Prove that $x(t)$ is bounded
$begingroup$
In this problem, I tried the following:
First I show that
$$x(t) = e^{tA}x_0 + intlimits_0^t e^{(t-s)A}f(s)ds$$
Then I take the norm for both sides
$$|x(t)|leq Ke^{-alpha t}|x_0|+left|intlimits_0^t e^{(t-s)A}f(s)dsright|$$
I have used Meiss’s Lemma ($|e^{tA}|leq K e^{-alpha t}|$ for some $K>0$, $alpha >0$).
Could you please help me in what is remaining, and if there is any mistake. Thanks.
ordinary-differential-equations eigenvalues-eigenvectors
asked Dec 7 '18 at 16:10
AhmedAhmed
1,247612
$endgroup$
|
show 1 more comment
$begingroup$
In this problem, I tried the following:
First I show that
$$x(t) = e^{tA}x_0 + intlimits_0^t e^{(t-s)A}f(s)ds$$
Then I take the norm for both sides
$$|x(t)|leq Ke^{-alpha t}|x_0|+left|intlimits_0^t e^{(t-s)A}f(s)dsright|$$
I have used Meiss’s Lemma ($|e^{tA}|leq K e^{-alpha t}|$ for some $K>0$, $alpha >0$).
Could you please help me in what is remaining, and if there is any mistake. Thanks.
ordinary-differential-equations eigenvalues-eigenvectors
asked Dec 7 '18 at 16:10
AhmedAhmed
1,247612
$endgroup$
2
$begingroup$
You just bring the norm inside the integral and you conclude
$endgroup$
– Federico
Dec 7 '18 at 16:13
1
$begingroup$
And see Norm bound on exponential matrix with eigenvalue negative real part to bound the matrix norm.
$endgroup$
– mathcounterexamples.net
Dec 7 '18 at 16:23
$begingroup$
It is not enough to bring the norm inside the integral, I think you have to take A into consideration.
$endgroup$
– Eduardo Elael
Dec 7 '18 at 16:30
$begingroup$
Ok ... My problem is: when we bring the norm inside what will happen to the integral as $t to infty$
$endgroup$
– Ahmed
Dec 8 '18 at 20:05
$begingroup$
Could you please check my comment @Federico
$endgroup$
– Ahmed
Dec 8 '18 at 20:23
|
show 1 more comment
$begingroup$
In this problem, I tried the following:
First I show that
$$x(t) = e^{tA}x_0 + intlimits_0^t e^{(t-s)A}f(s)ds$$
Then I take the norm for both sides
$$|x(t)|leq Ke^{-alpha t}|x_0|+left|intlimits_0^t e^{(t-s)A}f(s)dsright|$$
I have used Meiss’s Lemma ($|e^{tA}|leq K e^{-alpha t}|$ for some $K>0$, $alpha >0$).
Could you please help me in what is remaining, and if there is any mistake. Thanks.
ordinary-differential-equations eigenvalues-eigenvectors
asked Dec 7 '18 at 16:10
AhmedAhmed
1,247612
$endgroup$
In this problem, I tried the following:
First I show that
$$x(t) = e^{tA}x_0 + intlimits_0^t e^{(t-s)A}f(s)ds$$
Then I take the norm for both sides
$$|x(t)|leq Ke^{-alpha t}|x_0|+left|intlimits_0^t e^{(t-s)A}f(s)dsright|$$
I have used Meiss’s Lemma ($|e^{tA}|leq K e^{-alpha t}|$ for some $K>0$, $alpha >0$).
Could you please help me in what is remaining, and if there is any mistake. Thanks.
ordinary-differential-equations eigenvalues-eigenvectors
ordinary-differential-equations eigenvalues-eigenvectors
asked Dec 7 '18 at 16:10
AhmedAhmed
1,247612
asked Dec 7 '18 at 16:10
AhmedAhmed
1,247612
asked Dec 7 '18 at 16:10
AhmedAhmed
1,247612
asked Dec 7 '18 at 16:10
AhmedAhmed
1,247612
asked Dec 7 '18 at 16:10
AhmedAhmed
1,247612
1,247612
2
$begingroup$
You just bring the norm inside the integral and you conclude
$endgroup$
– Federico
Dec 7 '18 at 16:13
1
$begingroup$
And see Norm bound on exponential matrix with eigenvalue negative real part to bound the matrix norm.
$endgroup$
– mathcounterexamples.net
Dec 7 '18 at 16:23
$begingroup$
It is not enough to bring the norm inside the integral, I think you have to take A into consideration.
$endgroup$
– Eduardo Elael
Dec 7 '18 at 16:30
$begingroup$
Ok ... My problem is: when we bring the norm inside what will happen to the integral as $t to infty$
$endgroup$
– Ahmed
Dec 8 '18 at 20:05
$begingroup$
Could you please check my comment @Federico
$endgroup$
– Ahmed
Dec 8 '18 at 20:23
|
show 1 more comment
2
$begingroup$
You just bring the norm inside the integral and you conclude
$endgroup$
– Federico
Dec 7 '18 at 16:13
1
$begingroup$
And see Norm bound on exponential matrix with eigenvalue negative real part to bound the matrix norm.
$endgroup$
– mathcounterexamples.net
Dec 7 '18 at 16:23
$begingroup$
It is not enough to bring the norm inside the integral, I think you have to take A into consideration.
$endgroup$
– Eduardo Elael
Dec 7 '18 at 16:30
$begingroup$
Ok ... My problem is: when we bring the norm inside what will happen to the integral as $t to infty$
$endgroup$
– Ahmed
Dec 8 '18 at 20:05
$begingroup$
Could you please check my comment @Federico
$endgroup$
– Ahmed
Dec 8 '18 at 20:23
2
2
$begingroup$
You just bring the norm inside the integral and you conclude
$endgroup$
– Federico
Dec 7 '18 at 16:13
$begingroup$
You just bring the norm inside the integral and you conclude
$endgroup$
– Federico
Dec 7 '18 at 16:13
1
1
$begingroup$
And see Norm bound on exponential matrix with eigenvalue negative real part to bound the matrix norm.
$endgroup$
– mathcounterexamples.net
Dec 7 '18 at 16:23
$begingroup$
And see Norm bound on exponential matrix with eigenvalue negative real part to bound the matrix norm.
$endgroup$
– mathcounterexamples.net
Dec 7 '18 at 16:23
$begingroup$
It is not enough to bring the norm inside the integral, I think you have to take A into consideration.
$endgroup$
– Eduardo Elael
Dec 7 '18 at 16:30
$begingroup$
It is not enough to bring the norm inside the integral, I think you have to take A into consideration.
$endgroup$
– Eduardo Elael
Dec 7 '18 at 16:30
$begingroup$
Ok ... My problem is: when we bring the norm inside what will happen to the integral as $t to infty$
$endgroup$
– Ahmed
Dec 8 '18 at 20:05
$begingroup$
Ok ... My problem is: when we bring the norm inside what will happen to the integral as $t to infty$
$endgroup$
– Ahmed
Dec 8 '18 at 20:05
$begingroup$
Could you please check my comment @Federico
$endgroup$
– Ahmed
Dec 8 '18 at 20:23
$begingroup$
Could you please check my comment @Federico
$endgroup$
– Ahmed
Dec 8 '18 at 20:23
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
Following the already given hints,
$$
begin{split}
|x(t)|
&leq Ke^{-alpha t} |x_0| + left|int_0^t e^{(t-s)A}f(s),dsright| \
&leq Ke^{-alpha t} |x_0| + int_0^t bigl|e^{(t-s)A}f(s)bigr|,ds \
&leq Ke^{-alpha t} |x_0| + int_0^t bigl|e^{(t-s)A}bigr||f(s)|,ds \
&leq Ke^{-alpha t} |x_0| + K|f|_infty int_0^t e^{alpha(s-t)},ds \
&= Ke^{-alpha t} |x_0| + K|f|_infty frac{1-e^{-alpha t}}{alpha} \
&leq Ke^{-alpha t} |x_0| + frac{K|f|_infty}alpha .
end{split}
$$
answered Dec 10 '18 at 14:05
FedericoFederico
5,144514
$endgroup$
$begingroup$
Should we eliminate also $e^{-alpha t}$ as $ttoinfty$
$endgroup$
– Ahmed
Dec 11 '18 at 13:49
$begingroup$
What do you mean? I'm not taking any limits here. I only used $1-e^{-alpha t}leq 1$.
$endgroup$
– Federico
Dec 11 '18 at 14:00
1
$begingroup$
Oh, I see, you mean in the first term? Well then yes you can eliminate $Ke^{-alpha t}|x_0|leq K|x_0|$ because $e^{-alpha t}leq 1$. But you can also keep it because it provides a slightly better inequality
$endgroup$
– Federico
Dec 11 '18 at 14:01
add a comment |
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1 Answer
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1 Answer
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votes
$begingroup$
Following the already given hints,
$$
begin{split}
|x(t)|
&leq Ke^{-alpha t} |x_0| + left|int_0^t e^{(t-s)A}f(s),dsright| \
&leq Ke^{-alpha t} |x_0| + int_0^t bigl|e^{(t-s)A}f(s)bigr|,ds \
&leq Ke^{-alpha t} |x_0| + int_0^t bigl|e^{(t-s)A}bigr||f(s)|,ds \
&leq Ke^{-alpha t} |x_0| + K|f|_infty int_0^t e^{alpha(s-t)},ds \
&= Ke^{-alpha t} |x_0| + K|f|_infty frac{1-e^{-alpha t}}{alpha} \
&leq Ke^{-alpha t} |x_0| + frac{K|f|_infty}alpha .
end{split}
$$
answered Dec 10 '18 at 14:05
FedericoFederico
5,144514
$endgroup$
$begingroup$
Should we eliminate also $e^{-alpha t}$ as $ttoinfty$
$endgroup$
– Ahmed
Dec 11 '18 at 13:49
$begingroup$
What do you mean? I'm not taking any limits here. I only used $1-e^{-alpha t}leq 1$.
$endgroup$
– Federico
Dec 11 '18 at 14:00
1
$begingroup$
Oh, I see, you mean in the first term? Well then yes you can eliminate $Ke^{-alpha t}|x_0|leq K|x_0|$ because $e^{-alpha t}leq 1$. But you can also keep it because it provides a slightly better inequality
$endgroup$
– Federico
Dec 11 '18 at 14:01
add a comment |
$begingroup$
Following the already given hints,
$$
begin{split}
|x(t)|
&leq Ke^{-alpha t} |x_0| + left|int_0^t e^{(t-s)A}f(s),dsright| \
&leq Ke^{-alpha t} |x_0| + int_0^t bigl|e^{(t-s)A}f(s)bigr|,ds \
&leq Ke^{-alpha t} |x_0| + int_0^t bigl|e^{(t-s)A}bigr||f(s)|,ds \
&leq Ke^{-alpha t} |x_0| + K|f|_infty int_0^t e^{alpha(s-t)},ds \
&= Ke^{-alpha t} |x_0| + K|f|_infty frac{1-e^{-alpha t}}{alpha} \
&leq Ke^{-alpha t} |x_0| + frac{K|f|_infty}alpha .
end{split}
$$
answered Dec 10 '18 at 14:05
FedericoFederico
5,144514
$endgroup$
$begingroup$
Should we eliminate also $e^{-alpha t}$ as $ttoinfty$
$endgroup$
– Ahmed
Dec 11 '18 at 13:49
$begingroup$
What do you mean? I'm not taking any limits here. I only used $1-e^{-alpha t}leq 1$.
$endgroup$
– Federico
Dec 11 '18 at 14:00
1
$begingroup$
Oh, I see, you mean in the first term? Well then yes you can eliminate $Ke^{-alpha t}|x_0|leq K|x_0|$ because $e^{-alpha t}leq 1$. But you can also keep it because it provides a slightly better inequality
$endgroup$
– Federico
Dec 11 '18 at 14:01
add a comment |
$begingroup$
Following the already given hints,
$$
begin{split}
|x(t)|
&leq Ke^{-alpha t} |x_0| + left|int_0^t e^{(t-s)A}f(s),dsright| \
&leq Ke^{-alpha t} |x_0| + int_0^t bigl|e^{(t-s)A}f(s)bigr|,ds \
&leq Ke^{-alpha t} |x_0| + int_0^t bigl|e^{(t-s)A}bigr||f(s)|,ds \
&leq Ke^{-alpha t} |x_0| + K|f|_infty int_0^t e^{alpha(s-t)},ds \
&= Ke^{-alpha t} |x_0| + K|f|_infty frac{1-e^{-alpha t}}{alpha} \
&leq Ke^{-alpha t} |x_0| + frac{K|f|_infty}alpha .
end{split}
$$
answered Dec 10 '18 at 14:05
FedericoFederico
5,144514
$endgroup$
Following the already given hints,
$$
begin{split}
|x(t)|
&leq Ke^{-alpha t} |x_0| + left|int_0^t e^{(t-s)A}f(s),dsright| \
&leq Ke^{-alpha t} |x_0| + int_0^t bigl|e^{(t-s)A}f(s)bigr|,ds \
&leq Ke^{-alpha t} |x_0| + int_0^t bigl|e^{(t-s)A}bigr||f(s)|,ds \
&leq Ke^{-alpha t} |x_0| + K|f|_infty int_0^t e^{alpha(s-t)},ds \
&= Ke^{-alpha t} |x_0| + K|f|_infty frac{1-e^{-alpha t}}{alpha} \
&leq Ke^{-alpha t} |x_0| + frac{K|f|_infty}alpha .
end{split}
$$
answered Dec 10 '18 at 14:05
FedericoFederico
5,144514
answered Dec 10 '18 at 14:05
FedericoFederico
5,144514
answered Dec 10 '18 at 14:05
FedericoFederico
5,144514
answered Dec 10 '18 at 14:05
FedericoFederico
5,144514
5,144514
$begingroup$
Should we eliminate also $e^{-alpha t}$ as $ttoinfty$
$endgroup$
– Ahmed
Dec 11 '18 at 13:49
$begingroup$
What do you mean? I'm not taking any limits here. I only used $1-e^{-alpha t}leq 1$.
$endgroup$
– Federico
Dec 11 '18 at 14:00
1
$begingroup$
Oh, I see, you mean in the first term? Well then yes you can eliminate $Ke^{-alpha t}|x_0|leq K|x_0|$ because $e^{-alpha t}leq 1$. But you can also keep it because it provides a slightly better inequality
$endgroup$
– Federico
Dec 11 '18 at 14:01
add a comment |
$begingroup$
Should we eliminate also $e^{-alpha t}$ as $ttoinfty$
$endgroup$
– Ahmed
Dec 11 '18 at 13:49
$begingroup$
What do you mean? I'm not taking any limits here. I only used $1-e^{-alpha t}leq 1$.
$endgroup$
– Federico
Dec 11 '18 at 14:00
1
$begingroup$
Oh, I see, you mean in the first term? Well then yes you can eliminate $Ke^{-alpha t}|x_0|leq K|x_0|$ because $e^{-alpha t}leq 1$. But you can also keep it because it provides a slightly better inequality
$endgroup$
– Federico
Dec 11 '18 at 14:01
$begingroup$
Should we eliminate also $e^{-alpha t}$ as $ttoinfty$
$endgroup$
– Ahmed
Dec 11 '18 at 13:49
$begingroup$
Should we eliminate also $e^{-alpha t}$ as $ttoinfty$
$endgroup$
– Ahmed
Dec 11 '18 at 13:49
$begingroup$
What do you mean? I'm not taking any limits here. I only used $1-e^{-alpha t}leq 1$.
$endgroup$
– Federico
Dec 11 '18 at 14:00
$begingroup$
What do you mean? I'm not taking any limits here. I only used $1-e^{-alpha t}leq 1$.
$endgroup$
– Federico
Dec 11 '18 at 14:00
1
1
$begingroup$
Oh, I see, you mean in the first term? Well then yes you can eliminate $Ke^{-alpha t}|x_0|leq K|x_0|$ because $e^{-alpha t}leq 1$. But you can also keep it because it provides a slightly better inequality
$endgroup$
– Federico
Dec 11 '18 at 14:01
$begingroup$
Oh, I see, you mean in the first term? Well then yes you can eliminate $Ke^{-alpha t}|x_0|leq K|x_0|$ because $e^{-alpha t}leq 1$. But you can also keep it because it provides a slightly better inequality
$endgroup$
– Federico
Dec 11 '18 at 14:01
add a comment |
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2
$begingroup$
You just bring the norm inside the integral and you conclude
$endgroup$
– Federico
Dec 7 '18 at 16:13
1
$begingroup$
And see Norm bound on exponential matrix with eigenvalue negative real part to bound the matrix norm.
$endgroup$
– mathcounterexamples.net
Dec 7 '18 at 16:23
$begingroup$
It is not enough to bring the norm inside the integral, I think you have to take A into consideration.
$endgroup$
– Eduardo Elael
Dec 7 '18 at 16:30
$begingroup$
Ok ... My problem is: when we bring the norm inside what will happen to the integral as $t to infty$
$endgroup$
– Ahmed
Dec 8 '18 at 20:05
$begingroup$
Could you please check my comment @Federico
$endgroup$
– Ahmed
Dec 8 '18 at 20:23