Differential Equations

[Alexandru]

To be able to build a control [MATH] y_{1}{}'=y_1{}+y_{2} [/MATH]
[MATH]y_{2}{}'=y_2{}+u [/MATH]



[MATH]u \epsilon L^{2} (0,1)[/MATH]
for the care of the appropriate system solution [MATH]y_{1}(0)=y_{2}(0)=0[/MATH]
satisfy [MATH]y_{1}(1)=1 ,y_{2}(1)=0[/MATH]
Please kindly if you can help me
Discipline is Optimal Control


HELP!!!! i need to find control u




I am not cost functional, how to solve?

 

[HallsofIvy]

I do not know anything about "Optimal Control" and I don't know how you intend u to "control" anything. (And I certainly don't know what being "cost functional" means!)

But I do know a little about differential equations. The differential equation y2'= y2+ u has "associated homogeneous equation" y2'= dy2/dx= y2. That "separates" as dy2/y2= dx and, integrating both sides, ln(y2)= x+ C and then y2= e^{x+ C}= C'e^x. Whatever function of x u is we can look for a function that satisfies the entire equation of the form v(x)e^x (this method is called "variation of parameters" since we are treating the "parameter", C', as a variable). With y2= v(x)e^x. y2'= v'e^x+ ve^x= ve^x+ u. The "ve^x" terms cancel leaving v'e^x= u. v'= ue^{-x} and, integrating, v= int u(x)e^{-x}dx. So the general solution to the entire equation is y2(x)= C'e^x+ e^x int u(x)e^{-x}dx.

Since y1'= y1+y2= y1+ C'e^x+ e^x int u(x)e^{-x}dx you can solve that in a very similar way.