Prove that the solution of $x'(t)=Ax+frac{sin t}{t+1} (1,cdots,1)^T$ is bounded
I am trying to show that the solution $x(t)$ of $$x'(t)=Ax+frac{sin t}{t+1} (1,cdots,1)^T$$ is bounded on the interval $(0,infty)$,
where the eigenvalues of the $n times n$ matrix $A$ have negative real part. This is my trial:
I can show easily that
$$x(t) = e^{tA}left[x_0+intlimits_0^t e^{-As}frac{sin s}{s+1} (1,cdots,1)^T dsright]$$
Now,
$$|x(t)|leq left|e^{At}right|left[|x_0|+intlimits_0^t |e^{-As}| left|frac{sin s}{s+1}right| left|(1,cdots,1)^Tright| ds right]$$
Since the eigenvalues of $A$ have negative real part, then using Meiss's Lemma $|e^{At}|leq K e^{-alpha t}$ for some $K,alpha > 0$.
Also since $frac{sin s}{s+1}$ is bounded over $(0,infty)$ so $|frac{sin s}{s+1}|leq L$ and $|(1,cdots,1)^T|=N$, then
$$|x(t)|leq K e^{-alpha t}[|x_0| + MLt e^{-beta t}].$$
Therefore $x(t)$ is bounded.
differential-equations proof-verification eigenvalues-eigenvectors
edited Dec 9 at 15:14
Jakobian
2,489720
asked Dec 9 at 15:10
Ahmed
1,251512
add a comment |
I am trying to show that the solution $x(t)$ of $$x'(t)=Ax+frac{sin t}{t+1} (1,cdots,1)^T$$ is bounded on the interval $(0,infty)$,
where the eigenvalues of the $n times n$ matrix $A$ have negative real part. This is my trial:
I can show easily that
$$x(t) = e^{tA}left[x_0+intlimits_0^t e^{-As}frac{sin s}{s+1} (1,cdots,1)^T dsright]$$
Now,
$$|x(t)|leq left|e^{At}right|left[|x_0|+intlimits_0^t |e^{-As}| left|frac{sin s}{s+1}right| left|(1,cdots,1)^Tright| ds right]$$
Since the eigenvalues of $A$ have negative real part, then using Meiss's Lemma $|e^{At}|leq K e^{-alpha t}$ for some $K,alpha > 0$.
Also since $frac{sin s}{s+1}$ is bounded over $(0,infty)$ so $|frac{sin s}{s+1}|leq L$ and $|(1,cdots,1)^T|=N$, then
$$|x(t)|leq K e^{-alpha t}[|x_0| + MLt e^{-beta t}].$$
Therefore $x(t)$ is bounded.
differential-equations proof-verification eigenvalues-eigenvectors
edited Dec 9 at 15:14
Jakobian
2,489720
asked Dec 9 at 15:10
Ahmed
1,251512
This looks like a special case of an answer to your question of two days ago: Prove that x(t) is bounded.
– user539887
Dec 9 at 20:03
add a comment |
I am trying to show that the solution $x(t)$ of $$x'(t)=Ax+frac{sin t}{t+1} (1,cdots,1)^T$$ is bounded on the interval $(0,infty)$,
where the eigenvalues of the $n times n$ matrix $A$ have negative real part. This is my trial:
I can show easily that
$$x(t) = e^{tA}left[x_0+intlimits_0^t e^{-As}frac{sin s}{s+1} (1,cdots,1)^T dsright]$$
Now,
$$|x(t)|leq left|e^{At}right|left[|x_0|+intlimits_0^t |e^{-As}| left|frac{sin s}{s+1}right| left|(1,cdots,1)^Tright| ds right]$$
Since the eigenvalues of $A$ have negative real part, then using Meiss's Lemma $|e^{At}|leq K e^{-alpha t}$ for some $K,alpha > 0$.
Also since $frac{sin s}{s+1}$ is bounded over $(0,infty)$ so $|frac{sin s}{s+1}|leq L$ and $|(1,cdots,1)^T|=N$, then
$$|x(t)|leq K e^{-alpha t}[|x_0| + MLt e^{-beta t}].$$
Therefore $x(t)$ is bounded.
differential-equations proof-verification eigenvalues-eigenvectors
edited Dec 9 at 15:14
Jakobian
2,489720
asked Dec 9 at 15:10
Ahmed
1,251512
I am trying to show that the solution $x(t)$ of $$x'(t)=Ax+frac{sin t}{t+1} (1,cdots,1)^T$$ is bounded on the interval $(0,infty)$,
where the eigenvalues of the $n times n$ matrix $A$ have negative real part. This is my trial:
I can show easily that
$$x(t) = e^{tA}left[x_0+intlimits_0^t e^{-As}frac{sin s}{s+1} (1,cdots,1)^T dsright]$$
Now,
$$|x(t)|leq left|e^{At}right|left[|x_0|+intlimits_0^t |e^{-As}| left|frac{sin s}{s+1}right| left|(1,cdots,1)^Tright| ds right]$$
Since the eigenvalues of $A$ have negative real part, then using Meiss's Lemma $|e^{At}|leq K e^{-alpha t}$ for some $K,alpha > 0$.
Also since $frac{sin s}{s+1}$ is bounded over $(0,infty)$ so $|frac{sin s}{s+1}|leq L$ and $|(1,cdots,1)^T|=N$, then
$$|x(t)|leq K e^{-alpha t}[|x_0| + MLt e^{-beta t}].$$
Therefore $x(t)$ is bounded.
differential-equations proof-verification eigenvalues-eigenvectors
differential-equations proof-verification eigenvalues-eigenvectors
edited Dec 9 at 15:14
Jakobian
2,489720
asked Dec 9 at 15:10
Ahmed
1,251512
edited Dec 9 at 15:14
Jakobian
2,489720
asked Dec 9 at 15:10
Ahmed
1,251512
edited Dec 9 at 15:14
Jakobian
2,489720
edited Dec 9 at 15:14
Jakobian
2,489720
edited Dec 9 at 15:14
Jakobian
2,489720
2,489720
asked Dec 9 at 15:10
Ahmed
1,251512
asked Dec 9 at 15:10
Ahmed
1,251512
asked Dec 9 at 15:10
Ahmed
1,251512
1,251512
This looks like a special case of an answer to your question of two days ago: Prove that x(t) is bounded.
– user539887
Dec 9 at 20:03
add a comment |
This looks like a special case of an answer to your question of two days ago: Prove that x(t) is bounded.
– user539887
Dec 9 at 20:03
This looks like a special case of an answer to your question of two days ago: Prove that x(t) is bounded.
– user539887
Dec 9 at 20:03
This looks like a special case of an answer to your question of two days ago: Prove that x(t) is bounded.
– user539887
Dec 9 at 20:03
add a comment |
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This looks like a special case of an answer to your question of two days ago: Prove that x(t) is bounded.
– user539887
Dec 9 at 20:03