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What is the difference between linear stability and nonlinear stability of numerical schemes for the solution of time dependent PDEs?

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... of what? $\:$ – Ricky Demer Dec 5 '12 at 21:29
It is not clear what you want to know. Linear stability refers to stability of linear systems, $\dot u = A(t)u+b(t)$, in finite or infinite dimension. In the autonomous case, it reduces mostly to properties of the spectrum of $A$. Periodic linear systems may be partly reduced to the study of autonomous systemd. Nonlinear stability refers to stability of nonlinear systems, $\dot u=f(u)$. They are studied by means of Lyapunov functions, or by linarization in the sense of the Hartman-Grobmann theorem, &c. – Pietro Majer Dec 5 '12 at 21:38

Suppose you have a manifold $\mathcal{M}$ that represents "states of a system". $\mathcal{M}$, for example, could be a Euclidean space $\mathbb{R}^d$, or another example could be $L^p$ or even (sometimes only formally) $\mathcal{P}_2$. I give these possible examples because I don't know which one applies to your case.

Anyway, suppose you have a path $x(t)\in\mathcal{M}$ where $x: I \rightarrow \mathcal{M}$ where $I$ is some interval in $\mathbb{R}$, and say the path is $C^1$ for simplicity. This often occurs when you are looking at a "state" of a system that is evolving over time (hence the parameter $t$).

Now consider a steady-state $x_0 \in \mathcal{M}$ of an evolution. Suppose you have a path $x(t)$ that is generated by the evolution, and suppose T is such that $x(T)=x_0$. Since $x_0$ is a steady-state of the evolution, then by definition this means $\dot{x}(T)=0$.

Nonlinear Stability of a steady-state of an evolution is when, for a path generated by an evolution that starts at (say) $x(0)$: if $x(0)$ is "near" to $x_0$, then its evolution stays "near" to $x_0$ for all time. I use "near" here because it depends on the topology you're using, but if you are in a metric space then the statement can be written as: if the distance between $x(0)$ and $x_0$ is less than some $\delta>0$, then the distance between $x(t)$ and $x_0$ will be bounded by some $\varepsilon>0$ for all $t>0$.

Linear Stability is given by Pietro Majer in his comment. For example, suppose $\mathcal{M} \subset \mathbb{R}^d$ for simplicity, and consider an evolution is described by $\dot{x} = F(x)$. Since $x_0$ is a steady state we know that $F(x_0)=0$. This implies that first order expansion around $x_0$ gives gets $\dot{x} = DF(x_0)(x-x_0) + o(|x-x_0|^2)$. Then the steady-state $x_0$ is called Linearly Stable if $DF(x_0)$ is negative definite.

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The question has been sufficiently changed that my answer is no longer relevant (and I don't know the answer to the question as it now is). I'm new to mathoverflow, does this mean I should delete my answer now? – Robert L. Simione II Dec 7 '12 at 18:27

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