Nonlinear term in the KPZ equation
$begingroup$
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
asked Dec 27 '18 at 10:05
The Brainlet ExterminatorThe Brainlet Exterminator
340211
$endgroup$
|
show 2 more comments
$begingroup$
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
asked Dec 27 '18 at 10:05
The Brainlet ExterminatorThe Brainlet Exterminator
340211
$endgroup$
$begingroup$
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
|
show 2 more comments
$begingroup$
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
asked Dec 27 '18 at 10:05
The Brainlet ExterminatorThe Brainlet Exterminator
340211
$endgroup$
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
asked Dec 27 '18 at 10:05
The Brainlet ExterminatorThe Brainlet Exterminator
340211
asked Dec 27 '18 at 10:05
The Brainlet ExterminatorThe Brainlet Exterminator
340211
edited Dec 27 '18 at 10:27
The Brainlet Exterminator
asked Dec 27 '18 at 10:05
The Brainlet ExterminatorThe Brainlet Exterminator
340211
asked Dec 27 '18 at 10:05
The Brainlet ExterminatorThe Brainlet Exterminator
340211
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
|
show 2 more comments
$begingroup$
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
|
show 2 more comments
1 Answer
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active
oldest
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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.
answered Dec 27 '18 at 10:13
EddyEddy
939612
$endgroup$
$begingroup$
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
add a comment |
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$begingroup$
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.
answered Dec 27 '18 at 10:13
EddyEddy
939612
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Dec 27 '18 at 10:13
EddyEddy
939612
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Dec 27 '18 at 10:13
EddyEddy
939612
$endgroup$
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.
answered Dec 27 '18 at 10:13
EddyEddy
939612
answered Dec 27 '18 at 10:13
EddyEddy
939612
answered Dec 27 '18 at 10:13
EddyEddy
939612
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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