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3.2 Second exam

3.2.1 Questions

3.2.2 Key solution

3.2.3 My Expanded write-up of second exam

Problem 1

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.

Problem 2

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

$$\fbox{$x^k\left ( t\right ) =1+\frac{t^2}{2}$}$$

Problem 3

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

$$\fbox{$\left \Vert x\left ( 0\right ) \right \Vert =\frac{1}{7.389\,1}=0.135\,33$}$$

Problem 4

$$\begin{bmatrix} x_{1}^{\prime }\\ x_{2}^{\prime }\end{bmatrix} =\begin{bmatrix} 0 & 1\\ -1 & 0 \end{bmatrix}\begin{bmatrix} x_{1}\\ x_{2}\end{bmatrix}$$

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

$$\fbox{$x\left ( \frac{5\pi }{4}\right ) =\begin{bmatrix} \sqrt{2}\\ 0 \end{bmatrix} $}$$