Considering the following equation $$u_t + A u_x = 0,\quad t> 0$$ where $$ A = \begin{pmatrix} 1 & \sin t \\ \sin t & -1 \end{pmatrix} $$ Naturally, the system is hyperbolic if $A$ consist of 2 distinct eigenvalues and $A$ has 2 linearly independent (left) eigenvectors. This much is clear, but I run into problems when finding these eigenvectors, as one of them vanish completely or blows up completely.

For your convenience, here is the eigenvalues: $$ \lambda = \pm\sqrt{1+\sin^2(t) }$$

So my idea is to linearize $\sqrt{1+\sin^2(t)}$, but I am unsure on how to do this.

Any thoughts?

  • 1
    $\begingroup$ When you say 'linearize' you mean $t \sim 0$, right? What's wrong with $\sqrt{1 + \sin^2 t} \sim 1 + \frac{1}{2} \sin^2(t) \sim 1 + \frac{t^2}{2}$? $\endgroup$ May 22, 2013 at 10:20
  • $\begingroup$ Sorry, I see I forgot some details. It is correct now. I also see why linearizing is a bad idea in the first place. Thanks for your response though. $\endgroup$
    – Tom
    May 22, 2013 at 11:26

1 Answer 1


Clearly, $A(t) \in \mathbb{R}^{2\times 2}$ is symmetric which implies that it is always diagonalizable [1], [2] with real eigenvalues.

However, since you cannot guarantee for all times $t$ that your eigenvalues are distinct (consider $t = k \pi, k \in \mathbb{N}$) your system is not strictly hyperbolic.


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