1
$\begingroup$

I try to solve this differential system with delay:

DSolve[x'[t] == -0.12  x[t]  - 
   0.8 *  1.77   x[t] /(1 + x[t]^4)
      E^(-0.1  * 
     2.8    ) (1.6 * 1.77  1/(1 + x[t - 2.8]^4) +  0.01) x[
     t - 2.8] - (0.51/61)^-0.001  p[t]/(
    p[t] + 1) ((2.5*10^4 - 2.5*10^4 * 0.4 * 0.01)/(2.5*10^4)^0.99) x[
     t]^0.99, d'[t] == - 0.61  d[t] + 200 , 
 p'[t] == -0.0412 p[t] + 0.61 d[t], {x[t], d[t], p[t]}, t]

In the render I wasn't able to understand the error message.

Is there any charitable sool to help me out.

|improve this question
$\endgroup$
  • $\begingroup$ You need your three equations inside { and }. You have x'[t], y'[t] and p'[t] but are solving for x[t], d[t] and p[t]. There is still at least one more error I can't see at the moment. $\endgroup$ – Bill Feb 17 '15 at 7:22
  • $\begingroup$ DSolve might not recognize that x[t - 2.8] is just the time delay of x[t]. By the way, can DSolve solve time delay differential equation? $\endgroup$ – Harry Feb 17 '15 at 7:32
  • $\begingroup$ @Bill Excuse me. p[t] instead y[t], I give the correction. $\endgroup$ – Zbigniew Feb 17 '15 at 7:38
  • $\begingroup$ @Harry In mathematica 10 it is possible to sove an EDO with delay. $\endgroup$ – Zbigniew Feb 17 '15 at 7:39
  • $\begingroup$ @Zbigniew: First, you will have to enclose the equations in curly brackets, then you might need to specify the historic behavior, e.g. like x[t/;t<=0]==(*whatever*), and finally, you need to specify the range of t, e.g. {t,0,2}. $\endgroup$ – Jinxed Feb 17 '15 at 8:54
2
$\begingroup$

With the corrections suggested by Jinxed (with whatever set to 1) plus the additional correction of specifying the range of t (required for a delay differential equation), the code becomes

DSolve[{x'[t] == -0.12 x[t] - 0.8*1.77 x[t]/(1 + x[t]^4) E^(-0.1*2.8) (1.6*1.77 1/(1 + 
  x[t - 2.8]^4) + 0.01) x[t - 2.8] - (0.51/61)^-0.001 p[t]/(p[t] + 1) 
  ((2.5*10^4 - 2.5*10^4*0.4*0.01)/(2.5*10^4)^0.99) x[t]^0.99,
  d'[t] == -0.61 d[t] + 200, 
  p'[t] == -0.0412 p[t] + 0.61 d[t], 
  x[t /; t <= 0] == 1}, {x[t], d[t], p[t]}, ,{t, 0, 10}]

Since DSolve runs "forever", trying to solve this problem, I simplified the code by solving for d and p, which are independent of x.

ans1 = First@DSolve[{d'[t] == -0.61 d[t] + 200, p'[t] == -0.0412 p[t] + 0.61 d[t]}, {d[t],p[t]}, t]
(* {d[t] -> 327.869 + 1. E^(-0.61 t) C[1], 
    p[t] -> 1. E^(-0.0412 t) (5205.99 E^(0.0412 t) - 351.617 E^(0.61 t)) + 
    351.617 E^(-0.0412 t) (-1. E^(0.0412 t) + 1. E^(0.61 t)) + 
    1.07243 E^(-0.6512 t) (-1. E^(0.0412 t) + 1. E^(0.61 t)) C[1] + 1. E^(-0.0412 t) C[2]} *)

These then are substituted into the remaining differential equation

eq = Simplify[x'[t] == -0.12 x[t] - 0.8*1.77 x[t]/(1 + x[t]^4) E^(-0.1*2.8)
  (1.6*1.77 1/(1 + x[t - 2.8]^4) + 0.01) x[t - 2.8] - (0.51/61)^-0.001 p[t]/(p[t] + 1) 
  ((2.5*10^4 - 2.5*10^4*0.4*0.01)/(2.5*10^4)^0.99) x[t]^0.99 /. ans1]

The resulting expression, which is rather long to be displayed here, when inserted into DSolve

DSolve[{%, x[t /; t <= 0] == 1}, x[t], t]

still runs "forever", perhaps due to the x[t]^0.99 term.

This being the case, I used NDSolveValue. Doing so requires that the constants C[1] and C[2] be specified, and I chose 0. (Alternatively, one could use ParametricNDSolveValue with the two constants as parameters.)

Simplify[eq/.{C[1] -> 0, C[2] -> 0}]
sol = NDSolveValue[{%, x[t /; t <= 0] == 1}, x, {t, 0, 10}]
LogPlot[sol[t], {t, 0, 10}, AxesLabel -> {t, x}, PlotRange -> {10^-8, 1}]

enter image description here

|improve this answer
$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.