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In general, can I find an analytical solution for an equation of the form

$$ \dfrac{dV_{(t)}}{dt} = -\dfrac{V_{(t-1)}}{\tau} $$

where $t$ is time, $V$ is a variable that changes over time, and $\tau$ is a time constant. I think the result should be an exponential decay function, of the form

$$ V_0 \cdot e^{-\tau \cdot t} $$

where $V_0$ is some initial value for $V$.

Also, if anyone could recommend a beginner's book/resource for learning how to solve similar problems I would be most grateful.

Edit: Thanks for LutzL for pointing out this is not an ODE but rather a delay DE.

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  • $\begingroup$ This is a delay-differential equation. Usually there is no unique solution, every function on $[-1,0]$ can be extended to a solution over $t\ge 0$. $\endgroup$ Commented Jul 20, 2019 at 13:44
  • $\begingroup$ Thanks, LutzL. Does this mean my example DDE is possible to solve? $\endgroup$
    – Frank
    Commented Jul 20, 2019 at 14:04

1 Answer 1

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This is a delay-differential equation (DDE). Usually there is no unique solution, every function on $[−1,0]$ can be extended to a solution over $t≥0$ as $$ V(t)=V(0)-\int_{-1}^{t-1}\frac{V(s)}{τ}\,ds. $$


You can of course also look for solutions of the form $$V(t)=ae^{bt}.$$ Inserting you find $abe^{bt}=-\frac{a}{τ}e^{b(t-1)}\implies bτ=-e^{-b}\iff -\frac1{τ}=be^b$. This has a real solution only if $τ>e$, it can be expressed using the Lambert-W function as $b=W_0(-\frac1{τ})\in[-1,0]$ or $b=W_{-1}(-\frac1{τ})\in(-\infty,-1])$.

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  • $\begingroup$ So I think my solution is $ V_0 \cdot e^{- \dfrac{t}{\tau}} $. Is it correct? $\endgroup$
    – Frank
    Commented Jul 22, 2019 at 4:13
  • $\begingroup$ No, this is not a solution. $V_0e^{W_0(-\frac1τ)t}$ and $V_0e^{W_{-1}(-\frac1τ)t}$ are the solutions that are exponential functions. $\endgroup$ Commented Jul 22, 2019 at 5:15

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