This paper deals with the continuation of solutions to the generalized Camassa-Holm equation with higher-order nonlinearity beyond wave breaking. By introducing new variables, we transform the generalized Camassa-Holm equation to a semi-linear system and establish the global solutions to this semi-linear system, and by returning to the original variables, we obtain the existence of global conservative solutions to the original equation. We introduce a set of auxiliary variables tailored to a given conservative solution, which satisfy a suitable semi-linear system, and show that the solution for the semi-linear system is unique. Furthermore, it is obtained that the original equation has a unique global conservative solution. By Thom's transversality lemma, we prove that piecewise smooth solutions with only generic singularities are dense in the whole solution set, which means the generic regularity.
Citation: Li Yang, Chunlai Mu, Shouming Zhou, Xinyu Tu. The global conservative solutions for the generalized camassa-holm equation[J]. Electronic Research Archive, 2019, 27: 37-67. doi: 10.3934/era.2019009
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This paper deals with the continuation of solutions to the generalized Camassa-Holm equation with higher-order nonlinearity beyond wave breaking. By introducing new variables, we transform the generalized Camassa-Holm equation to a semi-linear system and establish the global solutions to this semi-linear system, and by returning to the original variables, we obtain the existence of global conservative solutions to the original equation. We introduce a set of auxiliary variables tailored to a given conservative solution, which satisfy a suitable semi-linear system, and show that the solution for the semi-linear system is unique. Furthermore, it is obtained that the original equation has a unique global conservative solution. By Thom's transversality lemma, we prove that piecewise smooth solutions with only generic singularities are dense in the whole solution set, which means the generic regularity.
In this paper we consider the continuation of solutions for the generalized Camassa-Holm (g-CH) equation
$ \begin{equation} \left\{ \begin{array}{llll} u_t-u_{xxt}+\frac{(m+2)(m+1)}{2}u^mu_x = \bigg(\frac{m}{2}u^{(m-1)}u_x^2+u^mu_{xx}\bigg)_x,&x\in \mathbb{R},t > 0,\\ u(x,0) = u_0(x), &x\in \mathbb{R}, \end{array} \right. \end{equation} $ | (1) |
where
The equation (1) was first proposed by Hakkaev and Kirchev in [18], and the local well-posedness of the Cauchy problem (1) was studied for the Sobolev spaces
In fact, the equation (1) is a natural generalization of the famous Camassa-Holm (CH) model
$ \begin{equation} u_t + u_{xxt} + 3uu_x = uu_{xxx}+2u_{x}u_{xx}. \end{equation} $ | (2) |
The Camassa-Holm equation first arisen in the context of hereditary symmetries was studied by Fokas and Fuchssteiner [14], but did not receive much attention until Camassa and Holm [9,7] derived it as a model of shallow water waves over a flat bottom. It has bi-Hamiltonian structure, infinitely many conservation laws and is completely integrable [8,7,14,19]. In addition, the stability of the smooth solitons and the orbital stability of the peaked solitons to (1) were established in [10] and [11] respectively. Particularly, the Camassa-Holm equation possesses solutions with presence of wave breaking (that is, the solution remains bounded while its slope becomes unbounded in finite time [6,12]). When these two waves collide at some time, the combined wave forms an infinite slope. After the collision, there are two things that happen: either two waves pass through each other with total energy preserved; or annihilate each other with a lose of energy. The solutions in the first case is called conservative, and the second case is called dissipative.
So far, the continuation of the solutions after wave breaking has been studied widely. Bressan and Constantin proved that the solution of the Camassa-Holm equation can be continued as either global conservative or global dissipative solutions [2,3]. Notice that, the conservative solutions are about preservation of the
In the case of a more general Camassa-Holm equation, the global existence and uniqueness to the solution are established in [22,25]. Moreover, for the Camassa-Holm equation with a forcing term
Similar to the well-known Camassa-Holm equation, the system (1) also models the peculiar wave breaking phenomena [24]. Therefore, in this paper, we still focus on the conservative case of equation (1.1) in
$ \begin{equation} \left\{ \begin{array}{llll} u_t+u^mu_x = -P_x,&x\in \mathbb{R},t > 0,\\ u(x,0) = u_0(x), &x\in \mathbb{R}, \end{array} \right. \end{equation} $ | (3) |
where
$ \begin{equation} P = G*\bigg(\frac{m}{2}u^{(m-1)}u_x^2+\frac{m(m+3)}{2(m+1)}\bigg). \end{equation} $ | (4) |
Motivated by [2,4,17], in this paper, we consider the global weak conservative solutions defined by as follows.
Definition 1.1. Let
(ⅰ) The map
(ⅱ) The solution
$ \begin{equation} \begin{split} &\int\int_\Omega\bigg( -u_x(\phi_t+u^m\phi_x)+(-\frac{m}{2}u^mu_x^2-\frac{m(m+3)}{2(m+1)}u^{m+1}+P) \phi\bigg)dxdt\\&\qquad+\int_{\mathbb{R}}u_{0x}\phi(0,x)dx = 0. \end{split} \end{equation} $ | (5) |
For a solution
There exists a family of Radon measure
$ \begin{equation} \begin{split} &\int_0^\infty \bigg(\int_\mathbb{R}(\phi_t+u^m\phi_x)d\mu_{(t)}+\int_\mathbb{R} 2u_x\bigg(\frac{m(m+3)}{2(m+1)}u^{m+1}-P\bigg)dx \bigg)dt\\&\qquad-\int_\mathbb{R}u_{0x}^2\phi(0,x)dx = 0. \end{split} \end{equation} $ | (6) |
Based on the characteristic, by introducing a new variables, we transform the equation (1) to a semi-linear system, and prove the semi-linear system has global solutions. Then by a reverse transformation, one can get the conservative solutions for equation (1). Our results are stated as follows.
Theorem 1.2. Let
(i)
(ii) The function
(iii) There exists a null set
(iv) The energy
$ E(t) = E(0)\quad \mathit{\text{for}} \; t\notin \mathcal{N},\quad E(t) < E(0)\; \mathit{\text{for}}\; t\in\mathcal{N}. $ |
(v) The continuous dependence of solutions to system (3) holds with the initial data belongs to
Theorem 1.3. Given any initial data
Remark 1. In fact, we know the process for the proof of the existence is an inverse, but the method here is an irreversible.
By virtue of the analysis of solutions along characteristic, we show that piecewise smooth solutions with only generic singularities are dense in the whole solution set. Using the Thom's transversality Lemma [1,15], we give the following generic regularity result.
Theorem 1.4. For any
Remark 2. The generic regularity is very interesting since it reflects the structure of singularities. Similar issue was first established for the variational wave equation[5], and later this method was applied to the Camassa-Holm equation [17].
This paper is organized as follows. In Section 2, we give the energy conservation laws and introduce a new set of independent and dependent variables. In Section 3, we first obtain a global conservative solution of the semi-linear system (27), and then by inverse transformation we prove the existence of the global conservative solution to equation (1). In Section 4, we establish the uniqueness of the characteristic curve through each initial point, and by considering the dynamics of a conservative solution along a characteristic, we obtain the proof of the uniqueness for the global conservation solution. In Section 5, the generic regularity of conservative solutions to equation (1) is investigated.
For smooth solutions, we claim that the total energy
$ \begin{equation} E(t) = \int_{\mathbb{R}}(u^2+u_x^2)dx \end{equation} $ | (7) |
is constant in time. In fact, by using
$ \begin{equation} u_{xt}+u^mu_{xx}+mu^{m-1}u_x^2 = \frac{m}{2}u^{m-1}u_x^2+\frac{m(m+3)}{2(m+1)}u^{m+1}-P. \end{equation} $ | (8) |
Multiplying (3) by
$ \begin{equation} (\frac{u^2}{2})_t+(\frac{1}{m+1}u^{m+1})_x+uP_x = 0, \end{equation} $ | (9) |
$ \begin{equation} (\frac{u_x^2}{2})_t+ (\frac{u^mu_x}{2})_x = \bigg(\frac{m(m+3)}{2(m+1)}u^{m+1}-P\bigg)u_x. \end{equation} $ | (10) |
It follows from (9)-(10) that
$ \begin{equation} \frac{d}{dt}E(t) = \frac{d}{dt}\int_{\mathbb{R}}(u^2+u_x^2)dx = 0. \end{equation} $ | (11) |
Therefore, the conservation law is given by
$ \begin{equation} E(t): = \int_{\mathbb{R}}(u^2+u_x^2)dx = E(0). \end{equation} $ | (12) |
Since
$ \begin{equation} \begin{split} &\|P(t)\|_{L^\infty},\; \|P_x(t)\|_{L^\infty}\\ &\quad\; \; \; \; \leq \|\frac{1}{2}e^{-|x|}\|_{L^\infty}\| \frac{m}{2}u^{m-1}u_x^2 +\frac{m(m+3)}{2(m+1)}u^{m+1}\|_{L^1}\leq \frac{m(m+3)}{4(m+1)}E(0)^\frac{m+1}{2}, \end{split} \end{equation} $ | (13) |
and
$ \begin{equation} \begin{split} &\|P(t)\|_{L^2},\; \|P_x(t)\|_{L^2}\\ &\quad\; \; \; \; \leq \|\frac{1}{2}e^{-|x|}\|_{L^2}\| \frac{m}{2}u^{m-1}u_x^2+\frac{m(m+3)}{2(m+1)}u^{m+1}\|_{L^1}\leq \frac{m(m+3)}{2(m+1)}E(0)^\frac{m+1}{2}. \end{split} \end{equation} $ | (14) |
Let
$ \begin{equation} \int_0^{\tilde{y}(\xi)}\big(1+\tilde{u}_x^2\big)dx = \xi. \end{equation} $ | (15) |
Then the characteristic map
$ \begin{equation} \frac{\partial}{\partial t}y(t,\xi) = u^{m}(t,y(t,\xi)), \quad y(0,\xi) = \tilde{y}(\xi). \end{equation} $ | (16) |
And the new variables
$ \begin{equation} \theta\doteq 2 \arctan u_x,\quad h \doteq 1+u_x^2\cdot\frac{\partial y}{\partial\xi}, \end{equation} $ | (17) |
$ \begin{equation} h(0,\xi)\equiv 1, \end{equation} $ | (18) |
$ \begin{equation} \frac{1}{1+u_x^2} = \cos^2\frac{\theta}{2},\quad \frac{u_x}{1+u_x^2} = \frac{1}{2}\sin \theta,\\ \frac{u_x^2}{1+u_x^2} = \sin^2\frac{\theta}{2}, \end{equation} $ | (19) |
$ \begin{equation} \frac{\partial y}{\partial \xi} = \frac{h}{(1+u_x^2)} = \cos^2\frac{\theta}{2}\cdot h, \end{equation} $ | (20) |
$ \begin{equation} y(t,\bar{\xi})-y(t,\xi) = \int_\xi^{\bar{\xi}} \cos^2\frac{\theta(t,s)}{2}\cdot h(t,s)ds. \end{equation} $ | (21) |
Furthermore, we get
$ \begin{equation} \begin{split} &P(t,\xi) = \frac{1}{2}\int_{-\infty}^\infty \exp\left\{-\left|y(t,\xi)-x\right|\right\}\left(\frac{m(m+3)}{2(m+1)}u^{m+1}+\frac{m}{2}u^{m-1}u_x^2\right)dx,\\ &P_x(t,\xi)\\& = \frac{1}{2}\big(\int_{y(t,\xi)}^\infty -\int_{-\infty}^{y(t,\xi)}\big)\exp\left\{-|y(t,\xi)-x|\right\}\left(\frac{m(m+3)}{2(m+1)}u^{m+1}+\frac{m}{2}u^{m-1}u_x^2\right)dx.\\ \end{split}\nonumber\ \end{equation} $ |
It follows from identities (18) to (21) that an expression for
$ \begin{equation} \begin{split} P(t,\xi) = &\frac{1}{2}\int_{-\infty}^\infty \exp\left\{-\left|\int_\xi^{\bar{\xi}}\left(h\cdot\cos^2\frac{\theta }{2}\right) (s)ds\right|\right\}\\ &\cdot\left[h \left(\frac{m(m+3)}{2(m+1)}u^{m+1} \cos^2\frac{\theta }{2} + \frac{m}{2}u^{m-1} \sin^2\frac{\theta }{2}\right)\right] (\bar{\xi})d\bar{\xi}, \end{split} \end{equation} $ | (22) |
$ \begin{equation} \begin{split} P_x(t,\xi) = &\frac{1}{2}\left(\int_{\xi}^\infty -\int_{-\infty}^{\xi}\right)\exp\left\{-\left|\int_\xi^{\bar{\xi}}\left(h\cdot\cos^2\frac{\theta }{2}\right) (s)ds\right|\right\}\\ &\cdot\left[h \left(\frac{m(m+3)}{2(m+1)}u^{m+1} \cos^2\frac{\theta }{2} + \frac{m}{2}u^{m-1} \sin^2\frac{\theta }{2}\right)\right] (\bar{\xi})d\bar{\xi}, \end{split} \end{equation} $ | (23) |
By (4) and (16), the evolution equation for
$ \begin{equation} \frac{\partial}{\partial t}u(t,\xi) = u_t+ u_y y_t = u_t+ u^{m}u_x = -P_x(t,\xi), \end{equation} $ | (24) |
where
By the definition of variable
$ \begin{equation} \int_{\xi^1}^{\xi^2}h(t,\xi)d\xi = \int_{y(t,\xi^1)}^{y(t,\xi^2)}\big(1+u_x^2(t,x)\big)dx.\nonumber\ \end{equation} $ |
By using (16) and
$ \begin{equation} \begin{split} \frac{d}{dt}\int_{\xi^1}^{\xi^2}h(t,\xi)d\xi& = \int_{y(t,\xi^1)}^{y(t,\xi^2)}\{[(1+u_x^2)]_t+ [u(1+u_x^2)]_x\}dx\\ & = \int_{y(t,\xi^1)}^{y(t,\xi^2)}\bigg(\frac{m(m+3)}{m+1}u^{m+1}+mu^{m-1}-2P\bigg)u_xdx. \end{split}\nonumber\ \end{equation} $ |
Differentiating with respect to
$ \begin{equation} \begin{split} \frac{\partial}{\partial t} h(t,\xi)& = \left(\frac{m(m+3)}{m+1}u^{m+1}+mu^{m-1}-2P\right)\frac{u_x}{1+u_x^2}\cdot h\\ & = \left(\frac{m(m+3)}{m+1}u^{2(m+1)}+\frac{m}{2}u^{m-1}-P\right)\sin \theta\cdot h. \end{split} \end{equation} $ | (25) |
Using (17) and (19), we see
$ \begin{equation} \begin{split} \frac{\partial}{\partial t} \theta(t,\xi)& = \frac{2}{1+u_x^2}(u_{xt} +u^mu_{xx})\\ & = \frac{2}{1+u_x^2}\left(-\frac{m}{2}u ^{m-1}(u_x)^2 + \frac{m(m+3)}{2(m+1)}u^{m+1} -P\right)\\ & = \bigg( \frac{m(m+3)}{m+1}u^{m+1} -2P\bigg)\cos^2\frac{\theta}{2}- mu^{m-1} \sin^2\frac{\theta}{2}, \end{split} \end{equation} $ | (26) |
where
According to (24)-(26), we obtain the following semi-linear system
$ \begin{equation} \left\{ \begin{array}{llll} u_t = -P_x,\\ \theta_t = \bigg( \frac{m(m+3)}{m+1}u^{m+1} -2P\bigg)\cos^2\frac{\theta}{2}- mu^{m-1} \sin^2\frac{\theta}{2},\\ h_t = \left(\frac{m(m+3)}{2(m+1)}u^{m+1}+\frac{m}{2}u^{m-1}-P\right)\sin \theta\cdot h \end{array} \right. \end{equation} $ | (27) |
with the initial data
$ \begin{equation} \left\{ \begin{array}{llll} u(0,\xi) = \tilde{u}(\tilde{y}(\xi)) \\ \theta(0,\xi) = 2\arctan \tilde{u}_x(\tilde{y}(\xi))\\ h(0,\xi) = 1,\\ \end{array} \right. \end{equation} $ | (28) |
where
$ \begin{equation} X\doteq H^1(\mathbb{R})\times[L^2(\mathbb{R})\cap L^\infty(\mathbb{R})]\times L^\infty(\mathbb{R}), \end{equation} $ | (29) |
with
In light of the standard theory of ODE in the Banach space, we can establish that all functions on the right-hand side of (27) are locally Lipschitz continuous, this implies the local existence of solutions to the system (27)-(28).
Lemma 3.1. Given initial data
Proof. Set any bounded domain
$ \begin{equation} \begin{split} \Lambda = (u,\theta,h) = \{ \|u\|_{H^1}\leq \gamma,\|\theta\|_{L^2}\leq \delta,\|\theta\|_{L^\infty}\leq \frac{3\pi}{2} ,h^{-} \leq h(x)\leq h^{+} a.e. x\in \mathbb{R}\}, \end{split} \end{equation} $ | (30) |
for any positive constants
$ \begin{equation} \|u\|_{L^\infty}\leq \|u\|_{H^1}, \end{equation} $ | (31) |
and the uniform boundedness of
$ \frac{m(m+3)}{m+1}u^{m+1}\cos^2\frac{\theta}{2},\; \; \; mu^{m-1} \sin^2\frac{\theta}{2},\; \; \; \left(\frac{m(m+3)}{m+1}u^{m+1}+\frac{m}{2}u^{m-1}\right)\sin \theta\cdot h $ |
are Lipschitz continuous from
$ \begin{equation} (u,\theta,h)\mapsto (P,P_x) \end{equation} $ | (32) |
are Lipschitz continuous from
$ \begin{equation*} \begin{split} &\text{measure}\left\{\xi\in \mathbb{R}; |\frac{\theta(\xi)}{2}|\geq \frac{\pi}{4}\right\}\leq\text{measure}\left\{\xi\in \mathbb{R};\sin^2 \frac{\theta(\xi)}{2}\geq \frac{1}{4}\right\}\\ &\; \; \; \; \; \; \; \; \; \leq 4\int_{\{\xi\in \mathbb{R};\sin^2 \frac{\theta(\xi)}{2}\geq \frac{1}{4}\}}\sin^2 \frac{\theta(\xi)}{2}d\xi\\ &\; \; \; \; \; \; \; \; \; \leq \frac{1}{4}\int_{\{\xi\in \mathbb{R};\sin^2 \frac{\theta(\xi)}{2}\geq \frac{1}{4}\}}\frac{\theta(\xi)}{2}d\xi \leq \frac{1}{4}\delta^2 \end{split} \end{equation*} $ |
for
$ \begin{equation} \int_{\xi}^{\bar{\xi}}\cos^2 \frac{\theta(\xi)}{2}h(\xi')d\xi\geq\int_{\{\xi\in [\xi,\bar{\xi}] \frac{\theta(\xi)}{2}\leq \frac{\pi}{4}\}}\frac{h^-}{2}d\xi\geq \frac{h^-}{2}(\bar{\xi}-\xi-\frac{1}{4}\delta^2)\,\,\,\text{for any}\,\,\, \xi\in\bar{\xi}, \end{equation} $ | (33) |
which guarantees that exponential term in the (22)-(23) for
$ \begin{equation} \Gamma(\epsilon)\doteq\min\bigg\{1,\exp{\bigg( \frac{1}{8}\delta^2 h^{-}- \frac{h^-}{2}|\epsilon|\bigg)}\bigg\}, \end{equation} $ | (34) |
we see
$ \begin{equation} \|\Gamma\|_{L^1} = \bigg(\int_{\epsilon\leq \frac{1}{4}\delta^2} +\int_{\epsilon\geq \ \frac{1}{4} \delta^2}\bigg) \Gamma(\epsilon)d\epsilon = \frac{1}{2}\delta^2+\frac{h^-}{4}. \end{equation} $ | (35) |
Next we show that
$ \begin{equation} P,\; \; \; P_\xi,\; \; \; P_x,\; \; \; P_{x\xi}\in L^2. \end{equation} $ | (36) |
Since the estimates for
$ \begin{equation} |P_x(\xi)|\leq \frac{h^+}{2}\bigg|\Gamma* \bigg(\frac{m(m+3)}{2(m+1)}u^{m+1} \cos^2\frac{\theta }{2} + \frac{m}{2}u^{m-1} \sin^2\frac{\theta }{2} \bigg) \bigg|. \end{equation} $ | (37) |
A standard properties of convolutions ensures that
$ \begin{equation} \begin{split} ||P_x||_{L^2}&\leq \frac{h^+}{2}||\Gamma||_{L^1}\bigg(\frac{m(m+3)}{2(m+1)}\|u^{m+1}\|_{L^2}+\frac{m}{8}||u^{m-1}\theta^2||_{L^2} \bigg)\\ &\leq \frac{Ch^+}{2}||\Gamma||_{L^1}\bigg(||u||^m_{L^\infty }||u||_{L^2}+||u||^{m-2}_{L^\infty}||u||_{L^2} ||\theta^2||_{L^2} \bigg) < \infty, \end{split} \end{equation} $ | (38) |
where
$ \begin{equation} \begin{split} P_{x\xi}& = \left[h \left(\frac{m(m+3)}{2(m+1)}u^{m+1} \cos^2\frac{\theta }{2} + \frac{m}{2}u^{m-1} \sin^2\frac{\theta }{2}\right)\right](\xi)\\ \quad&\; \; \; +\frac{1}{2}\left(\int_{\xi}^\infty -\int_{-\infty}^{\xi}\right)\exp\left\{-\left|\int_\xi^{\bar{\xi}}\left(h\cdot\cos^2\frac{\theta }{2}\right) (s)ds\right|\right\} \bigg[\cos^2\frac{\theta }{2}h(\xi)\bigg]\\ \quad&\; \; \; sign(\xi-\bar{\xi})\cdot\left[h \left(\frac{m(m+3)}{2(m+1)}u^{m+1} \cos^2\frac{\theta }{2} + \frac{m}{2}u^{m-1} \sin^2\frac{\theta }{2}\right)\right] (\bar{\xi})d\bar{\xi}. \end{split} \end{equation} $ | (39) |
Since
$ \begin{equation} \begin{split} |P_{x\xi}(\xi)|&\leq h^+\bigg|\frac{m(m+3)}{2(m+1)}u^{m+1}(\xi)+\frac{m}{8}u^{m-1}(\xi)\theta^2(\xi) \bigg|\\ \quad&\; \; \; \; \; \; \; +\frac{h^+}{2}\bigg|\Gamma* \bigg(\frac{m(m+3)}{2(m+1)}u^{m+1} \cos^2\frac{\theta }{2} + \frac{m}{2}u^{m-1} \sin^2\frac{\theta }{2} \bigg)(\xi)\bigg|, \end{split} \end{equation} $ | (40) |
this implies
$ \begin{equation} \begin{split} ||P_{x\xi}||_{L^2}& = h^+\bigg(\frac{m(m+3)}{2(m+1)}||u^{m+1}||_{L^2}+\frac{m}{8}||u^{m-1}\theta^2||_{L^2} \bigg)\\ &\; \; \; +\frac{h^+}{2}||\Gamma||_{L^1}\bigg(\frac{m(m+3)}{2(m+1)}u^{m+1}_{L^2}+\frac{m}{8}||u^{m-1}\theta^2||_{L^2} \bigg)\\ &\leq \frac{Ch^+}{2}||\Gamma||_{L^1}\bigg(||u||^m_{L^\infty }||u||_{L^2}+||u||^{m-2}_{L^\infty}||u||_{L^2} ||\theta^2||_{L^2} \bigg)\\ &\leq C(h^++\frac{h^+}{2})||\Gamma||_{L^1} < \infty. \end{split} \end{equation} $ | (41) |
A similar argument leads to
To show that the maps given in (32) are Lipschitz continuous. It suffices to verify that partial derivatives
$ \begin{equation} \frac{\partial_P}{\partial_u},\; \frac{\partial_P}{\partial_\theta},\; \frac{\partial_P}{\partial_h},\; \frac{\partial_{P_x}}{\partial_u},\; \frac{\partial_{P_x}}{\partial_\theta},\; \frac{\partial_{P_x}}{\partial_h}, \end{equation} $ | (42) |
are uniformly bounded for
For
$ \begin{equation} \begin{split} \bigg[\frac{\partial_{P_x}(u,\theta,h)}{\partial_u}u^*\bigg](\xi)& = \frac{1}{2}\left(\int_{\xi}^\infty -\int_{-\infty}^{\xi}\right)\exp\left\{-\left|\int_\xi^{\bar{\xi}}\left(h\cdot\cos^2\frac{\theta }{2}\right) (s)ds\right|\right\}\\ &\cdot\left[h u^*\left(\frac{m(m+3)}{2}u^{m} \cos^2\frac{\theta }{2} + \frac{m(m-1)}{2}u^{m-2} \sin^2\frac{\theta }{2}\right)\right] (\bar{\xi})d\bar{\xi} \end{split} \end{equation} $ | (43) |
and
$ \begin{equation} \begin{split} &\bigg[\frac{\partial(\partial_\xi\partial_{P_x})(u,\theta,h)}{\partial_u}u^*\bigg](\xi)\\& = -\left[h u^*\left(\frac{m(m+3)}{2}u^{m} \cos^2\frac{\theta }{2} + \frac{m(m-1)}{2}u^{m-2} \sin^2\frac{\theta }{2}\right)\right](\xi) \\ \quad&\; \; \; \; \; \; \; \; \; \; \; +\frac{1}{2}\left(\int_{\xi}^\infty -\int_{-\infty}^{\xi}\right)\exp\left\{-\left|\int_\xi^{\bar{\xi}}\left(h\cdot\cos^2\frac{\theta }{2}\right) (s)ds\right|\right\}\left(h\cdot\cos^2\frac{\theta }{2}\right) \\ \quad&\; \; \; \; \; \; \; \; \; \; \; sign(\bar{\xi}-\xi)\left(\frac{m(m+3)}{2}u^{m} \cos^2\frac{\theta }{2} + \frac{m(m-1)}{2}u^{m-2} \sin^2\frac{\theta }{2}\right)u^* (\bar{\xi})d\bar{\xi}. \end{split} \end{equation} $ | (44) |
In view of
$ \begin{equation} \begin{split} \bigg\|\frac{\partial_{P_x}}{\partial_u}u^*\bigg\|_{L^2}&\leq \frac{h^+}{2} \bigg\|\Gamma*\left(\frac{m(m+3)}{2}u^{m} \cos^2\frac{\theta }{2} + \frac{m(m-1)}{2}u^{m-2} \sin^2\frac{\theta }{2}\right) \bigg\|_{L^2} ||u^*||_{L^\infty}\\ &\leq \frac{h^+}{2}||\Gamma||_L^1\bigg(\frac{m(m+3)}{2}\|u^{m}\|_{L^2}+\frac{m(m-1)}{8}||u^{m-2}\theta^2||_{L^2} \bigg) ||u^*||_{H^1} \end{split} \end{equation} $ | (45) |
and
$ \begin{equation} \begin{split} &\bigg\|\frac{\partial(\partial_\xi\partial_{P_x})}{\partial_u}u^*\bigg\|_{L^2}\leq \frac{h^+}{2} \bigg\|\frac{m(m+3)}{2}u^{m} \cos^2\frac{\theta }{2} + \frac{m(m-1)}{2}u^{m-2} \sin^2\frac{\theta }{2} \bigg\|_{L^2} ||u^*||_{H^1}\\ &\quad\; \; \; \; \; \; \quad+\frac{(h^+)^2}{2}\bigg\|\Gamma*\left(\frac{m(m+3)}{2}u^{m} \cos^2\frac{\theta }{2} + \frac{m(m-1)}{2}u^{m-2} \sin^2\frac{\theta }{2}\right) \bigg\|_{L^2} ||u^*||_{H^1}. \end{split} \end{equation} $ | (46) |
By (45) and (46), we obtain
Next we show that the local solutions of the system (27) can be extended globally in time.
Lemma 3.2. Given initial data
Proof. To extend the local solutions of the system (27) to global solutions, we only need to prove that
$ \begin{equation} ||u||_{H^1}+||\theta||_{L^2}+||\theta||_{L^\infty}+||h||_{L^\infty}+\bigg\|\frac{1}{h}\bigg\|_{L^\infty} < \infty \end{equation} $ | (47) |
for all
$ \begin{equation} u_\xi = \frac{h}{2}\sin \theta. \end{equation} $ | (48) |
In fact, recalling (27), (22) and (23), one can get
$ \begin{equation} u_{\xi t} = h \left(\frac{m(m+3)}{2(m+1)}u^{m+1} \cos^2\frac{\theta }{2} + \frac{m}{2}u^{m-1} \sin^2\frac{\theta }{2}-P \cos^2\frac{\theta }{2}\right) = u_{t\xi}. \end{equation} $ | (49) |
Moreover, from (19)) and ((20) we have
$ u_\xi = \frac{u_x}{1+{u_x}^2} = \frac{1}{2}\sin \theta, \; \; \; h = 1 $ |
at
Next we prove
$ \begin{equation} \frac{d}{dt}\int_{\mathbb{R}}\bigg(u^2 \cos^2\frac{\theta }{2}+\sin^2\frac{\theta }{2}\bigg)h d\xi = 0. \end{equation} $ | (50) |
Using (27), a direct calculation yields that
$ \begin{equation} \begin{split} &\frac{d}{dt}\int_{\mathbb{R}}\bigg(u^2 \cos^2\frac{\theta }{2}+\sin^2\frac{\theta }{2}\bigg)h d\xi\\& = \int_{\mathbb{R}}h\{(u^2 \cos^2\frac{\theta }{2}+\sin^2\frac{\theta }{2})\left(\frac{m(m+3)}{m+1}u^{m+1}+mu^{m-1}-2P\right)\\ \quad&\; \; \; \cos\frac{\theta }{2}\sin\frac{\theta}{2}-2uP_x\cos^2\frac{\theta }{2}+\cos\frac{\theta }{2}\sin\frac{\theta}{2}(1-u^2)\\ \quad&\; \; \; \bigg[\bigg( \frac{m(m+3)}{m+1}u^{m+1} -2P\bigg)\cos^2\frac{\theta}{2}- mu^{m-1} \sin^2\frac{\theta}{2}\bigg]\}d\xi\\ & = \int_{\mathbb{R}}h \{\frac{2m(m+2)}{m+1}u^{m+1} \cos\frac{\theta }{2}\sin\frac{\theta}{2}-2P\\ \quad&\; \; \; \cos\frac{\theta }{2}\sin\frac{\theta}{2}-2uP_x\cos^2\frac{\theta }{2}\}d\xi. \end{split} \end{equation} $ | (51) |
By (20), we have
$ \begin{equation*} P_\xi = h P_x \cos^2\frac{\theta }{2}, \end{equation*} $ |
which implies
$ \begin{equation} (uP)_\xi = (P \sin\frac{\theta}{2}\cos\frac{\theta }{2}+uP_x\cos^2\frac{\theta }{2})h \end{equation} $ | (52) |
and
$ \begin{equation} \frac{2m(m+2)}{m+1}u^{m+1} \cos\frac{\theta }{2}\sin\frac{\theta}{2}h = \frac{2m(m+2)}{m+1}u^{m+1} u_\xi = \bigg(\frac{2m}{m+1}u^{m+2}\bigg)_\xi. \end{equation} $ | (53) |
From (52) and (53), one has
$ \begin{equation*} \begin{split} \frac{d}{dt}\int_{\mathbb{R}}\bigg(u^2 \cos^2\frac{\theta }{2}+\sin^2\frac{\theta }{2}\bigg) d\xi = 0, \end{split} \end{equation*} $ |
namely,
$ \begin{equation} \begin{split} \int_{\mathbb{R}}\bigg(u^2 \cos^2\frac{\theta }{2}+\sin^2\frac{\theta }{2}\bigg)d\xi = E(0) = E_0. \end{split} \end{equation} $ | (54) |
If the solution is well defined, we obtain a priori bound on
$ \begin{equation} \mathop {\sup}\limits_{\xi\in\mathbb{R}}|u^2(t,\xi)|\leq 2\int_\mathbb{R}|uu_\xi|d\xi\leq 2\int_\mathbb{R}|u|\cdot \left|\sin\frac{\theta}{2}\cos\frac{\theta}{2}\right|h d\xi\leq E_0. \end{equation} $ | (55) |
According to (22), (23) and (54) we know
$ \begin{equation} \begin{split} ||P(t)||_{L^\infty }, \; ||P_x(t)||_{L^\infty }&\leq \frac{1}{2}\bigg\|h \left(\frac{m(m+3)}{2(m+1)}u^{m+1} \cos^2\frac{\theta }{2} + \frac{m}{2}u^{m-1} \sin^2\frac{\theta }{2}\right)\bigg\|_{L^1} \\& \leq \frac{m(m+3)}{4(m+1)}E_0^{\frac{m+1}{2}}. \end{split} \end{equation} $ | (56) |
Using the third equation in (27), together with (55) and (56), we conclude
$ \begin{equation*} \begin{split} |h_t|\leq \bigg(\frac{3m(m+3)}{8(m+1)}E_0^{\frac{m+1}{2}}+\frac{m}{4}E_0^{\frac{m-1}{2}}\bigg)h. \end{split} \end{equation*} $ |
Therefore, it follows that
$ \begin{equation} \begin{split} e^{-\bigg(\frac{3m(m+3)}{8(m+1)}E_0^{\frac{m+1}{2}}+\frac{m}{4}E_0^{\frac{m-1}{2}}\bigg)t } \leq h(t)\leq e^{\bigg(\frac{3m(m+3)}{8(m+1)}E_0^{\frac{m+1}{2}}+\frac{m}{4}E_0^{\frac{m-1}{2}}\bigg)t}. \end{split} \end{equation} $ | (57) |
Also, we observe that there exists a suitable constant
$ \begin{equation} \begin{split} \|\theta\|_{L^\infty}\leq e^{At}. \end{split} \end{equation} $ | (58) |
Multiplying the first equation of (27) by
$ \begin{equation*} \bigg| \frac{d}{dt}\bigg(\frac{1}{2}{\|u(t)\|_{L^2}}^2\bigg)\bigg|\leq\|u(t)\|_{L^\infty}\|P_x\|_{L^1}. \end{equation*} $ |
Similarly, we deduce
$ \begin{equation*} \bigg|\frac{d}{dt}\bigg(\frac{1}{2}{\|u_\xi(t)\|_{L^2}}^2\bigg)\bigg|\leq\|u_\xi(t)\|_{L^\infty}\|\partial_\xi P_x\|_{L^1}. \end{equation*} $ |
Therefore, we obtain that
$ \begin{equation} \begin{split} \int_{\bar{\xi}}^\xi h \cos^2\frac{\theta }{2}(s)ds&\geq \int_{\{s\in [\bar {\xi},\xi],|\frac{\theta }{2}(\bar{\xi})|\geq\frac{\pi}{4}\}}h \cos^2\frac{\theta }{2}(s)ds\geq\int_{\{s\in [\bar {\xi},\xi],|\frac{\theta }{2}(\bar{\xi})|\geq\frac{\pi}{4}\}}\frac{h }{2}(s)ds\\ &\geq \frac{h^- }{2}(\xi-\bar{\xi})-\int_{\{s\in [\bar {\xi},\xi],|\frac{\theta }{2}(\bar{\xi})|\geq\frac{\pi}{4}\}}\frac{h }{2}(s)ds\geq \frac{h^- }{2}(\xi-\bar{\xi})\\ \quad&\; \; \; -\int_{\{s\in [\bar {\xi},\xi],|\frac{\theta }{2}(\bar{\xi})|\geq\frac{\pi}{4}\}}h \sin^2\frac{\theta }{2}(s)ds\geq \frac{h^- }{2}(\xi-\bar{\xi})-2E_0, \end{split} \end{equation} $ | (59) |
where
$ \begin{equation*} \Gamma_1(\epsilon)\doteq \min \bigg\{ 1, e^{(2E_0-\frac{h^- |\xi|}{2})}\bigg\}, \end{equation*} $ |
it follows from (39) that
$ \begin{equation*} \begin{split} \|\partial_\xi (\partial_xP(t))\|_{L^1}&\leq |P_{x\xi}(\xi)| = \bigg\| h \left(\frac{m(m+3)}{2(m+1)}u^{m+1} \cos^2\frac{\theta }{2} + \frac{m}{2}u^{m-1} \sin^2\frac{\theta }{2}\right)(\xi) \bigg\|_{L^1}\\ \quad&\; \; \; \; \; \; \; +\bigg\|\frac{h^+}{2}\bigg\|_{L^\infty}\bigg\|\Gamma_1* \bigg(\frac{m(m+3)}{2(m+1)}u^{m+1} \cos^2\frac{\theta }{2} + \frac{m}{2}u^{m-1} \sin^2\frac{\theta }{2} \bigg)(\xi)\bigg\|_{L^1}\\ &\leq\bigg(1+\frac{1}{2}\|h\|_{L^\infty}\|\Gamma_1\|_{L^1}\bigg)\bigg\| h \left(\frac{m(m+3)}{2(m+1)}u^{m+1} \cos^2\frac{\theta }{2} + \frac{m}{2}u^{m-1} \sin^2\frac{\theta }{2}\right)(\xi) \bigg\|_{L^1}\\ &\leq \frac{m(m+3)}{2(m+1)}[1+2e^{2Bt}(2E_0+1)]E_0^{\frac{m+1}{2}}, \end{split} \end{equation*} $ |
where
$ \begin{equation*} \|\Gamma_1\|_{L^1} = 4e^{Bt}(2E_0+1). \end{equation*} $ |
Multiplying the second equation (27) by
$ \begin{equation*} \begin{split} &\bigg| \frac{d}{dt}\bigg(\frac{1}{2}{\|\theta(t)\|_{L^2}}^2\bigg)\bigg|\leq \int_{\mathbb{R}}\bigg|\bigg( \frac{m(m+3)}{m+1}u^{m+1} -2P\bigg)\theta\bigg|d\xi+\int_{\mathbb{R}}\frac{m}{8}u^{m-1} |\theta|^3d\xi\\ &\; \; \; \; \; \; \; \; \; \; \; \leq\|\theta\|_{L^\infty}\bigg( \frac{m(m+3)}{m+1}\|u\|_{L^2}^2 \|u^{m-1}\|_{L^\infty}+2\|P\|_{L^1}\bigg)+ \frac{m}{8} \|u^{m-1}\|_{L^\infty}\|\theta\|_{L^\infty}\|\theta\|_{L^2}^2. \end{split} \end{equation*} $ |
From this, we prove that
Let
$ \begin{equation} y(t,\xi) \doteq \tilde{y}(\xi)+ \int_0^t u^m(\tau, \xi)d\tau. \end{equation} $ | (60) |
Then for each fixed
$ \begin{equation} \frac{\partial}{\partial t}y(t,\xi) = u^m(t,\xi), y(0,\xi) = \tilde{y}(\xi). \end{equation} $ | (61) |
Now we claim that
$ \begin{equation} u(t,x)\doteq u(t,\xi) \text{ if } y(t,\xi) = x, \end{equation} $ | (62) |
is a weak solutions of equation (3).
Proof of Theorem 1.2. The proof is divided into the following steps.
Step 1. It is clearly that we have the uniform bound
$ \begin{equation*} |u(t,\xi)|\leq E_0^{\frac{1}{2}}. \end{equation*} $ |
From (60), we have the estimate
$ \begin{equation*} \tilde{y}(\xi)-E_0^\frac{m}{2} t\leq y(t,\xi)\leq \tilde{y}(\xi)+E_0^\frac{m}{2} t, t\geq 0. \end{equation*} $ |
The definition of
$ \begin{equation*} \mathop {\lim }\limits_{\xi\rightarrow\pm\infty} y(t,\xi) = \pm\infty. \end{equation*} $ |
Hence the image of the map
Step 2. We check
$ \begin{equation} y_\xi = h\cos^2\frac{\theta}{2}\quad \text{ for all } t\geq 0 \text{ and a.e } \xi\in \mathbb{R}. \end{equation} $ | (63) |
Indeed, thanks to system (27), by a straightforward computation we have
$ \begin{align*} \frac{\partial}{\partial t}\left(h \cos^2 \frac{\theta}{2}\right)(t,\xi)& = -h \theta_t\sin \frac{\theta}{2}\cdot\cos\frac{\theta}{2}+h_t\cos^2\frac{v}{2}\\ & = mu^{m-1}h\sin \frac{\theta}{2}\cdot\cos\frac{\theta}{2}\\ & = (u^m)_\xi(t,\xi). \end{align*} $ |
On the other hand, (60) implies
$ \frac{\partial}{\partial t}y_\xi(t,\xi) = (u^m)_\xi(t,\xi). $ |
Since the function
$ \int_{\xi}^{\bar{\xi}} y_{\xi}(t,s)ds = \int_{\xi}^{\bar{\xi}} h(t,s)\cos^2\frac{\theta}{2}ds = 0. $ |
Hence
$ u(t,\bar{\xi})-u(t,\xi) = \int_{\xi}^ {\bar{\xi}}\frac{h(t,s)}{2}\sin \theta(t,s)\cos\frac{\theta}{2}ds = 0. $ |
This shows that the map
Step 3. Recalling the basic relations, we have
$ \begin{equation} \frac{\partial u(t,\xi)}{\partial \xi} = \frac{h(t,\xi)}{2}\sin\theta(t,\xi),\; \frac{\partial y(t,\xi)}{\partial \xi} = h(t,\xi)\cos^2\frac{\theta(t,\xi)}{2}. \end{equation} $ | (64) |
In addition, if
$ \begin{equation} u_x(t,x) = \frac{\sin \theta(t,\xi)}{1+\cos \theta(t,\xi)}, \end{equation} $ | (65) |
then for any time
$ \begin{equation} \begin{split} \int_\mathbb{R}&\left(u^2(t,x)+u^2_x(t,x)\right)dx\\ & = \int_{\mathbb{R}\cap\cos\frac{\theta}{2} \neq -1 }\left(u^2(t,\xi)\cos^2\frac{\theta(t,\xi)}{2}+\sin^2\frac{\theta(t,\xi)}{2}\right)h(t,\xi)d\xi\leq E_0, \end{split} \end{equation} $ | (66) |
which implies
$ \begin{equation*} \frac{d u(t,y(t,\xi))}{dt} = u_t\leq \infty. \end{equation*} $ |
On the basis of the Sobolev inequality,
Step 4. We are ready to show that the Lipschitz continuity of
$ \begin{equation} \begin{split} |u(\tau+s,x)-u(\tau,x)|&\leq |u(\tau+s,x)-u(\tau+s,y(\tau+s,\xi))|\\ &\quad+|u(\tau+s,y(\tau+s,\xi))-u(\tau,x)|\\ &\leq \mathop {\sup }\limits_{|y-x|\leq E_0^\frac{m}{2}s}\left|u(\tau+s,y)-u(\tau+s,x)\right|+\int_\tau^{\tau+s}|P_x(t,\xi)|dt. \end{split}\nonumber\ \end{equation} $ |
Integrating over
$ \begin{equation*} \begin{split} &\int_\mathbb{R}|u(\tau+s,y)-u(\tau,y)|^2dx \\ &\leq 2\int_\mathbb{R}\left(\int_{x-E_0^\frac{m}{2}s}^{x+E_0^\frac{m}{2}s}|u_x(\tau+s,y)|dy\right)^2dx\\&\qquad+ 2\int_\mathbb{R}\left(\int_{\tau}^{\tau+s}|P_x(t,\xi)|dt\right)^2h(\tau,\xi)\cos^2\frac{\theta(\tau,\xi)}{2}d\xi\\ &\leq 4E_0^\frac{m}{2}s\int_\mathbb{R}\int_{x-E_0^\frac{m}{2}s}^{x+E_0^\frac{m}{2}s}|u_x(\tau+s,y)|^2dydx+ 2s\|h\|_{L^\infty}\int_{\mathbb{R}}\int_{\tau}^{\tau+s}|P_x(t,\xi)|^2dtd\xi\\ &\leq 8E_0^m s^2\|u_x(\tau+s)||_{L^2(\mathbb{R})}^2+2s\|h\|_{L^\infty}||P_x(t)||_{L^2(\mathbb{R})}^2dt\leq Cs, \end{split} \end{equation*} $ |
where the constant
Step 5. Define
$ \begin{equation} \begin{split} 0& = \int\int_{\Omega}\bigg\{u_{\xi t}\phi_t+h\phi \left(\frac{m(m+3)}{2(m+1)}u^{m+1} \cos^2\frac{\theta }{2} + \frac{m}{2}u^{m-1} \sin^2\frac{\theta }{2}-P \cos^2\frac{\theta }{2}\right)\bigg\}\xi dt\\ & = \int\int_{\Omega}\bigg\{-u_\xi \phi_t\phi_t+h\phi \left(\frac{m(m+3)}{2(m+1)}u^{m+1} \cos^2\frac{\theta }{2} + \frac{m}{2}u^{m-1} \sin^2\frac{\theta }{2}-P \cos^2\frac{\theta }{2}\right)\bigg\}d\xi dt\\ & = \int\int_{\Omega'}\bigg\{-u_\xi \phi_t\phi_t+h\phi \left(\frac{m(m+3)}{2(m+1)}u^{m+1} \cos^2\frac{\theta }{2} + \frac{m}{2}u^{m-1} \sin^2\frac{\theta }{2}-P \cos^2\frac{\theta }{2}\right)\bigg\}\xi dt\\ & = \int\int_{\Omega}\bigg\{-u_x (\phi_t+u^m\phi_x)+\phi \left(-\frac{m(m+3)}{2(m+1)}u^{m+1}-\frac{m}{2}u^{m-1}u_x^2+P\right)\bigg\}dxdt, \end{split} \end{equation} $ | (67) |
which proves that (5) holds. Let us introduce the Radon measures
$ \mu_{t}(\mathcal{A}) = \int_{\mathcal{F}(\mathcal{A})}\sin^2\frac{\theta}{2}(t,\xi)d\xi. $ |
By (63), the measure
$ \begin{equation} \begin{split} -\int_{\mathbb{R}^+}\bigg\{\int(\phi_t+u^m\phi_x)d \nu_{(t)}\bigg\}dt& = -\int\int_{\Omega}\sin^2\frac{\theta}{2}h \phi_td\xi dt = \int\int_{\Omega}(\sin^2\frac{\theta}{2}h)_t \phi d\xi dt\\ & = \int\int_{\Omega}\bigg( \frac{m(m+3)}{m+1}u^{m+1} -2P\bigg)\sin\frac{\theta}{2}\cos\frac{\theta}{2} h \phi d\xi dt \end{split} \end{equation} $ | (68) |
$ \begin{equation*} \begin{split} & = \int\int_{\Omega}\bigg( \frac{m(m+3)}{m+1}u^{m+1} -2P\bigg)u_x \phi dx dt. \end{split} \end{equation*} $ |
Step 6. Ultimately, we show that for almost every
Let
$ \begin{equation} x(t,\beta)+\mu_{(t)}\{(-\infty,x)\}\leq \beta\leq x(t,\beta)+\mu_{(t)}\{(-\infty,x]\} \end{equation} $ | (69) |
for any time
$ \begin{equation} x(t,\beta)+\int_{-\infty}^{x(t,\beta)}u_x^2(t,\xi)d\xi = \beta. \end{equation} $ | (70) |
Next, we will give following Lemma which is helpful to prove the Lipschitz continuity of
Lemma 4.1. Assume that
Proof of Lemma 4.1. We split the proof into three steps.
Step 1. For any time
$ x\mapsto \beta(t,x) $ |
is right continuous and strictly increasing. Thus, the inverse
$ \begin{equation*} x(t,\beta_2)-x(t,\beta_1)+\mu_{(t)}\{(x(t,\beta_2),x(t,\beta_1))\}\leq \beta_2-\beta_1, \end{equation*} $ |
which implies
$ \begin{equation} x(t,\beta_2)-x(t,\beta_1)\leq \beta_2-\beta_1, \end{equation} $ | (71) |
and the map
Step 2. For the map
$ \begin{equation} \begin{split} |u(t,x(t,\beta_2))-u(t,x(t,\beta_1))|&\leq\int_{x(t,\beta_1)}^{x(t,\beta_2)}|u_x|dx\leq \int_{x(t,\beta_1)}^{x(t,\beta_2)}\frac{1}{2}(1+u_x^2)dy\\ &\leq \frac{1}{2}\big[x(t,\beta_2)-x(t,\beta_1)+\mu_{(t)}\{(x(t,\beta_2),x(t,\beta_1))\}\big]\\ &\leq \frac{1}{2}(\beta_2-\beta_1). \end{split} \end{equation} $ | (72) |
Hence, the map
Step 3. Now we claim the Lipschitz continuity of the map
$ \begin{equation} \begin{split} \|2u_x\bigg(\frac{m(m+3)}{2(m+1)}u^{m+1}-P\bigg)\|_{L^1}\leq 2\big(\frac{m(m+3)}{2(m+1)}\|u\|_{L^\infty}^m\|u\|_{L^2}+\|P\|_{L^2}\big)\|u_x\|_{L^2}\leq C_s. \end{split} \end{equation} $ | (73) |
For
$ \begin{equation*} \begin{split} \mu_{(t)}\{(-\infty,y-C_{\infty}(t-\tau))\}\leq \mu_{(\tau)}\{(-\infty,y)\}+C_s(t-\tau), \end{split} \end{equation*} $ |
where the constant
$ \begin{equation*} \begin{split} y^-(t)+\mu_{(t)}\{(-\infty,y^-(t)]\}&\leq y-(C_\infty+C_s)(t-\tau)+\mu_{(\tau)}\{(-\infty,y)\}+C_s(t-\tau)\\ &\leq y+\mu_{(\tau)}\{(-\infty,y)\}\leq \beta, \end{split} \end{equation*} $ |
which implies
$ \begin{equation*} x(t,\beta)\leq y^+(t)\doteq y+(C_\infty+C_s)(t-\tau). \end{equation*} $ |
This proves the uniform Lipschitz continuity of the map
Lemma 4.2. Let
$ \begin{equation} \begin{split} \frac{d}{dt}x(t) = u^m(t,x),\; \; \; \; \; \; x(0) = \tilde{y}, \end{split} \end{equation} $ | (74) |
and
$ \begin{equation} \begin{split} \frac{d}{dt}\int_{-\infty}^{x(t,\beta)}u_x^2 = \int_{-\infty}^{x(t,\beta)}2u_x\bigg(\frac{m(m+3)}{2(m+1)}u^{m+1}-P\bigg)dx. \end{split} \end{equation} $ | (75) |
Moreover, we have
$ \begin{equation} \begin{split} u(t,x(t))-u(\tau,x(\tau)) = -\int_{\tau}^tP_x(s,x(s))ds, \end{split} \end{equation} $ | (76) |
for any
Proof. Step 1. According to the adapted coordinates
$ \begin{equation} \begin{split} x(t)+&\int_{-\infty}^{x(t)}u_x^2(t,y)dy = \tilde{y}+\int_{-\infty}^{\tilde{y}}u_{0,x}^2dy\\ &+\int_0^t\bigg( u^m(s,x(s))+\int_{-\infty}^{x(s)} 2u_x\bigg(\frac{m(m+3)}{2(m+1)}u^{m+1}-P\bigg)(s,y)dy\bigg)ds. \end{split} \end{equation} $ | (77) |
For convenience, let
$ \begin{equation} G(t,\beta)\doteq \int_{-\infty}^{x(s)}2u_x\bigg(\frac{m(m+3)}{2(m+1)}u^{m+1}-P\bigg)dx \end{equation} $ | (78) |
and
$ \begin{equation} \tilde{\beta} = \tilde{y}+\int_{-\infty}^{\tilde{y}}u_{0,x}^2(y)dy. \end{equation} $ | (79) |
Therefore, we can rewrite the equation (77) as follow
$ \begin{equation} \beta(t) = \tilde{\beta}+\int_0^tG(s,\beta(s))ds, \; \; \; \; \; \text{for all}\; t > 0. \end{equation} $ | (80) |
Step 2. For every fix
$ \begin{equation} \begin{split} G_{\beta}& = 2u_x\bigg(\frac{m(m+3)}{2(m+1)}u^{m+1}-P\bigg)x_\beta\\ & = \frac{2u_x\bigg(\frac{m(m+3)}{2(m+1)}u^{m+1}-P\bigg)}{1+u_x^2}\in [-C,C] \end{split} \end{equation} $ | (81) |
for some constant
Step 3. Based on the Lipschitz continuity of the function
Step 4. Owing to the previous construction, we conclude that the map
$ \begin{equation} \dot{x}({\tau})\doteq u^m(\tau,x(\tau))+2\varepsilon_0 \end{equation} $ | (82) |
for some
$ \begin{equation} x^+(t)\doteq x(\tau)+(t-\tau)[ u^m(\tau,x(\tau))+\varepsilon_0] < x(t). \end{equation} $ | (83) |
We also observe that (6) still holds for any test
For any
$ \begin{equation*} \rho^\epsilon(s,y) = \left\{ \begin{array}{llll} 0\; \; \; \; \; \; \; \; \; \; \; \; & \text{if}\; y\leq -\epsilon^{-1},\\ (y+\epsilon^{-1})\; \; \; \; \; \; \; \; \; \; \; \; &\text{if}\; -\epsilon^{-1}\leq y\leq 1-\epsilon^{-1},\\ 1\; \; \; \; \; \; \; \; \; \; \; \; \; &\text{if}\; 1-\epsilon^{-1}\leq y\leq x^{+}(s),\\ 1-\epsilon^{-1}(y-x(s))\; \; \; \; \; \; \; \; \; \; &\text{if}\; x^{+}(s)\leq y\leq x^+(s)+\epsilon,\\ 0\; \; \; \; \; \; \; \; \; \; \; &\text{if}\; y\geq x^+(s)+\epsilon, \end{array} \right. \end{equation*} $ |
$ \begin{equation} \chi^\epsilon(s,y) = \left\{ \begin{array}{llll} 0\; \; \; \; \; \; \; \; \; \; \; \; & \text{if}\; s\leq \tau-\epsilon,\\ \epsilon^{-1}(s-\tau+\epsilon)\; \; \; \; \; \; \; \; \; \; \; \; &\text{if}\; \tau-\epsilon\leq s\leq \tau,\\ 1\; \; \; \; \; \; \; \; \; \; \; \; \; &\text{if}\; \tau\leq s\leq t,\\ 1-\epsilon^{-1}(s-t)\; \; \; \; \; \; \; \; \; \; &\text{if}\; t\leq s < t+\epsilon,\\ 0\; \; \; \; \; \; \; \; \; \; \; &\text{if}\; s\geq t+\epsilon. \end{array} \right. \end{equation} $ | (84) |
We define
$ \begin{equation} \psi^\epsilon (s,y)\doteq\min\{\rho^\epsilon(s,y),\chi^\epsilon(s)\}. \end{equation} $ | (85) |
Using
$ \begin{equation} \int\int \bigg[ u_x^2 \psi_t^\epsilon+ u^mu_x^2\psi_x^\epsilon+ 2u_x\bigg(\frac{m(m+3)}{2(m+1)}u^{m+1}-P\bigg) \psi^\epsilon \bigg]dxdt = 0. \end{equation} $ | (86) |
If
$ \begin{equation} \lim\limits_{\epsilon\rightarrow 0}\int_{\tau}^t\int_{x^+(s)-\epsilon}^{x^+(s)+\epsilon}u_x^2(\psi_t^\epsilon+ u^m\psi_x^\epsilon)dyds\geq 0. \end{equation} $ | (87) |
Actually, for
$ \begin{equation} 0 = \psi_t^\epsilon+[ u^m(\tau,x(\tau))+\varepsilon_0]\psi_x^\epsilon\leq \psi_t^\epsilon+ u^m(s,x)\psi_x^\epsilon, \end{equation} $ | (88) |
where we use the fact that
Due to the family of measures
$ \begin{equation*} \begin{split} 0& = \int_{-\infty}^{x(\tau)}u_x^2(\tau,y)dy-\int_{-\infty}^{x^+(t)}u_x^2(t,y)dy\\&\qquad+\int_{\tau}^t\int_{-\infty}^{x^{+}(s)} 2u_x\bigg(\frac{m(m+3)}{2(m+1)}u^{m+1}-P\bigg) dyds\\ &\quad\; \; \; \; \; \; +\lim\limits_{\epsilon\rightarrow 0}\int_{\tau}^t\int_{x^+(s)-\epsilon}^{x^+(s)+\epsilon}u_x^2(\psi_t^\epsilon+ u^m\psi_x^\epsilon)dyds\\ &\geq \int_{-\infty}^{x(\tau)}u_x^2(\tau,y)dy-\int_{-\infty}^{x^+(t)}u_x^2(t,y)dy\\&\qquad+\int_{\tau}^t\int_{-\infty}^{x^{+}(s)} 2u_x\bigg(\frac{m(m+3)}{2(m+1)}u^{m+1}-P\bigg) dyds, \end{split} \end{equation*} $ |
which yields
$ \begin{equation*} \begin{split} & \int_{-\infty}^{x^+(t)}u_x^2(t,y)dy\geq \int_{-\infty}^{x(\tau)}u_x^2(\tau,y)dy+\int_{\tau}^t\int_{-\infty}^{x^{+}(s)} 2u_x\bigg(\frac{m(m+3)}{2(m+1)}u^{m+1}-P\bigg)dyds\\ & = \int_{-\infty}^{x(\tau)}u_x^2(\tau,y)dy+\int_{\tau}^t\int_{-\infty}^{x(s)} 2u_x\bigg(\frac{m(m+3)}{2(m+1)}u^{m+1}-P\bigg) dyds+o_1(t-\tau). \end{split} \end{equation*} $ |
Note that the last term is higher order infinitesimal, satisfying
$ \begin{equation*} \begin{split} |o_1(t-\tau)|& = \bigg| \int_{\tau}^t\int_{x^+(s)}^{x^+(s)+\epsilon} 2u_x\bigg(\frac{m(m+3)}{2(m+1)}u^{m+1}-P\bigg)dxdyds\bigg|\\ &\leq \|2\bigg(\frac{m(m+3)}{2(m+1)}u^{m+1}-P\bigg)\|_{L^\infty} \int_{\tau}^t\int_{x^+(s)}^{x^+(s)+\epsilon}|u_x|dyds\\ &\leq\|2\bigg(\frac{m(m+3)}{2(m+1)}u^{m+1}-P\bigg)\|_{L^\infty}\int_{\tau}^t(x(s)-x^+(s))^\frac{1}{2}\|u_x(s,\cdot)\|_{L^2}ds\\ &\leq C\cdot (t-\tau)^\frac{3}{2}. \end{split} \end{equation*} $ |
On the other hand, together with (78) and (80), we see
$ \begin{equation} \beta(t) = \beta(\tau)+(t-\tau)\bigg[ u^m(\tau,x(\tau)+\int_{-\infty}^{x(\tau)}2u_x\bigg(\frac{m(m+3)}{2(m+1)}u^{m+1}-P\bigg) dy\bigg]+o_2(t-\tau), \end{equation} $ | (89) |
with
$ \begin{equation} \begin{split} \beta(t)& = x(t)+\int_{-\infty}^{x(t)}(u_x^2(t,y)dy\\ & > x(\tau)+(t-\tau)[u^m(\tau,x(\tau))+\varepsilon_0]+\int_{-\infty}^{x^+(t)}u_x^2(t,y)dy\\ &\geq x(\tau)+(t-\tau)[u^m(\tau,x(\tau))+\varepsilon_0]+\int_{-\infty}^{x(\tau)}u_x^2(\tau,y)dy\\ &\quad\; \; +\int_{\tau}^t\int_{-\infty}^{x(s)}2u_x\bigg(\frac{m(m+3)}{2(m+1)}u^{m+1}-P\bigg)dyds+o_1(t-\tau). \end{split} \end{equation} $ | (90) |
By (89) and (90), we have
$ \begin{equation} \begin{split} &\beta(t)+(t-\tau)\bigg[ u^m(\tau,x(\tau))+\int_{-\infty}^{x(\tau)} 2u_x\bigg(\frac{m(m+3)}{2(m+1)}u^{m+1}-P\bigg)dy \bigg]+o_2(t-\tau)\\ &\quad\; \; \; \geq\bigg[x(\tau)+\int_{-\infty}^{x(\tau)}u_x^2(\tau,y)dy\bigg]+(t-\tau)[ u^m(\tau,x(\tau))+\varepsilon_0]\\ &\quad\; \; \; \; \; \; +\int_{\tau}^t\int_{-\infty}^{x(s)}2u_x\bigg(\frac{m(m+3)}{2(m+1)}u^{m+1}-P\bigg )dyds+o_1(t-\tau). \end{split} \end{equation} $ | (91) |
Dividing both sides by
Step 5. Now we prove (75). By (3), one has
$ \begin{equation} \int_{0}^{\infty}\int\bigg[u\phi_t+\frac{u^{m+1}}{m+1}\phi_x+P_x\phi\bigg]dxdt+\int u_0(x)\phi(0,x)dx = 0, \end{equation} $ | (92) |
for every test function
$ \begin{equation} \int_{0}^{\infty}\int\bigg[u_x\psi_t+u^mu_x\psi_x+P_x\psi_x\bigg]dxdt+\int u_{0,x}(x)\psi(0,x)dx = 0. \end{equation} $ | (93) |
By an approximation argument, we find the identity (93) still holds for any test function
$ \begin{equation*} \eta(s,y) = \left\{ \begin{array}{llll} 0\; \; \; \; \; \; \; \; \; \; \; \; & if\; y\leq -\epsilon^{-1},\\ (y+\epsilon^{-1})\; \; \; \; \; \; \; \; \; \; \; \; &if\; -\epsilon^{-1}\leq y\leq 1-\epsilon^{-1},\\ 1\; \; \; \; \; \; \; \; \; \; \; \; \; &if\; 1-\epsilon^{-1}\leq y\leq x(s),\\ 1-\epsilon^{-1}(y-x(s))\; \; \; \; \; \; \; \; \; \; &if\; x(s)\leq y\leq x(s)+\epsilon,\\ 0\; \; \; \; \; \; \; \; \; \; \; &if\; y\geq x(s)+\epsilon, \end{array} \right. \end{equation*} $ |
for any
$ \begin{equation} \varphi^{\epsilon}(s,y) = \min\{\eta^\epsilon(s,y),\chi^\epsilon(s)\}, \end{equation} $ | (94) |
with
$ \begin{equation} \begin{split} \int_{-\infty}^{x(t)}u_x(t,y)dy& = \int_{-\infty}^{x(\tau)}u_x(\tau,y)dy-\int_{\tau}^t P_x(s,x(s))ds\\ &\quad\; \; \; +\lim\limits_{\epsilon\rightarrow 0}\int_{\tau-\epsilon}^{t+\epsilon}\int_{x(s)}^{x(s)+\epsilon}u_x^2(\varphi_t^\epsilon+ u^m\varphi_x^\epsilon)dyds. \end{split} \end{equation} $ | (95) |
It suffices to prove that the last term of the limit in (95) is zero. The Cauchy's inequality implies
$ \begin{equation} \begin{split} &\bigg|\int_{\tau}^t\int_{x(s)}^{x(s)+\epsilon}u_x(\varphi_t^\epsilon+u^m\varphi_x^\epsilon)dyds\bigg|\\ &\quad\; \; \; \leq\int_{\tau}^t\bigg(\int_{x(s)}^{x(s)+\epsilon}|u_x|^2dy\bigg)^\frac{1}{2} \bigg( \int_{x(s)}^{x(s)+\epsilon} (\varphi_t^\epsilon+ u^m\varphi_x^\epsilon)^2dy\bigg)^\frac{1}{2}ds, \end{split} \end{equation} $ | (96) |
where
$ \begin{equation} \varsigma_\epsilon(s)\doteq \big(\sup\limits_{x\in\mathbb{R}}\int _{x}^{x+\epsilon }u_x^2(s,y)dy\big)^\frac{1}{2}, \end{equation} $ | (97) |
we see that all functions
$ \begin{equation} \lim\limits_{\epsilon\rightarrow 0}\int_{\tau}^t\bigg(\int_{x(s)}^{x(s)+\epsilon}|u_x(s,y)|^2dx\bigg)^\frac{1}{2}ds \leq \lim\limits_{\epsilon\rightarrow 0}\int_{\tau}^t\varsigma_\epsilon(s)ds = 0. \end{equation} $ | (98) |
On the other hand, for every time
$ \varphi_x^\epsilon(s,y) = \epsilon^{-1},\; \; \; \varphi_t^\epsilon(s,y)+u^m(s,x(s))\varphi_x^\epsilon(s,y) = 0, $ |
for
$ \begin{equation} \begin{split} &\int_{x(s)}^{x(s)+\epsilon}|\varphi_t^\epsilon(s,y)+ u^m(s,x(s))\varphi_x^\epsilon(s,y)|^2dy\\& = \epsilon^{-2} \int_{x(s)}^{x(s)+\epsilon}|u^m(s,y)-u^m(s,x(s))|^2dy\\ &\leq\epsilon^{-1}\cdot \bigg( \max\limits_{x(s)\leq y\leq x(s)+\epsilon}{|u^m(s,y)-u^m(s,x(s))|}\bigg)^2\\ &\leq m^2\epsilon^{-1}\cdot\bigg(\int_{x(s)}^{x(s)+\epsilon}|u^{m-1}u_x(s,y)|dy\bigg)^2\\ &\leq\epsilon^{-1}m^2(\|u\|_{L^\infty}^{m-1}\epsilon^\frac{1}{2}\cdot\|u_x(s)\|_{L^2})^2\leq m^2\|u(s)\|_{H^1}^{2m}. \end{split} \end{equation} $ | (99) |
By (98) and (99), one has the integral in (96) approaches to zero as
$ \begin{equation} \begin{split} &\bigg|\bigg(\int_{\tau-\epsilon}^\tau+\int_t^{t+\epsilon}\bigg)\int_{x(s)}^{x(s)+k\epsilon} u_x(\psi_t^\epsilon+u^m\varphi_x^\epsilon)dxds\bigg|\\ &\quad\; \; \; \leq \bigg(\int_{\tau-\epsilon}^\tau+\int_t^{t+\epsilon}\bigg) \bigg(\int_{x(s)}^{x(s)+\epsilon}|u_x|^2dx\bigg)^\frac{1}{2}\bigg(\int_{x(s)}^{x(s)+\epsilon} (\psi_t^\epsilon+ u^m\varphi_x^\epsilon)^2dx\bigg)^\frac{1}{2}ds\\ &\quad\; \; \; \leq 2\epsilon\cdot \|u(s)\|_{H^1}\bigg(\int_{x(s)}^{x(s)+\epsilon}4\epsilon^{-2}\|u\|_{L^\infty})^{2m}dx \bigg)^\frac{1}{2}\leq C\epsilon^\frac{1}{2}\rightarrow 0 \end{split} \end{equation} $ | (100) |
as
$ \begin{equation} \lim\limits_{\epsilon\rightarrow 0}\int_{\tau-\epsilon}^{t+\epsilon}\int_{x(s)}^{x(s)+\epsilon}u_x^2(\psi_t^\epsilon+u^m u\varphi_x^\epsilon)dxds = 0. \end{equation} $ | (101) |
Therefore, using (95), we deduce (76).
Step 6. Finally, the uniqueness of the solution
Lemma 4.3. If
Proof. By (72), (76) and (80), we have
$ \begin{equation} \begin{split} &\bigg|u(t,x(t,\tilde{\beta}))-u(\tau,\tilde{\beta})\bigg|\\ &\quad\; \; \; \; \; \; \leq\big|u(t,x(t,\tilde{\beta}))-u(t,x(t,\beta(t)))\big|+\big|u(t,x(t,\beta(t)))-u(t,x(\tau,\beta(\tau)))\big|\\ &\quad\; \; \; \; \; \; \leq\frac{1}{2}|\beta(t)-\tilde{\beta}|+C(t-\tau)\leq C(t-\tau), \end{split} \end{equation} $ | (102) |
where
Lemma 4.4. Let
$ \begin{equation} \beta(t) = \tilde{\beta}+\int_{\tau}^t G(\tau,\beta(\tau))d\tau, \end{equation} $ | (103) |
where the
$ \begin{equation} |\beta(t;\tau,\tilde{\beta_1})- \beta(t;\tau,\tilde{\beta_2})|\leq e^{C|t-\tau|}|\tilde{\beta_1}-\tilde{\beta_2}|. \end{equation} $ | (104) |
Proof. Using the Lipschitz continuity of
Lemma 4.5. Suppose
$ \begin{equation} P_{xx} = P-\frac{m}{2}u^{m-1}u_x^2-\frac{m(m+3)}{2(m+1)}u^{m+1}. \end{equation} $ | (105) |
Proof. The function
$ D_x^2\phi = \phi-\delta_0. $ |
Here
$ D_x^2(\phi*f) = \phi*f-f. $ |
Choosing
In this subsection, we mainly prove the uniqueness of conservative solutions for equation (1).
Proof of Theorem 1.3. The proof is divided into following steps.
Step 1. It follows from Lemma 4.1 and Lemma 4.3 that the map
(GC) For a.e.
If the above condition is true, we say that
Step 2. We find an ODE to describe a change in these two quantities
$ \begin{equation} \frac{\partial\beta(t)}{\partial \tilde{\beta}} = 1+\int_{\tau}^t G_\beta(s,\beta(s;\tau,\tilde{\beta}))\cdot \frac{\partial} {\partial\tilde{\beta}}\beta(s;\tau,\tilde{\beta})ds. \end{equation} $ | (106) |
Next, differentiating w.r.t.
$ x(t,\beta(t;\tau,\tilde{\beta})) = x(\tau,\tilde{\beta})+\int_{\tau}^t u^m(s,x(s,\beta(t;\tau,\tilde{\beta})))ds, $ |
we have
$ \begin{equation} \begin{split} &x_\beta(t,\beta(t;\tau,\tilde{\beta}))\cdot\frac{\partial}{\partial \tilde{\beta}}\beta(t;\tau,\tilde{\beta})\\& = x_\beta(\tau,\tilde{\beta}) +\int_{\tau}^t u^m_\beta(s,\beta(s;\tau,\tilde{\beta}))\cdot \frac{\partial}{\partial \tilde{\beta}} \beta(t;s,\tilde{\beta})ds. \end{split} \end{equation} $ | (107) |
Finally, by (76), differentiating w.r.t.
$ \begin{equation} u_\beta(t,\beta(t;\tau,\tilde{\beta}))\cdot\frac{\partial}{\partial \tilde{\beta}} \beta(t;\tau,\tilde{\beta}) = u_\beta(\tau,\tilde{\beta})+\int_\tau^tP_{x,\beta}(s,\beta(s;\tau,\tilde{\beta})) \cdot \frac{\partial}{\partial \tilde{\beta}} \beta(t;s,\tilde{\beta})ds. \end{equation} $ | (108) |
Together with (106), (107) and (108) yield the following ODE system:
$ \begin{equation} \left\{ \begin{array}{llll} \frac{d}{dt}\big[\frac{\partial}{\partial \tilde{\beta}} \beta(t;\tau,\tilde{\beta})\big] = G_\beta(t,\beta(t;\tau,\tilde{\beta}))\cdot \frac{\partial}{\partial \tilde{\beta}} \beta(t;\tau,\tilde{\beta}),\\ \frac{d}{dt}\big[x_\beta(t,\beta(t;\tau,\tilde{\beta}))\cdot\frac{\partial}{\partial \tilde{\beta}} \beta(t;\tau,\tilde{\beta})\big] = (u^m)_\beta(t,\beta(t;\tau,\tilde{\beta}))\cdot \frac{\partial}{\partial \tilde{\beta}} \beta(t;\tau,\tilde{\beta}),\\ \frac{d}{dt}\big[u_\beta(t,\beta(t;\tau,\tilde{\beta}))\cdot\frac{\partial}{\partial \tilde{\beta}} \beta(t;\tau,\tilde{\beta})\big] = P_{x,\beta}(\beta(t;\tau,\tilde{\beta}))\cdot\beta(t;\tau,\tilde{\beta}). \end{array} \right. \end{equation} $ | (109) |
The quantities within square brackets on the left hand sides of (109) are absolutely continuous. By the above system and using Lemma 4.5, along a good characteristic, we deduce
$ \begin{equation} \left\{ \begin{array}{llll} \frac{d}{dt}x_\beta+G_\beta x_\beta& = mu^{m-1}u_\beta,\\ \frac{d}{dt}u_\beta+G_\beta u_\beta& = \bigg(\frac{m}{2}u^{m-1}u_x^2+\frac{m(m+3)}{2(m+1)}u^{m+1}-P\bigg)x_\beta\\ & = \bigg(\frac{m(m+3)}{2(m+1)}u^{m+1} +\frac{m}{2}u^{m-1}(\frac{1}{x_\beta}-1)-P\bigg)x_\beta\\ & = \bigg( \frac{m(m+3)}{2(m+1)}u^{m+1}-P-\frac{m}{2}u^{m-1} \bigg)x_\beta+\frac{m}{2}u^{m-1}. \end{array} \right. \end{equation} $ | (110) |
Step 3. Now we return to the original coordinates
Fixed a point
$ u_x^2(\tau,x) = \frac{1}{x_\beta(\tau,\tilde{\beta})}-1\geq 0,\; \; \; \; \; \; x_\beta(\tau,\tilde{\beta}) > 0. $ |
If
$ \begin{equation} u_x\bigg(t,x(t,\beta(t;\tau,\tilde{\beta}))\bigg) = \frac{u_\beta(t,\beta(t;\tau,\tilde{\beta}))} {x_\beta(t,\beta(t;\tau,\tilde{\beta}))}. \end{equation} $ | (111) |
From (110), we obtain that the map
$ \begin{equation} \begin{split} &\frac{d}{dt}u_x(t,x(t,\beta(t;\tau\tilde{\beta}))) = \frac{d}{dt}\big(\frac{u_\beta}{x_\beta}\big)\\ & = \frac{x_\beta\{ \bigg(\frac{m(m+3)}{2(m+1)}u^{m+1}-P-\frac{m}{2}u^{m-1} \bigg)x_\beta+\frac{m}{2}u^{m-1}x_\beta-u_\beta G_\beta \} -u_\beta\{mu^{m-1}u_\beta-x_\beta G_\beta\}}{x_\beta^2}\\ &\quad\; \; \; \; \; \; = \frac{m(m+3)}{2(m+1)}u^{m+1}-P-\frac{m}{2}u^{m-1} +\frac{mu^{m-1}}{2x_\beta}-\frac{u_\beta G_\beta}{x_\beta} -\frac{mu^{m-1}u_\beta^2}{x_\beta^2}+\frac{u_\beta G_\beta}{x_ \beta}\\ &\quad\; \; \; \; \; \; = \frac{m(m+3)}{2(m+1)}u^{m+1}-P-\frac{m}{2}u^{m-1} +\frac{mu^{m-1}}{2x_\beta} -\frac{mu^{m-1}u_\beta^2}{x_\beta^2}. \end{split} \end{equation} $ | (112) |
Thus, for
$ \begin{equation} \begin{split} &\frac{d}{dt}\arctan u_x(t,x(t,\beta(t;\tau\tilde{\beta}))) = \frac{1}{1+u_x^2}\cdot \frac{d}{dt}u_x\\ &\quad\; \; \; \; \; \; \; \; \; = \bigg(\frac{m(m+3)}{2(m+1)}u^{m+1}-P-\frac{m}{2}u^{m-1} +\frac{mu^{m-1}}{2x_\beta} -\frac{mu^{m-1}u_\beta^2}{x_\beta^2} \bigg)x_\beta\\ &\quad\; \; \; \; \; \; \; \; \; = \big(\frac{m(m+3)}{2(m+1)}u^{m+1}-P-\frac{m}{2}u^{m-1}\big)x_\beta+\frac{mu^{m-1}}{2} -\frac{mu^{m-1}u_\beta^2}{x_\beta}\\ &\quad\; \; \; \; \; \; \; \; \; = \big(\frac{m(m+3)}{2(m+1)}u^{m+1}-mu^{m-1}u_x^2-P-\frac{m}{2}u^{m-1}\big)x_\beta+\frac{mu^{m-1}}{2}. \end{split} \end{equation} $ | (113) |
Step 4. Introduce the function
$ \begin{equation} \theta\doteq\left\{ \begin{array}{llll} 2\arctan u_x\; \; \; \; \; \; \text{if} \; 0 < x_\beta\leq1,\\ \pi \; \; \; \; \; \; \text{if} \; x_\beta = 0. \end{array} \right. \end{equation} $ | (114) |
This yields
$ \begin{equation} x_\beta = \frac{1}{1+u_x^2} = \cos^2\frac{\theta}{2},\; \; \; \frac{u_x}{1+u_x^2} = \frac{1}{2}\sin \theta,\; \; \; \frac{u_x^2}{1+u_x^2} = \sin^2\frac{\theta}{2}. \end{equation} $ | (115) |
where
$ \begin{equation} \frac{d}{dt}\theta(t) = \theta_t = \bigg( \frac{m(m+3)}{m+1}u^{m+1} -2P\bigg)\cos^2\frac{\theta}{2}- \frac{m}{2}u^{m-1} \sin^2\frac{\theta}{2}, \end{equation} $ | (116) |
along each good characteristic. In fact, for simplicity, denote by
$ \begin{equation} u_x^2(t) = \frac{1-x_\beta(t)}{x_\beta(t)}, \end{equation} $ | (117) |
which is valid as long as
Step 5. If
$ \begin{equation} \left\{ \begin{array}{llll} \frac{d}{dt}\beta(t,\tilde{\beta}) = G(t,\beta(t,\tilde{\beta})),\\ \frac{d}{dt}x(t,\beta(t,\tilde{\beta})) = u^m(t,\beta(t,\tilde{\beta})),\\ \frac{d}{dt}u(t,\beta(t,\tilde{\beta})) = -P_x(t,\beta(t,\tilde{\beta})),\\ \frac{d}{dt}v(t,\beta(t,\tilde{\beta})) = \bigg( \frac{m(m+3)}{m+1}u^{m+1} -2P\bigg)\cos^2\frac{\theta}{2}- \frac{m}{2}u^{m-1} \sin^2\frac{\theta}{2}. \end{array} \right. \end{equation} $ | (118) |
Recalling the definition of
$ \begin{equation} \begin{split} &P(x(\beta)) = \frac{1}{2}\int_{-\infty}^{\infty}\exp\bigg\{-\bigg|\int_\beta^{\beta'}\cos^2\frac{v(s)}{2}ds\bigg|\bigg\}\\ &\quad\; \; \; \; \; \; \cdot\bigg[\bigg(\frac{m(m+3)}{2(m+1)}u^{m+1} \cos^2\frac{\theta }{2} + \frac{m}{2}u^{m-1} \sin^2\frac{\theta }{2}\bigg)(\beta') \bigg]d\beta', \end{split} \end{equation} $ | (119) |
$ \begin{equation} \begin{split} &P_x(x(\beta)) = \frac{1}{2}\bigg(\int_{\xi}^\infty-\int_{-\infty}^\xi\bigg) \exp\bigg\{-\bigg|\int_\xi^{\beta'}\cos^2\frac{v(s)}{2}ds\bigg|\bigg\}\\ &\quad\; \; \; \; \; \; \cdot\bigg[\bigg(\frac{m(m+3)}{2(m+1)}u^{m+1} \cos^2\frac{\theta }{2} + \frac{m}{2}u^{m-1} \sin^2\frac{\theta }{2}\bigg)(\beta') \bigg]d\beta'. \end{split} \end{equation} $ | (120) |
For every
$ \begin{equation} \left\{ \begin{array}{llll} \beta(0,\tilde{\beta}) = \tilde{\beta},\\ x(0,\tilde{\beta}) = x(0,\tilde{\beta}),\\ u(0,\tilde{\beta}) = u_0(x(0,\tilde{\beta})),\\ \theta(0,\tilde{\beta}) = 2\arctan u_{0,x}(x(0,\tilde{\beta})). \end{array} \right. \end{equation} $ | (121) |
Since the Lipschitz continuity of coefficients, the Cauchy problem (118), (121) has a unique solution with initial data condition (121), which is globally defined for all
Step 6. Suppose that
$ x(t,\beta) = \bar{x}(t,\beta),\; \; \; \; \; \; u(t,\beta) = \bar{u}(t,\beta). $ |
In turn, for a.e.
$ u(t,x) = u(t,\beta^*(t,x)) = \bar{u}(t,\bar{\beta}^*(t,x)) = \bar{u}(t,x). $ |
To prove the singularities of the solution for
Lemma 5.1. Let
(1) If
i) When
ii) At the point
$ rank\; D_\lambda(v^\lambda,\theta^\lambda_\xi,\theta^\lambda_{\xi\xi}) = 3. $ |
2) If
Proof. Let
$ \begin{equation} \begin{split} \frac{\partial}{\partial_t}\theta_\xi(t,\xi)& = \bigg(\frac{m(m+3)}{2}u^mu_\xi-P_\xi\bigg)(1+\cos \theta)\\ &-\bigg( \frac{m(m+3)}{2(m+1)} u^{m+1} -P+\frac{m}{2}u^{m-1} \bigg)\sin \theta \theta_\xi-\frac{m(m-1)}{2}u^{m-2} u_\xi(1-\cos \theta), \end{split} \end{equation} $ | (122) |
$ \begin{equation} \begin{split} \frac{\partial}{\partial_t}\theta_t(t,\xi)& = \bigg(\frac{m(m+3)}{2}u^mP_x-P_t\bigg)(1+\cos \theta)\\ &\quad\; \; \; -\bigg( \frac{m(m+3)}{2(m+1)} u^{m+1} -P+\frac{m}{2}u^{m-1} \bigg)\sin \theta \bigg[\bigg( \frac{m(m+3)}{m+1}u^{m+1} -2P\bigg) \end{split} \end{equation} $ | (123) |
$ \begin{equation*} \begin{split} &\quad\; \; \; \; \; \; \cos^2\frac{\theta}{2}- mu^{m-1} \sin^2\frac{\theta}{2}\bigg]+\frac{m(m-1)}{2}u^{m-2} P_x(1-\cos \theta), \end{split} \end{equation*} $ |
$ \begin{equation} \begin{split} \frac{\partial}{\partial_t}\theta_{\xi\xi}(t,\xi)& = \bigg(\frac{m^2(m+3)}{2}u^{m-1}u_\xi^2+\frac{m(m+3)}{2}u^mu_{\xi\xi} -P_{\xi\xi}\bigg)(1+\cos \theta)\\ &\quad\; \; \; -\bigg( m(m+3)u^mu_\xi-2P_\xi +\frac{m(m-1)}{2}u^{m-2}u_\xi\bigg)\sin \theta\theta_\xi\\ &\quad\; \; \; -\bigg(\frac{m(m+3)}{2(m+1)}u^{m+1}-P+\frac{m}{2}u^{m-1}(\cos\theta\theta_\xi^2+\sin \theta\theta_{\xi\xi})\bigg)\\ &-\frac{m(m-1)}{2}u^{m-2}u_\xi \sin \theta\theta_\xi -\frac{m(m-1)}{2}u^{m-2}u_\xi^2-\frac{m(m-1)(m-2)}{2}u^{m-3}u_\xi^2, \end{split} \end{equation} $ | (124) |
$ \begin{equation} \begin{split} \frac{\partial}{\partial_t}h_\xi(t,\xi)& = \bigg(\frac{m(m+3)}{2(m+1)}u^{m+1}+\frac{m}{2}u^{m-1}-P\bigg) \sin \theta h_\xi)\\ &\quad\; \; \; \bigg(\frac{m(m+3)}{2(m+1)}u^{m+1}+\frac{m}{2}u^{m-1}-P\bigg) \cos \theta \theta_\xi h\\ &\quad\; \; \; \bigg(\frac{m(m+3)}{2}u^mu_\xi+\frac{m(m-1)}{2}u^{m-2}u_\xi-P_\xi\bigg)\sin \theta h, \end{split} \end{equation} $ | (125) |
with
$ \begin{equation} u_\xi = \frac{1}{2}\sin \theta h,\quad u_{\xi\xi} = \frac{1}{2}\sin \theta h_\xi+\frac{1}{2}\cos \theta h \theta_\xi. \end{equation} $ | (126) |
We consider the families
$ \begin{equation} \bar{u}^\lambda = \bar{u}(\xi)+\sum\limits_{i = 1,2,3}\lambda_iU_i(\xi), \end{equation} $ | (127) |
$ \begin{equation} \bar{\theta}^\lambda = \bar{\theta}(\xi)+\sum\limits_{i = 1,2,3}\lambda_i\Theta_i(\xi), \end{equation} $ | (128) |
$ \begin{equation} \bar{h}^\lambda = \bar{h}(\xi)+\sum\limits_{i = 1,2,3}\lambda_iH_i(\xi). \end{equation} $ | (129) |
Together with (27), ((122), (124) and (125) form a complete system. Next, the following lemma will be used to get the rank which we desired.
Lemma 5.2 ([5]). Consider the following ODE system
$ \begin{equation} \begin{split} \frac{d}{d_t}u^\epsilon = g(u^\epsilon),\quad u^\epsilon(0) = u_0+\epsilon_1v_1+\cdot\cdot\cdot+\epsilon_k v_k, \end{split} \end{equation} $ | (130) |
where
$ \begin{equation} D_\epsilon u^\epsilon_0 = (v_1,v_2,\cdot\cdot\cdot,v_k)\in \mathbb{R}^{n\times k}, \end{equation} $ | (131) |
and the rank of this matrix is
$ \begin{equation} rank(D_\epsilon u^\epsilon_0) = l. \end{equation} $ | (132) |
Then
Proof Lemma 5.1. By (27), (122) and (124), we achieve an ODE system as follows,
$ \begin{equation} \begin{split} &\frac{\partial}{\partial_t}\left( \begin{array}{llll} u\\ \theta\\ h\\ \theta_\xi\\ \theta_{\xi\xi}\\ \end{array} \right) = \\& \left( \begin{array}{llll} -P_x\\ \bigg( \frac{m(m+3)}{m+1}u^{m+1} -2P\bigg)\cos^2\frac{\theta}{2}- mu^{m-1} \sin^2\frac{\theta}{2}\\ \left(\frac{m(m+3)}{2(m+1)}u^{m+1}+\frac{m}{2}u^{m-1}-P\right)\sin \theta\cdot h\\ \bigg(\frac{m(m+3)}{2}u^mu_\xi-P_\xi\bigg)(1+\cos \theta)\\ -\bigg( \frac{m(m+3)}{2(m+1)} u^{m+1} -P+\frac{m}{2}u^{m-1} \bigg)\sin \theta \theta_\xi\\ -\frac{m(m-1)}{2}u^{m-2} u_\xi(1-\cos \theta)\\ \bigg(\frac{m^2(m+3)}{2}u^{m-1}u_\xi^2+\frac{m(m+3)}{2}u^mu_{\xi\xi} -P_{\xi\xi}\bigg)(1+\cos \theta)\\ -\bigg( m(m+3)u^mu_\xi-2P_\xi +\frac{m(m-1)}{2}u^{m-2}u_\xi\bigg)\sin \theta\theta_\xi\\ -\bigg(\frac{m(m+3)}{2(m+1)}u^{m+1}-P+\frac{m}{2}u^{m-1}(\cos\theta\theta_\xi^2+\sin \theta\theta_{\xi\xi})\bigg)\\ -\frac{m(m-1)}{2}u^{m-2}u_\xi \sin \theta\theta_\xi -\frac{m(m-1)}{2}u^{m-2}u_\xi^2-\frac{m(m-1)(m-2)}{2}u^{m-3}u_\xi^2 \end{array} \right). \end{split} \end{equation} $ | (133) |
Then we establish a family of solutions
$ \begin{equation} \frac{\partial}{\partial_t}\left( \begin{array}{llll} D_\lambda \bar{u}^\lambda\\ D_\lambda \bar{\theta}^\lambda\\ D_\lambda \bar{h}^\lambda\\ D_\lambda \bar{\theta}_\xi^\lambda\\ D_\lambda \bar{\theta}_{\xi\xi}^\lambda\\ \end{array} \right) = \left( \begin{array}{llll} D_\lambda g^\lambda_1\\ D_\lambda g^\lambda_2\\ D_\lambda g^\lambda_3\\ D_\lambda g^\lambda_4\\ D_\lambda g^\lambda_5\\ \end{array} \right), \end{equation} $ | (134) |
where
$ \begin{equation} \begin{split} &\frac{\partial}{\partial_t}\left( \begin{array}{llll} D_\lambda\bar{u}^\lambda\\ D_\lambda \bar{\theta}^\lambda\\ D_\lambda \bar{h}^\lambda\\ D_\lambda \bar{\theta}_\xi^\lambda\\ D_\lambda \bar{\theta}_{\xi\xi}^\lambda\\ \end{array} \right)\\& = \left( \begin{array}{llll} D_u g^\lambda_1\; D_\theta g^\lambda_1\; D_h g^\lambda_1\; D_{\theta_\xi} g^\lambda_1\; D_{\theta_{\xi\xi}} g^\lambda_1\\ D_u g^\lambda_2\; D_\theta g^\lambda_2\; D_h g^\lambda_2\; D_{\theta_\xi} g^\lambda_2\; D_{\theta_{\xi\xi}} g^\lambda_2\\ D_u g^\lambda_3\; D_\theta g^\lambda_3\; D_h g^\lambda_3\; D_{\theta_\xi} g^\lambda_3\; D_{\theta_{\xi\xi}} g^\lambda_3\\ D_u g^\lambda_4\; D_\theta g^\lambda_4\; D_h g^\lambda_4\; D_{\theta_\xi} g^\lambda_4\; D_{\theta_{\xi\xi}} g^\lambda_4\\ D_u g^\lambda_5\; D_\theta g^\lambda_5\; D_h g^\lambda_5\; D_{\theta_\xi} g^\lambda_5\; D_{\theta_{\xi\xi}} g^\lambda_5\\ \end{array} \right)\cdot \left( \begin{array}{llll} D_{\lambda_1} \bar{u}^\lambda \; D_{\lambda_2} \bar{u}^\lambda\; D_{\lambda_3} \bar{u}^\lambda\\ D_{\lambda_1} \bar{\theta}^\lambda\; D_{\lambda_2} \bar{\theta}^\lambda\; D_{\lambda_3} \bar{\theta}^\lambda\\ D_{\lambda_1} \bar{h}^\lambda\; D_{\lambda_2} \bar{h}^\lambda\; D_{\lambda_3} \bar{h}^\lambda\\ D_{\lambda_1} \bar{\theta}_\xi^\lambda\; D_{\lambda_2} \bar{\theta}_\xi^\lambda\; D_{\lambda_3} \bar{\theta}_\xi^\lambda\\ D_{\lambda_1} \bar{\theta}_{\xi\xi}^\lambda\; D_{\lambda_2} \bar{\theta}_{\xi\xi}^\lambda\; D_{\lambda_3} \bar{\theta}_{\xi\xi}^\lambda\\ \end{array} \right). \end{split} \end{equation} $ | (135) |
Based on Lemma 5.2, we need to explain the Lipschitz continuity of
$ \begin{equation} rank D_\lambda \left( \begin{array}{llll} \theta\\ \theta_{\xi}\\ \theta_{\xi\xi} \end{array} \right) = 3. \end{equation} $ | (136) |
The system (27) combine with (122), (123) forms a complete system. By choosing suitable perturbation
$ \begin{equation} rank D_\lambda \left( \begin{array}{llll} \theta\\ \theta_{\xi}\\ \theta_{t} \end{array} \right) = 3, \end{equation} $ | (137) |
while
Next, we are going to investigate smooth solutions to the semi-linear system (27), and determine the generic structure of level sets
Lemma 5.3 ([5]). Given a compact domain
$ \mathcal{D}: = \{(t,\xi);0\leq t\leq T,|\xi|\leq M\}, $ |
let
$ \begin{equation} (\theta,\theta_\xi,\theta_{\xi\xi}) = (\pi,0,0),\; \; \; (\theta,\theta_\xi,\theta_t) = (\pi,0,0) \end{equation} $ | (138) |
cannot be obtained. Then
Proof of Theorem 1.4. Step 1. For convenience, we denote the space
$ \mathcal{M}: = C^3(\mathbb{R}^+)\cap H^1(\mathbb{R^+}), $ |
with the norm
$ \|u_0\|_{\mathcal{M}}: = \|u_0\|_{C^3}+\|u_0\|_{H^1}. $ |
Given a initial data
$ B_\delta: = \{u_0\in\mathcal{M};\|u_0-u^*_0\|_{\mathcal{M}} < \delta\}. $ |
By the definition of the space of
We define the subset
Step 2. We now claim the set
$ (\theta^n,\theta^n_\xi,\theta^n_{\xi\xi})(t^n,\theta^n) = (\pi,0,0), \; \; \; (t^n,x^n(t^n,\xi^n))\in\Delta $ |
for all
$ (\theta,\theta_\xi,\theta_{\xi\xi})(t,\xi = (\pi,0,0),\; \; \; (t,x(t,\xi))\in \Delta. $ |
This denotes
Step 3. We explain that
ⅰ) for every
ⅱ) The
$ \lim\limits_{n\rightarrow \infty}\|(u^n-u,\theta^n-\theta,h^n-h,x^n-x)\|_{C^k(I)} = 0, $ |
for every bounded set
$ \lim\limits_{n\rightarrow \infty}\|u_0^n-u_0\|_{C^k[a,b]} = 0 $ |
for every bounded set
Introducing a cutoff function
$ \begin{equation} p(x) = \left\{ \begin{array}{llll} 1,\; if |x|\leq r,\\ 0,\; if |x|\geq r+1, \end{array} \right. \end{equation} $ | (139) |
where
$ \tilde{u}_0^v: = p u_0^n+(1-p)u_0. $ |
We obtain
$ \lim\limits_{n\rightarrow \infty}\|\tilde{u}_0^n-u_0\|_{\mathcal{M}} = 0. $ |
Furthermore, choosing
$ \tilde{u}^n(t,x) = u^n(t,x). $ |
It is obvious that
Step 4. Finally, we prove that, for every initial data
Case Ⅰ.
Case Ⅱ.
$ \mathcal{W}^\theta: = \{(t,\xi)\in \Delta; \theta(t,\xi) = \pi\} $ |
is the union of finitely many
1. | Zhiying Meng, Zhaoyang Yin, Blow-up phenomena and the local well-posedness and ill-posedness of the generalized Camassa–Holm equation in critical Besov spaces, 2022, 0026-9255, 10.1007/s00605-022-01719-9 |