Mixing
Linear version of Gronwall's inequality, proof
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I am reading a proof of the following theorem:
Assume $phi$ is a continuous function in $[0,T]$ that satisfies $$phi(t) = alpha + int_0^t (beta phi(s) + gamma)ds, hspace{0.5mm} tin [0,T], $$ where $alpha, gamma in mathbb{R}$ and $beta > 0$. Then $$phi(t) leq alpha e^{beta t} + frac{gamma}{beta}(e^{beta t} - 1), hspace{0.5mm} tin [0,T]. $$
The proof begins by setting $$psi(t) = alpha + int_0^t (beta phi(s) + gamma)ds, $$ in order to obtain $$psi'(t) leq beta psi(t) + gamma implies psi'(t) - beta psi(t) leq gamma implies left(e^{-beta t} psi(t) right)' leq gamma e^{-beta t}. $$
After this, it says that this implies that $e^{-beta t} psi(t) - psi(0) leq frac{gamma}{beta}(1-e^{-beta t}), $ and this is what confuses me.
I think that $$left(e^{-beta t} psi(t) right)' leq gamma e^{-beta t} implies e^{-beta t} psi(t) leq -frac{gamma}{beta}e^{-beta t}, $$ but does subtracting $psi(0) = alpha$ on the left-hand side somehow correspond to adding $gamma / beta$ on the right-hand side? I don't see how it would.
UPDATE
Could it be that I have to integrate $gamma e^{-beta t} $ from $0$ to $t$? Because then that would be $frac{gamma}{beta}(1-e^{-beta t})$. But then I am still confused about what allows us to subtract $psi(0)$ from the left-hand side.
ordinary-differential-equations inequality integral-inequality
asked Jan 27 at 3:45
j.eeej.eee
787
$endgroup$
add a comment |
$begingroup$
I am reading a proof of the following theorem:
Assume $phi$ is a continuous function in $[0,T]$ that satisfies $$phi(t) = alpha + int_0^t (beta phi(s) + gamma)ds, hspace{0.5mm} tin [0,T], $$ where $alpha, gamma in mathbb{R}$ and $beta > 0$. Then $$phi(t) leq alpha e^{beta t} + frac{gamma}{beta}(e^{beta t} - 1), hspace{0.5mm} tin [0,T]. $$
The proof begins by setting $$psi(t) = alpha + int_0^t (beta phi(s) + gamma)ds, $$ in order to obtain $$psi'(t) leq beta psi(t) + gamma implies psi'(t) - beta psi(t) leq gamma implies left(e^{-beta t} psi(t) right)' leq gamma e^{-beta t}. $$
After this, it says that this implies that $e^{-beta t} psi(t) - psi(0) leq frac{gamma}{beta}(1-e^{-beta t}), $ and this is what confuses me.
I think that $$left(e^{-beta t} psi(t) right)' leq gamma e^{-beta t} implies e^{-beta t} psi(t) leq -frac{gamma}{beta}e^{-beta t}, $$ but does subtracting $psi(0) = alpha$ on the left-hand side somehow correspond to adding $gamma / beta$ on the right-hand side? I don't see how it would.
UPDATE
Could it be that I have to integrate $gamma e^{-beta t} $ from $0$ to $t$? Because then that would be $frac{gamma}{beta}(1-e^{-beta t})$. But then I am still confused about what allows us to subtract $psi(0)$ from the left-hand side.
ordinary-differential-equations inequality integral-inequality
asked Jan 27 at 3:45
j.eeej.eee
787
$endgroup$
add a comment |
$begingroup$
I am reading a proof of the following theorem:
Assume $phi$ is a continuous function in $[0,T]$ that satisfies $$phi(t) = alpha + int_0^t (beta phi(s) + gamma)ds, hspace{0.5mm} tin [0,T], $$ where $alpha, gamma in mathbb{R}$ and $beta > 0$. Then $$phi(t) leq alpha e^{beta t} + frac{gamma}{beta}(e^{beta t} - 1), hspace{0.5mm} tin [0,T]. $$
The proof begins by setting $$psi(t) = alpha + int_0^t (beta phi(s) + gamma)ds, $$ in order to obtain $$psi'(t) leq beta psi(t) + gamma implies psi'(t) - beta psi(t) leq gamma implies left(e^{-beta t} psi(t) right)' leq gamma e^{-beta t}. $$
After this, it says that this implies that $e^{-beta t} psi(t) - psi(0) leq frac{gamma}{beta}(1-e^{-beta t}), $ and this is what confuses me.
I think that $$left(e^{-beta t} psi(t) right)' leq gamma e^{-beta t} implies e^{-beta t} psi(t) leq -frac{gamma}{beta}e^{-beta t}, $$ but does subtracting $psi(0) = alpha$ on the left-hand side somehow correspond to adding $gamma / beta$ on the right-hand side? I don't see how it would.
UPDATE
Could it be that I have to integrate $gamma e^{-beta t} $ from $0$ to $t$? Because then that would be $frac{gamma}{beta}(1-e^{-beta t})$. But then I am still confused about what allows us to subtract $psi(0)$ from the left-hand side.
ordinary-differential-equations inequality integral-inequality
asked Jan 27 at 3:45
j.eeej.eee
787
$endgroup$
I am reading a proof of the following theorem:
Assume $phi$ is a continuous function in $[0,T]$ that satisfies $$phi(t) = alpha + int_0^t (beta phi(s) + gamma)ds, hspace{0.5mm} tin [0,T], $$ where $alpha, gamma in mathbb{R}$ and $beta > 0$. Then $$phi(t) leq alpha e^{beta t} + frac{gamma}{beta}(e^{beta t} - 1), hspace{0.5mm} tin [0,T]. $$
The proof begins by setting $$psi(t) = alpha + int_0^t (beta phi(s) + gamma)ds, $$ in order to obtain $$psi'(t) leq beta psi(t) + gamma implies psi'(t) - beta psi(t) leq gamma implies left(e^{-beta t} psi(t) right)' leq gamma e^{-beta t}. $$
After this, it says that this implies that $e^{-beta t} psi(t) - psi(0) leq frac{gamma}{beta}(1-e^{-beta t}), $ and this is what confuses me.
I think that $$left(e^{-beta t} psi(t) right)' leq gamma e^{-beta t} implies e^{-beta t} psi(t) leq -frac{gamma}{beta}e^{-beta t}, $$ but does subtracting $psi(0) = alpha$ on the left-hand side somehow correspond to adding $gamma / beta$ on the right-hand side? I don't see how it would.
UPDATE
Could it be that I have to integrate $gamma e^{-beta t} $ from $0$ to $t$? Because then that would be $frac{gamma}{beta}(1-e^{-beta t})$. But then I am still confused about what allows us to subtract $psi(0)$ from the left-hand side.
ordinary-differential-equations inequality integral-inequality
ordinary-differential-equations inequality integral-inequality
asked Jan 27 at 3:45
j.eeej.eee
787
asked Jan 27 at 3:45
j.eeej.eee
787
edited Jan 27 at 3:57
j.eee
asked Jan 27 at 3:45
j.eeej.eee
787
asked Jan 27 at 3:45
j.eeej.eee
787
asked Jan 27 at 3:45
j.eeej.eee
787
787
add a comment |
add a comment |
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