Ergodicity of Stochastic Burgers’ System with Dissipative Term

Theorem 8. With conditions in Lemma 2, when ϑ > 1/4, there exists an invariant measure for (2).

Proof. Multiplying (131) by v_τ and integrating on D, we get

For the third term on the left hand side of (133), we deduce that Substituting (134) into (133), we have For the third term on the right hand side of (135), we get by the Young inequality that For the last term on the right hand side of (135),

Since v_τ(t) is vector field, we denote it by $$, where $$ is real valued function, i = 1,2. For r₁, we have

Similarly for r₂, Analogously to r₁, we deduce that For r₄, we have Since {A^1/2W_A(t)} _t∈ℝ is a Gaussian process, we infer that Then, with the proof of Lemma 2, we know that $$ is continuous with respect to t. By (137)–(141), we have By (135), (136), and (143), we arrive at It is equivalent to Since ϑ > 1/4, let ɛ be small enough, such that Then, the above estimates can be changed into By the Gronwall inequality, we get Similarly to the argument of [26], we will prove that $$ has at most polynomial growth, when t → −∞ a.s. So, we conclude that Multiplying e^δt on both sides of (147) and integrating with respect to t, we have As by (149), we have By Theorem 7, we know that for problem (131) there exists unique mild solution, which has the following: Then, for any ζ ∈ (0, θ)∩(0, 1/4), where the θ is the parameter in Lemma 2, Since then, For z₁, we have In the following, we use Theorem 6.13 in chapter two of [27] to estimate them respectively as follows: the last inequality follows by Theorem A.8 in [25], where δ > 0,  $$, and $$. So, by Hölder inequality and interpolation inequality, we have For z_1,2, we have where Analogously to estimating z_1,1, we have Similarly, we can get the same estimates for z_1,3 and z_1,4. Therefore, Analogously to estimating z₁, we can get for z₂, z₃, and z₄ that So, by (163)–(164) and (156), we get For the third term on the right hand side of (154), we obtain since $$ and $$ are bounded for t, τ ∈ (−∞, T], the last inequality follows. For the first term on the right hand side of (154), we have Similar to [26], we can prove that $$ has at most polynomial growth when τ → ∞. For the reader convenience, we sketch a proof. By Lemma 2, we know that W(t) − W(s) is a D(A^θ/2) valued Brownian motion, for s ≤ t ≤ 0. So, by the law of iterated logarithm, we have Obviously, w_n is a i.i.d sequence. By the law of large numbers, there exists an integer-valued random variable n₀(w) > 0, when n ≥ n₀(w), we have This implies that for all n > 0. In other words, when s ≤ t ≤ [s] + 1. By the law of iterated logarithm, we have for some positive random variable. By Theorem 5.14 in [23], we know that So, we have that since s ≤ [t] − 1, the fourth inequality follows. By (167) and (174), we know that If we let ζ = 1/2 < θ, repeating the argument of (174), we can see that $$ also has at most polynomial growth, when t → −∞ a.s., since we have the Sobolev embedding H^3/2 ⊂ W^1,4. Consider the second term on the right hand side of (154), by (165), where the last inequality follows by (152). Analogously, we can prove that where we used (149) and (152) for the last inequality. By (154) and (175)–(177), we get for some positive random variable ξ(w). As H^1+δ ⊂ H¹ is compact, by Prohorov Theorem, we know that the family of laws for (v_τ(0)) _τ≥0 taking values in H¹ is tight. Since v_τ(0) = u_τ(0) − W_A(0), then so does the law of (u_τ(0)) _τ≥0 taking values in the same space. For t ≥ 0, set where $$. Following the arguments in [24], for all t₀ < s < t and all $$, by proving we can show that u is a Markov process. Here, ℱ_s is the σ-algebra generated by W(r) for r ≤ s. So, (P_t) _t≥0 is the Markov semigroup. Define a dual semigroup $$ in the space $$ of probability measures on $$ as follows: Let μ_τ be the law of u_τ(0), which is the solution of (2) with initial condition u(−τ) = 0. Then, we have where we use the fact that u(τ, ·; 0,0) and u_τ(0) have the same law, the second equality follows. Therefore, Since (μ_τ) _τ≥0 is tight, then by Prokhorov theorem, we know that (μ_τ) _τ≥0 is relatively compact. We can choose a subsequence of (μ_τ) _τ≥0 denoted by $$ such that for μ ∈ P(H^σ),

Theorem 9. Assume θ > 1/2 in Lemma 2 and ϑ > 1/4; then,

• (i)

  the stochastic Burgers equation (2) has a unique invariant measure μ;

• (ii)

  for all $$ and all Borel measurable functions φ, $$, such that $$,

  $$ ()

• (iii)

  for every Borel measure μ^* on $$, one has that

  $$ ()

where ∥·∥_TV stands for the total variation of a measure. In particularly, one has that for every Borel set $$(the Borel σ-algebra of $$).

Theorem 10. Assume that the probability measures $$, are all equivalent, in the sense that they are mutually absolutely continuous. Then, Theorem 9 holds true.

Lemma 11. Define Ψ(ϕ) = u(·, x, ϕ); then,

• (i)

  the mapping

  $$ ()

•  

  is continuous, where C₀([0, T]; B)≔{h ∈ C([0, T]; B); h(0) = 0} for Banach space B;

• (ii)

  for every x, y ∈ H^3/2 and T > 0 there exists $$ such that $$.

Proof. (i) is proved by (A.30) in the Appendix. To prove (ii), let x, y, ∈H^3/2 and T > 0, define $$ as

Obviously, $$. Define $$ as the solution of the following equation: with initial condition $$; then $$. Set $$; then it satisfies all the requirements of the lemma.

Proposition 12. With conditions in Theorem 9, the irreducibility property (I) is satisfied.

Proof. Let x ∈ H^3/2 and $$ be the same as (ii) in Lemma 11. By the above lemma, we have that for ɛ > 0, we can find δ > 0, such that

implies that If θ > 1/2 in Lemma 2, and denote z and $$ the corresponding Ornstein-Uhlenbeck process satisfying conditions in the lemma, then $$. Choose δ₁ > 0 such that δ₁ < δ and Then, for $$, we have that Recall now that the solution u of the stochastic Burgers equation is equal to Ψ(z), z being the Ornstein-Uhlenbeck process. Then, it remains to show that But this is obviously true. So far, we have proved that for for all t > 0, for  all  x, y ∈ H^3/2, for all ɛ > 0, Next, we will prove for all $$, the above inequality also holds. Indeed, for 0 < h < t, by Chapman-Kolmogorov equation, we have Since P(t − h, x₀, H^3/2) = 1, we will extend (204) to the case for all $$. If this is not true, there exists t₀ > 0,  $$, ɛ > 0 such that Then, we can choose y₁ ∈ H^3/2,  ɛ₁ > 0 such that B(y₁, ɛ₁) ⊂ B(y₀, ɛ). By (204), we have which is contrary to (205).

Proposition 13. There exists a unique mild solution $$ for (208) which is Markov process with the Feller property in $$, that is for every R > 0, t > 0, there exists a constant L = L(t, R) > 0 such that

holds for all $$, and all $$, where $$ is the transition probabilities corresponding to (204).

Proof. The proof of existence and uniqueness is similar to Section 2. Let ϕ₁ = ϕ₂ in (A.28), by the Gronwall inequality, we know that u_R is Lipschitz continuous with respect to initial value. Using the method in Proposition 4.3.3 in [24], we can prove that the solution is a Markov process. To prove the Fell property, we first consider the following Galerkin approximations of (208). Let P_n be the orthogonal projection in H defined as $$. Clearly, H_n≔P_nH for every n. Consider the equation in H_n as follows:

with initial condition $$. This is a finite-dimensional equation with globally Lipschitz nonlinear functions, so it has a unique progressively measurable solution with P-a.e. trajectory $$, which is also a Markov process in H_n with associated semigroup $$ defined as for all x ∈ H_n and ϕ ∈ C_b(H_n). For every R > 0, t > 0, we can prove that there exists a constant L = L(t, R) > 0 such that hold for all n ∈ ℕ, x, y ∈ H_n, and all ϕ ∈ C_b(H_n) with $$. Indeed, the following remarkable formula holds true for the differential in x of $$ [29]: for all h ∈ H_n, where β_n is a n-dimensional standard Wiener process with incremental covariance P_nQ and Q is the covariance operator of W(t). Obviously, Q is nonnegative, adjoint, Hilbert-Schmidt operator with inverse. Since the eigenvalues α_n of the Stokes operator A, in 2-space dimension, behave like n, let θ = 1/2 + ɛ for some ɛ > 0, in Lemma 2, we have D(A) ⊂ ℛ(Q) ⊂ D(A^3/4), where ℛ(Q) is the image of Q. Therefore, Since for y ∈ H_n it follows that where the last inequality follows by the Estimate 4 of the Appendix (note that C(R) is independent of x ∈ H_n and n ∈ ℕ). Indeed, $$ is given by v_n(t, x) + P_nz(t), where z is the Ornstein-Uhlenbeck process, and v_n is the solution of (A.2). Therefore, In the following step, we will let n → ∞ to get the Fell property for (208). Let $$ and $$ be given. From the Appendix, Remark A.1, we know that $$ converges to u^(R)(t) strongly in $$, p-a.s.. By the boundedness and continuous of ϕ as well as Lebesgue dominated convergence theorem, we have which implies that for some subsequence n_k, for a.e. t ∈ [0, T]. Take $$, by the previous argument, we can find a subsequence n_k such that the previous almost sure convergence in t ∈ [0, T] holds true both x and y.

Thus, from (212), we have

for a.e. t ∈ [0, T]. As u^(R)(t; x) has continuous trajectories with values in $$, the above inequality holds for all t ∈ [0, T].

Proposition 14. Under conditions of Theorem 9, (S) holds true.

Proof. Take $$ satisfying x_n → x in H¹. For every R > 0, we have that

as n → ∞ by Proposition 13. Then, where the inequality follows by the consistency of u(t; x) and u^(R)(t; x), when $$, and the limit follows by (A.21). Therefore, as n → ∞.