# Energy Conservation for 3D Tropical Climate Model in Periodic Domains

Authors: Hui Zhang
Journal: Journal of Mathematics
 (2021)

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## Abstract

In this paper, we prove the energy conservation for the weak solutions of the 3D tropical climate model under some sufficient conditions. Our results are similar to Onsager’s conjecture which is on energy conservation for weak solutions of Euler equations.


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## Body

## 1. Introduction

The incompressible Euler equations inℝ3 are read as follows:(1)∂tu+u⋅∇u+∇p=0,∇⋅u=0,ux,0=u0x,where u is the velocity and p is the pressure.Given that the solution to system (1) is sufficiently smooth, it is easy to see that the total kinetic energy of the flow is conserved. However, turbulent flows are highly irregular; one major character of turbulent fluids is the energy dissipation. In the context of hydrodynamics, this leads to the famous Onsager’s conjecture [1]. Specifically, Onsager’s conjecture that every weak solution to the Euler equations with Hölder continuity exponent α>1/3 conserves energy; and anomalous dissipation of energy occurs when α<1/3. The first part of the conjecture was proved by Eyink [2] and extended to Besov space in subsequent work of Constantin et al. [3], while significant progress has recently been made on the second part, for example, see [4, 5], and some prior work on the compressible Onsager’s conjecture can be found in the literature, for example, see [6, 7].The object of this study is to address the relation between the energy conservation and the degree of regularity of the solution for the tropical climate model inΩ=0,13 with periodic boundary conditions(2)∂tu+u⋅∇u+∇p+∇⋅v⊗v=0,∂tv+v⋅∇u+u⋅∇v+∇θ=0,∂tθ+u⋅∇θ+∇⋅v=0,∇⋅u=0,with initial data(3)ux,0=u0x,vx,0=v0x,θx,0=θ0x,where uandv denote the vector fields and θandp denote the temperature and the pressure, respectively.The system (2) is derived from the inviscid primitive equations by Frierson et al. [8]. For more details of the related background about the tropical climate model, for instance, see [9, 10] and the references therein. Here, it is worth particularly mentioning the solution of (2) with enough regularity to justify integration by parts, and it is clear that the following total energy is conserved:(4)total energy:Et=ux,tL2Ω2+vx,tL2Ω2+θx,tL2Ω2.So, it seems very reasonable to investigate the analogue of Onsager’s conjecture about system (2). To the best of our knowledge, there is no available result for the weak solutions of the tropical climate model. Our study is motivated by [3, 11]. However, we are unable to obtain the desired result using the same processes as [3, 12]. The obstacle is due to ∇⋅v≠0 and the nonlinear term u⋅∇θ. In fact, we require a stronger regularity about v due to the presence of the term u∇⋅v. Now, we are ready to state the main results.Theorem 1.
Letu∈L30,T;B3α,∞Ω∩C0,T,L2Ω, ∇⋅v∈L30,T;L3Ω, v∈L30,T;B3β,∞Ω∩C0,T,L2Ω, and θ∈L30,T;B3γ,∞Ω∩C0,T,L2Ω be a weak solution of systems (2) and (3), i.e.,(5)u,∂tφ+u⊗u,∇φ+p,∇φ+v⊗v,∇φ=0,v,∂tφ+u⊗v,∇φ+v⊗u,∇φ+u⋅∇v,φ+θ,∇⋅φ=0,θ,∂tϕ+uθ,∇ϕ+v,∇ϕ=0,where φ∈Cc∞Ω×R+ is a vector field and ϕ∈Cc∞Ω×R+ is a scalar field. If(6)α>13,α+2β>1,α+2γ>1,then we have(7)Et=E0,fort∈0,T.


## 2. Preliminary Remarks

In this section, we briefly discuss some properties of the Besov spaceB3α,∞Ω.(8)uB3α,∞Ω≔uL3Ω+supx∈Ωux+y−uxL3Ωyα<+∞.From the above definition, we have the following inequalities [3]:(9)ux+y−uxL3Ω≤CyαuB3α,∞Ω.Letη∈Cc∞ℝ3 be a standard mollifying kernel and set(10)ηϵ=1ϵ3ηxϵ.With the notation(11)uϵ=ηϵ∗u,x∈Ωϵ=x∈Ω:distx,∂Ω>ϵ,we have the following inequalities [3]:(12)u−uϵL3≤CϵαuB3α,∞Ω,(13)∇uϵL3≤Cϵα−1uB3α,∞Ω,(14)uϵL3≤u−uϵL3+uL3≤Cϵα+1uB3α,∞Ω.Now, we define(15)rϵu,vx=∫Ωηϵyux−y−ux⊗vx−y−vxdy,qϵu,vx=∫Ωηϵyux−y−ux∇⋅vx−y−vxdy,hϵu,θx=∫Ωηϵyux−y−uxθx−y−θxdy.Then, it is easy to check the following equalities, it is pointed out that the first equality was discovered in [3]:(16)u⊗vϵ=uϵ⊗vϵ+rϵu,v−u−uϵ⊗v−vϵ,u∇⋅vϵ=uϵ∇⋅vϵ+qϵu,v−u−uϵ∇⋅v−vϵ,uθϵ=uϵθϵ+hϵu,θ−u−uϵθ−θϵ.


## 3. Proof of the Main Theorem

For the sake of simplicity, we will proceed as if the solution is differentiable in time. The extra arguments need to be modifed in time are straightforward.Proof.
First, we apply the test functionφ=uϵϵ on the first equation and then we have(17)12ddt∫Ωuϵx,t2dx=∫Ωu⊗uϵ∇uϵdx+∫Ωv⊗vϵ∇uϵdx=I1+I2.
We point out that theI1 term can be estimated as same as that in [3], and we omit the details:(18)I1≤Cϵ3α−1uB3α,∞Ω3.
Now, we estimateI2:(19)I2=∫Ωv⊗vϵ∇uϵdx=∫Ωvϵ⊗vϵ+rϵv,v−v−vϵ⊗v−vϵ∇uϵdxdt=I21+I22+I23.
Using the following inequality and (12) and (13), we obtain(20)rϵv,vL3/2≤Cϵ2βvB3β,∞Ω2.
We can also obtain the following estimate through the above inequalities:(21)I22;I23≤Cϵα+2β−1vB3β,∞Ω2uB3α,∞Ω.
Because∇⋅v≠0, I21 cannot be zero, and we will estimate this term later.
We rewrite the second equation as follows:(22)∂tv+∇⋅u⊗v+∇⋅v⊗u−u∇⋅v+∇θ=0.
Second, we apply the test functionφ=vϵϵ on the above equation and then we have(23)12ddtvϵ22=∫Ωu⊗vϵ∇vϵdx+∫Ωv⊗uϵ∇vϵdx+∫Ωu∇⋅vϵvϵdx+∫Ωθϵ∇⋅vϵdx=H1+H2+H3+H4.
Using (12), one can easily obtain(24)rϵu,vL3/2≤Cϵα+βuB3α,∞ΩvB3β,∞Ω.
Then,(25)H1=∫Ωuϵ⊗vϵ+rϵu,v−u−uϵ⊗v−vϵ∇vϵdx=H11+H12+H13.
Similar to (21), we have(26)H12;H13≤Cϵα+2β−1vB3β,∞Ω2uB3α,∞Ω.
We can rewriteH11 as follows:(27)H11=−∫Ω∇⋅uϵ⊗vϵvϵdx=−∫Ωvϵ⋅∇uϵvϵdx−∫Ωuϵvϵ∇⋅vϵdx=−∫Ωvϵ⊗vϵ∇uϵdx−∫Ωuϵvϵ∇⋅vϵdx.
We can estimateH2 using a similar process as follows:(28)H2=∫Ωvϵ⊗uϵ+rϵv,u−v−vϵ⊗u−uϵ∇vϵdx=H21+H22+H23.
Then, we have the following estimate:(29)H22;H23≤Cϵα+2β−1vB3β,∞Ω2uB3α,∞Ω.
Here, we have used the following equality due to∇⋅u=0:(30)H21=−∫Ω∇⋅vϵ⊗uϵvϵdx=−∫Ωuϵ⋅∇vϵvϵdx−∫Ωuϵvϵ∇⋅uϵdx=0.
Now, we estimateH3 as follows:(31)H3=∫Ωuϵ∇⋅vϵ+qϵu,v−u−uϵ∇⋅v−vϵvϵdx=H31+H32+H33.
Using (12), one can obtain the following inequality:(32)qϵu,vL3/2≤CϵαuB3α,∞Ω∇⋅vL3Ω.
Then, we have(33)H32;H33≤CϵαCϵβ+1∇⋅vL3ΩuB3α,∞ΩvB3β,∞Ω.
From the above, we can easily find that(34)I21+H11+H31=0.
We will estimateH4 later. At last, we apply the test function ϕ=θϵϵ on the first equation and then we have(35)12ddtθϵL2Ω2=∫Ωuθϵ∇θϵdx−∫Ω∇⋅vϵθϵdx=N1+N2.
Now, we have(36)N1=∫Ωuϵθϵ+hϵu,θ−u−uϵθ−θϵ∇θϵdx=N11+N12+N13.
Similarly, we also have the following inequality:(37)hϵu,θL3/2≤Cϵα+γuB3α,∞ΩθB3γ,∞Ω.
Then, we have(38)N2;N3≤Cϵα+2γ−1uB3α,∞ΩθB3γ,∞Ω2.
Because∇⋅uϵ=0, we can easily find that(39)N11=0.
We can also see that(40)H4+N2=0.
Finally, we can establish the following inequality through the above estimates and condition (6):(41)Eϵt−Eϵ0≤Cϵ3α−1+Cϵα+2β−1+Cϵα+2γ−1⟶0,as ϵ⟶0, which completes the proof.



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*Source: 9922077-2021-08-09.xml*