Why in non autonomous ODE, the flow is $varphi ^{t,t_0}(x_0)$ (i.e. depend on $t_0$)?
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Let $dot x=f(x)$ an autonomous ODE. Then the flow is simply defined as $$varphi ^t(x_0)=x(t)$$ where $x_0=x(t_0)$. In particular, it's the trajectory of the solution that start at $x(t_0)$.
Now, if it's not autonomous, i.e. the ODE is $dot x(t)=f(x(t),t)$, then the flow is $$x(t)=varphi ^{t,t_0}(x_0).$$
Why do we need $t_0$ here ? It looks that $x(t)=varphi ^{t}(x_0)$ looks fine, isn't it ?
ordinary-differential-equations
asked Dec 16 '18 at 14:53
NewMathNewMath
4059
$endgroup$
add a comment |
$begingroup$
Let $dot x=f(x)$ an autonomous ODE. Then the flow is simply defined as $$varphi ^t(x_0)=x(t)$$ where $x_0=x(t_0)$. In particular, it's the trajectory of the solution that start at $x(t_0)$.
Now, if it's not autonomous, i.e. the ODE is $dot x(t)=f(x(t),t)$, then the flow is $$x(t)=varphi ^{t,t_0}(x_0).$$
Why do we need $t_0$ here ? It looks that $x(t)=varphi ^{t}(x_0)$ looks fine, isn't it ?
ordinary-differential-equations
asked Dec 16 '18 at 14:53
NewMathNewMath
4059
$endgroup$
1
$begingroup$
Try this for, e.g., $x'(t)=t+x$, $x_0=0$, for two trajectories starting respectively at $t_0=-1$ and $t_0=1$.
$endgroup$
– user10354138
Dec 16 '18 at 14:58
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In the autonomous case, you should take $t_0=0$, otherwise the definition doesn't make sense ($phi^t$ means “follow the flow for $t$ units of time”, not “for $t-t_0$ units of times where $t_0$ can be anything”).
$endgroup$
– Hans Lundmark
Dec 16 '18 at 15:08
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In the first case $x(t)=e^tint_1^t ue^{-u}du$ and in the second case $x(t)=e^tint_{-1}^t ue^{-u}du$... hummm very strange that changing the $t_0$ gives an other solution since I thought that given $x(t_0)$, the solution is unique (since $f(x,t)=x+t$ is Lipschtz wrt $x$). So the solution is not unique ?! @user10354138
$endgroup$
– NewMath
Dec 16 '18 at 15:12
add a comment |
$begingroup$
Let $dot x=f(x)$ an autonomous ODE. Then the flow is simply defined as $$varphi ^t(x_0)=x(t)$$ where $x_0=x(t_0)$. In particular, it's the trajectory of the solution that start at $x(t_0)$.
Now, if it's not autonomous, i.e. the ODE is $dot x(t)=f(x(t),t)$, then the flow is $$x(t)=varphi ^{t,t_0}(x_0).$$
Why do we need $t_0$ here ? It looks that $x(t)=varphi ^{t}(x_0)$ looks fine, isn't it ?
ordinary-differential-equations
asked Dec 16 '18 at 14:53
NewMathNewMath
4059
$endgroup$
Let $dot x=f(x)$ an autonomous ODE. Then the flow is simply defined as $$varphi ^t(x_0)=x(t)$$ where $x_0=x(t_0)$. In particular, it's the trajectory of the solution that start at $x(t_0)$.
Now, if it's not autonomous, i.e. the ODE is $dot x(t)=f(x(t),t)$, then the flow is $$x(t)=varphi ^{t,t_0}(x_0).$$
Why do we need $t_0$ here ? It looks that $x(t)=varphi ^{t}(x_0)$ looks fine, isn't it ?
ordinary-differential-equations
ordinary-differential-equations
asked Dec 16 '18 at 14:53
NewMathNewMath
4059
asked Dec 16 '18 at 14:53
NewMathNewMath
4059
asked Dec 16 '18 at 14:53
NewMathNewMath
4059
asked Dec 16 '18 at 14:53
NewMathNewMath
4059
asked Dec 16 '18 at 14:53
NewMathNewMath
4059
4059
1
$begingroup$
Try this for, e.g., $x'(t)=t+x$, $x_0=0$, for two trajectories starting respectively at $t_0=-1$ and $t_0=1$.
$endgroup$
– user10354138
Dec 16 '18 at 14:58
$begingroup$
In the autonomous case, you should take $t_0=0$, otherwise the definition doesn't make sense ($phi^t$ means “follow the flow for $t$ units of time”, not “for $t-t_0$ units of times where $t_0$ can be anything”).
$endgroup$
– Hans Lundmark
Dec 16 '18 at 15:08
$begingroup$
In the first case $x(t)=e^tint_1^t ue^{-u}du$ and in the second case $x(t)=e^tint_{-1}^t ue^{-u}du$... hummm very strange that changing the $t_0$ gives an other solution since I thought that given $x(t_0)$, the solution is unique (since $f(x,t)=x+t$ is Lipschtz wrt $x$). So the solution is not unique ?! @user10354138
$endgroup$
– NewMath
Dec 16 '18 at 15:12
add a comment |
1
$begingroup$
Try this for, e.g., $x'(t)=t+x$, $x_0=0$, for two trajectories starting respectively at $t_0=-1$ and $t_0=1$.
$endgroup$
– user10354138
Dec 16 '18 at 14:58
$begingroup$
In the autonomous case, you should take $t_0=0$, otherwise the definition doesn't make sense ($phi^t$ means “follow the flow for $t$ units of time”, not “for $t-t_0$ units of times where $t_0$ can be anything”).
$endgroup$
– Hans Lundmark
Dec 16 '18 at 15:08
$begingroup$
In the first case $x(t)=e^tint_1^t ue^{-u}du$ and in the second case $x(t)=e^tint_{-1}^t ue^{-u}du$... hummm very strange that changing the $t_0$ gives an other solution since I thought that given $x(t_0)$, the solution is unique (since $f(x,t)=x+t$ is Lipschtz wrt $x$). So the solution is not unique ?! @user10354138
$endgroup$
– NewMath
Dec 16 '18 at 15:12
1
1
$begingroup$
Try this for, e.g., $x'(t)=t+x$, $x_0=0$, for two trajectories starting respectively at $t_0=-1$ and $t_0=1$.
$endgroup$
– user10354138
Dec 16 '18 at 14:58
$begingroup$
Try this for, e.g., $x'(t)=t+x$, $x_0=0$, for two trajectories starting respectively at $t_0=-1$ and $t_0=1$.
$endgroup$
– user10354138
Dec 16 '18 at 14:58
$begingroup$
In the autonomous case, you should take $t_0=0$, otherwise the definition doesn't make sense ($phi^t$ means “follow the flow for $t$ units of time”, not “for $t-t_0$ units of times where $t_0$ can be anything”).
$endgroup$
– Hans Lundmark
Dec 16 '18 at 15:08
$begingroup$
In the autonomous case, you should take $t_0=0$, otherwise the definition doesn't make sense ($phi^t$ means “follow the flow for $t$ units of time”, not “for $t-t_0$ units of times where $t_0$ can be anything”).
$endgroup$
– Hans Lundmark
Dec 16 '18 at 15:08
$begingroup$
In the first case $x(t)=e^tint_1^t ue^{-u}du$ and in the second case $x(t)=e^tint_{-1}^t ue^{-u}du$... hummm very strange that changing the $t_0$ gives an other solution since I thought that given $x(t_0)$, the solution is unique (since $f(x,t)=x+t$ is Lipschtz wrt $x$). So the solution is not unique ?! @user10354138
$endgroup$
– NewMath
Dec 16 '18 at 15:12
$begingroup$
In the first case $x(t)=e^tint_1^t ue^{-u}du$ and in the second case $x(t)=e^tint_{-1}^t ue^{-u}du$... hummm very strange that changing the $t_0$ gives an other solution since I thought that given $x(t_0)$, the solution is unique (since $f(x,t)=x+t$ is Lipschtz wrt $x$). So the solution is not unique ?! @user10354138
$endgroup$
– NewMath
Dec 16 '18 at 15:12
add a comment |
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$begingroup$
Try this for, e.g., $x'(t)=t+x$, $x_0=0$, for two trajectories starting respectively at $t_0=-1$ and $t_0=1$.
$endgroup$
– user10354138
Dec 16 '18 at 14:58
$begingroup$
In the autonomous case, you should take $t_0=0$, otherwise the definition doesn't make sense ($phi^t$ means “follow the flow for $t$ units of time”, not “for $t-t_0$ units of times where $t_0$ can be anything”).
$endgroup$
– Hans Lundmark
Dec 16 '18 at 15:08
$begingroup$
In the first case $x(t)=e^tint_1^t ue^{-u}du$ and in the second case $x(t)=e^tint_{-1}^t ue^{-u}du$... hummm very strange that changing the $t_0$ gives an other solution since I thought that given $x(t_0)$, the solution is unique (since $f(x,t)=x+t$ is Lipschtz wrt $x$). So the solution is not unique ?! @user10354138
$endgroup$
– NewMath
Dec 16 '18 at 15:12