The reverse implication in the stability theory
$begingroup$
Let us consider the system of differential equation $dot x=Ax+h(t),$ where A is a constant matrix of dimension $ngeq2$ and $h$ is continuous on $[0,infty).$ All eigenvalues of $A$ have a negative real part. It is known, that $h(t)to 0$ for $ttoinfty$ implies that all solutions $x(t)$ of $dot x=Ax+h(t)$ converge to $0$ for $ttoinfty.$
Is the converse implication true too? (All solutions $x(t)to 0$ implies $h(t)to 0$).
Thank you.
differential-equations stability-in-odes
asked Nov 23 '18 at 18:27
RobertRobert
112
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add a comment |
$begingroup$
Let us consider the system of differential equation $dot x=Ax+h(t),$ where A is a constant matrix of dimension $ngeq2$ and $h$ is continuous on $[0,infty).$ All eigenvalues of $A$ have a negative real part. It is known, that $h(t)to 0$ for $ttoinfty$ implies that all solutions $x(t)$ of $dot x=Ax+h(t)$ converge to $0$ for $ttoinfty.$
Is the converse implication true too? (All solutions $x(t)to 0$ implies $h(t)to 0$).
Thank you.
differential-equations stability-in-odes
asked Nov 23 '18 at 18:27
RobertRobert
112
$endgroup$
add a comment |
$begingroup$
Let us consider the system of differential equation $dot x=Ax+h(t),$ where A is a constant matrix of dimension $ngeq2$ and $h$ is continuous on $[0,infty).$ All eigenvalues of $A$ have a negative real part. It is known, that $h(t)to 0$ for $ttoinfty$ implies that all solutions $x(t)$ of $dot x=Ax+h(t)$ converge to $0$ for $ttoinfty.$
Is the converse implication true too? (All solutions $x(t)to 0$ implies $h(t)to 0$).
Thank you.
differential-equations stability-in-odes
asked Nov 23 '18 at 18:27
RobertRobert
112
$endgroup$
Let us consider the system of differential equation $dot x=Ax+h(t),$ where A is a constant matrix of dimension $ngeq2$ and $h$ is continuous on $[0,infty).$ All eigenvalues of $A$ have a negative real part. It is known, that $h(t)to 0$ for $ttoinfty$ implies that all solutions $x(t)$ of $dot x=Ax+h(t)$ converge to $0$ for $ttoinfty.$
Is the converse implication true too? (All solutions $x(t)to 0$ implies $h(t)to 0$).
Thank you.
differential-equations stability-in-odes
differential-equations stability-in-odes
asked Nov 23 '18 at 18:27
RobertRobert
112
asked Nov 23 '18 at 18:27
RobertRobert
112
asked Nov 23 '18 at 18:27
RobertRobert
112
asked Nov 23 '18 at 18:27
RobertRobert
112
asked Nov 23 '18 at 18:27
RobertRobert
112
112
add a comment |
add a comment |
2 Answers
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More precisely, the equivalence holds if for some vector norm $leftVertcdotrightVert,$
begin{equation}
suplimits_{0leq uleq1}leftVertintlimits_{t}^{t+u}h(s)dsrightVertto 0
end{equation}
as $ttoinfty.$
For the details and (long) proof see Theorem B in:
Strauss, A.; Yorke, J. A., Perturbing uniform asymptotically stable nonlinear systems, J. Differ. Equations 6, 452-483 (1969). ZBL0182.12103.
answered Nov 24 '18 at 9:54
RobertRobert
112
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add a comment |
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No. Consider $h$ which is $0$ everywhere except near $n^2$ for $ninmathbb N$ where it has bumps of height $1$ and width $1/n$:
$$
h(t) = sum_{ninmathbb N} 1_{[n^2-1/n,n^2+1/n]}(t).
$$
Consider now for instance $x'(t)=-x(t)+h(t)$.
All trajectories become eventually positive. Then it's clear that the effect of $h$ perturbs $x$ by at most $2/n$ when $t$ passes the time $n$:
$$
x(n+1/n)-x(n-1/n) leq int_{n^2-1/n}^{n^2+1/n} h(t),dt=2/n.
$$
From this you should be able to conclude that $x$ goes to $0$ nonetheless, because in the interval $[n^2,(n+1)^2]$ of length $sim n$ it has plenty of time to become exponentially close to $0$.
answered Nov 23 '18 at 18:51
FedericoFederico
4,869514
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
More precisely, the equivalence holds if for some vector norm $leftVertcdotrightVert,$
begin{equation}
suplimits_{0leq uleq1}leftVertintlimits_{t}^{t+u}h(s)dsrightVertto 0
end{equation}
as $ttoinfty.$
For the details and (long) proof see Theorem B in:
Strauss, A.; Yorke, J. A., Perturbing uniform asymptotically stable nonlinear systems, J. Differ. Equations 6, 452-483 (1969). ZBL0182.12103.
answered Nov 24 '18 at 9:54
RobertRobert
112
$endgroup$
add a comment |
$begingroup$
More precisely, the equivalence holds if for some vector norm $leftVertcdotrightVert,$
begin{equation}
suplimits_{0leq uleq1}leftVertintlimits_{t}^{t+u}h(s)dsrightVertto 0
end{equation}
as $ttoinfty.$
For the details and (long) proof see Theorem B in:
Strauss, A.; Yorke, J. A., Perturbing uniform asymptotically stable nonlinear systems, J. Differ. Equations 6, 452-483 (1969). ZBL0182.12103.
answered Nov 24 '18 at 9:54
RobertRobert
112
$endgroup$
add a comment |
$begingroup$
More precisely, the equivalence holds if for some vector norm $leftVertcdotrightVert,$
begin{equation}
suplimits_{0leq uleq1}leftVertintlimits_{t}^{t+u}h(s)dsrightVertto 0
end{equation}
as $ttoinfty.$
For the details and (long) proof see Theorem B in:
Strauss, A.; Yorke, J. A., Perturbing uniform asymptotically stable nonlinear systems, J. Differ. Equations 6, 452-483 (1969). ZBL0182.12103.
answered Nov 24 '18 at 9:54
RobertRobert
112
$endgroup$
More precisely, the equivalence holds if for some vector norm $leftVertcdotrightVert,$
begin{equation}
suplimits_{0leq uleq1}leftVertintlimits_{t}^{t+u}h(s)dsrightVertto 0
end{equation}
as $ttoinfty.$
For the details and (long) proof see Theorem B in:
Strauss, A.; Yorke, J. A., Perturbing uniform asymptotically stable nonlinear systems, J. Differ. Equations 6, 452-483 (1969). ZBL0182.12103.
answered Nov 24 '18 at 9:54
RobertRobert
112
answered Nov 24 '18 at 9:54
RobertRobert
112
answered Nov 24 '18 at 9:54
RobertRobert
112
answered Nov 24 '18 at 9:54
RobertRobert
112
112
add a comment |
add a comment |
$begingroup$
No. Consider $h$ which is $0$ everywhere except near $n^2$ for $ninmathbb N$ where it has bumps of height $1$ and width $1/n$:
$$
h(t) = sum_{ninmathbb N} 1_{[n^2-1/n,n^2+1/n]}(t).
$$
Consider now for instance $x'(t)=-x(t)+h(t)$.
All trajectories become eventually positive. Then it's clear that the effect of $h$ perturbs $x$ by at most $2/n$ when $t$ passes the time $n$:
$$
x(n+1/n)-x(n-1/n) leq int_{n^2-1/n}^{n^2+1/n} h(t),dt=2/n.
$$
From this you should be able to conclude that $x$ goes to $0$ nonetheless, because in the interval $[n^2,(n+1)^2]$ of length $sim n$ it has plenty of time to become exponentially close to $0$.
answered Nov 23 '18 at 18:51
FedericoFederico
4,869514
$endgroup$
add a comment |
$begingroup$
No. Consider $h$ which is $0$ everywhere except near $n^2$ for $ninmathbb N$ where it has bumps of height $1$ and width $1/n$:
$$
h(t) = sum_{ninmathbb N} 1_{[n^2-1/n,n^2+1/n]}(t).
$$
Consider now for instance $x'(t)=-x(t)+h(t)$.
All trajectories become eventually positive. Then it's clear that the effect of $h$ perturbs $x$ by at most $2/n$ when $t$ passes the time $n$:
$$
x(n+1/n)-x(n-1/n) leq int_{n^2-1/n}^{n^2+1/n} h(t),dt=2/n.
$$
From this you should be able to conclude that $x$ goes to $0$ nonetheless, because in the interval $[n^2,(n+1)^2]$ of length $sim n$ it has plenty of time to become exponentially close to $0$.
answered Nov 23 '18 at 18:51
FedericoFederico
4,869514
$endgroup$
add a comment |
$begingroup$
No. Consider $h$ which is $0$ everywhere except near $n^2$ for $ninmathbb N$ where it has bumps of height $1$ and width $1/n$:
$$
h(t) = sum_{ninmathbb N} 1_{[n^2-1/n,n^2+1/n]}(t).
$$
Consider now for instance $x'(t)=-x(t)+h(t)$.
All trajectories become eventually positive. Then it's clear that the effect of $h$ perturbs $x$ by at most $2/n$ when $t$ passes the time $n$:
$$
x(n+1/n)-x(n-1/n) leq int_{n^2-1/n}^{n^2+1/n} h(t),dt=2/n.
$$
From this you should be able to conclude that $x$ goes to $0$ nonetheless, because in the interval $[n^2,(n+1)^2]$ of length $sim n$ it has plenty of time to become exponentially close to $0$.
answered Nov 23 '18 at 18:51
FedericoFederico
4,869514
$endgroup$
No. Consider $h$ which is $0$ everywhere except near $n^2$ for $ninmathbb N$ where it has bumps of height $1$ and width $1/n$:
$$
h(t) = sum_{ninmathbb N} 1_{[n^2-1/n,n^2+1/n]}(t).
$$
Consider now for instance $x'(t)=-x(t)+h(t)$.
All trajectories become eventually positive. Then it's clear that the effect of $h$ perturbs $x$ by at most $2/n$ when $t$ passes the time $n$:
$$
x(n+1/n)-x(n-1/n) leq int_{n^2-1/n}^{n^2+1/n} h(t),dt=2/n.
$$
From this you should be able to conclude that $x$ goes to $0$ nonetheless, because in the interval $[n^2,(n+1)^2]$ of length $sim n$ it has plenty of time to become exponentially close to $0$.
answered Nov 23 '18 at 18:51
FedericoFederico
4,869514
answered Nov 23 '18 at 18:51
FedericoFederico
4,869514
answered Nov 23 '18 at 18:51
FedericoFederico
4,869514
answered Nov 23 '18 at 18:51
FedericoFederico
4,869514
4,869514
add a comment |
add a comment |
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