When does gradient flow not converge?












25












$begingroup$


I've been thinking about gradient flows in the context of Morse theory, where we take a differentiable-enough function $f$ on some space (for now let's say a compact Riemannian manifold $M$) and use the trajectories of the gradient flow $x'(t) = - operatorname{grad} f(x(t))$ to analyse the space. In particular the (un)stable manifolds $$W^pm(p) = { x in M | x to p textrm{ under gradient flow as } t to pm infty}$$of critical points $p$ must fill up the whole space, which means that the gradient flow from each point must converge to a critical point of $f$.



Most references (I've been using Jost's Riemannian Geometry and Geometric Analysis) simply claim that when $f$ is Morse (has non-degenerate Hessian at all critical points) the gradient flow always converges and then move on, without any discussion of what can go wrong in the degenerate cases. For the purposes of Morse theory this is no problem (since there are a wealth of Morse functions anyway), but I'm curious about what the counterexamples look like.



I have no trouble proving that the flow converges in the Morse case, but I've had trouble finding an example of a function and an initial point for which the flow does not converge. In such a case the flow would necessarily be asymptotic to some subset of the critical set of $f$, just not necessarily to a single critical point. I discussed this with someone a while ago and was told that the flow would always converge in the real-analytic category, so any local examples would be given by non-analytic $f$. It was suggested to me that a function whose graph over a flat subset looked like an infinitely long "groove" cut out of a bump (something like
enter image description here
with the edges of the groove smoothed out) would do the job (with the flow proceeding down the groove forever); but upon further thought it seems such a function would fail to be differentiable at the accumulation circle. (Edit: in retrospect, I think this idea does work - if the depth of the groove decayed fast enough, such a function can be differentiable, just not analytic.)



Does anyone know any non-trivial (there are the obvious examples of unbounded functions on non-compact manifolds) examples of how the a trajectory of gradient flow can fail to converge? I'm particularly interested in the simple case of compact manifolds, but anything is welcome. Necessary conditions (or sufficient conditions less restrictive than Morse) for convergence might also be interesting.










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Maybe you should take a look at Morse-Bott functions first. This is further generalization of Morse functions that allows critical subsets to be closed manifolds other from discrete set of points. As far as I see, the provided example fits nice in that class. Also, "groovy" example can be simply modeled by function in cylindrical coordinates $(r, phi, z)$, like $z = e^{-r}cdot cos r$, which seems to be differentiable everywhere.
    $endgroup$
    – Evgeny
    Sep 23 '13 at 3:55












  • $begingroup$
    @Evgeny: thanks, I'll look in to Morse-Bott functions. I think my description of the groove example was lacking - I've updated with a more accurate plot, hopefully it shows what I'm trying to describe.
    $endgroup$
    – Anthony Carapetis
    Sep 23 '13 at 4:47










  • $begingroup$
    I think your "groove" example is fairly universal -- if any counter-example exists it would have to be more or less of that type. Have you ever heard someone claim there are counter-examples? Maybe there aren't.
    $endgroup$
    – Ryan Budney
    Sep 26 '13 at 19:28










  • $begingroup$
    @Ryan: I've come across an (alleged) example of non-convergence for geodesic heat flow, which is the gradient flow of the energy functional on the Hilbert manifold of $W^{1,2}$ loops in a Riemannian manifold. I was hoping for a more elementary example; but after going back to investigate the details just now it seems to come from a gradient flow on the manifold - I'll check it out. I'm still very much interested in the flat case - if a trajectory of the gradient flow of a function on $(mathbb R^n,delta)$ is bounded, must it converge?
    $endgroup$
    – Anthony Carapetis
    Sep 27 '13 at 1:42






  • 3




    $begingroup$
    Hi Anthony -- in my paper on generalised Helfrich flow of space curves I have a few examples of functionals that give rise to nonconvergence; I talk both about your groove possibility as well as some solutions of translation type. These are all gradient flows. (Unlike e.g. IMCF of a star-shaped hypersurface.)
    $endgroup$
    – Glen Wheeler
    Jul 17 '15 at 11:58
















25












$begingroup$


I've been thinking about gradient flows in the context of Morse theory, where we take a differentiable-enough function $f$ on some space (for now let's say a compact Riemannian manifold $M$) and use the trajectories of the gradient flow $x'(t) = - operatorname{grad} f(x(t))$ to analyse the space. In particular the (un)stable manifolds $$W^pm(p) = { x in M | x to p textrm{ under gradient flow as } t to pm infty}$$of critical points $p$ must fill up the whole space, which means that the gradient flow from each point must converge to a critical point of $f$.



Most references (I've been using Jost's Riemannian Geometry and Geometric Analysis) simply claim that when $f$ is Morse (has non-degenerate Hessian at all critical points) the gradient flow always converges and then move on, without any discussion of what can go wrong in the degenerate cases. For the purposes of Morse theory this is no problem (since there are a wealth of Morse functions anyway), but I'm curious about what the counterexamples look like.



I have no trouble proving that the flow converges in the Morse case, but I've had trouble finding an example of a function and an initial point for which the flow does not converge. In such a case the flow would necessarily be asymptotic to some subset of the critical set of $f$, just not necessarily to a single critical point. I discussed this with someone a while ago and was told that the flow would always converge in the real-analytic category, so any local examples would be given by non-analytic $f$. It was suggested to me that a function whose graph over a flat subset looked like an infinitely long "groove" cut out of a bump (something like
enter image description here
with the edges of the groove smoothed out) would do the job (with the flow proceeding down the groove forever); but upon further thought it seems such a function would fail to be differentiable at the accumulation circle. (Edit: in retrospect, I think this idea does work - if the depth of the groove decayed fast enough, such a function can be differentiable, just not analytic.)



Does anyone know any non-trivial (there are the obvious examples of unbounded functions on non-compact manifolds) examples of how the a trajectory of gradient flow can fail to converge? I'm particularly interested in the simple case of compact manifolds, but anything is welcome. Necessary conditions (or sufficient conditions less restrictive than Morse) for convergence might also be interesting.










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Maybe you should take a look at Morse-Bott functions first. This is further generalization of Morse functions that allows critical subsets to be closed manifolds other from discrete set of points. As far as I see, the provided example fits nice in that class. Also, "groovy" example can be simply modeled by function in cylindrical coordinates $(r, phi, z)$, like $z = e^{-r}cdot cos r$, which seems to be differentiable everywhere.
    $endgroup$
    – Evgeny
    Sep 23 '13 at 3:55












  • $begingroup$
    @Evgeny: thanks, I'll look in to Morse-Bott functions. I think my description of the groove example was lacking - I've updated with a more accurate plot, hopefully it shows what I'm trying to describe.
    $endgroup$
    – Anthony Carapetis
    Sep 23 '13 at 4:47










  • $begingroup$
    I think your "groove" example is fairly universal -- if any counter-example exists it would have to be more or less of that type. Have you ever heard someone claim there are counter-examples? Maybe there aren't.
    $endgroup$
    – Ryan Budney
    Sep 26 '13 at 19:28










  • $begingroup$
    @Ryan: I've come across an (alleged) example of non-convergence for geodesic heat flow, which is the gradient flow of the energy functional on the Hilbert manifold of $W^{1,2}$ loops in a Riemannian manifold. I was hoping for a more elementary example; but after going back to investigate the details just now it seems to come from a gradient flow on the manifold - I'll check it out. I'm still very much interested in the flat case - if a trajectory of the gradient flow of a function on $(mathbb R^n,delta)$ is bounded, must it converge?
    $endgroup$
    – Anthony Carapetis
    Sep 27 '13 at 1:42






  • 3




    $begingroup$
    Hi Anthony -- in my paper on generalised Helfrich flow of space curves I have a few examples of functionals that give rise to nonconvergence; I talk both about your groove possibility as well as some solutions of translation type. These are all gradient flows. (Unlike e.g. IMCF of a star-shaped hypersurface.)
    $endgroup$
    – Glen Wheeler
    Jul 17 '15 at 11:58














25












25








25


13



$begingroup$


I've been thinking about gradient flows in the context of Morse theory, where we take a differentiable-enough function $f$ on some space (for now let's say a compact Riemannian manifold $M$) and use the trajectories of the gradient flow $x'(t) = - operatorname{grad} f(x(t))$ to analyse the space. In particular the (un)stable manifolds $$W^pm(p) = { x in M | x to p textrm{ under gradient flow as } t to pm infty}$$of critical points $p$ must fill up the whole space, which means that the gradient flow from each point must converge to a critical point of $f$.



Most references (I've been using Jost's Riemannian Geometry and Geometric Analysis) simply claim that when $f$ is Morse (has non-degenerate Hessian at all critical points) the gradient flow always converges and then move on, without any discussion of what can go wrong in the degenerate cases. For the purposes of Morse theory this is no problem (since there are a wealth of Morse functions anyway), but I'm curious about what the counterexamples look like.



I have no trouble proving that the flow converges in the Morse case, but I've had trouble finding an example of a function and an initial point for which the flow does not converge. In such a case the flow would necessarily be asymptotic to some subset of the critical set of $f$, just not necessarily to a single critical point. I discussed this with someone a while ago and was told that the flow would always converge in the real-analytic category, so any local examples would be given by non-analytic $f$. It was suggested to me that a function whose graph over a flat subset looked like an infinitely long "groove" cut out of a bump (something like
enter image description here
with the edges of the groove smoothed out) would do the job (with the flow proceeding down the groove forever); but upon further thought it seems such a function would fail to be differentiable at the accumulation circle. (Edit: in retrospect, I think this idea does work - if the depth of the groove decayed fast enough, such a function can be differentiable, just not analytic.)



Does anyone know any non-trivial (there are the obvious examples of unbounded functions on non-compact manifolds) examples of how the a trajectory of gradient flow can fail to converge? I'm particularly interested in the simple case of compact manifolds, but anything is welcome. Necessary conditions (or sufficient conditions less restrictive than Morse) for convergence might also be interesting.










share|cite|improve this question











$endgroup$




I've been thinking about gradient flows in the context of Morse theory, where we take a differentiable-enough function $f$ on some space (for now let's say a compact Riemannian manifold $M$) and use the trajectories of the gradient flow $x'(t) = - operatorname{grad} f(x(t))$ to analyse the space. In particular the (un)stable manifolds $$W^pm(p) = { x in M | x to p textrm{ under gradient flow as } t to pm infty}$$of critical points $p$ must fill up the whole space, which means that the gradient flow from each point must converge to a critical point of $f$.



Most references (I've been using Jost's Riemannian Geometry and Geometric Analysis) simply claim that when $f$ is Morse (has non-degenerate Hessian at all critical points) the gradient flow always converges and then move on, without any discussion of what can go wrong in the degenerate cases. For the purposes of Morse theory this is no problem (since there are a wealth of Morse functions anyway), but I'm curious about what the counterexamples look like.



I have no trouble proving that the flow converges in the Morse case, but I've had trouble finding an example of a function and an initial point for which the flow does not converge. In such a case the flow would necessarily be asymptotic to some subset of the critical set of $f$, just not necessarily to a single critical point. I discussed this with someone a while ago and was told that the flow would always converge in the real-analytic category, so any local examples would be given by non-analytic $f$. It was suggested to me that a function whose graph over a flat subset looked like an infinitely long "groove" cut out of a bump (something like
enter image description here
with the edges of the groove smoothed out) would do the job (with the flow proceeding down the groove forever); but upon further thought it seems such a function would fail to be differentiable at the accumulation circle. (Edit: in retrospect, I think this idea does work - if the depth of the groove decayed fast enough, such a function can be differentiable, just not analytic.)



Does anyone know any non-trivial (there are the obvious examples of unbounded functions on non-compact manifolds) examples of how the a trajectory of gradient flow can fail to converge? I'm particularly interested in the simple case of compact manifolds, but anything is welcome. Necessary conditions (or sufficient conditions less restrictive than Morse) for convergence might also be interesting.







ordinary-differential-equations differential-geometry dynamical-systems morse-theory gradient-flows






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 12 '18 at 6:37







Anthony Carapetis

















asked Sep 22 '13 at 6:36









Anthony CarapetisAnthony Carapetis

27.3k32865




27.3k32865








  • 1




    $begingroup$
    Maybe you should take a look at Morse-Bott functions first. This is further generalization of Morse functions that allows critical subsets to be closed manifolds other from discrete set of points. As far as I see, the provided example fits nice in that class. Also, "groovy" example can be simply modeled by function in cylindrical coordinates $(r, phi, z)$, like $z = e^{-r}cdot cos r$, which seems to be differentiable everywhere.
    $endgroup$
    – Evgeny
    Sep 23 '13 at 3:55












  • $begingroup$
    @Evgeny: thanks, I'll look in to Morse-Bott functions. I think my description of the groove example was lacking - I've updated with a more accurate plot, hopefully it shows what I'm trying to describe.
    $endgroup$
    – Anthony Carapetis
    Sep 23 '13 at 4:47










  • $begingroup$
    I think your "groove" example is fairly universal -- if any counter-example exists it would have to be more or less of that type. Have you ever heard someone claim there are counter-examples? Maybe there aren't.
    $endgroup$
    – Ryan Budney
    Sep 26 '13 at 19:28










  • $begingroup$
    @Ryan: I've come across an (alleged) example of non-convergence for geodesic heat flow, which is the gradient flow of the energy functional on the Hilbert manifold of $W^{1,2}$ loops in a Riemannian manifold. I was hoping for a more elementary example; but after going back to investigate the details just now it seems to come from a gradient flow on the manifold - I'll check it out. I'm still very much interested in the flat case - if a trajectory of the gradient flow of a function on $(mathbb R^n,delta)$ is bounded, must it converge?
    $endgroup$
    – Anthony Carapetis
    Sep 27 '13 at 1:42






  • 3




    $begingroup$
    Hi Anthony -- in my paper on generalised Helfrich flow of space curves I have a few examples of functionals that give rise to nonconvergence; I talk both about your groove possibility as well as some solutions of translation type. These are all gradient flows. (Unlike e.g. IMCF of a star-shaped hypersurface.)
    $endgroup$
    – Glen Wheeler
    Jul 17 '15 at 11:58














  • 1




    $begingroup$
    Maybe you should take a look at Morse-Bott functions first. This is further generalization of Morse functions that allows critical subsets to be closed manifolds other from discrete set of points. As far as I see, the provided example fits nice in that class. Also, "groovy" example can be simply modeled by function in cylindrical coordinates $(r, phi, z)$, like $z = e^{-r}cdot cos r$, which seems to be differentiable everywhere.
    $endgroup$
    – Evgeny
    Sep 23 '13 at 3:55












  • $begingroup$
    @Evgeny: thanks, I'll look in to Morse-Bott functions. I think my description of the groove example was lacking - I've updated with a more accurate plot, hopefully it shows what I'm trying to describe.
    $endgroup$
    – Anthony Carapetis
    Sep 23 '13 at 4:47










  • $begingroup$
    I think your "groove" example is fairly universal -- if any counter-example exists it would have to be more or less of that type. Have you ever heard someone claim there are counter-examples? Maybe there aren't.
    $endgroup$
    – Ryan Budney
    Sep 26 '13 at 19:28










  • $begingroup$
    @Ryan: I've come across an (alleged) example of non-convergence for geodesic heat flow, which is the gradient flow of the energy functional on the Hilbert manifold of $W^{1,2}$ loops in a Riemannian manifold. I was hoping for a more elementary example; but after going back to investigate the details just now it seems to come from a gradient flow on the manifold - I'll check it out. I'm still very much interested in the flat case - if a trajectory of the gradient flow of a function on $(mathbb R^n,delta)$ is bounded, must it converge?
    $endgroup$
    – Anthony Carapetis
    Sep 27 '13 at 1:42






  • 3




    $begingroup$
    Hi Anthony -- in my paper on generalised Helfrich flow of space curves I have a few examples of functionals that give rise to nonconvergence; I talk both about your groove possibility as well as some solutions of translation type. These are all gradient flows. (Unlike e.g. IMCF of a star-shaped hypersurface.)
    $endgroup$
    – Glen Wheeler
    Jul 17 '15 at 11:58








1




1




$begingroup$
Maybe you should take a look at Morse-Bott functions first. This is further generalization of Morse functions that allows critical subsets to be closed manifolds other from discrete set of points. As far as I see, the provided example fits nice in that class. Also, "groovy" example can be simply modeled by function in cylindrical coordinates $(r, phi, z)$, like $z = e^{-r}cdot cos r$, which seems to be differentiable everywhere.
$endgroup$
– Evgeny
Sep 23 '13 at 3:55






$begingroup$
Maybe you should take a look at Morse-Bott functions first. This is further generalization of Morse functions that allows critical subsets to be closed manifolds other from discrete set of points. As far as I see, the provided example fits nice in that class. Also, "groovy" example can be simply modeled by function in cylindrical coordinates $(r, phi, z)$, like $z = e^{-r}cdot cos r$, which seems to be differentiable everywhere.
$endgroup$
– Evgeny
Sep 23 '13 at 3:55














$begingroup$
@Evgeny: thanks, I'll look in to Morse-Bott functions. I think my description of the groove example was lacking - I've updated with a more accurate plot, hopefully it shows what I'm trying to describe.
$endgroup$
– Anthony Carapetis
Sep 23 '13 at 4:47




$begingroup$
@Evgeny: thanks, I'll look in to Morse-Bott functions. I think my description of the groove example was lacking - I've updated with a more accurate plot, hopefully it shows what I'm trying to describe.
$endgroup$
– Anthony Carapetis
Sep 23 '13 at 4:47












$begingroup$
I think your "groove" example is fairly universal -- if any counter-example exists it would have to be more or less of that type. Have you ever heard someone claim there are counter-examples? Maybe there aren't.
$endgroup$
– Ryan Budney
Sep 26 '13 at 19:28




$begingroup$
I think your "groove" example is fairly universal -- if any counter-example exists it would have to be more or less of that type. Have you ever heard someone claim there are counter-examples? Maybe there aren't.
$endgroup$
– Ryan Budney
Sep 26 '13 at 19:28












$begingroup$
@Ryan: I've come across an (alleged) example of non-convergence for geodesic heat flow, which is the gradient flow of the energy functional on the Hilbert manifold of $W^{1,2}$ loops in a Riemannian manifold. I was hoping for a more elementary example; but after going back to investigate the details just now it seems to come from a gradient flow on the manifold - I'll check it out. I'm still very much interested in the flat case - if a trajectory of the gradient flow of a function on $(mathbb R^n,delta)$ is bounded, must it converge?
$endgroup$
– Anthony Carapetis
Sep 27 '13 at 1:42




$begingroup$
@Ryan: I've come across an (alleged) example of non-convergence for geodesic heat flow, which is the gradient flow of the energy functional on the Hilbert manifold of $W^{1,2}$ loops in a Riemannian manifold. I was hoping for a more elementary example; but after going back to investigate the details just now it seems to come from a gradient flow on the manifold - I'll check it out. I'm still very much interested in the flat case - if a trajectory of the gradient flow of a function on $(mathbb R^n,delta)$ is bounded, must it converge?
$endgroup$
– Anthony Carapetis
Sep 27 '13 at 1:42




3




3




$begingroup$
Hi Anthony -- in my paper on generalised Helfrich flow of space curves I have a few examples of functionals that give rise to nonconvergence; I talk both about your groove possibility as well as some solutions of translation type. These are all gradient flows. (Unlike e.g. IMCF of a star-shaped hypersurface.)
$endgroup$
– Glen Wheeler
Jul 17 '15 at 11:58




$begingroup$
Hi Anthony -- in my paper on generalised Helfrich flow of space curves I have a few examples of functionals that give rise to nonconvergence; I talk both about your groove possibility as well as some solutions of translation type. These are all gradient flows. (Unlike e.g. IMCF of a star-shaped hypersurface.)
$endgroup$
– Glen Wheeler
Jul 17 '15 at 11:58










1 Answer
1






active

oldest

votes


















7












$begingroup$

This is an adaptation (and slight correction) of the example of non-convergent geodesic heat flow found in Section 2 of http://www.math.msu.edu/~parker/ChoiParkerGeodesics.pdf.



All the details of interest
take place on a closed cylinder $left[0,frac{1}{2}right]times S^{1}$,
but we can use a bump function to embed the example nicely in a (compact) torus.
Let $M=mathbb{T}^{2}=left(mathbb{R}/2pimathbb{Z}right)^{2}$
with coordinates $u,v$ coming from the fundamental domain $[-pi,pi)^{2}$.
Let $eta$ be a bump function that is $1$ in $left[-1/2,1/2right]$
and $0$ outside $left[-1,1right]$. We will consider the function
$$
phileft(u,vright)=begin{cases}
etaleft(uright)e^{-1/u}left(1+sinleft(frac{1}{u}-vright)right) & textrm{if }uin(0,pi)\
0 & textrm{if }uinleft[-pi,0right]
end{cases}
$$
which one can check is smooth on the torus. We are hoping that we can make the curve
$v=frac{1}{u}$ for $uinleft(0,frac{1}{2}right)$ into a gradient flow trajectory. In this range
of $u$ we can use the coordinates $left(z,vright)$ for $z=frac{1}{u}-v$;
so we want to find a smooth metric such that the curve $left{ z=0right} $
is a gradient flow trajectory of $phi$. Note that in these coordinates
$phi$ takes the simple form $phileft(z,vright)=1+e^{-z-v}left(1+sin zright)$.
Thus we have
$$
dphi=-e^{-z-v}left(1+sin zright)dv+e^{-z-v}left(cos z-sin z-1right)dz;
$$
so when $z=0$ we have $dphi=-e^{-v}dv$. Thus a metric that has
the form $g=f^{2}dz^{2}+dv^{2}$ ($f$ a positive smooth function)
in this region would suffice, with gradient $nablaphi=-e^{-v}partial_{v}$
giving non-convergent solutions for all time of the form $left(u,vright)left(tright)=left(frac{1}{lnleft(c+tright)},lnleft(c+tright)right)$.
In the global coordinates such a metric would have the form



$$
g=u^{-4}f^{2}du^{2}+2u^{-2}f^{2}dudv+left(1+f^{2}right)dv^{2}
$$
in the region $uinleft(0,frac{1}{2}right)$. Choosing $f=u^{2}$
we have $g=du^{2}+2u^{2}dudv+left(1+u^{4}right)dv^{2}$ in this
region, which we can extend to a smooth non-degenerate metric on the
whole torus by interpolating between it and the flat metric with $eta$:
$$
g:=du^{2}+2etaleft(uright)u^{2}dudv+left(1+etaleft(uright)u^{4}right)dv^{2}.
$$
Since $eta$ is $1$ on the domain of interest this does not change
the gradient flow trajectory; so we have a smooth function on a compact
manifold with a gradient flow trajectory $left(u,vright)left(tright)=left(frac{1}{lnleft(c+tright)},lnleft(c+tright)right)$ ($c$ large enough that $1/ln(c) < 1/2$),
which has every point on the circle $left{ u=0right} $ as an
accumulation point.






share|cite|improve this answer









$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f501007%2fwhen-does-gradient-flow-not-converge%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    7












    $begingroup$

    This is an adaptation (and slight correction) of the example of non-convergent geodesic heat flow found in Section 2 of http://www.math.msu.edu/~parker/ChoiParkerGeodesics.pdf.



    All the details of interest
    take place on a closed cylinder $left[0,frac{1}{2}right]times S^{1}$,
    but we can use a bump function to embed the example nicely in a (compact) torus.
    Let $M=mathbb{T}^{2}=left(mathbb{R}/2pimathbb{Z}right)^{2}$
    with coordinates $u,v$ coming from the fundamental domain $[-pi,pi)^{2}$.
    Let $eta$ be a bump function that is $1$ in $left[-1/2,1/2right]$
    and $0$ outside $left[-1,1right]$. We will consider the function
    $$
    phileft(u,vright)=begin{cases}
    etaleft(uright)e^{-1/u}left(1+sinleft(frac{1}{u}-vright)right) & textrm{if }uin(0,pi)\
    0 & textrm{if }uinleft[-pi,0right]
    end{cases}
    $$
    which one can check is smooth on the torus. We are hoping that we can make the curve
    $v=frac{1}{u}$ for $uinleft(0,frac{1}{2}right)$ into a gradient flow trajectory. In this range
    of $u$ we can use the coordinates $left(z,vright)$ for $z=frac{1}{u}-v$;
    so we want to find a smooth metric such that the curve $left{ z=0right} $
    is a gradient flow trajectory of $phi$. Note that in these coordinates
    $phi$ takes the simple form $phileft(z,vright)=1+e^{-z-v}left(1+sin zright)$.
    Thus we have
    $$
    dphi=-e^{-z-v}left(1+sin zright)dv+e^{-z-v}left(cos z-sin z-1right)dz;
    $$
    so when $z=0$ we have $dphi=-e^{-v}dv$. Thus a metric that has
    the form $g=f^{2}dz^{2}+dv^{2}$ ($f$ a positive smooth function)
    in this region would suffice, with gradient $nablaphi=-e^{-v}partial_{v}$
    giving non-convergent solutions for all time of the form $left(u,vright)left(tright)=left(frac{1}{lnleft(c+tright)},lnleft(c+tright)right)$.
    In the global coordinates such a metric would have the form



    $$
    g=u^{-4}f^{2}du^{2}+2u^{-2}f^{2}dudv+left(1+f^{2}right)dv^{2}
    $$
    in the region $uinleft(0,frac{1}{2}right)$. Choosing $f=u^{2}$
    we have $g=du^{2}+2u^{2}dudv+left(1+u^{4}right)dv^{2}$ in this
    region, which we can extend to a smooth non-degenerate metric on the
    whole torus by interpolating between it and the flat metric with $eta$:
    $$
    g:=du^{2}+2etaleft(uright)u^{2}dudv+left(1+etaleft(uright)u^{4}right)dv^{2}.
    $$
    Since $eta$ is $1$ on the domain of interest this does not change
    the gradient flow trajectory; so we have a smooth function on a compact
    manifold with a gradient flow trajectory $left(u,vright)left(tright)=left(frac{1}{lnleft(c+tright)},lnleft(c+tright)right)$ ($c$ large enough that $1/ln(c) < 1/2$),
    which has every point on the circle $left{ u=0right} $ as an
    accumulation point.






    share|cite|improve this answer









    $endgroup$


















      7












      $begingroup$

      This is an adaptation (and slight correction) of the example of non-convergent geodesic heat flow found in Section 2 of http://www.math.msu.edu/~parker/ChoiParkerGeodesics.pdf.



      All the details of interest
      take place on a closed cylinder $left[0,frac{1}{2}right]times S^{1}$,
      but we can use a bump function to embed the example nicely in a (compact) torus.
      Let $M=mathbb{T}^{2}=left(mathbb{R}/2pimathbb{Z}right)^{2}$
      with coordinates $u,v$ coming from the fundamental domain $[-pi,pi)^{2}$.
      Let $eta$ be a bump function that is $1$ in $left[-1/2,1/2right]$
      and $0$ outside $left[-1,1right]$. We will consider the function
      $$
      phileft(u,vright)=begin{cases}
      etaleft(uright)e^{-1/u}left(1+sinleft(frac{1}{u}-vright)right) & textrm{if }uin(0,pi)\
      0 & textrm{if }uinleft[-pi,0right]
      end{cases}
      $$
      which one can check is smooth on the torus. We are hoping that we can make the curve
      $v=frac{1}{u}$ for $uinleft(0,frac{1}{2}right)$ into a gradient flow trajectory. In this range
      of $u$ we can use the coordinates $left(z,vright)$ for $z=frac{1}{u}-v$;
      so we want to find a smooth metric such that the curve $left{ z=0right} $
      is a gradient flow trajectory of $phi$. Note that in these coordinates
      $phi$ takes the simple form $phileft(z,vright)=1+e^{-z-v}left(1+sin zright)$.
      Thus we have
      $$
      dphi=-e^{-z-v}left(1+sin zright)dv+e^{-z-v}left(cos z-sin z-1right)dz;
      $$
      so when $z=0$ we have $dphi=-e^{-v}dv$. Thus a metric that has
      the form $g=f^{2}dz^{2}+dv^{2}$ ($f$ a positive smooth function)
      in this region would suffice, with gradient $nablaphi=-e^{-v}partial_{v}$
      giving non-convergent solutions for all time of the form $left(u,vright)left(tright)=left(frac{1}{lnleft(c+tright)},lnleft(c+tright)right)$.
      In the global coordinates such a metric would have the form



      $$
      g=u^{-4}f^{2}du^{2}+2u^{-2}f^{2}dudv+left(1+f^{2}right)dv^{2}
      $$
      in the region $uinleft(0,frac{1}{2}right)$. Choosing $f=u^{2}$
      we have $g=du^{2}+2u^{2}dudv+left(1+u^{4}right)dv^{2}$ in this
      region, which we can extend to a smooth non-degenerate metric on the
      whole torus by interpolating between it and the flat metric with $eta$:
      $$
      g:=du^{2}+2etaleft(uright)u^{2}dudv+left(1+etaleft(uright)u^{4}right)dv^{2}.
      $$
      Since $eta$ is $1$ on the domain of interest this does not change
      the gradient flow trajectory; so we have a smooth function on a compact
      manifold with a gradient flow trajectory $left(u,vright)left(tright)=left(frac{1}{lnleft(c+tright)},lnleft(c+tright)right)$ ($c$ large enough that $1/ln(c) < 1/2$),
      which has every point on the circle $left{ u=0right} $ as an
      accumulation point.






      share|cite|improve this answer









      $endgroup$
















        7












        7








        7





        $begingroup$

        This is an adaptation (and slight correction) of the example of non-convergent geodesic heat flow found in Section 2 of http://www.math.msu.edu/~parker/ChoiParkerGeodesics.pdf.



        All the details of interest
        take place on a closed cylinder $left[0,frac{1}{2}right]times S^{1}$,
        but we can use a bump function to embed the example nicely in a (compact) torus.
        Let $M=mathbb{T}^{2}=left(mathbb{R}/2pimathbb{Z}right)^{2}$
        with coordinates $u,v$ coming from the fundamental domain $[-pi,pi)^{2}$.
        Let $eta$ be a bump function that is $1$ in $left[-1/2,1/2right]$
        and $0$ outside $left[-1,1right]$. We will consider the function
        $$
        phileft(u,vright)=begin{cases}
        etaleft(uright)e^{-1/u}left(1+sinleft(frac{1}{u}-vright)right) & textrm{if }uin(0,pi)\
        0 & textrm{if }uinleft[-pi,0right]
        end{cases}
        $$
        which one can check is smooth on the torus. We are hoping that we can make the curve
        $v=frac{1}{u}$ for $uinleft(0,frac{1}{2}right)$ into a gradient flow trajectory. In this range
        of $u$ we can use the coordinates $left(z,vright)$ for $z=frac{1}{u}-v$;
        so we want to find a smooth metric such that the curve $left{ z=0right} $
        is a gradient flow trajectory of $phi$. Note that in these coordinates
        $phi$ takes the simple form $phileft(z,vright)=1+e^{-z-v}left(1+sin zright)$.
        Thus we have
        $$
        dphi=-e^{-z-v}left(1+sin zright)dv+e^{-z-v}left(cos z-sin z-1right)dz;
        $$
        so when $z=0$ we have $dphi=-e^{-v}dv$. Thus a metric that has
        the form $g=f^{2}dz^{2}+dv^{2}$ ($f$ a positive smooth function)
        in this region would suffice, with gradient $nablaphi=-e^{-v}partial_{v}$
        giving non-convergent solutions for all time of the form $left(u,vright)left(tright)=left(frac{1}{lnleft(c+tright)},lnleft(c+tright)right)$.
        In the global coordinates such a metric would have the form



        $$
        g=u^{-4}f^{2}du^{2}+2u^{-2}f^{2}dudv+left(1+f^{2}right)dv^{2}
        $$
        in the region $uinleft(0,frac{1}{2}right)$. Choosing $f=u^{2}$
        we have $g=du^{2}+2u^{2}dudv+left(1+u^{4}right)dv^{2}$ in this
        region, which we can extend to a smooth non-degenerate metric on the
        whole torus by interpolating between it and the flat metric with $eta$:
        $$
        g:=du^{2}+2etaleft(uright)u^{2}dudv+left(1+etaleft(uright)u^{4}right)dv^{2}.
        $$
        Since $eta$ is $1$ on the domain of interest this does not change
        the gradient flow trajectory; so we have a smooth function on a compact
        manifold with a gradient flow trajectory $left(u,vright)left(tright)=left(frac{1}{lnleft(c+tright)},lnleft(c+tright)right)$ ($c$ large enough that $1/ln(c) < 1/2$),
        which has every point on the circle $left{ u=0right} $ as an
        accumulation point.






        share|cite|improve this answer









        $endgroup$



        This is an adaptation (and slight correction) of the example of non-convergent geodesic heat flow found in Section 2 of http://www.math.msu.edu/~parker/ChoiParkerGeodesics.pdf.



        All the details of interest
        take place on a closed cylinder $left[0,frac{1}{2}right]times S^{1}$,
        but we can use a bump function to embed the example nicely in a (compact) torus.
        Let $M=mathbb{T}^{2}=left(mathbb{R}/2pimathbb{Z}right)^{2}$
        with coordinates $u,v$ coming from the fundamental domain $[-pi,pi)^{2}$.
        Let $eta$ be a bump function that is $1$ in $left[-1/2,1/2right]$
        and $0$ outside $left[-1,1right]$. We will consider the function
        $$
        phileft(u,vright)=begin{cases}
        etaleft(uright)e^{-1/u}left(1+sinleft(frac{1}{u}-vright)right) & textrm{if }uin(0,pi)\
        0 & textrm{if }uinleft[-pi,0right]
        end{cases}
        $$
        which one can check is smooth on the torus. We are hoping that we can make the curve
        $v=frac{1}{u}$ for $uinleft(0,frac{1}{2}right)$ into a gradient flow trajectory. In this range
        of $u$ we can use the coordinates $left(z,vright)$ for $z=frac{1}{u}-v$;
        so we want to find a smooth metric such that the curve $left{ z=0right} $
        is a gradient flow trajectory of $phi$. Note that in these coordinates
        $phi$ takes the simple form $phileft(z,vright)=1+e^{-z-v}left(1+sin zright)$.
        Thus we have
        $$
        dphi=-e^{-z-v}left(1+sin zright)dv+e^{-z-v}left(cos z-sin z-1right)dz;
        $$
        so when $z=0$ we have $dphi=-e^{-v}dv$. Thus a metric that has
        the form $g=f^{2}dz^{2}+dv^{2}$ ($f$ a positive smooth function)
        in this region would suffice, with gradient $nablaphi=-e^{-v}partial_{v}$
        giving non-convergent solutions for all time of the form $left(u,vright)left(tright)=left(frac{1}{lnleft(c+tright)},lnleft(c+tright)right)$.
        In the global coordinates such a metric would have the form



        $$
        g=u^{-4}f^{2}du^{2}+2u^{-2}f^{2}dudv+left(1+f^{2}right)dv^{2}
        $$
        in the region $uinleft(0,frac{1}{2}right)$. Choosing $f=u^{2}$
        we have $g=du^{2}+2u^{2}dudv+left(1+u^{4}right)dv^{2}$ in this
        region, which we can extend to a smooth non-degenerate metric on the
        whole torus by interpolating between it and the flat metric with $eta$:
        $$
        g:=du^{2}+2etaleft(uright)u^{2}dudv+left(1+etaleft(uright)u^{4}right)dv^{2}.
        $$
        Since $eta$ is $1$ on the domain of interest this does not change
        the gradient flow trajectory; so we have a smooth function on a compact
        manifold with a gradient flow trajectory $left(u,vright)left(tright)=left(frac{1}{lnleft(c+tright)},lnleft(c+tright)right)$ ($c$ large enough that $1/ln(c) < 1/2$),
        which has every point on the circle $left{ u=0right} $ as an
        accumulation point.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Oct 2 '13 at 4:31









        Anthony CarapetisAnthony Carapetis

        27.3k32865




        27.3k32865






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f501007%2fwhen-does-gradient-flow-not-converge%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            Popular posts from this blog

            Ellipse (mathématiques)

            Quarter-circle Tiles

            Mont Emei