Given the difference equation and the continuously differentiable function $g$: $$x(n+1)=x(n)+h\times g(x(n))$$ Determine conditions on $h$ for which an equilibrium point is asymptotically stable, respectively unstable.
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I just don't get how can you determine h without knowing g? – Anton Nov 29 '14 at 20:44
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Presumably, the conditions on $h$ will involve $g$ in some way. Most likely the derivative of $g$ at the equilibrium point (which will be a zero of $g$). – Harald Hanche-Olsen Nov 29 '14 at 20:47
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I have gotten to this point by myself but I just don't see any relation between g(x)=0 and g'(x)<0 – Anton Nov 29 '14 at 20:49
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What makes you think you need such a relation? – Harald Hanche-Olsen Nov 29 '14 at 20:50
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sorry f'(x)<1 where x is the equilibrium point – Anton Nov 29 '14 at 20:58
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Hint: From the comments, I infer that you are familiar with the requirements for stability of fixed points for the difference equation $x(n+1)=f(x(n))$. In the case you are asking about, you have $$ f(x)=x+hg(x) $$ for some function $g$ and a constant $h$. So you just need to insert this function $f$ into the criteria you already know, maybe pretty up the result a little, and you're done!
Harald Hanche-Olsen
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What relevant stability criteria do you know for the case $x(n+1)=f(x(n))$? Oh never mind, you have stability for $|f'(x_0)|<1$ and instability for $|f'(x_0)|>1$, where $x_0$ is the equilibrium point, right? Now insert $f'(x_0)=1+hg'(x_0)$. – Harald Hanche-Olsen Nov 29 '14 at 21:14
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With absolute values, yes. See my previous comment, which I amended as you were writing yours. – Harald Hanche-Olsen Nov 29 '14 at 21:19
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Sure, what else can you do? You don't know anything more about $g$, remember. Your problem seems to be that you think the question is asking more than is reasonable. – Harald Hanche-Olsen Nov 29 '14 at 21:21
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Ok, thanks. I had done the above reasoning myself but I thought that I had to take care of g as well. Anyway thanks for the help. – Anton Nov 29 '14 at 21:28