8

If $A$ is $n \times n$ with negative real parts of all eigenvalues, then there exists positive $K,\alpha$ such that

$$\|e^{At}\| \leq Ke^{-\alpha t}$$

Furthermore, if an eigenvalue has negative part zero, but with single multiplicity, then there exists $M > 0$ such that

$$\|e^{At}\| \leq M$$

Starting out: We have $e^{At} = Pe^{Jt}P^{-1}$, so $\|e^{At}\| \leq \|P\|\|e^{Jt}\|P^{-1}\| = K_1\|e^{Jt}\|$ since $P$ is invertible etc. I'm not sure how you bound the norm of the exponential Jordan matrix here. The book (Sze-Bi Hsu's ODE book) sort of just states that it's bounded somehow.

Benjamin Lindqvist
  • 3,112

1 Answers1

14

Clearly, if $A$ is diagonalizable then $J$ is diagonal and $\|e^{Jt}\|\le e^{-at}$ with $-a=\max{\text{Re}\lambda_i}$ (the slowest exponential).

If not then it is block diagonal and it is enough to bound each block. Let $J_k=\lambda I+N$ be a Jordan block of size $k$, $\lambda$ be the eigenvalue and $N$ is a matrix of all zeros except identities on the first superdiagonal. Then $$ e^{J_kt}=e^{(\lambda I+N)t}=[\text{$I$ and $N$ commute}]=e^{\lambda t}e^{Nt}= e^{\lambda t}\sum_{m=0}^k\frac{N^m}{m!}t^m. $$ The series for $e^{Nt}$ ends after $k$-th power since $N^k=0$. It gives for any small $\epsilon>0$ $$ \|e^{J_kt}\|\le e^{(\text{Re}\lambda+\epsilon)\cdot t}\left\|\sum_{m=0}^k\frac{N^m}{m!}t^me^{-\epsilon t}\right\|\le Ce^{(\text{Re}\lambda+\epsilon)\cdot t}. $$ Now pick $\epsilon>0$ small enough to get $\max\text{Re}\lambda_i+\epsilon=-\alpha<0$.

A.Γ.
  • 29,518