CUBO A Mathematical Journal Vol.12, No¯ 03, (13–32). October 2010 A Family of Stationary Solutions to the Euler Equations and Generalized Solutions JULIANA CONCEIÇÃO PRECIOSO Departamento de Matemática, Universidade Estadual Paulista, 15054-000, S.J.R. Preto, SP, Brazil email: precioso@ibilce.unesp.br ABSTRACT In this work, we present a interesting family of stationary solutions for the Euler equations, which behaves in the same way that the approximated solutions presented in [6]. RESUMEN En este trabajo, presentamos una familia interesante de soluciones estacionarias para las ecuaciones de Euler, que se comportan de la misma manera que las soluciones aproximadas presentadas en [6]. Key words and phrases: Euler equations, incompressible flows, generalized solutions. Math. Subj. Class.: 35D99. 14 Juliana C. Precioso CUBO 12, 3 (2010) 1 Introduction An ideal incompressible fluid moving inside D ⊂ Rn is classically described by a velocity field u(t, x) and a pressure field p(t, x), subject to the classical Euler equations: { ∂t u + (u ·∇)u +∇p = 0 ∇· u = 0, (1.1) with the boundary condition that u is tangent to ∂D. In classical continuous mechanics [1], the motion of an incompressible inviscid fluid in a compact domain D ⊂ Rn can be seen as a geodesic on the group of all diffeomorphisms of D with unit jacobian determinant, G(D). This set is included in S(D) the semigroup of all Borel maps h of D that satisfy ˆ D f (h(x))d x = ˆ D f (x)d x, ∀f ∈ C0(D). For more details see [1], [2] or [6]. We will denote V := { u : [0, T] × D −→ Rn such that u ∈ C0(Q), u(t,·) ∈ Li p(D) uniformly in 0 ≤ t ≤ T, div u = 0, u(t,·) · n̂∣∣ ∂D = 0 } . Note that the flow (t, x) 7→ g(t, x) describing the motion of fluid particles is defined by { ∂t g(t, x) = u(t, g(t, x)) g(0, x) = x. (1.2) By Cauchy-Lipschitz theorem, for each u ∈ V , there is a unique solution to (1.2) and for each time t the map g(t, x) = g(t,·) ∈ G(D). Then, by elementary calculations, the Euler equations can be replaced by the following equivalent equations: { ∂2tt g(t, x) +∇p(t, g(t, x)) = 0 det D x g(t, x) = 1. (1.3) Lets call (1.3) by the “Lagrangian formulation” of the Euler equations. From a geometrical point of view, different from the natural PDE point of view which consists in adressing the Euler equations as an evolution equation with prescribed initial CUBO 12, 3 (2010) A Family of Stationary Solutions ... 15 velocity field, it is natural to solve the problem to minimize the action A( g) = 1 2 ˆ T 0 ˆ D |∂t g(t, x)|2 d xd t, among all trajectories on G(D) connecting g(0,·) = I d and g(T,·) = h. The corresponding system of PDE’s is the lagrangian formulation of the Euler equations (1.3). Ebin and Marsden showed local existence and uniqueness for this problem, namely, if h and I are sufficiently close in a sufficiently high order Sobolev norm, then there is a unique geodesic connecting I d to h, see [9]. In the large, uniqueness can fail. However, in [12], A. I. Shnirelman shows that existence of minimal geodesics may fail to a class of data. To solve the problem to find minimal geodesics in a generalized sense, in particular for data h in Shnirelman class, was introduced suitable “Young measure", (see [15] and [18]) of different ways as in [4], [6], [12] and [13]. In [4], was used a concept that takes into account the dynamics of the particles. To each path t ∈ [0, T] 7−→ z(t) ∈ D, one associates the probability that it is followed by some material particle. More precisely, was proposed a notion of the generalized flow, as been a probability measure on set Ω = D[0,T] of all curves t ∈ [0, T] −→ z(t) ∈ D, namely, a Borel probability measure µ, on product space Ω = D[0,T], such that each projection µt for 0 ≤ t ≤ T is a Lebesgue measure on D. The action in this context is express by ˆ Ω ˆ T 0 1 2 |z′ (t)|2 dµt(z)d t. Brenier showed the existence of generalized solutions and, later in [5], the existence and uniqueness of the pressure gradient linked to them through a suitable Poisson equation, but did not obtain for them a complete set of equations beyond the classical Euler equations. However, in [6] it was possible. The problem to find minimal geodesics was reformulated in terms of a pair of measures associated to the field u, solution of the Euler equations, in the following way: Given a smooth trajectory t ∈ [0, T] 7→ g(t, x) ∈ G(D), we define the measures (respectively nonnegative and vector-valued) c(t, x, a) = δ(x − g(t, a)), m(t, x, a) = ∂t g(t, a)δ(x − g(t, a)), (1.4) defined on Q ′ = [0, T] × D × D. These measures satisfy ˆ D c(t, x, a)da = 1, (1.5) ∂t c +∇x · m = 0, (1.6) c(0, x, a) = δ(x − a); c(T, x, a) = δ(x − h(a)). (1.7) 16 Juliana C. Precioso CUBO 12, 3 (2010) Moreover, the measure m is absolutely continuous with respect to c, with a density v ∈ L2(Q ′ , d c), so that m = cv, and the action is given by A( g) = 1 2 ˆ 1 0 ˆ D |v(t, x, a)|2 c(t, x, a)d xda, or equivalently, A( g) = K [c, m], where K [c, m] := sup X {〈c, F〉+〈m,Φ〉}, (1.8) where (c, m) is of the form (1.4) and X = { (F,Φ) ∈ C0(Q′ ) × ( C0(Q ′ ) )n ; F(t, x, a) + 1 2 |Φ(t, x, a)|2 ≤ 0 } . Then, Brenier defined the relaxed problem, as the problem to look for pairs of measures (c, m) that minimize K [c, m] and are admissible in the sense of (1.5), (1.6) and (1.7), but do not necessarily satisfy (1.4). Also was showed that, for D = [0, 1]n and each data h ∈ S(D), the relaxed problem always has solutions (c, m) and that there is a unique locally bounded measure ∇x p in the interior of Q = [0, T] × D, depending only h, such that ∂t(cv) +∇x(cv ⊗ v) + c∇x p = 0, holds in the sense of distributions on the interior of Q ′ , where c is a extension of c for which the product c(t, x, a)∇x p(t, x) is well-defined. Moreover, was showed that for any h ∈ S([0, 1]3) of the form h(x1, x2, x3) = (H(x1, x2), x3), that , for any ε > 0, there is a uε ∈ V such that K (uε) + 1 2ε ||guε (T,·) − h||2L2(D) ≤ Iε(h) +ε, where Iε(h) = infu∈v { K (u) + 1 2ε ||gu(T,·) − h||2L2(D) } and k(uε) = 1 2 ˆ T 0 ˆ D |uε(t, x)|2 d xd t = 1 2 ˆ T 0 ˆ D |∂t gε(t, x)|2 d xd t = A( gε). In addition, the measures (cε, mε) associated with uε, through (1.4), converge, as ε → 0 to the generalized solutions of the relaxed problem. Moreover, the fields uε satisfy ∇x · uε = 0, ∂t uε + (uε ·∇) uε → −∇p, CUBO 12, 3 (2010) A Family of Stationary Solutions ... 17 weakly, as ε tends to zero. As observed in [6], with each solution (c, m) we may associated a measure-valued solution µ, in the sense of DiPerna and Majda, by setting ˆ Q×Rd f (t, x,ξ)dµ(t, x,ξ) = ˆ Q′ f (t, x, v(t, x, a))d c(t, x, a), for any continuous function f ∈ Q ×Rd with at most quadratic growth as ξ → ∞. For more details, see [6] and [7]. In [4], Brenier shows explicit examples of non trivial generalized solutions. A typical example is when D is the unit disk in 2D and h(x) = −x. We know that the problem of the minimal action has two trivial solutions g+(t, x) = eiπt x and g−(t, x) = e−iπt x with the same pressure field p(x) = π 2|x|2 2 . We have another (generalized) solution (c, m) to the same problem which is given by ˆ Q′ f (t, x, a)c(t, x, a)d td xda = ˆ [0,1]×D ˆ 1 0 f (t, G(t, a,θ), a)dθd tda, ˆ Q′ f (t, x, a)m(t, x, a)d td xda = ˆ [0,1]×D ˆ 1 0 ∂tG(t, a,θ) f (t, G(t, a,θ), a)dθd tda, for all continuous function f , where G(t,θ, a) = a cos(πt) + ( 1 −|a|2 ) 1 2 e2iπθ sin(πt) ∈ D. Note that each particle initially located at a ∈ D splits up along a circle of radius ( 1 −|a|2 ) 1 2 sin(πt), with center a cos(πt), that moves across the unit disk and shrinks down to the point −a when t = 1. In general, is very difficult to obtain explicit examples of non trivial generalized solu- tions and the explicit examples constructed by Brenier, are based on the model presented in [4], which takes in account a concept purely Lagrangian of Young measures, the so-called generalized flows. Beyond supplying an application of the model developed in [6], the results of this paper give an interesting information for the limit of a sequence of the stationary so- lutions, showing that they are associated with measures that satisfy the Euler equations in a specified weak sense. For another point of view, the results of the paper give example of as a sequence of highly oscillatory solutions still can have a limit that is solution in some sense. Namely, we exhibited a family uε (which behavior as the“approximated solutions" ar- gued above) such that when ε → 0, the velocity field gets more and more oscillatory, but the measures (cε, mε) associated to the field uε converges to a solution (c, m) of the equations ˆ D c(t, x, a)da = 1, ∂t c +∇x · m = 0, 18 Juliana C. Precioso CUBO 12, 3 (2010) and ∂t(cv) +∇x(cv ⊗ v) + c∇x p = 0. Equivalent vector fields to those ones treaties in this work have been studied as appli- cation of high order essentially no oscillatory (ENO) schemes for smooth solutions of Navier- Stokes and Euler equations, (see [14]), in problems involving the Taylor-Green vortex, (see [8], [3] and [10]) and to explore a discrete singular convolution algorithm (DSC) for solving certain mechanics problems, (see [16] and [17]). 2 A Family of (Stationary) Solutions In this work we consider the following family of stationary solutions to the Euler equations: un(x, y) = ( −cos(x) sin (n y) , 1 n sin(x) cos (n y) , 0 ) , pn(x, y) = − 1 4 ( cos(2x) + 1 n2 cos (2n y) ) . Note that, |D pn(x, y)| ≤ C, and when n goes to infinity the pressure field strongly con- verges to p(x, y) = − 1 4 cos(2x). Then, is easy verify that (un ·∇) un +∇p → 0, when n → ∞. For a moment, let us observe the behavior of family un. For n = 1 we have, u1(x, y) = (−cos(x) sin( y), sin(x) cos( y), 0) and { ẋ = −cos(x) sin( y) ẏ = sin(x) cos( y). (2.1) Then, we have (ẋ, ẏ) = (0, 0) ⇔ (x, y) = ( (2k + 1)π 2 , (2l + 1)π 2 ) or (kπ, lπ), where k, l ∈ Z, (see Figure 1). For n = 2 we have, (ẋ, ẏ) = (0, 0) ⇔ (x, y) = ( (2k + 1)π 2 , (2l + 1)π 4 ) or ( kπ, lπ 2 ) , where k, l ∈ Z, (see Figure 2). Thus, for n we have, (ẋ, ẏ) = (0, 0) ⇔ (x, y) = ( (2k + 1)π 2 , (2l + 1) π 2n ) or ( kπ, lπ n ) , where k, l ∈ Z. Note that, when n → ∞ the velocity field gets more and more oscillatory. In the next sec- tion we will show that the measures (cn, mn) defined by cn(t, x, a) = δ ( x − gun (t, a) ) , mn(t, x, a) = un(t, x)δ(x − gun (t, a)) converges to the solution (c, m) of the equations:ˆ D c(t, x, a)da = 1, (2.2) CUBO 12, 3 (2010) A Family of Stationary Solutions ... 19 –2 –1 0 1 2 y –2 –1 1 2 x Figure 1: Phase portrait of the velocity field u1(x, y) = (−cos(x) sin( y), sin(x) cos( y), 0) –2 –1 0 1 2 y –2 –1 1 2 x Figure 2: Phase portrait of the velocity field u2(x, y) = ( −cos(x) sin(2 y), 1 2 sin(x) cos(2 y), 0 ) ∂t c +∇x · m = 0 (2.3) ∂t(cv) +∇x · (cv ⊗ v) + c∇x p = 0, (2.4) By the consistency theorem in [6], or its generalization for variable density in [11], we know that if un is a solution of the Euler equations, then the pair of the measures (cn, mn) defined as below satisfy the equations (2.2), (2.3), (2.4). 20 Juliana C. Precioso CUBO 12, 3 (2010) 3 The Limite (c, m) In this section we build explicitly the limite (c, m). For this, in first we rewrite the field un(x, y) = ( −cos(x) sin(n y), 1 n sin(x) cos(n y), 0 ) as un(x, y) = ( u11(x, n y), 1 n u21(x, n y), 0 ) , where u11(x, n y) = −cos(x) sin(n y) and u21(x, n y) = sin(x) cos(n y). Of here in ahead, we will omit third coordinate of the fields u ′ ns. We also observe that the field has period 2π. Now, we define { x = x(t,γ,δ) y = y(t,γ,δ) the solution of    d x d t = u11(x, y) = cos(x) sin( y) d y d t = u21(x, y) = sin(x)cos( y) x(0,γ,δ) = γ y(0,γ,δ) = δ. (3.1) Let be 0 ≤ i ≤ n − 1, 0 ≤ α1 ≤ 2π, and 2πi n ≤ α2 ≤ 2π n (i + 1), where i, n ∈ N. This is: • for n = 1, i = 0 and we have 0 ≤ α2 ≤ 2π. • for n = 2, 0 ≤ i ≤ 1 and we have { 0 ≤ α2 ≤ π, if i = 0 π ≤ α2 ≤ 2π, if i = 1. • for n = k, 0 ≤ i ≤ k − 1 and we have    0 ≤ α2 ≤ 2π k , if i = 0 2π k ≤ α2 ≤ 4π k , if i = 1 . . . 2π(k − 1) k ≤ α2 ≤ 2π, if i = (k − 1). CUBO 12, 3 (2010) A Family of Stationary Solutions ... 21 α 2 α 2 α α1 1 0 0 2 π π2 2 π2 π n=1 π n=2 α 2 α 1 0 π2 2 π π n=k k 2π k 2π (k-1) Figure 3: Then, i counts the cells (in the vertical line) from 0 to 2π for each n, as we can observe in Figure 3. Now, we define    xin(t,α1,α2) := x ( t,α1, n ( α2 − 2πi n )) yin(t,α1,α2) := 1 n y ( t,α1, n ( α2 − 2πi n )) + 2πi n . (3.2) Note that, by the definition above • for n = 1, we have i = 0, thus { x01(t,α1,α2) := x(t,α1,α2), y01 (t,α1,α2) := y(t,α1,α2). • for n = 2, we have 0 ≤ i ≤ 1, thus    x02(t,α1,α2) := x(t,α1, 2α2) y02 (t,α1,α2) := 1 2 y(t,α1, 2α2) , if i = 0 and    x12(t,α1,α2) := x(t,α1, 2(α2 −π)) y12 (t,α1,α2) := 1 2 y(t,α1, 2(α2 −π)) +π , if i = 1. • for n = k, we have 0 ≤ i ≤ k − 1, thus    x0k(t,α1,α2) := x(t,α1, kα2) y0k(t,α1,α2) := 1 k y(t,α1, kα2) , if i = 0, 22 Juliana C. Precioso CUBO 12, 3 (2010) ...    xk−1k (t,α1,α2) := x ( t,α1, k ( α2 − 2(k − 1)π k )) yk−12 (t,α1,α2) := 1 k y ( t,α1, k ( α2 2(k − 1)π k )) + 2(k − 1)π k , if i = k − 1. Then, we conclude that 0 ≤ xin ≤ 2π, 2πi n ≤ yin ≤ 2π(i + 1) n and    xin(0,α1,α2) = x ( 0,α1, n ( α2 − 2πi n )) = α1 yin(0,α1,α2) = 1 n y ( 0,α1, n ( α2 − 2πi n )) + 2πi n = α2. Moreover, by (3.2) we have    d xin d t (t,α1,α2) = d x d t ( t,α1, n ( α2 − 2πi n )) d yin d t (t,α1,α2) = 1 n d y d t ( t,α1, n ( α2 − 2πi n )) (3.3) Therefore, by (3.1) and (3.3) d xin d t = u11 ( x ( t,α1, n ( α2 − 2πi n )) , y ( t,α1, n ( α2 − 2πi n ))) = u11 ( xin(t,α1,α2), n y i n(t,α1,α2) − 2πi ) = u11 ( xin(t,α1,α2), n y i n(t,α1,α2) ) = u1n ( xin(t,α1,α2), y i n(t,α1,α2) ) , and d yin d t = 1 n u21 ( x ( t,α1, n ( α2 − 2πi n )) , y ( t,α1, n ( α2 − 2πi n ))) = 1 n u21 ( xin(t,α1,α2), n y i n(t,α1,α2) − 2πi ) = 1 n u21 ( xin(t,α1,α2), n y i n(t,α1,α2) ) = u2n ( xin(t,α1,α2), y i n(t,α1,α2) ) . Now defining { xn(t,α1,α2) := xin(t,α1,α2) yn(t,α1,α2) := yin(t,α1,α2) if 2πi n ≤ α2 ≤ 2π(i + 1) n , CUBO 12, 3 (2010) A Family of Stationary Solutions ... 23 we conclude that    d xn d t (t,α1,α2) = u1n (xn(t,α1,α2), yn(t,α1,α2)) d yn d t (t,α1,α2) = u2n (xn(t,α1,α2), yn(t,α1,α2)) xn(0,α1,α2) = α1 yn(0,α1,α2) = α2. (3.4) In the remain of the work, for simplicity, we will use the following notation: D i = (0, 2π)i = (0, 2π)×···×(0, 2π), i times, where i = 1,··· , 4. Now, we are ready to show the following result: Theorem 3.1. Consider (xn, yn) solution of (3.4). Let { cn(t, x, y,α1,α2) = δ((x, y) − (xn(t,α1,α2), yn(t,α1,α2))) mn(t, x, y,α1,α2) = un(x, y)δ((x, y) − (xn(t,α1,α2), yn(t,α1,α2))) . Then, 〈ϕ, cn〉 → 1 2π ˆ D3 ˆ T 0 ϕ(x(t,α1,β2),γ,α1,γ, t)d tdα1 dβ2 dγ and 〈φ, mn〉 → 1 2π ˆ D3 ˆ T 0 φ 1(x(t,α1,β2),γ,α1,γ, t)u 1 1(x(t,α1,β2), y(t,α1,β2))d tdα1 dβ2 dγ, whenever n → ∞, for any ϕ ∈ C∞0 (D4 × (0, T)) and φ ∈ ( C∞0 (D4 × (0, T)) )2 . Proof. Let ϕ ∈ C∞0 (D4 × (0, T)) and cn(t, x, y,α1,α2) = δ((x, y) − (xn, yn)(t,α1,α2)) then, we have 〈ϕ, cn〉 = ˆ D2 ˆ T 0 ϕ(xn(t,α1,α2), yn(t,α1,α2),α1,α2, t)d tdα1 dα2 = n−1∑ i=0 ˆ 2π n (i+1) 2π n i ˆ 2π 0 ˆ T 0 ϕ ( xin(t,α1,α2), y i n (t,α1,α2) ,α1,α2, t ) d tdα1 dα2 = n−1∑ i=0 ˆ 2π n (i+1) 2π n i ˆ 2π 0 ˆ T 0 ϕ ( x ( t,α1, n ( α2 − 2πi n )) , 1 n y ( t,α1, n ( α2 − 2πi n )) + 2πi n ,α1,α2, t ) d tdα1 dα2 24 Juliana C. Precioso CUBO 12, 3 (2010) Now, make β2 = n ( α2 − 2πi n ) = nα2 − 2πi, namely, α2 = β2 n + 2πi n . Then, we have 〈ϕ, cn〉 = n−1∑ i=0 ˆ D2 ˆ T 0 ϕ ( x(t,α1,β2), 1 n y(t,α1,β2) + 2πi n , α1, β2 n + 2πi n , t ) d tdα1 dβ2 n = An + Bn, where An = 1 2π n−1∑ i=0 (ˆ D2 ˆ T 0 ϕ ( x(t,α1,β2), 2πi n ,α1, 2πi n , t ) d tdα1 dβ2 ) 2π n (3.5) and Bn = n−1∑ i=0 1 n ˆ D2 ˆ T 0 [ ϕ ( x(t,α1,β2), 1 n y(t,α1,β2) + 2πi n ,α1, β2 n + 2πi n , t ) −ϕ ( x(t,α1,β2), 2πi n ,α1, 2πi n , t )] d tdα1 dβ2. Note that, by Mean Value Theorem, we have ∣∣∣∣ϕ ( x(t,α1,β2), 1 n y(t,α1,β2) + 2πi n ,α1, β2 n + 2πi n , t ) − −ϕ ( x(t,α1,β2), 2πi n ,α1, 2πi n , t )∣∣∣∣ = ∣∣∣∣ ∂ϕ ∂y 1 n y(t,α1,α2) + ∂ϕ ∂β2 β2 n ∣∣∣∣ ≤ ≤ (∣∣∣∣ ∣∣∣∣ ∂ϕ ∂y ∣∣∣∣ ∣∣∣∣ L∞((0,2π)4×(0,T)) |y(t,α1,α2)| n + ∣∣∣∣ ∣∣∣∣ ∂ϕ ∂β2 ∣∣∣∣ ∣∣∣∣ L∞((0,2π)4×(0,T)) |β2| n ) ≤ ≤ 2π n (∣∣∣∣ ∣∣∣∣ ∂ϕ ∂y ∣∣∣∣ ∣∣∣∣ L∞ + ∣∣∣∣ ∣∣∣∣ ∂ϕ ∂β2 ∣∣∣∣ ∣∣∣∣ L∞ ) ≤ 2π n ∥ Dϕ ∥L∞ . Then, we obtain |Bn| ≤ n−1∑ i=0 1 n ˆ D2 ˆ T 0 ∣∣∣∣ [ ϕ ( x(t,α1,β2), 1 n y(t,α1,β2) + 2πi n ,α1, β2 n + 2πi n , t ) −ϕ ( x(t,α1,β2), 2πi n ,α1, 2πi n , t )]∣∣∣∣ d tdα1 dβ2 ≤ n−1∑ i=0 1 n 2π n ∥ Dϕ ∥L∞ ˆ 2π 0 ˆ 2π 0 ˆ T 0 d tdα1 dβ2 = 2π n ∥ Dϕ ∥L∞ 4π2 T − 2π n2 ∥ Dϕ ∥L∞ 4π2 T ≤ 2π n ∥ Dϕ ∥L∞ 4π2 T. CUBO 12, 3 (2010) A Family of Stationary Solutions ... 25 Therefore, Bn → 0, when n → ∞. Now, define the function ψ by ψ(γ) := ˆ D2 ˆ T 0 ϕ(x(t,α1,β2),γ,α1,γ, t)d tdα1 dβ2 thus, we rewrite (3.5) as An = 1 2π n−1∑ i=0 ψ ( 2πi n ) 2π n = 1 2π n−1∑ i=0 ψ ( 2πi n ) ( 2π(i + 1) n − 2πi n ) . By this form, if γi = 2πi n then, { γ0,γ1,··· ,γn } is a partition of the (0, 2π) and An = 1 2π n−1∑ i=0 ψ(γi ) ( γi+1 −γi ) is a Riemann sum. Therefore, An → 1 2π ˆ 2π 0 ψ(γ)dγ, when n → ∞. Then, we conclude that 〈ϕ, cn〉 → 1 2π ˆ D3 ˆ T 0 ϕ(x(t,α1,β2),γ,α1,γ, t)d tdα1 dβ2 dγ, when n → ∞, for any ϕ ∈ C∞0 (D4 × (0, T)) . Now, consider φ ∈ ( C∞0 (D4 × (0, T)) )2 and let mn = un(x, y)δ((x, y) − (xn(t,α1,α2), yn(t,α1,α2))) = ( u11(x, n y), 1 n u21(x, n y) ) δ((x, y) − (xn(t,α1,α2), yn(t,α1,α2))) . Then, we obtain 〈φ, mn〉 = ˆ D2 ˆ T 0 φ(xn(t,α1,α2), yn(t,α1,α2),α1,α2, t) [ u11(xn(t,α1,α2), n yn(t,α1,α2)), 1 n u21(xn(t,α1,α2), n yn(t,α1,α2)) ] d tdα1 dα2 = n−1∑ i=0 ˆ 2π n (i+1) 2π n i ˆ 2π 0 ˆ T 0 φ(xin(t,α1,α2), y i n(t,α1,α2),α1,α2, t) [ u11(x i n(t,α1,α2), n y i n(t,α1,α2)), 1 n u21(x i n(t,α1,α2), n yin(t,α1,α2)) ] d tdα1 dα2, 26 Juliana C. Precioso CUBO 12, 3 (2010) and therefore, making β2 = n ( α2 − 2πi n ) we have 〈φ, mn〉 = n−1∑ i=0 ˆ D2 ˆ T 0 1 n φ 1 ( x(t,α1,β2), 1 n y(t,α1,α2) + 2πi n ,α1, β2 n + 2πi n , t ) u11(x(t,α1,β2), y(t,α1,β2))d tdα1 dβ2 + + n−1∑ i=0 ˆ D2 ˆ T 0 1 n2 φ 2 ( x(t,α1,β2), 1 n y(t,α1,α2) + 2πi n ,α1, 2πi n , t ) u21(x(t,α1,β2), y(t,α1,β2))d tdα1 dβ2 = A1n + B1n + A2n + B2n, where, A1n = 1 2π n−1∑ i=0 (ˆ D2 ˆ T 0 1 n φ 1 ( x(t,α1,β2), 2πi n ,α1, 2πi n , t ) u11(x(t,α1,β2), y(t,α1,β2))d tdα1 dβ2 ) 2π, A2n = 1 2πn n−1∑ i=0 (ˆ D2 ˆ T 0 1 n φ 2 ( x(t,α1,β2), 2πi n ,α1, 2πi n , t ) u21(x(t,α1,β2), y(t,α1,β2))d tdα1 dβ2 ) 2π, B1n = n−1∑ i=0 ˆ D2 ˆ T 0 1 n [ φ 1 ( x(t,α1,β2), 1 n y(t,α1,β2) + 2πi n , α1, β2 n + 2πi n , t ) −φ1 ( x(t,α1,β2), 2πi n ,α1, 2πi n , t )] u11(x(t,α1,β2), y(t,α1,β2))d tdα1 dβ2, B2n = n−1∑ i=0 ˆ D2 ˆ T 0 1 n2 [ φ 2 ( x(t,α1,β2), 1 n y(t,α1,β2) + 2πi n ,α1, β2 n + 2πi n , t ) −φ2 ( x(t,α1,β2), 2πi n ,α1, 2πi n , t )] u21(x(t,α1,β2), y(t,α1,β2))d tdα1 dβ2, As we seen before, we conclude that ∣∣∣∣φ i ( x(t,α1,β2), 1 n y(t,α1,β2) + 2πi n ,α1, β2 n + 2πi n , t ) − −φi ( x(t,α1,β2), 2πi n ,α1, 2πi n , t )∣∣∣∣ ≤ 2π n ∥ Dφ ∥L∞ , i = 1, 2. CUBO 12, 3 (2010) A Family of Stationary Solutions ... 27 Thus, we have the estimate |B1n| ≤ n−1∑ i=0 1 n ˆ D2 ˆ T 0 ∣∣∣∣φ 1 ( x(t,α1,β2), 1 n y(t,α1,β2) + 2πi n ,α1, β2 n + 2πi n , t ) −φ1 ( x(t,α1,β2), 2πi n ,α1, 2πi n , t )∣∣∣∣|u 1 1(x(t,α1,β2), y(t,α1,β2))|d tdα1 dβ2, ≤ 2π n ∥ Dφ ∥L∞ 4π2 TC. Therefore, B1n → 0, when n → ∞. Now, we go to study the term B2n. |B2n| ≤ n−1∑ i=0 1 n2 ˆ D2 ˆ T 0 ∣∣∣∣φ 2 ( x(t,α1,β2), 1 n y(t,α1,β2) + 2πi n ,α1, β2 n + 2πi n , t ) −φ2 ( x(t,α1,β2), 2πi n ,α1, 2πi n , t )∣∣∣∣|u 2 1(x(t,α1,β2), y(t,α1,β2))|d tdα1 dβ2 ≤ 2π n2 ∥ Dφ ∥L∞ 4π2 TC, and, therefore, also B2n → 0, when n → ∞. Defining the function ψ by ψ i (γ) := ˆ D2 ˆ T 0 φ i (x(t,α1,β2),γ,α1,γ, t)u i 1(x(t,α1,β2), y(t,α1,β2))d tdα1 dβ2, i = 1, 2, we obtain, A1n = 1 2π n−1∑ i=0 ψ 1 ( 2πi n ) 2π n = 1 2π n−1∑ i=0 ψ 1 ( 2πi n ) ( 2π(i + 1) n − 2πi n ) . By this form, if γi = 2πi n then, { γ0,γ1,··· ,γn } is a partition of the (0, 2π) and A1n = 1 2π n−1∑ i=0 ψ 1(γi ) ( γi+1 −γi ) is a Riemann sum. Therefore, A1n → 1 2π ˆ 2π 0 ψ 1(γ)dγ, when n → ∞. For the last term, we have that A2n = 1 2πn n−1∑ i=0 ψ 2 ( 2πi n ) 2π n 28 Juliana C. Precioso CUBO 12, 3 (2010) = 1 2πn n−1∑ i=0 ψ 2 ( 2πi n ) ( 2π(i + 1) n − 2πi n ) , and therefore, A2n → 0, when n → ∞. Then, we conclude that 〈φ, mn〉 → 1 2π ˆ D3 ˆ T 0 φ 1(x(t,α1,β2),γ,α1,γ, t)u 1 1(x(t,α1,β2), y(t,α1,β2))d tdα1 dβ2 dγ, when n → ∞, for any φ ∈ ( C∞0 (D4 × (0, T)) )2 . By the last theorem we can conclude that 〈ϕ, cn〉 → 1 2π ˆ D3 ˆ T 0 ϕ(x(t,α1,β2),γ,α1,γ, t)d tdα1 dβ2 dγ and 〈φ, mn〉 → 1 2π ˆ D3 ˆ T 0 φ 1(x(t,α1,β2),γ,α1,γ, t)u 1 1(x(t,α1,β2), y(t,α1,β2))d tdα1 dβ2 dγ, whenever n → ∞, for any ϕ ∈ C∞0 (D4 × (0, T)) and φ ∈ ( C∞0 (D4 × (0, T)) )2 . Now, note that 2π〈ϕ, cn〉 → ˆ D3 ˆ T 0 ϕ(x(t,α1,β2),γ,α1,γ, t)d tdα1 dβ2 dγ is equivalent to ˆ 2π 0 〈ϕ, cn〉 dβ2 → ˆ D3 ˆ T 0 ϕ(x(t,α1,β2),γ,α1,γ, t)d tdα1 dγdβ2 and, therefore, ˆ 2π 0 [ 〈ϕ, cn〉− ˆ D2 ˆ T 0 ϕ(x(t,α1,β2),γ,α1,γ, t)d tdα1 dγ ] dβ2 → 0. Then, we conclude that 〈ϕ, cn〉 → ˆ D2 ˆ T 0 ϕ(x(t,α1,β2),γ,α1,γ, t)d tdα1 dγ. CUBO 12, 3 (2010) A Family of Stationary Solutions ... 29 Of completely analogous way, we also conclude that 〈φ, mn〉 → ˆ D2 ˆ T 0 φ 1(x(t,α1,β2),γ,α1,γ, t)u 1 1 ( x(t,α1,β2), y(t,α1,β2) ) d tdα1 dγ. Thus, the limite (c, m) is given by { c(x, y,α1,α2, t) = δ ( (x,α2) − (x(t,α1,β2), y) ) m(x, y,α1,α2, t) = δ ( (x,α2) − (x(t,α1,β2), y) ) ( u11(x, y(t,α1,β2)), 0 ) . (3.6) 4 Solution to the Relaxed Euler Equations In this section we will conclude our work showing that the pair (c, m), build in the before section, satisfy the relaxed Euler equations. Theorem 4.1. The pair of measures (c,m) defined in (3.6) satisfy the following equations ˆ D2 c(t, x, y,α1,α2)dα1 dα2 = 1, ∂t c +∇· m = 0, ∂t(cv) +∇· (cv ⊗ v) + c∇p = 0. in the sense of distributions. Proof. Note that 〈1, c〉 = ˆ D2 ˆ T 0 d tdα1 dα2 = ˆ D2 ˆ T 0 d td xd y. Then, we obtain ˆ D2 ˆ T 0 (ˆ D2 c(t, x, y,α1,α2)dα1 dα2 − 1 ) d td xd y = 0 and, therefore, ˆ D2 c(t, x, y,α1,α2)dα1 dα2 = 1. Now, we will show that pair (c, m) satisfy the equation ∂t c + ∇ · m = 0. Consider ϕ ∈ C∞0 (D4 × (0, T)), thus 〈ϕ(x, y,α1,α2, t),∂t c(t, x, y,α1,α2) +∇(x, y) · m(t, x, y,α1,α2)〉 = = − ˆ D2 ˆ T 0 ∂tϕ(x(t,α1,β2), y,α1, y, t)d tdα1 d y − ˆ D2 ˆ T 0 ∂xϕ(x(t,α1,β2), y,α1, y, t)u 1 1 ( x(t,α1,β2), y(t,α1,β2) ) d tdα1 d y 30 Juliana C. Precioso CUBO 12, 3 (2010) = − ˆ D2 ˆ T 0 ∂t ( ϕ(x(t,α1,β2), y,α1, y, t) ) d tdα1 d y = − ˆ D2 ( ϕ(α1, y,α1, y, T) −ϕ(β2, y,α1, y, 0) ) dα1 d y = 0. Finally, we will show that the pair (c, m) also satisfy the equation ∂t(cv)+∇·(cv⊗v)+c∇p = 0. First note that [ d iv(x, y)(u ⊗ u) ]i = ui d iv(x, y) u+u·∇(x, y) ui = u·∇(x, y) ui , because d iv(x, y) u = 0. Let ϕ ∈ ( C∞0 (D4 × (0, T)) )2 , then 〈ϕ1(x, y,α1,α2, t),∂t ( c(t, x, y,α1,α2)v1(x, y,α1,α2, t)〉+ +〈ϕ1(x, y,α1,α2, t),∇(x, y) · (c(t, x, y,α1,α2)v(x, y,α1,α2, t) ⊗ v(x, y,α1,α2, t))1〉 +〈ϕ1(x, y,α1,α2, t), c(x, y,α1,α2, t)∂x p(x, y)〉 = = − ˆ D2 ˆ T 0 u11(x(t,α1,β2), y(t,α1,β2))∂tϕ 1(x(t,α1,β2), y,α1, y, t)d tdα1 d y − ˆ D2 ˆ T 0 [ u11(x(t,α1,β2), y(t,α1,β2)) ]2 ∂xϕ 1(x(t,α1,β2), y,α1, y, t)d tdα1 d y + ˆ D2 ˆ T 0 ∂x p(x(t,α1,β2), y)ϕ 1(x(t,α1,β2), y,α1, y, t)d tdα1 d y = − ˆ D2 ˆ T 0 u11(x(t,α1,β2), y(t,α1,β2))∂t [ ϕ 1(x(t,α1,β2), y,α1, y, t) ] d tdα1 d y + ˆ D2 ˆ T 0 [ u11(x(t,α1,β2), y(t,α1,β2)) ]2 ∂xϕ 1(x(t,α1,β2), y,α1, y, t)d tdα1 d y − ˆ D2 ˆ T 0 [ u11(x(t,α1,β2), y(t,α1,β2)) ]2 ∂xϕ 1(x(t,α1,β2), y,α1, y, t)d tdα1 d y + ˆ D2 ˆ T 0 ∂x p(x(t,α1,β2), y)ϕ 1(x(t,α1,β2), y,α1, y, t)d tdα1 d y = ˆ D2 ˆ T 0 ∂t [ u11(x(t,α1,β2), y(t,α1,β2)) ] ϕ 1(x(t,α1,β2), y,α1, y, t)d tdα1 d y + ˆ D2 ˆ T 0 ∂x p(x(t,α1,β2), y)ϕ 1(x(t,α1,β2), y,α1, y, t)d tdα1 d y = ˆ D2 ˆ T 0 [ ∂x u 1 1(x(t,α1,β2), y(t,α1,β2))∂t x(t,α1,β2) +∂y u11(x(t,α1,β2), y(t,α1,β2))∂t y(t,α1,β2) ] ϕ 1(x(t,α1,β2), y,α1, y, t)d tdα1 d y + ˆ D2 ˆ T 0 ∂x p(x(t,α1,β2), y)ϕ 1(x(t,α1,β2), y,α1, y, t)d tdα1 d y = ˆ D2 ˆ T 0 ∇(x, y) u11(x(t,α1,β2), y(t,α1,β2)) CUBO 12, 3 (2010) A Family of Stationary Solutions ... 31 u11(x(t,α1,β2), y(t,α1,β2))ϕ 1(x(t,α1,β2), y,α1, y, t)d tdα1 d y + ˆ D2 ˆ T 0 ∂x p(x(t,α1,β2), y)ϕ 1(x(t,α1,β2), y,α1, y, t)d tdα1 d y. Since that p(x) = − 1 4 cos(2x) we have that ∂x p = 1 2 sin(2x) and then, ∂x p(x(t,α1,β2), y) = ∂x p(x(t,α1,β2), y(x(t,α1,β2)). Therefore, we conclude that 〈ϕ1(x, y,α1,α2, t),∂t ( c(t, x, y,α1,α2)v1(x, y,α1,α2, t)〉+ +〈ϕ1(x, y,α1,α2, t),∇(x, y) · (c(t, x, y,α1,α2)v(x, y,α1,α2, t) ⊗ v(x, y,α1,α2, t))1〉+ +〈ϕ1(x, y,α1,α2, t), c(x, y,α1,α2, t)∂x p(x, y)〉 = = ˆ D2 ˆ T 0 [ ∇(x, y) u11(x(t,α1,β2), y(t,α1,β2))u11(x(t,α1,β2), y(t,α1,β2)) +∂x p(x(t,α1,β2), y(t,α1,β2)) ] ϕ 1(x(t,α1,β2), y,α1, y, t)d tdα1 d y. Now, note that u11(x, n y) = −cos(x) sin(n y) and u21(x, n y) = sin(x) cos(n y), namely, u11(x, z) = −cos(x) sin(z) and u21(x, z) = sin(x) cos(z), where z = n y. Then, ∇u11 · u1 = −sin(x) cos(x) = − 1 2 sin(2x), and, therefore, ∇u11 · u1 +∂x p = − 1 2 sin(2x) + 1 2 sin(2x) = 0. Then, ∇(x, y) u11 ( x(t,α1,β2), y(t,α1,β2) ) · u1(x(t,α1,β2), y(t,α1,β2))+ +∂x p(x(t,α1,β2), y(t,α1,β2)) = 0 and we can conclude that the pair of measures (c, m = cv) satisfy ∂t(cv) +∇· (cv ⊗ v) + c∇p = 0. 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