Determining the maximum value for the solution of this delay differential equation?

Question

I am working on the following delay differential equation

$$\frac{df}{dt}=f-f^3-\alpha f(t-\delta)\tag{1},$$ where $\frac{1}{2}\leq\alpha\leq 1$ and $\delta\geq 1$.

I know that there are three equilibrium points to equation (1) by solving $$0=(1-\alpha)f-f^3\tag{2},$$ namely $f=0 \vee f=\pm\sqrt{1-\alpha}$.

I am however specifically interested in oscillatory solutions of this equation (close to and above the neutral curve).

My question is about the maximum value that $f$ can obtain for such a periodic solution, for given $\alpha$ and $\delta$. Basically I want to determine the amplitude of this oscillation. Is there a method (or approximation), that allows me to determine the maximum value of $f$?

Some of my thoughts:

I understand that the maximum value is at least $f>\sqrt{1-a}$, and from some numerical solutions I suspect that the maximum value is about twice this equilibrium value. However, I would like to have a more informed answer.

For example, for $\alpha=\frac{3}{4}$ and $\delta=2$, I numerically obtained the solution for different initial conditions $y_0$, and, as long as the initial condition is not the equilibrium point, it ends up in the same stable oscillation mode.

After finding the there is some bound present (see my incomplete answer below), I numerically investigated the behavior along the neutral curve:

$$ \delta=\frac{\arccos\left(\frac{3\alpha-2}{\alpha}\right)}{\sqrt{\alpha^2-(2-3\alpha)^2}},$$

and in the below figure I plotted the thus obtained $f_{max}$ as a function of $\alpha$. (Please be aware that the results are numerically obtained, and I did not carefully analyze the accuracy yet).

Apparently, I found for the maximum function value the following relation

$$f_{max}=\sqrt{2(1-\alpha^2)},$$

which fits perfectly between $\sqrt{1-\alpha}$ and $\sqrt{1+\alpha}$ on the given domain for $\alpha$. This cannot be a coindicent, can it?

Answer 1

Building on your own answer, one sees that the crucial part is getting some lower bound on $f(t_{max}-\delta)$.

The differential equation yields $$ \frac{df}{dt} = f-f^3-\alpha f(t-\delta) \leq (1+\alpha)f_{\max} $$ By Taylor expansion, $\exists \tau,\, f(t_{\max})-f(t_{max}-\delta) =\delta f'(\tau)$.

Thus, using your global bound, $f_{\max}-f(t_{max}-\delta) \leq \delta(1+\alpha)^{3/2}$. Multiplying by $\alpha$ and subtracting your eq. (1), we get $(\alpha-1)f_{\max} + f^3_{\max} \leq \alpha\delta(1+\alpha)^{3/2}$, which you can solve for a real upper bound on $f_{\max}$. This upper bound will be smaller for smaller $\delta$.

If it is not sharp enough yet for your use, then you can use the second derivative of $f$ for bounding $f_{\max}-f(t_{max}-\delta)$ with a Taylor expansion: $$\exists \tau,\, f(t_{\max})-f(t_{max}-\delta) =\delta f'(t_\max) -\frac{1}{2}\delta^2 f''(\tau) = -\frac{1}{2}\delta^2 f''(\tau) $$ Then, $f(t_{\max})-f(t_{max}-\delta) \leq -\frac{1}{2}\delta^2 \min_t f''(t)$. You must then bound $f''$ from below, a crude approach gives $$ f'' = f'-3f'f^2 - \alpha f'(t-\delta) \geq \min f' - \max f^2f' - \alpha\max f'\geq -(1+\alpha)^2f_\max - (1+\alpha)f_\max^3 $$ Following the above reasonning, one finds $(\alpha-1)f_{\max} + f^3_{\max} \leq \alpha\delta^2(1+\alpha)^{5/2}$.

Because you're in the case of finite $\delta$, it might be that the linear bound is sharper than the quadratic one for some values of $(\delta,\alpha)$.

Answer 2

This is not a complete answer! I don't know a full answer for all $\alpha,\delta$, but in this answer I describe how an absolute upper bound can be found, as it is too much for a comment.

At the peak of the solution $f(t)$, we obviously have $\frac{df}{dt}=0$, which means,

$$ 0=f_{max}-f_{max}^3-\alpha f(t_{max}-\delta)\tag{1}$$

At this stage, $f_{max}$ and $(t_{max}-\delta)$ are unknown.

However, we do know that $-f_{max}\leq f(t_{max}-\delta)\leq f_{max}$. So, we can test some situations

1. $f(t_{max}-\delta)=f_{max}$. This basically means $\delta=0$ and brings us to the equilibrium solution $f_{max}=\sqrt{1-\alpha}$.
2. $f(t_{max}-\delta)=-f_{max}$. Now, $f_{max}=\sqrt{1+\alpha}$.
3. $f(t_{max}-\delta)=0$. This is an intermediate solution with $f_{max}=1$.

It is my impression that the second case provides an absolute maximum to $f$, which is $\sqrt{1+\alpha}$. The third case provides for the parameters $\alpha=\frac{3}{4}$ and $\delta=2$, $f_{eq}=\frac{1}{2}$ and $f_{max}=1$.

How close this absolute maximum is reached, depends, for given $\alpha$, mainly on the value of $\delta$. From below figure it can be seen that for larger $\delta$, the extreme value $f_{max}=\sqrt{1+\alpha}$ (dashed line) is reached more closely.

Answer 3

a couple of remarks:

1/ It is not obvious to me that you necessarily get $t$ such that $f'(t)=0$. It looks like this is the case in your numerical examples, but analytically, I'm not sure how you would prove this.

2/ If indeed you have oscillatory solutions, then I agree with you, you need to solve the equations above. But as you have a parameter $\delta$ that makes $f(t-\delta)$ anything in $[-f_{max},f_{max}]$, I would suggest to consider it as a parameter and solve the 3rd degree equation. You will have between 1 and 3 solutions, all dependent on $f(t-\delta) \in [-f_{max},f_{max}]$. (It is not pretty in wolfram alpha)

3/ Next step, you can optimize the solutions you found above with respect to $f(t-\delta)$, which will ensure you have found a global extremum.

4/ Last step, you need to make sure everything is coherent, and potentially looking at your solutions in 2/ vs your optimization in 3/.

There probably is a way to use the fact that the solution is periodic, and that the minimum and maximum satisfy the same equation, but I can't really articulate it. Hope this is helpful.