How to solve this differential equation with integral?
$\omega = 1.2;$
$i1 = 25640;$
$ia = 4*10^6;$
$c2 = 1.4513*10^5;$
$t5 = 16000*\cos({\omega*t});$
$tpto = 2.8*10^6*{\theta'}(t);$
$cr(\omega)=\frac{1.8*10^7}{0.5}{\omega}^2\exp({-\omega})$
$R_{55}(t)=\frac{2}{\pi}{\int}_{-\infty}^{10}cr(\omega)\cos({\omega}t)d{\omega}$
$(i1+ia)*{\theta}^{''}(t)+{\int}_{-\infty}^{t}R_{55}(t-\tau){\theta}^{'}(\tau)d{\tau} +(c2)*{\theta}(t)=t5+tpto$
$(i1+ia)*{\theta}^{''}(t)+{\int}_{0}^{t}R_{55}(t-\tau){\theta}^{'}(\tau)d{\tau} +(c2)*{\theta}(t)=t5+tpto$
There is some error in my first edit and it should be the
${\int}_{- \infty}^{t}R_{55}(t-\tau){\theta}^{'}(\tau)d{\tau} $
And if when form $ -\infty $ to $0$ the ${\theta}(t) $ is undefined or is $0$ then
It is ${\int}_{0}^{t}R_{55}(t-\tau){\theta}^{'}(\tau)d{\tau} $
And also by using the FindFit
I get the similar r55
,a more simple form so is there a way to solve ?
Remove["Global`*"];
omega = 1.2;
i1 = 25640;
ia = 4*10^6;
c2 = 1.4513*10^5;
t5 = 16000*Cos[omega*t];
tpto = 2.8*10^6*Theta'[t];
(*r55=1/(Pi (1.` +t^2)^3) 2 \
(7.200000000000001`*^7-2.1600000000000006`*^8 \
t^2+(-199396.49151683348` -317073.10946119425` \
t^2-130751.79771595637` t^4) Cos[10 t]+(238622.0308316204` \
t+388986.59820497024` t^3+163439.7471449455` t^5) Sin[10 t]);*)
r55[t_]=(44559034.3038137*t-26314556.6322484)/(-0.507157643287066-t^\
4);
s = NDSolve[{(i1 + ia)*Theta''[t] +
Integrate[r55*Theta'[Tau], {Tau, 0, t}] + c2*Theta[t] ==
t5 + tpto, Theta'[0] == 1, Theta[0] == 0}, Theta, {t, 0, 10}]
(*NDSolve::rdelay: Delay Tau is not real valued. >>*)
(*NDSolve::rdelay: Delay Tau is not real valued. >>*)
s = NDSolve[{(i1 + ia)*Theta''[t] +
NIntegrate[r55*Theta'[Tau], {Tau, 0, t}] + c2*Theta[t] ==
t5 + tpto, Theta'[0] == 1, Theta[0] == 0}, Theta, {t, 0, 10}]
(*NIntegrate::nlim: Tau = t is not a valid limit of integration. >>*)
(*NIntegrate::nlim: Tau = t is not a valid limit of integration. >>*)
(*NIntegrate::nlim: Tau = #1 is not a valid limit of integration. >>*)
(*General::stop: Further output of NIntegrate::nlim will be suppressed during this calculation. >>*)
(*NDSolve::rdelay: Delay Tau is not real valued. >>*)
(*NDSolve::rdelay: Delay Tau is not real valued. >>*)
So is there a way that I can get the answer? Thanks!
Supplement
yes we can use the laplace transform to solve this question,bu the accuracy is the question,
r55[t_]=(44559034.3038137*t-26314556.6322484)/(-0.507157643287066-t^\
4);
i get the laplace transform of r55
,it is the function of p
,so we call it r55p
,but it is with Complex form,so i ge the fit of r55p,then i get
r55p = 13829522.310484 + -3538064.23662678/p + 63545.307621876*p^2 +
235429.331970918/p^2 - 1546395.29415638*p;
and finally i get the solution of theta,it is
Theta[t_] =
0.06541197518272371` E^(-33.8394608216844` t) -
0.38705195407627524` E^(-5.468703164047347` t) -
0.4851421230004468` E^(0.10884574410374667` t) +
0.8072987102672268` E^(
0.18393223395811042` t) - (0.0005166083732284499` + 0.` I) Cos[
1.2` t] + (0.000959271400198915` + 0.` I) Sin[1.2` t];
Plot[Theta[t], {t, 0, 10}]
and so we get the plot which is different from the answer 1
Theta[Tau]
is undefined forTau < 0
, andNIntegrate
should be used instead ofIntegrate
. However, even with these items corrected, it is not clear thatNDSolve
can solve such an integro-differential equation. By the way, I suggest you remove the first two blocks of code from the question, because they do nothing. I also suggest that you include the first few error messages that you receive. $\endgroup$DSolve
withIntegrate
can, in principle, solve such integro-differential equations, but not this one. $\endgroup$r55
, but as comments neither is available to the code that follows. Note also that the comments containr55[t]
and the subsequent lines of code justr55
, which is not the same. To obtain useful advice from readers, please display the code you actually used and what error messages, if any, you actually received. In the event of many consecutive error messages, including just the first few typically is sufficient. $\endgroup$R55
is not actually a function of tau, so you can factor out of the integral.. (Or should thet
in theR55
expression really bet-tau
? ) $\endgroup$