1
$\begingroup$

my question is as this and i want to get the solution of this ode x(t)

$ ({M_{33}}{\rm{ + }}{{\rm{m}}_{33}})x''(t) + \int\limits_{ - \infty }^t {x'(tau){K_{33}}(t - tau)d{\rm{ }}tau} = {F_3}(t)\\ {K_{33}}(t) = \frac{2}{\pi }\int\limits_0^{ + \infty } {{b_{33}}} (\omega )\cos (\omega t)d\omega \\ $

and the defined value is as this

${F_3}(t)= 31.4380978232711cos(3.9893240045584673t);\\ M_{33} = 1.95;\\ m_{33} = 2.5;$

codes asre as this

b33 = 2.52077405993092/(46.4663900509659 + omega^2 - 
11.8316929905336*omega);
k33 = 2/\[Pi]*Integrate[b33*Cos[omega*t], {omega, 0, +\[Infinity]}]

(*1.60477 If[
  t \[Element] 
   Reals, (0.0190724 - 
  0.0333163 I) ((3.33766 - 
     1.91069 I) Cosh[(3.38661 + 5.91585 I) Abs[
      t]] CosIntegral[(-5.91585 + 3.38661 I) Abs[
      t]] - (0.955346 + 
     1.66883 I) Sinh[(3.38661 + 5.91585 I) Abs[t]] (\[Pi] - 
     2. SinIntegral[(-5.91585 + 3.38661 I) Abs[
         t]])) + (0.0190724 + 
  0.0333163 I) ((3.33766 + 
     1.91069 I) Cos[(5.91585 + 3.38661 I) Abs[
      t]] CosIntegral[(-5.91585 - 3.38661 I) Abs[t]] + (1.66883 + 
     0.955346 I) Sin[(5.91585 + 3.38661 I) Abs[t]] (\[Pi] + 
     2 SinIntegral[(5.91585 + 3.38661 I) Abs[t]])), 
  Integrate[Cos[omega t]/(
   46.4664 - 11.8317 omega + omega^2), {omega, 0, \[Infinity]}, 
   Assumptions -> t \[NotElement] Reals]]*)

int2 = Integrate[x'[tau]*k33[t - tau], {tau, -[Infinity], t}]

it is x'[tau] rather than x'[t]

ft = 31.4380978232711` Cos[3.9893240045584673` t];
M33 = 1.95;
m33 = 2.5;


equation = (M33 + m33)*x''[t] + int2 == ft;
ts = 50;

s1 = NDSolve[{equation, x[0] == 0, x'[0] == 0}, x, {t, 0, ts}]
Plot[Evaluate[y[x] /. s], {x, 0, 30}, PlotRange -> All]

however it doesn't work so how to solve it?

updated3

now i use fit method to get the b33 and we can use this b33 to get the k33 but it seems that this k33 figure is queer from t=0-100`.

k33 in some essay is as this enter image description here

updated4

now the question changes to this ,how to solve the integral equation

k33[t_] := 
  20*(-1.01880661301011*
 Sin[-12.2868520317448*t]/(12.3729623105373*t + 
    31.6231386532942*t^2));

int2 = Integrate[x'[tau]*k33[t - tau], {tau, -\[Infinity], t}];

ft = 31.4380978232711` Cos[3.9893240045584673` t];
M33 = 1.95;
m33 = 2.5;

equation = (M33 + m33)*x''[t] + int2 == ft;
ts = 50;

s1 = NDSolve[{equation, x[0] == 1, x'[0] == 1}, x, {t, 0, ts}]
Plot[Evaluate[x[t] /. s1], {t, 0, ts}]
$\endgroup$
3
  • $\begingroup$ Let us continue this discussion in chat. $\endgroup$ – dcydhb Dec 13 '19 at 11:37
  • 1
    $\begingroup$ I'm voting to close this question as off-topic because the problem doesn't lie in Mathematica side, it's because the $B_z$ used by OP doesn't seem to be correct. $\endgroup$ – xzczd Dec 14 '19 at 3:06
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – Kuba Dec 14 '19 at 12:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.