On induction as you kindly requested (detailed)

[sarrah1]

Hi Evgeny

Thank you very much for your reply.

I will try to be more clear this time and in detail. The reason why it was not as clear, as it's lengthy. I have the following

$|{y}_{n}(x)|\le\int_{a}^{b} \,|k(x,s)| |{y}_{n-1}(s)| ds+\int_{a}^{x} \,|g(x,s)| |{y}_{n-1}(s)| ds$ , $x\in[a,b]$ $a,b$ real ... (1)

By defining operator $K$

$Ky=\int_{a}^{b} \,k(x,s) y(s) ds$

I define norms

$||K||={max}_{a<x<b}\int_{a}^{b} \,|k(x,s)|ds $ , $M={max}_{a<x<b}|g(x,s)|$ , $||y||={max}_{a<x<b}|y(x|$

From (1) $|{y}_{1}|\le||{y}_{0} ||K||+||{y}_{0}||M(x-a)$

Another iteration gives

$|{y}_{2}|\le||{y}_{0}|| {||K||}^{2}+||{y}_{0}|| ||K||M(x-a)+||{y}_{0}|| M||K|| (b-a)+||{y}_{0}||{M}^{2}\frac{{(x-a)}^{2}}{2!}$

In general for any $n$

$|{y}_{n}(x)|\le\left\{||K||\left(\sum_{j=0}^{n-1}{(||K||+M(b-a))}^{n-j-1}{M}^{j}\frac{{(x-a)}^{j}}{j!}\right)+{M}^{n}\frac{{(x-a)}^{n}}{n!}\right\}||{y}_{0}||$ ... (2)

MY PROBLEM STARTS HERE:

Instead of writing the above inequality for $n=3,4,$ etc.. I want to apply mathematical induction to prove (2). Assuming (2) is valid for $n$ I obtain using one further iteration

$|{y}_{n+1}(x)|\le\left\{||K||\left(\sum_{j=0}^{n-1}{(||K||+M(b-a))}^{n-j-1}{M}^{j}\left(||K||\frac{{(b-a)}^{j}}{j!}+M\frac{{(x-a)}^{j+1}}{j+1!}\right)\right)+{M}^{n}\left(||K||\frac{{(b-a)}^{n}}{n!}+M\frac{{(x-a)}^{n+1}}{n+1!}\right)\right\}||{y}_{0}||$ ... (3)

But since $(x-a)\le(b-a)$

$|{y}_{n+1}(x)|\le\left\{||K||\left(\sum_{j=0}^{n-1}{(||K||+M(b-a))}^{n-j}{M}^{j}\frac{{(b-a)}^{j}}{j!}\right)+{M}^{n}\left(||K||\frac{{(b-a)}^{n}}{n!}+M\frac{{(x-a)}^{n+1}}{n+1!}\right)\right\}||{y}_{0}||$ ... (4)

$\le\left\{||K||\left(\sum_{j=0}^{n}{(||K||+M(b-a))}^{n-j}{M}^{j}\frac{{{(b-a)}^{j}}}{j!}\right)+{M}^{n+1}\frac{{(x-a)}^{n+1}}{n+1!}\right\}||{y}_{0}||$ ... (5)

So whereas I need the form (2) in $x$ to carry the integration of ${(x-a)}^{j}$ , I ended up with (5) in which $(b-a)$ replaced $(x-a)$ in the first inside bracket. usually I must end in induction to exactly the same form except $n -> n+1$. Yet form (5) is really what I need except that I cannot start with $b-a$ instead because when integrated doesn't give $\frac{{(x-a)}^{n}}{n!}$ which what I need

I am really very grateful to sort out where is the problem
Sarrah

 

[Evgeny.Makarov]

You did not write how $y_{n+1}$ is obtained from $y_n$.

$M=\max_{a<x<b}|g(x,s)|$

This expression depends on $s$.

From (1) $|{y}_{1}|\le||{y}_{0} ||K||+||{y}_{0}||M(x-a)$

Is it correct that $|{y}_{n+1}|\le\|{y}_{n} \|K+\|{y}_{n}\|M(x-a)$ for all $n$? Do you mean $\|y_n\|$ instead of $|y_n|$? Also, how come $x$ is not used in the left-hand side of the inequality, but it is used in the right-hand side? Do you mean $|{y}_{1}(x)|\le||{y}_{0} \|K+\|{y}_{0}\|M(x-a)$ for all $x\in[a,b]$?

 

[sarrah1]

You did not write how $y_{n+1}$ is obtained from $y_n$.

This expression depends on $s$.

Is it correct that $|{y}_{n+1}|\le\|{y}_{n} \|K+\|{y}_{n}\|M(x-a)$ for all $n$? Do you mean $\|y_n\|$ instead of $|y_n|$? Also, how come $x$ is not used in the left-hand side of the inequality, but it is used in the right-hand side? Do you mean $|{y}_{1}(x)|\le||{y}_{0} \|K+\|{y}_{0}\|M(x-a)$ for all $x\in[a,b]$?

Thank you very much Evgeny for your prompt reply. I did some typing mistakes


$M={max}_{a\le s,x\le b}|g(x,s)|$

I have the recurrence

${y}_{n+1}(x)=\int_{a}^{b} \,k(x,s) {y}_{n}(s)ds +\int_{a}^{x} \,g(x,s) {y}_{n}(s) ds$ (i)

True $|{y}_{n+1}(x)|\le\|{y}_{n} \|K+\|{y}_{n}\|M(x-a)$ , but I WILL NOT use it since I need to integrate again the term $(x-a)$.

It is true that
$|{y}_{1}(x)|\le||{y}_{0} \|K+\|{y}_{0}\|M(x-a)$ , for all $x\in[a,b]$ (ii)
because
$|{y}_{1}(x)|\le\int_{a}^{b} \,|k(x,s)| |{y}_{0}(x)|ds +\int_{a}^{x} \,|g(x,s)| |{y}_{0}(x)|ds$
$\le{max}_{a \le s\le b}|{y}_{0}(s)|\int_{a}^{b} \,|k(x,s)|ds +{max}_{a \le s\le x}|{y}_{0}(s)|\int_{a}^{x} \,|g(x,s)|ds$
i.e. $|{y}_{1}(x)|\le||{y}_{0}||. ||K||+||{y}_{0}|| M(x-a)$

Upon substituting for $|{y}_{1}(s)|$ inside (i) I obtain $|{y}_{2}(x)|$ etc..
many thanks
Sarrah

 

[sarrah1]

Thank you very much Evgeny for your prompt reply. I did some typing mistakes


$M={max}_{a\le s,x\le b}|g(x,s)|$

I have the recurrence

${y}_{n+1}(x)=\int_{a}^{b} \,k(x,s) {y}_{n}(s)ds +\int_{a}^{x} \,g(x,s) {y}_{n}(s) ds$ (i)

True $|{y}_{n+1}(x)|\le\|{y}_{n} \|K+\|{y}_{n}\|M(x-a)$ , but I WILL NOT use it since I need to integrate again the term $(x-a)$.

It is true that
$|{y}_{1}(x)|\le||{y}_{0} \|K+\|{y}_{0}\|M(x-a)$ , for all $x\in[a,b]$ (ii)
because
$|{y}_{1}(x)|\le\int_{a}^{b} \,|k(x,s)| |{y}_{0}(x)|ds +\int_{a}^{x} \,|g(x,s)| |{y}_{0}(x)|ds$
$\le{max}_{a \le s\le b}|{y}_{0}(s)|\int_{a}^{b} \,|k(x,s)|ds +{max}_{a \le s\le x}|{y}_{0}(s)|\int_{a}^{x} \,|g(x,s)|ds$
i.e. $|{y}_{1}(x)|\le||{y}_{0}||. ||K||+||{y}_{0}|| M(x-a)$

Upon substituting for $|{y}_{1}(s)|$ inside (i) I obtain $|{y}_{2}(x)|$ etc..
many thanks
Sarrah

sorry I discovered a mistake, it's ${y}_{0}(s)$ and not ${y}_{0}(x)$ inside integral

$|{y}_{1}(x)|\le\int_{a}^{b} \,|k(x,s)| |{y}_{0}(s)|ds +\int_{a}^{x} \,|g(x,s)| |{y}_{0}(s)|ds$

thanks
sarrah