Given initial data Y Y Y

Given initial data (ψ0,u0)(ψ0,u0), it is standard to prove that (1.3) has a unique solution (ψ,u)(ψ,u) on [0,T][0,T] for some T>0T>0, so that there holds the first line of (6.9). While we get by taking div to the momentum equation of (1.3) that∇p=−2∇∂2ψ+∇(−Δ)−1div u⋅∇u+div(∇ψ⊗∇ψ) , which along with the first line of (6.9) implies that ∇p∈C([0,T];Hs−1(R2))∇p∈C([0,T];Hs−1(R2)). This proves (6.9).It remains to verify the blow-up criterion (6.10). Toward this, for any t<T?t<T?, we get by using a standard 5-BrdU estimate for (1.3) that12ddt(‖∇ψ(t)‖L22+‖u(t)‖L22)+‖∇u(t)‖L22=0, which implies that for any t<T?t<T?equation(6.11)‖∇ψ(t)‖L22+‖u(t)‖L22+‖∇u‖Lt2(L2)2≤‖∇ψ0‖L22+‖u0‖L22. While acting ΔjΔj to u1u1 equation of (1.3) and then taking the L2L2 inner product of the resulting equation with Δju1Δju1, it leads toequation(6.12)12ddt‖Δju1(t)‖L22+‖∇Δju1‖L22=−(Δj∂1(p+∂2ψ) Δju1)−(Δj(u⋅∇u1) Δju1)−(divΔj(∂1ψ∇ψ) Δju1). Similarly acting ΔjΔj to u2u2 equation of (1.3) and then taking the L2L2 inner product of the resulting equation with Δju2Δju2 leads toequation(6.13)12ddt‖Δju2(t)‖L22+‖∇Δju2‖L22+(ΔjΔψ Δju2)=−(Δj∂2(p+∂2ψ) Δju2)−(Δj(u⋅∇u2) Δju2)−(divΔj(∂2ψ∇ψ) Δju2). However by the transport equation of (1.3) and using integration by parts, one has(ΔjΔψ Δju2)=−(ΔjΔψ Δj(∂tψ+u⋅∇ψ))=12ddt‖∇Δjψ(t)‖L22+(Δj∇ψ ∇Δj(u⋅∇ψ)). Hence by combining (6.12) with (6.13) and using divu=0, we obtainequation(6.14)12ddt(‖Δju(t)‖L22+‖∇Δjψ(t)‖L22)+‖∇Δju‖L22=−(Δj(u⋅∇u) Δju)−(Δj∇ψ ∇Δj(u⋅∇ψ))−(divΔj(∇ψ⊗∇ψ) Δju).Next for s>0s>0, we claim thatequation(6.15) (Δj(u⋅∇b) Δjb) ?cj(t)22−2js(‖∇u(t)‖L∞‖b(t)‖H˙s2+‖∇b(t)‖L∞‖u(t)‖H˙s‖b(t)‖H˙s)or (Δj(u⋅∇b) Δjb) ?cj(t)22−2js(‖∇u(t)‖L∞‖b(t)‖H˙s2+‖b(t)‖L∞‖∇u(t)‖H˙s‖b(t)‖H˙s). Indeed applying Bony\'s decomposition (3.7) for u⋅∇bu⋅∇b and then using a sex chromosomes standard commutator\'s argument, we can write(Δj(u⋅∇b) Δjb)=∑ j−? ≤5([Δj;S?−1u]⋅∇Δ?b+(S?−1u−Sj−1u)⋅∇Δ?Δjb Δjb)+(Sj−1u⋅∇Δjb Δjb)−(Δj(R(u,∇b)) Δjb). It follows from the classical commutator\'s estimate (see [2] for instance) that∑ j−? ≤5 ([Δj;S?−1u]⋅∇Δ?b Δjb) ?∑ j−? ≤5‖∇S?−1u‖L∞‖Δ?b‖L2‖Δjb‖L2?cj(t)22−2js‖∇u(t)‖L∞‖b(t)‖H˙s2. The same estimate holds for ∑ j−? ≤5((S?−1u−Sj−1u)⋅∇Δ?Δjb Δjb) and (Sj−1u⋅∇Δjb Δjb).Whereas applying Lemma 3.1 gives‖ΔjR(u,∇b)‖L2?∑?≥j−N0‖Δ?u‖L2‖S?+2∇b‖L∞, which can be controlled by cj(t)2−js‖∇b(t)‖L∞‖u(t)‖H˙s or cj(t)2−js‖b(t)‖L∞‖∇u(t)‖H˙s as long as s>0s>0. This completes the proof of (6.15).Now we go back to (6.14). In fact, applying Lemma 3.4(i) and (6.15) to (6.14) gives12ddt(‖Δju(t)‖L22+‖∇Δjψ(t)‖L22)+‖∇Δju‖L22?cj(t)22−2js ‖∇u‖L∞(‖u‖H˙s2+‖∇ψ‖H˙s2)+‖∇ψ‖L∞‖∇u‖H˙s‖∇ψ‖H˙s fors>0. The above implies for any s>0s>0‖u(t)‖H˙s2+‖∇ψ(t)‖H˙s2+‖∇u‖Lt2(H˙s)2≤‖u0‖H˙s2+‖∇ψ0‖H˙s2+C∫0t(‖∇u(t′)‖L∞+‖∇ψ(t′)‖L∞2)(‖u(t′)‖H˙s2+‖∇ψ(t′)‖H˙s2)dt′. Applying Gronwall\'s inequality yields‖u(t)‖H˙s2+‖∇ψ(t)‖H˙s2+‖∇u‖Lt2(H˙s)2≤(‖u0‖H˙s2+‖∇ψ0‖H˙s2)exp? C∫0t(‖∇u(t′)‖L∞+‖∇ψ(t′)‖L∞2)dt′ fort<T?, which together with (6.11) implies (6.10). This completes the proof of Proposition 6.1.  □