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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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6  
Could you provide some more explanation? That equation is incomprehensible. –  Potato Dec 27 '11 at 20:08
    
Tried to fix up the equation. Rehana, please use LaTeX formatting in the future, as described here. –  Zev Chonoles Dec 27 '11 at 20:10
    
Thanks alot my friend for your attention, I edit my question by title' essential problem for homework". –  Rehana Dec 27 '11 at 21:08
1  
Please don't write \underset{X \in C^1([0,1])}{\inf}. I change this to \inf_{X \in C^1([0,1])}. In "display" mode, that causes the subscript to be directly under the "inf". In "inline" mode, the subscript will appear below and to the right of "inf". The whole point of having operator names like \inf, \det, \gcd, etc., instead of writing \mathrm{inf} or the like is that the standard conventions are built in. That also includes things like a space between "inf" and "A" when you write \inf A, and non-italicization of "inf". –  Michael Hardy Dec 27 '11 at 21:55
3  
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. –  Gerry Myerson Dec 28 '11 at 2:42

2 Answers 2

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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@Rehana, I'm not your dear friend, I'm a random stranger on the internet. Also, you have some funky unicode characters there. I have no idea what the mathematical meaning of U+2592 MEDIUM SHADE is even supposed to be. Please stick to LaTeX. –  Henning Makholm Dec 27 '11 at 22:02
    
,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〗 –  Rehana Dec 27 '11 at 22:07

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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@Rehana, it would be so much easier if you used latex to model your questions/comments. –  sxd Dec 27 '11 at 21:55
    
@DimitriSurinx, sorry, I don't have Latex, this software was damaged on my notebook. –  Rehana Dec 27 '11 at 22:06
    
@Patrick, 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〗 –  Rehana Dec 27 '11 at 22:08
    
@Rehana You don't have to have LaTeX installed on your machine in order to use it on this website. See the help page. If Javascript doesn't work on your computer then that's a more serious problem. –  Dylan Moreland Dec 27 '11 at 22:16
    
@Patrick, why did you say supremum is zero for above function? –  Rehana Dec 29 '11 at 20:36

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