Obtain phase portrait of this system
$begingroup$
I have a system which consists on a deposit with water, where the relation between the in and out fluxes with the height of the liquid is
$$
q_{in}(t) - q_{out}(t) = Adfrac{dh}{dt} quad text{,}
$$
knowing that $q_{in}(t)$ is constant and $q_{out}(t)=x(t)Ksqrt{h(t)}$, where $x(t)$ is the effect of a valve with values in the range of 0 to 1, so we can decide which fraction of $Ksqrt{h(t)}$ goes through the exit of the deposit.
I want to get the phase portrait of the evolution of $q_{out}(t)$ with changes in $x(t)$, which in my head seems to be composed of closed curves returning to the equilibrium point, for step changes on $x(t)$. But to get that phase portrait I think I have to get a equation with this form (maybe I'm wrong):
$$
dfrac{dq_{out}}{dx}=F(x)
$$
which I'm not able to get... I would appreciate any help with this.
ordinary-differential-equations dynamical-systems stability-theory
$endgroup$
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$begingroup$
I have a system which consists on a deposit with water, where the relation between the in and out fluxes with the height of the liquid is
$$
q_{in}(t) - q_{out}(t) = Adfrac{dh}{dt} quad text{,}
$$
knowing that $q_{in}(t)$ is constant and $q_{out}(t)=x(t)Ksqrt{h(t)}$, where $x(t)$ is the effect of a valve with values in the range of 0 to 1, so we can decide which fraction of $Ksqrt{h(t)}$ goes through the exit of the deposit.
I want to get the phase portrait of the evolution of $q_{out}(t)$ with changes in $x(t)$, which in my head seems to be composed of closed curves returning to the equilibrium point, for step changes on $x(t)$. But to get that phase portrait I think I have to get a equation with this form (maybe I'm wrong):
$$
dfrac{dq_{out}}{dx}=F(x)
$$
which I'm not able to get... I would appreciate any help with this.
ordinary-differential-equations dynamical-systems stability-theory
$endgroup$
add a comment |
$begingroup$
I have a system which consists on a deposit with water, where the relation between the in and out fluxes with the height of the liquid is
$$
q_{in}(t) - q_{out}(t) = Adfrac{dh}{dt} quad text{,}
$$
knowing that $q_{in}(t)$ is constant and $q_{out}(t)=x(t)Ksqrt{h(t)}$, where $x(t)$ is the effect of a valve with values in the range of 0 to 1, so we can decide which fraction of $Ksqrt{h(t)}$ goes through the exit of the deposit.
I want to get the phase portrait of the evolution of $q_{out}(t)$ with changes in $x(t)$, which in my head seems to be composed of closed curves returning to the equilibrium point, for step changes on $x(t)$. But to get that phase portrait I think I have to get a equation with this form (maybe I'm wrong):
$$
dfrac{dq_{out}}{dx}=F(x)
$$
which I'm not able to get... I would appreciate any help with this.
ordinary-differential-equations dynamical-systems stability-theory
$endgroup$
I have a system which consists on a deposit with water, where the relation between the in and out fluxes with the height of the liquid is
$$
q_{in}(t) - q_{out}(t) = Adfrac{dh}{dt} quad text{,}
$$
knowing that $q_{in}(t)$ is constant and $q_{out}(t)=x(t)Ksqrt{h(t)}$, where $x(t)$ is the effect of a valve with values in the range of 0 to 1, so we can decide which fraction of $Ksqrt{h(t)}$ goes through the exit of the deposit.
I want to get the phase portrait of the evolution of $q_{out}(t)$ with changes in $x(t)$, which in my head seems to be composed of closed curves returning to the equilibrium point, for step changes on $x(t)$. But to get that phase portrait I think I have to get a equation with this form (maybe I'm wrong):
$$
dfrac{dq_{out}}{dx}=F(x)
$$
which I'm not able to get... I would appreciate any help with this.
ordinary-differential-equations dynamical-systems stability-theory
ordinary-differential-equations dynamical-systems stability-theory
asked Dec 17 '18 at 11:55
Jaime_mc2Jaime_mc2
1017
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