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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...

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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!

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Question (a):

Seems okay to me.

Contraction Mapping Theorem:

Let $X$ be a complete, non-empty metric space with norm $\Vert \cdot \Vert$. Let $T:X\rightarrow X$ be a contraction map on $X$: that is, there exists constant $K \in [0,1)$ such that $$\Vert T(x)-T(y) \Vert ≤ K \Vert x-y \Vert$$ for all $x,y \in X$. Then there exists a unique fixed point in $X$.

Moreover, the fixed point is given by $$\lim_{n \rightarrow \infty} T^n(x_0)$$ where $x_0$ is an arbitrary element in $X$.


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:

$$\bigg \Vert \int_{t_0}^t \vec f(s,\vec x(s))ds \bigg \Vert ≤ \bigg| \int_{t_0}^t \big \Vert \vec f(s,\vec x(s)) \big \Vert ds \bigg |$$

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  • $\begingroup$ Thank you very much for your time. I really appreciate it. $\endgroup$ – MathIsHard May 22 '18 at 16:16
  • $\begingroup$ I think that the Lipschitz condition can be shown since the functions are all continuous. ie, $C^1$ implies locally Lipschitz as far as I understand... $\endgroup$ – MathIsHard May 22 '18 at 23:57
  • $\begingroup$ Seems reasonable $\endgroup$ – glowstonetrees May 23 '18 at 12:42
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    $\begingroup$ @MathIsHard careful that $C^1$ means continuously differentiable, not continuous. $\endgroup$ – Chee Han May 27 '18 at 8:16
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    $\begingroup$ Ah yes, you are right - fixed $\endgroup$ – glowstonetrees May 31 '18 at 0:00

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