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Research article Special Issues

The contribution of critical thinking to STEM disciplines at the time of generative intelligence


  • After more than two decades since its inception, the acronym STEM (Science, Technology, Engineering, and Mathematics), which designated the scientific disciplines to be strengthened in the U.S. to meet the challenges of the new millennium, is changing its orientation and representations. Furthermore, this is seen in Europe and Asia, a new approach to reading the contemporary world, which is based on a few key concepts: Interdisciplinarity and complexity. Philosophy, by its nature, plays a leading role in developing those skills that the Framework for the 21st Century Learning report has identified as indispensable and grouped under the 4Cs (Critical thinking, Communication, Collaboration, Creativity). Specifically, critical thinking, which originates with philosophy, helps to settle complex situations and problems and this discipline is also capable of bridging knowledge that seems distant from each other, bringing it into dialogue. In this paper, I aim to investigate, by literature review, the role of critical thinking in the STEM disciplines, which are closely connected to the development of technological knowledge, and thus of GAI (Generative Artificial Intelligence), to contribute to a discussion on how can offer a critical understanding of GAI and its uses. The result of this reflection, which does not yet seem to be outlined in the literature, but which hopefully will be more extensively addressed in the future, indicates that critical thinking, guided by philosophy, can play a crucial role in STEM, especially concerning the Post-Normal Science model, in which the construction of scientific knowledge leaves the academy. Moreover, GAI tools significantly modify the interactions between the different knowledge actors.

    Citation: Elena Guerra. The contribution of critical thinking to STEM disciplines at the time of generative intelligence[J]. STEM Education, 2024, 4(1): 71-81. doi: 10.3934/steme.2024005

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  • After more than two decades since its inception, the acronym STEM (Science, Technology, Engineering, and Mathematics), which designated the scientific disciplines to be strengthened in the U.S. to meet the challenges of the new millennium, is changing its orientation and representations. Furthermore, this is seen in Europe and Asia, a new approach to reading the contemporary world, which is based on a few key concepts: Interdisciplinarity and complexity. Philosophy, by its nature, plays a leading role in developing those skills that the Framework for the 21st Century Learning report has identified as indispensable and grouped under the 4Cs (Critical thinking, Communication, Collaboration, Creativity). Specifically, critical thinking, which originates with philosophy, helps to settle complex situations and problems and this discipline is also capable of bridging knowledge that seems distant from each other, bringing it into dialogue. In this paper, I aim to investigate, by literature review, the role of critical thinking in the STEM disciplines, which are closely connected to the development of technological knowledge, and thus of GAI (Generative Artificial Intelligence), to contribute to a discussion on how can offer a critical understanding of GAI and its uses. The result of this reflection, which does not yet seem to be outlined in the literature, but which hopefully will be more extensively addressed in the future, indicates that critical thinking, guided by philosophy, can play a crucial role in STEM, especially concerning the Post-Normal Science model, in which the construction of scientific knowledge leaves the academy. Moreover, GAI tools significantly modify the interactions between the different knowledge actors.



    In this paper, we consider the initial value problem for a high-order nonlinear dispersive wave equation

    ututxx+(k+2)ukux(k+1)uk1uxuxxukuxxx=0, (1.1)
    u(0,x)=u0(x), (1.2)

    where kN0 (N0 denotes the set of nonnegative integers) and u stands for the unknown function on the line R. Equation (1.1) was known as gkCH equation [18], which admits single peakon and multi-peakon traveling wave solutions, and possesses conserved laws

    R(u2+u2x)dx=R(u20+u20x)dx. (1.3)

    It is shown in [18] that this equation is well-posedness in Sobolev spaces Hs with s>3/2 on both the circle and the line in the sense of Hadamard by using a Galerkin-type approximation scheme. That is, the data-to-solution map is continuous. Furthermore, it is proved in [18] that this dependence is sharp by showing that the solution map is not uniformly continuous. The nonuniform dependence is proved using the method of approximate solutions and well-posedness estimates.

    For k=1, we obtain the integrable equation with quadratic nonlinearities

    ututxx+3uux2uxuxxuuxxx=0, (1.4)

    which was derived by Camassa and Holm [1] and by Fokas and Fuchssteiner [12]. It was called as Camassa-Holm equation. It describes the motion of shallow water waves and possesses a Lax pair, a bi-Hamiltonian structure and infinitely many conserved integrals [1], and it can be solved by the inverse scattering method. One of the remarkable features of the CH equation is that it has the single peakon solutions

    u(t,x)=ce|xct|,cR

    and the multi-peakon solutions

    u(t,x)=Ni=1pi(t)e|xqi(t)|,

    where pi(t), qi(t) satisfy the Hamilton system [1]

    dpidt=Hqi=ijpipjsign(qiqj)e|qiqj|, 
    dqidt=Hpi=jpje|qiqj|

    with the Hamiltonian H=12Ni,j=1pipje|qi|. It is shown that those peaked solitons were orbitally stable in the energy space [6]. Another remarkable feature of the CH equation is the so-called wave breaking phenomena, that is, the wave profile remains bounded while its slope becomes unbounded in finite time [7,8,9]. Hence, Eq (1.4) has attracted the attention of lots of mathematicians. The dynamic properties related to the equation can be found in [3,4,5,10,11,13,14,15,16,17,22,23,24,25,26,34,35,36] and the references therein.

    For k=2, we obtain the integrable equation with cubic nonlinearities

    ututxx+4u2ux3uuxuxxu2uxxx=0, (1.5)

    which was derived by Vladimir Novikov in a symmetry classification of nonlocal PDEs [29] and was known as the Novikov equation. It is shown in [29] that Eq (1.5) possesses soliton solutions, infinitely many conserved quantities, a Lax pair in matrix form and a bi-Hamiltonian structure. Equation (1.5) can be thought as a generalization of the Camassa-Holm equation. The conserved quantities

    H1[u(t)]=R(u2+u2x)dx

    and

    H2(t)=R(u4+2u2u2x13u4x)dx

    play an important role in the study of the dynamic properties related to the Eq (1.5). More information about the Novikov equation can be found in Himonas and Holliman [19], F. Tiglay [30], Ni and Zhou [28], Wu and Yin [31,32], Yan, Li and Zhang [33], Mi and Mu [27] and the references therein.

    Inspired by the reference cited above, the objective of this paper is to investigate the dynamic properties for the Problems (1.1) and (1.2). More precisely, we firstly establish two blow up criteria and derive a lower bound of the maximal existence time. Then for k=2p+1, p is nonnegative integer, we derive two blow-up phenomena under different initial data. Motivated by the idea from Chen etal's work[2], we apply the characteristic dynamics of P=2uux and Q=2u+ux to deduce the first blow-up phenomenon. For the Problems (1.1) and (1.2), in fact, the estimates of P and Q can be closed in the form of

    P(t)α(u)PQ+Θ1,Q(t)α(u)PQ+Θ2, (1.6)

    where α(u)0 and the nonlocal term Θi (i=1,2) can be bounded by the estimates from Problems (1.1) and (1.2). From (1.6) the monotonicity of P and Q can be established, and hence the finite-time blow-up follows. In blow-up analysis, one problematic issue is that we have to deal with high order nonlinear term uk2u3x to obtain accurate estimate. Luckily, we overcome the problem by finding a new L2k estimate uxL2kec0tu0xL2k+c0t (see Lemma 4.1). It is shown in [18] that Eq (1.1) has peakon travelling solution u(t,x)=c1ke|xct|. Follow the Definition 2.1, we show that the peakon solutions are global weak solutions.

    The rest of this paper is organized as follows. For the convenience, Section 2 give some preliminaries. Two blow-up criteria are established in Section 3. Section 4 give two blow-up phenomena. In Section 5, we prove that the peakon solutions are global weak solutions.

    We rewrite Problems (1.1) and (1.2) as follows

    ut+ukux=x(12x)1[uk+1+2k12uk1u2x](12x)1[k12uk2u3x], (2.1)
    u(0,x)=u0(x), (2.2)

    which is also equivalent to

    yt+(k+1)uk1uxy+ukyx=0, (2.3)
    y=uuxx, (2.4)
    u(0,x)=u0(x),y0=u0u0xx. (2.5)

    Recall that

    (12x)1f=Gf,whereG(x)=12e|x|

    and denotes the convolution product on R, defined by

    (fG)(x)=Rf(y)G(xy)dy. (2.6)

    Lemma 2.1. Given initial data u0Hs, s>32, the function u is said to be a weak solution to the initial-value Problem (1.1) and (1.2) if it satisfies the following identity

    T0Ruφt1k+1uk+1φxG(uk+1+2k12uk1u2x)φxG(k12uk2u3x)φdxdtRu0(x)φ(0,x)dx=0 (2.7)

    for any smooth test function φ(t,x)Cc([0,T)×R). If u is a weak solution on [0,T) for every T>0, then it is called a global weak solution.

    The characteristics q(t,x) relating to (2.3) is governed by

    qt(t,x)=uk(t,q(t,x)),t[0,T)
    q(0,x)=x,xR.

    Applying the classical results in the theory of ordinary differential equations, one can obtain that the characteristics q(t,x)C1([0,T)×R) with qx(t,x)>0 for all (t,x)[0,T)×R. Furthermore, it is shown in [21] that the potential y=uuxx satisfies

    y(t,q(t,x))q2x(t,x)=y0(x)et0(k1)uk1ux(τ,q(τ,x))dτ. (2.8)

    In this section, we investigate blow-up criteria and a lower bound of the maximal existence time. Now, we firstly give the first blow-up criterion.

    Theorem 3.1. Let u0Hs(R) with s>32. Let T>0 be the maximum existence time of the solution u to the Problem (1.1) and (1.2) with the initial data u0. Then the correspondingsolution u blows up in finite time if and only if

    limtTinfxRuk1ux=.

    Proof. Applying 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 Problem (1.1) and (1.2) with initial data u0H3(R). Due to y=uuxx, by direct computation, one has

    y2L2=R(uuxx)2dx=R(u2+2u2x+u2xx)dx. (3.1)

    So,

    u2H2≤∥y2L22u2H2. (3.2)

    Multiplying Eq (2.3) by 2y and integrating by parts, we obtain

    ddtRy2dx=2Ryytdx=2(k+1)Ruk1uxy2dxR2ukyyxdx=(k+2)Ruk1uxy2dx. (3.3)

    If there is a M>0 such that uk1ux>M, from (3.3) we deduce

    ddtRy2dx(k+2)MRy2dx. (3.4)

    By virtue of Gronwall's inequality, one has

    y2L2=Ry2dxe(k+2)Mty02L2. (3.5)

    This completes the proof of Theorem 3.1.

    Now we give the second blow-up criterion.

    Theorem 3.2. Let u0Hs(R) with s>32. Let T>0 be the maximum existence time of the solution u to the Problem (1.1) with the initial data u0. Then the correspondingsolution u blows up in finite time if and only if

    limtTinfxRux∣=.

    To complete the proof of Theorem 3.2, the following two lemmas is essential.

    Lemma 3.1. ([11]) The following estimates hold

    (i) For s0,

    fgHsC(fHsgL+fLgHs). (3.6)

    (ii) For s>0,

    fxgHsC(fHs+1gL+fLxgHs). (3.7)

    Lemma 3.2. ([20]) Let r>0. If uHrW1, and vHr1L, then

    [Λr,u]vL2C(uxLΛr1vL2+ΛruL2vL), (3.8)

    where Λ=(12x)12.

    Proof. Applying Λr with r1 to two sides of Eq (2.1) and multiplying by Λru and integrating on R

    12ddtR(Λru)2=RΛr(ukux)ΛrudxRΛrf(u)Λrudx. (3.9)

    where f(u)=x(12x)1[uk+1+2k12uk1u2x]+(12x)1[k12uk2u3x].

    Notice that

    RΛr(ukux)Λrudx=R[Λr,uk]uxΛrudx+RukΛruxΛrudx.≤∥[Λr,uk]uxL2ΛruL2+cuk1LuxLu2HrcuHr(uk1LuxLuHr+ukHruxL)+cuk1LuxLu2HrcuxLu2Hr, (3.10)

    where Lemmas 3.1, 3.2 and the inequality ukHrkuk1LuHr were used.

    In similar way, from Lemmas 3.1 and 3.2, we have

    RΛrf(u)Λrudx∣≤∥uHrΛrf(u)L2. (3.11)
    Λrf(u)L2≤∥Λrx(12x)1[uk+1+2k12uk1u2x]L2+Λr(12x)1[k12uk2u3x]L2c(uk+1Hr1+uk1u2xHr1+uk2u3xHr2). (3.12)

    Notice that

    uk1u2xHr1=∥Λr1uk1u2xL2≤∥[Λr1,uk1]u2xL2+uk1Λr1u2xL2c(ux2LuHr+uxLuHr) (3.13)

    and

    uk2u3xHr2=∥Λr2uk2u3xL2≤∥[Λr2,uk2]u3xL2+uk2Λr2u3xL2≤∥uk3LuxLu3xHr1+ux3Luk2Hr1+uk2Lu3xHr1c(ux3LuHr+cux2LuHr). (3.14)

    Thus, we obtain

    RΛrf(u)Λrudx∣≤cu2Hr(1+uxL+ux2L+ux3L). (3.15)

    It follows from (3.9), (3.10) and (3.15) that

    ddtu2Hrcu2Hr(1+uxL+ux2L+ux3L). (3.16)

    Therefore, if there exists a positive number M such that uxLM. The Gronwall' inequality gives rise to

    u2Hrcu02Hre(1+M+M2+M3)t. (3.17)

    This completes the proof of Theorem 3.2.

    Theorem 3.3. Assume that u0Hs with s>32, N=max{3,8k36}. Let ux(0)L<, T>0 be the maximum existence time of the solutionu to (2.1) and (2.2) with the initial data u0. Then T satisfies

    T12Nu0k2H1(u(0)L+ux(0)L)2.

    Proof. Notice that the Eq (2.1) is equivalent to the following equation

    ut+ukux+xG[uk+1+2k12uk1u2x]+G[k12uk2u3x]=0, (3.18)

    where G(x)=12e|x| is the Green function of (12x)1. Multiplying the above equation by u2n1 and integrating the resultant with respect to x, in view of Hölder's inequality, we obtain

    Ru2n1utdx=12nddtu2nL2n=∥u2n1L2nddtuL2n, (3.19)
    Ru2n1Gx[uk+1+2k12uk1u2x]dxcu2n1L2n(ukLuL2n+2k12ux2Luk2LuL2n), (3.20)

    and

    Ru2n1G[k12uk2u3x]dxk12u2n1L2nux2Luk2LuxL2n. (3.21)

    Combining (3.19)–(3.21), integrating over [0,t], it follows that

    uL2n≤∥u(0)L2n+t0uL2n(ukL+2k12uk2Lux2L)dt+k12t0ux2Luk2LuxL2ndt. (3.22)

    Letting n tend to infinity in the above inequality, we have

    uL≤∥u(0)L+t0uk2L(u3L+2k12uLux2L+k12ux3L)dx. (3.23)

    Differentiating (3.18) with respect to x, we obtain

    utx+kuk1u2x+ukuxxuk+12k12uk1u2x+G[uk+1+2k12uk1u2x]+Gx[k12uk2u3x]=0. (3.24)

    Multiplying the above equation by u2n1x and integrating the resultant with respect to x over R, still in view of Hölder's inequality, we get following estimates

    Ru2n1xutxdx=12nddtux2nL2n=∥ux2n1L2nddtuxL2n, (3.25)
    Rkuk1u2n+1xdx∣≤kux2Lux2n1L2nuk2LuL2n, (3.26)
    Ruku2n1xuxxdx∣≤k2nux2n1L2nux2Luk2LuL2n, (3.27)
    Ruk+1u2n1xdx∣≤∥ux2n1L2nukLuL2n, (3.28)
    R2k12uk1u2n+1xdx2k12ux2n1L2nux2Luk2LuL2n, (3.29)
    Ru2n1xG[uk+1+2k12uk1u2x]dx≤∥ux2n1L2n(ukLuL2n+2k12ux2Luk2LuL2n), (3.30)

    and

    Ru2n1xGx[k12uk2u3x]dxk12ux2n1L2nux2Luk2LuxL2n. (3.31)

    Combining (3.25)–(3.31), it gives rise to

    ddtuxL2n≤∥uk2L((k2n+3k1)ux2LuL2n+u2LuL2n+k12ux2LuxL2n). (3.32)

    Integrating (3.32) over [0,t] and letting n tend to infinity, it follows that

    uxL≤∥ux(0)L+t0uk2L(2u3L+(3k1)uLux2L+k12ux3L)dτ. (3.33)

    Combining (3.23) and (3.33), we deduce that

    uL+uxL≤∥u(0)L+ux(0)L+t0uk2L(3u3L+(8k3)2uLux2L+(k1)ux3L)dτ. (3.34)

    Define h(t)=∥uL+uxL and N=max{3,8k36}. Then h(0)=∥u(0)L+ux(0)L. One can obtain

    uxLh(t)h(0)+t0Nu0k2H1h3(t)dτ. (3.35)

    Solving (3.35), we get

    uxLh(t)h(0)12h2(0)Nu0k2H1t. (3.36)

    Therefore, let T=12Nu0k2H1(u(0)L+ux(0)L)2, for all tT, uxLh(t) holds.

    For the convenience, we firstly give several Lemmas.

    Lemma 4.1. (L2k estimate)Let u0Hs, s3/2 and u0xL2k<, k=2p+1, p is nonnegative integer. Let T be the lifespan of the solution to problem (1.1). The estimate

    uxL2kec0t(u0xL2k+c0t)

    holds for t[0,T).

    Proof. Differentiating the first equation of Problem (2.1) respect with to x, we have

    utx+12uk1u2x+ukuxx=uk+1(12x)1(uk+1+2k12uk1u2x)x(12x)1(k12uk2u3x). (4.1)

    Multiplying the above equation by 2ku2k1x and integrating the resultant over R, we obtain

    ddtRu2kxdx=ddtux2kL2k=2kux2k1L2kddtuxL2kc0ux2k1L2k+c0u2k1xL1u3xL1. (4.2)

    In view of Hölder's inequality, we derive the following estimates

    u2k1xL1(Ru2kxdx)12(Ru2k2xdx)12, (4.3)
    u2k1xL1(Ru2kxdx)12+14(Ru2k4xdx)14, (4.4)
    u2k1xL1(Ru2kxdx)12+14+18(Ru2k8xdx)18, (4.5)

    and

    u2k1xL1(Ru2kxdx)12+14+18+116(Ru2k16xdx)116, (4.6)

    On the other hand,

    u3xL1(Ru2xdx)12(Ru4xdx)12, (4.7)
    u3xL1(Ru2xdx)12+14(Ru6xdx)14, (4.8)
    u3xL1(Ru2xdx)12+14+18(Ru10xdx)18, (4.9)

    and

    u3xL1(Ru2xdx)12+14+18+116(Ru18xdx)116, (4.10)

    Therefore, if 2k2=2 (k=2), combining (4.3) and (4.7), we have

    u2k1xL1u3xL1≤∥u0H1(R)Ru2kxdx. (4.11)

    If 2k4=2 (k=3), combining (4.4) and (4.8), we have

    u2k1xL1u3xL1≤∥u0H1(R)Ru2kxdx. (4.12)

    If 2k8=2 (k=5), combining (4.5) and (4.9), we have

    u2k1xL1u3xL1≤∥u0H1(R)Ru2kxdx. (4.13)

    and if 2k16=2 (k=9), combining (4.6) and (4.10), we have

    u2k1xL1u3xL1≤∥u0H1(R)Ru2kxdx. (4.14)

    Therefore, if and only if k=2p+1, p is nonnegative integer, the following inequality follows

    2kux2k1L2kddtuxL2kc0ux2k1L2k+c0u2k1xL1u3xL1c0ux2k1L2k+c0ux2kL2k, (4.15)

    where c0=c0(u0H1(R)).

    ddtuxL2kc0+c0uxL2k. (4.16)

    In view of Gronwall's inequality, we have for t[0,T)

    uxL2kec0t(u0xL2k+c0t). (4.17)

    This completes the proof of Lemma 4.1.

    Remark. In case of p=0, then k=2, and we obtain

    uxL4ec0t(u0xL4+c0t), (4.18)

    which is exactly the same as the Lemma 3.2 in [23].

    Lemma 4.2. Given that u0Hs, s3/2. Let u0xL2k<, k=2p+1, p is nonnegative integer and T be the lifespan of the solution to Problem (1.1) and (1.2). The estimate

    R|uk2u3x|dxα:=(ec0T(u0xL2k+c0T))3u0k2H1

    holds.

    Proof. Applying Hölder'inequality, Lemma 4.1 and (1.3), we have

    R|uk2u3x|dx(R(u3x)2k3dx)32k(R(uk2)2k2k3dx)2k32k(R(u2kx)dx)32k(Ru2k(k2)2k3dx)2k32k(ec0tu0xL2k+c0t)3u(2k(k2)2k32)2k32kL(Ru2dx)2k32k(ec0tu0xL2k+c0t)3u0k2H1.

    Lemma 4.3. Given that u0Hs, s3. For k=2p+1 and p is nonnegative integer. Then

    u(t,q(t,x1))>0,for0<t<T1:=u0(x1)c0+cα.

    Proof. From the Young's inequality, Lemma 4.2 and (1.3), it follows that

    u(t)∣=∣x(12x)1(uk+1+2k12uk1u2x)+(12x)1(k12uk2u3x)≤∣x(12x)1(uk+1+2k12uk1u2x)+(12x)1(k12uk2u3x)c0+cα. (4.19)

    Therefore,

    c0cαu(t)c0+cα.

    Integrating over the time interval [0,t] yields

    u0(x1)[c0+cα]tu(t,q(t,x1))u0(x1)+[c0+cα]t. (4.20)

    So,

    u(t,q(t,x1))>0,for0<t<T1:=u0(x1)c0+cα.

    Theorem 4.1. Let u0xL2k<, k=2p+1, p is nonnegative integer and u0Hs(R) for s>32. Suppose that there exist some 0<λ<1 and x1R such that 12uk10(2u20u20x)+C20, 2u0<u0x, u0(x1)>0 and

    ln((λu0(x1))k12(2u20u20x)+2C(λu0(x1))k12(2u20u20x)2C)(1λ)u0(x1)C2(λu0(x1))k12.

    Then the corresponding solution u(t,x) blows up in finite time T with

    TT0=12(λu0(x1))k12Cln((λu0(x1))k12(2u20u20x)+2C(λu0(x1))k12(2u20u20x)2C),

    where C=cα+c0.

    Proof. We track the dynamics of P(t)=(2uux)(t,q(t,x1)) and Q(t)=(2u+ux)(t,q(t,x1)) along the characteristics

    P(t)=2(ut+uxqt)(utx+uxxqt)=12uk1(2u2u2x)2x(12x)1(uk+1+2k12uk1u2x)+2(12x)1(k12uk2u3x)+(12x)1(uk+1+2k12uk1u2x)x(12x)1(k12uk2u3x)12uk1PQC2 (4.21)

    and

    Q(t)=2(ut+uxqt)+(utx+uxxqt)=12uk1(2u2u2x)2x(12x)1(uk+1+2k12uk1u2x)+2(12x)1(k12uk2u3x)(12x)1(uk+1+2k12uk1u2x)+x(12x)1(k12uk2u3x)12uk1PQ+C2. (4.22)

    Then we obtain

    P(t)12uk1PQC2,Q(t)12uk1PQ+C2. (4.23)

    The expected monotonicity conditions on P and Q indicate that we would like to have

    12uk1PQ+C2<0. (4.24)

    It is shown from assumptions of Theorem 4.1 that the initial data satisfies

    12uk10(2u20u20x)+C2<0,2u0<u0x. (4.25)

    Therefore, along the characteristics emanating from x1, the following inequalities hold

    12uk10P(0)Q(0)+C2<0,P(0)>0,Q(0)<0 (4.26)

    and

    P(0)>0,Q(0)<0. (4.27)

    Therefore over the time of existence the following inequalities always hold

    P(t)>0,Q(t)<0. (4.28)

    Letting h(t)=PQ(t) and using the estimate QP2h(t), we have

    h(t)=PQ+PQ2PQ(12uk1PQ+C2)QP(12uk1PQ+C2)2PQ=(12uk1PQ+C2)(PQ)2PQ12uk1h2C2, (4.29)

    We focus on the time interval 0tT2:=(1λ)u0(x1)c0+cα, it implies that

    0<λu0(x1)u(t,q(t,x1))(2λ)u0(x1). (4.30)

    Solving for 0tT2

    h(t)12λk1u0(x1)k1h2C2, (4.31)

    we obtain

    ln(λu0(x1))k12h2C(λu0(x1))k12h+2Cln((λu0(x1))k12h02C(λu0(x1))k12h0+2C)+2(λu0(x1))k12Ct, (4.32)

    It is observed from assumption of Theorem 4.1 that T0<T2, (4.32) implies that h+ as tT

    TT0=12(λu0(x1))k12Cln((λu0(x1))k12h0+2C(λu0(x1))k12h02C). (4.33)

    Theorem 4.2. Let u0xL2k<, k=2p+1, p is nonnegative integer and u0Hs(R) for s>32. Suppose that there exist some 0<λ<1 and x2R such that u0(x2)>0, u0x(x2)<ba and

    ln(u0x(x2)+bau0x(x2)ba)2(1λ)u0(x2)ba, (4.34)

    where a=k2λk1uk10(x2) and b=c0+cα.Then the corresponding solution u(t,x) blows up in finite time T with

    TT4=12abln(u0x(x2)+bau0x(x2)ba), (4.35)

    Proof. Now, we prove the blow-up phenomenon along the characteristics q(t,x2). From (2.1), it follows that

    u(t)=x(12x)1(uk+1+2k12uk1u2x)(12x)1(k12uk2u3x), (4.36)

    and

    ux(t)=12uk1u2x+uk+1(12x)1(uk+1+2k12uk1u2x)x(12x)1(k12uk2u3x). (4.37)

    Setting M(t)=u(t,q(t,x1)) and using the Young's inequality, Lemmas 4.2 and 4.3 and (2.8) again, from (4.37), we get

    Mx(t)12Mk1M2x+b. (4.38)

    where b=c0+cα. Now, we focus on the time interval 0tT3:=(1λ)u0(x2)c0+cα, it implying that

    0<λu0(x2)u(t,q(t,x2))(2λ)u0(x2). (4.39)

    Therefore, for 0tT3, we deduce from (4.38) that

    Mx(t)aM2x+b, (4.40)

    where a=12λk1uk10(x2).

    It is observed from assumption of Theorem 4.2 that u0x(x2)<ba and T4<T3. Solving (4.40) results in

    MxastT, (4.41)

    where TT4=12abln(u0x(x2)bau0x(x2)+ba).

    In this section, we will turn our attention to peakon solution for the Problem (1.1).

    Theorem 5.1. The peakon function of the form

    u(t,x)=c1ke|xct|,c0isarbitaryconstant, (5.1)

    is a global weak solution to (1.1) and (1.2) in the sense of Definition 2.1.

    Proof. Let u=ae|xct| be peakon solution for the Problems (1.1) and (1.2), where a0 is an undetermined constant. We firstly claim that

    ut=asign(xct)u,ux=sign(xct)u. (5.2)

    Hence, using (2.6), (5.2) and integration by parts, we derive that

    T0Ruφt+1k+1uk+1φxdxdt+Ru0(x)φ(0,x)dx=T0Rφ(ut+ukux)dxdt=T0Rφsign(xct)(cuuk+1)dxdt. (5.3)

    On the other hand,

    T0RG(uk+1+2k12uk1u2x)φxG[k12uk2u3x]φdxdt=T0RφGx[2k12uk1u2x]φp[k12uk2u3x(k+1)ukux]dxdt. (5.4)

    Directly calculate

    k12uk2u3x+(k+1)ukux=(k+1)uk(sign(xct)u)+k12uk2(sign3(xct)u3)=[(k+1)+k12]sign(xct)uk+1=[k12(k+1)+1]x(uk+1). (5.5)

    Therefore, we obtain

    T0RG(uk+1+2k12uk1u2x)φxG[k12uk2u3x]φdxdt=T0RφGx[2k12uk1u2x(k12(k+1)+1)uk+1]dxdt=T0RφGx(k(k+2)k+1uk+1)dxdt. (5.6)

    Note that Gx=12sign(x)e|x|. For x>ct,

    Gx[(k(k+2)k+1uk+1]=12Rsign(xy)e|xy|(k(k+2)k+1ak+1e(k+1)|yct|dy=12(ct+xct+x)sign(xy)e|xy|k(k+2)k+1ak+1e(k+1)|yct|dy=I1+I2+I3. (5.7)

    We directly compute I1 as follows

    I1=12ctsign(xy)e|xy|k(k+2)k+1ak+1e(k+1)|yct|dy=12k(k+2)k+1ak+1ctex(k+1)cte(k+2)ydy=12k(k+2)k+1ak+1ex(k+1)ctcte(k+2)ydy=12(k+2)k(k+2)k+1ak+1ex+ct. (5.8)

    In a similar procedure, we obtain

    I2=12xctsign(xy)e|xy|k(k+2)k+1ak+1e(k+1)|yct|dy=12k(k+2)k+1ak+1xctex+(k+1)ctekydy=12k(k+2)k+1ak+1ex+(k+1)ctxctekydy=12kk(k+2)k+1ak+1(e(k+1)(xct)ex+ct). (5.9)

    and

    I3=12xsign(xy)e|xy|k(k+2)k+1ak+1e(k+1)|yct|dy=12k(k+2)k+1ak+1xex+(k+1)cte(k+2)ydy=12k(k+2)k+1ak+1xe(k+2)ydy=12(k+2)k(k+2)k+1ak+1e(k+1)(xct). (5.10)

    Substituting (5.8)–(5.10) into (5.7), we deduce that for x>ct

    Gx[k(k+2)k+1uk+1]=2(k+1)k(k+2)Ωex+ct2(k+1)k(k+2)Ωe(k+1)(xct)=ak+1ex+ct+ak+1e(k+1)(xct), (5.11)

    where Ω=12k(k+2)k+1ak+1.

    For x<ct,

    Gx[(k(k+2)k+1uk+1]=12Rsign(xy)e|xy|(k(k+2)k+1ak+1e(k+1)|yct|dy=12(x+ctx+ct)sign(xy)e|xy|k(k+2)k+1ak+1e(k+1)|yct|dy=Δ1+Δ2+Δ3. (5.12)

    We directly compute Δ1 as follows

    Δ1=12xsign(xy)e|xy|k(k+2)k+1ak+1e(k+1)|yct|dy=12k(k+2)k+1ak+1xex(k+1)cte(k+2)ydy=12k(k+2)k+1ak+1ex(k+1)ctxe(k+2)ydy=12(k+2)k(k+2)k+1ak+1e(k+1)(xct). (5.13)

    In a similar procedure, one has

    Δ2=12ctxsign(xy)e|xy|k(k+2)k+1ak+1e(k+1)|yct|dy=12k(k+2)k+1ak+1ctxex(k+1)ctekydy=12k(k+2)k+1ak+1ex(k+1)ctctxekydy=12kk(k+2)k+1ak+1(e(k+1)(xct)+exct). (5.14)

    and

    Δ3=12ctsign(xy)e|xy|k(k+2)k+1ak+1e(k+1)|yct|dy=12k(k+2)k+1ak+1ctex+(k+1)cte(k+2)ydy=12k(k+2)k+1ak+1ex+(k+1)ctcte(k+2)ydy=12(k+2)k(k+2)k+1ak+1exct. (5.15)

    Therefore, from (5.13)–(5.15), we deduce that for x<ct

    Gx[k(k+2)k+1uk+1]=2(k+1)k(k+2)Ωexct+2(k+1)k(k+2)Ωe(k+1)(xct), (5.16)

    where Ω=12k(k+2)k+1ak+1.

    Due to u=ae|xct|,

    sign(xct)(cuuk+1)={acex+ct+ak+1e(k+1)(xct),forx>ct,acexctak+1e(k+1)(xct),forxct.

    To ensure that u=ae|xct| is a global weak solution of (2.1) in the sense of Definition 2.1, we let

    ak+1=ac, (5.17)

    Solving (5.17), we get

    a=c1k,c>0. (5.18)

    From (5.18), we derive that

    u=c1ke|xct|,c>0, (5.19)

    which along with (5.11), (5.16) and (5.17) gives rise to

    asign(xct)(cuuk+1)(t,x)Gx(k(k+2)k+1uk+1(t,x)=0. (5.20)

    Therefore, we conclude that

    T0Ruφt+1k+1uk+1φx+G(uk+1+2k12uk1u2x)φx+G(k12uk2u3x)φdxdt+Ru0(x)φ(0,x)dx=0, (5.21)

    for every test function φ(t,x)Cc([0,+)×R), which completes the proof of Theorem 5.1.

    This work is supported by the GuiZhou Province Science and Technology Basic Project [Grant number QianKeHe Basic [2020]1Y011], Department of Guizhou Province Education project [grant number QianJiaoHe KY Zi [2019]124] and the GuiZhou Province Science and Technology Plan Project [Grant number QianKeHe Platform Talents [2018]5784-09].

    There are no conflict of interest.



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