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

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?

• Although I put an answer on my own question, I hope someone is able to construct a much better answer. My answer does not deserve a bounty in my opinion! – Bernhard Jul 22 '14 at 8:59
• Couldn't you get rid of $\delta$ by rescaling $t=\delta \tilde t$ ? – Joce Jul 23 '14 at 15:46
• Not really, I suppose, than I would have $f(\tilde{t}-1)$, being similarly annoying, right? Also, my derivative would be rescaled by $\delta$ – Bernhard Jul 23 '14 at 15:49

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.

• I still wonder if there are more informed guesses on the maximum value of the green line ($\delta=2$). – Bernhard Jul 18 '14 at 7:33
• It seems correct and complete to me. Especially, you may look for all possible couples of $f_{\max}$ and $f(t_{\max}-\delta)$ by solving $0 = x -x^3 - \alpha y$ for $-x<y<x$. You then have two symmetric sets of possible solutions (plus the trivial one), and $x$ is maximal if it is $-y$, your case 2. – Joce Jul 23 '14 at 19:44
• @Joce Well, it is correct for large $\delta$, but as you can see for the $\delta=2$ case, the maximum is clearly smaller than $\sqrt{1+\alpha}$ – Bernhard Jul 24 '14 at 7:01
• Allright, I hadn't realized that what you're looking for is a the maximum for a given $(\alpha,\delta)$. You can calculte an upper bound $f_{\max}-f(t_{\max}-\delta)\leq \delta(1+\alpha)^{3/2}+o(\delta)$ using your d.e., combined with (1) this will provide a $\delta$-dependent bound on $f_{\max}$ that in some cases will be smaller than $\sqrt{1+\alpha}$ (the expression is lengthy, as you've to solve a cubic equation). If this is not sharp enough yet, you can derive your d.e. to get the next order (the $o(\delta)$), and so on. – Joce Jul 24 '14 at 7:32
• @Joce This sounds really interesting! Do you have time to expand it to an answer? Is is not so obvious to me where the first expression in your comment comes from, so this might be an interesting lead for me! – Bernhard Jul 24 '14 at 7:45

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 $(\delta,\alpha)$.

• Thanks! I understand that $f-f^3-\alpha f(t-\delta) \leq (1+\alpha)^{3/2}$. I do not understand the step to $f_{max}-f(t_{max}-\delta)$. I am trying to grasp why the upper bound is linear in $\delta$. I think I do follow the next steps, but have to carefully work through that first. – Bernhard Jul 24 '14 at 8:57
• Thanks for your edit, helps me a lot. In your original comment on my answer, you meant to say $o(\delta^2)$ instead of $o(\delta)$, am I right? – Bernhard Jul 24 '14 at 9:07
• It is a $O(\delta^2)$ term, so $o(\delta)$. You can use any order of Taylor expansion, actually the gain may be interesting compared to the difficulty of the next order, as you get $\exists \tau,\, f(t_{\max})-f(t_{\max}-\delta) =\delta f'(t_\max) + \frac{\delta^2}{2} f''(\tau)$ and $f'(t_\max)=0$. Remains to check whether you get useful bounds for $f''$. – Joce Jul 24 '14 at 9:12
• In your derivation, you used that $f'(\tau)\leq (1+\alpha)^{3/2}$. However, as this is $f'(t_{max})$, we know that $f'(t_{max})=0$. Then, we would only have the second order term, and this gives me $(\alpha-1)f_{max}-f_{max}^3\leq\frac{1}{2}\alpha^2\delta^2(1+\alpha)^{3/2}$. What is your view on this? – Bernhard Jul 24 '14 at 12:16
• @Bernhard: I should have pointed a reference for this form of the remainder in Taylor expansion, it is the Lagrange form, en.wikipedia.org/wiki/… – Joce Jul 24 '14 at 13:55

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.

• I am not sure if I can follow your second bullet point. Do you mean to say: choose $Q=f(t-\delta)$ and solve $f'=f-f^3-Q=0$? (I do not know $f_{max}$ at this stage, obviously). Then, exactly how do I optimize? With respect to $Q$? – Bernhard Jul 23 '14 at 15:42
• Indeed. Solving $0=f-f^3-\alpha Q$ will give you between 1 and 3 solutions, $f_1,f_2,f_3$, all dependent on your parameter $Q$. Once you have those functions of $Q$, you can have take the derivative of $f_1(Q),f_2(Q),f_3(Q)$ and see how the function behaves with respect to $Q$. If you have a maximum, then you know you found a global maximum $f_{glob}$ (not necessarily attained though). And the last check is that $Q< f_{glob}$ to ensure consistency. – Matt B. Jul 23 '14 at 16:15
• I plotted those $f_i$ as a function of the $Q$ (which is basically just a rotated third-order-polynomial, so I think this does not give me a maximum. – Bernhard Jul 23 '14 at 16:18
• Ok, was worth a shot. – Matt B. Jul 23 '14 at 16:25