We have x^{\prime }\left ( t\right ) =A\left ( t\right ) x\left ( t\right ) ,x\left ( 0\right ) =x^{0} with Picard iterate x\left ( 0\right ) \equiv x^{0}\left ( t\right ) . Define x^{k+1}\left ( t\right ) =x^{0}+{\displaystyle \int \limits _{0}^{t}} A\left ( \tau \right ) x^{k}\left ( \tau \right ) d\tau for k=0,1,2,\cdots
This sequence lives in the bounded space B\left ( \left [ t_{0},t_{1}\right ] \mathbb{R} ^{n}\right ) . Lemma 1 from last lecture shows that x^{k}\left ( t\right ) is bounded. i.e. x^{k}\left ( t\right ) \in B. Now we go to lemma 2
lemma 2:
Convergence: The Picard iteration satisfy \overbrace{ \left \Vert x^{k+1}(t) -x^{k}(t) \right \Vert }\leq \frac{\left \Vert x^{0}\right \Vert \Pi ^{k+1}\left ( t\right ) }{\left ( k+1\right ) !} for k=0,1,2,\cdots . Notice the LHS is norm in \mathbb{R} ^{n} (pointwise convergence) since we used x\left ( t\right ) inside.
proof: By induction. For k=0,\begin{align*} \left \Vert x^{1}\left ( t\right ) -x^{0}\left ( t\right ) \right \Vert & =\left \Vert x^{0}+{\displaystyle \int \limits _{0}^{t}} A\left ( \tau \right ) x^{0}\left ( \tau \right ) d\tau -x^{0}\right \Vert \\ & =\left \Vert{\displaystyle \int \limits _{0}^{t}} A\left ( \tau \right ) x^{0}\left ( \tau \right ) d\tau \right \Vert \\ & \leq{\displaystyle \int \limits _{0}^{t}} \left \Vert A\left ( \tau \right ) x^{0}\left ( \tau \right ) \right \Vert d\tau \\ & \leq{\displaystyle \int \limits _{0}^{t}} \left \Vert A\left ( \tau \right ) \right \Vert \left \Vert x^{0}\left ( \tau \right ) \right \Vert d\tau \\ & =\left \Vert x^{0}\left ( \tau \right ) \right \Vert{\displaystyle \int \limits _{0}^{t}} \left \Vert A\left ( \tau \right ) \right \Vert d\tau \end{align*}
But \Pi \left ( t\right ) ={\displaystyle \int \limits _{0}^{t}} \left \Vert A\left ( \tau \right ) \right \Vert d\tau and \Pi \left ( t\right ) is non decreasing. It maximum is \Pi \left ( t_{1}\right ) Hence \left \Vert x^{1}\left ( t\right ) -x^{0}\left ( t\right ) \right \Vert \leq \left \Vert x^{0}\left ( \tau \right ) \right \Vert \Pi \left ( t_{1}\right ) So true for k=0. Now assume lemma is true for k and we need to show it is true for k+1. We form\begin{align*} \left \Vert x^{k+1}\left ( t\right ) -x^{k}\left ( t\right ) \right \Vert & =\left \Vert \left ( x^{0}+{\displaystyle \int \limits _{0}^{t}} A\left ( \tau \right ) x^{k}\left ( \tau \right ) d\tau \right ) -\left ( x^{0}+{\displaystyle \int \limits _{0}^{t}} A\left ( \tau \right ) x^{k-1}\left ( \tau \right ) d\tau \right ) \right \Vert \\ & =\left \Vert{\displaystyle \int \limits _{0}^{t}} A\left ( \tau \right ) x^{k}\left ( \tau \right ) d\tau -{\displaystyle \int \limits _{0}^{t}} A\left ( \tau \right ) x^{k-1}\left ( \tau \right ) d\tau \right \Vert \\ & =\left \Vert{\displaystyle \int \limits _{0}^{t}} A\left ( \tau \right ) \left ( x^{k}\left ( \tau \right ) -x^{k-1}\left ( \tau \right ) \right ) d\tau \right \Vert \\ & \leq{\displaystyle \int \limits _{0}^{t}} \left \Vert A\left ( \tau \right ) \left ( x^{k}\left ( \tau \right ) -x^{k-1}\left ( \tau \right ) \right ) \right \Vert d\tau \\ & \leq{\displaystyle \int \limits _{0}^{t}} \left \Vert A\left ( \tau \right ) \right \Vert \left \Vert \left ( x^{k}\left ( \tau \right ) -x^{k-1}\left ( \tau \right ) \right ) \right \Vert d\tau \end{align*}
Since we assumed it is true for k, i.e. \left \Vert \left ( x^{k}\left ( \tau \right ) -x^{k-1}\left ( \tau \right ) \right ) \right \Vert \leq \frac{\left \Vert x^{0}\right \Vert \Pi ^{k}\left ( t\right ) }{k!} is true by assumption. Then the above becomes\begin{align*} \left \Vert x^{k+1}\left ( t\right ) -x^{k}\left ( t\right ) \right \Vert & \leq{\displaystyle \int \limits _{0}^{t}} \left \Vert A\left ( \tau \right ) \right \Vert \frac{\left \Vert x^{0}\right \Vert \Pi ^{k}\left ( \tau \right ) }{k!}d\tau \\ & =\frac{\left \Vert x^{0}\right \Vert }{k!}{\displaystyle \int \limits _{0}^{t}} \left \Vert A\left ( \tau \right ) \right \Vert \Pi ^{k}\left ( \tau \right ) d\tau \end{align*}
But \frac{d}{d\tau }\Pi \left ( \tau \right ) =\left \Vert A\left ( \tau \right ) \right \Vert then d\Pi =\left \Vert A\left ( \tau \right ) \right \Vert d\tau and the above can be written as\begin{align*} \left \Vert x^{k+1}\left ( t\right ) -x^{k}\left ( t\right ) \right \Vert & \leq \frac{\left \Vert x^{0}\right \Vert }{k!}{\displaystyle \int \limits _{0}^{t}} \Pi ^{k}\left ( \tau \right ) d\Pi \\ & =\frac{\left \Vert x^{0}\right \Vert }{k!}\left ( \frac{\Pi ^{k+1}\left ( \tau \right ) }{k+1}\right ) _{0}^{t}\\ & =\frac{\left \Vert x^{0}\right \Vert }{k!}\left ( \frac{\Pi ^{k+1}\left ( t\right ) }{k+1}-\frac{\Pi ^{k+1}\left ( 0\right ) }{k+1}\right ) \end{align*}
But \Pi \left ( 0\right ) =0 from properties of \Pi , then the above reduces to \left \Vert x^{k+1}\left ( t\right ) -x^{k}\left ( t\right ) \right \Vert \leq \left \Vert x^{0}\right \Vert \frac{\Pi ^{k+1}\left ( t\right ) }{\left ( k+1\right ) !} and this proofs the lemma.
Lemma 3:
x^{k}\left ( t\right ) converges to the some limit. We need to show that x^{k}\left ( t\right ) converges uniformly to some x^{\ast }\left ( t\right ) \in B\left ( \left [ t_{0},t_{1}\right ] ,\mathbb{R} ^{n}\right ) . When we say a function converges in bounded space B, we always mean uniform convergence.
proof:
We need to generate a telescoping sequence, as in x^{\left ( 4\right ) }=\left ( x^{\left ( 4\right ) }-x^{\left ( 3\right ) }\right ) +\left ( x^{\left ( 3\right ) }-x^{\left ( 2\right ) }\right ) +\left ( x^{\left ( 2\right ) }-x^{\left ( 1\right ) }\right ) +\left ( x^{\left ( 1\right ) }-x^{\left ( 0\right ) }\right ) +x^{\left ( 0\right ) }, which mean x^{n}\left ( t\right ) =x^{0}\left ( t\right ) +\sum _{k=0}^{n-1}\left ( x^{k+1}\left ( t\right ) -x^{k}\left ( t\right ) \right ) We now need to use the M-test to bound \left \Vert x^{k+1}\left ( t\right ) -x^{k}\left ( t\right ) \right \Vert . From lemma 2 \left \Vert x^{k+1}-x^{k}\right \Vert _{I}\leq \sup _{0\leq t\leq t_{1}}\left \Vert x^{0}\right \Vert \frac{\Pi ^{k+1}\left ( t\right ) }{\left ( k+1\right ) !}=\left \Vert x^{0}\right \Vert \frac{\Pi ^{k+1}\left ( t_{1}\right ) }{\left ( k+1\right ) !} Since \Pi is non-decreasing, then we can bound the above from below by some M_{k}=\left \Vert x^{0}\right \Vert \frac{\Pi ^{k+1}\left ( t_{1}\right ) }{\left ( k+1\right ) !}, so now can use M-test\begin{align*} \sum _{k=0}^{\infty }M_{k} & =\left \Vert x^{0}\right \Vert \sum _{k=0}^{\infty }\frac{\Pi ^{k+1}\left ( t_{1}\right ) }{\left ( k+1\right ) !}\\ & =\left \Vert x^{0}\right \Vert \left ( e^{\Pi \left ( t_{1}\right ) }-1\right ) \end{align*}
Since \sum _{k=0}^{\infty }M_{k} is finite, then by the M-test we conclude that \sum _{k=0}^{n-1}\left ( x^{k+1}-x^{k}\right ) will converge to some limiting value, which implies x^{\left ( n\right ) } in the limit will also converge (uniformly) to some limit x^{\ast }
Lemma 4:
The x^{\ast } obtained from lemma 3 solves the state equation x^{\prime }=A\left ( t\right ) x\left ( t\right )
proof: We know that x^{k+1}=x^{0}+\int _{0}^{t}A\left ( \tau \right ) x^{k}\left ( \tau \right ) d\tau and we also know that x^{k+1}\left ( t\right ) will converge uniformly to some limit x^{\ast }\left ( t\right ) by lemma 3. Taking the limit of both sides of the Picard iteration formula above gives\begin{align*} \lim _{k\rightarrow \infty }x^{k+1}\left ( t\right ) & =\lim _{k\rightarrow \infty }\left ( x^{0}\left ( t\right ) +\int _{0}^{t}A\left ( \tau \right ) x^{k}\left ( \tau \right ) d\tau \right ) \\ x^{\ast }\left ( t\right ) & =x^{0}\left ( t\right ) +\lim _{k\rightarrow \infty }\int _{0}^{t}A\left ( \tau \right ) x^{k}\left ( \tau \right ) d\tau \end{align*}
To take the limit inside the integral, we need to first show that A\left ( \tau \right ) x^{k}\left ( \tau \right ) converges uniformly to A\left ( \tau \right ) x^{\ast }\left ( \tau \right )
\begin{align*} \left \Vert Ax^{k}-Ax^{\ast }\right \Vert _{I} & \leq \sup \left \Vert A\left ( t\right ) \right \Vert \overbrace{\left \Vert x^{k}(t) -x^{\ast }(t) \right \Vert }^{\text{converges uniformly to say z}}\\ & \leq \left \Vert A(t) \right \Vert z \end{align*}
But \left \Vert A\left ( t\right ) \right \Vert is bounded, hence A\left ( \tau \right ) x^{k}\left ( \tau \right ) converges uniformly and now we can take the limit inside the integral. x^{\ast }\left ( t\right ) =x^{0}+\int _{0}^{t}\lim _{k\rightarrow \infty }A\left ( \tau \right ) x^{k}\left ( \tau \right ) d\tau But \lim _{k\rightarrow \infty }x^{k}\left ( t\right ) =x^{\ast }\left ( t\right ) by lemma 3, hence the above becomes x^{\ast }\left ( t\right ) =x^{0}+\int _{0}^{t}A\left ( \tau \right ) x^{\ast }\left ( \tau \right ) d\tau This proofs the lemma.
We now need to establish uniqueness. Which means we need to show that x^{\ast } is the only solution to x^{\prime }=A\left ( t\right ) x\left ( t\right )
We will use what is called Granwall’s inequality.
If u\left ( t\right ) and \theta \left ( t\right ) are non-negative continuous functions on \left [ 0,t\right ] satisfying \theta \left ( t\right ) \leq{\displaystyle \int \limits _{0}^{t}} u\left ( \tau \right ) \theta \left ( \tau \right ) d\tau then \theta \left ( t\right ) =0 everywhere. Now we assume there are two solutions to state equation. x_{1}^{\prime }=Ax_{1} and x_{2}^{\prime }=Ax_{2}. Therefore\begin{align*} x_{1}\left ( t\right ) -x_{1}\left ( 0\right ) & ={\displaystyle \int \limits _{0}^{t}} A\left ( \tau \right ) x_{1}\left ( \tau \right ) d\tau \\ x_{2}\left ( t\right ) -x_{2}\left ( 0\right ) & ={\displaystyle \int \limits _{0}^{t}} A\left ( \tau \right ) x_{2}\left ( \tau \right ) d\tau \end{align*}
Hence\begin{align*} \left \Vert x_{1}\left ( t\right ) -x_{2}\left ( t\right ) \right \Vert & =\left \Vert{\displaystyle \int \limits _{0}^{t}} A\left ( \tau \right ) \left ( x_{1}\left ( \tau \right ) -x_{2}\left ( \tau \right ) \right ) d\tau \right \Vert \\ & \leq{\displaystyle \int \limits _{0}^{t}} \left \Vert A\left ( \tau \right ) \left ( x_{1}\left ( \tau \right ) -x_{2}\left ( \tau \right ) \right ) \right \Vert d\tau \\ & \leq{\displaystyle \int \limits _{0}^{t}} \left \Vert A\left ( \tau \right ) \right \Vert \left \Vert \left ( x_{1}\left ( \tau \right ) -x_{2}\left ( \tau \right ) \right ) \right \Vert d\tau \end{align*}
Let x_{1}\left ( t\right ) -x_{2}\left ( t\right ) \equiv \theta \left ( t\right ) and A\left ( t\right ) \equiv u\left ( t\right ) then by Granwall inequality x_{1}\left ( t\right ) -x_{2}\left ( t\right ) =0 or x_{1}=x_{2}. Therefore the solution to state space is unique.
Can we get unique solution to the state space problem? For large family of A\left ( t\right ) we can. We need to formulate the fundamental matrix.