Infinite superlinear growth of the gradient for the two-dimensional Euler equation

For two-dimensional Euler equation on the torus, we prove that the uniform norm of the gradient can grow superlinearly for some infinitely smooth initial data. We also show the exponential growth of the gradient for the finite time.


(Communicated by Roger Temam)
Abstract. For two-dimensional Euler equation on the torus, we prove that the L ∞ norm of the gradient can grow superlinearly for some infinitely smooth initial data. We also show the exponential growth of the gradient for finite time.

Introduction.
In this note, we are dealing with two-dimensional Euler equation. We will write the equation for vorticity in the following forṁ θ = ∇θ · u, u = ∇ ⊥ ζ = (ζ y , −ζ x ), ζ = ∆ −1 θ, θ(x, y, 0) = θ 0 (x, y) and θ is 2π-periodic in both x and y (e.g., the equation is considered on the torus T). We assume that θ 0 has zero average over T and then ∆ −1 is well-defined since the Euler flow is area-preserving and the average of θ(·, t) is zero as well.
The global existence of the smooth solution for smooth initial data is well-known [2]. It is also known that the gradient does not grow faster than the double exponential. On the other hand, the lower bounds for the gradient's norm are not studied well. There are some results on the infinite (not faster than linear) growth of the gradient for a domain with the boundary [4,5] or an annulus [3].
We will prove the following results. The proofs are inspired by the recent preprint [1].
That implies the infinite superlinear growth of the time average of the gradient.
That shows the possibility of the exponential growth over any (fixed) period of time.
2. Infinite superlinear growth. We start with several lemmas. We always assume that θ 0 (x, y) is infinitely smooth and its average over T is zero. The following result is borrowed from the recent preprint by A. Kiselev and F. Nazarov [1].
Proof. The following invariants are well-known on the Fourier side. Since outside the unit sphere, we have the statement of the lemma.
We also need the following elementary result on the preservation of some symmetries.
Proof. Assume that θ(x, y, t) is the solution. We need to show that ψ 1 (x, y, t) = θ(−x, −y, t) and ψ 2 (x, y, t) = θ(−y, x, t) both are solutions as well. Then, the uniqueness of the solution to Cauchy problem would yield the statement of the lemma. Notice thatψ as can be easily verified on the Fourier side. Therefore, ψ 1 solves Euler equation. For ψ 2 ,ψ e in1x+in2yψ 2 (n 1 , n 2 , t) n 2 1 + n 2
Obviously, the symmetries considered follow immediately from the symmetries of the multiplier (n 2 1 + n 2 2 ) −1 . Another symmetry preserved is θ(x, y, t) = −θ(y, x, t). The proof of this fact is similar but we are not going to use it. The following elementary lemma will be used later Before giving the proof to the theorem 1, we notice that the function θ * (x, y, t) = cos(x) + cos(y) has spectrum on the unit sphere. Therefore, ζ = θ * and θ * is a stationary solution. It is even and invariant with respect to rotation by π/2 degrees. The flow generated by it is and it is hyperbolic at the points D j = π(j 1 , j 2 ) where j 1 mod 2 = j 2 mod 2. One can consider, e.g., the points A 1 = (π, 0), A 2 = (0, π) and their 2π-translates. The proofs will be based on certain stability of θ * and perturbation theory around the hyperbolic scenario which (without nonlinear term) leads to infinite exponential growth of the gradient for suitably chosen initial data. The idea is this: we will show that away from the points D j the direction of the flow is basically the same as without nonlinearity. Therefore, if we place the bump at D j , the area around D j and inside the level set of this bump will be gradually exhausted by the flow. But since the total area is preserved, this will be manifested through narrowing of the "chanel of exhaustion" thus leading to collapse of two level sets and therefore growth of the gradient. Proof of theorem 1. Let δ > 0 be small and U δ be the disc of radius √ δ centered at origin. Denote its 2π-translates on R 2 beÛ δ . Consider the new orthogonal coordinate system with the origin at O 1 = A 1 = (π, 0) and axes ξ : y = x − π, 4 SERGEY A. DENISOV η : y = π − x, the orientation is positive. The relation to the original coordinates is then In this coordinate system, take the rectangle Π δ = {|ξ| < 0.1, |η| < δ}. Rotate it around the origin by π/2 degrees in the original coordinate system and denote the rectangle obtained by Π ′ δ . Consider all 2π-translates of Π δ and Π ′ δ and denote the collection of all rectangles obtained this way byΠ δ .
The existence of such θ 0 is obvious. Then, the solution θ(x, y, t) exists globally and is infinitely smooth. Due to the lemma 2.2, it is even, 2π-periodic, and is invariant under the π/2-rotation around the origin. Therefore, it must also be even with respect to all points D j . The function ζ(x, y, t) is therefore also even with respect to the origin and D j . If so, ∇ ⊥ ζ = 0 at the origin and at D j , so we have θ(O, t) = const, θ(D j , t) = const, and these points do not move under the flow.
Consider the following sets. Take the set of points inside the θ = 3 level curve at t = 0 (an ellipse). Consider its intersection with B 3ǫ = {|α| ≤ 3ǫ, |β| < 0.1}. Denote this set by S 0 . We write S 0 = S 1 0 ∪ S 2 0 where S 1 0 = S 0 ∩ B 2ǫ . Take E(0, 1)S 0 . It will be inside the tube |β| < 0.1. Intersect it with B 3ǫ and take the simply-connected component of this set containing the point O 1 (which does not move under the flow). Denote this set by S 1 . It has the following properties: (1) The boundary of S 1 consists of the part of θ(·, 1) = 3 level curve and parts of the vertical segments α = ±3ǫ. (2) The area |S 1 | ≤ |S 1 0 | = |S 0 | − |S 2 0 |. That simply follows from the lemma 2.4 since S 2 0 will be carried away from B 3ǫ by the flow. (3) S 1 contains the part of the level curve θ(·, 1) = 4 which is symmetric with respect to the origin and connects the following points: the origin and P 1 ± , where α-coordinate of P 1 ± is ±3ǫ, respectively (i.e., they lie on the left and right sides of S 1 ). Indeed, it follows from the fact that B 2ǫ -part of this level curve for t = 0 will stay inside the tube |β| < 0.1 and its edges (P 0 ± ) will be carried away from B 3ǫ at time 1. (4) From the previous property we get that in the decomposition S 1 = S 1 1 ∪ S 2 1 , the set S 2 1 is not empty and has a positive area. We then inductively define the sets S n for all times t n = n. All properties given above will hold true. In particular, For each n, S 2 n is symmetric with respect to O 1 . Consider its right part. The sides α = 2ǫ, α = 3ǫ contain the points of the level curve: θ = 4. Therefore, we have the trivial estimate where ξ n is some point inside S 2 n . The application of lemma 2.3 now yields The obvious modification of this argument gives (1). 2 Remark. By time scaling, we can show that the initial norm of the gradient can be taken as small as we like. We believe, though, that there is an exponential growth of the gradient in our scenario, not just superlinear. For instance, the Euler evolution of the bump in the exterior hyperbolic flow in R 2 allows much better estimates than superlinear. In that case, a simple multiscale argument allows to "cut out" more and more weight from the domain around zero. This is due to the fact that the L p norm of the solution around zero will decrease substantially in time. That we can not guarantee for the periodic case. Notice that the shear flow typically yields only linear growth of the gradient. We never succeeded in applying our method to perturbation of the shear flow (which is generated by another stable stationary solution, say, θ * (x, y) = cos x).
Analogous argument works for the family of equations where ζ = ∆ −γ θ and 1 > γ > 1/2 (although we do not know the global existence of solution). It is an interesting question to extend this proof to γ = 1/2 (the so-called SQG).

Exponential growth over finite time.
In this section, we will prove theorem 2. Proof of theorem 2. We need the following elementary perturbation lemma Lemma 3.1. Consider the following system of equations where f 1(2) , g 1(2) are C 1 -smooth and f 1(2) ∞ , g 1(2) ∞ < ǫ ≪ 1. Then, in every neighborhood of the origin there is a pair (α 0 , β 0 ) such that the solution to the corresponding Cauchy problem satisfies: Proof. Consider two sectors S 1 = {β > 2|α|} and S 2 = {β > |α|} and take any smooth curve γ 0 = γ(0) without self-intersections connecting the sides of S 1 and lying inside S 1 (see Picture 1). Let us control the evolution of this curve γ(t) under the flow given by (10). Clearly, γ(t) is smooth at any moment t. Take some point α 0 , β 0 on γ(0) and consider its trajectory α(t), β(t). The second equation easily implies that until this trajectory leaves the sector S 2 , we have Also, β(t) decreases. Next, take the endpoint on the curve γ(0) which lies on the right side of S 1 . Let it have coordinates (α 0 , 2α 0 ). The first equation of (10) shows that α(t)/2 <α(t) < 2α(t) until the corresponding trajectory (α(t), β(t)) is inside the sector Ω + = {α ≤ β ≤ 2α, α > 0}. Clearly, α(t) increases within this time interval and we have Therefore, we can conclude that (α(t), β(t)) ∈ Ω + for t ∈ (0, t 0 ), where t 0 = (2 ln 2)/3. Analogous inequalities hold true for the other endpoint of γ(0), the corresponding trajectory will not leave Ω − = {−α ≤ β ≤ −2α, α < 0} as long as t ∈ (0, t 0 ). It is also easy to see that all of γ(t) will be inside S 2 for this time interval. Therefore, we can take time t = t 0 and consider the part of the curve γ(t 0 ) which has no self-intersections, connects the opposite sides of S 1 , and lies inside S 1 . This is possible by simple topological argument since the endpoints of γ(t 0 ) are in the sectors Ω ± . Let us call this new curve γ 1 . Then, we consider the evolution of γ 1 repeating the same construction again and again. We will obtain the sequence of curves γ n which are all inside S 1 . Obviously, γ n is a part of γ(nt 0 ). Now, one can easily construct the solution with needed properties by the standard approximation argument. Indeed, for time t = nt 0 consider a point on the curve γ n with α = 0 (there might be many of those). Solve the equation backward obtaining the trajectory. Consider the functions α n (t), β n (t) given by this trajectory up to nt 0 and let α n (t) = 0, β n (t) = β n (nt 0 )e −(t−nt0)/2 for t > nt 0 (see Picture 2). The functions α n (t), β n (t), considered on [0, ∞), have uniformly bounded H 1 (R + ) norms. Indeed, by construction and (11), where the last inequality follows from the fact that the trajectory is inside the sector S 2 . The estimates on the derivative now easily follow from (10). By the Alaoglu theorem, there are α(t), β(t) ∈ H 1 (R + ), such that weakly in H 1 (R + ). The Sobolev embedding is compact and so the weak convergence in H 1 [0, b] implies the uniform convergence on [0, b] for any b > 0. We have β n k (s)[1 + g 1 (α n k (s), β n k (s), s)]ds + t 0 α n k (s)g 2 (α n k (s), β n k (s), s)ds where t ∈ [0, b] and k is large. Taking k → ∞, we see that α(t), β(t) satisfy the integral equations and therefore are solutions to (10). Obviously, we also have Picture 2 α β Remark. The lemma 3.1 is local, i.e. the functions f 1(2) , g 1(2) need to be defined and smooth only around the origin. Take any large T and consider the initial value where φ ǫ is supported around points D j and 2π-translates of O. Around each point D j it is given by (in α, β-coordinates) where φnonnegative spherically symmetric bump with φ(0, 0) = 1 and support inside the unit disc. Around the origin, φ ǫ is a similar bump chosen such that T θ(x, y, 0)dxdy = 0 Clearly, we can arrange for ∇θ(·, 0) ∞ < 10 We will chose ǫ(T ) later. The solution θ will always exist, will be even, invariant under π/2-rotation, and θ(·, t) = θ * (·) + ψ(·, t) where ψ(·, t) 2 ǫ 2 as follows form lemma 2.1 and symmetries of the solution. Assume now that the statement of the theorem is wrong and ∇ψ(·, t) ∞ < 0.1 exp(T /2) + 2 for all t ∈ [0, T ]. Since ψ is even with respect to all points D j , its gradient there is zero and we can write The analogous formula holds for the derivative in β. Let us estimate the second derivatives of ∆ −1 ψ. We consider, say, (∆ −1 ψ) αβ , the others can be treated similarly. Since ∆ −1 has a kernel with ln |z 1 − z 2 | singularity, we get The first term is not larger than Cρ −1 ψ 2 ρ −1 ǫ 2 in absolute value. By our assumption, the second term is dominated by Ce T /2 ρ. Thus, all second derivatives of ∆ −1 ψ are bounded by C(ρ −1 ǫ 2 +ρe T /2 ). By choosing ρ = 0.001C −1 e −T /2 , ǫ = 0.001 ρC −1 , we obtain for t ∈ [0, T ]. Then, the representation (12) allows to write equations for the flow (8) as α = cos β sin α + αf 1 + βf 2 β = − cos α sin β + αg 1 + βg 2 where f 1(2) (α, β, t), g 1(2) (α, β, t) are uniformly smaller than 0.01. The simple modification of the argument from lemma 3.1 shows existence of the flow trajectory that starts at the level set θ = ǫγ, γ < 1/2 and approaches the origin exponentially fast. We therefore have ∇ψ ∞ > 0.5e t/2 thus giving a contradiction at t = T . 2 Remark. Analogous argument shows the infinite exponential growth for the problems where ζ = ∆ −γ θ and γ > 1. For that case, it is also easy to prove that the gradient can not grow faster than the exponential, so the exponential growth is sharp.