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Let $a= t^0 \lt t^1 \lt … \lt t^N =b$ a random division of $[a,b]$ space where $h_n = t^{n+1} - t^n$ for $n=0,1,…,N-1$.

We consider the initial value problem $$ y'(t)=f(t,y(t))\text{ for } a \le t \le b \ \&\ y(a)= y_0 $$

Function $f \in ( C[a,b]×R )$ and satisfies the Lipschitz condition with constant $L>0$ for $a \le t \le b$

In addition, $M=\max|y''(t)|$ & $h=\max h_n $ for $0 \le n \le N-1$

Show that $| ε_n | \le M {h\over2L} [ e^{L(b-a)} -1]$

We define $ε_n =y( t^n )- y^n$ the error of Euler's method with variable step $h_n$ and by applying this method to the problem we conclude to this:

\begin{align} | ε_{n+1} | &\le (1+ h_n L)| ε_n | + h_n {h\over2} M \\ &\le M {h\over2} \Bigl[ h_n +(1+ h_n L) h_{n-1} +(1+ h_n L)(1+ h_{n-1} L) h_{n-2} + \\&\qquad\qquad…+(1+ h_n L)(1+ h_{n-1} L)\cdots(1+ h_1 L) h_0 \Bigr] \\ &\le M {h\over2} \Bigl[ h_n + e^{L(t^{n+1}-t^n)} ( t^n - t^{n-1} )+ e^{L(t^{n+1}-t^{n-1})} ( t^{n-1} - t^{n-2} )+ \\&\qquad\qquad…+ e^{L(t^{n+1}-t^1)} ( t^1 - t^0 )\Bigr] \\ &\le M {h\over2} \Bigl[ h_n + \int_{t^{n-1}}^{t^n} e^{L(t^{n+1}-s)} ds + \int_{t^{n-2}}^{t^{n-1}} e^{L(t^{n+1}-s)} ds + \\&\qquad\qquad…+ \int_{t^0}^{t^1} e^{L(t^{n+1}-s)} ds\Bigr] \\ &= M {h\over2} \Bigl[ h_n + \int_{t^0}^{t^n} e^{L(t^{n+1}-s)} ds\Bigr] \end{align}

At this point I have been stuck because I don't know how to get rid of the ' $h_n$ ' I would appreciate any help ..How could I continue in order to show this? Thanks in advance!!

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  • $\begingroup$ Please change your formula salad into readable mathjax/Latex using only one math environment per formula, using align environment for the large formula block at the bottom,... $\endgroup$ – Dr. Lutz Lehmann Mar 25 '16 at 15:44
  • $\begingroup$ I have to admit that you've been very polite but it was my first attempt to write in mathjax and certainly I didn't nail it... Although I can't see something unreadable here... so except "my formula salad" if you had something else to add about the problem, you 'd help me a lot! thanks $\endgroup$ – kaithkolesidou Mar 25 '16 at 16:10
  • $\begingroup$ See math.stackexchange.com/a/1696107/115115 for the idea for the standard treatment of this problem. $\endgroup$ – Dr. Lutz Lehmann Mar 25 '16 at 16:26
  • $\begingroup$ @kaithkolesidou - you don't need to split every single item into separate $ $ pairs. It really looks better and is a LOT easier to read if you just do an entire fomula instead of bits and pieces. For example: $$a=t^0 \lt t^1\lt \ldots \lt t^N =b$$ gives $$a=t^0 \lt t^1\lt \ldots \lt t^N =b$$ instead of "a=$t^0$ $\lt$ $t^1$ $\lt$ ......$\lt$ $t^N$ =b" $\endgroup$ – Paul Sinclair Mar 25 '16 at 16:26
  • $\begingroup$ @LutzL the ex. gives certain steps we should follow in order to achieve the final inequality, and the point i've stopped it's the last given step.. if i calculate the integral then i conclude to the above inequality BUT the hn is stil there... $\endgroup$ – kaithkolesidou Mar 25 '16 at 16:38
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Just use $$ h_n=e^{L(t^{n+1}-t^{n+1})}·(t^{n+1}-t^n) \le \int_{t^n}^{t^{n+1}}e^{L(t^{n+1}-s)} $$ to get to $$ \int_{t^0}^{t^{n+1}}e^{L(t^{n+1}-s)}ds=\frac1L\left(e^{L(t^{n+1}-t^0)}-1\right). $$

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  • $\begingroup$ yeeeees you're right... thank you very very much for your help and for your time!! $\endgroup$ – kaithkolesidou Mar 25 '16 at 16:52

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