I am solving a 1D advection problem of the the form $$dQ/dt=[A]Q$$ where {Q} is the vector of unknowns and [A] is the matrix of coefficients of spatial discretisation. I have worked out the eigenvalues of [A] to get an idea about the stability of the semi-discretised system and I am getting eigenvalues with zero real parts plus imaginary parts. Are there any good references for this?
2 Answers
Eigenvalues with zero eigenvalue correspond to purely oscillatory modes. You can see it by diagonalising the system. Your matrix $A$ can be written as \begin{equation} A = P \Sigma P^{-1} \end{equation} where $\Sigma$ is a diagonal matrix with entries $\lambda_n$ corresponding to the eigenvalues of $Q$. You can now transform your ODE to \begin{equation} dQ/dt = P \Sigma P^{-1} Q \Rightarrow d (P^{-1}Q)/dt = \Sigma \left( P^{-1} Q \right) \end{equation} since $A$ and thus $P$ do not depend on $t$. In the transformed coordinate $R:=P^{-1}Q$. the system decouples into \begin{equation} dR/dt = \Sigma R \end{equation} and thus, as $\Sigma$ is diagonal, into a number of independent ODEs \begin{equation} d r_n / dt = \lambda_n r_n \end{equation} with $r_n$ the n-th entry of $R$. This scalar ODE has the solution \begin{equation} r_n(t) = r_n(0) \exp(\lambda_n t) \end{equation} By Euler's formula, for $\lambda_n = i a_n$, i.e. zero real part, this becomes \begin{equation} r_n(t) = r_n(0) \left( \cos(a_n t) + i\sin(a_n t) \right). \end{equation}
In the 2-D case, this corresponds to an elliptic fixed point (an orbital, I believe).
You might look into Lyapunov stability, hopefully someone will be able to recommend a good resource on that.
I'll try to have another look later to see if I can find resources on this.
UPDATE
For related material, you might look into limit cycles, or the Poincaré-Bendixson theorem. I've managed to find a reference for the latter. You might also try searching on math.stackexchange for dynamical systems references.
Differential Equations, Dynamical Systems, and an Introduction to Chaos - Hirsch, Smale, Devaney
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1$\begingroup$ Daniel Ruprecht has provided a more rigorous, mathematical answer, but I'll leave this here for reference - sometimes it's helpful just knowing what resources to look for. $\endgroup$ Sep 24, 2015 at 5:52