Flow of ODE with monotone source Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?On properties of Wronskians of ODEPolar coordinates, bounded domain with $C^1$ boundaryLaplace problem with Robin boundary condition on a wedgeODE with Holder drift - Cauchy-Peano theoremQuantitative finite speed of propagation property for ODE (cone of dependence)The regularity of ODE with Zygmund coefficientsDerivative and Jacobian determinant of solution of ODEJacobian and Jacobian matrix of solutions of ODE with Sobolev vector fieldModulus of continuity of flow for non-Lipschitz vector fields satisfies Osgood conditionDifference quotient for solutions of ODE and Liouville equation
Flow of ODE with monotone source
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)
Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?On properties of Wronskians of ODEPolar coordinates, bounded domain with $C^1$ boundaryLaplace problem with Robin boundary condition on a wedgeODE with Holder drift - Cauchy-Peano theoremQuantitative finite speed of propagation property for ODE (cone of dependence)The regularity of ODE with Zygmund coefficientsDerivative and Jacobian determinant of solution of ODEJacobian and Jacobian matrix of solutions of ODE with Sobolev vector fieldModulus of continuity of flow for non-Lipschitz vector fields satisfies Osgood conditionDifference quotient for solutions of ODE and Liouville equation
$begingroup$
Let $Phi$ be the flow (defined as in page 6 of this paper) of the ODE
$$begincases
fracddtPhi(x,t) = f(Phi(x,t),t) quad t >0 \
Phi(x,0) = x quad x in mathbbR.
endcases$$
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
asked Apr 16 at 10:14
HiroHiro
697
$endgroup$
add a comment |
$begingroup$
Let $Phi$ be the flow (defined as in page 6 of this paper) of the ODE
$$begincases
fracddtPhi(x,t) = f(Phi(x,t),t) quad t >0 \
Phi(x,0) = x quad x in mathbbR.
endcases$$
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
asked Apr 16 at 10:14
HiroHiro
697
$endgroup$
add a comment |
$begingroup$
Let $Phi$ be the flow (defined as in page 6 of this paper) of the ODE
$$begincases
fracddtPhi(x,t) = f(Phi(x,t),t) quad t >0 \
Phi(x,0) = x quad x in mathbbR.
endcases$$
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
asked Apr 16 at 10:14
HiroHiro
697
$endgroup$
Let $Phi$ be the flow (defined as in page 6 of this paper) of the ODE
$$begincases
fracddtPhi(x,t) = f(Phi(x,t),t) quad t >0 \
Phi(x,0) = x quad x in mathbbR.
endcases$$
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 Apr 16 at 10:14
HiroHiro
697
asked Apr 16 at 10:14
HiroHiro
697
asked Apr 16 at 10:14
HiroHiro
697
asked Apr 16 at 10:14
HiroHiro
697
asked Apr 16 at 10:14
HiroHiro
697
697
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Suppose that $f$ is decreasing in $x$. Let $x(t)$, $y(t)$ be two solutions of the ode. Then
$$
dotx-doty= f(x,t)-f(y,t).
$$
Multiplying both sides by $x-y$ we deduce
$$
(dotx-doty)(x-y) =big(f(x,t)-f(y,t)big)(x-y)leq 0,
$$
where the last equality holds because $f$ is decreasing.
Hence
$$
frac12fracddtbig(x-y)^2leq 0.
$$
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.
answered Apr 16 at 10:44
Liviu NicolaescuLiviu Nicolaescu
26.2k260113
$endgroup$
$begingroup$
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
Apr 16 at 11:48
$begingroup$
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
Apr 16 at 12:05
$begingroup$
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
Apr 16 at 12:08
$begingroup$
In the scalar case all you need for existence and uniqueness is that $f$ is decreasing and the function $mathbbRni xmapsto f(x)-xinmathbbR$ is onto.
$endgroup$
– Liviu Nicolaescu
Apr 16 at 14:30
$begingroup$
Thank you. What if $f$ is increasing?
$endgroup$
– Hiro
Apr 16 at 23:48
|
show 4 more comments
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1 Answer
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1 Answer
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$begingroup$
Suppose that $f$ is decreasing in $x$. Let $x(t)$, $y(t)$ be two solutions of the ode. Then
$$
dotx-doty= f(x,t)-f(y,t).
$$
Multiplying both sides by $x-y$ we deduce
$$
(dotx-doty)(x-y) =big(f(x,t)-f(y,t)big)(x-y)leq 0,
$$
where the last equality holds because $f$ is decreasing.
Hence
$$
frac12fracddtbig(x-y)^2leq 0.
$$
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.
answered Apr 16 at 10:44
Liviu NicolaescuLiviu Nicolaescu
26.2k260113
$endgroup$
$begingroup$
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
Apr 16 at 11:48
$begingroup$
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
Apr 16 at 12:05
$begingroup$
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
Apr 16 at 12:08
$begingroup$
In the scalar case all you need for existence and uniqueness is that $f$ is decreasing and the function $mathbbRni xmapsto f(x)-xinmathbbR$ is onto.
$endgroup$
– Liviu Nicolaescu
Apr 16 at 14:30
$begingroup$
Thank you. What if $f$ is increasing?
$endgroup$
– Hiro
Apr 16 at 23:48
|
show 4 more comments
$begingroup$
Suppose that $f$ is decreasing in $x$. Let $x(t)$, $y(t)$ be two solutions of the ode. Then
$$
dotx-doty= f(x,t)-f(y,t).
$$
Multiplying both sides by $x-y$ we deduce
$$
(dotx-doty)(x-y) =big(f(x,t)-f(y,t)big)(x-y)leq 0,
$$
where the last equality holds because $f$ is decreasing.
Hence
$$
frac12fracddtbig(x-y)^2leq 0.
$$
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.
answered Apr 16 at 10:44
Liviu NicolaescuLiviu Nicolaescu
26.2k260113
$endgroup$
$begingroup$
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
Apr 16 at 11:48
$begingroup$
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
Apr 16 at 12:05
$begingroup$
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
Apr 16 at 12:08
$begingroup$
In the scalar case all you need for existence and uniqueness is that $f$ is decreasing and the function $mathbbRni xmapsto f(x)-xinmathbbR$ is onto.
$endgroup$
– Liviu Nicolaescu
Apr 16 at 14:30
$begingroup$
Thank you. What if $f$ is increasing?
$endgroup$
– Hiro
Apr 16 at 23:48
|
show 4 more comments
$begingroup$
Suppose that $f$ is decreasing in $x$. Let $x(t)$, $y(t)$ be two solutions of the ode. Then
$$
dotx-doty= f(x,t)-f(y,t).
$$
Multiplying both sides by $x-y$ we deduce
$$
(dotx-doty)(x-y) =big(f(x,t)-f(y,t)big)(x-y)leq 0,
$$
where the last equality holds because $f$ is decreasing.
Hence
$$
frac12fracddtbig(x-y)^2leq 0.
$$
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.
answered Apr 16 at 10:44
Liviu NicolaescuLiviu Nicolaescu
26.2k260113
$endgroup$
Suppose that $f$ is decreasing in $x$. Let $x(t)$, $y(t)$ be two solutions of the ode. Then
$$
dotx-doty= f(x,t)-f(y,t).
$$
Multiplying both sides by $x-y$ we deduce
$$
(dotx-doty)(x-y) =big(f(x,t)-f(y,t)big)(x-y)leq 0,
$$
where the last equality holds because $f$ is decreasing.
Hence
$$
frac12fracddtbig(x-y)^2leq 0.
$$
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.
answered Apr 16 at 10:44
Liviu NicolaescuLiviu Nicolaescu
26.2k260113
edited Apr 16 at 20:06
answered Apr 16 at 10:44
Liviu NicolaescuLiviu Nicolaescu
26.2k260113
answered Apr 16 at 10:44
Liviu NicolaescuLiviu Nicolaescu
26.2k260113
answered Apr 16 at 10:44
Liviu NicolaescuLiviu Nicolaescu
26.2k260113
26.2k260113
$begingroup$
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
Apr 16 at 11:48
$begingroup$
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
Apr 16 at 12:05
$begingroup$
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
Apr 16 at 12:08
$begingroup$
In the scalar case all you need for existence and uniqueness is that $f$ is decreasing and the function $mathbbRni xmapsto f(x)-xinmathbbR$ is onto.
$endgroup$
– Liviu Nicolaescu
Apr 16 at 14:30
$begingroup$
Thank you. What if $f$ is increasing?
$endgroup$
– Hiro
Apr 16 at 23:48
|
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
Apr 16 at 11:48
$begingroup$
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
Apr 16 at 12:05
$begingroup$
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
Apr 16 at 12:08
$begingroup$
In the scalar case all you need for existence and uniqueness is that $f$ is decreasing and the function $mathbbRni xmapsto f(x)-xinmathbbR$ is onto.
$endgroup$
– Liviu Nicolaescu
Apr 16 at 14:30
$begingroup$
Thank you. What if $f$ is increasing?
$endgroup$
– Hiro
Apr 16 at 23:48
$begingroup$
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
Apr 16 at 11:48
$begingroup$
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
Apr 16 at 11:48
$begingroup$
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
Apr 16 at 12:05
$begingroup$
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
Apr 16 at 12:05
$begingroup$
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
Apr 16 at 12:08
$begingroup$
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
Apr 16 at 12:08
$begingroup$
In the scalar case all you need for existence and uniqueness is that $f$ is decreasing and the function $mathbbRni xmapsto f(x)-xinmathbbR$ is onto.
$endgroup$
– Liviu Nicolaescu
Apr 16 at 14:30
$begingroup$
In the scalar case all you need for existence and uniqueness is that $f$ is decreasing and the function $mathbbRni xmapsto f(x)-xinmathbbR$ is onto.
$endgroup$
– Liviu Nicolaescu
Apr 16 at 14:30
$begingroup$
Thank you. What if $f$ is increasing?
$endgroup$
– Hiro
Apr 16 at 23:48
$begingroup$
Thank you. What if $f$ is increasing?
$endgroup$
– Hiro
Apr 16 at 23:48
|
show 4 more comments
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