
#1
Nov2312, 10:35 AM

P: 225

I have been thinking about numerical methods for ODEs, and the whole notion of stability confuses me.
Take Euler's method for solving an ODE: U_n+1 = U_n + h.A.U_n where U_n = U_n( t ), A is the Jacobian and h is step size. Rearrange: U_n+1 = ( 1 + hA ).U_n This method is only stable if (1 + hA) < 1 ( using the eigenvalues of A). But what does this mean!?? Every value of my function that I am numerically getting is less than the previous value. This seems rather useless, I don't get it? It appears to me that this method can only be used on functions that are strictly decreasing for all increasing t ??? 



#2
Nov2312, 09:48 PM

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Does this help http://courses.engr.illinois.edu/cs4...stability.pdf?




#3
Nov2412, 11:52 AM

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But it's a nice example of something that "obviosuly" look like a good idea, but turns out not to be. 



#4
Nov2912, 09:48 AM

P: 225

Stability of an ODE and Euler's method
Well, I'm still confused.
Say I have an ODE who's solution family y(t) is unstable. That is, for increasing t, the solution curves diverge from eachother. In this case, J = df(y, t)/dy < 0. So does this mean that ANY numerical method I use to solve this ODE will be unstable? With reference to http://courses.engr.illinois.edu/cs4...stability.pdf? there is a condition for all the methods, even the trapezoid rule etc. to be stable. And in each of these it implies that numerical values for each succesive value of y(t) are less than the previous, ie. y(t+h) < y(t). So, in essence, what I gather here is that unless an ODE has the property that the magnitude of each value of the function y is LESS than the previous value, then it CANNOT be solved with a numerical method accurately? Or, in another way, errors will always grow in solving an unstable ODE? All this seems rather strange to me then. We cannot solve an ODE accurately unless the function is monotonically decreasing? What rather tiny area of applicability then!! 


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