Page 421 Problem 15

Page 421 Problem 15

Problem 15a)

The problem makes use of uniqueness. The book uses this uniqueness
property on page 417 right after equation (27).

If we can show that both $\Phi(t)\Phi(s)$ and $\Phi(t+s)$ satisfy
the same initial value problem, then they are the same object. The
book's hint is telling us which initial value problem they satisfy.
Thus we simply have to show both satisfy $Z^{\prime}=AZ,\, Z(0)=\Phi(s)$.
Note that we are taking the derivative with respect to $t$, because
$s$ is fixed. And since they both satisfy the same initial value
problem for all $s$ they are the same.

$[\Phi(t)\Phi(s)]^{\prime}=\Phi^{\prime}(t)\Phi(s)=A\Phi(t)\Phi(s)$.
And $\Phi(0)\Phi(s)=I\Phi(s)=\Phi(s)$

I find the above to be rather clear. The next line might be more confusing:

$[\Phi(t+s)]^{\prime}=\Phi^{\prime}(t+s)=A\Phi(t+s)$. And $\Phi(0+s)=\Phi(s)$.

The first equality is chain rule. The derivative of $(t+s)$ with
respect to $t$ is $1$.

The second equality is from the assumption that $\Phi^{\prime}(x)=A\Phi(x)$.
In particular it is true for $x=t+s$.

Problem 15b)

Just let $s=-t$ in 15a) to get this result: $\Phi(t)\Phi(-t)=I$.
This holds for all $t$. So we have it for $t=u$ and $t=-u$ . Thus
$\Phi(u)\Phi(-u)=I=\Phi(-u)\Phi(u)$. And hence $\Phi(-u)=\Phi^{-1}(u)$.
(By definition $B=A^{-1}$ if $AB=BA=I$)

Problem 15c)

This is just a combination of 15a and 15b, Enjoy!