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Find the Local Truncation error in the form of $ \ o(h^k) \ $ of the multi-step method $ 2u_{i+3} = −3u_{i+2} + 6u_{i+1} − u_i + 6hf(t_{i+2}, u_{i+2}) $.

Also determine it is convergent or not.

Answer:

The scheme is $ \ 2u_{i+3} = −3u_{i+2} + 6u_{i+1} − u_i + 6hf(t_{i+2}, u_{i+2}) $.

We know that $ f(t_i,u_i )=\frac{u_{i+1}-u_i}{h} ,\ by \ Taylor \ series \ $

Then , $ \ 2u_{i+3} = −3u_{i+2} + 6u_{i+1} − u_i +6 u_{i+3}-6u_{i+2} $,

or,$ \ 4u_{i+3}-9u_{i+2}+6u_{i+1}-u_i=0 $

or, $ 4(u_{i+3}-u_{i+2})-5(u_{i+2}-u_{i+1})+(u_{i+1}-u_i)=0 $

Dividing by $ h \ $ ,

$ 4 .\frac{u_{i+3}-u_{i+2}}{h}-5. \frac{u_{i+2}-u_{i+1}}{h}+\frac{u_{i+1}-u_i}{h}=0 $ ,

or, $ 4 u'''+o(h^4)-5 u'' +o(h^2)+u'+o(h)=0, $

So the Local Truncation error is $ \ (h^4) \ $.

Am I right ? Is there any help ?

MAS
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  • First of all, what is it that you are trying to approximate? An ODE? If so, which ODE? There isn't much we can say about the 'error' if we don't know what we are approximating. – ekkilop Jul 30 '17 at 17:59

1 Answers1

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I'll assume this is approximating $y'(t) = f(t,y)$. The method is $$ 2y_{n+3} =-3 y_{n+2} + 6 y_{n+1} - y_n + 6 h f (t_{n+2}, y_{n+2}) $$ To find the (local) error, taylor expand each term: \begin{align*} 2 y_{n+3} &= 2\left( y_n + 3h y_n ' + \frac{(3h)^2}{2} y_n '' + \frac{(3h)^3}{6} y_n ''' + \frac{(3h)^4}{24} +O(h^5)\right) \\ -3 y_{n+2} &= -3 \left( y_n + 2h y_n ' + \frac{(2h)^2}{2} y_n '' + \frac{(2h)^3}{6} y_n ''' + \frac{(2h)^4}{24} + O(h^5) \right) \\ 6 y_{n+1} &= 6 \left( y_n + h y_n ' + \frac{h^2}{2} y_n '' + \frac{h^3}{6} y_n ''' + \frac{h^4}{24} + O(h^5) \right) \\ -y_n &= -y_n \\ 6 h f(t_{n+2}, y_{n+2}) &= 6h \left( y_n' + 2h y_n '' + \frac{(2h)^2}{2} y_n ''' + \frac{(2h)^3}{6} y_n '''' + O(h^4) \right) \end{align*}

and matching the coefficients of the scheme reveals truncation error $O(h^4)$.

To check if it converges, we look to the root condition (i.e. apply to the test equation $y'=f(t,y)=0$, and see if the scheme is zero-stable. $$ 2 y_{n+3} = -3 y_{n+2} + 6 y_{n+1} - y_n \implies 2 r^3 = -3 r^2 + 6r -1 $$

One of the roots is larger than 1 in absolute value, so by Dahlquist Equivalence Theorem, the scheme does not converge (since it doesn't satisfy the root condition)

user288742
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