Urgent: Existence and Uniqueness theorem

MT20
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Greating my friends,

I have just returned home today from heart surgery.

I still feeling the effects of the operation, because I'm affried the hospital send me home a bit to early. But I have to have these questions finished before tomorrow.

So therefore I would very much appreciate if somebody could help me answer these questions?

(a)

(I use the triangle inequality in (1) and (3)?)

Let I be a open interval and f: I \rightarrow \mathbb{R}^n be a continious function.

Let || \cdot || be a given norm on \mathbb{R}^n.

Show the following:

1) If there exists a C>0 then ||x|| \leq C ||C||_1; x \in \mathbb{R}^n, ||x|| _1 = \sum _{j=1} ^n |x_j|

2) The mapping I \ni t \rightarrow ||f(t)|| \in \mathbb{R} is continious on.

3) for all t1,t2 \in \ || \int_{t_1} ^{t_2} f(t) dt|| \leq | \int_{t_1} ^{t_2} ||f(t)||dt|


(b)

Looking at the system(*) of equations,

x1' = (a-bx2)* x1
x2' = (cx1 -d)*x2

open the open Quadrant K; here a,b,c and d er positive constants.

I need to show that the system can be integrated. Which means I need to show that there exist a C^1 -function, with the properties F:U \rightarrow \mathbb{R}, where U \subseteq K is open and close in K(close means that for every point in K, is a limit point for q sequence, whos elements belongs to K). Such that gradient F \notequal 0 for alle x \in U and such that F is constant for trajectories of system.

Finally I need to conclude that every max trajectory is contained in a compact subset of K, and that the system (*) field X: R \rightarrow \mathbb{R}^2.

Sincerley Yours
Maria Thomson 20

p.s. I'm so sorry for asking so much, but has been such a difficult couple of days for me.
 
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I'm just going to copy and past your post, adding in TeX tags so I can read your questions:

Greating my friends,

I have just returned home today from heart surgery.

I still feeling the effects of the operation, because I'm affried the hospital send me home a bit to early. But I have to have these questions finished before tomorrow.

So therefore I would very much appreciate if somebody could help me answer these questions?

(a)

(I use the triangle inequality in (1) and (3)?)

Let I be a open interval and f: I \rightarrow \mathbb{R}^n be a continious function.

Let || \cdot || be a given norm on \mathbb{R}^n.

Show the following:

1) If there exists a C>0 then ||x|| \leq C ||C||_1; x \in \mathbb{R}^n, ||x|| _1 = \sum _{j=1} ^n |x_j|

2) The mapping I \ni t \rightarrow ||f(t)|| \in \mathbb{R} is continious on.

3) for all t_1,t_2 \in \ || \int_{t_1} ^{t_2} f(t) dt|| \leq | \int_{t_1} ^{t_2} ||f(t)||dt|


(b)

Looking at the system(*) of equations,

x1' = (a-bx2)* x1
x2' = (cx1 -d)*x2

open the open Quadrant K; here a,b,c and d er positive constants.

I need to show that the system can be integrated. Which means I need to show that there exist a C^1 -function, with the properties F:U \rightarrow \mathbb{R}, where U \subseteq K is open and close in K(close means that for every point in K, is a limit point for q sequence, whos elements belongs to K). Such that \nabla F \notequal 0 for alle x \in U and such that F is constant for trajectories of system.

Finally I need to conclude that every max trajectory is contained in a compact subset of K, and that the system (*) field X: R \rightarrow \mathbb{R}^2.

Sincerley Yours
Maria Thomson 20

p.s. I'm so sorry for asking so much, but has been such a difficult couple of days for me.
 
Hi

I'm sorry, but I have these answered within the next 4 hours, so if there somebody here who can help me I would be very greatful.

Sincerely Yours and God bless

MT
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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