I am attempting to do this problem here for studying purposes for an exam I have in a couple months. I was hoping to get some help...
For part a) I was trying the following:
Given a contraction mapping K : C → X, K admits a unique fixed-point x* in X (i.e. K(x*) = x*). Furthermore, x* can be found as follows: start with an arbitrary element $x_0$ in X and define a sequence {$x_n$} by $x_n = T(x_n−1)$, then $x_n$ → x*. I was wondering if that is ok for a?
Also, I am really stuck on part b) and c). I don't know how to reformulate the problem to be a fixed point problem. I would really appreciate help on this.
Thanks!
Question (a):
Seems okay to me.
Question (b):
Let $\vec x (t) = \begin{pmatrix} x(t) \\ \dot x(t) \end{pmatrix}$. Then
$$\dfrac{d\vec x}{dt}(t)= \begin{pmatrix} \dot x(t) \\ \ddot x(t) \end{pmatrix} := \begin{pmatrix} f_1(t,x,\dot x) \\ f_2(t,x,\dot x) \end{pmatrix}= \begin{pmatrix} f_1(t,\vec x) \\ f_2(t,\vec x) \end{pmatrix}:=\vec f (t,\vec x)$$
where $f_1(t,\vec x) = \dot x$ and $f_2(t,\vec x) = \ddot x = r(t)-p(t) \dot x - q(t) x$.
Moreover, we have $$\vec x (t_0) = \begin{pmatrix} x(t_0) \\ \dot x (t_0) \end{pmatrix} = \begin{pmatrix} x_0 \\ x_1 \end{pmatrix} := \vec c$$
Hence, the initial value problem has become
$$\cdots \iff \frac{d \vec x(t)}{dt} = \vec f(t,\vec x(t)) \qquad \vec x(t_0) = \vec c$$
Moreover, we can transform this into an integral equation by integrating on both sides with respect to some dummy variable $s$, and fix the constant of integration using the intial value:
$$\cdots \iff \vec x(t) = \int_{t_0}^t \vec f(s,\vec x(s)) ds + \vec c \tag{*}$$
So, if we define the map
$$T\vec x (t) := \int_{t_0}^t \vec f(s,\vec x(s)) ds + \vec c$$
then $\vec x$ is a solution to $(*)$ if and only if $\vec x$ is a solution to $T\vec x(t)=\vec x(t)$, which is the required fixed point problem.
All of the $\iff$'s on the way guarantee that the two problems are equivalent.
Question (c):
Honestly, I was just studying for my exams too and I had practically the same question as this in a past paper, with the exception that I was given that $\vec f_2$ satisfies a Lipschitz condition
$$\Vert \vec f_2(t,\vec x) - \vec f_2(t,\vec y) \Vert≤ L\Vert \vec x - \vec y \Vert$$
So without this, I am not too sure either.
In any case, what you have to do in this part of the question is to prove that the map $T$ satisfies the hypotheses for the CMT, so that by the CMT there exists a unique fixed point.
Here, we define $X$ to be the space of continuous functions from $[t_0-h,t_0+h]$ to $B$ where
$$B = \{ \vec v \in \Bbb R ^2: \Vert \vec v - \vec c \Vert _1 ≤ k\}$$
where $\Vert (v_1,v_2)^T \Vert _1 := |v_1|+|v_2|$ is the $1$-norm. The $k$ and $h$ are constants that you need to choose later on. And define the norm of $X$ to be
$$\Vert \vec g(t) \Vert _\infty :=\sup_{t \in [t_0-h,t_0+h]} \Vert \vec g(t) \Vert _1$$
There are then two things you need to do:
(i) Show that $T:X \rightarrow X$
i.e. if $\vec g:[t_0-h,t_0+h] \rightarrow B$ is continuous, then we have $T \vec g:[t_0-h,t_0+h] \rightarrow B$ and that this is also continuous
(ii) Show that $T$ is indeed a contraction map
i.e. there exists constant $K \in [0,1)$ such that $\Vert T(x)-T(y) \Vert ≤ K \Vert x-y \Vert$ for all $x,y \in X$
Notes:
Note that $[t_0-h,t_0+h]$ and $B$ are compact, so that a continuous function defined on these sets are bounded. You will need this fact when fixing the constants $k$,$h$ and $K$.
Also, recall the triangle inequality for integrals:
$$\bigg \Vert \int_{t_0}^t \vec f(s,\vec x(s))ds \bigg \Vert ≤ \bigg| \int_{t_0}^t \big \Vert \vec f(s,\vec x(s)) \big \Vert ds \bigg |$$