Solving Inequality from the proof of First Order ODE.

Question

Let $x'=f(x)$ $x(0)=x_0$ assuming that $|f(x)-f(x)|\le L|x-y|$. We have $x(t)=x_0+\displaystyle\int_0^tf(k)dk$.

Let's say $$x_{n+1}=x_0+\int_0^tf(x_n(k))dk$$ and $$|x_{n+1}-x_n(t)|\le L\int_0^t|x_{n}(k)-x_{n-1}(k)|dk.$$ From that we know that $\max_{t\in[0,\frac{1}{2L}]}|x_{n+1}-x_n(t)|\le0.5\max_{t\in[0,\frac{1}{2L}]}|x_{n}(t)-x_{n-1}(t)|$.

show that max $\displaystyle\max_{t∈[0,\frac{1}{2L}]} |x_{n+1}(t) − x_n(t)|≤ \frac{1}{2^{n−1}} \displaystyle\max_{t∈[0,\frac{1}{2L}]} |x_1(t) − x_0(t)|$.

By writing $x_n(t) = [x_n(t) − x_{n−1}(t)] + [x_{n−1}(t) − x_{n−2}(t)] +···+[x_1(t) − x_0(t)] + x_0(t)$

deduce that

$\displaystyle\max_{t∈[0,\frac{1}{2L}]} |x_n(t) − x_m(t)|≤ \frac{1}{ 2^{N−2}} \displaystyle\max_{t∈[0,\frac{1}{2L}]} |x_1(t) − x_0(t)|$

for all $n,m ≥ N$.

It follows that $x_n(t)$ converges to some function $x_∞(t)$ as $n →∞$,and therefore taking limits in both side of $$x=x_0+\int_0^tf(x(k))dk$$ implies that $$x_{\infty}=x_0+\int_0^tf(x_\infty(k))dk$$. Thus $x_\infty(t)$ is the solution of differential equation.

Here is my problem:

I used induction to and got $\displaystyle\max_{t∈[0,\frac{1}{2L}]} |x_{n+1}(t) − x_n(t)|≤ \frac{1}{2^{n}} \displaystyle\max_{t∈[0,\frac{1}{2L}]} |x_1(t) − x_0(t)|$

instead of $\displaystyle\max_{t∈[0,\frac{1}{2L}]} |x_{n+1}(t) − x_n(t)|≤ \frac{1}{2^{n−1}} \displaystyle\max_{t∈[0,\frac{1}{2L}]} |x_1(t) − x_0(t)|$.

I don't know if I am wrong but I did check it very strictly. After deducing this inequality in wrong or right way I came to this one which is something I don't understand how to derive it from

$\displaystyle\max_{t∈[0,\frac{1}{2L}]} |x_n(t) − x_m(t)|≤ \frac{1}{ 2^{N−2}} \displaystyle\max_{t∈[0,\frac{1}{2L}]} |x_1(t) − x_0(t)|$

And I skipped it but I want this to be solved too so how do I solve it?

Also it says as $n$ approaches $\infty$ you will get to the solution my tought is the inequality's $N$ will become larger and larger so $x_n(t)$ will reach $x_m(t)$ may be it means this by it?

Can anyone explain how to solve this? If I have done any typos on the question pleaselet me know. I derived this from my memory. Thanks for taking time to read and Good luck!

Answer 1

First question : what do you want to show

In order to obtain the wanted inequality :

$$|x_{n+1}-x_n|=|\int_0^t f(x_n(t))-f(x_{n+1}(t))dt| $$

$$|x_{n+1}-x_n|\leq\int_0^t |f(x_n(t))-f(x_{n+1}(t))|\leq \int_0^tL|x_n-x_{n+1}| dt $$

Noting $$ s_n \triangleq = \displaystyle\max_{t∈[0,\frac{1}{2L}]} |x_{n+1}(t) − x_n(t)| $$

Then particularly restrictive $\forall t \in [0,\dfrac{1}{2L}] $

$$s_n \leq L\int_0^ts_{n-1}dt$$

Taking $t=\dfrac{1}{2L}$

$$s_n \leq L\int_0^{\frac{1}{2L}}s_{n-1}dt=\dfrac{s_{n-1}}{2}$$

By iteration

$$ \displaystyle\max_{t∈[0,\frac{1}{2L}]} |x_{n+1}(t) − x_n(t)|=s_n \leq \dfrac{s_0}{2^{n-1}}=\dfrac{1}{2^{n-1}}\displaystyle\max_{t∈[0,\frac{1}{2L}]} |x_1(t) − x_0(t)| $$

When you don't know how to cope with things write them :

$$\displaystyle\max_{t∈[0,\frac{1}{2L}]} |x_{n+1}(t) − x_n(t)|≤ \frac{1}{2^{n−1}} \displaystyle\max_{t∈[0,\frac{1}{2L}]} |x_1(t) − x_0(t)|$$.

Then let $N>0$, let $n,m>N$

$$ |x_{n+2}-x_n|\leq|x_{n+1}-x_n|+|x_{n+2}-x_{n+1}|$$ So

$$ |x_{n+2}-x_n|\leq\frac{1}{2^{n−1}} \displaystyle\max_{t∈[0,\frac{1}{2L}]} |x_1(t) − x_0(t)|+\frac{1}{2^n} \displaystyle\max_{t∈[0,\frac{1}{2L}]} |x_1(t) − x_0(t)|$$

So

$$ |x_{n+m}-x_n|\leq\sum_{i=0}^m\dfrac{1}{2^{n-1+k}}\displaystyle\max_{t∈[0,\frac{1}{2L}]} |x_1(t) − x_0(t)|=\dfrac{2^{m+1}-1}{2^{m}2^{n-1}}\displaystyle\max_{t∈[0,\frac{1}{2L}]} |x_1(t) − x_0(t)| $$

Consequently :

$$ \displaystyle\max_{t∈[0,\frac{1}{2L}]} |x_{n+m}(t) − x_n(t)|\leq(\frac{1}{2^{n-2}}-\dfrac{1}{2^{n+m-1}}) \displaystyle\max_{t∈[0,\frac{1}{2L}]} |x_1(t) − x_0(t)|\leq \dfrac{1}{2^{n-2}} \displaystyle\max_{t∈[0,\frac{1}{2L}]} |x_1(t) − x_0(t)| $$

So it can also be written because $n>N$:

$$\displaystyle\max_{t∈[0,\frac{1}{2L}]} |x_{n+m}(t) − x_n(t)|\leq\dfrac{1}{2^{N-2}} \displaystyle\max_{t∈[0,\frac{1}{2L}]} |x_{1}(t) − x_0(t)|$$

In others words : your sequences is Cauchy for the norm

$$|| \ ||:\to\displaystyle\max_{t∈[0,\frac{1}{2L}]} |f(t)| $$

and $$ (C^0[0,\dfrac{1}{2L}],|| \ ||) $$ is a Banach Space.

So your sequence is convergent because asbolutely convergent and it results that $$ x_{\infty} \triangleq \lim_{n \to \infty} x_n $$ is a solution to your equation on $[0,\dfrac{1}{2L}]$

Hence your results.

I hope it clarify your work don't hesitate for further questions.

Answer 2

Your first point is correct, if you insert $n=0$ or $n=1$ you want to recover the basic identity or inequality that was just shown. For the proof it makes no difference, it just makes the upper bound twice as large in a majorant sequence that converges to zero.

Then for the Cauchy property you get the same correction, if $n>m\ge N$, then $$ |x_n(t)-x_m(t)|\le\sum_{k=m}^{n-1}|x_{k+1}(t)-x_k(t)| \le \sum_{k=m}^{n-1}\frac1{2^k}\|x_1-x_0\|\le\sum_{k=N}^\infty\frac1{2^k}\|x_1-x_0\|=\frac1{2^{N-1}}\|x_1-x_0\|. $$ Such discrepancies can occur when in the course of writing the text the preference for the increment indices is shifted from $\|x_n-x_{n-1}\|$ to $\|x_{n+1}-x_n\|$ or back.