Citation: L. Galgani. Foundations of physics in Milan, Padua and Paris. Newtonian trajectories from celestial mechanics to atomic physics[J]. Mathematics in Engineering, 2021, 3(6): 1-24. doi: 10.3934/mine.2021045
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The equation:
{∂tu+∂xf(u)−β2∂2xu+δ∂3xu+κu+γ2|u|u=0,0<t<T,x∈R,u(0,x)=u0(x),x∈R, | (1.1) |
was originally derived in [14,17] with f(u)=au2 focusing on microbubbles coated by viscoelastic shells. These structures are crucial in ultrasound diagnosis using contrast agents, and the dynamics of individual coated bubbles are explored, taking into account nonlinear competition and dissipation factors such as dispersion, thermal effects, and drag force.
The coefficients β2, δ, κ, and γ2 are related to the dissipation, the dispersion, the thermal conduction dissipation, and to the drag force, repsctively.
If κ=γ=0, we obtain the Kudryashov-Sinelshchikov [18] Korteweg-de Vries-Burgers [3,20] equation
∂tu+a∂xu2−β2∂2xu+δ∂3xu=0, | (1.2) |
that models pressure waves in liquids with gas bubbles, taking into account heat transfer and viscosity. The mathematical results on Eq (1.2) are the following:
● analysis of exact solutions in [13],
● existence of the traveling waves in [2],
● well-posedness and asymptotic behavior in [7,11].
If β=0, we derive the Korteweg-de Vries equation:
∂tu+a∂xu2+δ∂3xu=0, | (1.3) |
which describes surface waves of small amplitude and long wavelength in shallow water. Here, u(t,x) represents the wave height above a flat bottom, x corresponds to the distance in the propagation direction, and t denotes the elapsed time. In [4,6,10,12,15,16], the completele integrability of Eq (1.3) and the existence of solitary wave solutions are proved.
Through the manuscript, we will assume
● on the coefficients
β,δ,κ,γ∈R,β,δ,γ≠0; | (1.4) |
● on the flux f, one of the following conditions:
f(u)=au2+bu3, | (1.5) |
f∈C1(R),|f′(u)|≤C0(1+|u|),u∈R, | (1.6) |
for some positive constant C0;
● on the initial value
u0∈H1(R). | (1.7) |
The main result of this paper is the following theorem.
Theorem 1.1. Assume Eqs (1.5)–(1.7). For fixed T>0, there exists a unique distributional solution u of Eq (1.1), such that
u∈L∞(0,T;H1(R))∩L4(0,T;W1,4(R))∩L6(0,T;W1,6(R))∂2xu∈L2((0,T)×R). | (1.8) |
Moreover, if u1 and u2 are solutions to Eq (1.1) corresponding to the initial conditions u1,0 and u2,0, respectively, it holds that:
‖u1(t,⋅)−u2(t,⋅)‖L2(R)≤eC(T)t‖u1,0−u2,0‖L2(R), | (1.9) |
for some suitable C(T)>0, and every, 0≤t≤T.
Observe that Theorem 1.1 gives the well-posedness of (1.1), without conditions on the constants. Moreover, the proof of Theorem 1.1 is based on the Aubin-Lions Lemma [5,21]. The analysis of Eq (1.1) is more delicate than the one of Eq (1.2) due to the presence of the nonlinear sources and the very general assumptions on the coefficients.
The structure of the paper is outlined as follows. Section 2 is dedicated to establishing several a priori estimates for a vanishing viscosity approximation of Eq (1.1). These estimates are crucial for proving our main result, which is presented in Section 3.
To establish existence, we utilize a vanishing viscosity approximation of equation (1.1), as discussed in [19]. Let 0<ε<1 be a small parameter, and denote by uε∈C∞([0,T)×R) the unique classical solution to the following problem [1,9]:
{∂tuε+∂xf(uε)−β2∂2xuε+δ∂3xuε+κu+γ2|u|u=−ε∂4xuε,0<t<T,x∈R,uε(0,x)=uε,0(x),x∈R, | (2.1) |
where uε,0 is a C∞ approximation of u0, such that
‖uε,0‖H1(R)≤‖u0‖H1(R). | (2.2) |
Let us prove some a priori estimates on uε, denoting with C0 constants which depend only on the initial data, and with C(T) the constants which depend also on T.
We begin by proving the following lemma:
Lemma 2.1. Let T>0 be fixed. There exists a constant C(T)>0, which does not depend on ε, such that
‖uε(t,⋅)‖2L2(R)+2γ2e|κ|t∫t0∫Re−|κ|su2ε|uε|dsdx+2β2e|κ|t∫t0e−|κ|s‖∂xuε(s,⋅)‖2L2(R)ds+2εe|κ|t∫t0e−|κ|s‖∂2xuε(s,⋅)‖2L2(R)≤C(T), | (2.3) |
for every 0≤t≤T.
Proof. For 0≤t≤T. Multiplying equations (2.1) by 2uε, and integrating over R yields
ddt‖uε(t,⋅)‖2L2(R)=2∫Ruε∂tuεdx=−2∫Ruεf′(uε)∂xuεdx⏟=0+2β2∫Ruε∂2xuεdx−2δ∫Ruε∂3xuεdx−κ‖uε(t,⋅)‖2L2(R)−2γ2∫R|uε|u2εdx−2ε∫Ruε∂4xuεdx=−2β2‖∂xuε(t,⋅)‖2L2(R)+2δ∫R∂xuε∂2xuεdx−κ‖uε(t,⋅)‖2L2(R)−2γ2∫R|uε|u2εdx+2ε∫R∂xuε∂3xuεdx=−2β2‖∂xuε(t,⋅)‖2L2(R)−κ‖uε(t,⋅)‖2L2(R)−2γ2∫R|uε|u2εdx−2ε‖∂2xuε(t,⋅)‖2L2(R). |
Thus, it follows that
ddt‖uε(t,⋅)‖2L2(R)+2β2‖∂xuε(t,⋅)‖2L2(R)+2γ2∫R|uε|u2εdx+2ε‖∂2xuε(t,⋅)‖2L2(R)=κ‖uε(t,⋅)‖2L2(R)≤|κ|‖uε(t,⋅)‖2L2(R). |
Therefore, applying the Gronwall's lemma and using Eq (2.2), we obtain
‖uε(t,⋅)‖2L2(R)+2β2e|κ|t∫t0e−|κ|s‖∂xuε(s,⋅)‖2L2(R)ds+2γ2e|κ|t∫t0∫Re−|κ|t|uε|u2εdsdx+2ε‖∂2xuε(t,⋅)‖2L2(R)+2εe|κ|t∫t0e−|κ|s‖∂2xuε(s,⋅)‖2L2(R)ds≤C0e|κ|t≤C(T), |
which gives Eq (2.3).
Lemma 2.2. Fix T>0 and assume (1.5). There exists a constant C(T)>0, independent of ε, such that
‖uε‖L∞((0,T)×R)≤C(T), | (2.4) |
‖∂xuε(t,⋅)‖2L2(R)+β2∫t0‖∂2xuε(s,⋅)‖2L2(R)ds | (2.5) |
+2ε∫t0‖∂3xuε(s,⋅)‖2L2(R)ds≤C(T),∫t0‖∂xuε(s,⋅)‖4L4(R)ds≤C(T), | (2.6) |
holds for every 0≤t≤T.
Proof. Let 0≤t≤T. Consider A,B as two real constants, which will be specified later. Thanks to Eq (1.5), multiplying Eq (2.1) by
−2∂2xuε+Au2ε+Bu3ε, |
we have that
(−2∂2xuε+Au2ε+Bu3ε)∂tuε+2a(−2∂2xuε+Au2ε+Bu3ε)uε∂xuε+3b(−2∂2xuε+Au2ε+Bu3ε)u2ε∂xuε−β2(−2∂2xuε+Au2ε+Bu3ε)∂2xuε+δ(−2∂2xuε+Au2ε+Bu3ε)∂3xuε+κ(−2∂2xuε+Au2ε+Bu3ε)uε+γ2(−2∂2xuε+Au2ε+Bu3ε)|uε|uε=−ε(−2∂2xuε+Au2ε+Bu3ε)∂4xuε. | (2.7) |
Observe that
∫R(−2∂2xuε+Au2ε+Bu3ε)∂tuεdx=ddt(‖∂xuε(t,⋅)‖2L2(R)+A3∫Ru3εdx+B4∫Ru4εdx),2a∫R(−2∂2xuε+Au2ε+Bu3ε)uε∂xuεdx=−4a∫Ruε∂xuε∂2xuεdx,3b∫R(−2∂2xuε+Au2ε+Bu3ε)u2ε∂xuεdx=−6b∫Ru2ε∂xuε∂2xuεdx,−β2∫R(−2∂2xuε+Au2ε+Bu3ε)∂2xuεdx=2β2‖∂2xuε(t,⋅)‖2L2(R)+2Aβ2∫Ruε(∂xuε)2dx+3Bβ2∫Ru2ε(∂xuε)2dx,δ∫R(−2∂2xuε+Au2ε+Bu3ε)∂3xuεdx=−2Aδ∫Ruε∂xuε∂2xuεdx−3Bδ∫Ru2ε∂xuε∂2xuεdx,κ∫R(−2∂2xuε+Au2ε+Bu3ε)uεdx=2κ‖∂xuε(t,⋅)‖2L2(R)+Aκ∫Ru3εdx+Bκ∫Ru4εdx,γ2∫R(−2∂2xuε+Au2ε+Bu3ε)|uε|uεdx=−2γ2∫R|uε|uε∂2xuεdx+Aγ2∫R|u|u3εdx+Bγ2∫R|uε|u4dx,−ε∫R(−2∂2xuε+Au2ε+Bu3ε)∂4xuεdx=−2ε‖∂3xuε(t,⋅)‖2L2(R)+2Aε∫Ruε∂xuε∂3xuεdx+3Bε∫Ru2ε∂xuε∂3xuεdx=−2ε‖∂3xuε(t,⋅)‖2L2(R)−Aε∫R(∂xuε)3dx−6Bε∫Ruε(∂xuε)2∂2xuεdx−3Bε‖uε(t,⋅)∂2xuε(t,⋅)‖2L2(R)=−2ε‖∂3xuε(t,⋅)‖2L2(R)−Aε∫R(∂xuε)3dx+2Bε∫R(∂xuε)4dx−3Bε‖uε(t,⋅)∂2xuε(t,⋅)‖2L2(R). |
Therefore, an integration on R gives
ddt(‖∂xuε(t,⋅)‖2L2(R)+A3∫Ru3εdx+B4∫Ru4εdx)+β2‖∂2xuε(t,⋅)‖2L2(R)+2ε‖∂3xuε(t,⋅)‖2L2(R)=−(4a+Aδ)∫Ruε∂xuε∂2xuεdx−3(2b+Bδ)∫Ru2ε∂xuε∂2xuεdx−2Aβ2∫Ruε(∂xuε)2dx−3Bβ2∫Ru2ε(∂xuε)2dx−κ‖∂xuε(t,⋅)‖2L2(R)−Aκ3∫Ru3εdx−Bκ4∫Ru4εdx+2γ2∫R|uε|uε∂2xuεdx−Aγ2∫R|uε|u3εdx−Bγ2∫R|uε|u4εdx−Aε∫R(∂xuε)3dx+2Bε∫R(∂xuε)4dx−3Bε‖uε(t,⋅)∂2xuε(t,⋅)‖2L2(R). |
Taking
(A,B)=(−4aδ,−2bδ), |
we get
ddt(‖∂xuε(t,⋅)‖2L2(R)−4a3δ∫Ru3εdx−bδ∫Ru4εdx)+2β2‖∂2xuε(t,⋅)‖2L2(R)+2ε‖∂3xuε(t,⋅)‖2L2(R)=8aβ2δ∫Ruε(∂xuε)2dx+6bβ2δ∫Ru2ε(∂xuε)2dx−κ‖∂xuε(t,⋅)‖2L2(R)+4aκ3δ∫Ru3εdx+bκ2∫Ru4εdx+2γ2∫R|uε|uε∂2xuεdx+4aγ2δ∫R|uε|u3εdx+2bγ2δ∫R|uε|u4εdx+4aεδ∫R(∂xuε)3dx−4bεδ∫R(∂xuε)4dx+6bεδ‖uε(t,⋅)∂2xuε(t,⋅)‖2L2(R). | (2.8) |
Since 0<ε<1, due to the Young inequality and (2.3),
8aβ2δ∫R|uε|(∂xuε)2dx≤4∫Ru2ε(∂xuε)2dx+4a2β4δ2‖∂xuε(t,⋅)‖2L2(R)≤4‖uε‖2L∞((0,T)×R)‖∂xuε(t,⋅)‖2L2(R)+4a2β4δ2‖∂xuε(t,⋅)‖2L2(R)≤C0(1+‖uε‖2L∞((0,T)×R))‖∂xuε(t,⋅)‖2L2(R),|6bβ2δ|∫Ru2ε(∂xuε)2dx≤|6bβ2δ|‖uε‖2L∞((0,T)×R)‖∂xuε(t,⋅)‖2L2(R),|4aκ3δ|∫R|uε|3dx≤|4aκ3δ|‖uε‖L∞((0,T)×R)‖uε(t,⋅)‖2L2(R)≤C(T)‖uε‖L∞((0,T)×R),|bκ2|∫Ru4εdx≤|bκ2|‖uε‖2L∞((0,T)×R)‖uε(t,⋅)‖2L2(R)≤C(T)‖uε‖2L∞((0,T)×R),2γ2∫R|uε|uε∂2xuεdx≤2∫R|γ2|uε|uεβ||β∂2xuε|dx≤γ4β2∫Ruε4dx+β2‖∂2xuε(t,⋅)‖2L2(R)≤γ4β2‖uε‖2L∞((0.T)×R)‖uε(t,⋅)‖2L2(R)+β2‖∂2xuε(t,⋅)‖2L2(R)≤C(T)‖uε‖2L∞((0,T)×R)+β2‖∂2xuε(t,⋅)‖2L2(R),|4aγ2δ|∫R|uε||uε|3dx=|4aγ2δ|∫Ru4εdx≤|4aγ2δ|‖uε‖2L∞((0,T)×R)‖uε(t,⋅)‖2L2(R)≤C(T)‖uε‖2L∞((0,T)×R),|2bγ2δ|∫R|uε|uε4dx≤|2bγ2δ|‖uε‖3L∞((0,T)×R)‖uε(t,⋅)‖2L2(R)≤C(T)‖uε‖3L∞((0,T)×R),|4aεδ|∫R|∂xuε|3dx≤|4aεδ|‖∂xuε(t,⋅)‖2L2(R)+|4aεδ|∫R(∂xuε)4dx≤|4aδ|‖∂xuε(t,⋅)‖2L2(R)+|4aεδ|∫R(∂xuε)4dx. |
It follows from Eq (2.8) that
ddt(‖∂xuε(t,⋅)‖2L2(R)−4a3δ∫Ru3εdx−bδ∫Ru4εdx)+β2‖∂2xuε(t,⋅)‖2L2(R)+2ε‖∂3xuε(t,⋅)‖2L2(R)≤C0(1+‖uε‖2L∞((0,T)×R))‖∂xuε(t,⋅)‖2L2(R)+C(T)‖uε‖L∞((0,T)×R)+C(T)‖uε‖2L∞((0,T)×R)+C(T)‖uε‖3L∞((0,T)×R)+C0ε∫R(∂xuε)4dx+C0ε‖uε(t,⋅)∂2xuε(t,⋅)‖2L2(R)+C0‖∂xuε(t,⋅)‖2L2(R). | (2.9) |
[8, Lemma 2.3] says that
∫R(∂xuε)4dx≤9∫Ru2ε(∂2xuε)2dx≤9‖uε‖2L∞((0,T)×R)‖∂2xuε(t,⋅)‖2L2(R). | (2.10) |
Moreover, we have that
‖uε(t,⋅)∂2xuε(t,⋅)‖2L2(R)=∫Ru2ε(∂2xuε)2dx≤‖uε‖2L∞((0,T)×R)‖∂2xuε(t,⋅)‖2L2(R). | (2.11) |
Consequentially, by Eqs (2.9)–(2.11), we have that
ddt(‖∂xuε(t,⋅)‖2L2(R)−4a3δ∫Ru3εdx−bδ∫Ru4εdx)+β2‖∂2xuε(t,⋅)‖2L2(R)+2ε‖∂3xuε(t,⋅)‖2L2(R)≤C0(1+‖uε‖2L∞((0,T)×R))‖∂xuε(t,⋅)‖2L2(R)+C(T)‖uε‖L∞((0,T)×R)+C(T)‖uε‖2L∞((0,T)×R)+C(T)‖uε‖3L∞((0,T)×R)+C0ε‖uε‖2L∞((0,T)×R)‖∂2xuε(t,⋅)‖2L2(R)+C0‖∂xuε(t,⋅)‖2L2(R). |
An integration on (0,t) and Eqs (2.2) and (2.3) give
‖∂xuε(t,⋅)‖2L2(R)−4a3δ∫Ru3εdx−bδ∫Ru4εdx+β2∫t0‖∂2xuε(s,⋅)‖2L2(R)ds+2ε∫t0‖∂3xuε(s,⋅)‖2L2(R)ds≤C0(1+‖uε‖2L∞((0,T)×R))∫t0‖∂xuε(s,⋅)‖2L2(R)ds+C(T)‖uε‖L∞((0,T)×R)t+C(T)‖uε‖2L∞((0,T)×R)t+C(T)‖uε‖3L∞((0,T)×R)t+C0ε‖uε‖2L∞((0,T)×R)∫t0‖∂2xuε(s,⋅)‖2L2(R)ds+C0∫t0‖∂xuε(s,⋅)‖2L2(R)ds≤C(T)(1+‖uε‖L∞((0,T)×R)+‖uε‖2L∞((0,T)×R)+‖uε‖3L∞((0,T)×R)). |
Therefore, by Eq (2.3),
‖∂xuε(t,⋅)‖2L2(R)+β2∫t0‖∂2xuε(s,⋅)‖2L2(R)ds+2ε∫t0‖∂3xuε(s,⋅)‖2L2(R)ds≤C(T)(1+‖uε‖L∞((0,T)×R)+‖uε‖2L∞((0,T)×R)+‖uε‖3L∞((0,T)×R))+4a3δ∫Ru3εdx+bδ∫Ru4εdx≤C(T)(1+‖uε‖L∞((0,T)×R)+‖uε‖2L∞((0,T)×R)+‖uε‖3L∞((0,T)×R))+|4a3δ|∫R|uε|3dx+|bδ|∫Ru4εdx≤C(T)(1+‖uε‖L∞((0,T)×R)+‖uε‖2L∞((0,T)×R)+‖uε‖3L∞((0,T)×R))+|4a3δ|‖uε‖L∞((0,T)×R)‖uε(t,⋅)‖2L2(R)+|bδ|‖uε‖2L∞((0,T)×R)‖uε(t,⋅)‖2L2(R)≤C(T)(1+‖uε‖L∞((0,T)×R)+‖uε‖2L∞((0,T)×R)+‖uε‖3L∞((0,T)×R)). | (2.12) |
We prove Eq (2.4). Thanks to the Hölder inequality,
u2ε(t,x)=2∫x−∞uε∂xuεdx≤2∫R|uε||∂xuε|dx≤2‖uε(t,⋅)‖L2(R)‖∂xuε(t,⋅)‖L2(R). |
Hence, we have that
‖uε(t,⋅)‖4L∞(R)≤4‖uε(t,⋅)‖2L2(R)‖∂xuε(t,⋅)‖2L2(R). | (2.13) |
Thanks to Eqs (2.3) and (2.12), we have that
‖uε‖4L∞((0,T)×R)≤C(T)(1+‖uε‖L∞((0,T)×R)+‖uε‖2L∞((0,T)×R)+‖uε‖3L∞((0,T)×R)). | (2.14) |
Due to the Young inequality,
C(T)‖uε‖3L∞((0,T)×R)≤12‖uε‖4L∞((0,T)×R)+C(T)‖uε‖2L∞((0,T)×R),C(T)‖uε‖L∞((0,T)×R)≤C(T)‖uε‖2L∞((0,T)×R)+C(T). |
By Eq (2.14), we have that
12‖uε‖4L∞((0,T)×R)−C(T)‖uε‖2L∞((0,T)×R)−C(T)≤0, |
which gives Eq (2.4).
Equation (2.5) follows from Eqs (2.4) and (2.12).
Finally, we prove Eq (2.6). We begin by observing that, from Eqs (2.4) and (2.10), we have
‖∂xuε(t,⋅)‖4L4(R)≤C(T)‖∂2xuε(t,⋅)‖2L2(R). |
An integration on (0,t) and Eqs (2.5) give Eq (2.6).
Lemma 2.3. Fix T>0 and assume (1.6). There exists a constant C(T)>0, independent of ε, such that Eq (2.4) holds. Moreover, we have Eqs (2.5) and (2.6).
Proof. Let 0≤t≤T. Multiplying Eq (2.1) by −2∂2xuε, an integration on R gives
ddt‖∂xuε(t,⋅)‖2L2(R)=−2∫R∂2xuε∂tuεdx=−2∫Rf′(uε)∂xuε∂2xuεdx−2β2‖∂2xuε(t,⋅)‖2L2(R)−2δ∫R∂2xuε∂3xuεdx−2κ∫Ruε∂2xuεdx−2γ2∫R|uε|uε∂2xuεdx+2ε∫R∂2xuε∂4xuεdx=−2∫Rf′(uε)∂xuε∂2xuεdx−2β2‖∂2xuε(t,⋅)‖2L2(R)+2κ‖∂xuε(t,⋅)‖2L2(R)+2γ2∫R|uε|uε∂2xuεdx−2ε‖∂3xuε(t,⋅)‖2L2(R). |
Therefore, we have that
ddt‖∂xuε(t,⋅)‖2L2(R)+2β2‖∂2xuε(t,⋅)‖2L2(R)+2ε‖∂3xuε(t,⋅)‖2L2(R)=−2∫Rf′(uε)∂xuε∂2xuεdx+2κ‖∂xuε(t,⋅)‖2L2(R)+2γ2∫R|uε|uε∂2xuεdx. | (2.15) |
Due Eqs (1.6) and (2.3) and the Young inequality,
2∫R|f′(uε)||∂xuε||∂2xuε|dx≤C0∫R|∂xuε∂2xuε|dx+C0∫R|uε∂xuε||∂2xuε|dx=2∫R|C0√3∂xuε2β||β∂2xuε√3|dx+2∫R|C0√3uε∂xuε2β||√3β∂2xuε|dx≤C0‖∂xuε(t,⋅)‖2L2(R)+C0∫Ru2ε(∂xuε)2dx+2β23‖∂2xuε(t,⋅)‖2L2(R)≤C0‖∂xuε(t,⋅)‖2L2(R)+C0‖uε‖2L∞((0,T)×R)‖∂xuε(t,⋅)‖2L2(R)+2β23‖∂2xuε(t,⋅)‖2L2(R)≤C0(1+‖uε‖2L∞((0,T)×R))‖∂xuε(t,⋅)‖2L2(R)+2β23‖∂2xuε(t,⋅)‖2L2(R),2γ2∫R|uε|uε∂2xuεdx≤2γ2∫Ru2ε|∂2xuε|dx=2∫R|√3γ2u2εβ||β∂2xuε√3|dx≤3γ4β2∫Ru4εdx+β23‖∂2xuε(t,⋅)‖2L2(R)≤3γ4β2‖uε‖2L∞((0,T)×R)‖uε(t,⋅)‖2L2(R)+β23‖∂2xuε(t,⋅)‖2L2(R)≤C(T)‖uε‖2L∞((0,T)×R)+β23‖∂2xuε(t,⋅)‖2L2(R). |
It follows from Eq (2.15) that
ddt‖∂xuε(t,⋅)‖2L2(R)+β2‖∂2xuε(t,⋅)‖2L2(R)+2ε‖∂3xuε(t,⋅)‖2L2(R)≤C0(1+‖uε‖2L∞((0,T)×R))‖∂xuε(t,⋅)‖2L2(R)+C(T)‖uε‖2L∞((0,T)×R). |
Integrating on (0,t), by Eq (2.3), we have that
‖∂xuε(t,⋅)‖2L2(R)+β2∫t0‖∂2xuε(s,⋅)‖2L2(R)ds+2ε∫t0‖∂3xuε(s,⋅)‖2L2(R)≤C0+C0(1+‖uε‖2L∞((0,T)×R))∫t0‖∂xuε(s,⋅)‖2L2(R)ds+C(T)‖uε‖2L∞((0,T)×R)t≤C(T)(1+‖uε‖2L∞((0,T)×R)). | (2.16) |
Thanks to Eqs (2.3), (2.13), and (2.16), we have that
‖uε‖4L∞((0,T)×R)≤C(T)(1+‖uε‖2L∞((0,T)×R)). |
Therefore,
‖uε‖4L∞((0,T)×R)−C(T)‖uε‖2L∞((0,T)×R)−C(T)≤0, |
which gives (2.4).
Equation (2.5) follows from (2.4) and (2.16), while, arguing as in Lemma 2.2, we have Eq (2.6).
Lemma 2.4. Fix T>0. There exists a constant C(T)>0, independent of ε, such that
∫t0‖∂xuε(s,⋅)‖6L6(R)ds≤C(T), | (2.17) |
for every 0≤t≤T.
Proof. Let 0≤t≤T. We begin by observing that,
∫R(∂xuε)6dx≤‖∂xuε(t,⋅)‖4L∞(R)‖∂xuε(t,⋅)‖2L2(R). | (2.18) |
Thanks to the Hölder inequality,
(∂xuε(t,x))2=2∫x−∞∂xuε∂2xuεdy≤2∫R|∂xuε||∂2xuε|dx≤2‖∂xuε(t,⋅)‖L2(R)‖∂2xuε(t,⋅)‖2L2(R). |
Hence,
‖u(t,⋅)‖4L∞(R)≤4‖∂xuε(t,⋅)‖2L2(R)‖∂2xuε(t,⋅)‖2L2(R). |
It follows from Eq (2.18) that
∫R(∂xuε)6dx≤4‖∂xuε(t,⋅)‖4L2(R)‖∂2xuε(t,⋅)‖2L2(R). |
Therefore, by Eq (2.5),
∫R(∂xuε)6dx≤C(T)‖∂2xuε(t,⋅)‖2L2(R). |
An integration on (0,t) and Eq (2.5) gives (2.17).
This section is devoted to the proof of Theorem 1.1.
We begin by proving the following result.
Lemma 3.1. Fix T>0. Then,
the family {uε}ε>0 is compact in L2loc((0,T)×R). | (3.1) |
Consequently, there exist a subsequence {uεk}k∈N and u∈L2loc((0,T)×R) such that
uεk→u in L2loc((0,T)×R) and a.e. in (0,T)×R. | (3.2) |
Moreover, u is a solution of Eq (1.1), satisfying Eq (1.8).
Proof. We begin by proving Eq (3.1). To prove Eq (3.1), we rely on the Aubin-Lions Lemma (see [5,21]). We recall that
H1loc(R)↪↪L2loc(R)↪H−1loc(R), |
where the first inclusion is compact and the second one is continuous. Owing to the Aubin-Lions Lemma [21], to prove Eq (3.1), it suffices to show that
{uε}ε>0 is uniformly bounded in L2(0,T;H1loc(R)), | (3.3) |
{∂tuε}ε>0 is uniformly bounded in L2(0,T;H−1loc(R)). | (3.4) |
We prove Eq (3.3). Thanks to Lemmas 2.1–2.3,
‖uε(t,⋅)‖2H1(R)=‖uε(t,⋅)‖2L2(R)+‖∂xuε(t,⋅)‖2L2(R)≤C(T). |
Therefore,
{uε}ε>0 is uniformly bounded in L∞(0,T;H1(R)), |
which gives Eq (3.3).
We prove Eq (3.4). Observe that, by Eq (2.1),
∂tuε=−∂x(G(uε))−f′(uε)∂xuε−κuε−γ2|uε|uε, |
where
G(uε)=β2∂xuε−δ∂2xuε−ε∂3xuε. | (3.5) |
Since 0<ε<1, thanks to Eq (2.5), we have that
β2‖∂xuε‖2L2((0,T)×R),δ2‖∂2xuε‖2L2((0,T)×R)≤C(T),ε2‖∂3xuε‖2L2((0,T)×R)≤C(T). | (3.6) |
Therefore, by Eqs (3.5) and (3.6), we have that
{∂x(G(uε))}ε>0 is bounded in L2(0,T;H−1(R)). | (3.7) |
We claim that
∫T0∫R(f′(uε))2(∂xuε)2dtdx≤C(T). | (3.8) |
Thanks to Eqs (2.4) and (2.5),
∫T0∫R(f′(uε))2(∂xuε)2dtdx≤‖f′‖2L∞(−C(T),C(T))∫T0‖∂xuε(t,⋅)‖2L2(R)dt≤C(T). |
Moreover, thanks to Eq (2.3),
|κ|∫T0∫R(uε)2dx≤C(T). | (3.9) |
We have that
γ2∫T0∫R(|uε|uε)2dsdx≤C(T). | (3.10) |
In fact, thanks to Eqs (2.3) and (2.4),
γ2∫T0∫R(|uε|uε)2dsdx≤γ2‖uε‖2L∞((0,T)×R)∫T0∫R(uε)2dsdx≤C(T)∫T0∫R(uε)2dsdx≤C(T). |
Therefore, Eq (3.4) follows from Eqs (3.7)–(3.10).
Thanks to the Aubin-Lions Lemma, Eqs (3.1) and (3.2) hold.
Consequently, arguing as in [5, Theorem 1.1], u is solution of Eq (1.1) and, thanks to Lemmas 2.1–2.3 and Eqs (2.4), (1.8) holds.
Proof of Theorem 1.1. Lemma 3.1 gives the existence of a solution of Eq (1.1).
We prove Eq (1.9). Let u1 and u2 be two solutions of Eq (1.1), which verify Eq (1.8), that is,
{∂tui+∂xf(ui)−β2∂2xui+δ∂3xui+κui+γ2|ui|ui=0,0<t<T,x∈R,ui(0,x)=ui,0(x),x∈R,i=1,2. |
Then, the function
ω(t,x)=u1(t,x)−u2(t,x), | (3.11) |
is the solution of the following Cauchy problem:
{∂tω+∂x(f(u1)−f(u2))−β2∂2xω+δ∂2xω+κω+γ2(|u1|u1−|u2|u2)=0,0<t<T,x∈R,ω(0,x)=u1,0(x)−u2,0(x),x∈R. | (3.12) |
Fixed T>0, since u1,u2∈H1(R), for every 0≤t≤T, we have that
‖u1‖L∞((0,T)×R),‖u2‖L∞((0,T)×R)≤C(T). | (3.13) |
We define
g=f(u1)−f(u2)ω | (3.14) |
and observe that, by Eq (3.13), we have that
|g|≤‖f′‖L∞(−C(T),C(T))≤C(T). | (3.15) |
Moreover, by Eq (3.11) we have that
||u1|−|u2||≤|u1−u2|=|ω|. | (3.16) |
Observe that thanks to Eq (3.11),
|u1|u1−|u2|u2=|u1|u1−|u1|u2+|u1|u2−|u2|u2=|u1|ω+u2(|u1|−|u2|). | (3.17) |
Thanks to Eqs (3.14) and (3.17), Equation (3.12) is equivalent to the following one:
∂tω+∂x(gω)−β2∂2xω+δ∂3xω+κω+γ2|u1|ω+γ2u2(|u1|−|u2|)=0. | (3.18) |
Multiplying Eq (3.18) by 2ω, an integration on R gives
dtdt‖ω(t,⋅)‖2L2(R)=2∫Rω∂tω=−2∫Rω∂x(gω)dx+2β2∫Rω∂2xωdx−2δ∫Rω∂3xωdx−2κ‖ω(t,⋅)‖2L2(R)−2γ2∫R|u1|ω2dx−2γ2∫Ru2(|u1|−|u2|)ωdx=2∫Rgω∂xωdx−2β2‖∂xω(t,⋅)‖2L2(R)+2δ∫R∂xω∂2xωdx−2κ‖ω(t,⋅)‖2L2(R)−2γ2∫R|u1|ω2dx−2γ2∫Ru2(|u1|−|u2|)ωdx=2∫Rgω∂xωdx−2β2‖∂xω(t,⋅)‖2L2(R)−2κ‖ω(t,⋅)‖2L2(R)−2γ2∫R|u1|ω2dx−2γ2∫Ru2(|u1|−|u2|)ωdx. |
Therefore, we have that
‖ω(t,⋅)‖2L2(R)+2β2‖∂xω(t,⋅)‖2L2(R)+2γ2∫R|u1|ω2dx=2∫Rgω∂xωdx−κ‖ω(t,⋅)‖2L2(R)−2γ2∫Ru2(|u1|−|u2|)ωdx. | (3.19) |
Due to Eqs (3.13), (3.15) and (3.16) and the Young inequality,
2∫R|g||ω||∂xω|dx≤2C(T)∫R|ω||∂xω|dx=2∫R|C(T)ωβ||β∂xω|dx≤C(T)‖ω(t,⋅)‖2L2(R)+β2‖∂xω(t,⋅)‖2L2(R),2γ2∫R|u2||(|u1|−|u2|)||ω|dx≤2γ2‖u2‖L∞((0,T)×R)∫R|(|u1|−|u2|)||ω|dx≤C(T)‖ω(t,⋅)‖2L2(R). |
It follows from Eq (3.19) that
‖ω(t,⋅)‖2L2(R)+β2‖∂xω(t,⋅)‖2L2(R)+2γ2∫R|u1|ω2dx≤C(T)‖ω(t,⋅)‖2L2(R). |
The Gronwall Lemma and Eq (3.12) give
‖ω(t,⋅)‖2L2(R)+β2eC(T)t∫t0e−C(T)s‖∂xω(s,⋅)‖2L2(R)ds+2γ2eC(T)t∫t0∫Re−C(T)s|u1|ω2dsdx≤eC(T)t‖ω0‖2L2(R). | (3.20) |
Equation (1.9) follows from Eqs (3.11) and (3.20).
Giuseppe Maria Coclite and Lorenzo Di Ruvo equally contributed to the methodologies, typesetting, and the development of the paper.
The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.
Giuseppe Maria Coclite is an editorial boardmember for [Networks and Heterogeneous Media] and was not involved inthe editorial review or the decision to publish this article.
GMC is member of the Gruppo Nazionale per l'Analisi Matematica, la Probabilità e le loro Applicazioni (GNAMPA) of the Istituto Nazionale di Alta Matematica (INdAM). GMC has been partially supported by the Project funded under the National Recovery and Resilience Plan (NRRP), Mission 4 Component 2 Investment 1.4 -Call for tender No. 3138 of 16/12/2021 of Italian Ministry of University and Research funded by the European Union -NextGenerationEUoAward Number: CN000023, Concession Decree No. 1033 of 17/06/2022 adopted by the Italian Ministry of University and Research, CUP: D93C22000410001, Centro Nazionale per la Mobilità Sostenibile, the Italian Ministry of Education, University and Research under the Programme Department of Excellence Legge 232/2016 (Grant No. CUP - D93C23000100001), and the Research Project of National Relevance "Evolution problems involving interacting scales" granted by the Italian Ministry of Education, University and Research (MIUR Prin 2022, project code 2022M9BKBC, Grant No. CUP D53D23005880006). GMC expresses its gratitude to the HIAS - Hamburg Institute for Advanced Study for their warm hospitality.
The authors declare there is no conflict of interest.
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