# Thread: Application of the Arzela-Ascoli Theorem

1. ## Application of the Arzela-Ascoli Theorem

Here is the problem as it is written in the text:
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Let $f(t,x)$ be defined and continuous for $a\leq t\leq b$ and $x\in\mathbb{R}^n$. The purpose of this exercise is to show that the problem $dx/dt=f(t,x)$, $x(a)=x_0$, has a solution on the interval $t\in[a,c]$ for some $c>a$. Perform the operations as follows: Divide $[a,b]$ into $n$ equal parts: $a=t_0,t_1,...,t_n=b$, and define a continuous function $x_n$ inductively by

$\bigg\{\begin{array}{l}x_n'(t)=f(t_i,x_n(t_i)),~~t _i

Put $\Delta_n(t)=x_n'(t)-f(t,x_n(t))$, so that

$x_n(t)=x_0+\int_a^t f(s,x_n(s))+\Delta_n(s)\,ds$

Use the Arzela-Ascoli Theorem to find a convergent subsequence of the $x_n$. Show that the limit satisfies $dx/dt=f(t,x)$ and $x(a)=x_0$.
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The version of the A-A Thm that seems the most applicable here is the following:

Let $A$ be a compact set. If $\mathcal{B}\subset\mathcal{C}(A,\mathbb{R}^n)$ is (uniformly) equicontinuous and pointwise bounded, then every sequence in $\mathcal{B}$ has a uniformly convergent subsequence.

Letting $x_n=\mathcal{B}$, I don't know how to show equicontinuity or pointwise boundedness. I know that each individual $x_n$ is bounded and uniformly continuous, but that doesn't mean that they're pointwise bounded or equicontinuous (because there are infinitely many $x_n$).

EDIT: I've figured out what to do after applying the A-A Theorem. However, I still don't know how to prove the criteria for using the A-A Theorem, so I still need help with that.

Also, I'm not sure where that constant $c$ came from. It pops up and then disappears almost immediately, never to be mentioned again. Maybe it's a typo.

2. Originally Posted by redsoxfan325
I'm not sure where that constant $c$ came from. It pops up and then disappears almost immediately, never to be mentioned again. Maybe it's a typo.
The constant c is an essential ingredient in this theorem. To see why it's important, look at this example: Let $f(t,x) = x^2$, defined for $0\leqslant t\leqslant 2$ and $x\in\mathbb R$. The initial value problem $dx/dt = f(t,x),\ x(0)=1$ has the solution $x(t) = 1/(1-t)$. But that is only defined on the interval [0,1), not on the interval [0,2] on which we defined f(x,t). Admittedly, the function f(t,x) does not explicitly involve t in that example, but it shows that the solution to the problem can blow up unexpectedly in the interval [a,b]. That is why it's going to be necessary to restrict to a shorter interval if you want the function x(t) to exist.

You can find a solution to the problem here (it's a theorem due to Peano). The proof is quite subtle. It starts by using the continuity of the function f(t,x) to get a neighbourhood of $(a,x_0)$ on which f is bounded.

3. Ugh, that's tough. The homework was due earlier today, and I submitted a proof that (based on what I understood of the correct proof in that textbook) should be good enough for half credit. (I had the general idea, but I apparently oversimplified a lot of the not-so-simple steps.) Thanks for explaining the $c$ though!