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Euler's Method - Global Error

  1. Nov 24, 2011 #1
    1. The problem statement, all variables and given/known data

    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]

    2. Relevant equations

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

    3. 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?
     
  2. jcsd
  3. Nov 24, 2011 #2
    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]
     
  4. Nov 25, 2011 #3
    Please, Math Gods, I beg thee...
     
  5. Nov 25, 2011 #4

    HallsofIvy

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    Yes, that's true.
     
  6. Nov 25, 2011 #5
    Thanks, it's T that I'm having trouble with... how would I find the upper bound for |y''(t)| ?
     
  7. Nov 26, 2011 #6
    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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