Stable shock formation near plane-symmetry for nonlinear wave equations

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1 Stable shock formation near plane-symmetry for nonlinear wave equations Willie WY Wong Michigan State University University of Minnesota 9 March

2 Based on: arxiv: , joint with Gustav Holzegel, Jonathan Luk, and Jared Speck arxiv: (to appear in JHDE), joint with Gustav Holzegel, Sergiu Klainerman, and Jared Speck 2

3 The set-up Cylindrical domain: R R T 3 (t,y, )=(x 0,x 1,x 2 ). Study behavior 2 t 2 y 2 = h(@ 2 (Der-QNLW) 2X i,j=0 1 p i p g (g 1 ) j = 0 (Geom-QNLW) 3

4 Derivative 2 t 2 y 2 = h(@ 2 Arises from Lagrangian field theories with Lagrangian density L(@ ) dvol Example: Irrotational Euler, with being the fluid potential. 4

5 Geometric QNLW where g( ) = 2X i,j=0 g = g( )= 1 p i p g (g 1 ) j =0 0 B@ O( ) CA 1 is a Lorentzian metric and g is the geometric wave operator for the metric g. 5

6 Conversion... In (Der-QNLW), of the equation we get that ~ solves g( ~) ~ = Q(@ ~,@~) where: g( ~) = diag( 1,1,1) + h( ~) ~ is treated as a scalar taking values in R 3 Q is quadratic form ~, with coe cients depending on ~ Key: Q is a strong null form 6

7 ... additional equations Our results apply to any equation that can be written in the form where Q is a strong null form. g( ) = Q(@,@ ) Strong null form: for every, the form Q is null relative to g( ). (Needs to hold even for cubic and higher terms!) Hence focus on (Geom-QNLW). 7

8 Di erent from Weak Null case Weak null 2 t 2 y 2 = h( 2. (Geom-QNLW) expands 2 t 2 y 2 = h( 2 + Q(@,@ )+h 0 The last term makes all the di erence when g( ). ( )g(0). Small data regime: define G LL = h 0 (0) (@ t y ) (@ t y ) measure failure of quadratic null condition. 8

9 Main results (abbreviated version) Let g be such that G LL, 0. Suppose solves (Geom-QNLW) with compactly supported smooth initial data ( 0 = t=0, 1 t t=0 ) satisfying the conditions 1. i = Small amplitude: 0 <. 3. Forward y 0 1 apple 0 and (g 1 ( )) (g 1 ( )) 01 1@ y 0 +(g 1 ( )) 11 (@ y 0 ) 2 =0. Then terminates in time (1 + O( )) 1 where := 1 2 supg LL@ y 0. in a shock singularity, Furthermore, the singularity formation is stable under su small, arbitrary perturbations. ciently 9

10 Remarks The first part of the theorem (singularity formation for -independent data) is essentially contained in the work of F. John from the 70s, which built on works of Lax and Glimm earlier. We will start with a quick review of this theory as applied to our simple setting. The second part of the theorem (stability) is new. The bulk of today s talk concerns this part. The conditions on small amplitude and forward propagating are not critical: they can be significantly relaxed and we assume them here for simplicity. 10

11 Part I: 1+1dimensional problem and importance of geometry 11

12 Reduction to 1+1dimensions Under the -independence assumption, (Geom-QNLW) reduces to the equation ḡ = 0 on R 2, where ḡ is the two-dimensional Lorentzian metric obtained by restricting to the t,y components of g dimensional geometric wave equations are conformally invariant. 12

13 Double null formulation ḡ( ) =0 v =0 where u,v are null/characteristic coordinates of R 2 relative to ḡ. u t y v t + y solve ḡ 1 (du,du)=ḡ 1 (dv,dv)=0. (A priori they depend non-locally on and hides the nonlinearity.) 13

14 Regular propagation Forward-propagating assumption =) = (u) is independent of v. Level sets of u are straight lines, along which is constant. Slope of these level sets determined at t = 0 by the initial data! 14

15 Shock formation Shock forms when two characteristics (level sets of u) intersect! This happens when wave speed decreases as y increases. Rate of change of wave speed governed by G y 0. 15

16 Time of existence When exactly does shock form? Let µ := (ḡ 1 (dt,du)) 1, the inverse foliation density of leaves of u. Shock forms when µ & 0. (We can re-parametrize u so that when t = 0, µ 1. ) 16

17 Time of existence The lines of constant u are geodesics of ḡ. Geodesic equation =) along lines of constant u, parametrized by t, we t µ = Ḡvv( )@ u. Here Ḡvv is the geometric version of G LL, and is equal to (1 + O( ))G LL. By propagation equation the RHS is constant along the level sets of u. Hence µ is linear in time. 17

18 Key structural conclusions u is the forward-going null coordinate. u are constant along level sets of u. µ, the inverse foliation density of u, evolves linearly in time t along level sets of u; its rate is governed by the (initial value) quantity G y 0. 18

19 Part II: structure of proof, heuristics, and obstructions 19

20 Structure of the 1+1proof In 3 steps 1. Rewrite equations in characteristic coordinates. (Here v u =0, v µ =0.) 2. Prove virtual global well-posedness in characteristic coordinates. (Trivial in ) 3. Show that initial data implies characteristic coordinates cannot be globally regular (µ! 0 in finite time) and hence solution exhibits shock. 20

21 What to do in higher dimensions? 1. Define a characteristic coordinate system so that the equations of motion appear to be perturbations of the case. 2. Prove, virtually globally, that smallness of the perturbation propagates under the equation. (Thereby proving virtual GWP.) (This is the hard part!) 3. Profit! v µ =small requires no modification.) 21

22 Coordinate choice Coordinate R R T by functions (t,u,!) 2 R R T. t is the same as the background rectilinear time.! is the angular variable. u is a null coordinate solving u t + y and g 1 (du,du) = 0. 22

23 Necessity of true characteristics One is tempted to simplify the construction of the characteristic coordinate function u by trying to replace it with either Projected null coordinates (that du restricted to constant surfaces are null in the induced geometry); The null coordinates for g(0) instead of g( ). Problem: In generic blows up at the shock point. When u is not a true characteristic, t to also blow-up, which would mean we become far from the perturbative regime that we want to be in! 23

24 Aside: why is one tempted? Solving the eikonal equation g 1 (du,du) = 0 loses derivatives... except in the case. Roughly speaking the eikonal equations can be re-written t where ru is R 3 -valued function representing rectangular derivatives of the coordinate function u. If we solve it as a transport equation this gives us regularity This is one derivative lower than g 1 ( )(du,du) = 0 may suggest. 24

25 Historical remark Importance of true characteristic geometry in shock problems first realized by S. Alinhac ca D. Christodoulou ca Both managed to deal with the derivative loss problem through essentially the same trick. (More on this later.) 25

26 Propagating smallness In use transport equation formulation. In higher dimensions, transport formulation (L 1 -L 1 estimates) loses derivative. Use energy estimate + Sobolev embedding. 26

27 Energy and anisotropy Problem: We are looking at shock formation! The solution is expected to be small. So g( ) expected to be close to Minkowskian. So volume forms are expected to be close to rectilinear. In particular: natural energy quantities will be similar to those of the linear wave equation, and is hence isotropic. However, expect solution to be anisotropic: derivatives tangent to u-level-sets are small, while transversal derivatives blow-up! 27

28 Solutions 1. To extract smallness: take derivatives of the equation in directions tangential to the u foliation. In the model case these derivatives are all vanishing. Hope to prove that in the perturbed case all these derivatives have very small energy. 2. To deal with the blow-up: instead of studying the natural energy (which blows up at shock formation), study the renormalized energy (µ times the natural energy). This quantity should remain bounded. 28

29 New problems 1. Commuting the equation with a derivative will generate error terms. Need to make sure the error terms don t themselves blow-up! To solve this require exposing the inherent null structures of the equations. (Here the strong null form assumption on the RHS plays a role.) 2. The the anisotropy of the expected evolution means that renormalized energy estimates loses coercivity near the blow-up time. To solve this use space-time integrated energy control that arises from the fact that the inverse foliation density is expected to be approximately monotonic along constant u slices. 29

30 Part III: aspects of the proof of higher dimensional stability theorem 30

31 Virtual Global Existence Prolonged time of existence in the wave zone compared to direct local existence argument. Standard local existence by energy method: solution exists at least up to time Expect that shock formation within time (1 + O( )) 1, where = 1 2 supg LL@ y 0. Will prove that as long as µ doesn t tend to zero, well-posedness holds up to time

32 Coordinates Let u be a solution to the eikonal equation g 1 (du,du) = 0 such that at t = 0 we u = 0 and µ = 1. Fix t to be the rectilinear time. Fix! so that at t = 0,! =. And furthermore that the lines {! =! 0,u = u 0 } are null geodesics. 32

33 Transformation map Let x µ, µ 2{0,1,2} be the rectilinear coordinates. Define the Jacobian components X µ t t (x µ ), X µ! (x µ ). (Don t need the u version: that can be solved from the metric and X µ t.) Recall µ = g 1 (dt,du) 1. Note t X µ! X µ t. Let g u! = g(@ u,@! ) and g!! = g(@!,@!). 33

34 Decomposition of metric In the coordinate system, we have! 2 g = 2µ du dt + µ 2 du 2 g + g u!!! du +d!. g!! And where (g 1 t t R + R = 1 u! g g!!! µ u. Let R = u. 34

35 The system: the main unknowns The function. The Jacobian components X t = µg 1 (du,dx ) and X! The inverse foliation density µ 35

36 The system: the main evolution equations Geodesic equation (transport): (G is Wave t µ = 1 2 G tt R( 1 ) 2 µ(g tt + G Rt )@ t t Xt = 1 1 X 2 g!g + O(1)@ t (Geod2)!! 0=µ g t [µ@ t +2 R( )] + µ 2!! 1 t g!! g!! R( ) + terms involving t (Wave) 36

37 The system: the main evolution equations In the -independent case, (Geod1) is unchanged (except that the RHS is constant in t). (Geod2) reduces t Xt = 0. (And Xt 0 = X2 t u level sets are flat.) = 0.) (This means that (Wave) becomes t pg!! [@ t +2 R( )]. 37

38 Commutation structure Observe that ]=0, t, R] [@!, R] This means that if we commute (Geod1), (Geod2), and (Wave) with u-tangential t,@!, we will not pick up extra R derivative terms. 38

39 Commutation structure This implies we can close a bootstrap argument using L 1 estimates for derivatives of R. L 2 estimates for derivatives of involving no more than three involving no more than one R. They are proved by Commuting (Geod1), (Geod2) t,@!, and up to two R and estimating as transport equation. Commuting (Wave) with t,@!, and doing energy estimates. 39

40 Energy estimates Let P denote a string t derivatives. (Wave) gives t [µ@ t P +2 R(P )] + µ 2!!P 1 t P g!! g!! R( ) + other error terms We focus on the one error term (in red), which can potentially lose derivatives, to illustrate the strategy. Some other error terms are similarly bad, but we ignore them for this presentation. 40

41 Energy estimates Multiply (Wave) t P + 2[µ@ t P + R(P )] and integrate by parts in a region {(t,u,!):t 2 [0,T),T <2 1,u 2 [0,U],U apple u } (By assumption the initial data has compact support, make the support contained in u 2 [0,u ].) u = 0 boundary gives 0 contribution by finite speed of propagation t = 0 boundary is initial data 41

42 Energy estimates t = T energy: E T,U (P )= Z U Z apple µ 2 (1 + 2µ)(@ tp ) 2 +2µ@ t P RP 0! + 2( RP ) 2 + µ 2 (1 + 2µ)(@!P ) 2 g!! p g!! d! du u = U flux: F T,U (P )= Z T 0 Z! " (1 + µ)(@ t P ) 2 + µ (@!P ) 2 # pg!! d! dt g!! 42

43 Energy estimates E T,U + F T,U = initial data + Z T Z U Z 0 0! Where Bulk includes, for example, Bulk p g!! d! du dt. From µ 2!!P, the t µ g!! (@! P ) 2. From 1 t P g!! g!! R, the term 1 t P g!! g!! ( R )( RP ). 43

44 Energy degeneracy When µ 1, E only controls RP, while F only t P. What P? Use the bulk term! 44

45 Point of no return Lemma µ< 1 4 tµ apple 1 4. So the t µ g!! (@! P ) 2 gives control of the time-integrated angular derivatives, which is su cient to close energy estimates. 45

46 Heuristic justification of lemma Recall that µ(t = 0) = 1. And we study the interval t 2 [0,2 1 ]. Furthermore we expect µ to be almost linear. So we expect that for µ to decay by 3/4 within time at most 2 1 means that the minimum slope t µ = 3 8 > 1 4. We leave some room for error terms. 46

47 Derivative / µ loss As mentioned before, the coe cients of (Wave) depend on u in a way that could potentially lead to derivative loss. We can see it t P g!!. g!! = g( ) X!X! So the term with the worst regularity t P g!! is To t X! we 2 t X t = 1 2 g X!@ t P X!. 1 g!! X!G 2!!

48 Derivatve / µ loss So the worst term t P g!! looks like (after some computations) Z t G tt@ 2!!P. This requires one higher degree of regularity than that appears in the energy (which is of ). 48

49 Derivative / µ dichotomy Solution: At non-top-level: control by Gronwall into higher energy. At top-level: use the equation! Replace 1 2!! = 1 t[µ@ t +2 R( )] +... So the problematic term becomes t P g!! ( R )( RP ) 2 g!! = 1 Gtt R (µ@ 4µ t P +2 RP )( RP )+l.o.t.(w/ t-integral) {z t µ 49

50 Top level End result: top-order energy estimate of the form E T,U (P ) E 0,U (P ) apple C Z T 0 sup t µ µ E s,u (P )ds +... Important: The number C is a universal structure constant, independent of the precise form of g. (In fact it is apple 12.) Assuming µ only depends on t and forgetting other error terms then we get E T,U (P ) apple µ C E 0,U (P ). Top level energy can blow up! 50

51 Below top level Suppose we know that E T,U (@! P the term that loses derivatives, ) apple µ C Data. We observe that Z t G tt@ 2!!P, can be controlled, after a L 2 spatial integration, by Z t µ C/2 1 2 Data. 0 51

52 Below top level Since we expect µ to be approximately linear, we in fact have Z t µ C/2 1/2 µ 1/2 C/2. 0 Noting that we have another time integration ( R t 0 G tt@ t P g!! ), we conclude finally that E T,U (@! P ) apple µ C =) E T,U (P ) apple µ C+2. 52

53 Descent scheme Sacrifice the top C/2 derivatives: their corresponding energies are allowed to blow up. Below that in the medium range we prove only energy estimates with no µ-related blow-up. Below that in the low range we gain (via Sobolev) pointwise estimates. 53

54 Thank you! 54

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