The Kuramoto-Sinelshchikov equation describes the evolution of a phase turbulence in reaction-diffusion systems or the evolution of the plane flame propagation, taking in account the combined influence of diffusion and thermal conduction of the gas on the stability of a plane flame front. In this paper, we prove the well-posedness of the classical solutions for the initial-boundary value problem for this equation, under appropriate boundary conditions.
Citation: Giuseppe Maria Coclite, Lorenzo di Ruvo. On the initial-boundary value problem for a Kuramoto-Sinelshchikov type equation[J]. Mathematics in Engineering, 2021, 3(4): 1-43. doi: 10.3934/mine.2021036
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The Kuramoto-Sinelshchikov equation describes the evolution of a phase turbulence in reaction-diffusion systems or the evolution of the plane flame propagation, taking in account the combined influence of diffusion and thermal conduction of the gas on the stability of a plane flame front. In this paper, we prove the well-posedness of the classical solutions for the initial-boundary value problem for this equation, under appropriate boundary conditions.
In this paper, we investigate the well-posedness of the classical solutions for the equation:
∂tu+κ∂xu2+q∂xu3+ν∂2xu+δ∂3xu+β2∂4xu=0, | (1.1) |
with κ,q,ν,δ,β∈R and β≠0.
We are interested in the initial-boundary value problem for this equation. More precisely we consider the following boundary conditions
{u(t,0)=g(t),t>0,∂xu(t,0)=h(t),t>0,g,h∈W1,∞(0,∞),g(0)=u0(0),q=0, | (1.2) |
{u(t,0)=g(t),t>0,∂2xu(t,0)=0,t>0,g∈W1,∞(0,∞),g(0)=u0(0),q=0, | (1.3) |
{∂xu(t,0)=0,t>0,∂3xu(t,0)=0,t>0,q=−a2,δ=0, | (1.4) |
{∂2xu(t,0)=0,t>0,∂3xu(t,0)=0,t>0,q=−a2≠0, | (1.5) |
{u(t,0)=0,t>0,∂xu(t,0)=0,t>0. | (1.6) |
Moreover, we augment (1.1) with the following initial datum:
u(0,x)=u0(x),x>0, | (1.7) |
on which assume
u0∈H2(0,∞). | (1.8) |
Assuming q=0, (1.1) reads
∂tu+κ∂xu2+ν∂2xu+δ∂3xu+β2∂4xu=0, | (1.9) |
(1.9) arises in interesting physical situations, for example as a model for long waves on a viscous fluid owing down an inclined plane [50]. and to derive drift waves in a plasma [20]. (1.9) was derived also independently by Kuramoto [30,31,32] as a model for phase turbulence in reaction-diffusion systems and by Sivashinsky [47] as a model for plane flame propagation, describing the combined influence of diffusion and thermal conduction of the gas on the stability of a plane flame front.
Equation (1.9) also describes incipient instabilities in a variety of physical and chemical systems [7,23,33]. Moreover, (1.9), which is also known as the Benney-Lin equation [4,41], was derived by Kuramoto in the study of phase turbulence in the Belousov-Zhabotinsky reaction [38].
The dynamical properties and the existence of exact solutions for (1.9) have been investigated in [21,26,28,43,44,51]. In [3,6,22], the control problem for (1.9) with periodic boundary conditions, and on a bounded interval are studied, respectively. In [8], the problem of global exponential stabilization of (1.9) with periodic boundary conditions is analyzed. In [24], it is proposed a generalization of optimal control theory for (1.9), while in [42] the problem of global boundary control of (1.9) is considered. In [45], the existence of solitonic solutions for (1.9) is proven. In [5,18,48], the well-posedness of the Cauchy problem for (1.9) is proven, using the energy space technique, a priori estimates together with an application of the Cauchy-Kovalevskaya and the fixed point method, respectively. In particular, in [18], the well-posedness of the Cauchy problem for (1.1) is proven. In [12], following [9,10,37,46] it is proven that, when ν,δ,β2 go to zero, the solution of (1.9) converges to the unique entropy one of the Burgers equation. Finally, the initial-boundary value problem for (1.9), under the conditions (1.2) is analyzed in [39,40], in a quarter plane and in a bounded domain, respectively, using the energy space technique, under appropriate assumptions on κ,ν,δ,β.
Taking q=ν=β=0 in (1.1), we have the Korteweg-de Vries equation [27]
∂tu+κ∂xu2+δ∂3xu=0, | (1.10) |
that has a very wide range of applications, such as magnetic fluid waves, ion sound waves, and longitudinal astigmatic waves.
From a mathematical point of view, in [16,19,25], the Cauchy problem for (1.10) is studied, while in [29], the author reviewed the travelling wave solutions for (1.10). Moreover, in [14,37,46], the convergence of the solution of (1.10) to the unique entropy one of the Burgers equation is proven.
Taking κ=r=δ=μ=β=0, (1.1) becomes
∂tu+q∂xu3+δ∂3xu=0, | (1.11) |
which is known as the modified Korteweg-de Vries equation.
[1,2,15,34,35,36] show that (1.11) is a non-slowly-varying envelope approximation model that describes the physics of few-cycle-pulse optical solitons. In [19,25], the Cauchy problem for (1.11) is studied, while, in [13,46], the convergence of the solution of (1.11) to the unique entropy solution of the following scalar conservation law
∂tu+q∂xu3=0. | (1.12) |
The main result of this paper is the following theorem.
Theorem 1.1. Fix T>0. The initial value problems (1.1)-(1.2)-(1.7), (1.1)-(1.3)-(1.7), (1.1)-(1.4)-(1.7), (1.1)-(1.5)-(1.7), (1.1)-(1.6)-(1.7), admit an unique solution
u∈H1((0,T)×(0,∞))∩L∞(0,T;H2(0,∞)). | (1.13) |
Moreover, if u1 and u2 are two solutions of the same initial-boundary value problem for (1.1), we have
‖u1(t,⋅)−u2(t,⋅)‖L2(0,∞)≤eC(T)t‖u1,0−u2,0‖L2(0,∞), | (1.14) |
for some suitable C(T)>0, and every ≤t≤T.
Compared to [39], Theorem 1.1 gives the well-posedness of the initial-boundary value problem (1.1)-(1.2)-(1.7), without any additional assumption on the constants. The proof of Theorem 1.1 relies on deriving suitable a priori estimates together with an application of the Cauchy-Kovalevskaya Theorem [49].
The paper is organized as follows. In Sections 2, 3, 4, 5, 6, we prove Theorem 1.1 for (1.1)-(1.2)-(1.7), (1.1)-(1.3)-(1.7), (1.1)-(1.4)-(1.7), (1.1)-(1.5)-(1.7), (1.1)-(1.6)-(1.7), respectively.
In this section, we prove Theorem 1.1 for (1.1)-(1.2)-(1.7).
Let us prove some a priori estimates on u, denoting with C0 the constants which depend only on the data, and with C(T), the constants which depend also on T.
Following [11,17], we introduce the auxiliary variable:
v(t,x)=u(t,x)−g(t)e−x−[g(t)+h(t)]xe−x. | (2.1) |
Observe that
∂tv(t,x)=∂tu(t,x)−g′(t)e−x−[g′(t)+h′(t)]xe−x,∂xv(t,x)=∂xu(t,x)−h(t)e−x+[g(t)+h(t)]xe−x,∂2xv(t,x)=∂2xu(t,x)+2h(t)e−x+g(t)e−x−[g(t)+h(t)]xe−x,∂3xv(t,x)=∂3xu(t,x)−3h(t)e−x−2g(t)e−x+[g(t)+h(t)]xe−x,∂4xv(t,x)=∂4xu(t,x)+4h(t)e−x+3g(t)e−x−[g(t)+h(t)]xe−x. | (2.2) |
In particular, thanks to (1.1)-(1.2)-(1.7), (2.1) and (2.2),
v(t,0)=u(t,0)−g(t)=0,∂xv(t,0)=∂xu(t,0)−h(t)=0. | (2.3) |
Moreover, thanks to (1.2) and (2.2), we have that
‖v0‖2L2(0,∞)≤‖u0‖2L2(0,∞). | (2.4) |
Again by (1.1)-(1.2)-(1.7), we have the following equation for v:
∂tv+2κv∂xv+ν∂2xv+δ∂3xv+β2∂4xv=−g′(t)e−x+[g′(t)+h′(t)]xe−x−2κh(t)e−xv+2κ[g(t)+h(t)]xe−xv−2κg(t)e−x∂xv−2κg(t)h(t)e−2x+2κg(t)[g(t)+h(t)]xe−2x−2κ[g(t)+h(t)]xe−x∂xv−2κh(t)[g(t)+h(t)]xe−2x+2κ[g(t)+h(t)]2x2e−2x+2h(t)e−x+g(t)e−x−[g(t)+h(t)]xe−x−3h(t)e−x−2g(t)e−x+[g(t)+h(t)]xe−x+4h(t)e−x+3g(t)e−x−[g(t)+h(t)]xe−x. | (2.5) |
Lemma 2.1. Fix T>0. There exists a constant C(T)>0, such that
‖v(t,⋅)‖2L2(0,∞)+β2eC0t2∫t0e−C0s‖∂2xv(s,⋅)‖2L2(0,∞)ds≤C(T), | (2.6) |
for every 0≤t≤T. In particular, we have that
∫t0‖∂xv(s,⋅)‖2L2(0,∞)ds≤C(T), | (2.7) |
for every 0≤t≤T.
Proof. Let 0≤t≤T. We begin by observing that, thanks to (2.3),
4κ∫∞0v2∂xvdx=0,2δ∫∞0v∂3xvdx=−2δ∫∞0∂xv∂2xv=0,2β∫∞0v∂4xvdx=−2β2∫∞0∂xv∂3xvdx=2β2‖∂2xv(t,⋅)‖2L2(0,∞). | (2.8) |
Therefore, by (2.8), multiplying (2.5) by 2v, an integration on (0,∞) gives
ddt‖v(t,⋅)‖2L2(0,∞)+2β2‖∂2xv(t,⋅)‖2L2(0,∞)=−2ν∫∞0v∂2xvdx+2g′(t)∫∞0e−xvdx+2[g′(t)+h′(t)]∫∞0xe−xvdx+4κ[g(t)+h(t)]∫∞0xe−xv2dx−4κg(t)∫Re−xv∂xvdx−4κ[g(t)+h(t)]∫∞0xe−xvdx−4κ[g(t)+h(t)]∫∞0xe−xv∂xvdx−4κh(t)[g(t)+h(t)]∫∞0xe−2xvdx+4κ[g(t)+h(t)]2∫∞0x2e−2xvdx+4h(t)∫∞0e−xvdx+2g(t)∫∞0e−xvdx−2[g(t)+h(t)]∫∞0xe−xvdx−6h(t)∫∞0e−xvdx−4g(t)∫∞0e−xvdx+2[g(t)+h(t)]∫∞0xe−xvdx+8h(t)∫∞0e−xvdx+6g(t)∫∞0e−xvdx−2[g(t)+h(t)]∫∞0xe−xvdx. | (2.9) |
Observe that, for each x∈(0,∞),
e−x≤1,xe−x≤e,∫∞0e−2xdx=12∫∞0x2e−4xdx=132,∫∞0x2e−2xdx=14,∫∞0x4e−4xdx=3128. | (2.10) |
Due (1.2), (2.10) and the Young inequality,
2|ν|∫∞0|v||∂2xv|dx=2∫∞0|νvβ||β∂2xv|dx≤ν2β2‖v(t,⋅)‖2L2(0,∞)+β2‖∂2xu(t,⋅)‖2L2(0,∞),2|g′(t)|∫∞0e−x|v|dx≤2C0∫∞0e−x|v|dx≤C0+C0‖v(t,⋅)‖2L2(0,∞),2|g′(t)+h′(t)|∫∞0xe−x|v|dx≤2C0∫∞0xe−x|v|dx≤C0+C0‖v(t,⋅)‖2L2(0,∞),4|κ||g(t)+h(t)|∫∞0xe−xv2dx≤C0∫∞0xe−xv2dx≤C0‖v(t,⋅)‖2L2(0,∞),4|κ||g(t)|∫∞0e−x|v||∂xv|dx≤2C0∫∞0e−x|v||∂xv|dx≤2C0∫∞0|v||∂xv|dx≤C0‖v(t,⋅)‖2L2(0,∞)+C0‖∂xv(t,⋅)‖2L2(0,∞),4|κ||g(t)+h(t)|∫∞0xe−x|v|dx≤2C0∫∞0xe−x|v|dx≤C0+C0‖v(t,⋅)‖2L2(0,∞),4κ|g(t)+h(t)|∫∞0xe−x|v||∂xv|dx≤2C0∫∞0xe−x|v||∂xv|dx≤C0‖v(t,⋅)‖2L2(0,∞)+C0‖∂xv(t,⋅)‖2L2(0,∞)4|κ||h(t)||g(t)+h(t)|∫∞0xe−2x|v|dx≤2C0∫∞0xe−2x|v|dx≤C0+C0‖v(t,⋅)‖2L2(0,∞),4|κ|[g(t)+h(t)]2∫∞0x2e−2x|v|dx≤2C0∫∞0x2e−2x|v|dx≤C0+C0‖v(t,⋅)‖2L2(0,∞),4|h(t)|∫∞0e−x|v|dx≤2C0∫∞0e−x|v|dx≤C0+C0‖v(t,⋅)‖2L2(0,∞),2|g(t)|∫∞0e−x|v|dx≤2C0∫∞0e−x|v|dx≤C0+C0‖v(t,⋅)‖2L2(0,∞),2|g(t)+h(t)|∫∞0xe−x|v|dx≤2C0∫∞0xe−x|v|dx≤C0+C0‖v(t,⋅)‖2L2(0,∞),6|h(t)|∫∞0e−x|v|dx≤2C0∫∞0e−x|v|dx≤C0+C0‖v(t,⋅)‖2L2(0,∞),4|g(t)|∫∞0e−x|v|dx≤2C0∫∞0e−x|v|dx≤C0+C0‖v(t,⋅)‖2L2(0,∞),2|g(t)+h(t)]|∫∞0xe−x|v|dx≤2C0∫∞0xe−x|v|dx≤C0+C0‖v(t,⋅)‖2L2(0,∞),8|h(t)|∫∞0e−x|v|dx≤2C0∫Re−x|v|dx≤C0+C0‖v(t,⋅)‖2L2(0,∞),6|g(t)|∫∞0e−x|v|dx≤2C0∫∞0e−x|v|dx≤C0+C0‖v(t,⋅)‖2L2(0,∞),2|g(t)+h(t)|∫∞0xe−x|v|dx≤2C0∫∞0xe−x|v|dx≤C0+C0‖v(t,⋅)‖2L2(0,∞). |
It follows from (2.9) that
ddt‖v(t,⋅)‖2L2(0,∞)+β2‖∂2xv(t,⋅)‖2L2(0,∞)≤C0‖v(t,⋅)‖2L2(0,∞)+C0‖∂xv(t,⋅)‖2L2(0,∞)+C0. | (2.11) |
Thanks to (2.3),
C0‖∂xv(t,⋅)‖2L2(0,∞)=C0∫∞0∂xv∂xvdx=−C0∫∞0v∂2xvdx. |
Therefore, by the Young inequality,
C0‖∂xv(t,⋅)‖2L2(0,∞)≤∫∞0|C0vβ||β∂2xv|dx≤C0‖v(t,⋅)‖2L2(0,∞)+β22‖∂2xv(t,⋅)‖2L2(0,∞). | (2.12) |
Consequently, by (2.11),
ddt‖v(t,⋅)‖2L2(0,∞)+β22‖∂2xv(t,⋅)‖2L2(0,∞)≤C0‖v(t,⋅)‖2L2(0,∞)+C0. |
By the Gronwall Lemma and (2.4), we have
‖v(t,⋅)‖2L2(0,∞)+β2eC0t2∫t0e−C0s‖∂2xv(s,⋅)‖2L2(0,∞)ds≤C0eC0t+C0t≤C(T), |
which gives (2.6).
Finally, we prove (2.7). By (2.6), and (2.12),
C0‖∂xv(t,⋅)‖2L2(0,∞)≤C(T)+β22‖∂2xv(t,⋅)‖2L2(0,∞). |
Integrating on (0,t), by (2.6), we have
C0∫t0‖∂xv(s,⋅)‖2L2(0,∞)ds≤C(T)t+β22∫t0‖∂2xv(s,⋅)‖2L2(0,∞)ds≤C(T), |
which gives (2.7).
Lemma 2.2. Fix T>0. There exists a constant C(T)>0, such that
‖u(t,⋅)‖L2(0,∞)≤C(T), | (2.13) |
∫t0‖∂xu(s,⋅)‖2L2(0,∞)ds≤C(T), | (2.14) |
∫t0‖∂2xu(s,⋅)‖2L2(0,∞)ds≤C(T), | (2.15) |
for every 0≤t≤T.
Proof. Let 0≤t≤T. We begin by proving (2.13). Observe that, by (2.1),
u2(t,x)=[v(t,x)+g(t)e−x+[g(t)+h(t)]xe−x]=v2(t,x)+g2(t)e−2x+[g(t)+h(t)]x2e−2x+2g(t)e−xv(t,x)+2[g(t)+h(t)]xe−xv+2g(t)[g(t)+h(t)]xe−2x. |
Consequently, an integration on (0,∞) gives
‖u(t,⋅)‖2L2(0,∞)=‖v(t,⋅)‖2L2(0,∞)+g2(t)∫∞0e−2xdx+[g(t)+h(t)]2∫∞0x2e−2xdx+2g(t)∫∞0e−xvdx+2[g(t)+h(t)]∫∞0xe−xvdx+2g(t)[g(t)+h(t)]∫x0xe−2xdx. | (2.16) |
Observe that
∫∞0xe−2x=14. | (2.17) |
Due to (1.2), (2.10), (2.17) and the Young inequality,
g2(t)∫∞0e−2xdx≤C0∫∞0e−2xdx≤C0,[g(t)+h(t)]2∫∞0x2e−2xdx≤C0∫∞0x2e−2xdx≤C0,2|g(t)|∫∞0e−x|v|dx≤2C0∫∞0e−x|v|dx≤C0+C0‖v(t,⋅)‖2L2(0,∞)2|g(t)+h(t)|∫∞0xe−x|v|dx≤2C0∫∞0xe−x|v|dx≤C0+C0‖v(t,⋅)‖2L2(0,∞)2|g(t)||g(t)+h(t)|]∫x0xe−2xdx≤C0∫x0xe−2xdx≤C0. |
Therefore, by (2.6) and (2.16),
‖u(t,⋅)‖2L2(0,∞)≤C0‖v(t,⋅)‖2L2(0,∞)+C0≤C(T), |
that is (2.13).
We prove (2.14). By (2.5), we have that
(∂xu(t,x))2=[∂xv(t,x)+h(t)e−x−[g(t)+h(t)]xe−x]2=(∂xv(t,x))2+h2(t)e−2x+[g(t)+h(t)]2x2e−2x+2h(t)e−x∂xv(t,x)−2[g(t)+h(t)]xe−x∂xv(t,x)−2h(t)[g(t)+h(t)]xe−2x. |
Therefore, an integration on (0,∞) gives
‖∂xu(t,⋅)‖2L2(0,∞)=‖∂xv(t,⋅)‖2L2(0,∞)+h2(t)∫∞0e−2xdx+[g(t)+h(t)]2∫∞0x2e−2xdx+2h(t)∫∞0e−x∂xvdx−2[g(t)+h(t)]∫∞0xe−x∂xvdx−2h(t)[g(t)+h(t)]∫∞0xe−2xdx. | (2.18) |
Due to (1.2), (2.10), (2.17) and the Young inequality,
h2(t)∫∞0e−2xdx≤C0∫∞0e−2xdx≤C0,[g(t)+h(t)]2∫∞0x2e−2xdx≤C0∫∞0x2e−2xdx≤C0,2|h(t)|∫∞0e−x|∂xv|dx≤2C0∫∞0e−x|∂xv|dx≤C0+C0‖∂xv(t,⋅)‖2L2(0,∞),2|g(t)+h(t)|∫∞0xe−x|∂xv|dx≤2C0∫∞0xe−x|∂xv|dx≤C0+C0‖∂xv(t,⋅)‖2L2(0,∞),2|h(t)||g(t)+h(t)|∫∞0xe−2xdx≤C0∫∞0xe−2xdx≤C0. |
It follows from (2.18) that
‖∂xu(t,⋅)‖2L2(0,∞)≤C0‖∂xv(t,⋅)‖2L2(0,∞)+C0. |
Integrating on (0,t), by (2.7), we have that
∫t0‖∂xu(s,⋅)‖2L2(0,∞)ds≤C0∫t0‖∂xv(s,⋅)‖2L2(0,∞)ds+C0t≤C(T), |
which gives (2.14).
Finally, we prove (2.15). By (2.2), we have that
(∂2xu(t,x))2=[∂2xv(t,x)−[2h(t)+g(t)]e−x+[g(t)+h(t)]xe−x]2=(∂2xv(t,x))2+[2h(t)+g(t)]2e−2x+[g(t)+h(t)]2x2e−2x−2[2h(t)+g(t)]e−x∂2xv+2[g(t)+h(t)]xe−x∂2xv−2[2h(t)+g(t)][g(t)+h(t)]xe−2x. |
An integration on (0,∞) gives
‖∂2xu(t,⋅)‖2L2(0,∞)=‖∂2xv(t,⋅)‖2L2(0,∞)+[2h(t)+g(t)]2∫∞0e−2xdx+[g(t)+h(t)]2∫∞0x2e−2xdx−2[2h(t)+g(t)]∫∞0e−x∂2xvdx+2[g(t)+h(t)]∫∞0xe−x∂2xvdx−2[2h(t)+g(t)][g(t)+h(t)]∫∞0xe−2xdx. | (2.19) |
Due to (1.2), (2.10), (2.17) and the Young inequality,
[2h(t)+g(t)]2∫∞0e−2xdx≤C0∫∞0e−2xdx≤C0,[g(t)+h(t)]2∫∞0x2e−2xdx≤C0∫∞0x2e−2xdx≤C0,2|2h(t)+g(t)|∫∞0e−x|∂2xv|dx≤2C0∫∞0e−x|∂2xv|dx≤C0+C0‖∂2xv(t,⋅)‖2L2(0,∞),2|g(t)+h(t)|∫∞0xe−x|∂2xv|dx≤2C0∫∞0xe−x|∂2xv|dx≤C0+C0‖∂2xv(t,⋅)‖2L2(0,∞),2|2h(t)+g(t)||g(t)+h(t)|∫∞0xe−2xdx≤C0∫∞0xe−2xdx≤C0. |
It follows from (2.19) that
‖∂2xu(t,⋅)‖2L2(0,∞)≤‖∂2xv(t,⋅)‖2L2(0,∞)+C0. |
Integrating on (0,t), by (2.6), we have that
∫t0‖∂2xu(s,⋅)‖2L2(0,∞)ds≤∫t0‖∂2xv(s,⋅)‖2L2(0,∞)ds+C0t≤C(T), |
which gives (2.15).
Lemma 2.3. Fix T>0. There exists a constant C(T)>0, such that
∫t0‖∂xu(s,⋅)‖2L∞(0,∞)ds≤C(T), | (2.20) |
∫t0‖u(s,⋅)∂xu(s,⋅)‖2L2(0,∞)ds≤C(T), | (2.21) |
for every 0≤t≤T.
Proof. Let 0≤t≤T. We begin by proving (2.20). Thanks to (1.2) and the Young inequality,
(∂xu(t,x))2=2∫x0∂xu∂2xudy+h2(t)≤2∫∞0|∂xu||∂2xu|dx+C0≤‖∂xu(t,⋅)‖2L2(0,∞)+‖∂2xu(t,⋅)‖2L2(0,∞)+C0. |
Hence,
‖∂xu(t,⋅)‖2L∞(0,∞)≤‖∂xu(t,⋅)‖2L2(0,∞)+‖∂2xu(t,⋅)‖2L2(0,∞)+C0 | (2.22) |
Integrating on (0,t), by (2.14) and (2.15), we have that
∫t0‖∂xu(s,⋅)‖2L∞(0,∞)ds≤∫t0‖∂xu(s,⋅)‖2L2(0,∞)ds+∫t0‖∂2xu(s,⋅)‖2L2(0,∞)ds+C0t≤C(T), |
which gives (2.20).
Finally, we prove (2.21). Thanks to (2.13), we have that
‖u(t,⋅)∂xu(s,⋅)‖2L2(0,∞)ds=∫∞0u2(∂xu)2dx≤‖∂xu(t,⋅)‖2L∞(0,∞)‖u(t,⋅)‖2L2(0,∞)≤C(T)‖∂xu(t,⋅)‖2L∞(0,∞). |
(2.21) follows from (2.20) and an integration on (0,t).
Lemma 2.4. Fix T>0. There exists a constant C(T)>0, such that
‖∂2xu(t,⋅)‖2L2(0,∞)+δ2+24β4‖∂xu(t,⋅)‖2L2(0,∞) | (2.23) |
+β22∫t0‖∂4xu(s,⋅)‖2L2(0,∞)ds+δ2+246β2∫0‖∂3xu(s,⋅)‖2L2(0,∞)ds≤C(T),∫t0(∂2xu(s,0))2ds≤C(T),∫t0(∂3xu(s,0))2ds≤C(T), | (2.24) |
‖∂xu‖L∞((0,T)×(0,∞))≤C(T), | (2.25) |
‖u‖L∞((0,T)×(0,∞))≤C(T), | (2.26) |
for every 0≤t≤T.
Proof. Let 0≤t≤T. Consider an positive constant A, which will be specified later. Multiplying (1.1)-(1.2)-(1.7) by
2∂4xu−2A∂2xu, |
we have that
(2∂4xu−2A∂2xu)∂tu+2κ(2∂4xu−2A∂2xu)u∂xu+ν(2∂4xu−2A∂2xu)∂2xu+δ(2∂4xu−2A∂2xu)∂3xu+β2(2∂4xu−2A∂2xu)∂4xu=0. | (2.27) |
Observe that, thanks to (1.1)-(1.2)-(1.7),
∫∞0(2∂4xu−2A∂2xu)∂tudx=−2∂3xu(t,0)∂tu(t,0)−2∫∞0∂3xu∂t∂xudx+2A∂xu(t,0)∂tu(t,0)+Addt‖∂xu(t,⋅)‖2L2(0,∞)=−2g′(t)∂3xu(t,0)+2h′(t)∂2xu(t,0)+2Ah(t)g(t)+ddt(‖∂2xu(t,⋅)‖2L2(0,∞)+A‖∂xu(t,⋅)‖2L2(0,∞)),−2Aβ2∫∞0∂2xu∂4xudx=2Aβ2∂2xu(t,0)∂3xu(t,0)+2Aβ2‖∂3xu(t,⋅)‖2L2(0,∞). | (2.28) |
Therefore, thanks to (2.28), an integration of (2.27) on (0,∞) gives
ddt(‖∂2xu(t,⋅)‖2L2(0,∞)+A‖∂xu(t,⋅)‖2L2(0,∞))+2β2‖∂4xu(t,⋅)‖2L2(0,∞)+2Aβ2‖∂3xu(t,⋅)‖2L2(0,∞)=−4κ∫∞0u∂xu∂4xudx+4Aκ∫∞0u∂xu∂2xudx−2ν∫∞0∂4xu∂2xudx−2Aν‖∂2xu(t,⋅)‖2L2(0,∞)−2δ∫∞0∂4xu∂3xudx+2Aδ∫∞0∂2xu∂3xudx+2g′(t)∂3xu(t,0)−2h′(t)∂2xu(t,0)−2Ah(t)g(t)−2Aβ2∂2xu(t,0)∂3xu(t,0). | (2.29) |
Due to the Young inequality,
4|κ|∫∞0|u∂xu||∂4xu|dx=2∫∞0|2κu∂xuβ√D1||β√D1∂4xu|dx≤4κ2β2D1‖u(t,⋅)∂xu(t,⋅)‖2L2(0,∞)+D1β2‖∂4xu(t,⋅)‖2L2(0,∞),4A|κ|∫∞0|u∂xu||∂2xu|dx=2∫∞0|2Aκu∂xu||∂2xu|dx≤4A2κ2‖u(t,⋅)∂xu(t,⋅)‖2L2(0,∞)+‖∂2xu(t,⋅)‖2L2(0,∞),2|ν|∫∞0|∂4xu||∂2xu|dx=2∫∞0|β√D1∂4xu||ν∂2xuβ√D1|dx≤D1β2‖∂4xu(t,⋅)‖2L2(0,∞)+ν2β2D1‖∂2xu(t,⋅)‖2L2(0,∞),2|δ|∫∞0|∂4xu||∂3xu|dx=2∫∞0|β√D1∂4xu||δ∂3xuβ√D1|dx≤D1β2‖∂4xu(t,⋅)‖2L2(0,∞)+δ2β2D1‖∂3xu(t,⋅)‖2L2(0,∞),2A|δ|∫∞0|∂2xu||∂3xu|dx=A∫∞0|δ∂2xuβ||β∂3xu|dx≤Aδ2β2‖∂2xu(t,⋅)‖2L2(0,∞)+Aβ2‖∂3xu(t,⋅)‖2L2(0,∞), |
where D1 is a positive constant, which will be specified later. It follows from (2.29) that
ddt(‖∂2xu(t,⋅)‖2L2(0,∞)+A‖∂xu(t,⋅)‖2L2(0,∞))+β2(2−3D1)‖∂4xu(t,⋅)‖2L2(0,∞)+(Aβ2−δ2β2D1)‖∂3xu(t,⋅)‖2L2(0,∞)≤4κ2(1β2D1+A2)‖u(t,⋅)∂xu(t,⋅)‖2L2(0,∞)+(1+ν2β2D1+2A|ν|+Aδ2β2)‖∂2xu(t,⋅)‖2L2(0,∞)+2|g′(t)||∂3xu(t,0)|+2|h′(t)||∂2xu(t,0)|+2A|h(t)||g(t)|+2Aβ2|∂2xu(t,0)||∂3xu(t,0)|. |
Choosing D1=3, we have that
ddt(‖∂2xu(t,⋅)‖2L2(0,∞)+A‖∂xu(t,⋅)‖2L2(0,∞))+β2‖∂4xu(t,⋅)‖2L2(0,∞)+(Aβ2−δ23β2)‖∂3xu(t,⋅)‖2L2(0,∞)≤4κ2(13β2+A2)‖u(t,⋅)∂xu(t,⋅)‖2L2(0,∞)+(1+ν23β2+2A|ν|+Aδ2β2)‖∂2xu(t,⋅)‖2L2(0,∞)+2|g′(t)||∂3xu(t,0)|+2|h′(t)||∂2xu(t,0)|+2A|h(t)||g(t)|+2Aβ2|∂2xu(t,0)||∂3xu(t,0)|. | (2.30) |
Due to (1.2) and the Young inequality,
2|g′(t)||∂3xu(t,0)|≤(g′(t))2+(∂3xu(t,0))2≤C0+(∂3xu(t,0))2,2|h′(t)||∂2xu(t,0)|≤(h′(t))2+(∂2xu(t,0))2≤C0+(∂2xu(t,0))2,2A|h(t)||g(t)|≤AC0,2Aβ2|∂2xu(t,0)||∂3xu(t,0)|≤A2β4(∂2xu(t,0))2+(∂3xu(t,0))2. |
Consequently, by (2.30),
ddt(‖∂2xu(t,⋅)‖2L2(0,∞)+A‖∂xu(t,⋅)‖2L2(0,∞))+β2‖∂4xu(t,⋅)‖2L2(0,∞)+(Aβ2−δ23β2)‖∂3xu(t,⋅)‖2L2(0,∞)≤4κ2(13β2+A2)‖u(t,⋅)∂xu(t,⋅)‖2L2(0,∞)+(1+ν23β2+2A|ν|+Aδ2β2)‖∂2xu(t,⋅)‖2L2(0,∞)+2(∂3xu(t,0))2+(1+A2β4)(∂2xu(t,0))2+C0(1+A). | (2.31) |
Thanks to the Young inequality,
2(∂3xu(t,0))2=−4∫∞0∂3xu∂4xudx≤∫∞0|4∂3xuβ||β∂4xu|dx≤8β2‖∂3xu(t,⋅)‖2L2(0,∞)+β22‖∂4xu(t,⋅)‖2L2(0,∞),(1+A2β4)(∂2xu(t,0))2=−2(1+A2β4)∫∞0∂2xu∂3xudx≤∫∞0|2(1+A2β4)∂2xuβ√A||β√A∂3xu|dx≤2(1+A2β4)2‖∂2xu(t,⋅)‖2L2(0,∞)+Aβ22‖∂3xu(t,⋅)‖2L2(0,∞). | (2.32) |
It follows from (2.31) that
ddt(‖∂2xu(t,⋅)‖2L2(0,∞)+A‖∂xu(t,⋅)‖2L2(0,∞))+β22‖∂4xu(t,⋅)‖2L2(0,∞)+(Aβ22−δ2+243β2)‖∂3xu(t,⋅)‖2L2(0,∞)≤4κ2(13β2+A2)‖u(t,⋅)∂xu(t,⋅)‖2L2(0,∞)+(1+ν23β2+2A|ν|+Aδ2β2+2(1+A2β4)2)‖∂2xu(t,⋅)‖2L2(0,∞)+C0(1+A). |
Choosing
A=δ2+24β4, | (2.33) |
we have that
ddt(‖∂2xu(t,⋅)‖2L2(0,∞)+δ2+24β4‖∂xu(t,⋅)‖2L2(0,∞))+β22‖∂4xu(t,⋅)‖2L2(0,∞)+δ2+246β2‖∂3xu(t,⋅)‖2L2(0,∞)≤C0‖u(t,⋅)∂xu(t,⋅)‖2L2(0,∞)+C0‖∂2xu(t,⋅)‖2L2(0,∞)+C0. |
Integrating on (0,t), by (1.8), (2.15) and (2.21), we get
‖∂2xu(t,⋅)‖2L2(0,∞)+δ2+24β4‖∂xu(t,⋅)‖2L2(0,∞)+β22∫t0‖∂4xu(s,⋅)‖2L2(0,∞)ds+δ2+246β2∫0‖∂3xu(s,⋅)‖2L2(0,∞)ds≤C0+C0∫t0‖u(s,⋅)∂xu(s,⋅)‖2L2(0,∞)ds+C0∫t0‖∂2xu(s,⋅)‖2L2(0,∞)ds+C0t≤C(T), |
which gives (2.23).
(2.24) follows from (2.15), (2.23), (2.33) and an integration of (2.32) on (0,t), while (2.22) and (2.23) give (2.25).
Finally, we prove (2.26). Due to (1.2), (2.13), (2.23) and the Hölder inequality,
u2(t,x)=2∫x0u∂xudy+g2(t)≤2∫∞0|u||∂xu|dx+C0≤‖u(t,⋅)‖L2(0,∞)‖∂xu(t,⋅)‖L2(0,∞)+C0≤C(T). |
Hence,
‖u‖2L∞((0,T)×(0,∞))≤C(T), |
which gives (2.26).
Lemma 2.5. Fix T>0. There exists a constant C(T)>0, such that
∫t0‖∂tu(s,⋅)‖2L2(0,∞)ds≤C(T), | (2.34) |
for every 0≤t≤T.
Proof. Let 0≤t≤T. Multiplying (1.1)-(1.2)-(1.7) by 2∂tu, an integration on (0,∞) gives
2‖∂tu(t,⋅)‖2L2(0,∞)=−4κ∫∞0u∂xu∂tudx−2ν∫∞0∂2xu∂tudx−2δ∫∞0∂3xu∂tudx−2β2∫∞0∂4xu∂tudx. | (2.35) |
Due to (2.23) and the Young inequality,
4|κ|∫∞0u∂xu∂tudx=2∫∞0|2κu∂xu√D2||√D2∂tu|dx≤4κ2D2‖u(t,⋅)∂xu(t,⋅)‖2L2(0,∞)+D2‖∂tu(t,⋅)‖2L2(0,∞),2|ν|∫∞0|∂2xu||∂tu|dx=2∫∞0|ν∂2xu√D2||√D2∂tu|dx≤ν2D2‖∂2xu(t,⋅)‖2L2(0,∞)+D2‖∂tu(t,⋅)‖2L2(0,∞)≤C(T)D2+D2‖∂tu(t,⋅)‖2L2(0,∞),2|δ|∫∞0|∂3xu||∂tu|dx=2∫∞0|δ∂3xu√D2||√D2∂tu|dx≤δ2D2‖∂3xu(t,⋅)‖2L2(0,∞)+D2‖∂tu(t,⋅)‖2L2(0,∞),2β2∫∞0|∂4xu||∂tu|dx=2∫∞0|β2∂4xu√D2||√D2∂tu|dx≤β4D2‖∂4xu(t,⋅)‖2L2(0,∞)+D2‖∂tu(t,⋅)‖2L2(0,∞). |
Consequently, by (2.35),
2(1−2D2)‖∂tu(t,⋅)‖2L2(0,∞)≤C(T)D2+4κ2D2‖u(t,⋅)∂xu(t,⋅)‖2L2(0,∞)+δ2D2‖∂3xu(t,⋅)‖2L2(0,∞)+β4D2‖∂4xu(t,⋅)‖2L2(0,∞). |
Choosing D2=14, we have that
‖∂tu(t,⋅)‖2L2(0,∞)≤C(T)+16κ2‖u(t,⋅)∂xu(t,⋅)‖2L2(0,∞)+4δ2‖∂3xu(t,⋅)‖2L2(0,∞)+4β4‖∂4xu(t,⋅)‖2L2(0,∞). |
Integrating on (0,t), by (2.21) and (2.23), we get
∫t0‖∂tu(s,⋅)‖2L2(0,∞)ds≤C(T)t+16κ2∫t0‖u(s,⋅)∂xu(t,⋅)‖2L2(0,∞)ds+4δ2∫t0‖∂3xu(s,⋅)‖2L2(0,∞)ds+4β4∫t0‖∂4xu(s,⋅)‖2L2(0,∞)ds≤C(T), |
which gives (2.34).
Now, we prove Theorem 1.1.
Proof. Fix T>0. Thanks to Lemmas 2.2, 2.4, 2.5 and the Cauchy-Kovalevskaya Theorem [49], we have that u is solution of (1.1)-(1.2)-(1.7) and (1.13) holds.
We prove (1.14). Let u1 and u2 be two solutions of (1.1)-(1.2)-(1.7), which verify (1.13), that is
{∂tu1+2κu1∂xu1+ν∂2xu1+δ∂3xu1+β2∂4xu1=0,t>0,x>0,u1(t,0)=g(t),t>0,∂xu1(t,0)=h(t),t>0,u1(0,x)=u1,0(x),x>0,{∂tu2+2κu2∂xu2+ν∂2xu2+δ∂3xu2+β2∂4xu2=0,t>0,x>0,u2(t,0)=g(t),t>0,∂xu2(t,0)=h(t),t>0,u2(0,x)=u2,0(x),x>0. |
Then, the function
ω=u1−u2 | (2.36) |
solves the following initial-boundary value problem:
{∂tω+2κ(u1∂xu1−u2∂xu2)+ν∂2xω+δ∂3xω+β2∂4xω=0,t>0,x>0,ω(t,0)=0,t>0,∂xω(t,0)=0,t>0,ω(0,x)=u1,0(x)−u2,0(x),x>0. | (2.37) |
Observe that, thanks to (2.36),
u1∂xu1−u2∂xu2=u1∂xu1−u1∂xu2+u1∂xu2−u2∂xu2=u1∂xω+∂xu2ω. |
Therefore, (2.37) is equivalent to the following equation:
∂tω+2κu1∂xω+2κ∂xu2ω+ν∂2xω+δ∂3xω+β2∂4xω=0. | (2.38) |
Moreover, since u1,u2∈L∞(0,T;H2(0,∞)), we have that
‖∂xu1‖L∞((0,T)×(0,∞)),‖∂xu2‖L∞((0,T)×(0,∞))≤C(T). | (2.39) |
Observe again that, thanks to (2.37),
4κ∫∞0u1ω∂xωdx=−2κ∫∞0∂xu1ω2dx,2δ∫∞0ω∂3xωdx=−2δ∫∞0∂xω∂2xωdx=0,2β2∫∞0ω∂4xωdx=−2β2∫∞0∂xω∂3xωdx=2β2‖∂2xω(t,⋅)‖2L2(0,∞). | (2.40) |
Therefore, multiplying (2.38) by 2ω, thanks to (2.40), an integration on (0,∞) gives
ddt‖∂xω(t,⋅)‖2L2(0,∞)+2β2‖∂2xω(t,⋅)‖2L2(0,∞)=2κ∫∞0(∂xu1−2∂xu2)ω2dx−2ν∫∞0ω∂2xωdx. | (2.41) |
Due to (2.39) and the Young inequality,
2|κ|∫∞0|∂xu1−2∂xu2|ω2dx≤C(T)‖ω(t,⋅)‖2L2(0,∞),2|ν|∫∞0|ω||∂2xω|dx=2∫∞0|νωβ||β∂2xω|dx≤ν2β2‖ω(t,⋅)‖2L2(0,∞)+β2‖∂2xω(t,⋅)‖2L2(0,∞). |
Therefore, by (2.41),
ddt‖∂xω(t,⋅)‖2L2(0,∞)+β2‖∂2xω(t,⋅)‖2L2(0,∞)≤C(T)‖ω(t,⋅)‖2L2(0,∞). |
The Gronwall Lemma and (2.37) gives
‖∂xω(t,⋅)‖2L2(0,∞)+β2eC(T)t∫t0e−C(T)s‖∂2xω(s,⋅)‖2L2(0,∞)ds≤eC(T)t‖ω0‖2L2(0,∞). | (2.42) |
(1.14) follows from (2.36) and (2.42).
In this section, we prove Theorem 1.1 for (1.1)-(1.3)-(1.7).
Let us prove some a priori estimates on u.
Following [11,17], we consider the following function:
v(t,x)=u(t,x)−g(t)e−x. | (3.1) |
Observe that
∂tv(t,x)=∂tu(t,x)−g′(t)e−x,∂xv(t,x)=∂xu(t,x)+g(t)e−x,∂2xv(t,x)=∂2xu(t,x)−g(t)e−x,∂3xv(t,x)=∂3xu(t,x)+g(t)e−x,∂4xv(t,x)=∂4xu−g(t)e−x. | (3.2) |
By (1.1)-(1.3)-(1.7) and (3.2),
v(t,0)=u(t,0)−g(t)=0,∂2xv(t,0)=−g(t), | (3.3) |
while, by (1.3) and (3.2), we have (2.4).
Again by (1.1)-(1.3)-(1.7) and (3.2), we have the following equation for v.
∂tv+2κv∂xv+ν∂2xv+δ∂3xv+β2∂4xv=−g′(t)e−x+2κg(t)e−xv−2κg(t)e−x∂xv+2κg2(t)e−2x−νg(t)e−x−δg(t)e−x−β2g(t)e−x. | (3.4) |
We prove the following result.
Lemma 3.1. Fix T>0. There exists a constant C(T)>0, such that
‖v(t,⋅)‖2L2(0,∞)+β2eC0t42∫t0e−C0s‖∂2xv(s,⋅)‖2L2(0,∞)ds≤C(T), | (3.5) |
for every 0≤t≤T. In particular, we have (2.7). Moreover,
∫t0(∂xv(s,0))2ds≤C(T), | (3.6) |
for every 0≤t≤T.
Proof. Let 0≤t≤T. We begin by observing that, thanks to (3.3),
2∫∞0v∂tvdx=ddt‖v(t,⋅)‖2L2(0,∞),2δ∫∞0v∂3xv=−2δ∫∞0∂xv∂2xvdx,2β2∫∞0v∂4xvdx=−2β2∫∞0∂xv∂3xvdx=2β2∂xv(t,0)∂2xv(t,0)+2β2‖∂2xv(t,⋅)‖2L2(0,∞)=2β2g(t)∂xv(t,0)+2β2‖∂2xv(t,⋅)‖2L2(0,∞). | (3.7) |
Consequently, multiplying (3.4) by 2v, thanks to (3.7), an integration of (3.7) on (0,∞) gives
ddt‖v(t,⋅)‖2L2(0,∞)+2β2‖∂2xv(t,⋅)‖2L2(0,∞)=−2g′(t)∫∞0e−xvdx+4κg(t)∫∞0e−xv2dx−4κg(t)∫∞0e−xv∂xvdx+4κg2(t)∫∞0e−2xvdx−2νg(t)∫∞0e−xvdx−2δg(t)∫∞0e−xvdx−β2g(t)∫∞0e−xvdx−2ν∫∞0v∂2xvdx+2δ∫∞0∂xv∂2xvdx−2β2g(t)∂xv(t,0). | (3.8) |
Since
∫∞0e−4xdx=14, | (3.9) |
thanks to (1.3), (2.10), (3.9) and the Young inequality,
2|g′(t)|∫∞0e−x|v|dx≤2C0∫∞0e−x|v|dx≤C0+C0‖v(t,⋅)‖2L2(0,∞),4|κ||g(t)|∫∞0e−xv2dx≤C0∫∞0e−xv2dx≤C0‖v(t,⋅)‖2L2(0,∞),4|κ||g(t)|∫∞0e−x|v||∂xv|dx≤2C0∫∞0e−x|v||∂xv|dx≤2C0∫∞0|v||∂xv|dx≤C0‖v(t,⋅)‖2L2(0,∞)+C0‖∂xv(t,⋅)‖2L2(0,∞),4|κ|g2(t)∫∞0e−2x|v|dx≤2C0∫∞0e−2x|v|dx≤C0+C0‖v(t,⋅)‖2L2(0,∞),2|ν+δ+β||g(t)|∫∞0e−x|v|dx≤2C0∫∞0e−x|v|dx≤C0+C0‖v(t,⋅)‖2L2(0,∞),2|ν|∫∞0v|∂2xv|dx=2∫∞0|νvβ||β∂2xv|dx≤ν2β2‖v(t,⋅)‖2L2(0,∞)+β2‖∂2xv(t,⋅)‖2L2(0,∞),2|δ|∫∞0|∂xv||∂2xv|dx=∫∞0|2δ∂xvβ||β∂2xv|dx≤2δ2β2‖∂xv(t,⋅)‖2L2(0,∞)+β22‖∂2xv(t,⋅)‖2L2(0,∞),2β2|g(t)||∂xv(t,0)|≤2C0|∂xv(t,0)|≤C0+C0(∂xv(t,0))2. |
It follows from (3.8) that
ddt‖v(t,⋅)‖2L2(0,∞)+β22‖∂2xv(t,⋅)‖2L2(0,∞)≤C0+C0‖v(t,⋅)‖2L2(0,∞)+C0‖∂xv(t,⋅)‖2L2(0,∞)+C0(∂xv(t,0))2. | (3.10) |
Observe that, by the Young inequality,
C0(∂xv(t,0))2=−2C0∫∞0∂xv∂2xvdx≤2C0∫∞0|∂xv||∂2xv|dx≤2∫∞0|√3C0∂xvβ||β√3∂2xv|dx≤C0‖∂xv(t,⋅)‖2L2(0,∞)+β23‖∂2xv(t,⋅)‖2L2(0,∞). | (3.11) |
Consequently, by (3.10),
ddt‖v(t,⋅)‖2L2(0,∞)+β26‖∂2xv(t,⋅)‖2L2(0,∞)≤C0+C0‖v(t,⋅)‖2L2(0,∞)+C0‖∂xv(t,⋅)‖2L2(0,∞). | (3.12) |
Observe that, thanks to (3.3),
C0‖∂xv(t,⋅)‖2L2(0,∞)=C0∫∞0∂xv∂xvdx=−C0∫∞0v∂xvdx. |
Therefore, by the Young inequality,
C0‖∂xv(t,⋅)‖2L2(0,∞)≤2∫∞0|√7C0v2β||β∂2xv√7|dx≤C0‖v(t,⋅)‖2L2(0,∞)+β27‖∂2xv(t,⋅)‖2L2(0,∞). | (3.13) |
It follows from (3.12) that
ddt‖v(t,⋅)‖2L2(0,∞)+β242‖∂2xv(t,⋅)‖2L2(0,∞)≤C0+C0‖v(t,⋅)‖2L2(0,∞). | (3.14) |
By (2.4) and the Gronwall Lemma, we get
‖v(t,⋅)‖2L2(0,∞)+β2eC0t42∫t0e−C0s‖∂2xv(s,⋅)‖2L2(0,∞)ds≤C0eC0t+C0eC0t∫t0e−C0sds≤C(T), |
which gives (3.5).
We prove (2.7). By (3.5) and (3.13), we have that
C0‖∂xv(t,⋅)‖2L2(0,∞)≤C(T)+β27‖∂2xv(t,⋅)‖2L2(0,∞). |
Integrating on (0,t), by (3.5), we have (2.7).
Finally, (3.6) follows from (2.7), (3.5), (3.11) and an integration on (0,t).
Lemma 3.2. Fix T>0. There exists a constant C(T)>0, such that
∫t0(∂xu(s,0))2ds≤C(T), | (3.15) |
for every 0≤t≤T. In particular, (2.13), (2.14), (2.15), (2.20), (2.21) hold.
Proof. Let 0≤t≤T. We prove (3.15). We begin by observing that, thanks (3.2),
∂xu(t,0)=∂xv(t,0)−g(t). |
Therefore, by (1.3) and the Young inequality,
(∂xu(t,0))2=(∂xv(t,0))2+g2(t)−2g(t)∂xv(t,0)≤(∂xv(t,0))2+g2(t)+2|g(t)||∂xv(t,0)|≤2(∂xv(t,0))2+2g2(t)≤2(∂xv(t,0))2+C0, |
Integrating on (0,t), by (3.6), we get
∫t0(∂xu(s,0))2ds≤2∫t0(∂xv(s,0))2ds+C0t≤C(T), |
which gives (3.15).
Arguing as in Lemma 2.2, we have (2.13), (2.14) and (2.15).
We prove (2.20). Thanks to the Young inequality,
(∂xu(t,x))2=2∫x0|∂xu||∂2xu|dx+2(∂xu(t,0))2≤2∫∞0|∂xu||∂2xu|dx+2(∂xu(t,0))2≤‖∂xu(t,⋅)‖2L2(0,∞)+‖∂2xu(t,⋅)‖2L2(0,∞)+2(∂xu(t,0))2. |
Therefore,
‖∂xu(t,⋅)‖2L∞(0,∞)≤‖∂xu(t,⋅)‖2L2(0,∞)+‖∂2xu(t,⋅)‖2L2(0,∞)+2(∂xu(t,0))2. | (3.16) |
Integrating on (0,t), by (2.14), (2.15) and (3.15), we have (2.20).
Finally, arguing as in Lemma 2.3, we have (2.20).
Lemma 3.3. Fix T>0. There exists a constant C(T)>0, such that
‖∂xu(t,⋅)‖2L2(0,∞)+β2∫t0‖∂3xu(s,⋅)‖2L2(0,∞)ds≤C(T), | (3.17) |
for every 0≤t≤T. In particular, (2.26) holds.
Proof. Let 0≤t≤T. We begin by observing that
−2∫∞0u∂2xudx=2∂xu(t,0)∂tu(t,0)+ddt‖∂xu(t,⋅)‖2L2(0,∞),−2δ∫∞0∂2xu∂3xudx=0,−2β2∫∞0∂2xu∂4xudx=2β2‖∂3xu(t,⋅)‖2L2(0,∞). | (3.18) |
Since, thanks (1.1)-(1.3)-(1.7), ∂tu(t,0)=g′(t), multiplying (1.1)-(1.3)-(1.7) by −2∂2xu, thanks to (3.18), an integration on (0,∞) gives
ddt‖∂xu(t,⋅)‖2L2(0,∞)+2β2‖∂3xu(t,⋅)‖2L2(0,∞)=−2g′(t)∂xu(t,0)+4κ∫∞0u∂xu∂2xudx+2ν‖∂2xu(t,⋅)‖2L2(0,∞). | (3.19) |
Due to (1.3) and the Young inequality,
2|g′(t)||∂xu(t,0)|≤2C0|∂xu(t,0)|≤C0+C0(∂xu(t,0))24|κ|∫∞0|u∂xu||∂2xu|dx≤2κ2‖u(t,⋅)∂xu(t,⋅)‖2L2(0,∞)+‖∂2xu(t,⋅)‖2L2(0,∞). |
Consequently, by (3.19),
ddt‖∂xu(t,⋅)‖2L2(0,∞)+2β2‖∂3xu(t,⋅)‖2L2(0,∞)≤C0+C0(∂xu(t,0))2+2κ2‖u(t,⋅)∂xu(t,⋅)‖2L2(0,∞)+(1+|ν|)‖∂2xu(t,⋅)‖2L2(0,∞). |
Integrating on (0,t), by (1.8), (2.15) and (2.21), we get
‖∂xu(t,⋅)‖2L2(0,∞)+2β2∫t0‖∂3xu(s,⋅)‖2L2(0,∞)ds≤C0+C0t+C0∫t0(∂xu(s,0))2ds+2κ2∫t0‖u(s,⋅)∂xu(s,⋅)‖2L2(0,∞)ds+(1+|ν|)∫t0‖∂2xu(s,⋅)‖2L2(0,∞)ds≤C(T), |
which gives (3.17).
Finally, arguing as in Lemma 2.4, we have (2.26).
Lemma 3.4. Fix T>0. There exists a constant C(T)>0, such that
‖∂2xu(t,⋅)‖2L2(0,∞)+β2∫t0‖∂4xu(s,⋅)‖2L2(0,∞)ds≤C(T), | (3.20) |
for ever 0≤t≤T. In particular, we have (2.25) and
∫t0(∂3xu(s,0))2ds≤C(T), | (3.21) |
for every 0≤t≤T. Moreover, (2.34) holds.
Proof. Let 0≤t≤T. We begin by observing that, thanks to (1.1)-(1.3)-(1.7),
2∫∞0∂4xu∂tudx=−2∂3xu(t,0)∂tu(t,0)−2∫∞0∂3xu∂t∂xudx=−2∂3xu(t,0)∂tu(t,0)+ddt‖∂2xu(t,⋅)‖2L2(0,∞),2ν∫∞0∂2xu∂4xudx=−2ν‖∂3xu(t,⋅)‖2L2(0,∞). | (3.22) |
Since, thanks to (1.1)-(1.3)-(1.7), ∂tu(t,0)=g′(t), multiplying (1.1)-(1.3)-(1.7) by 2∂4xu, thanks (3.22), an integration on (0,∞) gives
ddt‖∂2xu(t,⋅)‖2L2(0,∞)+2β2‖∂4xu(t,⋅)‖2L2(0,∞)=2g′(t)∂3xu(t,0)−4κ∫∞0u∂xu∂4xudx+2ν‖∂3xu(t,⋅)‖2L2(0,∞)−2δ∫∞0∂3xu∂4xudx. | (3.23) |
Due to (1.3) and the Young inequality,
2|g′(t)||∂3xu(t,0)|≤2C0|∂3xu(t,0)|≤C0+C0(∂3xu(t,0))2,4|κ|∫∞0|u∂xu||∂4xu|dx=2∫∞0|2κu∂xuβ||β∂4xu|dx≤4κ2β2‖u(t,⋅)∂xu(t,⋅)‖2L2(0,∞)+β2‖∂4xu(t,⋅)‖2L2(0,∞)2|δ|∫∞0|∂3xu||∂4xu|dx=∫∞0|2δ∂3xuβ||β∂4xu|dx≤2δ2β2‖∂3xu(t,⋅)‖2L2(0,∞)+β22‖∂4xu(t,⋅)‖2L2(0,∞). |
It follows from (3.23) that
ddt‖∂2xu(t,⋅)‖2L2(0,∞)+β22‖∂4xu(t,⋅)‖2L2(0,∞)≤C0+C0(∂3xu(t,0))2+4κ2β2‖u(t,⋅)∂xu(t,⋅)‖2L2(0,∞)+(2|ν|+2δ2β2)‖∂3xu(t,⋅)‖2L2(0,∞). | (3.24) |
Thanks to the Young inequality,
C0(∂3xu(t,0))2=−2C0∫∞0∂3xu∂4xudx≤2∫∞0|√3C0∂3xuβ||β∂4xu√3|dx≤C0‖∂3xu(t,⋅)‖2L2(0,∞)+β23‖∂4xu(t,⋅)‖2L2(0,∞). | (3.25) |
Consequently, by (3.24),
ddt‖∂2xu(t,⋅)‖2L2(0,∞)+β26‖∂4xu(t,⋅)‖2L2(0,∞)≤C0+4κ2β2‖u(t,⋅)∂xu(t,⋅)‖2L2(0,∞)+C0‖∂3xu(t,⋅)‖2L2(0,∞). |
Integrating on (0,t), by (1.8), (2.21) and (3.17), we get
‖∂2xu(t,⋅)‖2L2(0,∞)+β26∫t0‖∂4xu(s,⋅)‖2L2(0,∞)≤C0+C0t+4κ2β2∫t0‖u(s,⋅)∂xu(s,⋅)‖2L2(0,∞)ds+C0∫t0‖∂3xu(s,⋅)‖2L2(0,∞)ds≤C(T), |
which gives (3.20).
We prove (2.25). Thanks to the Hölder inequality,
(∂xu(t,0))2=−2∫∞0∂xu∂2xudx≤2∫∞0|∂xu||∂2xu|dx≤2‖∂xu(t,⋅)‖L2(0,∞)‖∂2xu(t,⋅)‖L2(0,∞). | (3.26) |
Therefore, by (3.17) and (3.20),
(∂xu(t,0))2≤C(T). | (3.27) |
By (3.16), (3.17) and (3.20), we obtain that
‖∂xu‖2L∞((0,T)×(0,∞))≤C(T), |
which gives (2.25).
Finally, (3.21) follows from (3.17), (3.20), (3.25) and an integration on (0,t), while arguing as in Lemma 2.5, we have (2.34).
Arguing as in Section 2, we have Theorem 1.1.
In this section, we prove Theorem 1.1 for (1.1)-(1.4)-(1.7).
Let us prove some a priori estimates on u.
We begin by proving the following lemma.
Lemma 4.1. Fix T>0. There exists a constant C(T)>0, such that
‖u(t,⋅)‖2L(0,∞)+β2eC0t∫t0e−C0s‖∂2xu(s,⋅)‖2L2(0,∞)ds+a2eC0t∫t0e−C0su4(s,0)ds≤C(T). | (4.1) |
for every 0≤t≤T. In particular, we have (2.14), (2.20), (2.21) and
∫t0u2(s,0)ds≤C(T), | (4.2) |
for every 0≤t≤T.
Proof. Let 0≤t≤T. Multiplying (1.1)-(1.4)-(1.7) by 2v, thanks to (1.1)-(1.4)-(1.7), an integration on (0,∞) give
ddt‖u(t,⋅)‖2L2(0,∞)=2∫∞0u∂tudx=−4κ∫∞0u2∂xudx+6a2∫∞0u3∂xudx−2ν∫∞0u∂2xudx−2β2∫∞0u∂4xudx=4κ3u3(t,0)−3a22u4(t,0)−2ν∫∞0u∂2xudx+2β2∫∞0∂xu∂3xudx=4κ3u3(t,0)−3a22u4(t,0)−2ν∫∞0u∂2xudx−2β2‖∂2xu(t,⋅)‖2L2(0,∞). |
Therefore, we have that
ddt‖u(t,⋅)‖2L2(0,∞)+2β2‖∂2xu(t,⋅)‖2L2(0,∞)+3a22u4(t,0)=4κ3u3(t,0)−2ν∫∞0u∂2xudx. | (4.3) |
Due to the Young inequality,
4|κ|3|u(t,0)|3=2(2|κu(t,0)|3√D3)(√D3u2)≤4κ29D3u2(t,0)+D3u4(t,0)=2(2κ29D3√D3)(√D3u2(t,0))+D3u4(t,0)≤4κ481D3+2D3u4(t,0),2|ν|∫∞0|u||∂2xu|dx≤2∫∞0|νuβ||β∂2xu|dx≤ν2β2‖u(t,⋅)‖2L2(0,∞)+β2‖∂2xu(t,⋅)‖2L2(0,∞), |
where D3 is a positive constant, which will be specified later. Consequently, by (4.3),
ddt‖u(t,⋅)‖2L2(0,∞)+β2‖∂2xu(t,⋅)‖2L2(0,∞)+(3a22−2D3)u4(t,0)≤4κ481D3+ν2β2‖u(t,⋅)‖2L2(0,∞). |
Choosing D3=a22, we have that
ddt‖u(t,⋅)‖2L2(0,∞)+2β2‖∂2xu(t,⋅)‖2L2(0,∞)+a2u4(t,0)≤C0+C0‖u(t,⋅)‖2L2(0,∞). |
By the Gronwall Lemma and (1.8), we get
‖u(t,⋅)‖2L2(0,∞)+2β2eC0t∫t0e−C0s‖∂2xu(s,⋅)‖2L2(0,∞)ds+a2eC0t∫t0e−C0su4(s,0)ds≤C0eC0t+C0eC0t∫t0e−C0sds≤C(T), |
which gives (4.1).
We prove (2.14). Thanks to (1.1)-(1.4)-(1.7), (4.1) and the Young inequality,
‖∂xu(t,⋅)‖2L2(0,∞)=∫∞0∂xu∂xudx=−∫∞0u∂2xudx≤∫∞0|u||∂2xu|dx≤12‖u(t,⋅)‖2L2(0,∞)+12‖∂2xu(t,⋅)‖2L2(0,∞)≤C(T)+12‖∂2xu(t,⋅)‖2L2(0,∞). |
(2.14) follows from (4.1) and an integration on (0,t).
We prove (2.20). By (3.16) and (1.1)-(1.4)-(1.7), we have that
‖∂xu(t,⋅)‖2L∞(0,∞)≤‖∂xu(t,⋅)‖2L2(0,∞)+‖∂2xu(t,⋅)‖2L2(0,∞) | (4.4) |
Therefore, an integration on (0,t), (2.14) and (4.1) gives (2.20).
Arguing as in Lemma 2.3, we have (2.21).
Finally, we prove (4.2). We begin by observing that, by the Young inequality,
u2(t,0)≤12+12u4(t,0). |
(4.2) follows from (4.1) and an integration on (0,t).
Lemma 4.2. Fix T>0. There exists a constant C(T)>0, such that (2.26) holds. In particular, we have that
‖∂xu(t,⋅)‖2L2(0,∞)+2β2∫t0‖∂3xu(s,⋅)‖2L2(0,∞)ds≤C(T), | (4.5) |
∫t0(∂2xu(s,0))2ds≤C(T), | (4.6) |
for every 0≤t≤T.
Proof. Let 0≤t≤T. Multiplying (1.1)-(1.4)-(1.7) by −2∂2xu, thanks to (1.1)-(1.4)-(1.7), an integration on (0,∞) gives
ddt‖∂xu(t,⋅)‖2L2(0,∞)=−2∫∞0∂2xu∂tudx=4κ∫∞0u∂xu∂4xudx−6a2∫∞0u2∂xu∂2xudx+2ν‖∂2xu(t,⋅)‖2L2(0,∞)+2β2∫∞0∂2xu∂4xudx=4κ∫∞0u∂xu∂4xudx−6a2∫∞0u2∂xu∂2xudx+2ν‖∂2xu(t,⋅)‖2L2(0,∞)−2β2‖∂3xu(t,⋅)‖2L2(0,∞). |
Therefore, we have that
ddt‖∂xu(t,⋅)‖2L2(0,∞)+2β2‖∂3xu(t,⋅)‖2L2(0,∞)=4κ∫∞0u∂xu∂2xudx−6a2∫∞0u2∂xu∂2xudx+2ν‖∂2xu(t,⋅)‖2L2(0,∞). | (4.7) |
Due to the Young inequality,
4|κ|∫∞0|u∂xu||∂4xu|dx≤2κ2‖u(t,⋅)∂xu(t,⋅)‖2L2(0,∞)+2‖∂2xu(t,⋅)‖2L2(0,∞),6a2∫∞0|u2∂xu||∂2xu|dx≤3a4∫∞0u4(∂xu)2dx+3‖∂2xu(t,⋅)‖2L2(0,∞)≤3a2‖u‖2L∞((0,T)×(0,∞))‖u(t,⋅)∂xu(t,⋅)‖2L2(0,∞)+3‖∂2xu(t,⋅)‖2L2(0,∞). |
It follows from (4.7) that
ddt‖∂xu(t,⋅)‖2L2(0,∞)+2β2‖∂3xu(t,⋅)‖2L2(0,∞)≤(2κ2+3a2‖u‖2L∞((0,T)×(0,∞)))‖u(t,⋅)∂xu(t,⋅)‖2L2(0,∞)+(2|ν|+4)‖∂2xu(t,⋅)‖2L2(0,∞). |
Integrating on (0,t), by (1.8), (2.21) and (4.1), we have that
‖∂xu(t,⋅)‖2L2(0,∞)+2β2∫t0‖∂3xu(s,⋅)‖2L2(0,∞)ds≤C0+(2κ2+3a2‖u‖2L∞((0,T)×(0,∞)))∫t0‖u(s,⋅)∂xu(s,⋅)‖2L2(0,∞)ds+(2|ν|+4)∫t0‖∂2xu(s,⋅)‖2L2(0,∞)ds≤C(T)(1+‖u‖2L∞((0,T)×(0,∞))). | (4.8) |
We prove (2.26). Thanks to (4.1), (4.8) and the Hölder inequality,
u2(t,x)=2∫x0u∂xudy−2∫∞0u∂xudx≤2∫∞0|u||∂xu|dx≤2‖u(t,⋅)‖L2(0,∞))‖∂xu(t,⋅)‖L2(0,∞)≤C(T)√(1+‖u‖2L∞((0,T)×(0,∞))). |
Therefore,
‖u‖4L∞((0,T)×(0,∞))−C(T)‖u‖2L∞((0,T)×(0,∞))−C(T)≤0, | (4.9) |
which gives (2.26).
(4.5) follows from (2.26) and (4.8).
Finally, we prove (4.6). Thanks to the Young inequality,
(∂2xu(t,0))2=−2∫∞0∂2xu∂3xudx≤2∫∞0|∂2xu||∂3xu|dx≤‖∂2xu(t,⋅)‖2L2(0,∞)+‖∂3xu(t,⋅)‖2L2(0,∞). |
(4.1), (4.5) and an integration on (0,t) give (4.6).
Lemma 4.3. Fix T>0. There exists a constant C(T)>0, such that
‖∂2xu(t,⋅)‖2L2(0,∞)+β2∫t0‖∂4xu(s,⋅)‖2L2(0,∞)ds≤C(T), | (4.10) |
for every 0≤t≤T. In particular, (2.25) holds. Moreover, we have (2.34).
Proof. Let 0≤t≤T. Since, by (1.1)-(1.4)-(1.7), ∂2txu(t,0)=0, multiplying (1.1)-(1.4)-(1.7) by 2∂4xu, thanks to (1.1)-(1.4)-(1.7), an integration on (0,t) gives
ddt‖∂2xu(t,⋅)‖2L2(0,∞)=2∫∞0∂4xu∂tudx=−4κ∫∞0u∂xu∂4xudx+6a2∫∞0u2∂xu∂4xudx−2ν∫∞0∂2xu∂4xudx−2β2‖∂4xu(t,⋅)‖2L2(0,∞)=−4κ∫∞0u∂xu∂4xudx+6a2∫∞0u2∂xu∂4xudx+2ν‖∂3xu(t,⋅)‖2L2(0,∞)−2β2‖∂4xu(t,⋅)‖2L2(0,∞). |
Therefore, we have
ddt‖∂2xu(t,⋅)‖2L2(0,∞)+2β2‖∂4xu(t,⋅)‖2L2(0,∞)=−4κ∫∞0u∂xu∂4xudx+6a2∫∞0u2∂xu∂4xudx+2ν‖∂3xu(t,⋅)‖2L2(0,∞). | (4.11) |
Due to (2.26), (4.5) and the Young inequality,
4|κ|∫∞0|u||∂xu||∂4xu|dx≤4|κ|‖u‖L∞((0,T)×(0,∞))∫∞0|∂xu||∂4xu|dx≤C(T)∫∞0|∂xu||∂4xu|dx=∫∞0|C(T)∂xuβ||β∂4xu|dx≤C(T)‖∂xu(t,⋅)‖2L2(0,∞)+β22‖∂4xu(t,⋅)‖2L2(0,∞)≤C(T)+β22‖∂4xu(t,⋅)‖2L2(0,∞),6a2∫∞0u2|∂xu||∂4xu|dx≤6a2‖u‖2L∞((0,T)×(0,∞))∫∞0|∂xu||∂4xu|dx≤C(T)∫∞0|∂xu||∂4xu|dx=∫∞0|C(T)∂xuβ||β∂4xu|dx≤C(T)‖∂xu(t,⋅)‖2L2(0,∞)+β22‖∂4xu(t,⋅)‖2L2(0,∞)≤C(T)+β22‖∂4xu(t,⋅)‖2L2(0,∞). |
It follows from (4.11) that
ddt‖∂2xu(t,⋅)‖2L2(0,∞)+β2‖∂4xu(t,⋅)‖2L2(0,∞)≤C(T)+2|ν|‖∂3xu(t,⋅)‖2L2(0,∞). |
Integrating on (0,t), by (1.8) and (4.5), we get
‖∂2xu(t,⋅)‖2L2(0,∞)+β2∫t0‖∂4xu(s,⋅)‖2L2(0,∞)ds≤C0C(T)t+2|ν|∫t0‖∂3xu(s,⋅)‖2L2(0,∞)ds≤C(T), |
which gives (4.10).
(2.25) follows from (4.4), (4.5) and (4.10).
Finally, we prove (2.34). Multiplying (1.1)-(1.4)-(1.7) by 2∂tu, an integration on (0,∞) give
2‖∂tu(t,⋅)‖2L2(0,∞)=−4κ∫∞0u∂xu∂tudx+6a2∫∞0u2∂xu∂tudx−2ν∫∞0∂2xu∂tudx−2β2∫∞0∂4xu∂tudx. | (4.12) |
Due to (2.26), (4.5), (4.10) and the Young inequality,
4|κ|∫∞0|u||∂xu||∂tu|dx≤4|κ|‖u‖L∞((0,T)×(0,∞))∫∞0|∂xu||∂tu|dx≤2C(T)∫∞0|∂xu||∂tu|dx=2∫∞0|C(T)∂xu√D4||√D4∂tu|dx≤C(T)D4‖∂xu(t,⋅)‖2L2(0,∞)+D4‖∂tu(t,⋅)‖2L2(0,∞)≤C(T)D4+D4‖∂tu(t,⋅)‖2L2(0,∞),6a2∫∞0u2|∂xu||∂tu|dx≤6a2‖u‖2L∞((0,T)×(0,∞))∫∞0|∂xu||∂tu|dx≤2C(T)∫∞0|∂xu||∂tu|dx=2∫∞0|C(T)∂xu√D4||√D4∂tu|dx≤C(T)D4‖∂xu(t,⋅)‖2L2(0,∞)+D4‖∂tu(t,⋅)‖2L2(0,∞)≤C(T)D4+D4‖∂tu(t,⋅)‖2L2(0,∞),2|ν|∫∞0|∂2xu||∂tu|dx=2∫∞0|ν∂2xu√D4||√D4∂tu|dx≤ν2D4‖∂2xu(t,⋅)‖2L2(0,∞)+D4‖∂tu(t,⋅)‖2L2(0,∞)≤C(T)D4+D4‖∂tu(t,⋅)‖2L2(0,∞),2β2∫∞0|∂4xu||∂tu|dx=2∫∞0|β2∂4xu√D2||√D4∂tu|dx≤β4D4‖∂4xu(t,⋅)‖2L2(0,∞)+D4‖∂tu(t,⋅)‖2L2(0,∞). |
Therefore, by (4.12),
2(1−2D4)‖∂tu(t,⋅)‖2L2(0,∞)≤C(T)D4+β4D4‖∂4xu(t,⋅)‖2L2(0,∞). |
Choosing D4=14, we have
‖∂tu(t,⋅)‖2L2(0,∞)≤C(T)+4β4‖∂4xu(t,⋅)‖2L2(0,∞). |
An integration on (0,t) and (4.10) give (2.34).
Arguing as in Section 2, we have Theorem 1.1.
In this section, we prove Theorem 1.1 for (1.1)-(1.5)-(1.7).
Let us prove some a priori estimates on u.
We begin by proving the following lemma.
Lemma 5.1. Fix T>0. There exists a constant C(T)>0, such that
‖u(t,⋅)‖2L(0,∞)+β2eC0t6∫t0e−C0s‖∂2xu(s,⋅)‖2L2(0,∞)ds+a2eC0t2∫t0e−C0su4(s,0)ds≤C(T). | (5.1) |
for every 0≤t≤T. In particular, we have (2.14), (2.20), (2.21) and (4.2).
Proof. Let 0≤t≤T. We begin by observing that, thanks to (1.1)-(1.5)-(1.7),
4κ∫∞0u2∂xudx=4κ3u3(t,0),−6a2∫∞0u2∂xudx=3a22u4(t,0),2δ∫x0u∂3xu=−2δ∫∞0∂xu∂2xudx,2β2∫∞0u∂4xudx=−2β2∫∞0u∂3xudx=2β2‖∂2xu(t,⋅)‖2L2(0,∞). | (5.2) |
Therefore, multiplying (1.1)-(1.5)-(1.7) by 2u, thanks to (5.2), an integration on (0,∞) gives
ddt‖u(t,⋅)‖2L2(0,∞)+2β2‖∂2xu(t,⋅)‖2L2(0,∞)+3a22u4(t,0)=4κ3u2(t,0)−2ν∫∞0u∂2xudx+2δ∫∞0∂xu∂2xudx. | (5.3) |
Due to the Young inequality,
4|κ|3u2(t,0)≤8κ29a2u2(t,0)+a22u4(t,0),2|ν|∫∞0|u||∂2xu|dx=∫∞0|2νuβ||β∂2xu|dx≤2ν2β2‖u(t,⋅)‖2L2(0,∞)+β22‖∂2xu(t,⋅)‖2L2(0,∞),2|δ|∫∞0|∂xu||∂2xu|dx=∫∞0|2δ∂xuβ||β∂4xu|dx≤2δ2β2‖∂xu(t,⋅)‖2L2(0,∞)+β22‖∂2xu(t,⋅)‖2L2(0,∞). |
It follows from (5.3) that
ddt‖u(t,⋅)‖2L2(0,∞)+β2‖∂2xu(t,⋅)‖2L2(0,∞)+a2u4(t,0)≤8κ29a2u2(t,0)+2ν2β2‖u(t,⋅)‖2L2(0,∞)+2δ2β2‖∂xu(t,⋅)‖2L2(0,∞)≤C0u2(t,0)+C0‖u(t,⋅)‖2L2(0,∞)+C0‖∂xu(t,⋅)‖2L2(0,∞). | (5.4) |
Observe that, by the Hölder inequality,
‖∂xu(t,⋅)‖2L2(0,∞)=∫∞0∂xu∂xudx=−u(t,0)∂xu(t,0)−∫∞0u∂2xudx≤|u(t,0)||∂xu(t,0)|+∫∞0|u||∂2xu|dx≤|u(t,0)||∂xu(t,0)|+‖u(t,⋅)‖L2(0,∞)‖∂2xu(t,⋅)‖L2(0,∞). | (5.5) |
Moreover, by the Young inequality,
|u(t,0)||∂xu(t,0)|≤12D5u2(t,0)+D52(∂xu(t,0))2, | (5.6) |
where D5 is a positive constant, which will be specified later. Observe that, by the Young inequality,
D52(∂xu(t,0))2=−D5∫∞0∂xu∂2xudx≤∫∞0|∂xu||D5∂2xu|dx≤12‖∂xu(t,⋅)‖2L2(0,∞)+D25‖∂2xu(t,⋅)‖2L2(0,∞). | (5.7) |
It follows from (5.5), (5.6) and (5.7) that
12‖∂xu(t,⋅)‖2L2(0,∞)≤12D5u2(t,0)+D25‖∂2xu(t,⋅)‖2L2(0,∞)+‖u(t,⋅)‖L2(0,∞)‖∂2xu(t,⋅)‖L2(0,∞), |
that is
‖∂xu(t,⋅)‖2L2(0,∞)≤1D5u2(t,0)+2D25‖∂2xu(t,⋅)‖2L2(0,∞)+2‖u(t,⋅)‖L2(0,∞)‖∂2xu(t,⋅)‖L2(0,∞). | (5.8) |
Due to the Young inequality,
2‖u(t,⋅)‖L2(0,∞)‖∂2xu(t,⋅)‖L2(0,∞)≤1D6‖u(t,⋅)‖2L2(0,∞)+D6‖∂2xu(t,⋅)‖2L2(0,∞), |
where D6 is a positive constant, which will be specified later. It follows from (5.8) that
‖∂xu(t,⋅)‖2L2(0,∞)≤1D5u2(t,0)+2D25‖∂2xu(t,⋅)‖2L2(0,∞)+1D6‖u(t,⋅)‖2L2(0,∞)+D6‖∂2xu(t,⋅)‖2L2(0,∞). | (5.9) |
Consequently, by (5.4) and (5.9),
ddt‖u(t,⋅)‖2L2(0,∞)+(β2−2D25−D6)‖∂2xu(t,⋅)‖2L2(0,∞)+a2u4(t,0)≤(C0+1D5)u2(t,0)+(C0+1D6)‖u(t,⋅)‖2L2(0,∞). |
Choosing
D5=|β|2,D6=13, | (5.10) |
we have that
ddt‖u(t,⋅)‖2L2(0,∞)+β26‖∂2xu(t,⋅)‖2L2(0,∞)+a2u4(t,0)≤C0u2(t,0)+C0‖u(t,⋅)‖2L2(0,∞). | (5.11) |
Due to the Young inequality,
C0u2(t,0)≤C0+a22u4(t,0). |
It follows from (5.11) that
ddt‖u(t,⋅)‖2L2(0,∞)+β26‖∂2xu(t,⋅)‖2L2(0,∞)+a22u4(t,0)≤C0+C0‖u(t,⋅)‖2L2(0,∞). |
By the Gronwall Lemma and (1.8), we get
‖u(t,⋅)‖2L2(0,∞)+β2eC0t6∫t0e−C0s‖∂2xu(s,⋅)‖2L2(0,∞)ds+a2eC0t2∫t0e−C0su4(s,0)ds≤C0+C0eC0t∫t0e−C0sds≤C(T), |
which gives (5.1).
Arguing as in Lemma 4.1, we have (4.2), while (4.2), (5.1), (5.9), (5.10) and an integration on (0,t) give (2.14).
We prove (2.20). By (3.16), (5.7) and (5.10), we have that
‖∂xu(t,⋅)‖2L∞(0,∞)≤C0‖∂xu(t,⋅)‖2L2(0,∞)+C0‖∂2xu(t,⋅)‖2L2(0,∞). |
(2.14), (5.1) and an integration on (0,t) give (2.20).
Arguing as in Lemma 2.3, we have (2.21), while (3.15) follows from (2.14), (5.1), (5.7), (5.10) and an integration on (0,t).
Lemma 5.2. Fix T>0. There exists a constant C(T)>0, such that (2.26) holds. In particular,
‖∂2xu(t,⋅)‖2L2(0,∞)+β22∫t0‖∂4xu(s,⋅)‖2L2(0,∞)ds≤C(T), | (5.12) |
for every 0≤t≤T. Moreover, we have (2.25),
‖∂xu(t,⋅)‖L2(0,∞)≤C(T), | (5.13) |
∫t0‖∂3xu(s,⋅)‖2L2(0,∞)ds≤C(T), | (5.14) |
for every 0≤t≤T.
Proof. Let 0≤t≤T. We begin by observing that, thanks (1.1)-(1.5)-(1.7),
2∫∞0∂4xu∂tudx=−2∫∞0∂3xu∂t∂xudx=ddt‖∂2xu(t,⋅)‖2L2(0,∞),2δ∫∞0∂3xu∂4xudx=0. | (5.15) |
Therefore, multiplying (1.1)-(1.5)-(1.7) by 2∂4xu, by (5.15) an integration on (0,∞ gives
ddt‖∂2xu(t,⋅)‖2L2(0,∞)+2β2‖∂4xu(t,⋅)‖2L2(0,∞)=−4κ∫∞0u∂xu∂4xudx+6a2∫∞0u2∂xu∂4xudx−2ν∫∞0∂2xu∂4xudx. | (5.16) |
Due to the Young inequality,
4|κ|∫∞0|u∂xu||∂4xu|dx=∫∞0|4κu∂xuβ||β∂4xu|dx≤8κ2β2‖u(t,⋅)∂xu(t,⋅)‖2L2(0,∞)+β22‖∂4xu(t,⋅)‖2L2(0,∞),6a2∫∞0|u2∂xu||∂4xu|dx=∫∞0|6a2u2∂xuβ||β∂4xu|dx≤18a4β2∫∞0u4(∂xu)2dx+β22‖∂4xu(t,⋅)‖2L2(0,∞)≤18a4β2‖u‖2L∞((0,∞)×(0infty))‖u(t,⋅)∂xu(t,⋅)‖2L2(0,∞)+β22‖∂4xu(t,⋅)‖2L2(0,∞),2|ν|∫∞0|∂2xu||∂4xu|dx=∫∞0|2ν∂2xuβ||β∂4xu|dx≤2ν2β2‖∂2xu(t,⋅)‖2L2(0,∞)+β22‖∂4xu(t,⋅)‖2L2(0,∞). |
It follows from (5.16) that
ddt‖∂2xu(t,⋅)‖2L2(0,∞)+β22‖∂4xu(t,⋅)‖2L2(0,∞)≤C0(1+‖u‖2L∞((0,∞)×(0,∞)))‖u(t,⋅)∂xu(t,⋅)‖2L2(0,∞)+C0‖∂2xu(t,⋅)‖2L2(0,∞). |
Integrating on (0,∞), by (1.8), (2.21) and (5.1), we have that
‖∂2xu(t,⋅)‖2L2(0,∞)+β22∫t0‖∂4xu(s,⋅)‖2L2(0,∞)ds≤C0+C0(1+‖u‖2L∞((0,∞)×(0,∞)))∫t0‖u(s,⋅)∂xu(s,⋅)‖2L2(0,∞)ds+C0∫t0‖∂2xu(s,⋅)‖2L2(0,∞)ds≤C(T)(1+‖u‖2L∞((0,∞)×(0,∞))). | (5.17) |
We prove that
‖∂xu(t,⋅)‖2L2(0,∞)≤C(T)(1+‖u‖2L∞((0,∞)×(0,∞))). | (5.18) |
We begin by observing that, by (5.1), (5.9) and (5.10),
‖∂xu(t,⋅)‖2L2(0,∞)≤C0u2(t,0)+C(T)(1+‖∂2xu(t,⋅)‖2L2(0,∞)). | (5.19) |
Due to (5.1) and the Young inequality,
C0u2(t,0)=−2C0∫∞0u∂xudx≤2C0∫∞0|u||∂xu|dx≤C0‖u(t,⋅)‖L2(0,∞)+12‖∂xu(t,⋅)‖2L2(0,∞)≤C(T)+12‖∂xu(t,⋅)‖2L2(0,∞). |
Therefore, by (5.18),
‖∂xu(t,⋅)‖2L2(0,∞)≤C(T)(1+‖∂2xu(t,⋅)‖2L2(0,∞)). | (5.20) |
(5.17) and (5.19) give (5.20).
We prove (2.26). Thanks to (5.1), (5.19) and the Hölder inequality, we have (4.9), which gives (2.26).
(5.12) and (5.13) follows from (2.26), (5.17) and (5.18), respectively, while (3.16), (5.7), (5.10) (5.12) and (5.13) gives (2.25).
Finally, we prove (5.14). We begin by observing that, thanks to (1.1)-(1.5)-(1.7),
‖∂3xu(t,⋅)‖2L2(0,∞)=∫∞0∂3xu∂3xudx=−∫∞0∂2xu∂3xudx. |
Therefore, by (5.12) and the Young inequality,
‖∂3xu(t,⋅)‖2L2(0,∞)≤∫∞0|∂2xu||∂3xu|dx≤12‖∂2xu(t,⋅)‖2L2(0,∞)+12‖∂4xu(t,⋅)‖2L2(0,∞)≤C(T)++12‖∂4xu(t,⋅)‖2L2(0,∞). |
(5.14) follows from (5.12) and an integration on (0,t).
Lemma 5.3. Fix T>0. There exists a constant C(T)>0, such that (2.34) holds.
Proof. Let 0≤t≤T. Arguing as in Lemma 4.3, we have that
‖∂tu(t,⋅)‖2L2(0,∞)≤C(T)+4β4‖∂4xu(t,⋅)‖2L2(0,∞)+2|δ|∫∞0|∂3xu|∂tu|dx. | (5.21) |
Due to the Young inequality,
2|δ|∫∞0|∂3xu|∂tu|dx≤2δ2‖∂3xu(t,⋅)‖2L2(0,∞)+12‖∂tu(t,⋅)‖2L2(0,∞). |
Therefore, by (5.21), we have
12‖∂tu(t,⋅)‖2L2(0,∞)≤C(T)+4β4‖∂4xu(t,⋅)‖2L2(0,∞)+2δ2‖∂3xu(t,⋅)‖2L2(0,∞). |
(2.34) follows from (5.12), (5.14) and an integration on (0,t).
Arguing as in Section 2, we have Theorem 1.1.
In this section, we prove Theorem 1.1 for (1.1)-(1.6)-(1.7).
Let us prove some a priori estimates on u.
We begin by proving the following lemma.
Lemma 6.1. Fix T>0. There exists a constant C(T)>0, such that
‖u(t,⋅)‖2L2(0,∞)+β2eC0t2∫t0e−C0s‖∂2xu(s,⋅)‖2L2(0,∞)ds≤C(T), | (6.1) |
for every 0≤t≤T. In particular, we have (2.14), (2.20) and (2.21).
Proof. Let 0≤t≤T. We begin by observing that, thanks to (1.1)-(1.6)-(1.7),
4κ∫∞0u2∂xudx=0,6q∫∞0u3∂xudx=0,2ν∫∞0u∂2xudx=−2ν‖∂xu(t,⋅)‖2L2(0,∞),2δ∫∞0u∂3xudx=−2δ∫∞0∂xu∂2xudx,2β2∫∞0u∂4xudx=−2β2∫∞0∂xu∂3xudx=2β2‖∂2xu(t,⋅)‖2L2(0,∞). | (6.2) |
Therefore, multiplying (1.1)-(1.6)-(1.7) by 2v, thanks to (6.2), an integration on 0,∞) gives
ddt‖u(t,⋅)‖2L2(0,∞)+2β2‖∂2xu(t,⋅)‖2L2(0,∞)=2ν‖∂xu(t,⋅)‖2L2(0,∞)−2δ∫∞0∂xu∂2xudx. | (6.3) |
Due to the Young inequality,
2|δ|∫∞0|∂xu||∂2xu|dx=2∫∞0|δ∂xuβ||β∂2xu|dx≤δ2β2‖∂xu(t,⋅)‖2L2(0,∞)+β2‖∂2xu(t,⋅)‖2L2(0,∞). |
Consequently, by (6.3),
ddt‖u(t,⋅)‖2L2(0,∞)+β2‖∂2xu(t,⋅)‖2L2(0,∞)≤C0‖∂xu(t,⋅)‖2L2(0,∞). | (6.4) |
Observe that, by (5.5) and (1.1)-(1.6)-(1.7),
C0‖∂xu(t,⋅)‖2L2(0,∞)≤C0‖u(t,⋅)‖L2(0,∞)‖∂2xu(t,⋅)‖2L2(0,∞) |
Therefore, by the Young inequality,
C0‖∂xu(t,⋅)‖2L2(0,∞)≤C0‖u(t,⋅)‖2L2(0,∞)+β22‖∂2xu(t,⋅)‖L2(0,∞). | (6.5) |
It follows from (6.4) that
ddt‖u(t,⋅)‖2L2(0,∞)+β22‖∂2xu(t,⋅)‖2L2(0,∞)≤C0‖u(t,⋅)‖2L2(0,∞). |
By the Gronwall Lemma and (1.8).
‖u(t,⋅)‖2L2(0,∞)+β2eC0t2∫t0e−C0s‖∂2xu(s,⋅)‖2L2(0,∞)ds≤C0eC0t≤C(T), |
which gives (6.1).
(2.14) follows from (6.1) and an integration on (0,t), while, arguing as in Lemma 2.3, we have (2.20) and (2.21).
Lemma 6.2. Fix T>0. There exists a constant C(T)>0, such that (2.26) holds. In particular,
‖∂2xu(t,⋅)‖2L2(0,∞)+4(4δ2+4)β4‖∂xu(t,⋅)‖2L2(0,∞)+β22∫t0‖∂4xu(s,⋅)‖2L2(0,∞)ds+4δ2+2β2∫t0‖∂3xu(s,⋅)‖2L2(0,∞)ds≤C(T), | (6.6) |
for every 0≤t≤T. Moreover, we have (2.24), (2.25) and (2.34).
Proof. Let 0≤t≤T. Consider an positive constant B, which will be specified later. Observing that, since, thanks to (1.1)-(1.6)-(1.7), ∂tu(t,0=∂t∂xu(t,0), we have that
2∫∞0∂4xu∂tudx=−2∫∞0∂3xu∂t∂xudx=ddt‖∂2xu(t,⋅)‖2L2(0,∞),−2B∫∞0∂2xu∂tudx=Bddt‖∂xu(t,⋅)‖2L2(0,∞),−2Bβ2∫0∂2xu∂4xudx=2Bβ2∂2xu(t,0)∂3xu(t,0)+2Bβ2‖∂3xu(t,⋅)‖2L2(0,∞). | (6.7) |
Therefore, multiplying (1.1)-(1.6)-(1.7) by 2∂4xu, thanks to (6.7), an integration on (0,∞) gives
ddt(‖∂2xu(t,⋅)‖2L2(0,∞)+B‖∂xu(t,⋅)‖2L2(0,∞))+2β2‖∂4xu(t,⋅)‖2L2(0,∞)+2Bβ2‖∂3xu(t,⋅)‖2L2(0,∞)=2Bβ2∂2xu(t,0)∂3xu(t,0)−4κ∫∞0u∂xu∂4xudx+4Bκ∫∞0u∂xu∂2xudx−6q∫∞0u2∂xu∂4xudx+6Bq∫∞0u2∂xu∂2xudx−2ν∫∞0∂2xu∂4xudx+2Bν‖∂2xu(t,⋅)‖2L2(0,∞)−2δ∫∞0∂3xu∂4xudx+2Bδ∫∞0∂3xu∂2xudx. | (6.8) |
Due to the Young inequality,
2Bβ2∂2xu(t,0)∂3xu(t,0)≤β4B2(∂2xu(t,0)2+(∂3xu(t,0))24|κ|∫∞0|u∂xu||∂4xu|dx=2∫∞0|2κu∂xuβ√D7||β√D7∂4xu|dx≤4κ2β2D7‖u(t,⋅)∂xu(t,⋅)‖2L2(0,∞)+D7β2‖∂4xu(t,⋅)‖2L2(0,∞),4B|κ|∫∞0|u∂xu||∂2xu|dx≤‖u(t,⋅)∂xu(t,⋅)‖2L2(0,∞)+4B2κ2‖∂2xu(t,⋅)‖2L2(0,∞),6|q|∫∞0|u2∂xu||∂4xu|dx=2∫∞0|3qu2∂xuβ√D7||β√D7∂4xu|dx≤9q2β2D7∫∞0u4(∂xu)2dx+D7β2‖∂4xu(t,⋅)‖2L2(0,∞)≤9q2β2D7‖u‖2L∞((0,T)×(0,∞))‖u(t,⋅)∂xu(t,⋅)‖2L2(0,∞)+D7β2‖∂4xu(t,⋅)‖2L2(0,∞)6B|q|∫∞0|u2∂xu||∂2xu|dx≤∫∞0u4(∂xu)2dx+9q2B2‖∂2xu(t,⋅)‖2L2(0,∞)≤‖u‖2L∞((0,T)×(0,∞))‖u(t,⋅)∂xu(t,⋅)‖2L2(0,∞)+9q2B2‖∂2xu(t,⋅)‖2L2(0,∞)2|ν|∫∞0|∂2xu||∂4xu|dx=2∫∞0|ν∂2xuβ√D7||β√D7∂4xu|dx≤ν2β2‖∂2xu(t,⋅)‖2L2(0,∞)+β2D7‖∂4xu(t,⋅)‖2L2(0,∞),2|δ|∫∞0|∂3xu||∂4xu|dx=2∫∞0|δ∂3xuβ√D7||β√D7∂4xu|dx≤δ2β2D7‖∂3xu(t,⋅)‖2L2(0,∞)+β2D7‖∂4xu(t,⋅)‖2L2(0,∞),2B|δ|∫∞0|∂3xu||∂2xu|dx=2B∫∞0|β∂3xu||δ∂2xuβ|dx≤Bβ2‖∂3xu(t,⋅)‖2L2(0,∞)+Bδ2β2‖∂2xu(t,⋅)‖2L2(0,∞), |
where D7 is a positive constant, which will be specified later. It follows from (6.8) that
ddt(‖∂2xu(t,⋅)‖2L2(0,∞)+B‖∂xu(t,⋅)‖2L2(0,∞))+(2−4D7)β2‖∂4xu(t,⋅)‖2L2(0,∞)+(Bβ2−δ2β2D7)‖∂3xu(t,⋅)‖2L2(0,∞)≤C0B2(∂2xu(t,0)2+(∂3xu(t,0))2+C0(1+14+B2+B)‖∂2xu(t,⋅)‖2L2(0,∞)+C0(1+1D7+(1+1D7)‖u‖2L∞((0,T)×(0,∞)))‖u(t,⋅)∂xu(t,⋅)‖2L2(0,∞). |
Choosing D7=14, we have that
ddt(‖∂2xu(t,⋅)‖2L2(0,∞)+B‖∂xu(t,⋅)‖2L2(0,∞))+β2‖∂4xu(t,⋅)‖2L2(0,∞)+(Bβ2−4δ2β2)‖∂3xu(t,⋅)‖2L2(0,∞)≤B2(∂2xu(t,0)2+(∂3xu(t,0))2+C0(1+B2+B)‖∂2xu(t,⋅)‖2L2(0,∞)+C0(1+‖u‖2L∞((0,T)×(0,∞)))‖u(t,⋅)∂xu(t,⋅)‖2L2(0,∞). | (6.9) |
Observe that by the Young inequality,
B2(∂2xu(t,0)2=−B2∫∞0∂2xu∂3xudx≤∫∞0|2B√B∂2xuβ||√Bβ∂3xu|dx≤B3β2‖∂2xu(t,⋅)‖2L2(0,∞)+Bβ22‖∂3xu(t,⋅)‖2L2(0,∞),(∂3xu(t,0))2=−2∫∞0∂3xu∂4xudx≤=∫∞0|2∂3xuβ||β∂4xu|dx≤2β2‖∂3xu(t,⋅)‖2L2(0,∞)+β22‖∂4xu(t,⋅)‖2L2(0,∞). |
Consequently, by (6.9),
ddt(‖∂2xu(t,⋅)‖2L2(0,∞)+B‖∂xu(t,⋅)‖2L2(0,∞))+β22‖∂4xu(t,⋅)‖2L2(0,∞)+(Bβ22−4δ2+2β2)‖∂3xu(t,⋅)‖2L2(0,∞)≤C0(1+‖u‖2L∞((0,T)×(0,∞)))‖u(t,⋅)∂xu(t,⋅)‖2L2(0,∞)+C0(1+B3+B2+B)‖∂2xu(t,⋅)‖2L2(0,∞). |
Choosing
B=4(4δ2+4)β4, |
we have that
ddt(‖∂2xu(t,⋅)‖2L2(0,∞)+4(4δ2+4)β4‖∂xu(t,⋅)‖2L2(0,∞))+β22‖∂4xu(t,⋅)‖2L2(0,∞)+4δ2+2β2‖∂3xu(t,⋅)‖2L2(0,∞)≤C0(1+‖u‖2L∞((0,T)×(0,∞)))‖u(t,⋅)∂xu(t,⋅)‖2L2(0,∞)+C0‖∂2xu(t,⋅)‖2L2(0,∞). |
Integrating on (0,t), by (1.8), (2.21) and (6.1), we get
‖∂2xu(t,⋅)‖2L2(0,∞)+4(4δ2+4)β4‖∂xu(t,⋅)‖2L2(0,∞)+β22∫t0‖∂4xu(s,⋅)‖2L2(0,∞)ds+4δ2+2β2∫t0‖∂3xu(s,⋅)‖2L2(0,∞)ds≤C0+C0(1+‖u‖2L∞((0,T)×(0,∞)))∫t0‖u(s,⋅)∂xu(s,⋅)‖2L2(0,∞)ds+C0∫t0‖∂2xu(s,⋅)‖2L2(0,∞)ds≤C(T)(1+‖u‖2L∞((0,T)×(0,∞))). | (6.10) |
We prove (2.26). Thanks to (6.1), (6.10) and the Hölder inequality, we have (4.9), which gives (2.26).
(6.6) follows from (2.26) and (6.10).
Finally, arguing as in Lemma 2.4, we have (2.24) and (2.25), while arguing as in Lemma 5.3, (2.34) holds.
Arguing as in Section 2, we have Theorem 1.1.
The authors are members of the Gruppo Nazionale per l'Analisi Matematica, la Probabilità e le loro Applicazioni (GNAMPA) of the Istituto Nazionale di Alta Matematica (INdAM).
The authors declare that there are no conflict of interest.
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