May 8th, 2018, 04:17 PM  #1 
Member Joined: Nov 2016 From: Kansas Posts: 73 Thanks: 1  Absolute stability
f(t,u) = iu where i$^{2}$ = 1 find theta for which the function has a range of absolute stability 
May 8th, 2018, 06:44 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,511 Thanks: 585 
There is no theta in the expression. What do you mean by absolute stability? 
May 9th, 2018, 04:41 AM  #3 
Member Joined: Nov 2016 From: Kansas Posts: 73 Thanks: 1 
Absolute stability: the numerical method/function is absolutely stable if all its roots lie in a unit circle. u(t$_{n+1}$) = u(t$_{n}$) + h[(1theta) f(t$_{n}$,u(t$_{n}$) ) + (theta)f(t$_{n+1}$,u(t$_{n+1}$) )] 
May 9th, 2018, 01:42 PM  #4 
Global Moderator Joined: May 2007 Posts: 6,511 Thanks: 585 
It would be helpful if you could explain your notation. What is $t_n$? What is $u(t_n)$?

May 9th, 2018, 04:16 PM  #5  
Math Team Joined: May 2013 From: The Astral plane Posts: 1,797 Thanks: 715 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
If f(t, u) = i u then why include the t in the notation? You have $\displaystyle u( t_{n+1} ) = u( t_{n} ) + h [ ( 1  \theta) f( t_{n} , u( t_{n} ) ) + \theta f( t_{n+1},u( t_{n+1}) ) ] $ $\displaystyle u( t_{n+1} ) = u( t_{n} ) + h [ ( 1  \theta) ( i u( t_{n} ) + \theta ( i u( t_{n+1}) ) ] $ $\displaystyle u( t_{n+1} ) = u( t_{n} ) + h [ i u( t_{n} ) + i \theta \left \{ u( t_{n})  ~ u( t_{n+1}) \right \} ] $ That's as far as I can help until you define the h[ ] thing. Dan Last edited by topsquark; May 9th, 2018 at 04:32 PM.  
May 9th, 2018, 06:45 PM  #6 
Member Joined: Nov 2016 From: Kansas Posts: 73 Thanks: 1 
h is a constant. u(t$_{n}$) is technically u$_{n}$ General definition of this notation is under theta method (in numerical analysis) while calculating error 
May 10th, 2018, 12:41 PM  #7 
Global Moderator Joined: May 2007 Posts: 6,511 Thanks: 585  

Tags 
absolute, convergence, numerical analysis, numerical method, stability 
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