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
ut−utxx+(k+2)ukux−(k+1)uk−1uxuxx−ukuxxx=0, | (1.1) |
u(0,x)=u0(x), | (1.2) |
where k∈N0 (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
ut−utxx+3uux−2uxuxx−uuxxx=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−|x−ct|,c∈R |
and the multi-peakon solutions
u(t,x)=N∑i=1pi(t)e−|x−qi(t)|, |
where pi(t), qi(t) satisfy the Hamilton system [1]
dpidt=−∂H∂qi=∑i≠jpipjsign(qi−qj)e|qi−qj|, |
dqidt=−∂H∂pi=∑jpje|qi−qj| |
with the Hamiltonian H=12∑Ni,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
ut−utxx+4u2ux−3uuxuxx−u2uxxx=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+2u2u2x−13u4x)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=√2u−ux 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 uk−2u3x to obtain accurate estimate. Luckily, we overcome the problem by finding a new L2k estimate ∥ux∥L2k≤ec0t∥u0x∥L2k+c0t (see Lemma 4.1). It is shown in [18] that Eq (1.1) has peakon travelling solution u(t,x)=c1ke−|x−ct|. 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(1−∂2x)−1[uk+1+2k−12uk−1u2x]−(1−∂2x)−1[k−12uk−2u3x], | (2.1) |
u(0,x)=u0(x), | (2.2) |
which is also equivalent to
yt+(k+1)uk−1uxy+ukyx=0, | (2.3) |
y=u−uxx, | (2.4) |
u(0,x)=u0(x),y0=u0−u0xx. | (2.5) |
Recall that
(1−∂2x)−1f=G∗f,whereG(x)=12e−|x| |
and ∗ denotes the convolution product on R, defined by
(f∗G)(x)=∫Rf(y)G(x−y)dy. | (2.6) |
Lemma 2.1. Given initial data u0∈Hs, 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
∫T0∫Ruφt−1k+1uk+1φx−G∗(uk+1+2k−12uk−1u2x)φx−G∗(k−12uk−2u3x)φdxdt−∫Ru0(x)φ(0,x)dx=0 | (2.7) |
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.
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,x∈R. |
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=u−uxx satisfies
y(t,q(t,x))q2x(t,x)=y0(x)e∫t0(k−1)uk−1ux(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 u0∈Hs(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
limt→T−infx∈Ruk−1ux=−∞. |
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 u0∈H3(R). Due to y=u−uxx, by direct computation, one has
∥y∥2L2=∫R(u−uxx)2dx=∫R(u2+2u2x+u2xx)dx. | (3.1) |
So,
∥u∥2H2≤∥y∥2L2≤2∥u∥2H2. | (3.2) |
Multiplying Eq (2.3) by 2y and integrating by parts, we obtain
ddt∫Ry2dx=2∫Ryytdx=−2(k+1)∫Ruk−1uxy2dx−∫R2ukyyxdx=−(k+2)∫Ruk−1uxy2dx. | (3.3) |
If there is a M>0 such that uk−1ux>−M, from (3.3) we deduce
ddt∫Ry2dx≤(k+2)M∫Ry2dx. | (3.4) |
By virtue of Gronwall's inequality, one has
∥y∥2L2=∫Ry2dx≤e(k+2)Mt∥y0∥2L2. | (3.5) |
This completes the proof of Theorem 3.1.
Now we give the second blow-up criterion.
Theorem 3.2. Let u0∈Hs(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
limt→T−infx∈R∣ux∣=∞. |
To complete the proof of Theorem 3.2, the following two lemmas is essential.
Lemma 3.1. ([11]) The following estimates hold
(i) For s≥0,
∥fg∥Hs≤C(∥f∥Hs∥g∥L∞+∥f∥L∞∥g∥Hs). | (3.6) |
(ii) For s>0,
∥f∂xg∥Hs≤C(∥f∥Hs+1∥g∥L∞+∥f∥L∞∥∂xg∥Hs). | (3.7) |
Lemma 3.2. ([20]) Let r>0. If u∈Hr∩W1,∞ and v∈Hr−1∪L∞, then
∥[Λr,u]v∥L2≤C(∥ux∥L∞∥Λr−1v∥L2+∥Λru∥L2∥v∥L∞), | (3.8) |
where Λ=(1−∂2x)12.
Proof. Applying Λr with r≥1 to two sides of Eq (2.1) and multiplying by Λru and integrating on R
12ddt∫R(Λru)2=−∫RΛr(ukux)Λrudx−∫RΛrf(u)Λrudx. | (3.9) |
where f(u)=∂x(1−∂2x)−1[uk+1+2k−12uk−1u2x]+(1−∂2x)−1[k−12uk−2u3x].
Notice that
∫RΛr(ukux)Λrudx=∫R[Λr,uk]uxΛrudx+∫RukΛruxΛrudx.≤∥[Λr,uk]ux∥L2∥Λru∥L2+c∥u∥k−1L∞∥ux∥L∞∥u∥2Hr≤c∥u∥Hr(∥u∥k−1L∞∥ux∥L∞∥u∥Hr+∥uk∥Hr∥ux∥L∞)+c∥u∥k−1L∞∥ux∥L∞∥u∥2Hr≤c∥ux∥L∞∥u∥2Hr, | (3.10) |
where Lemmas 3.1, 3.2 and the inequality ∥uk∥Hr≤k∥u∥k−1L∞∥u∥Hr were used.
In similar way, from Lemmas 3.1 and 3.2, we have
∣∫RΛrf(u)Λrudx∣≤∥u∥Hr∥Λrf(u)∥L2. | (3.11) |
∥Λrf(u)∥L2≤∥Λr∂x(1−∂2x)−1[uk+1+2k−12uk−1u2x]∥L2+∥Λr(1−∂2x)−1[k−12uk−2u3x]∥L2≤c(∥uk+1∥Hr−1+∥uk−1u2x∥Hr−1+∥uk−2u3x∥Hr−2). | (3.12) |
Notice that
∥uk−1u2x∥Hr−1=∥Λr−1uk−1u2x∥L2≤∥[Λr−1,uk−1]u2x∥L2+∥uk−1Λr−1u2x∥L2≤c(∥ux∥2L∞∥u∥Hr+∥ux∥L∞∥u∥Hr) | (3.13) |
and
∥uk−2u3x∥Hr−2=∥Λr−2uk−2u3x∥L2≤∥[Λr−2,uk−2]u3x∥L2+∥uk−2Λr−2u3x∥L2≤∥u∥k−3L∞∥ux∥L∞∥u3x∥Hr−1+∥ux∥3L∞∥uk−2∥Hr−1+∥uk−2∥L∞∥u3x∥Hr−1≤c(∥ux∥3L∞∥u∥Hr+c∥ux∥2L∞∥u∥Hr). | (3.14) |
Thus, we obtain
∣∫RΛrf(u)Λrudx∣≤c∥u∥2Hr(1+∥ux∥L∞+∥ux∥2L∞+∥ux∥3L∞). | (3.15) |
It follows from (3.9), (3.10) and (3.15) that
ddt∥u∥2Hr≤c∥u∥2Hr(1+∥ux∥L∞+∥ux∥2L∞+∥ux∥3L∞). | (3.16) |
Therefore, if there exists a positive number M such that ∥ux∥L∞≤M. The Gronwall' inequality gives rise to
∥u∥2Hr≤c∥u0∥2Hre(1+M+M2+M3)t. | (3.17) |
This completes the proof of Theorem 3.2.
Theorem 3.3. Assume that u0∈Hs with s>32, N=max{3,8k−36}. 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
T≤12N∥u0∥k−2H1(∥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+2k−12uk−1u2x]+G∗[k−12uk−2u3x]=0, | (3.18) |
where G(x)=12e−|x| is the Green function of (1−∂2x)−1. Multiplying the above equation by u2n−1 and integrating the resultant with respect to x, in view of Hölder's inequality, we obtain
∫Ru2n−1utdx=12nddt∥u∥2nL2n=∥u∥2n−1L2nddt∥u∥L2n, | (3.19) |
∫Ru2n−1Gx∗[uk+1+2k−12uk−1u2x]dx≤c∥u∥2n−1L2n(∥u∥kL∞∥u∥L2n+2k−12∥ux∥2L∞∥u∥k−2L∞∥u∥L2n), | (3.20) |
and
∫Ru2n−1G∗[k−12uk−2u3x]dx≤k−12∥u∥2n−1L2n∥ux∥2L∞∥u∥k−2L∞∥ux∥L2n. | (3.21) |
Combining (3.19)–(3.21), integrating over [0,t], it follows that
∥u∥L2n≤∥u(0)∥L2n+∫t0∥u∥L2n(∥u∥kL∞+2k−12∥u∥k−2L∞∥ux∥2L∞)dt+k−12∫t0∥ux∥2L∞∥u∥k−2L∞∥ux∥L2ndt. | (3.22) |
Letting n tend to infinity in the above inequality, we have
∥u∥L∞≤∥u(0)∥L∞+∫t0∥u∥k−2L∞(∥u∥3L∞+2k−12∥u∥L∞∥ux∥2L∞+k−12∥ux∥3L∞)dx. | (3.23) |
Differentiating (3.18) with respect to x, we obtain
utx+kuk−1u2x+ukuxx−uk+1−2k−12uk−1u2x+G∗[uk+1+2k−12uk−1u2x]+Gx∗[k−12uk−2u3x]=0. | (3.24) |
Multiplying the above equation by u2n−1x and integrating the resultant with respect to x over R, still in view of Hölder's inequality, we get following estimates
∫Ru2n−1xutxdx=12nddt∥ux∥2nL2n=∥ux∥2n−1L2nddt∥ux∥L2n, | (3.25) |
∣∫Rkuk−1u2n+1xdx∣≤k∥ux∥2L∞∥ux∥2n−1L2n∥u∥k−2L∞∥u∥L2n, | (3.26) |
∣∫Ruku2n−1xuxxdx∣≤k2n∥ux∥2n−1L2n∥ux∥2L∞∥u∥k−2L∞∥u∥L2n, | (3.27) |
∣∫Ruk+1u2n−1xdx∣≤∥ux∥2n−1L2n∥u∥kL∞∥u∥L2n, | (3.28) |
∣∫R2k−12uk−1u2n+1xdx∣≤2k−12∥ux∥2n−1L2n∥ux∥2L∞∥u∥k−2L∞∥u∥L2n, | (3.29) |
∣∫Ru2n−1xG∗[uk+1+2k−12uk−1u2x]dx∣≤∥ux∥2n−1L2n(∥u∥kL∞∥u∥L2n+2k−12∥ux∥2L∞∥u∥k−2L∞∥u∥L2n), | (3.30) |
and
∣∫Ru2n−1xGx∗[k−12uk−2u3x]dx∣≤k−12∥ux∥2n−1L2n∥ux∥2L∞∥u∥k−2L∞∥ux∥L2n. | (3.31) |
Combining (3.25)–(3.31), it gives rise to
ddt∥ux∥L2n≤∥u∥k−2L∞((k2n+3k−1)∥ux∥2L∞∥u∥L2n+∥u∥2L∞∥u∥L2n+k−12∥ux∥2L∞∥ux∥L2n). | (3.32) |
Integrating (3.32) over [0,t] and letting n tend to infinity, it follows that
∥ux∥L∞≤∥ux(0)∥L∞+∫t0∥u∥k−2L∞(2∥u∥3L∞+(3k−1)∥u∥L∞∥ux∥2L∞+k−12∥ux∥3L∞)dτ. | (3.33) |
Combining (3.23) and (3.33), we deduce that
∥u∥L∞+∥ux∥L∞≤∥u(0)∥L∞+∥ux(0)∥L∞+∫t0∥u∥k−2L∞(3∥u∥3L∞+(8k−3)2∥u∥L∞∥ux∥2L∞+(k−1)∥ux∥3L∞)dτ. | (3.34) |
Define h(t)=∥u∥L∞+∥ux∥L∞ and N=max{3,8k−36}. Then h(0)=∥u(0)∥L∞+∥ux(0)∥L∞. One can obtain
∥ux∥L∞≤h(t)≤h(0)+∫t0N∥u0∥k−2H1h3(t)dτ. | (3.35) |
Solving (3.35), we get
∥ux∥L∞≤h(t)≤h(0)√1−2h2(0)N∥u0∥k−2H1t. | (3.36) |
Therefore, let T=12N∥u0∥k−2H1(∥u(0)∥L∞+∥ux(0)∥L∞)2, for all t≤T, ∥ux∥L∞≤h(t) holds.
For the convenience, we firstly give several Lemmas.
Lemma 4.1. (L2k estimate)Let u0∈Hs, s≥3/2 and ∥u0x∥L2k<∞, k=2p+1, p is nonnegative integer. Let T be the lifespan of the solution to problem (1.1). The estimate
∥ux∥L2k≤ec0t(∥u0x∥L2k+c0t) |
holds for t∈[0,T).
Proof. Differentiating the first equation of Problem (2.1) respect with to x, we have
utx+12uk−1u2x+ukuxx=uk+1−(1−∂2x)−1(uk+1+2k−12uk−1u2x)−∂x(1−∂2x)−1(k−12uk−2u3x). | (4.1) |
Multiplying the above equation by 2ku2k−1x and integrating the resultant over R, we obtain
ddt∫Ru2kxdx=ddt∥ux∥2kL2k=2k∥ux∥2k−1L2kddt∥ux∥L2k≤c0∥ux∥2k−1L2k+c0∥u2k−1x∥L1∥u3x∥L1. | (4.2) |
In view of Hölder's inequality, we derive the following estimates
∥u2k−1x∥L1≤(∫Ru2kxdx)12(∫Ru2k−2xdx)12, | (4.3) |
∥u2k−1x∥L1≤(∫Ru2kxdx)12+14(∫Ru2k−4xdx)14, | (4.4) |
∥u2k−1x∥L1≤(∫Ru2kxdx)12+14+18(∫Ru2k−8xdx)18, | (4.5) |
and
∥u2k−1x∥L1≤(∫Ru2kxdx)12+14+18+116(∫Ru2k−16xdx)116, | (4.6) |
⋅⋅⋅⋅⋅⋅
On the other hand,
∥u3x∥L1≤(∫Ru2xdx)12(∫Ru4xdx)12, | (4.7) |
∥u3x∥L1≤(∫Ru2xdx)12+14(∫Ru6xdx)14, | (4.8) |
∥u3x∥L1≤(∫Ru2xdx)12+14+18(∫Ru10xdx)18, | (4.9) |
and
∥u3x∥L1≤(∫Ru2xdx)12+14+18+116(∫Ru18xdx)116, | (4.10) |
⋅⋅⋅⋅⋅⋅
Therefore, if 2k−2=2 (k=2), combining (4.3) and (4.7), we have
∥u2k−1x∥L1∥u3x∥L1≤∥u0∥H1(R)∫Ru2kxdx. | (4.11) |
If 2k−4=2 (k=3), combining (4.4) and (4.8), we have
∥u2k−1x∥L1∥u3x∥L1≤∥u0∥H1(R)∫Ru2kxdx. | (4.12) |
If 2k−8=2 (k=5), combining (4.5) and (4.9), we have
∥u2k−1x∥L1∥u3x∥L1≤∥u0∥H1(R)∫Ru2kxdx. | (4.13) |
and if 2k−16=2 (k=9), combining (4.6) and (4.10), we have
∥u2k−1x∥L1∥u3x∥L1≤∥u0∥H1(R)∫Ru2kxdx. | (4.14) |
⋅⋅⋅⋅⋅⋅
Therefore, if and only if k=2p+1, p is nonnegative integer, the following inequality follows
2k∥ux∥2k−1L2kddt∥ux∥L2k≤c0∥ux∥2k−1L2k+c0∥u2k−1x∥L1∥u3x∥L1≤c0∥ux∥2k−1L2k+c0∥ux∥2kL2k, | (4.15) |
where c0=c0(∥u0∥H1(R)).
ddt∥ux∥L2k≤c0+c0∥ux∥L2k. | (4.16) |
In view of Gronwall's inequality, we have for t∈[0,T)
∥ux∥L2k≤ec0t(∥u0x∥L2k+c0t). | (4.17) |
This completes the proof of Lemma 4.1.
Remark. In case of p=0, then k=2, and we obtain
∥ux∥L4≤ec0t(∥u0x∥L4+c0t), | (4.18) |
which is exactly the same as the Lemma 3.2 in [23].
Lemma 4.2. Given that u0∈Hs, s≥3/2. Let ∥u0x∥L2k<∞, 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|uk−2u3x|dx≤α:=(ec0T(∥u0x∥L2k+c0T))3∥u0∥k−2H1 |
holds.
Proof. Applying Hölder'inequality, Lemma 4.1 and (1.3), we have
∫R|uk−2u3x|dx≤(∫R(u3x)2k3dx)32k(∫R(uk−2)2k2k−3dx)2k−32k≤(∫R(u2kx)dx)32k(∫Ru2k(k−2)2k−3dx)2k−32k≤(ec0t∥u0x∥L2k+c0t)3∥u∥(2k(k−2)2k−3−2)2k−32kL∞(∫Ru2dx)2k−32k≤(ec0t∥u0x∥L2k+c0t)3∥u0∥k−2H1. |
Lemma 4.3. Given that u0∈Hs, s≥3. 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(1−∂2x)−1(uk+1+2k−12uk−1u2x)+(1−∂2x)−1(k−12uk−2u3x)∣≤∣−∂x(1−∂2x)−1(uk+1+2k−12uk−1u2x)∣+∣(1−∂2x)−1(k−12uk−2u3x)∣≤c0+cα. | (4.19) |
Therefore,
−c0−cα≤u′(t)≤c0+cα. |
Integrating over the time interval [0,t] yields
u0(x1)−[c0+cα]t≤u(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 ∥u0x∥L2k<∞, k=2p+1, p is nonnegative integer and u0∈Hs(R) for s>32. Suppose that there exist some 0<λ<1 and x1∈R such that 12uk−10(2u20−u20x)+C2≤0, √2u0<−u0x, u0(x1)>0 and
ln((λu0(x1))k−12(2u20−u20x)+√2C(λu0(x1))k−12(2u20−u20x)−√2C)≤(1−λ)u0(x1)√C√2(λu0(x1))k−12. |
Then the corresponding solution u(t,x) blows up in finite time T∗ with
T∗≤T0=1√2(λu0(x1))k−12Cln((λu0(x1))k−12(2u20−u20x)+√2C(λu0(x1))k−12(2u20−u20x)−√2C), |
where C=√cα+c0.
Proof. We track the dynamics of P(t)=(√2u−ux)(t,q(t,x1)) and Q(t)=(√2u+ux)(t,q(t,x1)) along the characteristics
P′(t)=√2(ut+uxqt)−(utx+uxxqt)=−12uk−1(2u2−u2x)−√2∂x(1−∂2x)−1(uk+1+2k−12uk−1u2x)+√2(1−∂2x)−1(k−12uk−2u3x)+(1−∂2x)−1(uk+1+2k−12uk−1u2x)−∂x(1−∂2x)−1(k−12uk−2u3x)≥−12uk−1PQ−C2 | (4.21) |
and
Q′(t)=√2(ut+uxqt)+(utx+uxxqt)=12uk−1(2u2−u2x)−√2∂x(1−∂2x)−1(uk+1+2k−12uk−1u2x)+√2(1−∂2x)−1(k−12uk−2u3x)−(1−∂2x)−1(uk+1+2k−12uk−1u2x)+∂x(1−∂2x)−1(k−12uk−2u3x)≤12uk−1PQ+C2. | (4.22) |
Then we obtain
P′(t)≥−12uk−1PQ−C2,Q′(t)≤12uk−1PQ+C2. | (4.23) |
The expected monotonicity conditions on P and Q indicate that we would like to have
12uk−1PQ+C2<0. | (4.24) |
It is shown from assumptions of Theorem 4.1 that the initial data satisfies
12uk−10(2u20−u20x)+C2<0,√2u0<−u0x. | (4.25) |
Therefore, along the characteristics emanating from x1, the following inequalities hold
12uk−10P(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 Q−P2≥h(t), we have
h′(t)=−P′Q+PQ′2√−PQ≥(12uk−1PQ+C2)Q−P(12uk−1PQ+C2)2√−PQ=−(12uk−1PQ+C2)(P−Q)2√−PQ≥12uk−1h2−C2, | (4.29) |
We focus on the time interval 0≤t≤T2:=(1−λ)u0(x1)c0+cα, it implies that
0<λu0(x1)≤u(t,q(t,x1))≤(2−λ)u0(x1). | (4.30) |
Solving for 0≤t≤T2
h′(t)≥12λk−1u0(x1)k−1h2−C2, | (4.31) |
we obtain
ln(λu0(x1))k−12h−√2C(λu0(x1))k−12h+√2C≥ln((λu0(x1))k−12h0−√2C(λu0(x1))k−12h0+√2C)+√2(λu0(x1))k−12Ct, | (4.32) |
It is observed from assumption of Theorem 4.1 that T0<T2, (4.32) implies that h→+∞ as t→T∗
T∗≤T0=1√2(λu0(x1))k−12Cln((λu0(x1))k−12h0+√2C(λu0(x1))k−12h0−√2C). | (4.33) |
Theorem 4.2. Let ∥u0x∥L2k<∞, k=2p+1, p is nonnegative integer and u0∈Hs(R) for s>32. Suppose that there exist some 0<λ<1 and x2∈R such that u0(x2)>0, u0x(x2)<−√ba and
ln(−u0x(x2)+√ba−u0x(x2)−√ba)≤2(1−λ)u0(x2)√ba, | (4.34) |
where a=k2λk−1uk−10(x2) and b=c0+cα.Then the corresponding solution u(t,x) blows up in finite time T∗∗ with
T∗∗≤T4=12√abln(−u0x(x2)+√ba−u0x(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(1−∂2x)−1(uk+1+2k−12uk−1u2x)−(1−∂2x)−1(k−12uk−2u3x), | (4.36) |
and
u′x(t)=−12uk−1u2x+uk+1−(1−∂2x)−1(uk+1+2k−12uk−1u2x)−∂x(1−∂2x)−1(k−12uk−2u3x). | (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
M′x(t)≤−12Mk−1M2x+b. | (4.38) |
where b=c0+cα. Now, we focus on the time interval 0≤t≤T3:=(1−λ)u0(x2)c0+cα, it implying that
0<λu0(x2)≤u(t,q(t,x2))≤(2−λ)u0(x2). | (4.39) |
Therefore, for 0≤t≤T3, we deduce from (4.38) that
M′x(t)≤−aM2x+b, | (4.40) |
where a=12λk−1uk−10(x2).
It is observed from assumption of Theorem 4.2 that u0x(x2)<−√ba and T4<T3. Solving (4.40) results in
Mx→−∞ast→T∗∗, | (4.41) |
where T∗∗≤T4=12√abln(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−|x−ct|,c≠0isarbitaryconstant, | (5.1) |
is a global weak solution to (1.1) and (1.2) in the sense of Definition 2.1.
Proof. Let u=ae−|x−ct| be peakon solution for the Problems (1.1) and (1.2), where a≠0 is an undetermined constant. We firstly claim that
ut=asign(x−ct)u,ux=−sign(x−ct)u. | (5.2) |
Hence, using (2.6), (5.2) and integration by parts, we derive that
∫T0∫Ruφt+1k+1uk+1φxdxdt+∫Ru0(x)φ(0,x)dx=−∫T0∫Rφ(ut+ukux)dxdt=−∫T0∫Rφsign(x−ct)(cu−uk+1)dxdt. | (5.3) |
On the other hand,
∫T0∫RG∗(uk+1+2k−12uk−1u2x)φx−G∗[k−12uk−2u3x]φdxdt=∫T0∫R−φGx∗[2k−12uk−1u2x]−φp∗[k−12uk−2u3x−(k+1)ukux]dxdt. | (5.4) |
Directly calculate
k−12uk−2u3x+(k+1)ukux=(k+1)uk(−sign(x−ct)u)+k−12uk−2(−sign3(x−ct)u3)=−[(k+1)+k−12]sign(x−ct)uk+1=[k−12(k+1)+1]∂x(uk+1). | (5.5) |
Therefore, we obtain
∫T0∫RG∗(uk+1+2k−12uk−1u2x)φx−G∗[k−12uk−2u3x]φdxdt=∫T0∫RφGx∗[−2k−12uk−1u2x−(k−12(k+1)+1)uk+1]dxdt=−∫T0∫RφGx∗(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]=−12∫Rsign(x−y)e−|x−y|(k(k+2)k+1ak+1e−(k+1)|y−ct|dy=−12(∫ct−∞+∫xct+∫∞x)sign(x−y)e−|x−y|k(k+2)k+1ak+1e−(k+1)|y−ct|dy=I1+I2+I3. | (5.7) |
We directly compute I1 as follows
I1=−12∫ct−∞sign(x−y)e−|x−y|k(k+2)k+1ak+1e−(k+1)|y−ct|dy=−12k(k+2)k+1ak+1∫ct−∞e−x−(k+1)cte(k+2)ydy=−12k(k+2)k+1ak+1e−x−(k+1)ct∫ct−∞e(k+2)ydy=−12(k+2)k(k+2)k+1ak+1e−x+ct. | (5.8) |
In a similar procedure, we obtain
I2=−12∫xctsign(x−y)e−|x−y|k(k+2)k+1ak+1e−(k+1)|y−ct|dy=−12k(k+2)k+1ak+1∫xcte−x+(k+1)cte−kydy=−12k(k+2)k+1ak+1e−x+(k+1)ct∫xcte−kydy=12kk(k+2)k+1ak+1(e−(k+1)(x−ct)−e−x+ct). | (5.9) |
and
I3=−12∫∞xsign(x−y)e−|x−y|k(k+2)k+1ak+1e−(k+1)|y−ct|dy=−12k(k+2)k+1ak+1∫∞xex+(k+1)cte−(k+2)ydy=−12k(k+2)k+1ak+1∫∞xe−(k+2)ydy=12(k+2)k(k+2)k+1ak+1e−(k+1)(x−ct). | (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)Ωe−x+ct−2(k+1)k(k+2)Ωe−(k+1)(x−ct)=−ak+1e−x+ct+ak+1e−(k+1)(x−ct), | (5.11) |
where Ω=−12k(k+2)k+1ak+1.
For x<ct,
Gx∗[(k(k+2)k+1uk+1]=−12∫Rsign(x−y)e−|x−y|(k(k+2)k+1ak+1e−(k+1)|y−ct|dy=−12(∫x−∞+∫ctx+∫∞ct)sign(x−y)e−|x−y|k(k+2)k+1ak+1e−(k+1)|y−ct|dy=Δ1+Δ2+Δ3. | (5.12) |
We directly compute Δ1 as follows
Δ1=−12∫x−∞sign(x−y)e−|x−y|k(k+2)k+1ak+1e−(k+1)|y−ct|dy=−12k(k+2)k+1ak+1∫x−∞e−x−(k+1)cte(k+2)ydy=−12k(k+2)k+1ak+1e−x−(k+1)ct∫x−∞e(k+2)ydy=−12(k+2)k(k+2)k+1ak+1e(k+1)(x−ct). | (5.13) |
In a similar procedure, one has
Δ2=−12∫ctxsign(x−y)e−|x−y|k(k+2)k+1ak+1e−(k+1)|y−ct|dy=12k(k+2)k+1ak+1∫ctxex−(k+1)ctekydy=12k(k+2)k+1ak+1ex−(k+1)ct∫ctxekydy=12kk(k+2)k+1ak+1(−e(k+1)(x−ct)+ex−ct). | (5.14) |
and
Δ3=−12∫∞ctsign(x−y)e−|x−y|k(k+2)k+1ak+1e−(k+1)|y−ct|dy=12k(k+2)k+1ak+1∫∞ctex+(k+1)cte−(k+2)ydy=12k(k+2)k+1ak+1ex+(k+1)ct∫∞cte−(k+2)ydy=12(k+2)k(k+2)k+1ak+1ex−ct. | (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)Ωex−ct+2(k+1)k(k+2)Ωe(k+1)(x−ct), | (5.16) |
where Ω=12k(k+2)k+1ak+1.
Due to u=ae−|x−ct|,
sign(x−ct)(cu−uk+1)={−ace−x+ct+ak+1e−(k+1)(x−ct),forx>ct,acex−ct−ak+1e(k+1)(x−ct),forx≤ct. |
To ensure that u=ae−|x−ct| 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−|x−ct|,c>0, | (5.19) |
which along with (5.11), (5.16) and (5.17) gives rise to
asign(x−ct)(cu−uk+1)(t,x)−Gx∗(k(k+2)k+1uk+1(t,x)=0. | (5.20) |
Therefore, we conclude that
∫T0∫Ruφt+1k+1uk+1φx+G∗(uk+1+2k−12uk−1u2x)φx+G∗(k−12uk−2u3x)φdxdt+∫Ru0(x)φ(0,x)dx=0, | (5.21) |
for every test function φ(t,x)∈C∞c([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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