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I've been working with this kind of delay differential equation

\begin{equation} \begin{split} &f(x)' = f(x-1)^2 - f(x) \cdot (f(x+1) + 1), \\ &f(x) = f_0>0, \,\,\, 0\leq x < 1. \end{split} \end{equation}

I've already tried using Laplace transform methods, but it seems that they are not so helpful when it comes with product of general functions. Also, common tricks for DDE seem not working or be to really messy in my case, like step by step solution for every interval $[n, n+1]$.

Do you have any idea how to solve this equation? There are theorems that assure at least existence/uniqueness for this kind of non linear and not homogeneous equations?

Even partial solutions or comments are really appreciated.

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  • $\begingroup$ My first reaction is to express the "unknown" as a function of the known data, i.e.express $f(x+1)=...$ $\endgroup$
    – Jean Marie
    Dec 28, 2018 at 14:43
  • $\begingroup$ Your "initial condition" should be not just at one point but on an interval. $\endgroup$
    – Robert Israel
    Dec 28, 2018 at 15:22
  • $\begingroup$ I have added a new tag "functional equation" $\endgroup$
    – Jean Marie
    Dec 28, 2018 at 15:58

2 Answers 2

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Strictly speaking, this is not a solution to your problem but to an approximation of it (see figure).

I have been working with a function $f$ defined on $[0,2]$ whose curve is close to its horizontal asymptote, thus not very far from a constant ; I have attempted to find a solution that is continuous at $t=2$, which is the less we can desire for a solution to this kind of issue.

Let us consider $f$ defined by :

$$f(t)=\frac{at+b}{ct+d} \ \ \ \text{for} \ \ \ 0\leq t \leq 2$$

with

$$f'(t)=\frac{ad-bc}{(ct+d)^2}. $$

Taking into account the expression given by @Robert Israel, slightly modified under the form :

$$ f(t+2) = \frac{f(t)^2 - f'(t+1)}{f(t+1)} - 1 $$

Continuity in $t=2$ is warranted by condition : $$f(2)=\frac{f(0)^2-f'(1)}{f(1)}-1$$

giving a constraint on parameters $a,b,c,d$ :

$$\frac{2a + b}{2c + d} = - ((c + d)((ad - bc)/(c + d)^2 - b^2/d^2))/(a + b) - 1$$

Among the infinite set of solutions, here is a simple one, corresponding to the image below :

$$(a=0,b=16/45,c=3,d=2),$$

giving $f(t):=\frac{(16/45)}{3t+2}$.

enter image description here

Take note that 1) the first part is almost a constant 2) continuity is obtained in $t=2$ but not in $t=4$.

Remarks :

1) We can manage to have $c$ and $d$ such that the restriction of $f$ to $[0,2]$ is arbitrarily close to a constant. For that, we need to take $c=1$ and $d$ an arbitrarily large positive number.

2) Take $a=c=1;d=20;b= 156.386$ to see a slightly decreasing function, approximating a solution to the exact problem which is almost continuous at $t=4$ too.

3) Had we taken $$ (a=-4,b=24/11,c=0,d=1)$$ we would have had the plain affine function given by: $f(t):=-4t+24/11.$

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    $\begingroup$ Really interesting. I'll do some calculation and come back to your solution. It's not a complete solution of the problem (if any exists) but it gives very useful partial information. $\endgroup$
    – Jane Doe
    Dec 29, 2018 at 15:11
  • $\begingroup$ I just had an idea : why not working backwards with :$f(x-1)=\sqrt{f'(x)+f(x)(f(x+1)+1)}$ ? (or with $-\sqrt{\cdots}$). $\endgroup$
    – Jean Marie
    Dec 30, 2018 at 13:38
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Write the equation as $$ f(t) = \frac{f(t-2)^2 - f'(t-1)}{f(t-1)} - 1 $$

Given a solution on an interval of length $2$, say $(-2,0]$, this will determine it on the next interval $(0,1]$. Continuity and differentiability are not guaranteed, though, and if ever $f(x) = 0$ you'll likely have a singularity at $x+1$.

EDIT: If $f(x) = f_0$ on $[-1,1)$, so $f'(x) = 0$ on $(-1,1)$, then you get $f(1) = f_0 - 1$, so there's a discontinuity at $x=1$.

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