Lecture 16. Tuesday October 21 2014

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

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