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
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)?
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
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_{n}e^{-B} - A(1-e^{-e-n}) > v_{n}e^{-B} - Ae^{-n} So v_{n} > w_{n} where w_{n+1} = w_{n}e^{-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.