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Differential Equations

Alexandru

New member
Jun 6, 2020
1
To be able to build a control \(\displaystyle
y_{1}{}'=y_1{}+y_{2}

\)

\(\displaystyle y_{2}{}'=y_2{}+u \)




\(\displaystyle u \epsilon L^{2} (0,1)\)

for the care of the appropriate system solution \(\displaystyle y_{1}(0)=y_{2}(0)=0\)

satisfy \(\displaystyle y_{1}(1)=1 ,y_{2}(1)=0\)

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?
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
362
I'm not really up on "control" problems and I have no idea what being "cost functional" means but this looks to me like solving a system of differential equations. I am going to write the independent variable as "t".

If you are used to dealing with "control" problems, you might be used to setting a system of differential equations up as a matrix equation. For this problem that would be


$\begin{pmatrix}y_1 \\ y_2 \end{pmatrix}'= \begin{pmatrix}1 & 1 \\ 0 & 1 \end{pmatrix}$$\begin{pmatrix}y_1 \\ y_2 \end{pmatrix}$$+ \begin{pmatrix} 0 \\ u \end{pmatrix}$
and then observe that the matrix $\begin{pmatrix}1 & 1 \\ 0 & 1 \end{pmatrix}$ has 1 as a double eigenvalue.

Personally I wouldn't get that "sophisticated". I would, rather, note that the second equation, $y_2'= y_2+ u$ doesn't involve $y_1$ so can be solved directly. The "associated homogeneous equation" is $y_2'= y_2$ which has the general solution $y_2= Ce^t$ for C any constant. Using the "variation of parameters" method, we seek a solution of the form $y_2= v(t)e^t$ (thus "varying the parameter"- letting C be a function). $y_2'= v'e^t+ ve^t= ve^t+ u$ so that $v'e^t= u$, $v'= e^{-t}u$ and $v= \int_1^t e^{-z}u(z)dz$ (I have taken 1 as the lower limit in order to simplify using $y_2(1)= 0$).

So $y_2(t)= Ce^t+ e^t\int_1^t e^{-z}u(z)dz$. Setting t= 1, $y_2(1)= Ce= 0$ so $C= 0$. $y_2(t)= e^t\int_1^t e^{-z}u(z)dz$.

Now turn to the first equation. $y_1'= y_1+ y_2= y_1+ e^t\int_1^t e^{-z}u(z)dz$. The "associated homogeneous equation" for this is $y_1'= y_1$ which has general solution $y_1(t)= De^t$. Again, to use "variation of parameters" we seek a solution of the form $y_1(t)= w(t)e^t$. $y_1'= w'e^t+ we^t= we^t+ e^t\int_1^t e^{-z}u(z)dz$ so $w' e^t= e^t\int_1^t e^{-z}u(z)dz$. $w'(t)= \int_1^z e^{-z}u(z)dz$ and $w(t)= \int_1^t\int_1^s e^{-z}u(z)dz ds$.
Then $y_1(t)= De^t+ e^t\int_1^t\int_1^s e^{-z}u(z)dz ds$.
Since $y_1(1)= D= 0$, $y_1(t)= e^t\int_1^t\int_1^s e^{-z}u(z)dz ds$.

Now that we have solved the differential equations in terms of "u" what would make u a "control"?
 
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