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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 $G_0,\Lambda,\omega,\mu,\tau$ 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

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    $\begingroup$ Why do you think it has a closed-form solution? $\endgroup$ – GEdgar Apr 27 '18 at 16:49
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You solve the equation on intervals of length $\tau$. Let's find the solution on $[0,\tau]$. If $t\in[0,\tau]$, then $t-\tau\in[-\tau,0], G(t-\tau)=G_0$ and the the equation becomes $$ G'(t)=\Lambda +\omega\,G_0-\mu\,G(t), $$ with initial condition $G(0)=G_0$. It is a linear equation, whose solution is, if I have made no errors, $$ G(t)=\frac1\mu\bigl((\mu-\omega)G_0-\Lambda)\,e^{-\mu t}+\Lambda+\omega\,G_0\bigr). $$ Knowing $G$ on $[0,\tau]$, proceed now to find $G$ on $[\tau,2\,\tau]$, and so on.

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  • $\begingroup$ Thank you for your suggestion and help! $\endgroup$ – Ana May 3 '18 at 8:15
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You can try the Laplace Transform. Assuming null initial conditions you have after transformation

$$ sG(s)=\frac{\Lambda}{s}+\omega G(s)e^{-s\tau}-\mu G(s) $$

and then

$$ G(s) = \frac{\Lambda}{s(\mu+s-\omega e^{-s\tau})} $$

Now you can approximate $e^{-s\tau}$ by a Padé expansion.

NOTE

According to order 3 Padé expansion

$$ e^{-s\tau}\approx \frac{-\frac{1}{120} t^3 \tau ^3+\frac{t^2 \tau ^2}{10}-\frac{t \tau }{2}+1}{\frac{t^3 \tau ^3}{120}+\frac{t^2 \tau ^2}{10}+\frac{t \tau }{2}+1} $$

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  • $\begingroup$ I can't assume that $G_0=0$, because $G_0>0$. I forgot to write this detail... Sorry! $\endgroup$ – Ana Apr 27 '18 at 17:11
  • $\begingroup$ Thank you for your suggestion and help! $\endgroup$ – Ana May 3 '18 at 8:15

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