Stability of a linear equation
$begingroup$
If $A$ is a matrix, then $e^{At} leq C e^{-lambda t}$ if and only if the spectrum of $A$ consists of eigenvalues with negative real parts.
Is there a similar result, relating stability to the spectrum, when $A$ is a non-selfadjoint operator, when $e^{-At}$ is interpreted as the solution to
$$frac{partial u}{partial t} = - A u,$$
with appropriate boundary conditions? I assume that there is, but I haven't been able to find a reference.
operator-theory stability-theory
asked Jan 12 at 18:24
Roberto RastapopoulosRoberto Rastapopoulos
896424
$endgroup$
add a comment |
$begingroup$
If $A$ is a matrix, then $e^{At} leq C e^{-lambda t}$ if and only if the spectrum of $A$ consists of eigenvalues with negative real parts.
Is there a similar result, relating stability to the spectrum, when $A$ is a non-selfadjoint operator, when $e^{-At}$ is interpreted as the solution to
$$frac{partial u}{partial t} = - A u,$$
with appropriate boundary conditions? I assume that there is, but I haven't been able to find a reference.
operator-theory stability-theory
asked Jan 12 at 18:24
Roberto RastapopoulosRoberto Rastapopoulos
896424
$endgroup$
add a comment |
$begingroup$
If $A$ is a matrix, then $e^{At} leq C e^{-lambda t}$ if and only if the spectrum of $A$ consists of eigenvalues with negative real parts.
Is there a similar result, relating stability to the spectrum, when $A$ is a non-selfadjoint operator, when $e^{-At}$ is interpreted as the solution to
$$frac{partial u}{partial t} = - A u,$$
with appropriate boundary conditions? I assume that there is, but I haven't been able to find a reference.
operator-theory stability-theory
asked Jan 12 at 18:24
Roberto RastapopoulosRoberto Rastapopoulos
896424
$endgroup$
If $A$ is a matrix, then $e^{At} leq C e^{-lambda t}$ if and only if the spectrum of $A$ consists of eigenvalues with negative real parts.
Is there a similar result, relating stability to the spectrum, when $A$ is a non-selfadjoint operator, when $e^{-At}$ is interpreted as the solution to
$$frac{partial u}{partial t} = - A u,$$
with appropriate boundary conditions? I assume that there is, but I haven't been able to find a reference.
operator-theory stability-theory
operator-theory stability-theory
asked Jan 12 at 18:24
Roberto RastapopoulosRoberto Rastapopoulos
896424
asked Jan 12 at 18:24
Roberto RastapopoulosRoberto Rastapopoulos
896424
asked Jan 12 at 18:24
Roberto RastapopoulosRoberto Rastapopoulos
896424
asked Jan 12 at 18:24
Roberto RastapopoulosRoberto Rastapopoulos
896424
asked Jan 12 at 18:24
Roberto RastapopoulosRoberto Rastapopoulos
896424
896424
add a comment |
add a comment |
1 Answer
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You need some assumptions on $A$ to have a unique solution of the evolution equation (and one should also talk about the type of solution). Anyway, I think the answer to your question lies in the Hille-Yosida theorem:
If $A$ is the generator of the strongly continuous semigroup $T$, then $|T(t)|leq M e^{omega t}$ if and only if the spectrum of $A$ is contained in the half-plane ${zinmathbb{C}mid operatorname{Re}zleq omega}$.
In this case $u(t)=T(t)x$ is the unique mild solution of the initial value problem
begin{align*}
dot u(t)&=Au(t),\
u(0)&=x.
end{align*}
The Hille-Yosida theorem also gives a characterization of the operator that generate strongly continuous semigroups.
answered Jan 15 at 13:01
MaoWaoMaoWao
2,853617
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1 Answer
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1 Answer
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oldest
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votes
$begingroup$
You need some assumptions on $A$ to have a unique solution of the evolution equation (and one should also talk about the type of solution). Anyway, I think the answer to your question lies in the Hille-Yosida theorem:
If $A$ is the generator of the strongly continuous semigroup $T$, then $|T(t)|leq M e^{omega t}$ if and only if the spectrum of $A$ is contained in the half-plane ${zinmathbb{C}mid operatorname{Re}zleq omega}$.
In this case $u(t)=T(t)x$ is the unique mild solution of the initial value problem
begin{align*}
dot u(t)&=Au(t),\
u(0)&=x.
end{align*}
The Hille-Yosida theorem also gives a characterization of the operator that generate strongly continuous semigroups.
answered Jan 15 at 13:01
MaoWaoMaoWao
2,853617
$endgroup$
add a comment |
$begingroup$
You need some assumptions on $A$ to have a unique solution of the evolution equation (and one should also talk about the type of solution). Anyway, I think the answer to your question lies in the Hille-Yosida theorem:
If $A$ is the generator of the strongly continuous semigroup $T$, then $|T(t)|leq M e^{omega t}$ if and only if the spectrum of $A$ is contained in the half-plane ${zinmathbb{C}mid operatorname{Re}zleq omega}$.
In this case $u(t)=T(t)x$ is the unique mild solution of the initial value problem
begin{align*}
dot u(t)&=Au(t),\
u(0)&=x.
end{align*}
The Hille-Yosida theorem also gives a characterization of the operator that generate strongly continuous semigroups.
answered Jan 15 at 13:01
MaoWaoMaoWao
2,853617
$endgroup$
add a comment |
$begingroup$
You need some assumptions on $A$ to have a unique solution of the evolution equation (and one should also talk about the type of solution). Anyway, I think the answer to your question lies in the Hille-Yosida theorem:
If $A$ is the generator of the strongly continuous semigroup $T$, then $|T(t)|leq M e^{omega t}$ if and only if the spectrum of $A$ is contained in the half-plane ${zinmathbb{C}mid operatorname{Re}zleq omega}$.
In this case $u(t)=T(t)x$ is the unique mild solution of the initial value problem
begin{align*}
dot u(t)&=Au(t),\
u(0)&=x.
end{align*}
The Hille-Yosida theorem also gives a characterization of the operator that generate strongly continuous semigroups.
answered Jan 15 at 13:01
MaoWaoMaoWao
2,853617
$endgroup$
You need some assumptions on $A$ to have a unique solution of the evolution equation (and one should also talk about the type of solution). Anyway, I think the answer to your question lies in the Hille-Yosida theorem:
If $A$ is the generator of the strongly continuous semigroup $T$, then $|T(t)|leq M e^{omega t}$ if and only if the spectrum of $A$ is contained in the half-plane ${zinmathbb{C}mid operatorname{Re}zleq omega}$.
In this case $u(t)=T(t)x$ is the unique mild solution of the initial value problem
begin{align*}
dot u(t)&=Au(t),\
u(0)&=x.
end{align*}
The Hille-Yosida theorem also gives a characterization of the operator that generate strongly continuous semigroups.
answered Jan 15 at 13:01
MaoWaoMaoWao
2,853617
answered Jan 15 at 13:01
MaoWaoMaoWao
2,853617
answered Jan 15 at 13:01
MaoWaoMaoWao
2,853617
answered Jan 15 at 13:01
MaoWaoMaoWao
2,853617
2,853617
add a comment |
add a comment |
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