Linear control ODE - exponential convergence?

[haluza]

Hello,

I'm having hard times with the following simple linear ODE coming from a control problem:
$$u(t)' \leq \alpha(t) - u(t)\,,\quad u(0) = u_0 > 0$$
with a given smooth α(t) satisfying
$$0 \leq \alpha(t) \leq u(t) \quad\mbox{for all } t\geq 0.$$
My intuition is that $$\lim_{t\to\infty} u(t) - \alpha(t) = 0,$$
and that the convergence is exponential, i.e., $$|u(t) - \alpha(t)| = u(t) - \alpha(t) \leq c_1 e^{-c_2 t}.$$
For instance, if α was a constant, then the exponential convergence clearly holds (just solve the related ODE and use a "maximum principle").
Do you see a simple proof for time-dependent α (could not prove neither of the "statements" - probably I'm missing something very elementary); or is my intuition wrong?

Many thanks, Peter

 

[haruspex]

For a u(t) to exist satisfying those constraints puts constraints on α(t). Not exactly that it is monotonic non-increasing, but something approaching that. Do you know of such a constraint (beyond that implied)?

 

[nvir477]

For the fundamental solution set S={ex,e2x,e3x} can we construct a linear ODE with constant coefficients?

I have verified that the solution set is linearly independent via wronskian. I have got the annihilators as (D-1),(D-2),(D-3). However after that I'm not sure how to proceed. What do I do to get the ODE?

Thanks

 

[haruspex]

Consider α(t) as follows:
In the n^th period of time B, α(t) = A > 0, except for the last e^-n, where it is 0. If u'(t) = α(t) - u(t) and v_n = u(B_n) - A,
v_n+1 = v_ne^-B - A(1-e^-e-n)
> v_ne^-B - Ae^-n
So v_n > w_n where
w_n+1 = w_ne^-B - Ae^-n
which I believe gives:
w_n = Ce^-Bn - Ae^-n/(e^-1-e^-B)
w_n tends to 0 as n goes to infinity, not going negative. Hence u(t) converges to A, and the difference between u and α exceeds A on occasions beyond any specified t.

 

[blue_raver22]

Hello Peter,

Thank you for sharing your question and thoughts on this linear control ODE. It seems that your intuition is correct - the convergence to zero is indeed exponential. This can be proven by using the fact that the solution to this ODE can be written as $$u(t) = u_0e^{-t} + \int_0^t\alpha(s)e^{-(t-s)}ds.$$
By substituting this into the ODE, we get
$$u'(t) = \alpha(t)e^{-t} \leq \alpha(t)e^{-t} - u(t)e^{-t} = (\alpha(t) - u(t))e^{-t}.$$
Since we know that $$0 \leq \alpha(t) \leq u(t)$$
for all t, we can conclude that $$u'(t) \leq 0$$
for all t. This means that the solution u(t) is decreasing, and since it starts at a positive value, it must converge to a limit as t goes to infinity. We can also use the fact that $$u(t) \geq \alpha(t)$$
to show that the limit must be zero.

To prove the exponential convergence, we can use the Gronwall's inequality. Let $$v(t) = u(t) - \alpha(t).$$
Then we have $$v'(t) = u'(t) - \alpha'(t) \leq -v(t)e^{-t}$$
and since v(0) = u(0) - α(0) = u0 > 0, we can apply Gronwall's inequality to get
$$v(t) \leq u_0e^{-t}\int_0^te^{s}ds = u_0(e^t-1) \leq c_1e^{c_2t}$$
for some constants $$c_1, c_2 > 0.$$
Therefore, we have $$|u(t) - \alpha(t)| = v(t) \leq c_1e^{c_2t},$$
which shows that the convergence is indeed exponential.

I hope this helps and clarifies your intuition. Let me know if you have any further questions. Keep up the good work!

Best,