

My colleagues are correct, but I think there is more that can be said. First, to your actual question. The output of NDSolve is a list of the form
{{x[t]->InterpolatingFunction[...]}, {x[t]->InterpolatingFunction[...]}, ...}
where the second and subsequent replacement rules are only there if more than one solution is present. I have never encountered a case using NDSolve where that is true, but it makes the answer consistent with Solve, where multiple solutions is not uncommon. Therefor, with only one solution, you have a double list, i.e.
{{x[t]->InterpolatingFunction[...]}}
As per Mr. Wizard, you can use First, or you can use Part, i.e.
NDSolve[ ... ][[ 1 ]]
which is my preferred method, although it is slightly more difficult to read and may obscure your intent. You should be aware that the InterpolatingFunction that NDSolve returns is a function, and it will accept variables directly. So, the variables on the left hand side of the declarations
x[t_] = x[t] /. s
and from Belisarius
xr[u_] := ((x[t] /. s[[1]]) /. t -> u)
are superfluous at best, and the second one requires the replacement to occur every time xr is used. Instead, you can declare
x = x[t] /. s
and then writing x[t] afterwards will return IntepolatingFunction[t], exactly like you want. Then, as Belisarius points out, you can use it, or its derivative, in Plot directly, instead of first building a table of values and feeding them into ListPlot.
Edit: when I first posted this, I didn't notice a quirk with NDSolve. If you explicitly solve for x[t] not x, then NDSolve returns InterpolatingFunction[...][t], but if you just solve for x you get what I posted. This quirk allows both the OP's and Belisarius's solutions to function, otherwise the replacement shouldn't occur.

Answer 3

It is most likely that x'[t] is returning something of the form {x_i}. Try replacing the data=Table... line with this
data = Table[Evaluate[{First[x'[t]], .5}], {t, 0, 1000, tstep}]
An alternative would be to do
data=data /. {{x_}, y_} :> {x, y};
which uses ReplaceAll (/.) to replace every occurrence of {{x_i},y_i} with {x_i,y_i}
Example:

Answer 4

There are arguably better ways to accomplish what you are doing, but that is not what you asked.
To remove the extra {} recognize this comes from the result of NDSolve, and therefore use:
s = First # NDSolve[{x''[t] + b x'[t] - x[t] + x[t]^3 - .5 Cos[w t] == 0,
x'[0] == 0, x[0] == 0}, x[t], {t, 0, 1000}, MaxSteps -> Infinity];