Jtdylktuy

I wish to show that if all eigenvalues have real part negative or zero and if those eigenvalues with zero real part are simple, there exists a constant $K>0$ such that $|exp(tA)| leq K$, $(0<t<infty)$, and hence every solution of $y' = Ay$ is bounded on $(0 leq t < infty)$.

We can write our matrix $A$ as $A = PBP^{-1}$ and since all eigenvalues are simple, $B$ is diagonal so $A^n = PB^nP^{-1}$. If $B$ is nilpotent then our sum has finite number of terms and stops at the $n-1$ term.

EDIT:

In this answer, I will focus on showing that $Vert e^{At} Vert$ is bounded where $Vert cdot Vert$ is the standard operator norm,

$Vert A Vert = sup {Vert Ax Vert, ; Vert x Vert = 1 } = inf {C mid Vert Ax Vert le C Vert x Vert, ; forall x }, tag 0$

where $Vert x Vert$ is the standard Hermitian norm for vectors $x in Bbb C^n$, that is, $Vert x Vert$ is derived from the Hermitian inner product

$langle y, z rangle = displaystyle sum_1^n bar y_i z_i, y = (y_1, y_2, ldots, y_n)^T in Bbb C^n, ; text{etc,} tag{0.1}$

by

$Vert x Vert^2 = langle x, x rangle = displaystyle sum_1^n bar x_i x_i. tag{0.2}$

We resort to extending the normal real Euclidean inner product on $Bbb R^n$ to the Heritian product (0.1) to facilitate addressing and handling situations in which some of the eigenvalues of $A$ are non-real complex numbers.

We want to show that

$forall t in Bbb R, ; 0 le t < infty, Vert e^{At} Vert < K tag 1$

for some $0 < K in Bbb R$.

We know that $A$ may be cast into Jordan canonical form by a similarity transformation

$A to PAP^{-1} tag 2$

for some non-singular matrix $P$; we thus first deal with the case of $A$ a Jordan matrix; indeed, we will first address the situation when $A$ is a single Jordan block, that is, a matrix of the form

$A = lambda I_m + N, tag 3$

here $I_m$ is the $m times m$ identity matrix, $lambda$ is an eigenvalue of $A$ and $N$ is the $m times m$ nilpotent matrix consisting of $m - 1$ $1$s on the superdiagonal and zeroes everywhere else. Since $I_m$ commutes with $N_m$, that is,

$I_m N_m = N_m = N_m I_m, tag 4$

it follows that

$e^{At} = e^{(lambda I_m + N_m)t} = e^{lambda I_m t + N_m t} = e^{lambda I_m t} e^{N_m t}; tag 5$

now it is both well-known and easy to see from the matrix power series of $exp$ that

$e^{lambda I_m t} = displaystyle sum_0^infty dfrac{(lambda I_m t)^k}{k!} = sum_0^infty dfrac{lambda^k t^k I_m }{k!} = sum_0^infty dfrac{lambda^k t^k}{k!} I_m = e^{lambda t} I_m; tag 6$

also, since the nilpotent matrix $N$ satisfies

$N^m = 0, tag 7$

we have

$e^{N_m t} = displaystyle sum_0^{m - 1} dfrac{N_m^k t^k}{k!}, tag 8$

which is an $m times m$ matrix whose entries are polynomials in $t$ of degree at most $m - 1$; it follows from (5), (6) and (8) that $e^{At}$ takes the form

$e^{At} = e^{lambda t} I_m displaystyle sum_0^{m - 1} dfrac{N_m^k t^k}{k!} = e^{lambda t} sum_0^{m - 1} dfrac{N_m^k t^k}{k!}, tag 9$

and if

$lambda = sigma + i omega, ; sigma < 0, tag{10}$

we may further decompose $e^{At}$ as

$e^{At} = e^{iomega t} e^{sigma t} displaystyle sum_0^{m - 1} dfrac{N_m^k t^k}{k!}, tag{11}$

whence

$Vert e^{At} Vert = left Vert e^{iomega t} e^{sigma t} displaystyle sum_0^{m - 1} dfrac{N_m^k t^k}{k!} right Vert = vert e^{iomega t} vert left Vert e^{sigma t} displaystyle sum_0^{m - 1} dfrac{N_m^k t^k}{k!} right Vert = left Vert e^{sigma t} displaystyle sum_0^{m - 1} dfrac{N_m^k t^k}{k!} right Vert. tag{12}$

Now it is well-known that for $sigma < 0$ the expression on the right of (12), being dominated by the exponential $e^{sigma t}$, eventually decreases to $0$ as $t to infty$:

$displaystyle lim_{t to infty} Vert e^{At} Vert = lim_{t to infty} left Vert e^{sigma t} displaystyle sum_0^{m - 1} dfrac{N_m^k t^k}{k!} right Vert = 0; tag{13}$

since

$displaystyle Vert e^{At} Vert = left Vert e^{sigma t} displaystyle sum_0^{m - 1} dfrac{N_m^k t^k}{k!} right Vert. tag{14}$

is continuous as a function of $t$, it is bounded on any compact interval $[0, tau]$; by choosing $tau$ sufficiently large we may, in the light of the limit (13), assume $Vert e^{At} Vert < epsilon$, $0 < epsilon in Bbb R$, for $t ge tau$; therefore (14) is bounded by some $0 < K in Bbb R$ for all $t in [0, infty)$. We note that in the case $sigma = 0$, we have $m = 1$ by our hypotheses on the matrix $A$, and the Jordan block reduces to $e^{iomega t}$, which is manifestly bounded; indeed, as long as the block size is $1$, the block reduces to $e^{(sigma + i omega)t}$ with $sigma le 0$ and is thus easily seen to be bounded in norm.

The preceding discussion shows that $e^{At}$ is bounded for any Jordan block (3) of $A$ as long as $sigma = Re(lambda) le 0$; if $A$ is comprised of multiple Jordan blocks, then there is a bound for each and hence, since the number of Jordan blocks is finite, a bound over all the Jordan blocks of $A$ is obtained by taking the greatest such bound.

The preceding discussion covers the situation when $A$ is in Jordan form; it result may be extended to any $A$ by inverting the similarity transformation (2), an operation which preserves the boundedness of $Vert e^{At} Vert$ (though it may alter a particular bound) since $P$ does not depend upon $t$. We conclude that, in all specified/required cases, there is some $K > 0$ with

$Vert e^{At} Vert < K, ; t in [0, infty). tag{15}$

And we are done.