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Linear control ODE - exponential convergence?

  1. Jul 3, 2012 #1

    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
  2. jcsd
  3. Jul 5, 2012 #2


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    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)?
  4. Jul 5, 2012 #3
    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?

    Last edited by a moderator: Jul 5, 2012
  5. Jul 5, 2012 #4


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    Consider α(t) as follows:
    In the nth 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 vn = u(Bn) - A,
    vn+1 = vne-B - A(1-e-e-n)
    > vne-B - Ae-n
    So vn > wn where
    wn+1 = wne-B - Ae-n
    which I believe gives:
    wn = Ce-Bn - Ae-n/(e-1-e-B)
    wn 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.
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