 Okay so okay so let me start. First of all I'd like to thank the organizers for the invitation. So today I'm going to talk about the bubbling of what we call Q curvature equations of all manifolds in a new case. So this is a joint work with the home job from Thao China Normal University in China. Okay let me go straight to the motivation of my work. So I assume that M is a compact remanian for manifold without boundary and suppose that f is a non-trivial smooth function on M. So I will talk about the equation 1f which is the prescribing Q curvature equation. So in this equation on the left hand side we have a differential operator called the Panette operator. This is a force auto operator and on the right hand side we have a function f times exponential of 4u. And the reason why we want to study equation 1f because if you have a solution used to that equation then you have a conformal metric whose Q curvature is f. So I will talk about what is Q curvature but let's start with 1f. So in this case we have two simple but remarkable results. The first one is the non-existent result which says that if f is non-positive then the equation has no solution. But if we increase the function f a little bit so we say f less than or equal to small number epsilon which is small then we have some solution. So we conduct this research project because we realize that if we let epsilon goes to zero then solution to the equation 1f with f less than epsilon will escape to infinity because when epsilon equals zero we have no solution. So we put that observation as follows. We fix some function f zero which is non-positive and has some maximum at zero and then we perturb f zero by some small lambda. So we conduct f zero plus lambda equals f lambda and then we decrease lambda goes to zero and we want to understand the behavior of solution u lambda to the equation 1f with f replaced by f lambda. Okay so the outline of the talk is as follows. So first of all I talk about the Panette operator is Q curvature and then the PDE. Then I will talk about two equations. One is the Q curvature equation. The other is the Q curvature flow equation and then I will talk about some perturbation or the Q curvature equation. So the perturbation is it comes from the fact that I want to unify both equation and the flow equations and I want to study this equation at the same time. And then for the last session I will talk about the main results. So the main result is to show that there is some bubbling solution to the particular equation and then I give some remarks. Okay so to start with the Panette operators let me remind you the second order case first. So we know that the Laplace-Beltram operator is given by delta zero equals some summation and then if we make you a conformal train in g0 which is 2e2u g0 then the Laplace operator enjoy the conformal train. I show here. Furthermore we also have what we call Gaussian curvature equation which relates the Gaussian curvature of the metric g0 and the metric g0. So for higher order case we have what we call the Panette operator denoted by capital P. So this operator discovering 1983 by Panette on four manifold. So it's given by this formula and the leading term of the Panette operator is Laplace square. So this is by Laplace and we have lower order term here. So in the case of R4 the Panette operator is exactly the by Laplace and on S4 the four sphere it is a product of between Laplace and Laplace minus 2. Okay so the previous two differential operator Laplace Laplace and the Panette operator actually belongs to a wider class of operator. We call this is a conformally covariate operators by degree AB. So these operator enjoy this conformal trade rule. So if you want to compute with respect to the conformal metric g0 then it is just exponential minus du times the operator with respect to the background metric of some some multiplication of of fine here. Okay so in the case of Laplace operator it is of by degree zero two. Okay and in the case of Panette operator it is of by degree zero four. Okay so for convenience I also put a table showing that in fact for other dimensions three four or five and so we can modify the operator the Laplaceian and then the Panette operator to have a set of conformal conformal transformation operators. So in the case of two manifolds we have Laplaceian and then for higher dimension we have conformal Laplaceian. So it is different from Laplaceian by another factor involving the scale coefficient. So you see that for the two dimension this term drops because we have n minus 2 here and for for the case of a Panette operator for dimension three and five and above we have what we usually call the Panette Branson operator. So it is almost the same as a Panette operator but it plus some q curvature there. So in the case of the four dimension that curvature disappear because we have a coefficient n minus four. Okay oops I'm sorry I have trouble with the keyboard. Okay so so now I want to talk about the q curvature so this is due to Branson and his co-author. So like the Gaussian curvatures on the two manifolds we have a dehydrated order underlocked for four manifolds called the q curvature. So the q curvature is given by this formula. Here's the scalar curvature here and then here is this curvature. So if we train the metric g0 by multiplied by is 2u same then we have a conformal transformation rule for q gu. So it is almost the same as the transformation of the Gaussian curvature under this conformal chain. Okay so so now I want to talk about the prescribing q curvature problem. So the problem is stated as follow. So given the smooth function f on m is there a metric in the conformal class of the background matrix g0 such that the q curvature is exactly f. So in fact there's some there is one restriction here a metric must be in the conformal class of the g0. So without looking for a metric in this conformal conformal class of g0 so the problem is determined because we only want equation which is a q curvature equals f but we have times m minus 1 over 2 variables. So this is an entertainment problem. So going back to the conformal chain of the q curvature we arrive at what I just call the q curvature equation. So this equation is nothing but the p u plus q equals f times e for u. So geometrically if you can solve this equation to have some solution u then you can construct a conformal metric with the prescribing q curvature f. Okay so why should we interest in the conformal chain? So there's one reason why we are interested in the conformal chain of the metric because on one hand it equalizes the number of equations and the number of variables so that the problem is now determined. The other is we can start with any metric in the conformal class. So I put here a diagram. Okay so for this direction this is the conformal class of g0. So here's the g0 itself and this is another metric in the conformal class. So I denote it by e to v0. So if you start with g0 then you have a q curvature q g0 and you have the q curvature equation computed with respect to this metric and if you can solve for u and then you can construct a metric and then you have the q curvature metric is f. Okay but if you start with the other metric e to v0 then you have a corresponding q curvature equation and if you start you can solve this equation for w and then again you can construct a conformal metric and you have the q curvature which is f and the relation between the solution of these two q curvature equations is that you have w equals u minus v or u equals w plus v. So this diagram shows that well you can start any metric in this conformal class. So usually people start with some metric such that the q curvature is constant. So you turn this equation into a simpler one. So this scalar is now constant, it's not a function anymore. So this leads to the prescribing constant q curvature problem. So this is this study by Alice Giannepolniani in 1990, 1992 and others. Okay so well we talked about the finite operator and q curvature just a high order analog of the Laplacian and the Gaussian curvature. So why why we should study these curvatures? So well in fact I do not know. I try to google this and then there's not much information about why we should study q curvature equation except that the q curvature enjoy the Jones-Gauss Bonnet formula. So on the left hand side you have the total integral of q curvature plus one-quarter of the integral of the y tensor square and on the right hand side you have 8 pi square times Euler characteristic. So so perhaps we can say that the q curvature enjoy this formula so there's some topological information here. And now one key information coming from this formula is that the total integral of the q curvature is a topological invariance. This is because the integral of y tensor square here is also a topological invariance. So so well when you to start with you can see the matrix on the m that you can use the the side of the value of this number. So so for this region when starting q curvature equation we usually split the problem into three cases. The first one is corresponding to the positive invariant. The second is the zero invariant and the third one is the negative one. So so in my talk I'm interested in the second case. So I call this the known case that means the total integral of the q curvature is zero. And in fact because of the the results of Alice Chen and Paul Yan we can choose the background matrix zero such that the q curvature is zero. So so we don't have the scalar q zero in the q curvature equation anymore. Okay so so back to the three cases here. So we have some obvious necessary condition when we started the q curvature equation. So I put here the necessary condition. So for the known case the function f is side-changing. Okay okay so so in order to to study the q curvature equation we need to have some some looks and some we need to realize some the difference between the lower case and the higher case. So at least here are some major differences between the two operators. So first of all the panacea operator is of all the four and the the Laplacian is only two. Okay the second is the panacea operator is not necessarily positive. So here's some some results on the positivity of the panacea operators. Okay so the third difference is the panacea operators need not satisfy the maximum principle. So this is a typical feature of a high order operators. So usually we we should not expect the maximum principle to hold. Okay but remarkably for some very small assumption on the the matrix so the Gershky and McCullis and Hang and Po Yang can prove the maximum principle for dimension three and dimension five level. Okay and then the last difference between the two operators is the kernel of the panacea operator is not necessarily trivial. So you know that the kernel of Laplacian is is what but the kernel of the panacea operator is not necessarily trivial. Okay so now I so for the moments I will talk about two types of equation. So the first one is the Q curvature equation and then this the second one will be the Q curvature flow equation. So let me start with the Q curvature equation in the no case. So there's a very nice result. So the very nice existing result. So assuming that the panacea operator is a positive with a kernel consisting of constant functions then under this two condition the first one is the function f is positive somewhere and the total integral of f is negative. Then we have one solution we have at least one solution used to the equation. Okay and the proof is quite as straightforward. You just construct one energy's functionals and then you look at the minimizer of that energy and they use Lagrangian to prove that the solution to the equation. So okay so now let me let me comment on the the result of the current energy. The the assumption on the the positive of maximum f is necessary as I said before but here we need extra condition on the integral of f. Okay and compared to the case of the prescribing Gaussian curvature equation in the no case the condition on the side of integral f is necessary but in the case of Q curvature equation this extra condition is not necessary. So but we cannot we don't know how to remove this that's fine as I know. Okay so to support these comments there's a paper by Beth and others in 2006. They construct a metric on the four torus and then they compute the the Q curvature and then they prove that in that case the integral of Q curvature is positive. So limiting to the the negative total integral is not necessary. So but if we through or away that's the condition then in fact we can prove the existence of solution up to a sign. So we have a plus or minus here and then we cannot determine which one is the correct one. Okay so now I go to the Q curvature flow. So instead of using the direct method to look for a solution I use the heat flow. So I fix a metric as usual and then I try to I try to form less an evolution of the time dependent metric g lambda t which is conformal to the background metric d0 by the e to u lambda and then at the initial time we can put any initial conformal factor u lambda 0 and then for time the positive time the evolution is is in this screen. So the time derivative of the metric is equal to minus twice of the Q curvature minus some alpha f and the lambda. So if you replace g lambda by e to u lambda g0 then you can transfer this equation for the metric to the equation for the conformal factor. So we will get the time derivative of u lambda equals alpha lambda f lambda minus the Q curvature of the time dependent metric. Okay so the way we construct this evolution is to ensure that this is a negative gradient flow. Okay so using the previous energies the time derivative is not the positive. Okay so so far alpha lambda is just a constant the time dependent constant. So we need to specify what is the alpha lambda. So we choose the alpha lambda in such a way that the integral of alpha lambda is constant. So if we use this then we can prove that in fact along the flow the volume is also constant. So using this evolution we have two constraints. One is the constant of the integral of alpha lambda and the other is the volume. Okay so with Zhang Hong we can prove some existing solution to the flow one. Okay so before I talk about that for example let me let me show why should we study that flow. Okay so suppose that the flow converges namely we have u lambda t goes to u lambda infinity in some sense. Then from the flow equation for u lambda okay so this equation and we pass t to infinity and then we end up at alpha lambda infinity f lambda equals q of g lambda infinity. So that means we have a metric g lambda infinity which has a q curvature proportional to f lambda. So we just scale u lambda infinity a little bit then we have some conformal metric with the q curvature f lambda. Okay so now let me talk about the these results of conversions. So we prove that the flow has solution u t defy all time and then there's a sequence of times some constant of lambda infinity and we have some conversion up to that time sequence. Okay so initially we expected to have a uniform convergence but we failed to prove this and then we tried several ways but we cannot. Therefore in principle we shouldn't expect to have uniform convergence and perhaps this is the reason why we work on the particular equation that I'm going to talk later. Okay so back to the flow equation if we use the conformal chain of a q curvature apply to the q curvature of a time dependence metric. Okay here's I remind you the conformal chain of the q curvature and you rewrite this equation and then you end up p u lambda is equal to f lambda e for u lambda and we have an extra term minus the time derivative of u lambda e for u lambda. So therefore compare the q curvature equation and this new form of q curvature flow equation they are more the same except that there are some extra term here. Okay so putting the q curvature equation and this flow equation together we propose to study a particular q curvature equation. So that particular q curvature equation will help us to handle the q curvature equation and this flow equation at the same time. Okay so here's okay so I try to I try to summarize what I'm trying to to speak. Okay so on the left hand side we have the equation so this is a q curvature equation in our case and on the right hand side we have a flow equation so they are all the same. Okay so so in view of these two equations I try to consider the the red equation here p u equals f e for u plus h e for u and for some reason we expect that there are some blow up a curve. Okay so to to inspire a work of Struve in 2009 for the Gaussian curvature flow in the north case so we propose to study the same problem but for the fourth order case and we try to consider some similar scale like we particularly you know function by a little bit by a little constant lambda and we try to understand what happened if we lose lambda to zero. Okay okay so so the configuration of the problem as I said at the beginning of talk if we fix some some function f zero which is non-positive we perturbed by small lambda and we can see the class of solution u lambda followed by using the direct method so just say just minimize the energy function over some constraint and then we have some solution u lambda so u lambda is solution to equation two lambda corresponding to the f lambda. Okay so obviously uh two zero has no solution of course right because now the function f zero is non-positive and uh and if uh if lambda is sufficiently small but positive then we have some minimal some solution u lambda followed by the Gaussian results I talked earlier so so if we let lambda goes to zero so the solution u lambda will fly away right so we are interested in the behavior of u lambda. Okay so we can prove that the limit of lambda for lambda is bounded formable by 64 pi square and also as a consequence we have a limit of integral of positive part of q curfew of g lambda we also bounded formable by 64 pi square okay so the proof is quite involved so I know okay so so um uh so for for for our case uh we uh we would consider the equation of three k and for some sequence lambda k to go to zero to be determined later and we want to understand the blow up behavior of w k s k goes to infinity so uh okay just we go back to previous slide so so in in view of the the first important fact uh there will be some sequence lambda goes to zero such that lambda of lambda is bounded from both by 64 pi square so I can assume that the limit of lambda k of alpha k is bounded from both by 64 pi square okay and for the term involving the perturbation h k um so uh I make another assumption because that h k we can freely freely uh a piecewise so uh my assumption is uh the L2 norm of h k is little over one okay so uh with this uh behavior this is the behavior of h k with k large uh we can we can apply for the q-coach equation in this case k is zero and for the q-coach flow equation okay uh and in in in the case of flows equation you can we can take x to be the time derivative of u and then I form the analysis of flow equation uh we we know that ut satisfy this condition so we can we can make u this assumption okay so um so uh in in in our work we need to perform uh three uh so the first one is to realize the the blow-off so we need to prove that well the blow-off must occur because solution u lambda goes in phase and somewhere must blow-off and then a second one is to characterize the behavior of uh w k away from any bubble uh any blow-off points and then the the the last one is to guess the right behavior of the w k near the blow-off point okay so in order to realize that the blow-off must occur we need to prove something like we need to realize that the blow-off must occur is in fact equivalent to uh showing that there's some concentration of the curvature okay so uh the concentration of curvature is simply uh you integrate the the positive part of the the q curvature uh some around some more then that one is bounded from below by some constant so in our case the constant is eight pi square so if you can prove this then we can talk about we have some concentration of curvature and and okay then that the blow-off must occur so um so proof is uh quite involved so the idea is to turn uh uh to prove some some some some bounded of the uh the volume and then we use a lot of standard elliptic estimate and then we arise with some contradiction and I say try so so therefore the we have some concentration of curvature okay so I don't want to talk in detail here okay so um so for the first talks the main result is there will be some capital i and and finally many points x infinity i i from one to capital i says that uh the the integral of the positive part of q curvature um around the ball uh over the ball of the center x infinity i is is bounded from below by eight pi square here the radius is is arbitrary okay so uh and uh another the fact is um the x infinity i is the maximum point of the function epsilon okay and uh and because of the um the integral of the positive part of q curvature is bounded from above by 64 pi square from the second uh important uh facts then uh we have an upper bound for the the number of blow point so in this case is eight so okay so so no matter how many uh a maximum point of f zero um the number of blow point in our case is always at most eight okay and uh okay so for the second talks uh we want to understand wk away from all blow up point x infinity i so the theorem is is as follow uh so away from this point um wk the solution of the particular equation goes to minus infinity locally in a uniformly and of course at each x infinity i uh wk is going to plus infinity okay um so so the proof of this theorem uh follows standard uh a blow up analysis away from this point you have a upper bound for uh the deposit part of the two curvature this equation less than eight pi square here of course if we have the reverse inequality then we have another concentration point so uh we must have the less than so for the less than then uh then we have a loss of uh we make you a loss of uh elliptic estimate and some functional inequalities and eventually we can prove that wk must goes to minus infinity so uh so this this condition uh this finding is uh unreasonable because uh at a capital i point x infinity i uh the solution wk is going to infinity so in order to maintain the uh constant volume uh the integral of e4 wk equals one uh the other other point the wk must go to minus infinity otherwise uh we cannot have this uh constraint okay okay so uh so now we we we work on the last part of uh uh the problem so we want to understand the behavior of the wk near blow points so um okay so um so we we describe the behavior of wk near the x infinity i as well that's two scenarios first one is a fast blow up and the second is the slow blow up so uh so at each uh blow point infinity i uh if you pull back um pull back uh the equation uh sorry uh the the the the solution wk to the tangent space at x infinity i and then do some scaling in a dilation then uh the the projected solution uh wk two times will will um con will converge to some limit profile so in the fast blow up case we can write down precisely the uh the form of the blow profile you choose the w infinity hat which is a log of some some quotient and for the slow blow up uh uh we we cannot we just uh we can we just say that we have some blow up for five w infinity hat we show some um force all the equation of exponential non-minority here okay so to prove of this theorem uh we'll uh it also based on the standard analysis of standard blow analysis uh with some some circumstances due to the fact that we are we are handling a higher order case but the idea is to to turn the um concentration of curvature into the concentration of a volume and then if you have a concentration of a volume then you can select a sequence xk i going to x infinity i such that the volume also concentrated at xk i okay so uh once you have the sequence of xk i then you just project wk onto the tangent space at x infinity i and then appropriate uh appropriate scaling and dilation uh you can prove that wk delta uh belongs to some sort of space and then you have some equation of force order and then you use some known classification for a force order equation with the exponential non-linearity and you can write down the blow up for five okay okay so uh before a complete thing in my talk i would like to give some remarks so um so first remark our work is totally inspired by a similar work by strouf recently appeared in his work he considered the bubbling of the gaussian the perturbation of gaussian curvature equation uh of the same form so instead of uh an x operator we have laplacian on the left hand side so the rest is the same okay and uh he he he obtained uh he obtained the the bubbling when uh we take h lambda goes to zero okay of course there are some uh some difference in between some differences between the the two problem uh okay i i don't want to talk in detail here okay um one one one major difference between the finding of these two work is that uh for the for the case of the gaussian curvature equation uh strouver uh can prove that there's no slow blow up so so any block is fast okay but in our case we cannot in fact uh in in principle uh either yes or no can occur okay so uh we we limit ourselves to the case of an odd case but in the net of the case uh the problem was studied by galler betty in 2017 but he purely worked on the the q curvature equation in that case so without uh the extra perturbation here so uh we can we can repeat our argument for this negative case and therefore can give some slight improvement of the galler betty result okay but uh for the positive case uh uh no result is no so far okay so uh thank you so much yeah i finished my talk okay thank you quok for this lovely talk we thank you thanks um very nice talk um well uh now uh we are open to questions if anybody wants to make some questions can unmute yourself or uh write in the chat okay can i ask a question and the browski from sisa yes of course so i have a few questions but let's um first question is uh so you said that these solutions go to infinity but you maybe i mean is understood go to infinity in space or they become singular that's what in some finite concentration points is this what you have in mind okay so the value of the solution goes to infinity at some point okay that's what i mean okay my second question is related to okay so as you know uh the is this finite operator has been discussed only so for scalar functions for functions i understand right has it been discussed for spinor fields also like similar there is a in relation to the dirac operator ford power or something like that or you never know uh i think even for the second on a case uh screw will also rise uh raise the same question for for the question you just talked so no i i say no no result okay okay okay let me just in the moment stop okay thanks okay sure thanks thank you any other questions or comments uh can i ask a question yes of course all right so is there some statement about uniqueness of the solution um what was the solution uh you are talking about the two curvature division yeah for you i mean i mean you have some condition on existence right but what about um oh okay um well uh you mean um this the this results uniqueness yes is it what yeah what does this one is there only one or yeah yeah yeah so uh so this one is i mean at least one not just one yeah but but in general i mean is it known like is there no no no solutions or infinite no unique this result at the moment okay yeah so it's it's not even known if there is finite number of solutions or there can be like family of solutions oh well i haven't seen any result this thing yeah i i haven't seen such a kind of results perhaps there's a lot of uh solution yeah but that's a very good question yeah yeah well i also got kind of a naive question so this uh so you have a you you say that there is a requirement that f should vanish somewhere right on a four manifold necessarily sorry um can can you repeat your question so there was some there was a some as far as the suit there was a necessary condition at some point that f should vanish f should vanish somewhere on your four manifold right um which case because um well um if if f is the mood and yeah in this case then of course f will so my question is i mean suppose there is some nice kind of a nice uh situation this zero locus of f forms some three manifold inside four manifold so does it does this three manifold has some sort of meaning i don't know okay okay thanks okay you're welcome well i have just a question to do but just uh for curiosities maybe it's a silly questions i don't know if you have sense but it is possible to to think this kind of problem to solve this equation in higher dimensions or there is something that maybe um but well um so uh so i was talking about two problems i mean uh in fact one problem two approaches one approaches is a direct method and the other is a variation method so flow method so um for for the the prescribing q curvature problem uh in fact we can we can consider um similar problem for higher dimension manifold or even higher dimension operator so um okay so uh if you continue this this order so we also have a similar operator order six or the eight so and so forth so what will we call these operator the the g uh so the the g okay so gsm p operators okay okay so yeah so for for for other dimension you can start uh talking about uh the prescribing uh with the pannet prance and operator so people also consider this problem uh as well so you can easily find a lot of uh results in the literature okay thank you any other question or comment i i still have dabrowski i still have some small naive questions so i understand this pannet operator is elliptic operator of course right principal symbol starts from the fourth power of right correct so my my question is about the what is the interpretation of the index of this operator well um yeah uh i have no knowledge about uh that part but uh yeah so so yeah so and another related on special manifolds like i don't know complex manifolds or color manifold is something more known about these operators and solutions as i know uh people also consider for cr manifolds but for color manifolds i haven't seen uh yeah i haven't seen thanks for speech to see our manifolds we have yes they call two pram uh uh curvy sure thanks okay well um i think any other question or comment if if not well if not we are going to thanks again quok for this very nice talk and thanks for being here uh today with us thank you so much yeah i want to add a comment yes okay sure yeah sir as i said earlier so this is a lunar new just eve so happy new year oh thank you happy new year all right okay well thank you so much yeah thank you for having me