Flow of ODE with monotone source Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)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
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)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 18 hours ago
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 18 hours ago
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 18 hours ago
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 18 hours ago
HiroHiro
697
asked 18 hours ago
HiroHiro
697
asked 18 hours ago
HiroHiro
697
asked 18 hours ago
HiroHiro
697
asked 18 hours ago
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 17 hours ago
Liviu NicolaescuLiviu Nicolaescu
26.1k260112
$endgroup$
$begingroup$
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
16 hours ago
$begingroup$
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
16 hours ago
$begingroup$
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
16 hours ago
$begingroup$
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
14 hours ago
$begingroup$
Thank you. What if $f$ is increasing?
$endgroup$
– Hiro
4 hours ago
add a comment |
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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 17 hours ago
Liviu NicolaescuLiviu Nicolaescu
26.1k260112
$endgroup$
$begingroup$
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
16 hours ago
$begingroup$
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
16 hours ago
$begingroup$
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
16 hours ago
$begingroup$
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
14 hours ago
$begingroup$
Thank you. What if $f$ is increasing?
$endgroup$
– Hiro
4 hours ago
add a comment |
$begingroup$
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 17 hours ago
Liviu NicolaescuLiviu Nicolaescu
26.1k260112
$endgroup$
$begingroup$
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
16 hours ago
$begingroup$
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
16 hours ago
$begingroup$
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
16 hours ago
$begingroup$
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
14 hours ago
$begingroup$
Thank you. What if $f$ is increasing?
$endgroup$
– Hiro
4 hours ago
add a comment |
$begingroup$
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 17 hours ago
Liviu NicolaescuLiviu Nicolaescu
26.1k260112
$endgroup$
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 17 hours ago
Liviu NicolaescuLiviu Nicolaescu
26.1k260112
edited 8 hours ago
answered 17 hours ago
Liviu NicolaescuLiviu Nicolaescu
26.1k260112
answered 17 hours ago
Liviu NicolaescuLiviu Nicolaescu
26.1k260112
answered 17 hours ago
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
16 hours ago
$begingroup$
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
16 hours ago
$begingroup$
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
16 hours ago
$begingroup$
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
14 hours ago
$begingroup$
Thank you. What if $f$ is increasing?
$endgroup$
– Hiro
4 hours ago
add a comment |
$begingroup$
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
16 hours ago
$begingroup$
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
16 hours ago
$begingroup$
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
16 hours ago
$begingroup$
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
14 hours ago
$begingroup$
Thank you. What if $f$ is increasing?
$endgroup$
– Hiro
4 hours ago
$begingroup$
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
16 hours ago
$begingroup$
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
16 hours ago
$begingroup$
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
16 hours ago
$begingroup$
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
16 hours ago
$begingroup$
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
16 hours ago
$begingroup$
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
16 hours ago
$begingroup$
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
14 hours ago
$begingroup$
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
14 hours ago
$begingroup$
Thank you. What if $f$ is increasing?
$endgroup$
– Hiro
4 hours ago
$begingroup$
Thank you. What if $f$ is increasing?
$endgroup$
– Hiro
4 hours ago
add a comment |
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