Mixing
I have $y'(t) = e^{-[g(y)]^2}$
$begingroup$
I have $y'(t) = e^{-[g(y)]^2}$, with initial value $y(0) = y_0$, $g: mathbb{R} times mathbb{R} to mathbb{R}$ and g $in C^1$. I want to find the limits of $t to pm infty$ taking into account the possible values of $y_0$.
Now, I know that there exists a unique local solution because f(y) = $e^{-[g(y)]^2}$ is locally Lipschzit. the integral of an exponential is always exponential times something defined on all $mathbb{R}$ so there exists unique a global solution which is maximal because defined on all $mathbb{R}$.
That means that I can actually take those limits to plus and minus infinity. However I am stuck on how to actually calculate those. What theorem should I use?
ordinary-differential-equations initial-value-problems
asked Jan 12 at 15:55
qcc101qcc101
617213
$endgroup$
add a comment |
$begingroup$
I have $y'(t) = e^{-[g(y)]^2}$, with initial value $y(0) = y_0$, $g: mathbb{R} times mathbb{R} to mathbb{R}$ and g $in C^1$. I want to find the limits of $t to pm infty$ taking into account the possible values of $y_0$.
Now, I know that there exists a unique local solution because f(y) = $e^{-[g(y)]^2}$ is locally Lipschzit. the integral of an exponential is always exponential times something defined on all $mathbb{R}$ so there exists unique a global solution which is maximal because defined on all $mathbb{R}$.
That means that I can actually take those limits to plus and minus infinity. However I am stuck on how to actually calculate those. What theorem should I use?
ordinary-differential-equations initial-value-problems
asked Jan 12 at 15:55
qcc101qcc101
617213
$endgroup$
$begingroup$
Because the ODE is autonomous you can get an implicit representation for $y(t)$, $t > 0$: $int limits_{y_0}^{y(t)} e^{g(y)^2} , mathrm{d}y = t$.
$endgroup$
– Christoph
Jan 12 at 17:08
add a comment |
$begingroup$
I have $y'(t) = e^{-[g(y)]^2}$, with initial value $y(0) = y_0$, $g: mathbb{R} times mathbb{R} to mathbb{R}$ and g $in C^1$. I want to find the limits of $t to pm infty$ taking into account the possible values of $y_0$.
Now, I know that there exists a unique local solution because f(y) = $e^{-[g(y)]^2}$ is locally Lipschzit. the integral of an exponential is always exponential times something defined on all $mathbb{R}$ so there exists unique a global solution which is maximal because defined on all $mathbb{R}$.
That means that I can actually take those limits to plus and minus infinity. However I am stuck on how to actually calculate those. What theorem should I use?
ordinary-differential-equations initial-value-problems
asked Jan 12 at 15:55
qcc101qcc101
617213
$endgroup$
I have $y'(t) = e^{-[g(y)]^2}$, with initial value $y(0) = y_0$, $g: mathbb{R} times mathbb{R} to mathbb{R}$ and g $in C^1$. I want to find the limits of $t to pm infty$ taking into account the possible values of $y_0$.
Now, I know that there exists a unique local solution because f(y) = $e^{-[g(y)]^2}$ is locally Lipschzit. the integral of an exponential is always exponential times something defined on all $mathbb{R}$ so there exists unique a global solution which is maximal because defined on all $mathbb{R}$.
That means that I can actually take those limits to plus and minus infinity. However I am stuck on how to actually calculate those. What theorem should I use?
ordinary-differential-equations initial-value-problems
ordinary-differential-equations initial-value-problems
asked Jan 12 at 15:55
qcc101qcc101
617213
asked Jan 12 at 15:55
qcc101qcc101
617213
asked Jan 12 at 15:55
qcc101qcc101
617213
asked Jan 12 at 15:55
qcc101qcc101
617213
asked Jan 12 at 15:55
qcc101qcc101
617213
617213
$begingroup$
Because the ODE is autonomous you can get an implicit representation for $y(t)$, $t > 0$: $int limits_{y_0}^{y(t)} e^{g(y)^2} , mathrm{d}y = t$.
$endgroup$
– Christoph
Jan 12 at 17:08
add a comment |
$begingroup$
Because the ODE is autonomous you can get an implicit representation for $y(t)$, $t > 0$: $int limits_{y_0}^{y(t)} e^{g(y)^2} , mathrm{d}y = t$.
$endgroup$
– Christoph
Jan 12 at 17:08
$begingroup$
Because the ODE is autonomous you can get an implicit representation for $y(t)$, $t > 0$: $int limits_{y_0}^{y(t)} e^{g(y)^2} , mathrm{d}y = t$.
$endgroup$
– Christoph
Jan 12 at 17:08
$begingroup$
Because the ODE is autonomous you can get an implicit representation for $y(t)$, $t > 0$: $int limits_{y_0}^{y(t)} e^{g(y)^2} , mathrm{d}y = t$.
$endgroup$
– Christoph
Jan 12 at 17:08
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
I will consider the case $tto+infty$, the other being similar.
Since $0<e^{-g(y)^2}le1$, solutions are strictly increasing, global, and satisfy the inequality
$$
y(t)le y_0+t,quad tge0.
$$
Claim: $lim_{tto+infty}y(t)=+infty$.
Proof: Suppose that $y$ is bounded, that is, $y_0le y(t)le M$ for some $M$. Since $g$ is regular, $e^{-g(y)^2}$ is bounded below on $[y_0,M]$, that is, $e^{-g(y(t))^2}ge c$ for some constant $c>0$ and all $tge0$. This implies that $y(y)ge y_0+c,t$, in contradiction with the assumption that $y$ is bounded.
answered Jan 12 at 19:21
Julián AguirreJulián Aguirre
68.8k24096
$endgroup$
add a comment |
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1 Answer
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active
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1 Answer
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$begingroup$
I will consider the case $tto+infty$, the other being similar.
Since $0<e^{-g(y)^2}le1$, solutions are strictly increasing, global, and satisfy the inequality
$$
y(t)le y_0+t,quad tge0.
$$
Claim: $lim_{tto+infty}y(t)=+infty$.
Proof: Suppose that $y$ is bounded, that is, $y_0le y(t)le M$ for some $M$. Since $g$ is regular, $e^{-g(y)^2}$ is bounded below on $[y_0,M]$, that is, $e^{-g(y(t))^2}ge c$ for some constant $c>0$ and all $tge0$. This implies that $y(y)ge y_0+c,t$, in contradiction with the assumption that $y$ is bounded.
answered Jan 12 at 19:21
Julián AguirreJulián Aguirre
68.8k24096
$endgroup$
add a comment |
$begingroup$
I will consider the case $tto+infty$, the other being similar.
Since $0<e^{-g(y)^2}le1$, solutions are strictly increasing, global, and satisfy the inequality
$$
y(t)le y_0+t,quad tge0.
$$
Claim: $lim_{tto+infty}y(t)=+infty$.
Proof: Suppose that $y$ is bounded, that is, $y_0le y(t)le M$ for some $M$. Since $g$ is regular, $e^{-g(y)^2}$ is bounded below on $[y_0,M]$, that is, $e^{-g(y(t))^2}ge c$ for some constant $c>0$ and all $tge0$. This implies that $y(y)ge y_0+c,t$, in contradiction with the assumption that $y$ is bounded.
answered Jan 12 at 19:21
Julián AguirreJulián Aguirre
68.8k24096
$endgroup$
add a comment |
$begingroup$
I will consider the case $tto+infty$, the other being similar.
Since $0<e^{-g(y)^2}le1$, solutions are strictly increasing, global, and satisfy the inequality
$$
y(t)le y_0+t,quad tge0.
$$
Claim: $lim_{tto+infty}y(t)=+infty$.
Proof: Suppose that $y$ is bounded, that is, $y_0le y(t)le M$ for some $M$. Since $g$ is regular, $e^{-g(y)^2}$ is bounded below on $[y_0,M]$, that is, $e^{-g(y(t))^2}ge c$ for some constant $c>0$ and all $tge0$. This implies that $y(y)ge y_0+c,t$, in contradiction with the assumption that $y$ is bounded.
answered Jan 12 at 19:21
Julián AguirreJulián Aguirre
68.8k24096
$endgroup$
I will consider the case $tto+infty$, the other being similar.
Since $0<e^{-g(y)^2}le1$, solutions are strictly increasing, global, and satisfy the inequality
$$
y(t)le y_0+t,quad tge0.
$$
Claim: $lim_{tto+infty}y(t)=+infty$.
Proof: Suppose that $y$ is bounded, that is, $y_0le y(t)le M$ for some $M$. Since $g$ is regular, $e^{-g(y)^2}$ is bounded below on $[y_0,M]$, that is, $e^{-g(y(t))^2}ge c$ for some constant $c>0$ and all $tge0$. This implies that $y(y)ge y_0+c,t$, in contradiction with the assumption that $y$ is bounded.
answered Jan 12 at 19:21
Julián AguirreJulián Aguirre
68.8k24096
answered Jan 12 at 19:21
Julián AguirreJulián Aguirre
68.8k24096
answered Jan 12 at 19:21
Julián AguirreJulián Aguirre
68.8k24096
answered Jan 12 at 19:21
Julián AguirreJulián Aguirre
68.8k24096
68.8k24096
add a comment |
add a comment |
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$begingroup$
Because the ODE is autonomous you can get an implicit representation for $y(t)$, $t > 0$: $int limits_{y_0}^{y(t)} e^{g(y)^2} , mathrm{d}y = t$.
$endgroup$
– Christoph
Jan 12 at 17:08