Nonlinear term in the KPZ equation












2












$begingroup$


I'm reading up on the KPZ equation through the article by Bertini and Giacomin from 1997 and some lecture notes by Jeremy Quastel, the equation in 1+1 dimensions is stated as (for $h_t$ the height of the surface and $mathcal{W}_t$ space-time white noise)



$$ partial_t h_t = frac{1}{2} Delta h_t - frac{1}{2} (nabla h_t)^2 + mathcal{W}_t. $$



As indicated in the notes here by Quastel (https://math.arizona.edu/~mathphys/school_2012/IntroKPZ-Arizona.pdf, section 1.4 to 1.6) the nonlinear term is a problem that would actually make the equation ill-posed, but I don't really see how the nonlinear term would give the problems here. The only thing I can think of is integrability issues. The second derivative term requires more regularity, and would that not give more problems?



The lecture notes say that in order for the equation to make sense, we would need an "infinite renormalization", that is



$$ partial_t h_t = frac{1}{2} Delta h_t - frac{1}{2} ( (nabla h_t)^2 - infty) + mathcal{W}_t. $$



Now this is only formally, but I don't see how the equation as it is (without the $infty$) would make any less sense than the PDE without the nonlinear term. What's the issue with simply squaring a term that makes the entire PDE ill-posed? Could someone help me out?



Cheers.






pde stochastic-processes stochastic-calculus sde







share|cite|improve this question











$endgroup$












  • $begingroup$
    As to why the second derivative term doesn't give more problems, I would advise looking into the literature about the stochastic heat equation.
    $endgroup$
    – Eddy
    Dec 27 '18 at 11:17






  • 1




    $begingroup$
    The solution $h_t$ of the SPDE above is not a function, it is a (Schwartz) distribution. Note that, the white noise is itself a distribution. Since you cannot square a distribution (or more generally, multiply them), KPZ is mathematically ill-posed.
    $endgroup$
    – Sayantan
    Dec 27 '18 at 12:30






  • 1




    $begingroup$
    @TheBrainletExterminator I don't know much about renormlization. But do take a look at these notes.
    $endgroup$
    – Sayantan
    Dec 27 '18 at 22:31








  • 1




    $begingroup$
    A clarification: What I said about $h_t$ above is partly incorrect. You can define $h_t$ to be a function (e.g. in the Hopf-Cole sense), but it will be too rough to have pointwise derivative. Hence $nabla h_t$ must be interpreted as a distribution.
    $endgroup$
    – Sayantan
    Dec 28 '18 at 0:33






  • 1




    $begingroup$
    @TheBrainletExterminator You are correct, there is no classical meaning to $Delta h_t$ as well and we have to interpret it in distributional sense (integrating against test functions).
    $endgroup$
    – Sayantan
    Dec 28 '18 at 12:51
















2












$begingroup$


I'm reading up on the KPZ equation through the article by Bertini and Giacomin from 1997 and some lecture notes by Jeremy Quastel, the equation in 1+1 dimensions is stated as (for $h_t$ the height of the surface and $mathcal{W}_t$ space-time white noise)



$$ partial_t h_t = frac{1}{2} Delta h_t - frac{1}{2} (nabla h_t)^2 + mathcal{W}_t. $$



As indicated in the notes here by Quastel (https://math.arizona.edu/~mathphys/school_2012/IntroKPZ-Arizona.pdf, section 1.4 to 1.6) the nonlinear term is a problem that would actually make the equation ill-posed, but I don't really see how the nonlinear term would give the problems here. The only thing I can think of is integrability issues. The second derivative term requires more regularity, and would that not give more problems?



The lecture notes say that in order for the equation to make sense, we would need an "infinite renormalization", that is



$$ partial_t h_t = frac{1}{2} Delta h_t - frac{1}{2} ( (nabla h_t)^2 - infty) + mathcal{W}_t. $$



Now this is only formally, but I don't see how the equation as it is (without the $infty$) would make any less sense than the PDE without the nonlinear term. What's the issue with simply squaring a term that makes the entire PDE ill-posed? Could someone help me out?



Cheers.






pde stochastic-processes stochastic-calculus sde







share|cite|improve this question











$endgroup$












  • $begingroup$
    As to why the second derivative term doesn't give more problems, I would advise looking into the literature about the stochastic heat equation.
    $endgroup$
    – Eddy
    Dec 27 '18 at 11:17






  • 1




    $begingroup$
    The solution $h_t$ of the SPDE above is not a function, it is a (Schwartz) distribution. Note that, the white noise is itself a distribution. Since you cannot square a distribution (or more generally, multiply them), KPZ is mathematically ill-posed.
    $endgroup$
    – Sayantan
    Dec 27 '18 at 12:30






  • 1




    $begingroup$
    @TheBrainletExterminator I don't know much about renormlization. But do take a look at these notes.
    $endgroup$
    – Sayantan
    Dec 27 '18 at 22:31








  • 1




    $begingroup$
    A clarification: What I said about $h_t$ above is partly incorrect. You can define $h_t$ to be a function (e.g. in the Hopf-Cole sense), but it will be too rough to have pointwise derivative. Hence $nabla h_t$ must be interpreted as a distribution.
    $endgroup$
    – Sayantan
    Dec 28 '18 at 0:33






  • 1




    $begingroup$
    @TheBrainletExterminator You are correct, there is no classical meaning to $Delta h_t$ as well and we have to interpret it in distributional sense (integrating against test functions).
    $endgroup$
    – Sayantan
    Dec 28 '18 at 12:51














2












2








2


2



$begingroup$


I'm reading up on the KPZ equation through the article by Bertini and Giacomin from 1997 and some lecture notes by Jeremy Quastel, the equation in 1+1 dimensions is stated as (for $h_t$ the height of the surface and $mathcal{W}_t$ space-time white noise)



$$ partial_t h_t = frac{1}{2} Delta h_t - frac{1}{2} (nabla h_t)^2 + mathcal{W}_t. $$



As indicated in the notes here by Quastel (https://math.arizona.edu/~mathphys/school_2012/IntroKPZ-Arizona.pdf, section 1.4 to 1.6) the nonlinear term is a problem that would actually make the equation ill-posed, but I don't really see how the nonlinear term would give the problems here. The only thing I can think of is integrability issues. The second derivative term requires more regularity, and would that not give more problems?



The lecture notes say that in order for the equation to make sense, we would need an "infinite renormalization", that is



$$ partial_t h_t = frac{1}{2} Delta h_t - frac{1}{2} ( (nabla h_t)^2 - infty) + mathcal{W}_t. $$



Now this is only formally, but I don't see how the equation as it is (without the $infty$) would make any less sense than the PDE without the nonlinear term. What's the issue with simply squaring a term that makes the entire PDE ill-posed? Could someone help me out?



Cheers.






pde stochastic-processes stochastic-calculus sde







share|cite|improve this question











$endgroup$




I'm reading up on the KPZ equation through the article by Bertini and Giacomin from 1997 and some lecture notes by Jeremy Quastel, the equation in 1+1 dimensions is stated as (for $h_t$ the height of the surface and $mathcal{W}_t$ space-time white noise)



$$ partial_t h_t = frac{1}{2} Delta h_t - frac{1}{2} (nabla h_t)^2 + mathcal{W}_t. $$



As indicated in the notes here by Quastel (https://math.arizona.edu/~mathphys/school_2012/IntroKPZ-Arizona.pdf, section 1.4 to 1.6) the nonlinear term is a problem that would actually make the equation ill-posed, but I don't really see how the nonlinear term would give the problems here. The only thing I can think of is integrability issues. The second derivative term requires more regularity, and would that not give more problems?



The lecture notes say that in order for the equation to make sense, we would need an "infinite renormalization", that is



$$ partial_t h_t = frac{1}{2} Delta h_t - frac{1}{2} ( (nabla h_t)^2 - infty) + mathcal{W}_t. $$



Now this is only formally, but I don't see how the equation as it is (without the $infty$) would make any less sense than the PDE without the nonlinear term. What's the issue with simply squaring a term that makes the entire PDE ill-posed? Could someone help me out?



Cheers.






pde stochastic-processes stochastic-calculus sde




pde stochastic-processes stochastic-calculus sde






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 27 '18 at 10:27







The Brainlet Exterminator

















asked Dec 27 '18 at 10:05









The Brainlet ExterminatorThe Brainlet Exterminator

340211




340211












  • $begingroup$
    As to why the second derivative term doesn't give more problems, I would advise looking into the literature about the stochastic heat equation.
    $endgroup$
    – Eddy
    Dec 27 '18 at 11:17






  • 1




    $begingroup$
    The solution $h_t$ of the SPDE above is not a function, it is a (Schwartz) distribution. Note that, the white noise is itself a distribution. Since you cannot square a distribution (or more generally, multiply them), KPZ is mathematically ill-posed.
    $endgroup$
    – Sayantan
    Dec 27 '18 at 12:30






  • 1




    $begingroup$
    @TheBrainletExterminator I don't know much about renormlization. But do take a look at these notes.
    $endgroup$
    – Sayantan
    Dec 27 '18 at 22:31








  • 1




    $begingroup$
    A clarification: What I said about $h_t$ above is partly incorrect. You can define $h_t$ to be a function (e.g. in the Hopf-Cole sense), but it will be too rough to have pointwise derivative. Hence $nabla h_t$ must be interpreted as a distribution.
    $endgroup$
    – Sayantan
    Dec 28 '18 at 0:33






  • 1




    $begingroup$
    @TheBrainletExterminator You are correct, there is no classical meaning to $Delta h_t$ as well and we have to interpret it in distributional sense (integrating against test functions).
    $endgroup$
    – Sayantan
    Dec 28 '18 at 12:51


















  • $begingroup$
    As to why the second derivative term doesn't give more problems, I would advise looking into the literature about the stochastic heat equation.
    $endgroup$
    – Eddy
    Dec 27 '18 at 11:17






  • 1




    $begingroup$
    The solution $h_t$ of the SPDE above is not a function, it is a (Schwartz) distribution. Note that, the white noise is itself a distribution. Since you cannot square a distribution (or more generally, multiply them), KPZ is mathematically ill-posed.
    $endgroup$
    – Sayantan
    Dec 27 '18 at 12:30






  • 1




    $begingroup$
    @TheBrainletExterminator I don't know much about renormlization. But do take a look at these notes.
    $endgroup$
    – Sayantan
    Dec 27 '18 at 22:31








  • 1




    $begingroup$
    A clarification: What I said about $h_t$ above is partly incorrect. You can define $h_t$ to be a function (e.g. in the Hopf-Cole sense), but it will be too rough to have pointwise derivative. Hence $nabla h_t$ must be interpreted as a distribution.
    $endgroup$
    – Sayantan
    Dec 28 '18 at 0:33






  • 1




    $begingroup$
    @TheBrainletExterminator You are correct, there is no classical meaning to $Delta h_t$ as well and we have to interpret it in distributional sense (integrating against test functions).
    $endgroup$
    – Sayantan
    Dec 28 '18 at 12:51
















$begingroup$
As to why the second derivative term doesn't give more problems, I would advise looking into the literature about the stochastic heat equation.
$endgroup$
– Eddy
Dec 27 '18 at 11:17




$begingroup$
As to why the second derivative term doesn't give more problems, I would advise looking into the literature about the stochastic heat equation.
$endgroup$
– Eddy
Dec 27 '18 at 11:17




1




1




$begingroup$
The solution $h_t$ of the SPDE above is not a function, it is a (Schwartz) distribution. Note that, the white noise is itself a distribution. Since you cannot square a distribution (or more generally, multiply them), KPZ is mathematically ill-posed.
$endgroup$
– Sayantan
Dec 27 '18 at 12:30




$begingroup$
The solution $h_t$ of the SPDE above is not a function, it is a (Schwartz) distribution. Note that, the white noise is itself a distribution. Since you cannot square a distribution (or more generally, multiply them), KPZ is mathematically ill-posed.
$endgroup$
– Sayantan
Dec 27 '18 at 12:30




1




1




$begingroup$
@TheBrainletExterminator I don't know much about renormlization. But do take a look at these notes.
$endgroup$
– Sayantan
Dec 27 '18 at 22:31






$begingroup$
@TheBrainletExterminator I don't know much about renormlization. But do take a look at these notes.
$endgroup$
– Sayantan
Dec 27 '18 at 22:31






1




1




$begingroup$
A clarification: What I said about $h_t$ above is partly incorrect. You can define $h_t$ to be a function (e.g. in the Hopf-Cole sense), but it will be too rough to have pointwise derivative. Hence $nabla h_t$ must be interpreted as a distribution.
$endgroup$
– Sayantan
Dec 28 '18 at 0:33




$begingroup$
A clarification: What I said about $h_t$ above is partly incorrect. You can define $h_t$ to be a function (e.g. in the Hopf-Cole sense), but it will be too rough to have pointwise derivative. Hence $nabla h_t$ must be interpreted as a distribution.
$endgroup$
– Sayantan
Dec 28 '18 at 0:33




1




1




$begingroup$
@TheBrainletExterminator You are correct, there is no classical meaning to $Delta h_t$ as well and we have to interpret it in distributional sense (integrating against test functions).
$endgroup$
– Sayantan
Dec 28 '18 at 12:51




$begingroup$
@TheBrainletExterminator You are correct, there is no classical meaning to $Delta h_t$ as well and we have to interpret it in distributional sense (integrating against test functions).
$endgroup$
– Sayantan
Dec 28 '18 at 12:51










1 Answer
1






active

oldest

votes


















0












$begingroup$

The 'problems' they are referring to are problems in solving the equation. There are many, many methods for solving linear PDEs, both for specific and general cases. For example, if we were to remove the squaring so that term was linear, then we could solve on a rectangular domain using seperation of variables. Nonlinearity in the equation (or boundary conditions) is almost always the reason why equations occurring in physics are hard to solve exactly, and in many cases people resort to numerics.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Hi, I should have been more clear. It's not just a problem, the notes and the articles I've read about it state that the equation is actually ill-posed and would need an "infinite renormalization", i.e. $ partial_t h_t =frac{1}{2} Delta h_t - frac{1}{2} ( (nabla h_t)^2 - infty) + mathcal{W}_t$ to make sense. I will edit the question.
    $endgroup$
    – The Brainlet Exterminator
    Dec 27 '18 at 10:22













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%2f3053765%2fnonlinear-term-in-the-kpz-equation%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









0












$begingroup$

The 'problems' they are referring to are problems in solving the equation. There are many, many methods for solving linear PDEs, both for specific and general cases. For example, if we were to remove the squaring so that term was linear, then we could solve on a rectangular domain using seperation of variables. Nonlinearity in the equation (or boundary conditions) is almost always the reason why equations occurring in physics are hard to solve exactly, and in many cases people resort to numerics.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Hi, I should have been more clear. It's not just a problem, the notes and the articles I've read about it state that the equation is actually ill-posed and would need an "infinite renormalization", i.e. $ partial_t h_t =frac{1}{2} Delta h_t - frac{1}{2} ( (nabla h_t)^2 - infty) + mathcal{W}_t$ to make sense. I will edit the question.
    $endgroup$
    – The Brainlet Exterminator
    Dec 27 '18 at 10:22


















0












$begingroup$

The 'problems' they are referring to are problems in solving the equation. There are many, many methods for solving linear PDEs, both for specific and general cases. For example, if we were to remove the squaring so that term was linear, then we could solve on a rectangular domain using seperation of variables. Nonlinearity in the equation (or boundary conditions) is almost always the reason why equations occurring in physics are hard to solve exactly, and in many cases people resort to numerics.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Hi, I should have been more clear. It's not just a problem, the notes and the articles I've read about it state that the equation is actually ill-posed and would need an "infinite renormalization", i.e. $ partial_t h_t =frac{1}{2} Delta h_t - frac{1}{2} ( (nabla h_t)^2 - infty) + mathcal{W}_t$ to make sense. I will edit the question.
    $endgroup$
    – The Brainlet Exterminator
    Dec 27 '18 at 10:22
















0












0








0





$begingroup$

The 'problems' they are referring to are problems in solving the equation. There are many, many methods for solving linear PDEs, both for specific and general cases. For example, if we were to remove the squaring so that term was linear, then we could solve on a rectangular domain using seperation of variables. Nonlinearity in the equation (or boundary conditions) is almost always the reason why equations occurring in physics are hard to solve exactly, and in many cases people resort to numerics.






share|cite|improve this answer









$endgroup$



The 'problems' they are referring to are problems in solving the equation. There are many, many methods for solving linear PDEs, both for specific and general cases. For example, if we were to remove the squaring so that term was linear, then we could solve on a rectangular domain using seperation of variables. Nonlinearity in the equation (or boundary conditions) is almost always the reason why equations occurring in physics are hard to solve exactly, and in many cases people resort to numerics.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 27 '18 at 10:13









EddyEddy

939612




939612












  • $begingroup$
    Hi, I should have been more clear. It's not just a problem, the notes and the articles I've read about it state that the equation is actually ill-posed and would need an "infinite renormalization", i.e. $ partial_t h_t =frac{1}{2} Delta h_t - frac{1}{2} ( (nabla h_t)^2 - infty) + mathcal{W}_t$ to make sense. I will edit the question.
    $endgroup$
    – The Brainlet Exterminator
    Dec 27 '18 at 10:22




















  • $begingroup$
    Hi, I should have been more clear. It's not just a problem, the notes and the articles I've read about it state that the equation is actually ill-posed and would need an "infinite renormalization", i.e. $ partial_t h_t =frac{1}{2} Delta h_t - frac{1}{2} ( (nabla h_t)^2 - infty) + mathcal{W}_t$ to make sense. I will edit the question.
    $endgroup$
    – The Brainlet Exterminator
    Dec 27 '18 at 10:22


















$begingroup$
Hi, I should have been more clear. It's not just a problem, the notes and the articles I've read about it state that the equation is actually ill-posed and would need an "infinite renormalization", i.e. $ partial_t h_t =frac{1}{2} Delta h_t - frac{1}{2} ( (nabla h_t)^2 - infty) + mathcal{W}_t$ to make sense. I will edit the question.
$endgroup$
– The Brainlet Exterminator
Dec 27 '18 at 10:22






$begingroup$
Hi, I should have been more clear. It's not just a problem, the notes and the articles I've read about it state that the equation is actually ill-posed and would need an "infinite renormalization", i.e. $ partial_t h_t =frac{1}{2} Delta h_t - frac{1}{2} ( (nabla h_t)^2 - infty) + mathcal{W}_t$ to make sense. I will edit the question.
$endgroup$
– The Brainlet Exterminator
Dec 27 '18 at 10:22




















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%2f3053765%2fnonlinear-term-in-the-kpz-equation%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)

Image
Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
Read more

Quarter-circle Tiles

Image
up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
Read more

Mont Emei

Image
Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
Read more