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.