# Bound of solutions of autonomous linear ODEs

Given the linear system $\dot{z} = Az$.

(a) Assume all the eigenvalues of $A$ have negative real part. Give a counterexample to this statement: every solution of $\dot{z} = Az$ satisfies $|z(t)|\leq |z(s)| \ \forall\ t> s$

(b) Assume A is symmetric and all the eigenvalues of $A$ have negative real part. Prove that every solution satisfies $|z(t)|\leq |z(s)|\ \forall\ t> s$.

My thought: I was trying many different types of matrices (the ones which has $2$ real eigenvalues with $2$ negative real parts or $2$ complex eigenvalues with $2$ negative real part), but they all satisfy the inequality. Can someone please help me with an example?

For part (b), I think of using $A$ as symmetric must have all real eigenvalues. Thus, by applying Lemma A, which is following:

Consider $\dot{x} = Ax + f(t,x) + f_0(t)$ with $A,\ f$ and $f_0$ are continuous and $f(t,0) = 0 \ \forall\ t\in R$. Assume there exists constants $K\geq 1$, $M,L\geq 0$ and $\theta > \lambda + KL$ ($\lambda < 0$ is the negative real part). Then if we have:

(a) $||e^{At}||\leq Ke^{\lambda t}$ for $t\geq 0$

(b) $||f(t,x) - f(t,y)||\leq L||x-y||$ for all $L,x,y\in R$.

(c) $||f_0(t)||\leq Me^{\theta t}$ for $t\geq \tau$ ($\tau$ is some fixed constant).

Then if $u(t)$ is a solution, then for all $t\geq \tau$, we have: $||u(t)||e^{-\theta t}\leq K||u(\tau)||e^{-\theta \tau} + \frac{KM}{\theta - \lambda - KL}$.

Applying the result above into part (b) with sufficiently large $K\geq 1,\ M=L=\theta = 0$ (since $\lambda < 0$), we have: $||u(t)||\leq K||u(\tau)||$. But how do we "cancel" the constant $K$ here?

The only way to get an initial increase is to have eigenvalues with multiplicity greater than 1. So try $$A=\begin{bmatrix}-1 & N\\ 0 & -1\end{bmatrix}$$ and make $N$ large enough (positive or negative) to produce that initial increase.

One can probably also construct examples where the eigenvectors are not "orthogonal" for the vector norm used. Then in the initial point there could be some cancellation that is undone by the different decay velocities, resulting in an initial growth.

Both those principles are not true for symmetric matrices using the euclidean norm. (Why else use symmetric matrices.) Then one can use the Pythagorean theorem to separate the dimensions.

• Thank you so much for your great example in part (a)! My testing matrices were so close:(( But how do you prove part (b) without using the principles I mentioned (why you said it's wrong though?? The only thing I can see is that it might not be strong enough to solve part (b), but it's not wrong) – user177196 Jan 16 '16 at 6:18
• I'm quite sure that you need the Euclidean norm for (b), since only for that the symmetric matrices are self-adjoint. Compute the solution in an eigenvector basis and draw conclusions from that. – LutzL Jan 16 '16 at 12:51
• The principle I mentioned above works for any norm since we are in finite-dimensional. I will try your suggested approach, and will let you know for help if I still get stucked. Thank you for your patience:) – user177196 Jan 16 '16 at 19:02
• Yes, but then you can not guarantee that $K=1$. Any norm gives you just some constant $K\ge 1$. – LutzL Jan 16 '16 at 19:12
• That can not happen. Normal matrices always only have simple eigenvalues. There is always an orthonormal basis of eigenvectors. – LutzL Jan 18 '16 at 9:02