1. Introduction
In this paper, we address the relation between symmetric waves and traveling waves for a Camassa-Holm (CH) type equation with higher-order nonlinearity 1.1 ) has the nonlinearity of degree k + 1 and includes three integrable members, namely the CH equation (k = 1, b = 2)
$\begin{eqnarray}{u}_{t}-{u}_{txx}+(b+1){u}^{k}{u}_{x}-b{u}^{k-1}{u}_{x}{u}_{xx}-{u}^{k}{u}_{xxx}=0,\end{eqnarray}$
which was proposed by Grayshan and Himonas in [1]. Here $k\in {{\mathbb{N}}}^{+}$, $b\in {\mathbb{R}}$. The physical background of this relation is that of irrotational two-dimensional gravity water waves for a perfect fluid over a flat bottom [2]. This equation possesses peakon solutions [1] of the form $u(t,x)={c}^{\frac{1}{k}}{{\rm{e}}}^{-| x-ct| }$, c > 0. The local well-posedness, ill-posedness, blow-up criterions, global existence of strong and weak solutions, global analyticity, persistence properties, and asymptotic behavior of solutions to this equation have been investigated in [3-11]. Equation ( $\begin{eqnarray}{u}_{t}-{u}_{txx}+3u{u}_{x}-2{u}_{x}{u}_{xx}-u{u}_{xxx}=0,\end{eqnarray}$
the Degasperis-Procesi (DP) equation (k = 1, b = 3) $\begin{eqnarray}{u}_{t}-{u}_{txx}+4u{u}_{x}-3{u}_{x}{u}_{xx}-u{u}_{xxx}=0,\end{eqnarray}$
and the Novikov equation (k = 2, b = 3) $\begin{eqnarray}{u}_{t}-{u}_{txx}+4{u}^{2}{u}_{x}-3u{u}_{x}{u}_{xx}-{u}^{2}{u}_{xxx}=0.\end{eqnarray}$
The CH equation (1.2 ) was originally obtained by Fuchssteiner and Fokas [12] as an abstract bi-Hamiltonian equation with an infinite sequence of conservation laws, and subsequently re-obtained as an approximation of the incompressible Euler equations by Camassa and Holm [13] from physical principles (see [14] for a rigorous derivation). Moreover, the CH equation is a re-expression of the geodesic flow on the diffeomorphism group with the kinetic energy metric induced by the H1-norm [15], which indicates that it possesses the H1-norm conservation law. The CH equation has received much attention, and there have been numerous results on the well-posedness, blow-up via wave breaking, orbital stability and asymptotic stability of the peakon solutions, persistence properties, symmetry of traveling wave solutions, analyticity, Gevrey regularity, etc. [16-29].
The DP equation (1.3 ) was discovered by Degasperis and Procesi [30] via the method of asymptotic integrability. The DP equation has a similar structure and shares a lot of properties with the famous CH equation. It is also an approximation to the incompressible Euler equations, and its asymptotic accuracy is the same as for the CH equation [14]. It is completely integrable and admits the same explicit peakon solutions [31]. Both CH and DP equations can exhibit the wave-breaking phenomenon [19, 20, 32]. However, in contrast to the CH equation, the DP equation has shock peaked waves [33] of the form
$\begin{eqnarray*}u(t,x)=-\frac{1}{t+k}\mathrm{sgn}(x){{\rm{e}}}^{-| x| },\quad k\gt 0.\end{eqnarray*}$
Another important difference between these two equations is that the CH equation conserves the H1-norm of solutions while the DP equation does not [32].The Novikov equation (1.4 ) is an integrable peakon model with cubic nonlinearity and can be viewed as a generalization of the CH equation or the DP equation. This equation was first derived by Novikov [34] from the classification of generalized CH-type equations having infinite hierarchies of higher symmetries. Recently it was re-derived in the context of shallow water waves [35]. After its derivation, the qualitative properties of solutions to the Novikov equation have been investigated intensively, such as the local well-posedness [36-39], global existence of strong and weak solutions [38, 40-42], blow-up [43, 44], orbital stability and asymptotic stability of the peakon solutions [45, 46].
Inspired by the work in [26], the purpose of this paper is to prove that horizontally symmetric weak solutions of equation (1.1 ) are traveling wave solutions. A key step of the proof is to give the form of a weak solution of (1.1 ), cf. Lemma 2.1 . The detailed proof is given in the following section.
2. Symmetry of traveling wave solutions
In this section, we shall consider a unique x-symmetric weak solution of equation (1.1 ). Following [26], we will prove that such a solution must be a traveling wave. First, we recall the definitions of an x-symmetric solution and a weak solution of (1.1 ).
[26] A solution u is x-symmetric if there exists a function $a(t)\in {C}^{1}({{\mathbb{R}}}_{+})$ such that
$\begin{eqnarray*}u(t,x)=u(t,2a(t)-x),\quad \forall \,t\in (0,\infty ),\end{eqnarray*}$
for almost every $x\in {\mathbb{R}}$. We call a(t) the symmetric axis of u. We say u is a weak solution of equation (
$\begin{eqnarray}\begin{array}{l}\displaystyle \int {\displaystyle \int }_{{{\mathbb{R}}}_{+}\times {\mathbb{R}}}\left[u(1-{\partial }_{x}^{2}){\phi }_{t}-\frac{(b-k)(k-1)}{2}{u}^{k-2}{u}_{x}^{3}\phi \right.\\ \quad +\left(\frac{b+1}{k+1}{u}^{k+1}+\frac{3k-b}{2}{u}^{k-1}{u}_{x}^{2}\right){\phi }_{x}\\ \quad \,\left.-\frac{1}{k+1}{u}^{k+1}{\phi }_{xxx}\right]{\rm{d}}t{\rm{d}}x=0,\end{array}\end{eqnarray}$
for all $\phi \in {C}_{0}^{\infty }({{\mathbb{R}}}_{+}\times {\mathbb{R}})$.Further introducing the notation $\langle$ ·,· $\rangle$ for distributions, (2.1 ) becomes
$\begin{eqnarray}\begin{array}{l}\left\langle u,(1-{\partial }_{x}^{2}){\phi }_{t}\right\rangle -\left\langle \frac{(b-k)(k-1)}{2}{u}^{k-2}{u}_{x}^{3},\phi \right\rangle \\ +\left\langle \frac{b+1}{k+1}{u}^{k+1}+\frac{3k-b}{2}{u}^{k-1}{u}_{x}^{2},{\phi }_{x}\right\rangle \\ -\left\langle \frac{1}{k+1}{u}^{k+1},{\phi }_{xxx}\right\rangle =0.\end{array}\end{eqnarray}$
Now we prove the following lemma.
Assume that $U(x)\in {H}^{1}({\mathbb{R}})$ and it satisfies
$\begin{eqnarray}\begin{array}{l}{\displaystyle \int }_{{\mathbb{R}}}\left[-cU(1-{\partial }_{x}^{2}){\varphi }_{x}-\frac{(b-k)(k-1)}{2}{U}^{k-2}{U}_{x}^{3}\varphi \right.\\ +\left(\frac{b+1}{k+1}{U}^{k+1}+\frac{3k-b}{2}{U}^{k-1}{U}_{x}^{2}\right){\varphi }_{x}\\ \quad \,\left.-\frac{1}{k+1}{U}^{k+1}{\varphi }_{xxx}\right]{\rm{d}}x=0,\end{array}\end{eqnarray}$
for all $\varphi \in {C}_{0}^{\infty }({\mathbb{R}})$. Then we obtain a weak solution of ( $\begin{eqnarray}u(t,x)=U(x-c(t-{t}_{0})),\end{eqnarray}$
for a fixed ${t}_{0}\in {\mathbb{R}}$.We set without loss of generality that t0 = 0. We follow the arguments in [26] to get $u\in C({\mathbb{R}},{H}^{1}({\mathbb{R}}))$. Next we define
$\begin{eqnarray*}{\phi }_{c}(t,x)=\phi (t,x+ct),\end{eqnarray*}$
where $\phi \in {C}_{0}^{\infty }({{\mathbb{R}}}_{+}\times {\mathbb{R}})$. Then one has $\begin{eqnarray}{\partial }_{x}({\phi }_{c})={({\phi }_{x})}_{c},\quad {\partial }_{t}({\phi }_{c})={({\phi }_{t})}_{c}+c{({\phi }_{x})}_{c}.\end{eqnarray}$
Suppose u(t, x) = U(x - ct), we have $\begin{eqnarray}\begin{array}{rcl}\langle u,\phi \rangle & = & \langle U,{\phi }_{c}\rangle ,\quad \left\langle {u}^{k+1},\phi \right\rangle =\left\langle {U}^{k+1},{\phi }_{c}\right\rangle ,\\ \left\langle {u}^{k-2}{u}_{x}^{3},\phi \right\rangle & = & \left\langle {U}^{k-2}{U}_{x}^{3},{\phi }_{c}\right\rangle ,\\ \left\langle {u}^{k-1}{u}_{x}^{2},\phi \right\rangle & = & \left\langle {U}^{k-1}{U}_{x}^{2},{\phi }_{c}\right\rangle ,\end{array}\end{eqnarray}$
where U = U(x). From ( $\begin{eqnarray}\begin{array}{l}\left\langle u,(1-{\partial }_{x}^{2}){\phi }_{t}\right\rangle =\left\langle U,{\left((1-{\partial }_{x}^{2}){\phi }_{t}\right)}_{c}\right\rangle \\ =\left\langle U,(1-{\partial }_{x}^{2})({\partial }_{t}{\phi }_{c}-c{\partial }_{x}{\phi }_{c})\right\rangle ,\\ \left\langle \frac{(b-k)(k-1)}{2}{u}^{k-2}{u}_{x}^{3},\phi \right\rangle \\ =\left\langle \frac{(b-k)(k-1)}{2}{U}^{k-2}{U}_{x}^{3},{\phi }_{c}\right\rangle ,\\ \left\langle \frac{b+1}{k+1}{u}^{k+1}+\frac{3k-b}{2}{u}^{k-1}{u}_{x}^{2},{\phi }_{x}\right\rangle \\ =\left\langle \frac{b+1}{k+1}{U}^{k+1}+\frac{3k-b}{2}{U}^{k-1}{U}_{x}^{2},{\partial }_{x}{\phi }_{c}\right\rangle ,\\ \left\langle \frac{1}{k+1}{u}^{k+1},{\phi }_{xxx}\right\rangle =\left\langle \frac{1}{k+1}{U}^{k+1},{({\phi }_{xxx})}_{c}\right\rangle \\ =\left\langle \frac{1}{k+1}{U}^{k+1},{\partial }_{x}^{3}{\phi }_{c}\right\rangle .\end{array}\end{eqnarray}$
Since U is independent of time, so it does not belong to the support of Φc for sufficiently large T. This implies $\begin{eqnarray}\begin{array}{rcl}\left\langle U,(1-{\partial }_{x}^{2}){\partial }_{t}{\phi }_{c}\right\rangle & = & {\displaystyle \int }_{{\mathbb{R}}}U(x){\displaystyle \int }_{{{\mathbb{R}}}_{+}}{\partial }_{t}(1-{\partial }_{x}^{2}){\phi }_{c}\,{\rm{d}}t{\rm{d}}x\\ & & ={\displaystyle \int }_{{\mathbb{R}}}U(x)\left[(1-{\partial }_{x}^{2}){\phi }_{c}(T,x)\right.\\ & & \left.-(1-{\partial }_{x}^{2}){\phi }_{c}(0,x)\right]\,{\rm{d}}x=0,\\ \langle U,{\partial }_{t}{\phi }_{c}\rangle & = & {\displaystyle \int }_{{\mathbb{R}}}U(x){\displaystyle \int }_{{{\mathbb{R}}}_{+}}{\partial }_{t}{\phi }_{c}\,{\rm{d}}t{\rm{d}}x\\ & = & {\displaystyle \int }_{{\mathbb{R}}}U(x)\left[{\phi }_{c}(T,x)-{\phi }_{c}(0,x)\right]\,{\rm{d}}x=0.\end{array}\end{eqnarray}$
Then, ( $\begin{eqnarray*}\begin{array}{l}\,\,\left\langle u,(1-{\partial }_{x}^{2}){\phi }_{t}\right\rangle -\left\langle \frac{(b-k)(k-1)}{2}{u}^{k-2}{u}_{x}^{3},\phi \right\rangle \\ +\left\langle \frac{b+1}{k+1}{u}^{k+1}+\frac{3k-b}{2}{u}^{k-1}{u}_{x}^{2},{\phi }_{x}\right\rangle \\ -\left\langle \frac{1}{k+1}{u}^{k+1},{\phi }_{xxx}\right\rangle \\ =\left\langle U,-c(1-{\partial }_{x}^{2}){\partial }_{x}{\phi }_{c}\right\rangle -\left\langle \frac{(b-k)(k-1)}{2}{U}^{k-2}{U}_{x}^{3},{\phi }_{c}\right\rangle \\ +\left\langle \frac{b+1}{k+1}{U}^{k+1}+\frac{3k-b}{2}{U}^{k-1}{U}_{x}^{2},{\partial }_{x}{\phi }_{c}\right\rangle \\ -\left\langle \frac{1}{k+1}{U}^{k+1},{\partial }_{x}^{3}{\phi }_{c}\right\rangle ={\displaystyle \int }_{{{\mathbb{R}}}_{+}}{\displaystyle \int }_{{\mathbb{R}}}\left[-cU(1-{\partial }_{x}^{2}){\partial }_{x}{\phi }_{c}\right.\\ -\frac{(b-k)(k-1)}{2}{U}^{k-2}{U}_{x}^{3}{\phi }_{c}\\ +\left(\frac{b+1}{k+1}{U}^{k+1}+\frac{3k-b}{2}{U}^{k-1}{U}_{x}^{2}\right){\partial }_{x}{\phi }_{c}\\ \left.-\frac{1}{k+1}{U}^{k+1}{\partial }_{x}^{3}{\phi }_{c}\right]{\rm{d}}x{\rm{d}}t=0,\end{array}\end{eqnarray*}$
where we have used (Next, we give our main result.
Let u(t, x) be x-symmetric. If u(t, x) is a unique weak solution of equation (
Since ${C}_{0}^{\infty }({{\mathbb{R}}}_{+}\times {\mathbb{R}})$ is dense in ${C}_{0}^{1}({{\mathbb{R}}}_{+},{C}_{0}^{3}({\mathbb{R}}))$, we only consider the test function Φ in Definition
$\begin{eqnarray*}{\phi }_{a}(t,x)=\phi (t,2a(t)-x),\quad a(t)\in {C}^{1}({\mathbb{R}}).\end{eqnarray*}$
It is easy to see that ${({\phi }_{a})}_{a}=\phi $ and $\begin{eqnarray}\begin{array}{l}{\partial }_{x}{u}_{a}=-{({\partial }_{x}u)}_{a},\quad {\partial }_{x}{\phi }_{a}=-{({\partial }_{x}\phi )}_{a},\\ {\partial }_{t}{\phi }_{a}={({\partial }_{t}\phi )}_{a}+2\dot{a}{({\partial }_{x}\phi )}_{a},\end{array}\end{eqnarray}$
where $\dot{a}$ represents the time derivative of a(t). Moreover, we deduce that $\begin{eqnarray}\begin{array}{l}\langle {u}_{a},\phi \rangle =\langle u,{\phi }_{a}\rangle ,\quad \left\langle {u}_{a}^{k-2}{({\partial }_{x}{u}_{a})}^{3},\phi \right\rangle =-\left\langle {u}^{k-2}{u}_{x}^{3},{\phi }_{a}\right\rangle ,\\ \left\langle {u}_{a}^{k+1},\phi \right\rangle =\left\langle {u}^{k+1},{\phi }_{a}\right\rangle ,\quad \left\langle {u}_{a}^{k-1}{({\partial }_{x}{u}_{a})}^{2},\phi \right\rangle =\left\langle {u}^{k-1}{u}_{x}^{2},{\phi }_{a}\right\rangle .\end{array}\end{eqnarray}$
Since u is x-symmetric, it follows from ( $\begin{eqnarray}\begin{array}{l}\left\langle u,(1-{\partial }_{x}^{2}){\phi }_{t}\right\rangle =\left\langle u,{\left((1-{\partial }_{x}^{2}){\partial }_{t}\phi \right)}_{a}\right\rangle \\ =\left\langle u,(1-{\partial }_{x}^{2})({\partial }_{t}{\phi }_{a}+2\dot{a}{\partial }_{x}{\phi }_{a})\right\rangle ,\\ \left\langle \frac{(b-k)(k-1)}{2}{u}^{k-2}{u}_{x}^{3},\phi \right\rangle \\ =-\left\langle \frac{(b-k)(k-1)}{2}{u}^{k-2}{u}_{x}^{3},{\phi }_{a}\right\rangle ,\\ \left\langle \frac{b+1}{k+1}{u}^{k+1}+\frac{3k-b}{2}{u}^{k-1}{u}_{x}^{2},{\phi }_{x}\right\rangle \\ =-\left\langle \frac{b+1}{k+1}{u}^{k+1}+\frac{3k-b}{2}{u}^{k-1}{u}_{x}^{2},{\partial }_{x}{\phi }_{a}\right\rangle ,\\ \left\langle \frac{1}{k+1}{u}^{k+1},{\phi }_{xxx}\right\rangle =-\left\langle \frac{1}{k+1}{u}^{k+1},{\partial }_{x}^{3}{\phi }_{a}\right\rangle .\end{array}\end{eqnarray}$
From ( $\begin{eqnarray}\begin{array}{l}\left\langle u,(1-{\partial }_{x}^{2}){\phi }_{t}\right\rangle -\left\langle \frac{(b-k)(k-1)}{2}{u}^{k-2}{u}_{x}^{3},\phi \right\rangle \\ +\left\langle \frac{b+1}{k+1}{u}^{k+1}+\frac{3k-b}{2}{u}^{k-1}{u}_{x}^{2},{\phi }_{x}\right\rangle -\left\langle \frac{1}{k+1}{u}^{k+1},{\phi }_{xxx}\right\rangle \\ =\left\langle u,(1-{\partial }_{x}^{2})({\partial }_{t}{\phi }_{a}+2\dot{a}{\partial }_{x}{\phi }_{a})\right\rangle +\left\langle \frac{(b-k)(k-1)}{2}{u}^{k-2}{u}_{x}^{3},{\phi }_{a}\right\rangle \\ -\left\langle \frac{b+1}{k+1}{u}^{k+1}+\frac{3k-b}{2}{u}^{k-1}{u}_{x}^{2},{\partial }_{x}{\phi }_{a}\right\rangle \\ +\left\langle \frac{1}{k+1}{u}^{k+1},{\partial }_{x}^{3}{\phi }_{a}\right\rangle =0.\end{array}\end{eqnarray}$
Owing to ${({\phi }_{a})}_{a}=\phi $, we take Φ = Φa in ( $\begin{eqnarray*}\begin{array}{l}\left\langle u,(1-{\partial }_{x}^{2})({\partial }_{t}\phi +2\dot{a}{\partial }_{x}\phi )\right\rangle +\left\langle \frac{(b-k)(k-1)}{2}{u}^{k-2}{u}_{x}^{3},\phi \right\rangle \\ \quad -\left\langle \frac{b+1}{k+1}{u}^{k+1}+\frac{3k-b}{2}{u}^{k-1}{u}_{x}^{2},{\partial }_{x}\phi \right\rangle \\ \quad +\left\langle \frac{1}{k+1}{u}^{k+1},{\partial }_{x}^{3}\phi \right\rangle =0,\end{array}\end{eqnarray*}$
which together with ( $\begin{eqnarray}\begin{array}{l}\left\langle u,\dot{a}(1-{\partial }_{x}^{2}){\partial }_{x}\phi \right\rangle +\left\langle \frac{(b-k)(k-1)}{2}{u}^{k-2}{u}_{x}^{3},\phi \right\rangle \\ \quad \,-\left\langle \frac{b+1}{k+1}{u}^{k+1}+\frac{3k-b}{2}{u}^{k-1}{u}_{x}^{2},{\partial }_{x}\phi \right\rangle \\ \quad \,+\left\langle \frac{1}{k+1}{u}^{k+1},{\partial }_{x}^{3}\phi \right\rangle =0.\end{array}\end{eqnarray}$
We introduce a mollifier ${\rho }_{\epsilon }\in {C}_{0}^{\infty }({{\mathbb{R}}}_{+})$ with ρε → δ(t - t0), the Dirac mass at a fixed t0, as ε → 0. Then we define the following test function $\begin{eqnarray*}{\phi }_{\epsilon }(t,x)=\varphi (x){\rho }_{\epsilon }(t),\quad {\rm{forany}}\,\varphi \in {C}_{0}^{\infty }({\mathbb{R}}).\end{eqnarray*}$
Substituting the above test function into ( $\begin{eqnarray}\begin{array}{l}{\displaystyle \int }_{{\mathbb{R}}}\left[(1-{\partial }_{x}^{2}){\partial }_{x}\varphi {\displaystyle \int }_{{{\mathbb{R}}}_{+}}\dot{a}u{\rho }_{\epsilon }(t)\,{\rm{d}}t\right]{\rm{d}}x\\ +{\displaystyle \int }_{{\mathbb{R}}}\left[\varphi {\displaystyle \int }_{{{\mathbb{R}}}_{+}}\frac{(b-k)(k-1)}{2}{u}^{k-2}{u}_{x}^{3}{\rho }_{\epsilon }(t)\,{\rm{d}}t\right]{\rm{d}}x\\ \quad -{\displaystyle \int }_{{\mathbb{R}}}\left[{\partial }_{x}\varphi {\displaystyle \int }_{{{\mathbb{R}}}_{+}}\left(\frac{b+1}{k+1}{u}^{k+1}+\frac{3k-b}{2}{u}^{k-1}{u}_{x}^{2}\right){\rho }_{\epsilon }(t)\,{\rm{d}}t\right]{\rm{d}}x\\ +\frac{1}{k+1}{\displaystyle \int }_{{\mathbb{R}}}\left({\partial }_{x}^{3}\varphi {\displaystyle \int }_{{{\mathbb{R}}}_{+}}{u}^{k+1}{\rho }_{\epsilon }(t)\,{\rm{d}}t\right){\rm{d}}x=0.\end{array}\end{eqnarray}$
Note that $\begin{eqnarray*}\mathop{\mathrm{lim}}\limits_{\epsilon \to 0}{\int }_{{{\mathbb{R}}}_{+}}\dot{a}u{\rho }_{\epsilon }(t)\,{\rm{d}}t=\dot{a}({t}_{0})u({t}_{0},x),\quad {\rm{in}}\,{L}^{2}({\mathbb{R}}),\end{eqnarray*}$
and $\begin{eqnarray*}\begin{array}{l}\mathop{\mathrm{lim}}\limits_{\epsilon \to 0}{\displaystyle \int }_{{{\mathbb{R}}}_{+}}{u}^{k-2}{u}_{x}^{3}{\rho }_{\epsilon }(t)\,{\rm{d}}t={u}^{k-2}({t}_{0},x){u}_{x}^{3}({t}_{0},x),\\ \mathop{\mathrm{lim}}\limits_{\epsilon \to 0}{\displaystyle \int }_{{{\mathbb{R}}}_{+}}\left(\frac{b+1}{k+1}{u}^{k+1}+\frac{3k-b}{2}{u}^{k-1}{u}_{x}^{2}\right){\rho }_{\epsilon }(t)\,{\rm{d}}t\\ \quad =\frac{b+1}{k+1}{u}^{k+1}({t}_{0},x)+\frac{3k-b}{2}{u}^{k-1}({t}_{0},x){u}_{x}^{2}({t}_{0},x),\\ \mathop{\mathrm{lim}}\limits_{\epsilon \to 0}{\displaystyle \int }_{{{\mathbb{R}}}_{+}}{u}^{k+1}{\rho }_{\epsilon }(t)\,{\rm{d}}t={u}^{k+1}({t}_{0},x),\end{array}\end{eqnarray*}$
in ${L}^{1}({\mathbb{R}})$. So, letting ε → 0 in ( $\begin{eqnarray*}\begin{array}{l}{\displaystyle \int }_{{\mathbb{R}}}\dot{a}({t}_{0})u({t}_{0},x)(1-{\partial }_{x}^{2}){\partial }_{x}\varphi \,{\rm{d}}x+\frac{(b-k)(k-1)}{2}\\ \quad \,{\displaystyle \int }_{{\mathbb{R}}}{u}^{k-2}({t}_{0},x){u}_{x}^{3}({t}_{0},x)\varphi \,{\rm{d}}x\\ \quad -{\displaystyle \int }_{{\mathbb{R}}}\left(\frac{b+1}{k+1}{u}^{k+1}({t}_{0},x)+\frac{3k-b}{2}{u}^{k-1}({t}_{0},x){u}_{x}^{2}({t}_{0},x)\right){\partial }_{x}\varphi \,{\rm{d}}x\\ \quad \,+\frac{1}{k+1}{\displaystyle \int }_{{\mathbb{R}}}{u}^{k+1}({t}_{0},x){\partial }_{x}^{3}\varphi \,{\rm{d}}x=0.\end{array}\end{eqnarray*}$
It is clear that u(t0, x) satisfies ( Theorem