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Linear control ODE  exponential convergence? 
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#1
Jul312, 07:27 PM

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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 timedependent α (could not prove neither of the "statements"  probably I'm missing something very elementary); or is my intuition wrong? Many thanks, Peter 


#2
Jul512, 12:17 AM

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



#3
Jul512, 12:36 AM

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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 (D1),(D2),(D3). However after that I'm not sure how to proceed. What do I do to get the ODE? Thanks 


#4
Jul512, 03:36 AM

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Linear control ODE  exponential convergence?
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(1e^{en}) > 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. 


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