Integrating over an interval

Question

I have a function I want to integrate, which is $v(t)$.

solv = NDSolve[{v'[t] + (v[t])^2 + f == 0, v[0] == -I*k55}, 
v[t], {t, 0, 150}]

Also, I have a range tRange=[0,150,0.001]. Now, what I want to do is to integrate this function v=v[t]/.solv using the range I proposed. For example, say x1 and x2 are in the range, and xmin=0, xmax=150; how do I integrate v from xmin to x_1, then x_1 to x_2, then so on until xmax with 0.001 of difference between each step. At the end, I want to put everything in a table. Is there any way how? Or is it possible to make a For[] loop to get the results?

sol = NDSolve[{x''[t] + 3 x'[t] Sqrt[(x'[t])^2/6 + (1 - Exp[-x[t] 
Sqrt[2/3]])^2/4] + 
Sqrt[3/2] Exp[-x[t] Sqrt[2/3]] (1 - Exp[-x[t] Sqrt[2/3]]) == 0, 
a'[t]/a[t] == Sqrt[(x'[t])^2/6 + (1 - Exp[-x[t] Sqrt[2/3]])^2/4],
τ'[t] == 1/a[t],
x'[0] == -0.008226306418212731,
x[0] == 5.630991866033891,
a[0] == 1,
τ[149.4517772937791] == 0},
{x, τ, a},
{t, 0, 500}]
a = a[t] /. sol;
\[Tau] = \[Tau][t] /. sol;
x = x[t] /. sol;
xt = x'[t] /. sol;
att = a''[t] /. sol;
tEnd = t /. FindRoot[(att == 0), {t, 0, 150}]
aEnd = a /. t -> tEnd;
H = Sqrt[(xt)^2/6 + (1 - Exp[-x*Sqrt[2/3]])^2/4];
V = 3/4 (1 - Exp[-Sqrt[2/3] x])^2;
Vt = Sqrt[3/2] Exp[-x Sqrt[2/3]] (1 - Exp[-x Sqrt[2/3]]);
Vtt = -Exp[-2 Sqrt[2/3]*x]*(Exp[Sqrt[2/3]*x] - 2);
t55 = t /. FindRoot[(aEnd/a == E^55), {t, 0, 150}];
\[Tau]55 = \[Tau] /. t -> t55;
a55 = a /. t -> t55;
eps1 = (1/2) (Vt/V)^2;
eta = Vtt/V;
eps2 = -4 eps1 + 2 eta;
k55 = a55*(H /. t -> t55);
\[Nu] = 3/2 + eps1 + 1/2*eps2;
f = k55^2 - (\[Nu]^2 - 1/4)/(\[Tau]^2);

Answer 1

NOT AN ANSWER
(Too long for a comment, and I'm not sure to understand your aim.)

It would be simpler to call NDSolve with a maximum step size option of 0.001, and let NDSolve decide itself where the steps are.

Here is an example :

I don't copy k55 and f that are defined in your large block of code.

x1 = 10;
x2 = 20;
x3 = 30;
xmax = 150;

stateData = 
  First[NDSolve`ProcessEquations[{v'[t] + (v[t])^2 + f == 0, 
     v[0] == -I*k55[[1]] (* and not k55 *)}, v[t], t, MaxStepSize -> 0.001]];

res = Table[NDSolve`Iterate[stateData, t];
  varAndDerivativesValues = 
   NDSolve`ProcessSolutions[stateData, "Forward"];
  stateData = 
   First@NDSolve`Reinitialize[stateData, 
     Equal @@@ Most[varAndDerivativesValues]];
  varAndDerivativesValues, {t, {x1, x2, x3, xmax}}]  

The integration stops at t=138 due to another problem not related with the question :

Related

Answer 2

The block which evaluates k55, f can only be executed once without an error message. Strange, don't know why.

Change the last codelines to

k55 = (a55*(H /. t -> t55))[[1]];
\[Nu] = 3/2 + eps1 + 1/2*eps2;
f = (k55^2 - (\[Nu]^2 - 1/4)/(\[Tau]^2))[[1]];

Introduce a second variable intv[t] in NDSolveValue which defines the integral of v[t]

NDSolveValue[{v'[t] + (v[t])^2 + f == 0, intv'[t] == v[t], intv[0] == 0 + 0 I, v[0] == -I*k55}, {v, intv} , {t, 0, 150}]

NDSolve claims reduced simulation range {t,0,149}

Check result

Plot[ReIm[intV[t]], {t, 0, 139}, PlotRange -> Automatic,Evaluated -> True]

Answer 3

[H]ow do I integrate v from xmin to x_1, then x_1 to x_2, then so on until xmax with 0.001 of difference between each step. At the end, I want to put everything in a table.

The answer is Method -> {"FixedStep", Method -> ...} and StartingStepSize -> 0.001. To put all 150,000 steps in a table seems overkill — for the site, if not in the actual use-case. Here's an example of 20 steps:

solv = NDSolveValue[{v'[t] + (v[t])^2 + 1 == 0, v[0] == 1}, 
   v, {t, 0, 0.02}, 
   Method -> {"FixedStep", Method -> "ExplicitRungeKutta"}, 
   StartingStepSize -> 0.001];

TableForm[
 Transpose@{Flatten@solv@"Grid", solv@"ValuesOnGrid"},
 TableHeadings -> {Range[0, 20], {"Time", v[t] // TraditionalForm}}]