Proof. We shall complete the proof of the theorem by an inductive argument with respect to the index s. Let us first give upper bound for ‖ρ‖L∞(S).
From (31), we derive(34)ρL∞=ρ0L∞e∫0tuxs,-ηs,xds≤ρ0L∞e∫0tuxs,·L∞ds.Next, we split four steps to finish the proof of Theorem 10.
Step 1. For s∈(2,3), applying Lemma 3 to the second equation of system (23), we have(35)ρHs-2S≤ρ0Hs-2+C∫0tρuxHs-2dτ+C∫0tρHs-2u+2γ2L∞+∂xuL∞dτ.Using (6), we get(36)uxρHs-2S≤CuxHs-2ρL∞+uxL∞ρHs-2=Cu+γ1xHs-2ρL∞+uxL∞ρHs-2.Therefore, it yields(37)ρHs-2S≤ρ0Hs-2+C∫0tu+γ1Hs-1ρL∞dτ+C∫0tρHs-2uL∞+1+uxL∞dτ.
Differentiating once the second equation of system (23) with respect to x, we have(38)∂tρx-u+2γ2∂xρx=2uxρx+uxxρ.Using Lemma 3, we get(39)ρxHs-2S≤ρ0xHs-2+C∫0t2ρxux+ρuxxHs-2dτ+C∫0tρxHs-2u+2γ2L∞+∂xuL∞dτ.From (7), we have(40)uxρxHs-2≤CuxHs-1ρL∞+uxL∞ρxHs-2=Cu+γ1xHs-1ρL∞+uxL∞ρxHs-2and(41)ρ∂xuxHs-2≤CρHs-1uxL∞+ρL∞uxxHs-2=CρHs-1uxL∞+ρL∞u+γ1xxHs-2.Therefore, we have(42)ρxHs-2S≤ρ0xHs-2+C∫0tu+γ1HsρL∞dτ+C∫0tρHs-1uL∞+1+uxL∞dτ.From (37) and (42), it has(43)ρHs-1S≤ρ0Hs-1+C∫0tu+γ1HsρL∞dτ+C∫0tρHs-1uL∞+1+uxL∞dτ.
On the other hand, the first equation of system (23) is equivalent to(44)∂t-u-γ1+-u-γ1∂x-u-γ1=-A-1∂x2μ0u+γ1+12u+γ1x2+12ρ2.
Therefore, using Lemma 2, we get from (44)(45)u+γ1HsS≤C∫0tA-1∂x2μ0u+γ1+12u+γ1x2+12ρ2Hsdτ+u0+γ1Hs+C∫0tu+γ1Hs∂xuτL∞dτ.Note that(46)A-∂xfHsS≤A-∂xfL2S+A-∂x2fHs-1S≤gx∗fL2S+-f+∫Sf dxHs-1S≤gxL1SfL2S+2fHs-1S≤CfHs-1S,where we have used (27), (28), and Young’s inequality.
Using (6) and (46), one has(47)A-1∂x2μ0u+γ1+12u+γ1x2+12ρ2Hs≤C2μ0u+γ1+12u+γ1x2+12ρ2Hs-1≤Cu+γ1Hs-1+u+γ1xHs-1uxL∞+ρHs-1ρL∞.
Thus, we reach(48)u+γ1HsS≤u0+γ1Hs+C∫0tu+γ1Hs1+uxL∞dτ+C∫0tρτHs-1ρτL∞dτ,which together with (43) reaches(49)u+γ1HsS+ρtHs-1S≤u0+γ1Hs+ρ0Hs-1+C∫0tu+γ1Hs+ρtHs-1×uxL∞+1+uL∞+ρL∞dτ.
Using Gronwall’s inequality, we have(50)ut+γ1HsS+ρtHs-1S≤u0+γ1Hs+ρ0Hs-1eC∫0tuL∞+uxL∞+1+ρL∞dτ.
From (22) and (34), we get(51)ut+γ1HsS+ρtHs-1S≤u0+γ1Hs+ρ0Hs-1eC1t+1exp∫0tuxL∞dτ,where C1=C1(|μ0|,μ1,‖ρ0‖L∞).
Hence, if the maximal existence time T<∞ satisfies ∫0t‖ux‖L∞dτ<∞, we obtain from (51) that(52)lim sup t→Tut+γ1HsS+ρtHs-1S<∞,
which contradicts the assumption on the maximal existence time T<∞. It completes the Theorem 10 for s∈(2,3).
Step 2. For s∈[2,5/2), applying Lemma 2 to the second equation of system (23), we get(53)ρHs-1S≤ρ0Hs-1+C∫0tuxρHs-1dτ+C∫0tρHs-1∂xu+γ1L∞∩H1/2dτ.From (6), we get(54)ρuxHs-1S≤CuxHs-1ρL∞+ρHs-1uxL∞=Cu+γ1xHs-1ρL∞+ρHs-1uxL∞,which ensures that(55)ρHs-1S≤ρ0Hs-1+C∫0tu+γ1HsρL∞dτ+C∫0tρHs-1uxL∞+∂xu+γ1L∞∩H1/2dτ,
which together with (48) gives rise to(56)ut+γ1Hs+ρtHs-1≤u0+γ1Hs+ρ0Hs-1+C∫0tu+γ1Hs+ρtHs-1×uxL∞+1+u+γ1xH1/2∩L∞+ρL∞dτ.
Using Gronwall’s inequality, we have(57)ut+γ1Hs+ρtHs-1≤u0+γ1Hs+ρ0Hs-1eC∫0t1+u+γ1H3/2+ε+ρL∞dτ,where ε∈(0,1/2) and we used the fact that H1/2+ε↪L∞∩H1/2.
Therefore, using Step 1 and arguing by induction assumption, we get that(58)u+γ1H3/2+ε+ρL∞is uniformly bounded. We obtain from (57) that(59)lim supt→T ut+γ1Hs+ρtHs-1<∞,
which contradicts the assumption on the maximal existence time T<∞. It completes the Theorem 10 for s∈[2,5/2).
Step 3. For s=k∈N, k≥3. Differentiating k-2 times the second equation of system (23) with respect to x, we obtain(60)∂t∂xk-2ρ-u+2γ2∂x∂xk-2ρ+∑l1+l2=k-3,l1,l2≥0Cl1,l2∂xl1+1u∂xl2+1ρ+ρ∂x∂xk-2u=0.Using Lemma 2, we get from (60) that(61)∂xk-2ρH1≤∂xk-2ρ0H1+C∫0t∂xk-2ρH1∂xuH1/2∩L∞dτ+C∫0t∑l1+l2=k-3,l1,l2≥0Cl1,l2∂xl1+1u∂xl2+1ρ+ρ∂xk-1uH1dτ.Since H1 is an algebra, we have(62)ρ∂xk-1uH1≤CρH1∂xk-1uH1≤CρH1u+γ1Hsand(63)∑l1+l2=k-3,l1,l2≥0Cl1,l2∂xl1+1u∂xl2+1ρH1≤CρHs-1u+γ1Hs-1.Therefore, we have(64)∂xk-2ρH1≤∂xk-2ρ0H1+C∫0tu+γ1Hs+ρHs-1×u+γ1Hs-1+ρH1dτ.Applying Lemma 3 to the second equation of system (23) yields(65)ρHσS≤ρ0Hσ+C∫0tu+γ1Hσ+1ρL∞dτ+C∫0tρHσuL∞+∂xuL∞+1dτ,where we used Lemma 1.
From (64) and (65), it yields(66)ρHs-1≤ρ0Hs-1+C∫0tu+γ1Hs+ρHs-1×u+γ1Hs-1+ρH1dτ,where we used the Gagliardo-Nirenberg inequality ‖ρ‖Hs-1≤C(‖ρ‖Hσ+‖∂xk-2ρ‖H1) for σ∈(0,1).
Using (48) implies that(67)ut+γ1Hs+ρtHs-1≤Cu0+γ1Hs+ρ0Hs-1+C∫0tu+γ1Hs+ρtHs-1×u+γ1Hs-1+ρH1+1dτ.Using Gronwall’s inequality, we get(68)ut+γ1Hs+ρtHs-1≤Cu0+γ1Hs+ρ0Hs-1eC∫0tu+γ1Hs-1+ρH1+1dτ.Therefore, if the maximal existence time T<∞ satisfies ∫0T‖ux‖L∞dτ<∞, using Step 2, we get that(69)u+γ1Hs-1+ρH1is uniformly bounded by the induction assumption. From (66), we get(70)lim supt→T ut+γ1Hs+ρtHs-1<∞,which contradicts the assumption that T<∞ is the maximal existence time. This completes the proof of Theorem 10 for s=k∈N and k≥3.
Step 4. For s∈(k,k+1), k∈N and k≥3. Differentiating k-1 times the second equation of system (23) with respect to x, we obtain(71)∂t∂xk-1ρ-u+2γ2∂x∂xk-1ρ+∑l1+l2=k-2,l1,l2≥0Cl1,l2∂xl1+1u∂xl2+1ρ+ρ∂x∂xk-1u=0.Applying Lemma 3 to (71) and noting s-k∈(0,1), we get(72)∂xk-1ρHs-k≤C∫0t∑l1+l2=k-2,l1,l2≥0Cl1,l2∂xl1+1u∂xl2+1ρ+ρ∂xkuHs-kdτ+C∫0t∂xk-1ρHs-kuL∞+∂xuL∞+1dτ+∂xk-1ρ0Hs-k.For each ε∈(0,1/2), using (7) and the fact that H1/2+ε(R)↪L∞(R), we have(73)ρ∂xkuHs-k≤CρHs-k+1∂xk-1uL∞+∂xkuHs-kρL∞≤CρHs-k+1u+γ1Hk-1/2+ε+u+γ1Hs-1ρL∞and(74)∑l1+l2=k-2,l1,l2≥0Cl1,l2∂xl1+1u∂xl2+1ρHs-k≤C∑l1+l2=k-2,l1,l2≥0Cl1,l2∂xl1+1uHs-k+1∂xl2ρL∞+∂xl1+1uL∞∂xl2+1ρHs-k≤Cu+γ1HsρHk-3/2+ε+u+γ1Hk-1/2+ερHs-1,where we used Lemma 1.
Therefore, from (72), (73), and (74), we get(75)∂xk-1ρHs-k≤C∫0tu+γ1Hs+ρHs-1u+γ1Hk-1/2+ε+ρHk-3/2+ε+1dτ+∂xk-1ρ0Hs-k,which together with (48) and (37) (where s-2 is replaced by s-k) gives rise to(76)ut+γ1Hs+ρtHs-1≤Cu0+γ1Hs+ρ0Hs-1+C∫0tu+γ1Hs+ρtHs-1×u+γ1Hk-1/2+ε+ρHk-3/2+ε+1dτ,from which we have(77)ut+γ1Hs+ρtHs-1≤Cu0+γ1Hs+ρ0Hs-1eC∫0tu+γ1Hk-1/2+ε+ρHk-3/2+ε+1dτ.Noting that k-1/2+ε<k, k-3/2+ε<k-1 and k≥3. Therefore, if the maximal existence time T<∞ satisfies ∫0T‖ux‖L∞ dτ<∞, using Step 3, we get that(78)u+γ1Hk-1/2+ε+ρHk-3/2+εis uniformly bounded by the induction assumption. From (77), we get(79)lim supt→T ut+γ1Hs+ρtHs-1<∞,which contradicts the assumption that T<∞ is the maximal existence time. This completes the proof of Theorem 10 for s=k∈N and k≥3.
Therefore, from Steps 1–4, we complete the proof of Theorem 10.