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I want to find one method or approach or idea which compute following statement: $ \sup_{t \in [0,1]} \left( \inf_{X \in C^1([0,1])} \left\| \frac{dX(t)}{dt} - A(t)X(t) - F(t) \right\| \right) $ Thanks alot.

I want to find another idea for solving:

$\displaystyle\min\int_0^1\left(\left\|\frac{dX(t)}{dt}-A(t)X(t)-F(t)\right\|\mathrm{d}t\right) ,X(0)=X_0,X(t)\in K\subseteq\mathbb{R}^n\;,$

where $K$ is a closed interval.

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    Can I ask someone who understands the question to edit the title into something more informative than "essential problem", please? The best I can come up with is "supremum of something involving a derivative," but surely someone can do better than that.2011-12-28

2 Answers 2

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Since $X$ is allowed depend on $t$, you can just choose $X$ to be $u\mapsto (u-t)F(t)$. Then $X(t)=0$ takes care of the $A(t)X(t)$ term, and $-F(t)$ is cancelled out by the derivative of $X$.

Therefore the $\inf$ is $0$ for every $t$, and thus the result of the entire expression is $0$, too.

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    ,Thanks , It is good idea, would you mind to give me another idea which is similar to sup inf for solve (dX(t))/dt=A(t)X(t)+F(t) by min∫_0^1▒‖dX(t)/dt-A(t)X(t)-F(t) ‖dt X(0)=X_0 X(t)∈K⊆R^(n ),K is close interval〗2011-12-27
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We can do the following. Let $X(t) = Y(t) \, \mathrm{exp}\left( \left( \int_0^t A(z) \, dz \right)t \right)$ that I will write (for short) $Y \, e^{t \int A }$. Then the differential equation $ \frac{dX}{dt} - A(t)X(t) - F(t) = 0 $ becomes \frac{d(Y \, e^{t \int A })}{dt} - A Y \, e^{t \int A } - F(t) = Y'e^{t \int A } + AY \, e^{t \int A } -A Y \, e^{t \int A } - F(t) = Y' \, e^{t \int A } - F(t) = 0 which means that $ X(t) = Y(t) \, e^{t \int A }= e^{t \int_0^t A} \int_0^t e^{-u \int_0^u A} F(u) \, du. $ Therefore your supremum is zero whenever this function is in $C^1([0,1])$. When $A$ and $F$ are continuous we can see that this is the case.

Hope that helps,

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    @Patrick, why did you say supremum is zero $f$or above function?2011-12-29