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Euler's Method - Global Error

  • Thread starter spitz
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  • #1
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Homework Statement



Use Euler's method with [tex]h = 1/2[/tex] to estimate [tex]y(1)[/tex] for the IVP:

[tex]y(0)=1[/tex] [tex]y'(t)=t^2-y(t)[/tex]

Assuming that [tex]|y(t)| \le 1[/tex] for [tex]0 \le t \le 1[/tex] determine the value of n needed to ensure that [tex]|E_n| \le 10^{-2}[/tex]

Homework Equations



[tex]|E_n| \le \frac{T}{L}(e^{L(t_n-t_0)-1})[/tex]

The Attempt at a Solution



The first part is easy enough:

[tex]y_1=y_0+f(t_0,y_0)h=1+f(0,1)(1/2)=1/2[/tex]
[tex]y_2=y_1+f(x_1,y_1)h=1/2-1/8=3/8[/tex]
[tex]\Rightarrow y(1)=3/8[/tex]

I'm having trouble with the second part. Could somebody help me out?
 

Answers and Replies

  • #2
60
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Is this correct for L:

[tex]|f(t,u)-f(t,v)|=|t^2-u-t^2+v|=|u-v|[/tex]

Lipschitz with [tex]L=1[/tex]
 
  • #3
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Please, Math Gods, I beg thee...
 
  • #4
HallsofIvy
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Is this correct for L:

[tex]|f(t,u)-f(t,v)|=|t^2-u-t^2+v|=|u-v|[/tex]

Lipschitz with [tex]L=1[/tex]
Yes, that's true.
 
  • #5
60
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Thanks, it's T that I'm having trouble with... how would I find the upper bound for |y''(t)| ?
 
  • #6
60
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Is this right?

[tex]y''(t)=2t-t^2+y(t)[/tex]

So, [tex]T=(1/2)*(1/2)*y''(1)=(1/4)(2(1)-1^2-3/8)=0.15625[/tex]

and

[tex]E_n\le|0.15625(e-1)|\Rightarrow E_n\le 0.26848[/tex]

Then set [tex]E_n=0.01[/tex]

[tex]\Rightarrow 0.01 = h(0.53696)[/tex]
[tex]h=0.018623361[/tex]
[tex]\frac{1-0}{n}=0.018623361[/tex]
[tex]\Rightarrow n ≈ 54[/tex]

Anyone? Anyone?
 

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