Part(a)
The first step is to find the pointwise limit for \(f_{k}\left ( t\right ) \) over \(\left [ 0,1\right ] \) \[ f^{\ast }\left ( t\right ) =\lim _{k\rightarrow \infty }f_{k}\left ( t\right ) =\lim _{k\rightarrow \infty }-\frac{1}{k}t^{2}+\left ( 1+\frac{1}{k^{2}}\right ) t+1=1+t \]
Hence \begin{align*} \left \Vert f_{k}-f^{\ast }\right \Vert & =\sup _{t\in \left [ 0,1\right ] }\left \vert f_{k}\left ( t\right ) -f^{\ast }\left ( t\right ) \right \vert \\ & =\sup _{t\in \left [ 0,1\right ] }\left \vert -\frac{1}{k}t^{2}+\left ( 1+\frac{1}{k^{2}}\right ) t+1-\left ( 1+t\right ) \right \vert \\ & =\sup _{t\in \left [ 0,1\right ] }\left \vert -\frac{1}{k}t^{2}+\frac{t}{k^{2}}\right \vert \end{align*}
At \(t=0\,,\left \Vert f_{k}-f^{\ast }\right \Vert =0\) and at \(t=1\) then \(\left \Vert f_{k}-f^{\ast }\right \Vert =\sup _{t\in \left [ 0,1\right ] }\left \vert -\frac{1}{k}+\frac{1}{k^{2}}\right \vert =0\) in the limit as \(k\rightarrow \infty \).
For \(0<t<1\), the maximum is found by taking derivative1 and solving for \(t\)\begin{align*} \frac{d}{dt}\left ( -\frac{1}{k}t^{2}+\frac{t}{k^{2}}\right ) & =0\\ \frac{-2t}{k}+\frac{1}{k^{2}} & =0\\ t & =\frac{1}{2k} \end{align*}
So the maximum occurs at \(t=\frac{1}{2k}\), Hence \[ \sup f_{k}\left ( t\right ) =\left \vert -\frac{1}{k}t^{2}+\frac{t}{k^{2}}\right \vert _{t=\frac{1}{2k}}=\left \vert -\frac{1}{k}\left ( \frac{1}{2k}\right ) ^{2}+\frac{1}{k^{2}}\frac{1}{2k}\right \vert =\left \vert -\frac{1}{4k^{3}}+\frac{1}{2k^{3}}\right \vert =\left \vert \frac{1}{4k^{3}}\right \vert \] Hence in the limit as \(k\rightarrow \infty \), \(\left \Vert f_{k}-f^{\ast }\right \Vert =\lim _{k\rightarrow \infty }\frac{1}{4k^{3}}\rightarrow 0\) for \(t\in \left [ 0,1\right ] \). Therefore the answer is true. It does converge uniformly.
Part(b)
\begin{align} \frac{\partial \Phi \left ( t,\tau \right ) }{\partial \tau } & =\frac{\partial }{\partial \tau }\left ( \Psi \left ( t\right ) \Psi ^{-1}\left ( \tau \right ) \right ) \nonumber \\ & =\Psi \left ( t\right ) \frac{d}{d\tau }\Psi ^{-1}\left ( \tau \right ) \tag{1} \end{align}
But \[ \frac{d}{d\tau }\Psi ^{-1}\left ( \tau \right ) =-\Psi ^{-1}\left ( \tau \right ) \left ( \frac{d}{dt}\Psi \left ( \tau \right ) \right ) \Psi ^{-1}\left ( \tau \right ) \] Hence (1) becomes\begin{align*} \frac{\partial \Phi \left ( t,\tau \right ) }{\partial \tau } & =\Psi \left ( t\right ) \left ( -\Psi ^{-1}\left ( \tau \right ) \left ( \frac{d}{dt}\Psi \left ( \tau \right ) \right ) \Psi ^{-1}\left ( \tau \right ) \right ) \\ & =-\overset{\Phi \left ( t,\tau \right ) }{\overbrace{\Psi \left ( t\right ) \Psi ^{-1}\left ( \tau \right ) }}\left ( \frac{d}{dt}\Psi \left ( \tau \right ) \right ) \Psi ^{-1}\left ( \tau \right ) \\ & =-\Phi \left ( t,\tau \right ) \left ( \frac{d}{dt}\Psi \left ( \tau \right ) \right ) \Psi ^{-1}\left ( \tau \right ) \end{align*}
but \(\frac{d}{dt}\Psi \left ( \tau \right ) =A\left ( \tau \right ) \Psi \left ( \tau \right ) \) since \(\Psi \left ( \tau \right ) \) satisfies the state equation. Hence the above becomes\begin{align*} \frac{\partial \Phi \left ( t,\tau \right ) }{\partial \tau } & =-\Phi \left ( t,\tau \right ) A\left ( \tau \right ) \overset{I}{\overbrace{\Psi \left ( \tau \right ) \Psi ^{-1}\left ( \tau \right ) }}\\ & =-\Phi \left ( t,\tau \right ) A\left ( \tau \right ) \end{align*}
Hence the answer is true.
Part(c)
Let the pointwise limit be \(f\left ( t\right ) \), then since\[ \left \vert f_{k}^{2}\left ( t\right ) -f^{2}\left ( t\right ) \right \vert =\left \vert f_{k}\left ( t\right ) -f\left ( t\right ) \right \vert \times \left \vert f_{k}\left ( t\right ) +f\left ( t\right ) \right \vert \] Since \(\left \Vert f_{k} \right \Vert \leq 1\), then \[ \left \vert f_{k}^{2}\left ( t\right ) -f^{2}\left ( t\right ) \right \vert \leq 2\left \vert f_{k}\left ( t\right ) -f\left ( t\right ) \right \vert \] Looking at the above in the sup norm results in\[ \left \Vert f_{k}^{2}-f^{2}\right \Vert \leq 2\left \Vert f_{k}-f\right \Vert \] Since we are told that \(f_{k}(t)\) converges uniformly, then \(\left \Vert f_{k}-f\right \Vert \rightarrow 0\) and so this implies that \(\left \Vert f_{k}^{2}-f^{2}\right \Vert \rightarrow 0\), which means that \(f_{k}^{2}\left ( t\right ) \) converges uniformly. Now we can take the limit inside the integral, hence\[ \lim _{k\rightarrow \infty }{\displaystyle \int \limits _{0}^{1}} f_{k}^{2}\left ( t\right ) dt={\displaystyle \int \limits _{0}^{1}} \lim _{k\rightarrow \infty }f_{k}^{2}\left ( t\right ) dt={\displaystyle \int \limits _{0}^{1}} f^{2}\left ( t\right ) dt \]
Hence the answer is true.
Let \(x^{0}\left ( t\right ) =x\left ( 0\right ) =1\), hence the first iteration is
\begin{align*} x^{1}\left ( t\right ) & =x^{0}\left ( t\right ) +{\displaystyle \int \limits _{0}^{t}} f\left ( x^{0}\left ( \tau \right ) ,\tau \right ) d\tau \\ & =1+{\displaystyle \int \limits _{0}^{t}} \min \left \{ x^{0}\left ( \tau \right ) ,\tau \right \} d\tau \\ & =1+{\displaystyle \int \limits _{0}^{1}} \min \left \{ 1,\tau \right \} d\tau +{\displaystyle \int \limits _{1}^{t}} \min \left \{ 1,\tau \right \} d\tau \\ & =1+{\displaystyle \int \limits _{0}^{1}} \tau d\tau +{\displaystyle \int \limits _{1}^{t}} 1d\tau \end{align*}
Hence \[ x^{1}\left ( t\right ) =\left \{ \begin{array} [c]{cc}1+{\displaystyle \int \limits _{0}^{1}} \tau d\tau =1+\frac{t^{2}}{2} & 0<t\leq 1\\ 1+{\displaystyle \int \limits _{0}^{1}} \tau d\tau +{\displaystyle \int \limits _{1}^{t}} 1d\tau =\overset{\left . 1+\frac{t^{2}}{2}\right \vert _{t=1}}{\overbrace{1+\frac{1}{2}}}+\left ( t-1\right ) =\frac{1}{2}+t & t>1 \end{array} \right . \]
Now we do the second iteration\begin{align*} x^{2}\left ( t\right ) & =x^{0}\left ( t\right ) +{\displaystyle \int \limits _{0}^{t}} f\left ( x^{1}\left ( \tau \right ) ,\tau \right ) d\tau \\ & =1+{\displaystyle \int \limits _{0}^{t}} \overbrace{\min \left \{ x^{1}\left ( \tau \right ) ,\tau \right \} }^{\tau }d\tau \\ & =1+{\displaystyle \int \limits _{0}^{t}} \tau d\tau =1+\frac{t^{2}}{2} \end{align*}
Hence\[ \fbox{$x^k\left ( t\right ) =1+\frac{t^2}{2}$}\]
Since \(A\) commutes with itself, which means \(A\left ( t_{1}\right ) A\left ( t_{2}\right ) =A\left ( t_{2}\right ) A\left ( t_{1}\right ) \) we can use a short cut to find \(\Phi \left ( t,\tau \right ) \). See HW 5, problem 2, part (b). Hence
\begin{align*} \Phi \left ( t,\tau \right ) & =\exp \left ({\displaystyle \int \limits _{\tau }^{t}} A\left ( \xi \right ) d\xi \right ) =\exp \left ({\displaystyle \int \limits _{\tau }^{t}} \begin{bmatrix} \sin \xi & \cos \xi \\ \cos \xi & \sin \xi \end{bmatrix} d\xi \right ) =\exp \left ( \begin{bmatrix}{\displaystyle \int \limits _{\tau }^{t}} \sin \xi d\xi &{\displaystyle \int \limits _{\tau }^{t}} \cos \xi d\xi \\{\displaystyle \int \limits _{\tau }^{t}} \cos \xi d\xi &{\displaystyle \int \limits _{\tau }^{t}} \sin \xi d\xi \end{bmatrix} \right ) \\ & =\exp \left ( \begin{bmatrix} -\cos t+\cos \tau & \sin t-\sin \tau \\ \sin t-\sin \tau & -\cos t+\cos \tau \end{bmatrix} \right ) \end{align*}
Since \[ x\left ( t\right ) =\Phi \left ( t,0\right ) x\left ( 0\right ) \]
Then applying this to \(x\left ( \pi \right ) =1\) gives\begin{align*} x\left ( \pi \right ) & =\Phi \left ( \pi ,0\right ) x\left ( 0\right ) \\ x\left ( \pi \right ) & =\exp \left ( \begin{bmatrix} -\cos \pi +\cos 0 & \sin \pi -\sin 0\\ \sin \pi -\sin 0 & -\cos \pi +\cos 0 \end{bmatrix} \right ) x\left ( 0\right ) \\ & =\exp \left ( \begin{bmatrix} 2 & 0\\ 0 & 2 \end{bmatrix} \right ) x\left ( 0\right ) \end{align*}
Since \(\begin{bmatrix} 2 & 0\\ 0 & 2 \end{bmatrix} \) is diagonal, then the above simplifies to\[ x\left ( \pi \right ) =\begin{bmatrix} e^{2} & 0\\ 0 & e^{2}\end{bmatrix} x\left ( 0\right ) \]
Now since \(\left \Vert x\left ( \pi \right ) \right \Vert =1\) then\[ 1=\left \Vert \begin{bmatrix} e^{2} & 0\\ 0 & e^{2}\end{bmatrix} x\left ( 0\right ) \right \Vert =\left \Vert \begin{bmatrix} e^{2} & 0\\ 0 & e^{2}\end{bmatrix} \right \Vert \left \Vert x\left ( 0\right ) \right \Vert \]
Using spectral norm (sqrt of maximum eigenvalue of \(A^{T}A)\) we find that \(\left \Vert \begin{bmatrix} e^{2} & 0\\ 0 & e^{2}\end{bmatrix} \right \Vert _{2}=\sqrt{e^{4}}=e^{2}=7.389\,1\), hence\[ \fbox{$\left \Vert x\left ( 0\right ) \right \Vert =\frac{1}{7.389\,1}=0.135\,33$}\]
\[\begin{bmatrix} x_{1}^{\prime }\\ x_{2}^{\prime }\end{bmatrix} =\begin{bmatrix} 0 & 1\\ -1 & 0 \end{bmatrix}\begin{bmatrix} x_{1}\\ x_{2}\end{bmatrix} \]
The eigenvalues of \(A\) are \(\left \{ i,-i\right \} \), For \(\lambda =i\) the eigenvector is \(\begin{bmatrix} 1\\ i \end{bmatrix} \) and for \(\lambda =-i\) the eigenvector is \(\begin{bmatrix} 1\\ -i \end{bmatrix} \). Therefore\[ V=\begin{bmatrix} 1 & 1\\ i & -i \end{bmatrix} \]
Hence \begin{align*} V^{-1}AV & =\Lambda \\ A & =V\Lambda V^{-1}\\ & =\begin{bmatrix} 1 & 1\\ i & -i \end{bmatrix}\begin{bmatrix} i & 0\\ 0 & -i \end{bmatrix}\begin{bmatrix} 1 & 1\\ i & -i \end{bmatrix} ^{-1} \end{align*}
Therefore
\begin{align*} e^{At} & =\begin{bmatrix} 1 & 1\\ i & -i \end{bmatrix}\begin{bmatrix} e^{it} & 0\\ 0 & e^{-it}\end{bmatrix}\begin{bmatrix} 1 & 1\\ i & -i \end{bmatrix} ^{-1}\\ & =\begin{bmatrix} \frac{e^{it}+e^{-it}}{2} & \frac{e^{it}-e^{-it}}{2i}\\ \frac{-e^{it}+e^{-it}}{2i} & \frac{e^{it}+e^{-it}}{2}\end{bmatrix} \\ & =\begin{bmatrix} \cos t & \sin t\\ -\sin t & \cos t \end{bmatrix} \end{align*}
Now \[ x\left ( t\right ) =e^{A\left ( t-\tau \right ) }x\left ( \tau \right ) \]
Given that \(x\left ( \tau \right ) =x\left ( \pi \right ) =\begin{bmatrix} 1\\ 1 \end{bmatrix} \), then
\begin{align*} x\left ( \frac{5\pi }{4}\right ) & =e^{A\left ( \frac{5\pi }{4}-\pi \right ) }\begin{bmatrix} 1\\ 1 \end{bmatrix} \\ & =e^{A\left ( \frac{\pi }{4}\right ) }\begin{bmatrix} 1\\ 1 \end{bmatrix} \\ & =\begin{bmatrix} \cos \frac{\pi }{4} & \sin \frac{\pi }{4}\\ -\sin \frac{\pi }{4} & \cos \frac{\pi }{4}\end{bmatrix}\begin{bmatrix} 1\\ 1 \end{bmatrix} \end{align*}
Hence\[ \fbox{$x\left ( \frac{5\pi }{4}\right ) =\begin{bmatrix} \sqrt{2}\\ 0 \end{bmatrix} $}\]