Processing math: 100%

1.18 Lecture 16. Tuesday October 21 2014

Properties of $\Phi \left ( t,\tau \right )$.

We developed Picard for solution of linear time varying $x^{\prime }=A\left ( t\right ) x$ in the last two lectures. Established: That solution exist and the solution is unique. Some disadvantages of Picard method are

1.

Each time the initial conditions $x^{0}$ changes, we have to run the method again to find the solution.

2.

No closed form solution, so we lose insight by not being able to do some qualitative analysis on the solution if it were analytical solution.

3.

LTI system always have closed for solution, and for many LTV, there is also closed form solution, so we should try to find closed form solution.

4.

If input $u\left ( t\right )$ changes, we have to run Picard method again

To find closed form solution, we need to obtain what is called the fundamental matrix.  Let $X^{01},X^{02},\cdots ,X^{0n}$ be $n$ linearly independent initial conditions for a system with $n$ states. For example, for $n=3$, always take these as $X^{01}=\begin{pmatrix} 1\\ 0\\ 0 \end{pmatrix} ,X^{02}=\begin{pmatrix} 0\\ 1\\ 0 \end{pmatrix} ,X^{03}=\begin{pmatrix} 0\\ 0\\ 1 \end{pmatrix}$ and so on for more states. For each one of these $X^{0i}$, let $\Psi ^{i}$ be the corresponding solution of $x^{\prime }=A\left ( t\right ) x$. i.e. $\Psi ^{1}\left ( 0\right ) =X^{01},\Psi ^{2}\left ( 0\right ) =X^{02},\Psi ^{3}\left ( 0\right ) =X^{03}$. i.e. $$\dot{\Psi }^{i}\left ( t\right ) =A\left ( t\right ) \Psi ^{i}\left ( t\right )$$ Now form the fundamental matrix solution $$\Psi \left ( t\right ) =\begin{pmatrix} \Psi ^{1}\left ( t\right ) & \Psi ^{2}\left ( t\right ) & \cdots & \Psi ^{n}\left ( t\right ) \end{pmatrix}$$ Each $\Psi ^{i}$ is $n\times 1$, and there are $n$ such columns, hence $\Psi \left ( t\right )$ is $n\times n$ matrix. Any solution can now be found with the help of this $\Psi \left ( t\right )$, for any initial conditions. Remark: Matrix $\Psi \left ( t\right )$ satisfies the state equation. $$\Psi ^{\prime }\left ( t\right ) =A\left ( t\right ) \Psi \left ( t\right )$$ At $t=0$,$\Psi \left ( 0\right )$ has $n$ linearly independent columns by construction. What about for $t>0$?

Reader: Show that $\Psi \left ( t\right )$ has $n$ linearly independent columns for $t>0.$ Proof: By contradiction. Assume at $t^{\ast }$ , $\Psi \left ( t^{\ast }\right )$ no longer has linearly independent columns. Then there exist vector $\vec{x}\left ( t^{\ast }\right )$ not zero s.t. $\Psi \left ( t^{\ast }\right ) \vec{x}\left ( t^{\ast }\right ) =\vec{0}$.  This implies that $x^{\prime }\left ( t^{\ast }\right ) =0$, which means that $x\left ( t\right ) =0$, hence contradiction.

Example: Let $x_{1}^{\prime }=x_{1}+tx_{2},x_{2}^{\prime }=x_{2}$. Hence $x^{\prime }=\begin{pmatrix} 1 & t\\ 0 & 1 \end{pmatrix}\begin{pmatrix} x_{1}\\ x_{2}\end{pmatrix}$. Now let $X^{01}=\begin{pmatrix} 1\\ 0 \end{pmatrix}$ to be one linearly independent initial conditions. We use this to solve the state equation. Next to use the second $X^{02}=\begin{pmatrix} 0\\ 1 \end{pmatrix}$ and repeat the process. So we end up with two solutions. These make up $\Psi$ matrix. Using $X^{01}$, we see that $x_{1}\left ( 0\right ) =1,x_{2}\left ( 0\right ) =0$. Now we solve the state equation. $x_{1}^{\prime }=x_{1}+tx_{2},x_{2}^{\prime }=x_{2}\left ( t\right )$. This results in $\Psi ^{1}\left ( t\right ) =\begin{pmatrix} e^{t}\\ 0 \end{pmatrix}$. Now using initial conditions $x_{1}\left ( 0\right ) =0,x_{2}\left ( 0\right ) =1$ we solve the same state equation again, this results in $\Psi ^{2}\left ( t\right ) =\begin{pmatrix} \frac{1}{2}t^{2}e^{t}\\ e^{t}\end{pmatrix}$, hence $$\Psi \left ( t\right ) =\begin{pmatrix} e^{t} & \frac{1}{2}t^{2}e^{t}\\ 0 & e^{t}\end{pmatrix}$$

DSolve[{x1'[t] == x1[t] + t x2[t], x2'[t] == x2[t], x1[0] == 1, x2[0] == 0},  
{x1[t], x2[t]}, t]  
{{x1[t] -> E^t, x2[t] -> 0}}  
DSolve[{x1'[t] == x1[t] + t x2[t], x2'[t] == x2[t], x1[0] == 0, x2[0] == 1},  
{x1[t], x2[t]}, t]  
{{x1[t] -> (E^t t^2)/2, x2[t] -> E^t}}

Now that we have found $\Psi \left ( t\right )$ we need to find the general solution to $x^{\prime }=A\left ( t\right ) x\left ( t\right ) +B\left ( t\right ) u$ with given any $x\left ( 0\right )$ (this initial condition has nothing to do with $X^{0i}$ used to find $\Psi \left ( t\right )$, this is the actual initial condition for the problem itself.

Assume the general solution is $$x\left ( t\right ) =\Psi \left ( t\right ) \theta \left ( t\right )$$ where $\theta \left ( t\right )$ is some function to be found. Plugging this solution into the state space equation, we obtain $$\Psi ^{\prime }\left ( t\right ) \theta \left ( t\right ) +\Psi \left ( t\right ) \theta ^{\prime }\left ( t\right ) =A\left ( t\right ) \Psi \left ( t\right ) \theta \left ( t\right ) +B\left ( t\right ) u$$ But $\Psi ^{\prime }\left ( t\right ) =A\left ( t\right ) \Psi \left ( t\right )$, so the above simplifies to $$\begin{align*} \Psi \left ( t\right ) \theta ^{\prime }\left ( t\right ) & =B\left ( t\right ) u\\ \theta ^{\prime }\left ( t\right ) & =\Psi ^{-1}\left ( t\right ) B\left ( t\right ) u \end{align*}$$

Integrating $$\theta \left ( t\right ) -\theta \left ( 0\right ) ={\displaystyle \int \limits _{0}^{t}} \Psi ^{-1}\left ( \tau \right ) B\left ( \tau \right ) u\left ( \tau \right ) d\tau$$ But $\theta \left ( 0\right ) =\Psi ^{-1}\left ( 0\right ) X\left ( 0\right )$ where $X\left ( 0\right )$ is the initial conditions. (why??). Hence the above becomes $$\theta \left ( t\right ) =\Psi ^{-1}\left ( 0\right ) X\left ( 0\right ) +{\displaystyle \int \limits _{0}^{t}} \Psi ^{-1}\left ( \tau \right ) B\left ( \tau \right ) u\left ( \tau \right ) d\tau$$ Therefore, since $x\left ( t\right ) =\Psi \left ( t\right ) \theta \left ( t\right ) \,.$ Then $$\begin{align*} x\left ( t\right ) & =\Psi \left ( t\right ) \left ( \Psi ^{-1}\left ( 0\right ) X\left ( 0\right ) +{\displaystyle \int \limits _{0}^{t}} \Psi ^{-1}\left ( \tau \right ) B\left ( \tau \right ) u\left ( \tau \right ) d\tau \right ) \\ & =\Psi \left ( t\right ) \Psi ^{-1}\left ( 0\right ) X\left ( 0\right ) +\Psi \left ( t\right ){\displaystyle \int \limits _{0}^{t}} \Psi ^{-1}\left ( \tau \right ) B\left ( \tau \right ) u\left ( \tau \right ) d\tau \\ & =\Psi \left ( t\right ) \Psi ^{-1}\left ( 0\right ) X\left ( 0\right ) +{\displaystyle \int \limits _{0}^{t}} \Psi \left ( t\right ) \Psi ^{-1}\left ( \tau \right ) B\left ( \tau \right ) u\left ( \tau \right ) d\tau \end{align*}$$

Let $$\Psi \left ( t\right ) \Psi ^{-1}\left ( \tau \right ) =\Phi \left ( t,\tau \right )$$ called the transition matrix, then the above becomes $$\fbox{$x\left ( t\right ) =\Phi \left ( t,0\right ) X\left ( 0\right ) +{\displaystyle \int \limits _0^t} \Phi \left ( t,\tau \right ) B\left ( \tau \right ) u\left ( \tau \right ) d\tau $}$$ Reader: Find $\Phi \left ( t,\tau \right )$ for the last example, and then find $x\left ( t\right )$ for unit step $u\left ( t\right )$.

Properties of $\Phi (t,\tau )$:

1.

$\Phi \left ( 0,0\right ) =I$. Note that $\Psi \left ( t\right )$ does not depend on the actual initial conditions for the problem. (these are eigenfunctions of the system).

2.

$\Phi \left ( t_{3},t_{1}\right ) =\Phi \left ( t_{3},t_{2}\right ) \Phi \left ( t_{2},t_{1}\right )$. Proof: $\Psi \left ( 3\right ) \Psi ^{-1}\left ( 1\right ) =\Psi \left ( 3\right ) \Psi ^{-1}\left ( 2\right ) \Psi \left ( 2\right ) \Psi ^{-1}\left ( 1\right ) =\Psi \left ( 3\right ) I\Psi ^{-1}\left ( 1\right ) =\Psi \left ( 3\right ) \Psi ^{-1}\left ( 1\right )$

3.

$\Phi \left ( t_{m},t_{1}\right ) =\Phi \left ( t_{m},t_{m-1}\right ) \Phi \left ( t_{m-1},t_{m-2}\right ) \cdots \Phi \left ( t_{2},t_{1}\right )$

4.

$\Phi \left ( t_{m},t_{m}\right ) =I$

5.

$\Phi \left ( t,t_{0}\right ) =\Phi ^{-1}\left ( t_{0},t\right )$

6.

$\Phi \left ( t,\tau \right )$ satisfies the state equation under appropriate conditions. proof: $\frac{\partial \Phi \left ( t,\tau \right ) }{\partial t}=\frac{\partial \Psi \left ( t\right ) \Psi ^{-1}\left ( \tau \right ) }{\partial t}=\overset{A\left ( t\right ) \Psi \left ( t\right ) }{\overbrace{\frac{\partial \Psi \left ( t\right ) }{\partial t}}}\Psi ^{-1}\left ( \tau \right ) =A\left ( t\right ) \Psi \left ( t\right ) \Psi ^{-1}\left ( \tau \right ) =A\left ( t\right ) \Phi \left ( t,\tau \right )$. But we can’t differentiate w.r.t. $\tau$ in the above. Reader: Think about $\frac{\partial \Phi \left ( t,\tau \right ) }{\partial \tau }$

side note: Remember this $\frac{\partial }{\partial t}{\displaystyle \int \limits _{t_{0}}^{t}} f\left ( t,\tau \right ) d\tau ={\displaystyle \int \limits _{t_{0}}^{t}} \frac{\partial }{\partial t}f\left ( t,\tau \right ) d\tau +\left . f\left ( t,\tau \right ) \right \vert _{\tau =t}$