Citation: Sizhao Li, Xinyu Han, Dapeng Lang, Songsong Dai. On the stability of two functional equations for (S,N)-implications[J]. AIMS Mathematics, 2021, 6(2): 1822-1832. doi: 10.3934/math.2021110
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In this paper, we are concerned with the Cauchy problem for the generalized Camassa-Holm equation
(1−∂2x)ut+(k+1)(k+2)2ukux−k(k−1)2uk−2u3x−2kuk−1uxuxx−ukuxxx=0,k∈N+, | (1.1) |
u(0,x)=u0(x), | (1.2) |
which is introduced by S. Hakkaev and K. Kirchev [14,15]. In Eq (1.1), u=u(t,x) stands for the fluid velocity at time t>0 in the spatial direction. Equation (1.1) admits following conservative laws
E=∫R12(uk+2+uku2x)dx, | (1.3) |
F=∫R12(u2+u2x)dx, | (1.4) |
and
H=∫Rudx. | (1.5) |
An important feature of Eq (1.1) is the existence of traveling solitary waves, interacting like solitons, also called "peakons"
u(t,x)=c1ke−∣x−ct∣. | (1.6) |
Since then, Eq (1.1) attracted lots of the attentions in the last few years. The well-posedness of the solutions for Eq (1.1) is obtained by parabolic regularization method [14], Kato's semigroup approach [20] and by classical Friedrichs's regularization method [21], respectively. In [17], Lai and Wu obtained a sufficient condition for the existence of weak solutions of Eq (1.1) in lower order Sobolev space Hs(R) with 1<s≤3/2. Yan [23] proved that Eq (1.1) does not depend uniformly continuously on the initial data in Hs(R) with s<3/2 and that the Cauchy problem for the generalized Camassa-Holm equation is locally well-posedness in B3/22,1. In [24], Zhou focused on the persistence property in weighted Lp spaces.
When k=1, Eq (1.1) reduced to the well-known Camassa-Holm equation
(1−∂2x)ut=−3uux+2uxuxx+uuxxx, | (1.7) |
which was derived by Camassa and Holm [1] and by Fokas and Fuchssteiner [11]. It describes the motion of shallow water waves and possesses soliton solutions, a Lax pair, a bi-Hamiltonian structures and infinitely many conserved integrals [1,32], and it can be solved by the Inverse Scattering Method. The dynamic properties related the equation can be found in [3,4,5,10,12,13,14,15,16,18,19,20,22,25] and the references therein. It is well-known that a major interest in water waves is the existence of breaking waves (solutions that remain bounded but whose slope becomes unbounded in finite time[7]). Comparing with KdV equation, another important feature of Camassa-Holm equation is that it possesses breaking wave[6,7,8,9].
To our best knowledge, blow-up, analyticity and analytical solutions have not been investigated yet for the problems (1.1) and (1.2). Inspired by the ideas from [7], the objective of this paper is to investigate the blow-up phenomenon, analyticity and analytical solutions for the problems (1.1) and (1.2). In our blow-up phenomenon analysis, the quantity ∫R((uk)x)3dx plays a key role. Taking advantage of complicated calculation, we obtain the Riccati inequality of quantity ∫R((uk)x)3dx to arrive at a new blow-up result. In addition, we present some analytical solutions for the problems (1.1) and (1.2). Finally, we prove the analyticity. The results we obtained complements earlier results in this direction.
Notations. The space of all infinitely differentiable functions ϕ(t,x) with compact support in [0,+∞)×R is denoted by C∞0. Let Lp=Lp(R)(1≤p<+∞) be the space of all measurable functions h such that ∥h∥PLP=∫R|h(t,x)|pdx<∞. We define L∞=L∞(R) with the standard norm ∥h∥L∞=infm(e)=0supx∈R∖e|h(t,x)|. For any real number s, Hs=Hs(R) denotes the Sobolev space with the norm defined by
∥h∥Hs=(∫R(1+|ξ|2)s|ˆh(t,ξ)|2dξ)12<∞, |
where ˆh(t,ξ)=∫Re−ixξh(t,x)dx.
We denote by ∗ the convolution, using the green function G(x)=12e−∣x∣, we have (1−∂2x)−1f=G(x)∗f for all f∈L2, and p∗(u−uxx)=u. For T>0 and nonnegative number s, C([0,T);Hs(R)) denotes the Frechet space of all continuous Hs-valued functions on [0,T). For simplicity, throughout this article, we let c denote any positive constant
The Cauchy problems (1.1) and (1.2) is equivalent to
ut+ukux=−∂x(1−∂2x)−1(k(k+3)2(k+1)uk+1+k2uk−1u2x), | (1.8) |
u(0,x)=u0(x), | (1.9) |
which is also equivalent to
yt+ukyx+2kuk−1uxy+k(k−1)2(ukux−uk−2u3x)=0, | (1.10) |
y=u−uxx,u(0,x)=u0(x). | (1.11) |
The rest sections are organized as follows. In the second section, we give a blow up criterion and a new blow up phenomenon. Existence of weak solution (CH-type peakon) and analytical solutions are studied in third section. In the forth section, we proved analyticity of strong solutions..
We firstly give some useful Lemmas.
Lemma 2.1. Given u(x,0)=u0∈Hs(R),s>3/2, then there exist a maximal T=T(u0) and a unique solution u to the problems (1.1) and (1.2) such that
u=u(⋅,u0)∈C([0,T);Hs(R))⋂C1([0,T);Hs−1(R)). |
Moreover, the solution depends continuously on the initial data, i.e., the mapping u0→u(⋅,u0):Hs→C([0,T);Hs(R))⋂C1([0,T);Hs−1(R)) is continuous.
Proof. Using the Kato's theorem[27], we can prove the above theorem. Because there exist some similarities, here we omit the proof of Lemma 2.1, a detailed proof can be found in [26].
Lemma 2.2. ([2]) Let f∈C1(R), a>0, b>0 and f(0)>√ba. If f′(t)≥af2(t)−b, then
f(t)→+∞ast→T=12√ablog(f(0)+√baf(0)−√ba). | (2.1) |
Lemma 2.3. (see [23]) Let u0∈B3/22,1 and u be the corresponding solution to (1.1). Assume that T is the maximal time of existence of the solution to the problems (1.1) and (1.2). If T<∞, then
∫T0∥ux∥L∞dτ=+∞. | (2.2) |
Remark 1. For s>32, it is well known that Hs↪B3/22,1, so we have the following result:
Let u0∈Hs with s>32 and u be the corresponding solution to (1.1). Assume that T is the maximal time of existence of the solution to the problems (1.1) and (1.2). If T<∞, then
∫T0∥ux∥L∞dτ=+∞. | (2.3) |
Lemma 2.4. Let u0∈Hs(R) with s>32. Let T>0 be the maximum existence time of the solution u to the problems (1.1) and (1.2) with the initial data u0. Then the corresponding solution u blows up in finite time if and only if
limt→T−∥uk−1ux∥L∞=+∞. |
Proof. Applying Lemma 2.1 and a simple density argument, it suffices to consider the case s=3. Let T>0 be the maximal time of existence of solution u to the problems (1.1) and (1.2) with initial data u0∈H3(R). From Lemma 2.1 we know that u∈C([0,T);H3(R))⋂C1([0,T);H2(R)). Due to y=u−uxx, by direct computation, one has
∥y∥2L2=∫R(u−uxx)2dx=∫R(u2+2u2x+u2xx)dx. | (2.4) |
So,
∥u∥2H2≤∥y∥2L2≤2∥u∥2H2. | (2.5) |
Multiplying equation (1.10) by 2y and integrating by parts and using the interpolation ∥ux∥2L∞≤C∥u∥H1∥y∥L2, we obtain
ddt∥y∥2L2=2∫Ryytdx=−3k∫Ruk−1uxy2dx−k(k−1)∫Rukuxydx+k(k−1)∫Ruk−2u3xydx≤c(∥uk−1ux∥L∞∥y∥2L2+∥uk∥L∞∥y∥L2∥ux∥L2+∥uk−2∥L∞∥u2x∥L∞∥y∥L2∥ux∥L2)≤c(∥uk−1ux∥L∞+C1)∥y∥2L2. | (2.6) |
where C1=C1(∥u0∥H1).
If there exists a constant M>0 such that ∥uk−1ux∥L∞<M, from (2.6) we deduce that
ddt∥y∥2L2≤c(M+C1)∥y∥2L2. | (2.7) |
By virtue of Gronwall's inequality, one has
∥y∥2L2≤∥y0∥2L2ec(M+C1)t. | (2.8) |
On the other hand, due to u=p∗y and ux=px∗y, then
∥uk−1ux∥L∞≤∥u∥k−1L∞∥ux∥L∞≤∥p∥k−1L2∥px∥L2∥y∥kL2 |
This completes the proof of Lemma 2.4.
Now, we present the blow-up phenomenon.
Theorem 2.5. Let u0∈Hs(R) for s>32. Suppose that u(t,x) be corresponding solution of problems (1.1) and (1.2) with the initial datum u0. If the slope of uk0 satisfies
∫R(uk0)3x<−KK1, | (2.9) |
where K=3k(6k2+7k−2)4(k+1)∥u0∥2kH1 and K1=12ck∥u0∥kH1. Then there exists the lifespan T<∞ such that the corresponding solution u(t,x) blows up in finite time T with
T=12KK1log(K1h(0)−KK1h(0)+K). | (2.10) |
Proof. Defining g(t)=uk(t,x), h(t)=∫Rg3xdx, it follows that
gt+ggx=kuk−1Q, | (2.11) |
where Q=−∂x(1−∂2x)−1(k(k+2)2(k+1)uk+1+k2uk−1u2x).
Differentiating the above Eq (2.11) with respect x yields
gtx+ggxx=−12g2x+k(k−1)uk−2uxQ+k2(k+2)2(k+1)g2−k2(k+2)2(k+1)uk−1(1−∂2x)−1(uk+1)−k22uk−1(1−∂2x)−1(uk−1u2x). | (2.12) |
Multiplying 3g2x both sides of (2.12) and integrating with respect x over R, one has
ddt∫Rg3xdx=−12∫Rg4xdx+3k(k−1)∫Ruk−2uxg2xQdx+3k2(k+2)2(k+1)∫Rg2g2xdx−3k2(k+2)2(k+1)∫Rg2xuk−1(1−∂2x)−1(uk+1)dx−3k22∫Rg2xuk−1(1−∂2x)−1(uk−1u2x)dx=Γ1+Γ2+Γ3+Γ4+Γ5. | (2.13) |
Using the Hölder's inequality, Yong's inequality and (1.4), from (2.13) we get
Γ2=3k(k−1)∫Ruk−2uxg2xQdx≤3k(k−1)∥Q∥L∞(∫R(uk−2ux)2dx)12(∫R(gx)2dx)12≤3k(k−1)2∥u0∥2kH1(∫R(gx)4dx)12≤3k(k−1)2(∥u0∥4kH12ϵ+ϵ∫R(gx)4dx2), | (2.14) |
Γ3=3k2(k+2)2(k+1)∫Rg2g2xdx≤3k2(k+2)2(k+1)(∫Rg4dx)12(∫R(gx)4dx)12≤3k2(k+2)2(k+1)∥u0∥2kH1(∫R(gx)4dx)12≤3k2(k+2)2(k+1)(∥u0∥4kH12ϵ+ϵ∫R(gx)4dx2), | (2.15) |
Γ4=3k2(k+2)2(k+1)∫Rg2xuk−1(1−∂2x)−1(uk+1)dx≤3k2(k+2)2(k+1)∥(1−∂2x)−1(uk+1)∥L∞(∫Ru2k−2dx)12(∫R(gx)4dx)12≤3k2(k+2)2(k+1)∥u0∥2kH1(∫R(gx)4dx)12≤3k2(k+2)2(k+1)(∥u0∥4kH12ϵ+ϵ∫R(gx)4dx2), | (2.16) |
and
Γ5=3k22∫Rg2xuk−1(1−∂2x)−1(uk−1u2x)dx≤3k22∥(1−∂2x)−1(uk−1u2x)∥L∞(∫Ru2k−2dx)12(∫R(gx)4dx)12≤3k22∥u0∥2kH1(∫R(gx)4dx)12≤3k22(∥u0∥4kH12ϵ+ϵ∫R(gx)4dx2). | (2.17) |
Combining the above inequalities (2.14)–(2.17), we obtain
∣Γ2∣+∣Γ3∣+∣Γ4∣+∣Γ5∣≤3k(6k2+7k−2)8ϵ(k+1)∥u0∥4kH1+ϵ23k(6k2+7k−2)4(k+1)∫R(gx)4dx. | (2.18) |
Choosing ϵ=2(k+1)3k(6k2+7k−2), which results in
ddt∫Rg3xdx≤−14∫Rg4xdx+K2. | (2.19) |
in which K2=9k2(6k2+7k−2)216(k+1)2∥u0∥4kH1. Using the Hölder's inequality, we get
(∫Rg3xdx)2≤c∫Rg2xdx∫Rg4xdx≤ck2∥u0∥2kH1∫Rg4xdx. | (2.20) |
Combining (2.19) and (2.20), we have
ddth(t)≤−K21h2(t)+K2, | (2.21) |
where K21=14ck2∥u0∥2kH1.
It is observed from assumption of Theorem that h(0)<−KK1, the continuity argument ensures that h(t)<h(0). Lemma 2.1 (a=K21 and b=K2) implies that h(t)→−∞ as t→T=12K1KlogK1h(0)−KK1h(0)+K.
On the other hand, by using the fact that
∣∫Rg3xdx∣≤∫R∣g3x∣dx≤k3∥uk−1ux(t,x)∥L∞∫Rg2xdx=k3∥uk−1ux(t,x)∥L∞∥u0∥2kH1. | (2.22) |
Lemma 2.4 implies that the Theorem 2.5 is true. This completes the proof of Theorem 2.5.
The solitons do not belong to the spaces Hs(R) with s>32 [28,29], so it motivates us to carry out the study of analytical solutions to problems (1.1) and (1.2).
Definition 3.1. Given initial data u0∈Hs, s>32, the function u is said to be a weak solution to the initial-value problems (1.8) and (1.9) if it satisfies the following identity
∫T0∫Ruφt+1k+1uk+1φx+G∗(k(k+3)2(k+1)uk+1+k2uk−1u2x)φxdxdt+∫Ru0(x)φ(0,x)dx=0 | (3.1) |
for any smooth test function φ(t,x)∈C∞c([0,T)×R). If u is a weak solution on [0,T) for every T>0, then it is called a global weak solution.
Theorem 3.2 The peakon function of the form
u(t,x)=p(t)e−|x−q(t)| | (3.2) |
is a global weak solution to problems (1.1) and (1.2) in the sense of Definition 3.1. Assumed that the functions p(t) and q(t) satisfy
pk+1(t)−p(t)q′(t)−p′(t)=0, |
and
pk+1(t)−p(t)q′(t)+p′(t)=0. |
where ′ denotes differentiation.
Proof. We firstly claim that
u=p(t)e−|x−q(t)| | (3.3) |
is a peakon solution of (1.1) and
ut=p′(t)e−|x−q(t)|+p(t)sign(x−q(t))q′(t)e−|x−q(t)|,ux=−p(t)sign(x−q(t))e−|x−q(t)|. | (3.4) |
Hence, using (3.1), (3.4) and integration by parts, we derive that
∫T0∫Ruφt+1k+1uk+1φxdxdt+∫Ru0(x)φ(0,x)dx=−∫T0∫Rφ(ut+ukux)dxdt=−∫T0∫Rφ[p′(t)e−|x−q(t)|+sign(x−q(t))(p(t)q′(t)e−|x−q(t)|−pk+1e−(k+1)|x−q(t)|)]dxdt. | (3.5) |
On the other hand, using (3.4), we obtain
∫T0∫RG∗(k(k+3)2(k+1)uk+1+k2uk−1u2x)φxdxdt=∫T0∫R−φGx∗[k2uk−1u2x+k(k+3)2(k+1)uk+1]dxdt=∫T0∫R−φGx∗[k(k+2)k+1uk+1]dxdt. | (3.6) |
Note that Gx=−12sign(x)e−|x|. For x>q(t), directly calculate
Gx∗[k(k+2)k+1uk+1]=−12∫Rsign(x−y)e−|x−y|k(k+2)k+1pk+1(t)e−(k+1)|y−q(t)|dy=−12(∫q(t)−∞+∫xq(t)+∫∞x)sign(x−y)e−|x−y|k(k+2)k+1pk+1(t)e−(k+1)|y−q(t)|dy=I1+I2+I3. | (3.7) |
We directly compute I1 as follows
I1=−12∫q(t)−∞sign(x−y)e−|x−y|k(k+2)k+1pk+1(t)e−(k+1)|y−q(t)|dy=−12k(k+2)k+1pk+1(t)∫q(t)−∞e−x−(k+1)q(t)e(k+2)ydy=−12k(k+2)k+1pk+1(t)e−x−(k+1)q(t)∫q(t)−∞e(k+2)ydy=−12(k+2)k(k+2)k+1pk+1(t)e−x+q(t). | (3.8) |
In a similar procedure,
I2=−12∫xq(t)sign(x−y)e−|x−y|k(k+2)k+1ak+1e−(k+1)|y−q(t)|dy=−12k(k+2)k+1pk+1∫xq(t)e−x+(k+1)q(t)e−kydy=−12k(k+2)k+1pk+1e−x+(k+1)q(t)∫xq(t)e−kydy=12kk(k+2)k+1pk+1(e−(k+1)(x−q(t))−e−x+q(t)), | (3.9) |
and
I3=−12∫∞xsign(x−y)e−|x−y|k(k+2)k+1pk+1e−(k+1)|y−q(t)|dy=12k(k+2)k+1pk+1∫∞xex+(k+1)q(t)e−(k+2)ydy=12k(k+2)k+1pk+1ex+(k+1)q(t)∫∞xe−(k+2)ydy=12(k+2)k(k+2)k+1pk+1e−(k+1)(x−q(t)). | (3.10) |
Substituting (3.8)–(3.10) into (3.7), we deduce that for x>q(t)
Gx∗[k(k+2)k+1uk+1]=2(k+1)k(k+2)Ωe−x+q(t)−2(k+1)k(k+2)Ωe−(k+1)(x−q(t))=−pk+1(t)e−x+q(t)+pk+1(t)e−(k+1)(x−q(t)), | (3.11) |
where Ω=−12k(k+2)k+1pk+1(t).
For x<q(t),
Gx∗[(k(k+2)k+1uk+1]=−12∫Rsign(x−y)e−|x−y|(k(k+2)k+1pk+1(t)e−(k+1)|y−q(t)|dy=−12(∫x−∞+∫q(t)x+∫∞q(t))sign(x−y)e−|x−y|k(k+2)k+1pk+1(t)e−(k+1)|y−q(t)|dy=Δ1+Δ2+Δ3. | (3.12) |
We directly compute Δ1 as follows
Δ1=−12∫x−∞sign(x−y)e−|x−y|k(k+2)k+1pk+1e−(k+1)|y−q(t)|dy=−12k(k+2)k+1pk+1(t)∫x−∞e−x−(k+1)q(t)e(k+2)ydy=−12k(k+2)k+1pk+1(t)e−x−(k+1)q(t)∫x−∞e(k+2)ydy=−12(k+2)k(k+2)k+1pk+1(t)e(k+1)(x−q(t)). | (3.13) |
In a similar procedure,
Δ2=−12∫ctxsign(x−y)e−|x−y|k(k+2)k+1pk+1e−(k+1)|y−q(t)|dy=12k(k+2)k+1pk+1(t)∫q(t)xex−(k+1)q(t)ekydy=12k(k+2)k+1pk+1(t)ex−(k+1)q(t)∫q(t)xekydy=12kk(k+2)k+1pk+1(t)(−e(k+1)(x−q(t))+ex−q(t)), | (3.14) |
and
Δ3=−12∫∞q(t)sign(x−y)e−|x−y|k(k+2)k+1pk+1(t)e−(k+1)|y−q(t)|dy=12k(k+2)k+1pk+1(t)∫∞q(t)ex+(k+1)q(t)e−(k+2)ydy=12k(k+2)k+1pk+1(t)ex+(k+1)q(t)∫∞q(t)e−(k+2)ydy=12(k+2)k(k+2)k+1pk+1(t)ex−q(t). | (3.15) |
Therefore, from (3.8)–(3.10), we deduce that for x<q(t)
Gx∗[k(k+2)k+1uk+1]=2(k+1)k(k+2)Θex−q(t)−2(k+1)k(k+2)Θe(k+1)(x−q(t))=pk+1(t)ex−q(t)−pk+1(t)e(k+1)(x−q(t)), | (3.16) |
where Θ=12k(k+2)k+1pk+1(t).
Recalling u=p(t)e−|x−q(t)|, we have
p′(t)e−|x−q(t)|+sign(x−q(t))(p(t)q′(t)e−|x−q(t)|−pk+1e−(k+1)|x−q(t)|)={p′(t)e−x+q(t)+p(t)q′(t)e−x+q(t)−pk+1e−(k+1)(x−q(t),forx>q(t),p′(t)ex−q(t)−p(t)q′(t)ex−q(t)+pk+1e(k+1)(x−q(t)),forx≤q(t). |
To ensure that u=p(t)e−|x−q(t)| is a global weak solution of (1.1) in the sense of Definition 3.1, we infer that
pk+1(t)−p(t)q′(t)−p′(t)=0 | (3.17) |
and
pk+1(t)−p(t)q′(t)+p′(t)=0 | (3.18) |
hold.
It completes the proof of Theorem 3.2.
Remark 2. Solving Eqs (3.17) and (3.18), we get
p(t)=c1k,andq(t)=ct+x0c>0. | (3.19) |
Therefore, we conclude that peakon solution for problems (1.1) and (1.2)
u=c1ke−|x−ct−x0|,c>0. | (3.20) |
Remark 3. For x>q(t), the solution of problems (1.1) and (1.2) is of following form
u=p(t)e−x+q(t), | (3.21) |
where p(t) and q(t) satisfy
pk+1(t)−p(t)q′(t)−p′(t)=0. | (3.22) |
For x<q(t), the solution of problems (1.1) and (1.2) is of following form
u=p(t)ex−q(t), | (3.23) |
where p(t) and q(t) satisfy
pk+1(t)−p(t)q′(t)+p′(t)=0. | (3.24) |
Example. For x>q(t), letting q(t)=√t+c,c>0, from (3.17) we derive that
p′+12√tp−pk+1=0. | (3.25) |
(3.25) implies that
p=(2√t+2k)−1k. | (3.26) |
Hence, we obtain from (3.3) the solution of (1.1) for x>q(t).
u=(2√t+2k)−1ke−x+√t+c. | (3.27) |
For x<q(t), letting q(t)=√t+c,c>0, from (3.18) we derive that
p′−12√tp+pk+1=0. | (3.28) |
(3.28) implies that
p=(2√t−2k)−1k. | (3.29) |
Therefore, we obtain from (3.3) the solution of (1.1) for x<q(t).
u=(2√t−2k)−1kex−√t−c. | (3.30) |
In this section, we focus on the analyticity of the Cauchy problems (1.1) and (1.2) based on a contraction type argument in a suitably chosen scale of the Banach spaces. In order to state the main result, we will need a suitable scale of the Banach spaces as follows. For any s>0, we set
Es={u∈C∞(R):|||u|||s=supk∈N0sk||∂ku||H2k!/(k+1)2<∞}, |
where H2(R) is the Sobolev space of order two on the real line and N0 is the set nonnegative integers. One can easily verify that Es equipped with the norm |||⋅|||s is a Banach space and that, for any 0<s′<s, Es is continuously embedded in Es′ with
|||u|||s′≤|||u|||s. |
Another simple consequence of the definition is that any u in Es is a real analytic function on R. Our main theorem is stated as follows.
Theorem 4.1. If the initial data u0 is analytic and belongs to a space Es0, for some 0<s0≤1, then there exist an ε>0 and a unique solution u(t,x) to the Cauchy problems (1.1) and (1.2) that is analytic on (−ε,ε)×R.
For the proof of Theorem 4.1, we need the following lemmas
Lemma 4.2. ([30]) Let 0<s<1. There is a constant C>0, independent of s, such that for any u and v in Es we have
|||uv|||s≤C|||u|||s|||v|||s. |
Lemma 4.3. ([30]) There is a constant C>0 such that for any 0<s′<s<1, we have |||∂xu|||s′≤Cs−s′, and |||(1−∂2x)−1u|||s≤|||u|||s, |||∂x(1−∂2x)−1u|||s≤|||u|||s.
Lemma 4.4. ([31]) Let {Xs}0<s<1 be a scale of decreasing Banach spaces, namely for any s′<s we have Xs⊂Xs′ and |||⋅|||s′≤|||⋅|||s. Consider the Cauchy problem
dudt=F(t,u(t)), | (4.1) |
u(0,x)=0. | (4.2) |
Let T, R and C be positive constants and assume that F satisfies the following conditions:
(1) If for 0<s′<s<1 the function t↦u(t) is holomorphic in |t|<T and continuous on |t|≤T with values in Xs and
sup|t|≤T|||u(t)|||s<R, |
then t↦F(t,u(t)) is a holomorphic function on |t|<T with values in Xs′.
(2) For any 0<s′<s<1 and only u,v∈Xs with |||u|||s<R, |||v|||s<R,
sup|t|≤T|||F(t,u)−F(t,v)|||s′≤Cs−s′|||u−v|||s. |
(3) There exists M>0 such that for any 0<s<1,
sup|t|≤T|||F(t,0)|||s≤M1−s, |
then there exist a T0∈(0,T) and a unique function u(t), which for every s∈(0,1) is holomorphic in |t|<(1−s)T0 with values in Xs, and is a solution to the Cauchy problems (1.1) and (1.2).
Let u1=u and u2=ux, then the problems (1.1) and (1.2) can be written as a system for u1 and u2.
u1t=−uk1u2−∂x(1−∂2x)−1(k(k−3)2(k−1)uk+11+k2uk−11u22)=F1(u1,u2), | (4.3) |
u2t=−kuk−11u22−uk1u2x−∂2x(1−∂2x)−1(k(k−3)2(k−1)uk+11+k2uk−11u22)=F2(u1,u2), | (4.4) |
u1(x,0)=u(x,0)=u0(x),u2(x,0)=ux(x,0)=u0x(x). | (4.5) |
To apply Lemma 4.4 to prove Theorem 4.1, we rewrite the system (4.3)-(3.28) as
dUdt=F(u1,u2), | (4.6) |
U(0)=(u0,u′0), | (4.7) |
where U=(u1,u2) and F(t,U)=F(u1,u2)=(F1(u1,u2),F2(u1,u2)).
Proof of Theorem 4.1. Theorem 4.1 is a straightforward consequence of the Cauchy-Kowalevski theorem [31]. We only need verify the conditions (1)–(3) in the statement of the abstract Cauchy-Kowalevski theorem (see Lemma 4.4) for both F1(u1,u2) and F2(u1,u2) in the systems (4.3)–(3.28), since neither F1 nor F2 depends on t explicitly. For 0<s′<s<1, we derive from Lemmas 4.2 and 4.3 that
|||F1(u1,u2)|||s′≤|||u1|||ks|||u2|||s+C|||u1|||k+1s+C|||u1|||k−1s|||u2|||2s |
|||F2(u1,u2)|||s′≤C|||u1|||k−1s|||u2|||2s+Cs−s′|||u1|||ks|||u2|||s+Cs−s′|||u1|||k+1s+Cs−s′|||u1|||k−1s|||u2|||2s, |
where the constant C depends only on R, so condition (1) holds.
Notice that to verify the second condition it is sufficient to estimate
|||F(u1,u2)−F(v1,v2)|||s′≤|||F1(u1,u2)−F1(v1,v2)|||s′+|||F2(u1,u2)−F2(v1,v2)|||s′≤C|||uk1u2−vk1v2|||s′+C|||∂x(1−∂2x)−1(uk+11−vk+11)|||s′+C|||∂x(1−∂2x)−1(uk−11u22−vk−11v22)|||s′+C|||uk−11u22−vk−11v22|||s′+C|||uk1u2x−vk1v2x|||s′+C|||∂2x(1−∂2x)−1(uk+11−vk+11)|||s′+C|||∂2x(1−∂2x)−1(uk−11u22−vk−11v22)|||s′, |
Using Lemmas 4.2 and 4.3, we get the following estimates
|||uk1u2−vk1v2|||s′≤|||uk1u2−uk1v2|||s′+|||uk1v2−vk1v2|||s′≤C|||u2−v2|||s|||u1|||ks+|||u1−v1|||s|||v2|||s(k−1∑0(|||u1|||k−1−is|||v1|||is)), |
|||∂x(1−∂2x)−1(uk+11−vk+11)|||s′≤|||uk+11−vk+11|||s′≤|||u1−v1|||s(k∑0(|||u1|||k−is|||v1|||is)), |
|||uk−11u22−vk−11v22|||s′≤|||uk−11u22−uk−11v22|||s′+|||uk−11v22−vk−11v22|||s′≤C|||u2−v2|||s(|||u2+v2|||s)|||u1|||k−1s+|||u1−v1|||s|||v2|||2s(k−2∑0(|||u1|||k−2−is|||v1|||is)), |
and
|||∂x(1−∂2x)−1(uk−11u22−vk−11v22)|||s′≤C|||u2−v2|||s(|||u2+v2|||s)|||u1|||k−1s+|||u1−v1|||s|||v2|||2s(k−2∑0(|||u1|||k−2−is|||v1|||is)), |
|||uk1u2x−vk1v2x|||s′≤|||uk1u2−uk1v2x|||s′+|||uk1v2x−vk1v2|||s′≤Cs−s′|||u2−v2|||s|||u1|||ks+Cs−s′|||u1−v1|||s|||v2|||s(k−1∑0(|||u1|||k−1−is|||v1|||is)), |
|||∂2x(1−∂2x)−1(uk+11−vk+11)|||s′≤Cs−s′|||uk+11−vk+11|||s′≤Cs−s′|||u1−v1|||s(k∑0(|||u1|||k−is|||v1|||is)), |
|||∂2x(1−∂2x)−1(uk−11u22−vk−11v22)|||s′≤Cs−s′|||u2−v2|||s(|||u2+v2|||s)|||u1|||k−1s+Cs−s′|||u1−v1|||s|||v2|||2s(k−2∑0(|||u1|||k−2−is|||v1|||is)). |
Therefore, we arrive at
|||F(u1,u2)−F(v1,v2)|||s′≤C|||u2−v2|||s|||u1|||ks+|||u1−v1|||s|||v2|||s(k−1∑0(|||u1|||k−1−is|||v1|||is))+C|||u1−v1|||s(k∑0(|||u1|||k−is|||v1|||is))+Cs−s′|||u1−v1|||s(k∑0(|||u1|||k−is|||v1|||is))+2C|||u2−v2|||s|||u2+v2|||s|||u1|||k−1s+|||u1−v1|||s|||v2|||2s(k−2∑0(|||u1|||k−2−is|||v1|||is))+Cs−s′|||u2−v2|||s|||u1|||ks+Cs−s′|||u1−v1|||s|||v2|||s(k−1∑0(|||u1|||k−1−is|||v1|||is))+Cs−s′|||u2−v2|||s|||u2+v2|||s|||u1|||k−1s+Cs−s′|||u1−v1|||s|||v2|||2s(k−2∑0(|||u1|||k−2−is|||v1|||is)), |
where the constant C depends only on R and k. The conditions (1)–(3) above are easily verified once our system is transformed into a new system with zero initial data as (4.1) and (4.2). So, we have completed the proof of Theorem 4.1.
In this paper, we focus on several dynamic properties of the Cauchy problems (1.1) and (1.2). We first establish a new blow-up criterion and a blow-up phenomenon for the problem, then we study analytical solutions for the equation by using a new method, here, we present two analytical solutions for the problems (1.1) and (1.2) for the first time. Finally, we study the analyticity in a suitable scale of the Banach spaces. The properties of the problems (1.1) and (1.2) not only present fundamental importance from mathematical point of view but also are of great physical interest.
This research was funded by the Guizhou Province Science and Technology Basic Project (Grant No. QianKeHe Basic [2020]1Y011).
There is no conflict of interest.
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