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Associations of CYP1A1 gene polymorphisms and risk of breast cancer in Indian women: a meta-analysis

  • Reported associations of CYP1A1 polymorphisms with breast cancer have been inconsistent. In this meta-analysis examining breast cancer associations of three CYP1A1 polymorphisms (M1, M2 and M4) among Indian women may yield information that may be of clinical and epidemiological use for this particular demography. We searched MEDLINE using PubMed and Embase for association studies. From seven published case-control studies, we estimated overall associations and applied subgroup analysis to explore differential effects. All three polymorphisms exhibited overall increased risk, significant in M1 (OR 1.61-1.65, p = 0.04) and M4 (OR 2.02-3.92, p = 0.02-0.04). Differential effects were observed only in the M1 polymorphism where M1 effects were significant in South Indians (OR 2.20-4.34, p < 0.0001) but not the North population, who were at reduced risk (OR 0.64-0.77, p = 0.03-0.55). These populations were not materially different in regard to M2 and M4 as did the women stratified by menopausal status. In this meta-analysis, M1 and M4 effects may render Indian women susceptible, but may be limited by heterogeneity of the studies. Differential effects of the M1 polymorphism in breast cancer render South Indians susceptible compared to those in the North.

    Citation: Noel Pabalan, Neetu Singh, Eloisa Singian, Caio Parente Barbosa, Bianca Bianco, Hamdi Jarjanazi. Associations of CYP1A1 gene polymorphisms and risk of breast cancer in Indian women: a meta-analysis[J]. AIMS Genetics, 2015, 2(4): 250-262. doi: 10.3934/genet.2015.4.250

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  • Reported associations of CYP1A1 polymorphisms with breast cancer have been inconsistent. In this meta-analysis examining breast cancer associations of three CYP1A1 polymorphisms (M1, M2 and M4) among Indian women may yield information that may be of clinical and epidemiological use for this particular demography. We searched MEDLINE using PubMed and Embase for association studies. From seven published case-control studies, we estimated overall associations and applied subgroup analysis to explore differential effects. All three polymorphisms exhibited overall increased risk, significant in M1 (OR 1.61-1.65, p = 0.04) and M4 (OR 2.02-3.92, p = 0.02-0.04). Differential effects were observed only in the M1 polymorphism where M1 effects were significant in South Indians (OR 2.20-4.34, p < 0.0001) but not the North population, who were at reduced risk (OR 0.64-0.77, p = 0.03-0.55). These populations were not materially different in regard to M2 and M4 as did the women stratified by menopausal status. In this meta-analysis, M1 and M4 effects may render Indian women susceptible, but may be limited by heterogeneity of the studies. Differential effects of the M1 polymorphism in breast cancer render South Indians susceptible compared to those in the North.


    In this paper, we are concerned with the Cauchy problem for the generalized Camassa-Holm equation

    $ (12x)ut+(k+1)(k+2)2ukuxk(k1)2uk2u3x2kuk1uxuxxukuxxx=0,kN+,
    $
    (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)=c1kexct.
    $
    (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 $ H^{s}(\mathbb{R}) $ with $ 1 < s\leq3/2 $. Yan [23] proved that Eq (1.1) does not depend uniformly continuously on the initial data in $ H^{s}(\mathbb{R}) $ with $ s < 3/2 $ and that the Cauchy problem for the generalized Camassa-Holm equation is locally well-posedness in $ B_{2, 1}^{3/2} $. In [24], Zhou focused on the persistence property in weighted $ L^{p} $ spaces.

    When $ k = 1 $, Eq (1.1) reduced to the well-known Camassa-Holm equation

    $ (12x)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 $ \int_{\mathbb{R}}((u^{k})_{x})^{3}dx $ plays a key role. Taking advantage of complicated calculation, we obtain the Riccati inequality of quantity $ \int_{\mathbb{R}}((u^{k})_{x})^{3}dx $ 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 $ \phi(t, x) $ with compact support in $ [0, +\infty)\times \mathbb{R} $ is denoted by $ C^\infty_0 $. Let $ L^p = L^p(\mathbb{R})(1\leq p < +\infty) $ be the space of all measurable functions $ h $ such that $ \parallel h\parallel^{P}_{L^{P}} = \int_{\mathbb{R}}|h(t, x)|^{p} dx < \infty $. We define $ L^\infty = L^{\infty}(\mathbb{R}) $ with the standard norm $ \parallel h\parallel_{L^{\infty}} = {\rm{inf}}_{m(e) = 0} {\rm{sup}}_{x\in R\setminus e}| h(t, x)| $. For any real number $ s $, $ H^{s} = H^{s}(\mathbb{R}) $ denotes the Sobolev space with the norm defined by

    $ hHs=(R(1+|ξ|2)s|ˆh(t,ξ)|2dξ)12<,
    $

    where $ \hat{h}(t, \xi) = \int_\mathbb{R}e^{-ix\xi}h(t, x)dx $.

    We denote by $ \ast $ the convolution, using the green function $ G(x) = \frac{1}{2}e^{-\mid x\mid} $, we have $ (1-\partial_{x}^{2})^{-1}f = G(x)*f $ for all $ f\in L^{2} $, and $ p*(u-u_{xx}) = u $. For $ T > 0 $ and nonnegative number $ s $, $ C([0, T); H^{s}(\mathbb{R})) $ denotes the Frechet space of all continuous $ H^{s} $-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(12x)1(k(k+3)2(k+1)uk+1+k2uk1u2x),
    $
    (1.8)
    $ u(0,x)=u0(x),
    $
    (1.9)

    which is also equivalent to

    $ yt+ukyx+2kuk1uxy+k(k1)2(ukuxuk2u3x)=0,
    $
    (1.10)
    $ y=uuxx,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) = u_{0}\in H^{s}(\mathbb{R}), s > 3/2 $, then there exist a maximal $ T = T(u_{0}) $ and a unique solution $ u $ to the problems (1.1) and (1.2) such that

    $ u = u(\cdot, u_{0})\in C([0, T);H^{s}(\mathbb{R}))\bigcap C^{1}([0, T);H^{s-1}(\mathbb{R})). $

    Moreover, the solution depends continuously on the initial data, i.e., the mapping $ u_{0}\rightarrow u(\cdot, u_{0}):H^{s}\rightarrow C([0, T); H^{s}(\mathbb{R}))\bigcap C^{1}([0, T); H^{s-1}(\mathbb{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\in C^{1}(\mathbb{R}) $, $ a > 0 $, $ b > 0 $ and $ f(0) > \sqrt{\frac{b}{a}} $. If $ f'(t)\geq af^{2}(t)-b $, then

    $ f(t)+astT=12ablog(f(0)+baf(0)ba).
    $
    (2.1)

    Lemma 2.3. (see [23]) Let $ u_{0}\in B_{2, 1}^{3/2} $ 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 < \infty $, then

    $ T0uxLdτ=+.
    $
    (2.2)

    Remark 1. For $ s > \frac{3}{2} $, it is well known that $ H^{s}\hookrightarrow B_{2, 1}^{3/2} $, so we have the following result:

    Let $ u_{0}\in H^{s} $ with $ s > \frac{3}{2} $ 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 < \infty $, then

    $ T0uxLdτ=+.
    $
    (2.3)

    Lemma 2.4. Let $ u_{0}\in H^{s}(\mathbb{R}) $ with $ s > \frac{3}{2} $. Let $ T > 0 $ be the maximum existence time of the solution $ u $ to the problems (1.1) and (1.2) with the initial data $ u_{0} $. Then the corresponding solution $ u $ blows up in finite time if and only if

    $ limtTuk1uxL=+.
    $

    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 $ u_{0}\in H^{3}(\mathbb{R}) $. From Lemma 2.1 we know that $ u\in C([0, T); H^{3}(\mathbb{R}))\bigcap C^{1}([0, T); H^{2}(\mathbb{R})) $. Due to $ y = u-u_{xx} $, by direct computation, one has

    $ y2L2=R(uuxx)2dx=R(u2+2u2x+u2xx)dx.
    $
    (2.4)

    So,

    $ u2H2≤∥y2L22u2H2.
    $
    (2.5)

    Multiplying equation (1.10) by $ 2y $ and integrating by parts and using the interpolation $ \parallel u_{x}\parallel^{2}_{L^{\infty}}\leq C \parallel u\parallel_{H^{1}}\parallel y\parallel_{L^{2}} $, we obtain

    $ ddty2L2=2Ryytdx=3kRuk1uxy2dxk(k1)Rukuxydx+k(k1)Ruk2u3xydxc(uk1uxLy2L2+ukLyL2uxL2+uk2Lu2xLyL2uxL2)c(uk1uxL+C1)y2L2.
    $
    (2.6)

    where $ C_{1} = C_{1}(\parallel u_{0}\parallel_{H^{1}}) $.

    If there exists a constant $ M > 0 $ such that $ \parallel u^{k-1}u_{x}\parallel_{L^{\infty}} < M $, from (2.6) we deduce that

    $ ddty2L2c(M+C1)y2L2.
    $
    (2.7)

    By virtue of Gronwall's inequality, one has

    $ y2L2≤∥y02L2ec(M+C1)t.
    $
    (2.8)

    On the other hand, due to $ u = p\ast y $ and $ u_{x} = p_{x}\ast y $, then

    $ uk1uxL≤∥uk1LuxL≤∥pk1L2pxL2ykL2
    $

    This completes the proof of Lemma 2.4.

    Now, we present the blow-up phenomenon.

    Theorem 2.5. Let $ u_{0}\in H^{s}(\mathbb{R}) $ for $ s > \frac{3}{2} $. Suppose that $ u(t, x) $ be corresponding solution of problems (1.1) and (1.2) with the initial datum $ u_{0} $. If the slope of $ u^{k}_{0} $ satisfies

    $ R(uk0)3x<KK1,
    $
    (2.9)

    where $ K = \frac{3k(6k^{2}+7k-2)}{4(k+1)}\parallel u_{0}\parallel^{2k}_{H^{1}} $ and $ K_{1} = \frac{1}{2ck\parallel u_{0}\parallel^{k}_{H^{1}}} $. Then there exists the lifespan $ T < \infty $ 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) = u^{k}(t, x) $, $ h(t) = \int_{\mathbb{R}}g^{3}_{x}dx $, it follows that

    $ gt+ggx=kuk1Q,
    $
    (2.11)

    where $ Q = -\partial_{x}(1-\partial^{2}_{x})^{-1}(\frac{k(k+2)}{2(k+1)}u^{k+1}+\frac{k}{2}u^{k-1}u^{2}_{x}) $.

    Differentiating the above Eq (2.11) with respect $ x $ yields

    $ gtx+ggxx=12g2x+k(k1)uk2uxQ+k2(k+2)2(k+1)g2k2(k+2)2(k+1)uk1(12x)1(uk+1)k22uk1(12x)1(uk1u2x).
    $
    (2.12)

    Multiplying $ 3g^{2}_{x} $ both sides of (2.12) and integrating with respect $ x $ over $ \mathbb{R} $, one has

    $ ddtRg3xdx=12Rg4xdx+3k(k1)Ruk2uxg2xQdx+3k2(k+2)2(k+1)Rg2g2xdx3k2(k+2)2(k+1)Rg2xuk1(12x)1(uk+1)dx3k22Rg2xuk1(12x)1(uk1u2x)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(k1)Ruk2uxg2xQdx3k(k1)QL(R(uk2ux)2dx)12(R(gx)2dx)123k(k1)2u02kH1(R(gx)4dx)123k(k1)2(u04kH12ϵ+ϵR(gx)4dx2),
    $
    (2.14)
    $ Γ3=3k2(k+2)2(k+1)Rg2g2xdx3k2(k+2)2(k+1)(Rg4dx)12(R(gx)4dx)123k2(k+2)2(k+1)u02kH1(R(gx)4dx)123k2(k+2)2(k+1)(u04kH12ϵ+ϵR(gx)4dx2),
    $
    (2.15)
    $ Γ4=3k2(k+2)2(k+1)Rg2xuk1(12x)1(uk+1)dx3k2(k+2)2(k+1)(12x)1(uk+1)L(Ru2k2dx)12(R(gx)4dx)123k2(k+2)2(k+1)u02kH1(R(gx)4dx)123k2(k+2)2(k+1)(u04kH12ϵ+ϵR(gx)4dx2),
    $
    (2.16)

    and

    $ Γ5=3k22Rg2xuk1(12x)1(uk1u2x)dx3k22(12x)1(uk1u2x)L(Ru2k2dx)12(R(gx)4dx)123k22u02kH1(R(gx)4dx)123k22(u04kH12ϵ+ϵR(gx)4dx2).
    $
    (2.17)

    Combining the above inequalities (2.14)–(2.17), we obtain

    $ Γ2+Γ3+Γ4+Γ5∣≤3k(6k2+7k2)8ϵ(k+1)u04kH1+ϵ23k(6k2+7k2)4(k+1)R(gx)4dx.
    $
    (2.18)

    Choosing $ \epsilon = \frac{2(k+1)}{3k(6k^{2}+7k-2)} $, which results in

    $ ddtRg3xdx14Rg4xdx+K2.
    $
    (2.19)

    in which $ K^{2} = \frac{9k^{2}(6k^{2}+7k-2)^{2}}{16(k+1)^{2}}\parallel u_{0}\parallel^{4k}_{H^{1}} $. Using the Hölder's inequality, we get

    $ (Rg3xdx)2cRg2xdxRg4xdxck2u02kH1Rg4xdx.
    $
    (2.20)

    Combining (2.19) and (2.20), we have

    $ ddth(t)K21h2(t)+K2,
    $
    (2.21)

    where $ K^{2}_{1} = \frac{1}{4ck^{2}\parallel u_{0}\parallel^{2k}_{H^{1}}} $.

    It is observed from assumption of Theorem that $ h(0) < -\frac{K}{K_{1}} $, the continuity argument ensures that $ h(t) < h(0) $. Lemma 2.1 ($ a = K^{2}_{1} $ and $ b = K^{2} $) implies that $ h(t)\rightarrow -\infty $ as $ t\rightarrow T = \frac{1}{2K_{1}K}\log\frac{K_{1}h(0)-K}{K_{1}h(0)+K} $.

    On the other hand, by using the fact that

    $ Rg3xdx∣≤Rg3xdxk3uk1ux(t,x)LRg2xdx=k3uk1ux(t,x)Lu02kH1.
    $
    (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 $ H^{s}(\mathbb{R}) $ with $ s > \frac{3}{2} $ [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 $ u_{0}\in H^{s} $, $ s > \frac{3}{2} $, 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

    $ T0Ruφt+1k+1uk+1φx+G(k(k+3)2(k+1)uk+1+k2uk1u2x)φxdxdt+Ru0(x)φ(0,x)dx=0
    $
    (3.1)

    for any smooth test function $ \varphi(t, x)\in C^{\infty}_{c}([0, T)\times \mathbb{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|xq(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|xq(t)|
    $
    (3.3)

    is a peakon solution of (1.1) and

    $ ut=p(t)e|xq(t)|+p(t)sign(xq(t))q(t)e|xq(t)|,ux=p(t)sign(xq(t))e|xq(t)|.
    $
    (3.4)

    Hence, using (3.1), (3.4) and integration by parts, we derive that

    $ T0Ruφt+1k+1uk+1φxdxdt+Ru0(x)φ(0,x)dx=T0Rφ(ut+ukux)dxdt=T0Rφ[p(t)e|xq(t)|+sign(xq(t))(p(t)q(t)e|xq(t)|pk+1e(k+1)|xq(t)|)]dxdt.
    $
    (3.5)

    On the other hand, using (3.4), we obtain

    $ T0RG(k(k+3)2(k+1)uk+1+k2uk1u2x)φxdxdt=T0RφGx[k2uk1u2x+k(k+3)2(k+1)uk+1]dxdt=T0RφGx[k(k+2)k+1uk+1]dxdt.
    $
    (3.6)

    Note that $ G_{x} = -\frac{1}{2}sign(x)e^{-|x|} $. For $ x > q(t) $, directly calculate

    $ Gx[k(k+2)k+1uk+1]=12Rsign(xy)e|xy|k(k+2)k+1pk+1(t)e(k+1)|yq(t)|dy=12(q(t)+xq(t)+x)sign(xy)e|xy|k(k+2)k+1pk+1(t)e(k+1)|yq(t)|dy=I1+I2+I3.
    $
    (3.7)

    We directly compute $ I_{1} $ as follows

    $ I1=12q(t)sign(xy)e|xy|k(k+2)k+1pk+1(t)e(k+1)|yq(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).
    $
    (3.8)

    In a similar procedure,

    $ I2=12xq(t)sign(xy)e|xy|k(k+2)k+1ak+1e(k+1)|yq(t)|dy=12k(k+2)k+1pk+1xq(t)ex+(k+1)q(t)ekydy=12k(k+2)k+1pk+1ex+(k+1)q(t)xq(t)ekydy=12kk(k+2)k+1pk+1(e(k+1)(xq(t))ex+q(t)),
    $
    (3.9)

    and

    $ I3=12xsign(xy)e|xy|k(k+2)k+1pk+1e(k+1)|yq(t)|dy=12k(k+2)k+1pk+1xex+(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)(xq(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)Ωex+q(t)2(k+1)k(k+2)Ωe(k+1)(xq(t))=pk+1(t)ex+q(t)+pk+1(t)e(k+1)(xq(t)),
    $
    (3.11)

    where $ \Omega = -\frac{1}{2}\frac{k(k+2)}{k+1}p^{k+1}(t) $.

    For $ x < q(t) $,

    $ Gx[(k(k+2)k+1uk+1]=12Rsign(xy)e|xy|(k(k+2)k+1pk+1(t)e(k+1)|yq(t)|dy=12(x+q(t)x+q(t))sign(xy)e|xy|k(k+2)k+1pk+1(t)e(k+1)|yq(t)|dy=Δ1+Δ2+Δ3.
    $
    (3.12)

    We directly compute $ \Delta_{1} $ as follows

    $ Δ1=12xsign(xy)e|xy|k(k+2)k+1pk+1e(k+1)|yq(t)|dy=12k(k+2)k+1pk+1(t)xex(k+1)q(t)e(k+2)ydy=12k(k+2)k+1pk+1(t)ex(k+1)q(t)xe(k+2)ydy=12(k+2)k(k+2)k+1pk+1(t)e(k+1)(xq(t)).
    $
    (3.13)

    In a similar procedure,

    $ Δ2=12ctxsign(xy)e|xy|k(k+2)k+1pk+1e(k+1)|yq(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)(xq(t))+exq(t)),
    $
    (3.14)

    and

    $ Δ3=12q(t)sign(xy)e|xy|k(k+2)k+1pk+1(t)e(k+1)|yq(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)exq(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)Θexq(t)2(k+1)k(k+2)Θe(k+1)(xq(t))=pk+1(t)exq(t)pk+1(t)e(k+1)(xq(t)),
    $
    (3.16)

    where $ \Theta = \frac{1}{2}\frac{k(k+2)}{k+1}p^{k+1}(t) $.

    Recalling $ u = p(t)e^{-|x-q(t)|} $, we have

    $ p(t)e|xq(t)|+sign(xq(t))(p(t)q(t)e|xq(t)|pk+1e(k+1)|xq(t)|)={p(t)ex+q(t)+p(t)q(t)ex+q(t)pk+1e(k+1)(xq(t),forx>q(t),p(t)exq(t)p(t)q(t)exq(t)+pk+1e(k+1)(xq(t)),forxq(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|xctx0|,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)ex+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)exq(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) = \sqrt{t}+c, c > 0 $, from (3.17) we derive that

    $ p+12tppk+1=0.
    $
    (3.25)

    (3.25) implies that

    $ p=(2t+2k)1k.
    $
    (3.26)

    Hence, we obtain from (3.3) the solution of (1.1) for $ x > q(t) $.

    $ u=(2t+2k)1kex+t+c.
    $
    (3.27)

    For $ x < q(t) $, letting $ q(t) = \sqrt{t}+c, c > 0 $, from (3.18) we derive that

    $ p12tp+pk+1=0.
    $
    (3.28)

    (3.28) implies that

    $ p=(2t2k)1k.
    $
    (3.29)

    Therefore, we obtain from (3.3) the solution of (1.1) for $ x < q(t) $.

    $ u=(2t2k)1kextc.
    $
    (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={uC(R):|||u|||s=supkN0sk||ku||H2k!/(k+1)2<},
    $

    where $ H^{2}(\mathbb{R}) $ is the Sobolev space of order two on the real line and $ N_{0} $ is the set nonnegative integers. One can easily verify that $ E_{s} $ equipped with the norm $ |||\cdot|||_{s} $ is a Banach space and that, for any $ 0 < s' < s $, $ E_{s} $ is continuously embedded in $ E_{s'} $ with

    $ |||u|||_{s'}\leq|||u|||_{s}. $

    Another simple consequence of the definition is that any $ u $ in $ E_{s} $ is a real analytic function on $ \mathbb{R} $. Our main theorem is stated as follows.

    Theorem 4.1. If the initial data $ u_{0} $ is analytic and belongs to a space $ E_{s_{0}} $, for some $ 0 < s_{0}\leq1 $, then there exist an $ \varepsilon > 0 $ and a unique solution $ u(t, x) $ to the Cauchy problems (1.1) and (1.2) that is analytic on $ (-\varepsilon, \varepsilon)\times \mathbb{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 $ E_{s} $ we have

    $ |||uv|||_{s}\leq 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 $ |||\partial_{x}u|||_{s'}\leq \frac{C}{s-s'} $, and $ |||(1-\partial^{2}_{x})^{-1}u|||_{s}\leq|||u|||_{s} $, $ |||\partial_{x}(1-\partial^{2}_{x})^{-1}u|||_{s}\leq|||u|||_{s}. $

    Lemma 4.4. ([31]) Let $ \{X_{s}\}_{0 < s < 1} $ be a scale of decreasing Banach spaces, namely for any $ s' < s $ we have $ X_{s}\subset X_{s'} $ and $ |||\cdot|||_{s'}\leq|||\cdot|||_{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\mapsto u(t) $ is holomorphic in $ |t| < T $ and continuous on $ |t|\leq T $ with values in $ X_{s} $ and

    $ \sup\limits_{|t|\leq T}|||u(t)|||_{s} < R, $

    then $ t\mapsto F(t, u(t)) $ is a holomorphic function on $ |t| < T $ with values in $ X_{s'} $.

    $ (2) $ For any $ 0 < s' < s < 1 $ and only $ u, v\in X_{s} $ with $ |||u|||_{s} < R $, $ |||v|||_{s} < R $,

    $ \sup\limits_{|t|\leq T}|||F(t, u)-F(t, v)|||_{s'}\leq \frac{C}{s-s'}|||u-v|||_{s}. $

    $ (3) $ There exists $ M > 0 $ such that for any $ 0 < s < 1 $,

    $ \sup\limits_{|t|\leq T}|||F(t, 0)|||_{s}\leq\frac{M}{1-s}, $

    then there exist a $ T_{0}\in (0, T) $ and a unique function $ u(t) $, which for every $ s\in (0, 1) $ is holomorphic in $ |t| < (1-s)T_{0} $ with values in $ X_{s} $, and is a solution to the Cauchy problems (1.1) and (1.2).

    Let $ u_{1} = u $ and $ u_{2} = u_{x} $, then the problems (1.1) and (1.2) can be written as a system for $ u_{1} $ and $ u_{2} $.

    $ u1t=uk1u2x(12x)1(k(k3)2(k1)uk+11+k2uk11u22)=F1(u1,u2),
    $
    (4.3)
    $ u2t=kuk11u22uk1u2x2x(12x)1(k(k3)2(k1)uk+11+k2uk11u22)=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,u0),
    $
    (4.7)

    where $ U = (u_{1}, u_{2}) $ and $ F(t, U) = F(u_{1}, u_{2}) = (F_{1}(u_{1}, u_{2}), F_{2}(u_{1}, u_{2})) $.

    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 $ F_{1}(u_{1}, u_{2}) $ and $ F_{2}(u_{1}, u_{2}) $ in the systems (4.3)–(3.28), since neither $ F_{1} $ nor $ F_{2} $ 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|||k1s|||u2|||2s
    $
    $ |||F2(u1,u2)|||sC|||u1|||k1s|||u2|||2s+Css|||u1|||ks|||u2|||s+Css|||u1|||k+1s+Css|||u1|||k1s|||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)|||sC|||uk1u2vk1v2|||s+C|||x(12x)1(uk+11vk+11)|||s+C|||x(12x)1(uk11u22vk11v22)|||s+C|||uk11u22vk11v22|||s+C|||uk1u2xvk1v2x|||s+C|||2x(12x)1(uk+11vk+11)|||s+C|||2x(12x)1(uk11u22vk11v22)|||s,
    $

    Using Lemmas 4.2 and 4.3, we get the following estimates

    $ |||uk1u2vk1v2|||s|||uk1u2uk1v2|||s+|||uk1v2vk1v2|||sC|||u2v2|||s|||u1|||ks+|||u1v1|||s|||v2|||s(k10(|||u1|||k1is|||v1|||is)),
    $
    $ |||x(12x)1(uk+11vk+11)|||s|||uk+11vk+11|||s|||u1v1|||s(k0(|||u1|||kis|||v1|||is)),
    $
    $ |||uk11u22vk11v22|||s|||uk11u22uk11v22|||s+|||uk11v22vk11v22|||sC|||u2v2|||s(|||u2+v2|||s)|||u1|||k1s+|||u1v1|||s|||v2|||2s(k20(|||u1|||k2is|||v1|||is)),
    $

    and

    $ |||x(12x)1(uk11u22vk11v22)|||sC|||u2v2|||s(|||u2+v2|||s)|||u1|||k1s+|||u1v1|||s|||v2|||2s(k20(|||u1|||k2is|||v1|||is)),
    $
    $ |||uk1u2xvk1v2x|||s|||uk1u2uk1v2x|||s+|||uk1v2xvk1v2|||sCss|||u2v2|||s|||u1|||ks+Css|||u1v1|||s|||v2|||s(k10(|||u1|||k1is|||v1|||is)),
    $
    $ |||2x(12x)1(uk+11vk+11)|||sCss|||uk+11vk+11|||sCss|||u1v1|||s(k0(|||u1|||kis|||v1|||is)),
    $
    $ |||2x(12x)1(uk11u22vk11v22)|||sCss|||u2v2|||s(|||u2+v2|||s)|||u1|||k1s+Css|||u1v1|||s|||v2|||2s(k20(|||u1|||k2is|||v1|||is)).
    $

    Therefore, we arrive at

    $ |||F(u1,u2)F(v1,v2)|||sC|||u2v2|||s|||u1|||ks+|||u1v1|||s|||v2|||s(k10(|||u1|||k1is|||v1|||is))+C|||u1v1|||s(k0(|||u1|||kis|||v1|||is))+Css|||u1v1|||s(k0(|||u1|||kis|||v1|||is))+2C|||u2v2|||s|||u2+v2|||s|||u1|||k1s+|||u1v1|||s|||v2|||2s(k20(|||u1|||k2is|||v1|||is))+Css|||u2v2|||s|||u1|||ks+Css|||u1v1|||s|||v2|||s(k10(|||u1|||k1is|||v1|||is))+Css|||u2v2|||s|||u2+v2|||s|||u1|||k1s+Css|||u1v1|||s|||v2|||2s(k20(|||u1|||k2is|||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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