Prove that $x(t)$ is bounded












1














enter image description here



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.






differential-equations eigenvalues-eigenvectors







share|cite|improve this question


















  • 2




    You just bring the norm inside the integral and you conclude
    – Federico
    Dec 7 at 16:13






  • 1




    And see Norm bound on exponential matrix with eigenvalue negative real part to bound the matrix norm.
    – mathcounterexamples.net
    Dec 7 at 16:23












  • It is not enough to bring the norm inside the integral, I think you have to take A into consideration.
    – Eduardo Elael
    Dec 7 at 16:30










  • Ok ... My problem is: when we bring the norm inside what will happen to the integral as $t to infty$
    – Ahmed
    Dec 8 at 20:05












  • Could you please check my comment @Federico
    – Ahmed
    Dec 8 at 20:23
















1














enter image description here



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.






differential-equations eigenvalues-eigenvectors







share|cite|improve this question


















  • 2




    You just bring the norm inside the integral and you conclude
    – Federico
    Dec 7 at 16:13






  • 1




    And see Norm bound on exponential matrix with eigenvalue negative real part to bound the matrix norm.
    – mathcounterexamples.net
    Dec 7 at 16:23












  • It is not enough to bring the norm inside the integral, I think you have to take A into consideration.
    – Eduardo Elael
    Dec 7 at 16:30










  • Ok ... My problem is: when we bring the norm inside what will happen to the integral as $t to infty$
    – Ahmed
    Dec 8 at 20:05












  • Could you please check my comment @Federico
    – Ahmed
    Dec 8 at 20:23














1












1








1


1





enter image description here



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.






differential-equations eigenvalues-eigenvectors







share|cite|improve this question













enter image description here



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.






differential-equations eigenvalues-eigenvectors




differential-equations eigenvalues-eigenvectors






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 7 at 16:10









Ahmed

1,251512




1,251512








  • 2




    You just bring the norm inside the integral and you conclude
    – Federico
    Dec 7 at 16:13






  • 1




    And see Norm bound on exponential matrix with eigenvalue negative real part to bound the matrix norm.
    – mathcounterexamples.net
    Dec 7 at 16:23












  • It is not enough to bring the norm inside the integral, I think you have to take A into consideration.
    – Eduardo Elael
    Dec 7 at 16:30










  • Ok ... My problem is: when we bring the norm inside what will happen to the integral as $t to infty$
    – Ahmed
    Dec 8 at 20:05












  • Could you please check my comment @Federico
    – Ahmed
    Dec 8 at 20:23














  • 2




    You just bring the norm inside the integral and you conclude
    – Federico
    Dec 7 at 16:13






  • 1




    And see Norm bound on exponential matrix with eigenvalue negative real part to bound the matrix norm.
    – mathcounterexamples.net
    Dec 7 at 16:23












  • It is not enough to bring the norm inside the integral, I think you have to take A into consideration.
    – Eduardo Elael
    Dec 7 at 16:30










  • Ok ... My problem is: when we bring the norm inside what will happen to the integral as $t to infty$
    – Ahmed
    Dec 8 at 20:05












  • Could you please check my comment @Federico
    – Ahmed
    Dec 8 at 20:23








2




2




You just bring the norm inside the integral and you conclude
– Federico
Dec 7 at 16:13




You just bring the norm inside the integral and you conclude
– Federico
Dec 7 at 16:13




1




1




And see Norm bound on exponential matrix with eigenvalue negative real part to bound the matrix norm.
– mathcounterexamples.net
Dec 7 at 16:23






And see Norm bound on exponential matrix with eigenvalue negative real part to bound the matrix norm.
– mathcounterexamples.net
Dec 7 at 16:23














It is not enough to bring the norm inside the integral, I think you have to take A into consideration.
– Eduardo Elael
Dec 7 at 16:30




It is not enough to bring the norm inside the integral, I think you have to take A into consideration.
– Eduardo Elael
Dec 7 at 16:30












Ok ... My problem is: when we bring the norm inside what will happen to the integral as $t to infty$
– Ahmed
Dec 8 at 20:05






Ok ... My problem is: when we bring the norm inside what will happen to the integral as $t to infty$
– Ahmed
Dec 8 at 20:05














Could you please check my comment @Federico
– Ahmed
Dec 8 at 20:23




Could you please check my comment @Federico
– Ahmed
Dec 8 at 20:23










1 Answer
1






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oldest

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1














Following the already given hints,
$$
begin{split}
|x(t)|
&leq Ke^{-alpha t} |x_0| + left|int_0^t e^{(t-s)A}f(s),dsright| \
&leq Ke^{-alpha t} |x_0| + int_0^t bigl|e^{(t-s)A}f(s)bigr|,ds \
&leq Ke^{-alpha t} |x_0| + int_0^t bigl|e^{(t-s)A}bigr||f(s)|,ds \
&leq Ke^{-alpha t} |x_0| + K|f|_infty int_0^t e^{alpha(s-t)},ds \
&= Ke^{-alpha t} |x_0| + K|f|_infty frac{1-e^{-alpha t}}{alpha} \
&leq Ke^{-alpha t} |x_0| + frac{K|f|_infty}alpha .
end{split}
$$






share|cite|improve this answer





















  • Should we eliminate also $e^{-alpha t}$ as $ttoinfty$
    – Ahmed
    Dec 11 at 13:49










  • What do you mean? I'm not taking any limits here. I only used $1-e^{-alpha t}leq 1$.
    – Federico
    Dec 11 at 14:00






  • 1




    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
    – Federico
    Dec 11 at 14:01













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1














Following the already given hints,
$$
begin{split}
|x(t)|
&leq Ke^{-alpha t} |x_0| + left|int_0^t e^{(t-s)A}f(s),dsright| \
&leq Ke^{-alpha t} |x_0| + int_0^t bigl|e^{(t-s)A}f(s)bigr|,ds \
&leq Ke^{-alpha t} |x_0| + int_0^t bigl|e^{(t-s)A}bigr||f(s)|,ds \
&leq Ke^{-alpha t} |x_0| + K|f|_infty int_0^t e^{alpha(s-t)},ds \
&= Ke^{-alpha t} |x_0| + K|f|_infty frac{1-e^{-alpha t}}{alpha} \
&leq Ke^{-alpha t} |x_0| + frac{K|f|_infty}alpha .
end{split}
$$






share|cite|improve this answer





















  • Should we eliminate also $e^{-alpha t}$ as $ttoinfty$
    – Ahmed
    Dec 11 at 13:49










  • What do you mean? I'm not taking any limits here. I only used $1-e^{-alpha t}leq 1$.
    – Federico
    Dec 11 at 14:00






  • 1




    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
    – Federico
    Dec 11 at 14:01


















1














Following the already given hints,
$$
begin{split}
|x(t)|
&leq Ke^{-alpha t} |x_0| + left|int_0^t e^{(t-s)A}f(s),dsright| \
&leq Ke^{-alpha t} |x_0| + int_0^t bigl|e^{(t-s)A}f(s)bigr|,ds \
&leq Ke^{-alpha t} |x_0| + int_0^t bigl|e^{(t-s)A}bigr||f(s)|,ds \
&leq Ke^{-alpha t} |x_0| + K|f|_infty int_0^t e^{alpha(s-t)},ds \
&= Ke^{-alpha t} |x_0| + K|f|_infty frac{1-e^{-alpha t}}{alpha} \
&leq Ke^{-alpha t} |x_0| + frac{K|f|_infty}alpha .
end{split}
$$






share|cite|improve this answer





















  • Should we eliminate also $e^{-alpha t}$ as $ttoinfty$
    – Ahmed
    Dec 11 at 13:49










  • What do you mean? I'm not taking any limits here. I only used $1-e^{-alpha t}leq 1$.
    – Federico
    Dec 11 at 14:00






  • 1




    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
    – Federico
    Dec 11 at 14:01
















1












1








1






Following the already given hints,
$$
begin{split}
|x(t)|
&leq Ke^{-alpha t} |x_0| + left|int_0^t e^{(t-s)A}f(s),dsright| \
&leq Ke^{-alpha t} |x_0| + int_0^t bigl|e^{(t-s)A}f(s)bigr|,ds \
&leq Ke^{-alpha t} |x_0| + int_0^t bigl|e^{(t-s)A}bigr||f(s)|,ds \
&leq Ke^{-alpha t} |x_0| + K|f|_infty int_0^t e^{alpha(s-t)},ds \
&= Ke^{-alpha t} |x_0| + K|f|_infty frac{1-e^{-alpha t}}{alpha} \
&leq Ke^{-alpha t} |x_0| + frac{K|f|_infty}alpha .
end{split}
$$






share|cite|improve this answer












Following the already given hints,
$$
begin{split}
|x(t)|
&leq Ke^{-alpha t} |x_0| + left|int_0^t e^{(t-s)A}f(s),dsright| \
&leq Ke^{-alpha t} |x_0| + int_0^t bigl|e^{(t-s)A}f(s)bigr|,ds \
&leq Ke^{-alpha t} |x_0| + int_0^t bigl|e^{(t-s)A}bigr||f(s)|,ds \
&leq Ke^{-alpha t} |x_0| + K|f|_infty int_0^t e^{alpha(s-t)},ds \
&= Ke^{-alpha t} |x_0| + K|f|_infty frac{1-e^{-alpha t}}{alpha} \
&leq Ke^{-alpha t} |x_0| + frac{K|f|_infty}alpha .
end{split}
$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 10 at 14:05









Federico

4,459512




4,459512












  • Should we eliminate also $e^{-alpha t}$ as $ttoinfty$
    – Ahmed
    Dec 11 at 13:49










  • What do you mean? I'm not taking any limits here. I only used $1-e^{-alpha t}leq 1$.
    – Federico
    Dec 11 at 14:00






  • 1




    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
    – Federico
    Dec 11 at 14:01




















  • Should we eliminate also $e^{-alpha t}$ as $ttoinfty$
    – Ahmed
    Dec 11 at 13:49










  • What do you mean? I'm not taking any limits here. I only used $1-e^{-alpha t}leq 1$.
    – Federico
    Dec 11 at 14:00






  • 1




    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
    – Federico
    Dec 11 at 14:01


















Should we eliminate also $e^{-alpha t}$ as $ttoinfty$
– Ahmed
Dec 11 at 13:49




Should we eliminate also $e^{-alpha t}$ as $ttoinfty$
– Ahmed
Dec 11 at 13:49












What do you mean? I'm not taking any limits here. I only used $1-e^{-alpha t}leq 1$.
– Federico
Dec 11 at 14:00




What do you mean? I'm not taking any limits here. I only used $1-e^{-alpha t}leq 1$.
– Federico
Dec 11 at 14:00




1




1




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
– Federico
Dec 11 at 14:01






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
– Federico
Dec 11 at 14:01




















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