# Delayed differential equations - conditions for a steady-state solution

Is there a way of finding the condition required for a delayed differential equation to have a steady state solution? By 'steady state' I mean that $$\frac{dx}{dt}=0$$, or alternatively, $$x(t)$$ approaching some constant as $$t \rightarrow\infty$$.

If we have an ordinary differential equation:

$$\frac{d}{dt}x(t) = ax(t) + c$$

then we can solve it using the integrating factor method to get the following solution:

$$x(t) = e^{at}x(0) + -\frac{c}{a}[1-e^{at}]$$

assuming we have the initial condition $$x(0)$$. Call this equation (*). We can now see that a steady state solution exists only if $$a<0$$, so all of the exponential terms become $$0$$ as $$t \rightarrow \infty$$.

My question is this: is there an equivalent condition for a steady state solution to exist for a delayed differential equation of the form:

$$\frac{d}{dt}x(t) = a x(t) + bx(t-\tau) + c$$ ?

My hand-waving solution is that the condition is $$(a +b)<0$$ for steady state to exist, since when $$t \rightarrow\infty$$, there isn't a difference between $$t$$ and $$t-\tau$$. This also recovers the correct answer in the case where $$\tau \rightarrow 0$$. I was hoping someone might be able to confirm or dis-confirm this in a slightly more rigourous way. Thanks!

• Do you know the Laplace transform? With this you could get more clues about the stability of the differential equation. – Cesareo May 8 at 16:15
• thanks, yes, I do know the Laplace transform. I can use it to get a solution for $x(s)$ in the frequency domain, but am not sure how to use this to find the condition for a steady state solution. – asph May 8 at 17:07
• I prepared a post because as a comment would we too long. – Cesareo May 8 at 17:45

## 2 Answers

To have a steady state response we need a stable system. Now calling $$u(t) = x(t)-\frac{c}{a+b}$$ (assuming $$a +b \ne 0$$ etc.) we have

$$\dot{u}(t)=au(t)+bu(t-\tau)$$

and after Laplace transforming

$$U(s) = \frac{u(0)}{s-a-be^{-\tau s}}$$

so $$u(t)$$ is stable as far as the poles for $$U(s)$$ are left complex plane included.

As $$s-a-be^{-\tau s} = x-a-b e^{-\tau x}\cos(\tau y) + i\left(y+be^{-\tau x}\sin(\tau y)\right)$$ we can construct some graphics to show the poles dependence on $$a,b,\tau$$

Follows a plot showing the case $$a = 1, b = -2, \tau = 1$$

and the case with $$a = -2, b = 1, \tau = 1$$

The poles are at the intersection between the blue and red curves. In the first case we can verify that the response is unstable and in the second case the response is stable. In delayed DE appears infinite poles and we should assure that no one invades the positive complex plane.

It seems that for this differential equation it is sufficient that $$a + b <0$$, to be stable although this would have to be demonstrated.

"Steady-state solution" means a constant solution, not a solution that approaches a constant as $$t \to \infty$$. But if there is a solution that approaches a constant as $$t \to \infty$$, that constant must also be a solution.

If $$a+b \ne 0$$ there is a steady-state solution, namely $$x(t) = -c/(a+b)$$. Also if $$c = 0$$ there is a steady-state solution, namely $$x(t) = 0$$. Conversely, if $$a+b = 0$$ and $$c \ne 0$$ there is no steady state solution because for $$x(t) = k$$ the right side would be $$c$$ and the left side $$0$$.