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I have been trying to model a specific type of situation using the following formula, which I have come to learn is a "Delay Differential Equation" (DDE)

$$f'(t) = cf(t-\tau)$$

I understand that it is quite trivial to "simulate" this DDE to get an accurate approximation, but is it possible to find an exact solution for a differential equation in terms of $c$ and $\tau$ this form given an initial condition? In particular, the initial condition that I am interested in is $f(0) =1$. If this is possible, what steps would I take to solve it? Otherwise, is it impossible to obtain an equation, or simply intractable.

Thank you

user1135299
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  • It is not clear how to simulate this DDE. Perhaps you can do it if you were given an interval of initial conditions $f(t)$ for all $t \in [-\tau, 0]$. – Michael Dec 30 '22 at 01:11

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One solution is $ f(t) = e^{rt}$ for a certain value of $r$ to be defined later. This meets the initial condition $f(0)=1$. Also, we get from this function \begin{align} &f'(t) = re^{rt} \\ &cf(t-\tau) = ce^{r(t-\tau)} = ce^{-r\tau}e^{rt} \end{align} Equating the two gives $$ r = ce^{-r\tau}$$ Given $c>0, \tau>0$, you can always find $r>0$ that solves this (since $r$ starts at 0 and increases linearly with $r$, and $ce^{-r\tau}$ starts at $c>0$ and decreases continuously with $r$). For example if $c=1$ and $\tau=1$ then $r\approx 0.567143$.

There may be other solutions. DDEs do not have the same uniqueness properties as ODEs.

In the case $c=-1/4$ and $\tau=1$ the equation becomes $$ r = (-1/4)e^{-r}$$ which has two solutions $r \approx -.357403$ and $r\approx -2.15329$. So this is an example of non-uniqueness of the DDE solution (two different solutions with same initial condition $f(0)=1$).

Michael
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