EXPONENTIAL GROWTH OF THE VORTIITY GRADIENT FOR THE EULER EQUATION ON THE TORUS ANDREJ ZLATOŠ Abstract. We prove that there are solutions to the Euler equation on the torus with 1,α vorticity and smooth except at one point such that the vorticity gradient grows in L at least exponentially as t. The same result is shown to hold for the vorticity Hessian and smooth solutions. Our proofs use a version of a recent result by Kiselev and Šverák [5]. 1. Introduction Let (T = [ 1, 1 be the two-dimensional torus (i.e., we identify opposite sides of the square and consider the Euler equation on (T, in vorticity formulation: ω t + u ω =, ω(, = ω. (1.1 The velocity u is found from the vorticity ω via the Biot-Savart law u(t, x = 1 (x y n, + y 1 + n 1 ω(t, ydy, [ 1,1] x y n n Z obtained by taking the Biot-Savart kernel K(x = 1 (x, x on R and extending ω periodically. Initial data ω will here be 1 and odd in both and x, hence the latter property will hold for all t, as well as ω(t, L = ω L. Global regularity of bounded solutions to (1.1 was first proved by Wolibner [8] and Hölder [4]. We consider here the question of how fast the gradient of ω can grow in L as t. The well known upper bound is double-exponential ω(t, L e et but it has been a long standing open question whether this is attainable. The best infinite time result in the plane or on the torus (i.e., domains without a boundary so far has been the proof of the possibility of super-linear growth for smooth solutions by Denisov [1]. He also proved that doubleexponential growth is possible on arbitrarily long finite time intervals [], and constructed patch solutions to the D Euler equation with a regular prescribed stirring for which the distance between two approaching patches decreases double-exponentially in time [3]. For domains with boundaries (and no flow boundary condition, Yudovich [9, 1] and Nadirashvili [7] provided examples with unbounded growth and at least linear growth, respectively. These results have been dramatically improved in a striking recent work by Kiselev and Šverák [5], who proved the possibility of infinite time double-exponential growth of the vorticity gradient in a disc, thereby answering in the affirmative the above open question in 1
ANDREJ ZLATOŠ this setting. The boundary is crucial in [5] and the double-exponential growth is proved to occur on it as well. Related to this, we note that the double-exponential upper bound is only known for D domains with regular boundaries. In fact, Kiselev and the author [6] proved that there are domains whose boundary is smooth except at two points, and on which some solutions to the Euler equation with smooth initial data blow up in finite time. We refer the reader to [5] for more history and further references related to (1.1. In the present paper we prove that on the torus, at least exponential growth of the vorticity gradient happens for some 1,α initial data, as well as that such growth is possible for the vorticity Hessian for smooth initial data. Our proof uses a sharper version of a key result from [5] (see Lemma.1 below, applied on the torus instead of the disc. Theorem 1.1. (i For any α (, 1 and A <, there is ω 1,α ((T with ω L 1 and there is T such that the solution of (1.1 satisfies for all T T, sup ω(t, L ([, exp( AT ] e AT. t T (ii For any A <, there is ω ((T with ω L 1 and there is T such that the solution of (1.1 satisfies for all T T, sup D ω(t, L ([, exp( AT ] e AT. t T Our ω will be very simple. For instance, in (i it can be smooth except at the origin, with ω (r, φ = r 1+α sin(φ (in polar coordinates near the origin. Then ω remains in 1,α, and since u(t, = for all t > by symmetry (oddness of ω in, x, it follows that both u, ω are smooth except at the origin at all times. We also note that we will have ω L = 1, but as in [5], this can be arbitrarily small. Acknowledgements. The author thanks Sergey Denisov and Alexander Kiselev for useful discussions and comments. He also acknowledges partial support by NSF grants DMS- 15637 and DMS-1159133.. Proof of Theorem 1.1 For x [, 1] we let x := (, x, x := (, x, and Q(x := [, 1] [x, 1]. In what follows, will always be some universal constant which may change between inequalities. Lemma.1. Let ω(t, L ((T be odd in both and x. If, x [, 1 ], then ( 4 u j (t, x = ( 1 j y 1 y y ω(t, ydy + B j(t, x x 4 j (j = 1, (.1 Q(x where, with some universal, ( { ( B 1 (t, x ω(t, L 1 + min log 1 + x ( { ( B (t, x ω(t, L 1 + min log 1 + x } ω(t, L, x ([,x ] ω(t, L, ω(t, L ([, ] ω(t, L, (. }. (.3
EXPONENTIAL GROWTH FOR EULER EQUATIONS 3 Remark. If c x (resp. cx for some c <, then the min can obviously be dropped in (. (resp. (.3 as long as = (c. This version of these formulas was proved in [5] on the disc. In that case Q(x can be replaced by Q(x as well, which is done in [5]. Proof. Let us only consider j = 1 because j = follows by symmetry: K shows that if ω(t, x := ω(t, x,, then ũ(t, x = (u (t, x,, u 1 (t, x,. The Biot-Savart law gives u 1 (t, x = [ y1 ( n 1 (x y n x (y + n x (ỹ + n y ] 1( n 1 (x + y n ω(t, ydy, x (ȳ + n x ( y + n n Z [,1] (.4 by using the symmetries ω(t, ỹ = ω(t, y and then ω(t, ȳ = ω(t, y to express the integral over [ 1, 1] via that over [, 1]. Let us first consider the right-hand side of (.4 with the term n = (, removed. The first term in the integral equals (recall that, x [, 1 ] y 1 n 1 (x y n x (y + n x (ỹ + n + O( n 3. We combine it with the same term for ñ = ( n 1, n to obtain 3y 1 n 1 (x y n [n 1 (x y n + x 1 y 1 + 4n 1] x (y + n x (ỹ + n x (y + ñ x (ỹ + ñ + O( n 3 = O( n 3. This means that n (, y 1 ( n 1 (x y n ω(t, ydy [,1] x (y + n x (ỹ + n ω(t, L. An identical argument proves this also for the second term in (.4. We therefore only need to consider the term with n = (,, which is times [ y1 (x y x y x ỹ y ] 1(x + y ω(t, ydy. (.5 x ȳ x + y Q(x [,1] We will show that this equals to 4 times the integral in (.1, plus an error controlled by the right-hand side of (., thus proving (.1. We will again only consider the first term in the integral since the second will be handled in the same way. We separate the integral into either 3 or 4 regions. If x, then these regions will be Q(x, [, ] [, 1], and [, 1] [, x ]. If < x, we also split the last region into [, x ] [, x ] and [x, 1] [, x ]. The 3 region case is parallel to the treatment in [5] (where the domain is a disc. In the 4 region case we need to obtain an extra estimate for the integral over [, x ] [, x ] (this is not necessary for the second term in (.5. We start with Q(x, where we have y 1 x x y x ỹ dy x 1 Q(x y dy x dr. 3 x r
4 ANDREJ ZLATOŠ In Q(x also y 1 y x y x ỹ y 1y y = O( x y 5 = O( x y 3, 4 y 8 so the integral of the absolute value of this difference over Q(x is also bounded by 1 x x dr. r Hence integration of the first term in (.5 over Q(x gives ( times the integral in (.1, plus an error bounded by ω(t, L. (Integration of the second term in (.5 gives the same, whence the factor of 4 in (.1. Next integrate over [, ] [, 1]. After substituting z j := y j x j, the absolute value of the integral of the first term in (.5 can be bounded by ω(t, L times x1 1 z (z 1 + z (x 1 + z dz dz 1 = 1 x 1 + z x 1 + z arctan z dz arctan 1. Finally, the corresponding integral over [, 1] [, x ] is bounded by ω(t, L 1 x z 1 z x ( (z1 + z dz z dz 1 log 1 + x dz log x 1 + x x 1 times This gives the first term in the min in (.. If x, then we are done because the min is 1 and can be absorbed in the 1 in (.. So let us assume < x, and perform the above integration over [x, 1] [, x ] instead of [, 1] [, x ]. We see that the integral is bounded by ω(t, L log(1 + x x = ω(t, L log, and therefore it only remains to consider the integral over the remaining set [, x ] [, x ]. Let us denote M := ω(t, L ([,x ] and write ω(t, y = v(t, y 1 + w(t, y, where v(t, y 1 := ω(t, y 1, x and so w(t, y = ω(t, y 1, y ω(t, y 1, x M x y. For y 1 [, x ] we have with z := y x, x y 1 (x y x y x ỹ v(t, y 1dy = x x and we also have x x y 1 (x y w(t, y x y x ỹ dy x dy 1 M. y 1 z [( y 1 + z ][( + y 1 + z ] v(t, y 1dz = x y 1 z x (y1 + z dzdy 1 M This gives the second term in the min in (.. We note that for the second term in (.5, one can always integrate over [, 1] [, x ] because the substitution z 1 := y 1 and z := y + x yields 1 3x z 1 z x (z1 + z dz dz 1 3x z x x 1 + z dz log x 1 + 3x x 1 + x. z x 1 + z dz Mx.
EXPONENTIAL GROWTH FOR EULER EQUATIONS 5 Proof of Theorem 1.1. (i Given α and A, pick a function ω : (T [ 1, 1] which is odd in both and x, non-negative on [, 1] and equal to 1 on a subset of [, 1] of measure 1 δ (for some δ (, 1 to be chosen later, with ω 1 1,α ((T loc ((T \ {} and ω (s, s = s 1+α for s [, δ]. For instance, on B δ ( we could have in polar coordinates ω (r, φ = (r/ 1+α sin(φ. Take any T T := 1 log δ so that e AT δ, let X(t solve X (t = u(t, X(t with A X( = (e aat, e aat for some a > 1 to be chosen later, and let T := min{t, T }, with T the exit time of X from the square [, e AT ]. Obviously, for all t. Let us also assume that ω(t, X(t = ω (X( = e a(1+αat (.6 sup ω(t, L ([, exp( AT ] e AT (.7 t T because otherwise we are done. Since X(t [, e AT ] for t T, x ω(t, L ([,x ] 1 (.8 in (. when t T and x = X(t (and the same estimate applies to (.3. We then have (.1 with B j (t, X(t for t T (recall that ω(t, L = ω L = 1. A crucial observation of [5] is that ω on [, 1] and ω = 1 on a subset of [, 1] of measure 1 δ (along with the distribution function of ω(t, being the same for all t guarantees that the integral in (.1 is no less than 1 1 log δ when δ < and x [, 1 δ], for some universal >. (If instead we had odd ω : (T [ ε, ε] equal to ε on a subset of [, 1] of measure 1 δ, then this would be ε log δ, and our proof would be unchanged. So if we denote by k(t the value of the integral in (.1 for x = X(t, multiplied by 4, then k(t 1 log δ for t T. Hence we have for t T, ( 1 u 1 (t, X(t log δ ( 1 u (t, X(t log δ X 1 (t, (.9 X (t. (.1 If we take δ < e, it follows that X 1 (T < e AT and T (a 1AT ( 1 log δ 1. We will in fact pick δ e (a 1A so that also (a 1A( 1 log δ 1 < 1. Hence T = T < T and X (T = e AT. In addition, d dt [log X 1(t + log X (t] (.11 for t T by Lemma.1 and (.8. Therefore, log X 1 (T log X 1 ( log X (T + log X ( + T [ aa + A + ]T. (.1 But this, (.6, and ω(t,, e AT = give log ω(t, L ([,exp( AT ] log ω(t, X(T X 1 (T [a(1 αa A ]T,
6 ANDREJ ZLATOŠ which equals AT if we pick a := A + (1 αa and then δ as above. The proof of (i is finished because T T. (ii Given A, pick ω : (T [ 1, 1] which is odd in both and x, non-negative on [, 1] and equal to 1 on a subset of [, 1] of measure 1 δ (for some δ (, 1 to be chosen 1 later, smooth, and with ω (, x = sin 3 ( sin(x when min{, x } δ. 4 Take any T T := 1 log δ so that e AT δ, let X(t solve A 4 4 X (t = u(t, X(t with X( = (e AT, e (a 1AT for some a > 1 to be chosen later, and let T := min{t, T }, with T the exit time of X from the square [, e AT ]. Obviously, ω(t, X(t = ω (X( = sin 3 (e AT sin(e (a 1AT e (a+at (.13 for all t. Let us also assume (.7 (because otherwise we are done, using ω(t, =. As above, we obtain (.1 with B j (t, X(t for t T. We thus again have (.9 (.11 for t T, as well as X (T = e AT > X 1 (T and T = T < T, provided we pick δ < e (a A. So (.1 follows as well and then by (.13, log sup s [,exp([ aa+a+]t ] ω x1 (T, s, e AT log ω(t, X(T X 1 (T [3A + ]T. (.14 The result will follow if we can show that ω x1 (T,, e AT =, because this and (.14 imply provided a 1 + A Let v(t, x := ω x1 (t, x. Then log D ω(t, L ([,exp( AT ] [aa 4A ]T, (so that aa + A + A. Then we only need to pick a := 5A + A. v t + u 1 v x1 + u v x + (u 1 x1 v + (u x1 ω x =. We have u 1 (t,, x = = ω(t,, x by symmetry, so also ω x (t,, x =. This shows that if we denote w(t, s := v(t,, s for s T, then for (t, s R + T, w t + u (t,, sw s + (u 1 x1 (t,, sw =. Since w(, s = (ω x1 (, s = and u is smooth, it follows that w. Thus we indeed obtain ω x1 (T,, e AT =, and the proof of (ii is finished. References [1] S. Denisov, Infinite superlinear growth of the gradient for the two-dimensional Euler equation. Discrete ontin. Dyn. Syst. A 3 (9, 755 764. [] S. Denisov, Double-exponential growth of the vorticity gradient for the two-dimensional Euler equation Proceedings of the AMS, to appear. [3] S. Denisov, The sharp corner formation in D Euler dynamics of patches: infinite double exponential rate of merging, Arch. Rational Mech. Anal., to appear.
EXPONENTIAL GROWTH FOR EULER EQUATIONS 7 [4] E. Hölder, Über die unbeschränkte Fortsetzbarkeit einer stetigen ebenen Bewegung in einer unbegrentzten inkompressiblen Flüssigkeit (German, Math. Z. 37 (1933, 77 738. [5] A. Kiselev and V. Šverák, Small scale creation for solutions of the incompressible two dimensional Euler equation, preprint. [6] A. Kiselev and A. Zlatoš, Blow up for the D Euler equation on some bounded domains, preprint. [7] N. S. Nadirashvili, Wandering solutions of the two-dimensional Euler equation (Russian, Funktsional. Anal. i Prilozhen. 5 (1991, 7 71; translation in Funct. Anal. Appl. 5 (1991, 1. [8] W. Wolibner, Un theorème sur l existence du mouvement plan d un fluide parfait, homogène, incompressible, pendant un temps infiniment long (French, Mat. Z. 37 (1933, 698 76. [9] V. I. Judovič, The loss of smoothness of the solutions of Euler equations with time (Russian, Dinamika Splošn. Sredy 16 (1974, 71 78. [1] V. I. Yudovich, On the loss of smoothness of the solutions of the Euler equations and the inherent instability of flows of an ideal fluid, haos 1 (, 75 719. Department of Mathematics, University of Wisconsin, Madison, WI 5376, USA Email: zlatos@math.wisc.edu