# Can we approximate delayed-differential equations with higher-order-ordinary-differential-equations?

I noticed that the most simple numerical approximation of a higher order-differential equation has the same form as the numerical approximation of a delayed first-order differential equation. This leads me to the following hypothesis:

Hypothesis: Delayed first-order differential equations can be approximated by higher-order ordinary differential equations.

I wanted to investigate to what extent this is true so I came up with a method to find approximations of delayed differential equations. However, my initial tests seem to indicate that this approach doesn't work. I hope someone can explain why, and if there is a better, similar approach.

Here is my approach:

Take the simple delayed first-order differential equation for arbitrary function $f$: $$\dot x(t+1)=f(x(t))$$ By taking $\Delta x =1$, this is numerically approximated by the difference equation: $$x_{t+2}-x_{t+1}=f(x_t)$$ adding to both sides $x_t-x_{t+1}$, gives: $$x_{t+2}-2x_{t+1}+x_t=f(x_t) - (x_{t+1}-x_t)$$ Which, if we again take $\Delta x =1$, is the numerical approximation of $$\ddot x (t)=f(x(t))-\dot x(t)$$

Hence we might approximate a simple first-order delayed equation by a higher order non-delayed equation: $$\dot x(t+1)=f(x(t)), \quad \text {is approximated by}\quad \ddot x(t)+\dot x(t)=f(x(t))$$ By the same approach we could show that $$\dot x(t+2)=f(x(t)), \quad \text {is approximated by}\quad \dddot x(t)+2\ddot x(t) +\dot x(t)=f(x(t))$$ $$\dot x(t+3)=f(x(t)), \quad \text {is approximated by}\quad \ddddot x(t)+3\dddot x(t) -3\ddot x(t)+\dot x(t)=f(x(t))$$ And so forth...

However, I've been using $\Delta x =1$ here. If we take $\Delta x=1/n$, and gradually increment $n$ upwards, a similar pattern to the one for larger delays occurs: $$\dot x(t+1)=f(x(t))$$ is approximated for $n=2$ by: $$2\cdot (x(t+1+\frac {1}{2})-x(t+1))=f(x(t))$$ using the same approach as above, but $\Delta x = \frac{1}{2}$ instead of $\Delta x = 1$, and a lot of tedious algebra, one can show that this is equivalent to the approximation of
$$2^{-2}\dddot x(t)+2^{-1}\cdot 2 \ddot x(t)+2^{-0}\dot x(t)$$

Hence we might approach a closer and closer approximation of $\dot x(t+1)=f(x(t))$, by taking increasingly larger $n$, and smaller $\Delta x=\frac{1}{n}$, as follows:

using $n=2$ (resulting in $\Delta x=\frac {1}{2}$): $$\dot x(t+1)=f(x(t)), \quad \text {is approximated by}\quad n^{-n}\cdot \dddot x(t)+n^{-n+1}\cdot 2 \ddot x(t)+n^{-n+2} \cdot \dot x(t)$$ and using $n=3$: $$\dot x(t+1)=f(x(t)), \quad \text {is approximated by}\quad$$ $$n^{-n}\cdot \ddddot x(t) +n^{-n+1}\cdot 3\dddot x(t)+n^{-n+2} \cdot 3\ddot x(t)+ n^{-n+3}\cdot \dot x(t)$$

And so forth...

The form of this approximation is the same as for $\Delta x =1$, but with higher delays, except for the coefficients of $n$.

The most important point I noticed about this approximation is that as we increase $n$, increasingly higher order derivatives are added, but the coefficients of those derivatives seem to decrease hyperexponentially (i.e. $n^{-n}$). This made me hope that perhaps the approximations would converge quickly to the delayed equation as $\Delta x$ decreases.

However, I did some tests on the delayed equation $\dot x(t)=x(t-1)$, and $\dot x(t) = x^2(t-1)$, where my approach fails miserably. increasing $n$ actually deteriorates the approximation.

On the one hand (in hindsight) this makes sense to me, since increasing the order of the derivative should increase the rate of growth of $x$ for large $t$, but it still bugs me that it doesn't work, despite the fact that the difference equations for the two are so similar.

So my question is: Why doesn't my approach work, given that the difference-equation approximations of delayed-differential-equations, and higher-order-differential equations have the same form? and more importantly, is there a different but similar approach to approximate delayed-differential-equations?

• A delay system can sometimes be reduced to a higher ODE system, thus I think it can also be approximated by a higher ODE system. But I am not an expert in the computational aspect of your approach. Sound great though, good luck! :) Oct 22, 2016 at 4:09

If looking for a solution for $\dot x(t+1)=f(x(t))$ on $[0,\infty]$, take any continuous "seed" function $x:[-1,0]\to\Bbb R$ with $x(0)=x_0$ and define the solution as $$x(t)=x_0+\int_0^t f(x(s-1))ds$$ The values on $[0,1]$ are well-defined using the values on $[-1,0]$, the values on $[1,2]$ by the values on $[0,1]$ and consequently $[-1,1]$ etc. By construction this function is continuous and by the integration it is differentiable on $(0,\infty)$.
A delay can be written as a differential $$e^Dx(t)=x(t+1)$$. So it might make sense to approximate $$\dot{x}(t+1)=f(x(t))$$ by cutting off the following series at some $$n$$, $$e^DDx(t)=f(x(t))$$ $$(\cdots+\frac{D^n}{(n-1)!}+\cdots+D^2+D)x(t)=f(x(t))$$ This ties in with Lutz Lehmann's answer in that an infinite number of initial data is required to specify the solution.
For example, approximate $$\dot{x}(t+1)=f(x)$$ by $$(\frac{D^5}{4!}+\cdots+D^2+D)x(t)=f\circ x(t)$$.