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When solving non-linear equations via Newton's method, load increments are often used to improve convergence. In mechanics for example, if the final load in 90N, one could choose 3 load steps of 30N each. At each load step several Newton iteration are used until convergence, and the final converged result is used as the guess for the next load step. What I do not understand is why at the intermediate load steps, convergence is needed and not just for the final load step? Why not take just one or two iterations at each load step (where the change in the solution is significant) and use that as the initial guess for the next one.

Thanks!

Uri

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To answer completely, one should know what is the application that you have in mind and if there is any specific convergence result for the Newton method.

But, in general, I think that you are right: solving an inexact model to full precision is not needed. If you are solving an inexact problem $P'$ whose solution $x'$ is at distance $d$ from the solution $x$ to the true problem $P$, then it is useless to determine $x'$ more precisely than within a radius $O(d)$. However, a good deal of error analysis is required to determine precisely how much imprecise you can be in solving the intermediate problems.

Of course this is only needed if you wish to prove things in a rigorous setting (and if your problem allows it): if, as it often happens in engineering, the approach is "run Newton, cross fingers and it will converge to the solution", then all these strategies are just heuristics, too, and nothing more precise can be proved.

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  • $\begingroup$ Thanks Fredrico. I am just pretty sure that I have seen somewhere that the convergence in the intermediate stages is important (and I think this is done in practice too), and I could not find a good explaition. Regarding the application, well it is related to structural mechanics, where loads often need to be applied incremetally to get convergence. $\endgroup$
    – Uri
    Commented Jul 27, 2012 at 20:33
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Could I ask one question, what's the meaning of load increment? I have used Wiki to search load increment, but there's no answer. Can you offer some derivations and explanations of this terminology?

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  • $\begingroup$ I'll try to give an example: Say you have a fishing rod, and you want to see how it bends as a function of the weight if the fish. So, if the fish weights 10 kg, you would start the analysis with a 1 kg weight, get a converged solution, and use that as an initial guess for a 2kg weight (i.e. 1 kg increments). You proceed with this procedure until you reach 10 kgs. $\endgroup$
    – Uri
    Commented Jul 27, 2012 at 11:40

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