I want to solve the following delayed differential equation

$$G'(t)=\Lambda +\omega G(t-\tau)-\mu G(t),$$

when $G(t)=G_0>0$ for $t\in[-\tau,0]$ and $\Lambda,\omega, \mu, \tau>0$. Note that $G_0, \Lambda, \omega, \tau, \mu$ and final time $T_f$ are arbitrary.

I tried to solve it using Wolfram Mathematica 11.0 with code

sol=DSolve[{G'[t]==a+b*G[t-d]-u*G[t], G[t/;t<=0]==G0}, G[t], {t, 0, Tf}]

and I didn't receive the solution.

Can somebody help me?

Thank you very much! Ana


1 Answer 1



d = 1;
Tf = 1;
DSolve[{G'[t] == a + b*G[t - d] - u*G[t], G[t /; t <= 0] == G0}, G[t], {t, 0, Tf}];

(* {{G[t] -> Piecewise[{{G0, t <= 0},
{(-a + a*E^(t*u) - b*G0 + b*E^(t*u)*G0 + G0*u)/(E^(t*u)*u), 
  Inequality[0, Less, t, LessEqual, 1]}}, Indeterminate]}}*)

DSolve needs a value for d(better exact) and Tf to solve. If you don't gives then MMA can't solve or gives a Error messages.

EDITED -27.04.2018.

Using Laplace transfrom to eq:

eq = G'[t] == a + b*G[t - d] - u*G[t];
eq2 = LaplaceTransform[eq, t, s] 

(* -G[0] + s LaplaceTransform[G[t], t, s] == 
a/s - u LaplaceTransform[G[t], t, s] + 
b LaplaceTransform[G[-d + t], t, s] *)

in the last expression MMA can't find transform of LaplaceTransform(G[t - d], t, s) see page 120 (it's weakness of this function) and puttting initial condition G[0]=G0

eq3 = -G[0] + s LaplaceTransform[G[t], t, s] == a/s - u LaplaceTransform[G[t], t, s] + 
b *Exp[-d*s]*LaplaceTransform[G[t], t, s] /. G[0] -> G0

sol = LaplaceTransform[G[t], t, s] /. Solve[eq3, LaplaceTransform[G[t], t, s]][[1]]
(* (E^(d s) (a + G0 s))/(s (-b + E^(d s) s + E^(d s) u)) *)

Inverse Laplace Transform:

Total@(InverseLaplaceTransform[#, s, t] & /@ {(sol // Apart)[[1]], (sol // Apart)[[2]]})

(* a/u + (E^(-t u) (-a + G0 u))/u + 
b InverseLaplaceTransform[(a + G0 s)/(
s (s + u) (-b + E^(d s) s + E^(d s) u)), s, t]*)

MMA can only find for first experssion.

  • $\begingroup$ I understood your suggestion, but I want to solve the differential equation for an arbitrary $\tau>0$ and for all final time $T_f>0$... $\endgroup$
    – Ana
    Apr 26, 2018 at 14:53
  • $\begingroup$ As I said, DSolve can't solve for arbitrary d and Tf.DSolve have some limitations. Putt your questions to math.stackexchange.com maybe will help you there. $\endgroup$ Apr 26, 2018 at 15:31
  • $\begingroup$ I followed your suggestion, putting my question in math.stackexchange.com.I'm going to analyze your previous edition (27-04-2018). Thank you! $\endgroup$
    – Ana
    Apr 27, 2018 at 16:57

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