Flow of ODE with monotone source
$begingroup$
Let $Phi$ be the flow (defined as in page 6 of this paper) of the ODE
$$begin{cases}
frac{d}{dt}Phi(x,t) = f(Phi(x,t),t) quad t >0 \
Phi(x,0) = x quad x in mathbb{R}.
end{cases}$$
Is it true that if $f$ is monotone in the first variable then $Phi$ is Lipschitz?
reference-request ca.classical-analysis-and-odes measure-theory geometric-measure-theory
$endgroup$
add a comment |
$begingroup$
Let $Phi$ be the flow (defined as in page 6 of this paper) of the ODE
$$begin{cases}
frac{d}{dt}Phi(x,t) = f(Phi(x,t),t) quad t >0 \
Phi(x,0) = x quad x in mathbb{R}.
end{cases}$$
Is it true that if $f$ is monotone in the first variable then $Phi$ is Lipschitz?
reference-request ca.classical-analysis-and-odes measure-theory geometric-measure-theory
$endgroup$
add a comment |
$begingroup$
Let $Phi$ be the flow (defined as in page 6 of this paper) of the ODE
$$begin{cases}
frac{d}{dt}Phi(x,t) = f(Phi(x,t),t) quad t >0 \
Phi(x,0) = x quad x in mathbb{R}.
end{cases}$$
Is it true that if $f$ is monotone in the first variable then $Phi$ is Lipschitz?
reference-request ca.classical-analysis-and-odes measure-theory geometric-measure-theory
$endgroup$
Let $Phi$ be the flow (defined as in page 6 of this paper) of the ODE
$$begin{cases}
frac{d}{dt}Phi(x,t) = f(Phi(x,t),t) quad t >0 \
Phi(x,0) = x quad x in mathbb{R}.
end{cases}$$
Is it true that if $f$ is monotone in the first variable then $Phi$ is Lipschitz?
reference-request ca.classical-analysis-and-odes measure-theory geometric-measure-theory
reference-request ca.classical-analysis-and-odes measure-theory geometric-measure-theory
asked yesterday
HiroHiro
697
697
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Suppose that $f$ is decreasing in $x$. Let $x(t)$, $y(t)$ be two solutions of the ode. Then
$$
dot{x}-dot{y}= f(x,t)-f(y,t).
$$
Multiplying both sides by $x-y$ we deduce
$$
(dot{x}-dot{y})(x-y) =big(f(x,t)-f(y,t)big)(x-y)leq 0,
$$
where the last equality holds because $f$ is decreasing.
Hence
$$
frac{1}{2}frac{d}{dt}big(x-y)^2leq 0.
$$
Thus the function $tmapsto big( x(t)-y(t)big)^2 $ is decreasing so
$$
big(x(t)-y(t)big)^2leq big( x(0)-y(0)big)^2,;;forall tgeq 0,
$$
i.e.,
$$
Big(Phi(x_0,t)-Phi(y_0,t)Big)^2leq Big(x_0-y_0Big)^2,;;forall tgeq 0.
$$
In other words, for $tgeq 0$, $Phi(x,t)$ is Lipschitz in $x$ with Lipschitz constant $1$ if $f$ is decreasing.
$endgroup$
$begingroup$
Thank you. How can the argument be made rigorous even when $f$ is not smooth and $Phi$ is not a classical solution but a regular Lagrangian flow?
$endgroup$
– Hiro
yesterday
$begingroup$
The function $f$ coud even be multivalued, and you can work in an infinite dimensional Hilbert space as well This is a special case of the general theory of maximal monotone operators and the associated differential equations. Perhaps the friendliest introduction is Brezis' book Operateurs maximaux monotones et semi-groupes de contractions dans les espaces de Hilbert. The ultimate reference is however V. Barbu's book Nonlinear semigroups and differen tial equations in Banach spaces
$endgroup$
– Liviu Nicolaescu
yesterday
$begingroup$
The finite dimensional case is discussed in V. Barbu's recent book Differential Equations Springer 2016, Example 2.4 and Sec. 2.7.
$endgroup$
– Liviu Nicolaescu
yesterday
$begingroup$
In the scalar case all you need for existence and uniqueness is that $f$ is decreasing and the function $mathbb{R}ni xmapsto f(x)-xinmathbb{R}$ is onto.
$endgroup$
– Liviu Nicolaescu
yesterday
$begingroup$
Thank you. What if $f$ is increasing?
$endgroup$
– Hiro
yesterday
|
show 4 more comments
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1 Answer
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active
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1 Answer
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active
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$begingroup$
Suppose that $f$ is decreasing in $x$. Let $x(t)$, $y(t)$ be two solutions of the ode. Then
$$
dot{x}-dot{y}= f(x,t)-f(y,t).
$$
Multiplying both sides by $x-y$ we deduce
$$
(dot{x}-dot{y})(x-y) =big(f(x,t)-f(y,t)big)(x-y)leq 0,
$$
where the last equality holds because $f$ is decreasing.
Hence
$$
frac{1}{2}frac{d}{dt}big(x-y)^2leq 0.
$$
Thus the function $tmapsto big( x(t)-y(t)big)^2 $ is decreasing so
$$
big(x(t)-y(t)big)^2leq big( x(0)-y(0)big)^2,;;forall tgeq 0,
$$
i.e.,
$$
Big(Phi(x_0,t)-Phi(y_0,t)Big)^2leq Big(x_0-y_0Big)^2,;;forall tgeq 0.
$$
In other words, for $tgeq 0$, $Phi(x,t)$ is Lipschitz in $x$ with Lipschitz constant $1$ if $f$ is decreasing.
$endgroup$
$begingroup$
Thank you. How can the argument be made rigorous even when $f$ is not smooth and $Phi$ is not a classical solution but a regular Lagrangian flow?
$endgroup$
– Hiro
yesterday
$begingroup$
The function $f$ coud even be multivalued, and you can work in an infinite dimensional Hilbert space as well This is a special case of the general theory of maximal monotone operators and the associated differential equations. Perhaps the friendliest introduction is Brezis' book Operateurs maximaux monotones et semi-groupes de contractions dans les espaces de Hilbert. The ultimate reference is however V. Barbu's book Nonlinear semigroups and differen tial equations in Banach spaces
$endgroup$
– Liviu Nicolaescu
yesterday
$begingroup$
The finite dimensional case is discussed in V. Barbu's recent book Differential Equations Springer 2016, Example 2.4 and Sec. 2.7.
$endgroup$
– Liviu Nicolaescu
yesterday
$begingroup$
In the scalar case all you need for existence and uniqueness is that $f$ is decreasing and the function $mathbb{R}ni xmapsto f(x)-xinmathbb{R}$ is onto.
$endgroup$
– Liviu Nicolaescu
yesterday
$begingroup$
Thank you. What if $f$ is increasing?
$endgroup$
– Hiro
yesterday
|
show 4 more comments
$begingroup$
Suppose that $f$ is decreasing in $x$. Let $x(t)$, $y(t)$ be two solutions of the ode. Then
$$
dot{x}-dot{y}= f(x,t)-f(y,t).
$$
Multiplying both sides by $x-y$ we deduce
$$
(dot{x}-dot{y})(x-y) =big(f(x,t)-f(y,t)big)(x-y)leq 0,
$$
where the last equality holds because $f$ is decreasing.
Hence
$$
frac{1}{2}frac{d}{dt}big(x-y)^2leq 0.
$$
Thus the function $tmapsto big( x(t)-y(t)big)^2 $ is decreasing so
$$
big(x(t)-y(t)big)^2leq big( x(0)-y(0)big)^2,;;forall tgeq 0,
$$
i.e.,
$$
Big(Phi(x_0,t)-Phi(y_0,t)Big)^2leq Big(x_0-y_0Big)^2,;;forall tgeq 0.
$$
In other words, for $tgeq 0$, $Phi(x,t)$ is Lipschitz in $x$ with Lipschitz constant $1$ if $f$ is decreasing.
$endgroup$
$begingroup$
Thank you. How can the argument be made rigorous even when $f$ is not smooth and $Phi$ is not a classical solution but a regular Lagrangian flow?
$endgroup$
– Hiro
yesterday
$begingroup$
The function $f$ coud even be multivalued, and you can work in an infinite dimensional Hilbert space as well This is a special case of the general theory of maximal monotone operators and the associated differential equations. Perhaps the friendliest introduction is Brezis' book Operateurs maximaux monotones et semi-groupes de contractions dans les espaces de Hilbert. The ultimate reference is however V. Barbu's book Nonlinear semigroups and differen tial equations in Banach spaces
$endgroup$
– Liviu Nicolaescu
yesterday
$begingroup$
The finite dimensional case is discussed in V. Barbu's recent book Differential Equations Springer 2016, Example 2.4 and Sec. 2.7.
$endgroup$
– Liviu Nicolaescu
yesterday
$begingroup$
In the scalar case all you need for existence and uniqueness is that $f$ is decreasing and the function $mathbb{R}ni xmapsto f(x)-xinmathbb{R}$ is onto.
$endgroup$
– Liviu Nicolaescu
yesterday
$begingroup$
Thank you. What if $f$ is increasing?
$endgroup$
– Hiro
yesterday
|
show 4 more comments
$begingroup$
Suppose that $f$ is decreasing in $x$. Let $x(t)$, $y(t)$ be two solutions of the ode. Then
$$
dot{x}-dot{y}= f(x,t)-f(y,t).
$$
Multiplying both sides by $x-y$ we deduce
$$
(dot{x}-dot{y})(x-y) =big(f(x,t)-f(y,t)big)(x-y)leq 0,
$$
where the last equality holds because $f$ is decreasing.
Hence
$$
frac{1}{2}frac{d}{dt}big(x-y)^2leq 0.
$$
Thus the function $tmapsto big( x(t)-y(t)big)^2 $ is decreasing so
$$
big(x(t)-y(t)big)^2leq big( x(0)-y(0)big)^2,;;forall tgeq 0,
$$
i.e.,
$$
Big(Phi(x_0,t)-Phi(y_0,t)Big)^2leq Big(x_0-y_0Big)^2,;;forall tgeq 0.
$$
In other words, for $tgeq 0$, $Phi(x,t)$ is Lipschitz in $x$ with Lipschitz constant $1$ if $f$ is decreasing.
$endgroup$
Suppose that $f$ is decreasing in $x$. Let $x(t)$, $y(t)$ be two solutions of the ode. Then
$$
dot{x}-dot{y}= f(x,t)-f(y,t).
$$
Multiplying both sides by $x-y$ we deduce
$$
(dot{x}-dot{y})(x-y) =big(f(x,t)-f(y,t)big)(x-y)leq 0,
$$
where the last equality holds because $f$ is decreasing.
Hence
$$
frac{1}{2}frac{d}{dt}big(x-y)^2leq 0.
$$
Thus the function $tmapsto big( x(t)-y(t)big)^2 $ is decreasing so
$$
big(x(t)-y(t)big)^2leq big( x(0)-y(0)big)^2,;;forall tgeq 0,
$$
i.e.,
$$
Big(Phi(x_0,t)-Phi(y_0,t)Big)^2leq Big(x_0-y_0Big)^2,;;forall tgeq 0.
$$
In other words, for $tgeq 0$, $Phi(x,t)$ is Lipschitz in $x$ with Lipschitz constant $1$ if $f$ is decreasing.
edited yesterday
answered yesterday
Liviu NicolaescuLiviu Nicolaescu
26.1k260112
26.1k260112
$begingroup$
Thank you. How can the argument be made rigorous even when $f$ is not smooth and $Phi$ is not a classical solution but a regular Lagrangian flow?
$endgroup$
– Hiro
yesterday
$begingroup$
The function $f$ coud even be multivalued, and you can work in an infinite dimensional Hilbert space as well This is a special case of the general theory of maximal monotone operators and the associated differential equations. Perhaps the friendliest introduction is Brezis' book Operateurs maximaux monotones et semi-groupes de contractions dans les espaces de Hilbert. The ultimate reference is however V. Barbu's book Nonlinear semigroups and differen tial equations in Banach spaces
$endgroup$
– Liviu Nicolaescu
yesterday
$begingroup$
The finite dimensional case is discussed in V. Barbu's recent book Differential Equations Springer 2016, Example 2.4 and Sec. 2.7.
$endgroup$
– Liviu Nicolaescu
yesterday
$begingroup$
In the scalar case all you need for existence and uniqueness is that $f$ is decreasing and the function $mathbb{R}ni xmapsto f(x)-xinmathbb{R}$ is onto.
$endgroup$
– Liviu Nicolaescu
yesterday
$begingroup$
Thank you. What if $f$ is increasing?
$endgroup$
– Hiro
yesterday
|
show 4 more comments
$begingroup$
Thank you. How can the argument be made rigorous even when $f$ is not smooth and $Phi$ is not a classical solution but a regular Lagrangian flow?
$endgroup$
– Hiro
yesterday
$begingroup$
The function $f$ coud even be multivalued, and you can work in an infinite dimensional Hilbert space as well This is a special case of the general theory of maximal monotone operators and the associated differential equations. Perhaps the friendliest introduction is Brezis' book Operateurs maximaux monotones et semi-groupes de contractions dans les espaces de Hilbert. The ultimate reference is however V. Barbu's book Nonlinear semigroups and differen tial equations in Banach spaces
$endgroup$
– Liviu Nicolaescu
yesterday
$begingroup$
The finite dimensional case is discussed in V. Barbu's recent book Differential Equations Springer 2016, Example 2.4 and Sec. 2.7.
$endgroup$
– Liviu Nicolaescu
yesterday
$begingroup$
In the scalar case all you need for existence and uniqueness is that $f$ is decreasing and the function $mathbb{R}ni xmapsto f(x)-xinmathbb{R}$ is onto.
$endgroup$
– Liviu Nicolaescu
yesterday
$begingroup$
Thank you. What if $f$ is increasing?
$endgroup$
– Hiro
yesterday
$begingroup$
Thank you. How can the argument be made rigorous even when $f$ is not smooth and $Phi$ is not a classical solution but a regular Lagrangian flow?
$endgroup$
– Hiro
yesterday
$begingroup$
Thank you. How can the argument be made rigorous even when $f$ is not smooth and $Phi$ is not a classical solution but a regular Lagrangian flow?
$endgroup$
– Hiro
yesterday
$begingroup$
The function $f$ coud even be multivalued, and you can work in an infinite dimensional Hilbert space as well This is a special case of the general theory of maximal monotone operators and the associated differential equations. Perhaps the friendliest introduction is Brezis' book Operateurs maximaux monotones et semi-groupes de contractions dans les espaces de Hilbert. The ultimate reference is however V. Barbu's book Nonlinear semigroups and differen tial equations in Banach spaces
$endgroup$
– Liviu Nicolaescu
yesterday
$begingroup$
The function $f$ coud even be multivalued, and you can work in an infinite dimensional Hilbert space as well This is a special case of the general theory of maximal monotone operators and the associated differential equations. Perhaps the friendliest introduction is Brezis' book Operateurs maximaux monotones et semi-groupes de contractions dans les espaces de Hilbert. The ultimate reference is however V. Barbu's book Nonlinear semigroups and differen tial equations in Banach spaces
$endgroup$
– Liviu Nicolaescu
yesterday
$begingroup$
The finite dimensional case is discussed in V. Barbu's recent book Differential Equations Springer 2016, Example 2.4 and Sec. 2.7.
$endgroup$
– Liviu Nicolaescu
yesterday
$begingroup$
The finite dimensional case is discussed in V. Barbu's recent book Differential Equations Springer 2016, Example 2.4 and Sec. 2.7.
$endgroup$
– Liviu Nicolaescu
yesterday
$begingroup$
In the scalar case all you need for existence and uniqueness is that $f$ is decreasing and the function $mathbb{R}ni xmapsto f(x)-xinmathbb{R}$ is onto.
$endgroup$
– Liviu Nicolaescu
yesterday
$begingroup$
In the scalar case all you need for existence and uniqueness is that $f$ is decreasing and the function $mathbb{R}ni xmapsto f(x)-xinmathbb{R}$ is onto.
$endgroup$
– Liviu Nicolaescu
yesterday
$begingroup$
Thank you. What if $f$ is increasing?
$endgroup$
– Hiro
yesterday
$begingroup$
Thank you. What if $f$ is increasing?
$endgroup$
– Hiro
yesterday
|
show 4 more comments
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