# what do zero real parts of eigenvalues mean? Any good references?

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?

Eigenvalues with zero eigenvalue correspond to purely oscillatory modes. You can see it by diagonalising the system. Your matrix $A$ can be written as $$A = P \Sigma P^{-1}$$ where $\Sigma$ is a diagonal matrix with entries $\lambda_n$ corresponding to the eigenvalues of $Q$. You can now transform your ODE to $$dQ/dt = P \Sigma P^{-1} Q \Rightarrow d (P^{-1}Q)/dt = \Sigma \left( P^{-1} Q \right)$$ since $A$ and thus $P$ do not depend on $t$. In the transformed coordinate $R:=P^{-1}Q$. the system decouples into $$dR/dt = \Sigma R$$ and thus, as $\Sigma$ is diagonal, into a number of independent ODEs $$d r_n / dt = \lambda_n r_n$$ with $r_n$ the n-th entry of $R$. This scalar ODE has the solution $$r_n(t) = r_n(0) \exp(\lambda_n t)$$ By Euler's formula, for $\lambda_n = i a_n$, i.e. zero real part, this becomes $$r_n(t) = r_n(0) \left( \cos(a_n t) + i\sin(a_n t) \right).$$

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

• Thanks Roland, I think I've found good book that explains what I was after: link but please let me know if you found anything else. – melody Sep 22 '15 at 9:12
• 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. – Roland Heath Sep 24 '15 at 5:52