\documentclass[reqno]{amsart}
\usepackage{hyperref}
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\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2014 (2014), No. 69, pp. 1--16.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2014 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2014/69\hfil Local solvability and blow-up]
{Local solvability and blow-up for Benjamin-Bona-Mahony-Burgers,
Rosenau-Burgers and Korteweg-de Vries-Benjamin-Bona-Mahony equations}


\author[M. O. Korpusov, E. V. Yushkov \hfil EJDE-2014/69\hfilneg]
{Maxim O. Korpusov, Egor V. Yushkov }  % in alphabetical order

\address{Maxim O. Korpusov \newline
Chair of Mathematics, Physics Faculty,
M.~V.~Lomonosov Moscow State University,
Moscow, Russia}
\email{korpusov@gmail.com}

\address{Egor V. Yushkov \newline
Chair of Mathematics, Physics Faculty,
M.~V.~Lomonosov Moscow State University,
Moscow, Russia}
\email{yushkov.msu@mail.ru}

\thanks{Submitted August 6, 2013. Published March 15, 2014.}
\subjclass[2000]{35B44}
\keywords{Local solvability; blow-up; Benjamin-Bona-Mahony-Burgers;
 \hfill\break\indent  Rosenau-Burgers; Korteweg-de Vries-Benjamin-Bona-Mahony}

\begin{abstract}
 In this article some well-known problems in mathematical physics for
 Benjamin-Bona-Mahony-Burgers, Rosenau-Burgers and
 Korteweg-de Vries-Benjamin-Bona-Mahony equations are considered.
 These equations describe important processes in different fields of physics,
 particulary in hydro- and electrodynamics. We study initial-boundary problems
 with natural physical boundary conditions. Sufficient conditions of local
 solvability and blow-up in finite time are obtained. For this the methods
 of contraction mapping and nonlinear capacity, developed by  Galaktionov,
 Pokhozhaev and Mitidieri, are used.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}[theorem]{Remark}
\allowdisplaybreaks

\section{Introduction}

The Kortweg-de Vries equation is well known in different fields of science
and technology,
\begin{equation}
u_t+uu_x+u_{xxx}=0.\label{eKdv}
\end{equation}
Recently  Pokhozaev obtained  sufficient conditions for the finite time
blow-up of solutions of initial-boundary problems for KdV equation
\cite{P1,P2,P3,P4}. He used the  powerful method of nonlinear capacity,
developed in \cite{Po}. Note that in these papers both classical, and
weak solutions of the problems were considered.

By the method of nonlinear capacity, we study the following three equations
which are important in different physical applications such as
waves on shallow water, and  processes in semiconductors with negative
differential conductivity:
Benjamin-Bona-Mahony-Burgers equation
\begin{equation}
\frac{\partial}{\partial t}(u_{xx}-u)+u_{xx}-uu_x=0,\label{eBBMB}
\end{equation}
Rosenau-Burgers equation:
\begin{equation}
\frac{\partial}{\partial t}(u_{xxxx}+u)+u_{xx}-uu_x=0,\label{eRB}
\end{equation}
Korteweg-de Vries-Benjamin-Bona-Mahony equation:
\begin{equation}
\frac{\partial}{\partial t}(u_{xx}-u)-u_{xxx}-uu_x=0.\label{eKdv-BBM}
\end{equation}
The literature on these equations is very extensive. Among them, we mention
the classical papers \cite{Albert-1989}--\cite{Rosenau1}.
Also we mention our papers \cite{Korpusov1,Korpusov2,Yushkov}, 
where sufficient conditions for blow-up in
one- and multidimensional \eqref{eBBMB}, \eqref{eRB} and
\eqref{eKdv-BBM} equations were obtained, but for other boundary conditions
and without solving of the local solvability question.

Finally, for completeness we recall that there are three main methods
for  studying blow-up: The method of nonlinear capacity, developed by
 Pokhozhaev and  Mitidieri  \cite{Mitidieri,Po,Pohozaev1};
the energy method developed by
Levine \cite{Al,Kal,Levine6,Levine1,Levine3,Levine4,Levine5};
 and the method, based on maximum principle, proposed by
 Samarskii,  Galaktionov,  Kurdyumov and  Mikhailov
\cite{Galact, Samar}.

\section{Blow-up for equation \eqref{eBBMB}}

Let us consider the initial-boundary problem \eqref{eBBMB} and
\begin{gather}\label{2.1}
\frac{\partial}{\partial t}(u_{xx}-u)+u_{xx}-uu_x=0,\quad x\in(0,L),\; t>0, \\
\label{2.2}
u(0,t)=u_x(0,t)=0,\quad u(x,0)=u_0(x),\quad x\in[0,L],\quad
t\geqslant 0,
\end{gather}
where $L\in(0,+\infty)$. We are looking for the solution of
 problem \eqref{2.1}--\eqref{2.2} such that
\begin{equation}\label{2.3}
u(x)(t)\in\mathbb{C}^{(1)}([0,T];\mathbb{C}_0^{(2)}([0,L]))
\end{equation}
for some $T>0$. By definition, if $f\in \mathbb{C}_0^{(2)}$, then
$f\in \mathbb{C}^{(2)}$ and $f(0)=f_x(0)=0$.
In this section we show the following result.

\begin{theorem} \label{thm1}
For any $u_0\in\mathbb{C}_0^{(2)}([0,L])$ there exists one and only one
solution of problem \eqref{2.1}, \eqref{2.2} such that
$$
u(x)(t)\in\mathbb{C}^{(1)}([0,T_0);\mathbb{C}_0^{(2)}([0,L])),
$$
where either $T_0=+\infty$ or $T_0<+\infty$ and the limiting equality holds
\begin{equation}\label{2.3-1}
\limsup_{t\uparrow T_0}\sup_{x\in[0,L]}|u(x)(t)|=+\infty.
\end{equation}
\end{theorem}

\begin{proof}
We begin with some notation. For any $f\in\mathbb{C}[0,L]$ we use
the  Volterra operator
$$
G\ast f=\int_0^x\,dy\int_0^yf(z)\,dz
=\int_0^x(x-z)f(z)\,dz\quad\text{for } x\in[0,L].
$$
By \eqref{2.3} we can rewrite \eqref{2.1} as
\begin{equation}\label{2.4}
\frac{\partial}{\partial t}[\mathbb{I}-G\ast]u+u
=\frac{1}{2}\int_0^xu^2\,dx.
\end{equation}
The spectral radius of the Volterra operator is zero.
Thus multiplying both sides \eqref{2.4} by $[\mathbb{I}-G\ast]^{-1}$,
we obtain
\begin{equation}\label{2.5}
\frac{\partial u}{\partial t}+u=
-\sum_{k=1}^{+\infty}[G\ast]^ku+\frac{1}{2}\sum_{k=0}^{+\infty}[G\ast]^k
\int_0^xu^2\,dy.
\end{equation}
With the new variables $w=e^tu$,
integrating  \eqref{2.5}, we have
\begin{equation}\label{2.6}
w=\mathbb{F}[w]\equiv u_0+\int_0^tds B[w](s),\quad
B[w]=B_1[w]-B_2[w],
\end{equation}
 where
$$
B_1[w]=-\sum_{k=1}^{+\infty}[G\ast]^kw,\quad
B_2[w]=-\frac{e^{-t}}{2}\sum_{k=0}^{+\infty}[G\ast]^k\int_0^xw^2\,dy.
$$
Let us consider the following closed, bounded and convex subset in the Banach space
$\mathbb{C}([0,T]\times[0,L])$:
$$
\mathbb{B}_R\equiv\big\{w\in\mathbb{C}([0,T]\times[0,L]):\,\lVert
w\rVert\equiv\sup_{(t,x)\in[0,T]\times[0,L]}|w(x,t)|\leqslant R\big\}.
$$
We prove that the operator $\mathbb{F}[w]:\mathbb{B}_R\to \mathbb{B}_R$ f
or large enough $R>0$ and small $T>0$.
 Indeed, suppose that $(t_1,x_1)$ and
$(t_2,x_2)$ belong to $[0,L]\times[0,T]$. Without loss of generality we
suppose that $t_2\leqslant t_1$. Then the
following chain of inequalities holds
\begin{equation} \label{2.7}
\begin{aligned}
&\lVert\int_0^{t_1}ds B_1[w](s,x_1)-\int_0^{t_2}ds B_1[w](s,x_2)\rVert \\
&\leqslant \int_{t_1}^{t_2}\lVert B_1[w](s,x_1)\rVert\,ds+\int_0^{t_1}\lVert
B_1[w](s,x_1)-B_1[w](s,x_2)\rVert \\
&\leqslant |t_2-t_1|\sum_{k=1}^{+\infty}\frac{L^{2k}}{(2k)!}\lVert
w\rVert+T|x_1-x_2|\sum_{k=1}^{+\infty}\frac{L^{2k-1}}{(2k-1)!}\lVert
w\rVert.
\end{aligned}
\end{equation}
In the same way, for $B_2[w]$ we have the estimate
\begin{equation} \label{2.8}
\begin{aligned}
&\lVert\int_0^{t_1}ds B_2[w](s,x_1)
-\int_0^{t_2}ds B_2[w](s,x_2) \rVert \\
&\leqslant \int_{t_1}^{t_2}\lVert B_2[w](s,x_1)\rVert\,ds
+\int_0^{t_1}\lVert B_2[w](s,x_1)-B_2[w](s,x_2)\rVert\\
&\leqslant |t_2-t_1|\frac{1}{2}\sum_{k=0}^{+\infty}\frac{L^{2k+1}}{(2k+1)!}\lVert
w\rVert^2+T|x_1-x_2|\frac{1}{2}\sum_{k=0}^{+\infty}\frac{L^{2k}}{(2k)!}\lVert
w\rVert^2.
\end{aligned}
\end{equation}
This yields that the operator $\mathbb{F}$ is a mapping from
$\mathbb{C}([0,T]\times[0,L])$ to $\mathbb{C}([0,T]\times[0,L])$.
Now show that the operator is a mapping from $\mathbb{B}_R$ to $\mathbb{B}_R$.
It is clear that
\begin{equation} \label{2.9}
\begin{aligned}
\lVert\mathbb{F}[w]\rVert
&\leqslant\lVert u_0\rVert+\int_0^Tds\lVert B[w](s)\rVert\\
&\leqslant \lVert u_0\rVert+T\sum_{k=1}^{+\infty}\frac{L^{2k}}{(2k)!}\lVert
w\rVert+\frac{T}{2}\sum_{k=0}^{+\infty}\frac{L^{2k+1}}{(2k+1)!}\lVert
w\rVert^2.
\end{aligned}
\end{equation}
From \eqref{2.9} we get that for large enough $R>0$,
$$
\lVert u_0\rVert\leqslant\frac{R}{2},
$$
and for small enough $T>0$,
$$
T\sum_{k=1}^{+\infty}\frac{L^{2k}}{(2k)!}R
+\frac{T}{2}\sum_{k=0}^{+\infty}\frac{L^{2k+1}}{(2k+1)!}R^2\leqslant\frac{R}{2}\,.
$$
The result of Theorem 1 is true.
Now let us demonstrate the contraction mapping $\mathbb{F}[w]$ on $\mathbb{B}_R$.
It is not hard to show that the following chain of inequalities holds
\begin{equation}\label{2.10}
\begin{aligned}
\lVert\mathbb{F}[w_1]-\mathbb{F}[w_2]\rVert
&\leqslant T\lVert B[w_1]-B[w_2]\rVert\\
&\leqslant T\Big[\sum_{k=1}^{+\infty}\frac{L^{2k}}{(2k)!}
 +\sum_{k=0}^{+\infty}\frac{L^{2k+1}}{(2k+1)!}R\Big]\lVert w_1-w_2\rVert,
\end{aligned}
\end{equation}
then from \eqref{2.10}, for any
$$
T\big[\sum_{k=1}^{+\infty}\frac{L^{2k}}{(2k)!}+\sum_{k=0}^{+\infty}
\frac{L^{2k+1}}{(2k+1)!}R\big]\leqslant\frac{1}{2}
$$
we have a contraction $\mathbb{F}[w]$ on $\mathbb{B}_R$.
This means that the unique, local in time, solvability of the
integral equation \eqref{2.6} is proved.

We claim that the solution $w$ of \eqref{2.6} belongs to the class
$\mathbb{C}([0,T_0)\times[0,L])$;
moreover, either $T_0=+\infty$, or $T_0<+\infty$ and
$\lim_{t\uparrow T_0}\lVert w\rVert=+\infty$.
Assuming the converse: $T_0<+\infty$, then we have
$$
\sup_{t\in[0,T_0]}\lVert w\rVert<+\infty.
$$
From  equation \eqref{2.6} we get the estimate
$$
\lVert w'\rVert\leqslant
c(T_0)<+\infty\quad\text{for all } t\in[0,T_0].
$$
This implies that
$$
|w(x)(t_1)-w(x)(t_2)|\leqslant\int_{t_1}^{t_2}|w'(x)(\tau)|\,d\tau\leqslant
c(T_0)|t_2-t_1|,
$$
where the constant $c(T_0)<+\infty$ does not depend on $x\in[0,L]$.
Whence, for all $x\in[0,L]$ we can define $w(x)(T_0)$. Furthemore,
if we take supremum of both sides of the last
inequality, then we obtain
$$
w(x)(T_0)\in\mathbb{B}[0,L],
$$
where $\mathbb{B}[0,L]$ is the set of bounded functions on the line
segment $[0,L]$. We shall see that
$$
w(x)(T_0)\in\mathbb{C}[0,L].
$$
Obviously, for all $\varepsilon>0$ there is  $\delta(\varepsilon)>0$
such that for all $x_1, x_2\in[0,L]$,
$$
|x_1-x_2|\leqslant\delta(\varepsilon)/2
$$
and we can choose  $t_1,t_2\in[0,T_0)$ such that
\[
|T_0-t_1|\leqslant\delta(\varepsilon)/4,\quad
|T_0-t_2|\leqslant\delta(\varepsilon)/4
\]
imply
\[
|t_1-t_2|\leqslant|T_0-t_1|+|T_0-t_2|\leqslant\delta(\varepsilon)/2,
|t_1-t_2|+|x_1-x_2|\leqslant\delta(\varepsilon),
\]
so the following chain of inequalities holds:
\begin{equation}\label{2.10-1}
\begin{aligned}
&|w(x_1)(T_0)-w(x_2)(T_0)|\\
&\leqslant|w(x_1)(T_0)-w(x_1)(t_1)|+|w(x_1)(t_1)-w(x_2)(t_2)|
 +|w(x_2)(T_0)-w(x_2)(t_2)| \\
&\leqslant c(T_0)|T_0-t_1|+|w(x_1)(t_1)-w(x_2)(t_2)|+c(T_0)|T_0-t_2| \\
&\leqslant c(T_0)(\frac{\varepsilon}{4}+\frac{\varepsilon}{2}
 +\frac{\varepsilon}{4})=c(T_0)\varepsilon.
\end{aligned}
\end{equation}
Therefore, we can define
$$
w(x)(T_0)\in\mathbb{C}([0,L])
$$
and can extend the classical solution over the time moment $T_0>0$,
taking the following initial function $w(x)(t)$ for $t=T_0$.
This contradiction concludes the result of existance of a maximal solution.
Finally, using so called "bootstrap" method for the integral equation
\eqref{2.6} under the additional condition
$u_0(x)\in\mathbb{C}_0^{(2)}([0,L])$, we get that
$$
w(x)(t)\in\mathbb{C}^{(1)}([0,T_0);\mathbb{C}_0^{(2)}([0,L])).
$$
proof is complete.
\end{proof}


Now we shall obtain a blow-up result.

\begin{theorem} \label{thm2}
Suppose that for some positive $\lambda\geqslant 3$, the initial function
satisfies the  condition
\begin{equation}\label{bb.7}
J(0)>\frac{m}{k},
\end{equation}
where
$$
J(t)=\int_0^L(L-x)^{\lambda-3}[\lambda(\lambda-1)-(L-x)^2]((L-x)u-(\lambda-1))\,dx,
$$
or
\begin{align*}
&\int_0^L(L-x)^{\lambda-2}[\lambda(\lambda-1)-(L-x)^2]u_0\,dx\\
&> (\lambda-1)L^{\lambda-1}
\Big|\frac{\lambda(\lambda-1)^2}{\lambda-2}-2\frac{1}{\lambda}(\lambda-1)L^2+
\frac{\lambda L^{4}}{\lambda(\lambda+2)}\Big|^{1/2}\\
&\quad + \frac{\lambda(\lambda-1)^2}{\lambda-2}L^{\lambda-2}
- \frac {\lambda-1}{\lambda}L^{\lambda},
\end{align*}
then the classical solution of problem {\rm\eqref{2.1}--\eqref{2.2}}
does not exist globally in time.
Furthermore, the following estimate holds
\begin{equation}\label{bb.8}
J(t)\geqslant\frac{m}{k}\frac{(kJ(0)+m)+(kJ(0)-m)
\exp(2mkt)}{(kJ(0)+m)-(kJ(0)-m)\exp(2mkt)},
\end{equation}
consequently,
\begin{equation}\label{bb.9}
\lim_{t\to T_b}J(t)=+\infty,\quad
T_b\leqslant -\frac{1}{2mk}\ln(\frac{kJ(0)-m}{kJ(0)+m}),
\end{equation}
$$
m^2=\frac{(\lambda-1)^2L^{\lambda}}{2},\quad
k^2=\frac{L^{2-\lambda}}{2}\Big|\frac{\lambda(\lambda-1)^2}{\lambda-2}
-2\frac{1}{\lambda}(\lambda-1)L^2+
\frac{\lambda L^{4}}{\lambda(\lambda+2)}\Big|^{-1}.
$$
\end{theorem}

\begin{proof}
To obtain sufficient conditions of the blow-up we use the method of
nonlinear capacity. Let us take the  test function
$$
\varphi(x)=(L-x)^{\lambda},\quad \lambda\geqslant 3
$$
to remove boundary conditions on the right side $(x=L)$.
Multiplying both sides of  \eqref{eBBMB}
by $\varphi(x)$ and integrating by parts over $[0,L]$, we use the following equalities
\begin{gather*}
\int_0^L (L-x)^{\lambda}u_{xx}(x,t)\, dx
=\lambda(\lambda-1)\int_0^L (L-x)^{\lambda-2} u\, dx,\\
\int_0^L (L-x)^{\lambda}uu_x(x,t)\, dx=
\frac{\lambda}{2}\int_0^L (L-x)^{\lambda-1} u^2\, dx,
\end{gather*}
and get the ordinary differential equation
\begin{equation} \label{bb.2}
\begin{aligned}
&\frac{d}{dt}\int_0^L(L-x)^{\lambda-2}[\lambda(\lambda-1)-(L-x)^2]u\,dx\\
&= -\lambda(\lambda-1)\int_0^L(L-x)^{\lambda-2}u\,dx+
\frac{\lambda}{2}\int_0^L u^2(L-x)^{\lambda-1}\,dx.
\end{aligned}
\end{equation}
Constructing the perfect square in the right side of \eqref{bb.2}
 and adding in the left side the time-independent function,
we obtain
\begin{equation} \label{bb.3}
\begin{aligned}
&\frac{d}{dt}\int_0^L(L-x)^{\lambda-3}[\lambda(\lambda-1)-(L-x)^2]((L-x)u
-(\lambda-1))\,dx\\
&= \frac{\lambda}{2}\int_0^L [u(L-x)-(\lambda-1)]^2(L-x)^{\lambda-3}\,dx
- \frac{\lambda(\lambda-1)^2}{2}\int_0^L(L-x)^{\lambda-1}\,dx.
\end{aligned}
\end{equation}
By Holder's inequality
\begin{equation}\label{bb.zep}
\big|\int ab\big|\leqslant \big|\int a^2\big|^{1/2}\big|\int b^2\big|^{1/2}
\end{equation}
we can substitute
\begin{gather*}
a^2=[u (L-x)-(\lambda-1)]^2(L-x)^{\lambda-3}, \\
b^2=[\lambda(\lambda-1)-(L-x)^2]^2(L-x)^{\lambda-3},
\end{gather*}
and integrate over the line segment $[0,L]$.
Then \eqref{bb.zep} gives the following estimate
$$
\int_0^L [u (L-x)-(\lambda-1)]^2(L-x)^{\lambda-3}\,dx\geqslant
J^2|\int_0^L b^2 dx|^{-1}.
$$
From  equation \eqref{bb.3} we get the  ODE
\begin{equation}\label{bb.4}
\frac{dJ(t)}{dt}\geqslant \frac{\lambda
J^2}{2}\big|\int_0^L b^2 dx\big|^{-1}
-\frac{\lambda(\lambda-1)^2}{2}\int_0^L(L-x)^{\lambda-1}\,dx,
\end{equation}
 where
\[
J(t)=\int_0^L(L-x)^{\lambda-3}[\lambda(\lambda-1)-(L-x)^2]((L-x)u-(\lambda-1))\,dx.
\]
The final inequality \eqref{bb.4} can be written as
\begin{equation}\label{bb.5}
\frac{dJ(t)}{dt}\geqslant k^2J^2-m^2,
\end{equation}
where
\begin{gather*} %\label{bb.6}
m^2=\frac{\lambda(\lambda-1)^2}{2}\int_0^L(L-x)^{\lambda-1}\,dx
=\frac{(\lambda-1)^2L^{\lambda}}{2},\\
\begin{aligned}
k^2&=\frac{\lambda}{2}\big|\int_0^L b^2 dx\big|^{-1}\\
&= \frac{\lambda}{2}\big|\int_0^L [\lambda^2(\lambda-1)^2-2\lambda(\lambda-1)(L-x)^2
+(L-x)^4](L-x)^{\lambda-3}\,dx \big|^{-1}\\
&= \frac{L^{2-\lambda}}{2}\big|\frac{\lambda(\lambda-1)^2}{\lambda-2}
-2\frac{1}{\lambda}(\lambda-1)L^2+
\frac{ L^{4}}{\lambda(\lambda+2)}\big|^{-1}.
\end{aligned}
\end{gather*}
Integrating the inequality \eqref{bb.5}, we complete the proof.
\end{proof}

Note that the Theorem \ref{thm2} is proved for any fixed $L\in(0,+\infty)$.
But we should stress that the time moment $T_0>0$ is the function of $L>0$
such that the following situation is possible:
$$
T_0\to +0\quad\text{as } L\to +\infty.
$$

\begin{theorem} \label{thm3}
Suppose that the initial data satisfies the inequalities
\begin{equation}\label{bb.31}
J(0)>1,\quad\lambda>1,
\end{equation}
where
$$
J(t)=\int_0^{+\infty}(u-\lambda)e^{-\lambda x}\,dx,
$$
or, which is the same,
$$
\int_0^{+\infty}u_0(x) e^{-\lambda x}\,dx>2\,.
$$
Then the global in time solution of the problem \eqref{2.1}, \eqref{2.2}
 does not exist in the half-space ($L=\infty$).
Furthermore,
\begin{equation}\label{bb.32}
J(t)\geqslant \frac{1+C\exp(\lambda^2t/(\lambda^2-1))}
{1-C\exp(\lambda^2t/(\lambda^2-1))},
\end{equation}
where
\begin{equation}\label{bb.33}
\lim_{t\to T_b}J(t)=+\infty,\quad
T_b\leqslant -\frac{\lambda^2-1}{\lambda^2}\ln C,\quad
C=\frac{J(0)-1}{J(0)+1}.
\end{equation}
\end{theorem}

\begin{proof}
To remove conditions for $L=+\infty$ we use the test function
$\varphi(x)=\exp (-\lambda x)$, $\lambda>1$. Multiplying both sides
of  \eqref{eBBMB} by $\varphi(x)$, we integrate by parts over $[0,\infty)$.
By the boundary conditions we have the following equalities
\begin{gather*}
\int_0^{+\infty} e^{-\lambda x} u_{xx}(x,t)\, dx
=\lambda^2 \int_0^{+\infty}  e^{-\lambda x} u(x,t)\, dx,
\\
\int_0^{+\infty} e^{-\lambda x}uu_x(x,t)\, dx=
\frac{\lambda}{2}\int_0^{+\infty} e^{-\lambda x}
u^2(x,t)\, dx.
\end{gather*}
So we can rewrite the equation \eqref{2.1} as
\begin{equation}\label{bb.27}
\frac{d}{dt}\int_0^{\infty} e^{-\lambda
x}[\lambda^2-1]u\,dx= -\lambda^2 \int_0^{\infty}e^{-\lambda
x}u\,dx+ \frac{\lambda}{2}\int_0^{\infty} e^{-\lambda
x}u^2\,dx.
\end{equation}
As above, we make the perfect square in the right side of \eqref{bb.27}
and add the time-independent function in the left
side, we obtain
\begin{equation}\label{bb.28}
\frac{d}{dt}\int_0^{+\infty}e^{-\lambda
x}[\lambda^2-1](u-\lambda)\,dx=
\frac{\lambda}{2}\int_0^{+\infty} (u- \lambda)^2
e^{-\lambda x}\,dx- \frac{\lambda^3}{2}\int_0^{+\infty}
e^{-\lambda x}\,dx.
\end{equation}
It is not hard to prove the  inequality
\begin{equation}\label{bb.29}
\begin{aligned}
\big|\int_0^{+\infty}(u-\lambda)e^{-\lambda x}\,dx \big|^2
&\leqslant \int_0^{+\infty} (u-\lambda)^2 e^{-\lambda x}\,dx
 \int_0^{+\infty} e^{-\lambda x}\,dx\\
&= \frac{1}{\lambda}\int_0^{+\infty} (u- \lambda)^2 e^{-\lambda x}\,dx.
\end{aligned}
\end{equation}
Thus from \eqref{bb.28} we get the ODE
\begin{equation}\label{bb.30}
\frac{dJ(t)}{dt}\geqslant \frac{\lambda^2 }{2(\lambda^2-1)}J^2
-\frac{\lambda^2}{2(\lambda^2-1)},
\end{equation}
 where
\[
J(t)=\int_0^{+\infty}(u-\lambda)e^{-\lambda x}\,dx.
\]
Integrating  inequality \eqref{bb.30}, we obtain the stated result.
\end{proof}

\section{Blow-up for the Rosenau-Burgers equation}

Let us consider \eqref{eRB} with the initial-boundary conditions
\begin{gather}\label{2.11}
\frac{\partial}{\partial
t}(u_{xxxx}+u)+u_{xx}-uu_x=0,\quad x\in(0,L),\quad t>0,
\\
\label{2.12}
u(0,t)=u_x(0,t)=u_{xx}(0,t)=u_{xxx}(0,t)=0,\quad
u(x,0)=u_0(x),
\end{gather}
where $L\in(0,+\infty)$. Assume that there is such $T>0$,
that classical solution $u$ of the problem \eqref{2.11}--\eqref{2.12}
exists and satisfies
\begin{equation}\label{2.13}
u(x)(t)\in\mathbb{C}^{(1)}([0,T];\mathbb{C}_0^{(4)}([0,L]))\,.
\end{equation}
By definition, if $f\in \mathbb{C}_0^{(4)}$, then $f\in \mathbb{C}^{(4)}$
and $f(0)=f_x(0)=f_{xx}(0)=f_{xxx}(0)=0$.
In this section we prove the following result.

\begin{theorem} \label{thm4}
For any initial data $u_0\in\mathbb{C}_0^{(4)}([0,L])$ there exists only one
solution $u$ of  problem \eqref{2.11}, \eqref{2.12} such that
$$
u(x)(t)\in\mathbb{C}^{(1)}([0,T_0);\mathbb{C}_0^{(4)}([0,L])),
$$
for some $T_0>0$ such that either $T_0=+\infty$ or $T_0<+\infty$ and
\begin{equation}\label{2.14}
\limsup_{t\uparrow T_0}\sup_{x\in[0,L]}|u(x)(t)|=+\infty.
\end{equation}
\end{theorem}

The proof of the above theorem is the same as that for the Theorem \ref{thm1}.
So we omit it.  Let us prove now a blow-up result.

\begin{theorem} \label{thm5}
Suppose that there exists such $\lambda\geqslant 7$ that initial function
satisfies the inequality
\begin{equation}\label{bb.15}
J(0)>\frac{m}{k},
\end{equation}
where
$$
J(t)=\int_0^L(L-x)^{\lambda-5}[\lambda(\lambda-1)(\lambda-2)(\lambda-3)
+(L-x)^4]((L-x)u-(\lambda-1))\,dx.
$$
then the the classical solution of  \eqref{2.11} does not exist globally in time.
Moreover, for
$m$ and $k$ from \eqref{2.12} the following estimate holds
\begin{equation}\label{bb.16}
J(t)\geqslant\frac{m}{k}\frac{(kJ(0)+m)+(kJ(0)-m)\exp(2mkt)}
{(kJ(0)+m)-(kJ(0)-m)\exp(2mkt)},
\end{equation}
and, consequently,
\begin{equation}\label{bb.17}
\lim_{t\to T_b}J(t)=+\infty,
\end{equation}
where
\begin{gather*}
T_b\leqslant
-\frac{1}{2mk}\ln(\frac{kJ(0)-m}{kJ(0)+m}),\quad
m^2=\frac{(\lambda-1)^2L^{\lambda}}{2},\\
k^2=\frac{L^{6-\lambda}}{2}\left|\frac{\lambda(\lambda-1)^2(\lambda-2)^2(\lambda-3)^2}{(\lambda-6)}+
2(\lambda-1)(\lambda-3)L^{4}+\frac{L^{8}}{\lambda(\lambda+2)}\right|^{-1}
\end{gather*}
\end{theorem}

\begin{proof}
Multiplying both sides \eqref{2.11} by  $\varphi(x)=(L-x)^{\lambda}$, we integrate
by parts. By the boundary conditions, we have
$$
\int_0^L (L-x)^{\lambda}u_{xxxx}(x,t)\, dx=
\lambda(\lambda-1)(\lambda-2)(\lambda-3)\int_0^L
(L-x)^{\lambda-4} u(x,t)\, dx;
$$
By this equality we can rewrite \eqref{eRB}  as the  integro-differential expression
\begin{equation} \label{bb.11}
\begin{aligned}
&\frac{d}{dt}\int_0^L(L-x)^{\lambda-4}[\lambda(\lambda-1)(\lambda-2)(\lambda-3)
+(L-x)^4]u\,dx\\
&= -\lambda(\lambda-1)\int_0^L(L-x)^{\lambda-2}u\,dx+
\frac{\lambda}{2}\int_0^L u^2(L-x)^{\lambda-1}\,dx.
\end{aligned}
\end{equation}
As before, in the right-hand side of \eqref{bb.11} we complete the square and
in the left-hand side  add the time-independent function:
\begin{equation} \label{bb.12}
\begin{aligned}
&\frac{d}{dt}\int_0^L(L-x)^{\lambda-5}[\lambda(\lambda-1)(\lambda-2)(\lambda-3)
+(L-x)^4]((L-x)u-(\lambda-1))\,dx\\
&= \frac{\lambda}{2}\int_0^L [u(L-x)-(\lambda-1)]^2(L-x)^{\lambda-3}\,dx
-\frac{\lambda(\lambda-1)^2}{2}\int_0^L(L-x)^{\lambda-1}\,dx.
\end{aligned}
\end{equation}
Substituting
\begin{gather*}
a^2=\left[u (L-x)-(\lambda-1)\right]^2(L-x)^{\lambda-3},\\
b^2=[\lambda(\lambda-1)(\lambda-2)(\lambda-3)-(L-x)^4]^2(L-x)^{\lambda-7}
\end{gather*}
in \eqref{bb.zep} and integrating over the segment $[0,L]$, we obtain
\begin{equation} \label{bb.13}
\begin{aligned}
&\int_0^L [u(L-x)-(\lambda-1)]^2(L-x)^{\lambda-3}\,dx\leqslant
J^2\Big|\int_0^L b^2 dx\Big|^{-1}\\
&= J^2\Big|\frac{\lambda^2(\lambda-1)^2(\lambda-2)^2(\lambda-3)^2}{(\lambda-6)}
L^{\lambda-6}+ 2\lambda(\lambda-1)(\lambda-3)L^{\lambda-2}
+\frac{L^{\lambda+2}}{\lambda+2}\Big|^{-1}.
\end{aligned}
\end{equation}
Finally, from  \eqref{bb.12} we obtain the ordinary differential inequality
\begin{equation}\label{bb.14}
\frac{dJ(t)}{dt}\geqslant k^2J^2-m^2,
\end{equation}
where $m^2=\frac{(\lambda-1)^2L^{\lambda}}{2}$,
\begin{gather*}
k^2=\frac{L^{6-\lambda}}{2}\Big|\frac{\lambda(\lambda-1)^2(\lambda-2)^2(\lambda-3)^2}{(\lambda-6)}+
2(\lambda-1)(\lambda-3)L^{4}+\frac{L^{8}}{\lambda(\lambda+2)}\Big|^{-1},
\\
J(t)=\int_0^L(L-x)^{\lambda-5}[\lambda(\lambda-1)(\lambda-2)(\lambda-3)
+(L-x)^4]((L-x)u-(\lambda-1))\,dx.
\end{gather*}
Integrating  \eqref{bb.14}, we complete the proof.
\end{proof}

As for equation \eqref{eBBMB}, we can obtain results for the
Rosenau-Burgers in the unbounded domain.
For  problem \eqref{2.11}-\eqref{2.12} on the half-line $[0,+\infty)$
we shall prove the following result.

\begin{theorem} \label{thm6}
Suppose that the initial data satisfies the inequality
$$
\int_0^{+\infty}u_0(x) e^{-\lambda x}\,dx>2,
$$
then the classical solution of the problem for Rosenau-Burgers on the half
line does not exist globally in time.
Furthermore, there exists the lower estimate
\begin{equation}\label{rb.32}
J(t)\geqslant \frac{1+C\exp(\lambda^2t/(\lambda^4+1))}
{1-C\exp(\lambda^2t/(\lambda^4+1))},\quad
J(t)=\int_0^{+\infty}(u-\lambda)e^{-\lambda x}\,dx,
\end{equation}
and, consequently,
\begin{equation}\label{rb.33}
\lim_{t\to T_b}J(t)=+\infty,\quad
T_b\leqslant -\frac{\lambda^4+1}{\lambda^2}\ln C,\quad
C=\frac{J(0)-1}{J(0)+1}.
\end{equation}
\end{theorem}

\begin{proof}
The proof is similar to the Theorem \ref{thm3}; so we present on the main points.
First let us multiply both sides of \eqref{eRB}  by $\varphi(x)=\exp (-\lambda x)$,
with $\lambda>1$, and integrate by parts over $[0,L)$:
\begin{equation}\label{rb.27}
\frac{d}{dt}\int_0^{\infty} e^{-\lambda
x}[\lambda^4+1]u\,dx= -\lambda^2 \int_0^{\infty}e^{-\lambda
x}u\,dx+ \frac{\lambda}{2}\int_0^{\infty} e^{-\lambda x}u^2\,dx.
\end{equation}
Then extracting perfect square in the side part \eqref{rb.27} and adding
 a time-independed function to the left-hand side,
we obtain the equality
\begin{equation}\label{rb.28}
\frac{d}{dt}\int_0^{+\infty}e^{-\lambda
x}[\lambda^4+1](u-\lambda)\,dx=
\frac{\lambda}{2}\int_0^{+\infty} (u- \lambda)^2
e^{-\lambda x}\,dx- \frac{\lambda^3}{2}\int_0^{+\infty}
e^{-\lambda x}\,dx.
\end{equation}
By  \eqref{bb.29} we can rewrite \eqref{rb.28} as
\begin{equation}\label{rb.30}
\frac{dJ(t)}{dt}\geqslant \frac{\lambda^2 }{2(\lambda^4+1)}J^2
-\frac{\lambda^2}{2(\lambda^4+1)},
\end{equation}
where
\[
J(t)=\int_0^{+\infty}(u-\lambda)e^{-\lambda x}\,dx.
\]
Integrating  \eqref{rb.30}, we obtain the result of the theorem.
\end{proof}

\section{Blow-up for equation \eqref{eKdv-BBM}}

It is readily seen that the method of nonlinear capacity is usable
for Sobolev type equations also with additional terms
like higher-order derivatives $u^{(n)}$. As an example, we consider the
problem for the equation, which describes processes
in cristalline semiconductors and has the third-order derivative:
\begin{gather}\label{bb.18}
(u_{xx}-u)_t+u_{xxx}-uu_x=0,\quad t>0,\; x\in(0,L),\;L>0, \\
\label{bb.18-1}
u(0,t)=u_x(0,t)=u_{xx}(0,t)=0,\quad u(x,0)=u_0(x),\quad
x\in[0,L],\quad t\geqslant 0.
\end{gather}
The physical model for problem \eqref{bb.18}--\eqref{bb.18-1} is presented
in \cite{Yushkov2}.
Let us prove the solvability result.

\begin{theorem} \label{thm7}
For any function $\tilde{u}_0(x)$ such that
$$
\tilde{u}_0(x)=\begin{cases} u_0(x),&\text{for }x\geqslant 0;\\
0,&\text{for }x<0,
\end{cases}\quad
\tilde{u}_0(x)\in\mathbb{C}^{(4)}(-\infty,L],
$$
there exists only one solution $u$ of \eqref{bb.18}, \eqref{bb.18-1} such that
$$
u(x)(t)\in\mathbb{C}^{(1)}([0,T_0);\mathbb{C}_0^{(3)}([0,L])),
$$
for some $T_0>0$. By definition, if $f\in \mathbb{C}_0^{(3)}$, then
$f\in \mathbb{C}^{(3)}$ and $f(0)=f_x(0)=f_{xx}(0)=0$.
Moreover, it can be proved that either $T_0=+\infty$ or $T_0<+\infty$
and the following limiting equality holds
$$
\sup_{t\uparrow T_0}\sup_{x\in[0,L]}|u(x)(t)|=+\infty.
$$
\end{theorem}

\begin{remark} \label{rmk1} \rm
By the boundary conditions \eqref{bb.18-1} and natural matching condition
of initial and boundary data it is easily shown that
$\tilde{u}_0(x)\in\mathbb{C}_0^{(3)}(-\infty,L]$, however,
larger  smoothness is not indispensable. In addition, it is clear that
necessary and sufficient conditions for
$\tilde{u}_0(x)\in\mathbb{C}^{(4)}(-\infty,L]$ are the following
$$
u_0(x)\in\mathbb{C}^{(4)}([0,L])\quad\text{and}\quad
(u_{0})_{xxx}(0+0)=(u_{0})_{xxxx}(0+0)=0.
$$
\end{remark}

\begin{proof}[Proof of Theorem \ref{thm7}]
As above for equation \eqref{eBBMB}, taking into account the boundary conditions,
integrating with respect to $x$ and inverting the operator
$$
(\mathbb{I}-G\ast),\quad\text{where}\quad
G\ast f=\int_0^xdy\int_0^yf(z)\,dz,
$$
we obtain the  equation
\begin{equation}\label{2.15}
\frac{\partial u}{\partial t} +\frac{\partial u}{\partial x}
=\mathbb{Q}[u](x,t)\equiv\sum_{k=0}^{+\infty}[G\ast]^k\int_0^xu\,dy+
\frac{1}{2}\sum_{k=0}^{+\infty}[G\ast]^k\int_0^xu^2\,dy,
\end{equation}
where we use that
$\int_0^xu_y\,dy=u(x)$, and, thus,
$$
\sum_{k=1}^{+\infty}[G\ast]^ku_x=\sum_{k=0}^{+\infty}[G\ast]^k\int_0^xu\,dy.
$$
From the definition of operator $\mathbb{Q}[u](x,t)$ it follows that
$\mathbb{Q}[u](0,t)=0$, moreover, $u(0,t)=0$.
Then we can continue the function $\mathbb{Q}[u](x,t)$ by zero
for $x<0$, furthermore, this function $\mathbb{Q}[u](x,t)$ is smooth
function with respect to the first input.
Corresponding to this continuation we define
$$
f(x,t)\equiv\widetilde{\mathbb{Q}}[u](x,t)\quad\text{and}\quad\tilde{u}(x,t).
$$
Now consider the differential problem
\begin{equation}\label{2.16}
\frac{\partial v}{\partial t}+\frac{\partial v}{\partial x}=f(x,t),\quad
v(x,0)=v_0(x),\quad v(x,t)\equiv\tilde{u}(x,t).
\end{equation}
The solution of  \eqref{2.15} we can be written as
\begin{equation}\label{2.17}
u(x,t)=\tilde{u}_0(x-t)+\int_0^t f(x-t+\tau,\tau)\,d\tau,\quad
x\in[0,L],\; t\in[0,T],
\end{equation}
where $L>0$ is fixed, and $T>0$ is small enough.

To prove the local solvability of the integral equation
\eqref{2.17} we use the contraction mapping method.
In the notation of the previous section we can rewrite
the integral equation \eqref{2.17} in the  abstract form
\begin{equation}\label{2.18}
u(x,t)=\mathbb{F}[u](x,t)\equiv
\tilde{u}_0(x-t)+\int_0^t\widetilde{\mathbb{Q}}[u](x-t+\tau,\tau)\,d\tau.
\end{equation}
As before, it can easily be checked that, if $u_0(x)\in\mathbb{C}_0^{(3)}([0,L])$,
then
$$
\mathbb{F}[\cdot]:\mathbb{C}([0,T]\times[0,L])\to \mathbb{C}([0,T]\times[0,L]).
$$
Let us prove now that $\mathbb{F}[\cdot]$ is an operator between
$\mathbb{B}_R$ and $\mathbb{B}_R$ for large enough $R>0$ and small $T>0$.
Indeed, we can fix $R>0$ such large that
$$
\lVert u_0\rVert\leqslant\frac{R}{2}.
$$
On the other hand, we have the inequality
\begin{equation}\label{2.19}
\lVert u\rVert\leqslant\lVert
u_0\rVert+T\sum_{k=0}^{+\infty}\frac{(L^{\ast})^{2k+1}}{(2k+1)!}\big[\lVert
u\rVert+\frac{1}{2}\lVert u\rVert^2\big],\quad
L^{\ast}=\max\{L,|L-T|\}.
\end{equation}
From \eqref{2.19}, for small enough $T>0$,
$$
T\sum_{k=0}^{+\infty}\frac{(L^{\ast})^{2k+1}}{(2k+1)!}
[R+\frac{R^2}{2}]\leqslant\frac{R}{2},
$$
and we obtain our result.
As in the second section it is easily shown that $\mathbb{F}$ is contraction
mapping on a space $\mathbb{B}_R$, if we change $L$ to $L^{\ast}$.
Thus, we complete the proof of local in time solvability of the
integral equation \eqref{2.17}.

Now we can continue the solution in time. Assume that
$u_0(x)\in\mathbb{C}_0^{(3)}([0,L])$, then
from the necessary matching condition of initial and boundary data we obtain that
$\tilde{u}_0(x)\in\mathbb{C}^{(4)}((-\infty,L])$. Moreover,
 it can easily be checked that
$$
u(x)(t)\in\mathbb{C}_0^{(3)}([0,T]\times[0,L]).
$$
Let us denote
$$
w(x)(t)\equiv\tilde{u}_0(x-t)+\int_0^t\widetilde{\mathbb{Q}}[u]
(x-t+\tau,\tau)\,d\tau,
$$
then under the condition $T_0<+\infty$, the following inequality holds
$$
\sup_{t\in[0,T_0]}\lVert u\rVert<+\infty.
$$
By a standard way we obtain
$w(x)(T_0)\in\mathbb{C}([0,L])$.
At the same time we can write
$$
w(x)(T_0)=\tilde{u}_0(x-T_0)+\int_0^{T_0}\widetilde{\mathbb{Q}}[u]
(x-T_0+\tau,\tau)\,d\tau\in\mathbb{C}^{(1)}([0,L]).
$$
Therefore, in \eqref{2.15}, taking  $u(x)(t)$ for $t=T_0$ as initial data,
we can continue the solution over the time moment $T_0$.
This contradiction concludes the proof of existence of the
maximum solution.

Using the ``bootstrap'' method, it is possible to show that,
if $\tilde{u}_0(x)\in\mathbb{C}^{(4)}(-\infty,L]$,
then the solution of  \eqref{2.17} belong to the class
$\mathbb{C}^{(1)}([0,T_0);\mathbb{C}_0^{(3)}([0,L]))$.
\end{proof}

As above, to obtain sufficient conditions of blow-up we use the method of
nonlinear capacity for the power test function
$\varphi(x)=(L-x)^{5}$. Suppose that the solution $u(x,t)$ of  problem
\eqref{eKdv-BBM}, \eqref{bb.18-1} is classical:
$$
u(x)(t)\in \mathbb{C}^{(1)}([0,T];\mathbb{C}^{(3)}([0;L]))\quad\text{for some }T>0.
$$
Multiplying both sides of \eqref{eKdv-BBM} by the test function $\varphi(x)$
and integrating by parts, we obtain the equality:
\begin{equation}\label{bb.19}
\frac{d}{dt}\int_0^L(L-x)^3[20-(L-x)^2]u\,dx=
\frac{5}{2}\int_0^L u^2(L-x)^4\,dx+ 60\int_0^L
u(L-x)^2\,dx.
\end{equation}
As before, let us make in the right-hnad side \eqref{bb.2} the perfect square
and in the left add the time-independent function
\begin{equation}\label{bb.20}
\frac{d}{dt}\int_0^L (L-x) [20-(L-x)^2]((L-x)^2u+12)\,dx=
\frac{5}{2}\int_0^L [(L-x)^2 u+12]^2\,dx-360 L.
\end{equation}
The following inequalities can be easily proved:
\begin{equation} \label{bb.21}
\begin{aligned}
&\Big|\int_0^L (L-x) [20-(L-x)^2]((L-x)^2u+12)\,dx\Big|^2 \\
&\leqslant \Big|\int_0^L (L-x)^2 [20-(L-x)^2]^2\,dx\Big|\,
\Big|\int_0^L ((L-x)^2u+12)^2\,dx\Big|\\
&= \frac{5}{2 k^2}\int_0^L ((L-x)^2u+12)^2\,dx,
\end{aligned}
\end{equation}
where
\[
k^{2}=\Big|\frac{160L^3}{3}-\frac{16L^5}{5}+\frac{2L^7}{35}\Big|^{-1}.
\]
From  equation \eqref{bb.20} and estimate \eqref{bb.21} we obtain the ordinary
differential inequality
\begin{equation}\label{bb.22}
\frac{dJ(t)}{dt}\geqslant k^2J^2-360L,
\end{equation}
 where
\[
J(t)=\int_0^L (L-x) [20-(L-x)^2]((L-x)^2u+12)\,dx.
\]
Integrating \eqref{bb.22}, we obtain the following result.

\begin{theorem} \label{thm8}
 Let the initial function satisfies the condition
\begin{equation}\label{bb.23}
J(0)=\int_0^L (L-x)
[20-(L-x)^2]((L-x)^2u_0(x)+12)\,dx>\sqrt{\frac{360L}{k^{2}}}
\end{equation}
or
$$
\int_0^L (L-x)^3
[20-(L-x)^2]u_0(x)\,dx>120L^2\sqrt{\frac{4}{3}
-\frac{2L^2}{25}+\frac{L^4}{700}}-120L^2+3L^4,
$$
then there is no global in time classical solutions of \eqref{eKdv-BBM}.
Moreover, the following lower bound holds
\begin{equation}\label{bb.24}
J(t)\geqslant\frac{6\sqrt{10L}}{k}
\frac{1+C\exp(12\sqrt{10L}kt)}{1-C\exp(12\sqrt{10L}kt)},
\end{equation}
and, consequently,
$\lim_{t\to T_b}J(t)=+\infty$,
\begin{gather*} 
T\leqslant T_b\leqslant -\frac{1}{12\sqrt{10L}k}\ln C,\\
C=\frac{kJ(0)-6\sqrt{10L}}{kJ(0)+6\sqrt{10L}},\quad
k^{2}=\Big|\frac{160L^3}{3}-\frac{16L^5}{5}+\frac{2L^7}{35}\Big|^{-1}.
\end{gather*}
\end{theorem}

Using the  nonlinearity capacity method with the test function
$\varphi(x)=\exp (-\lambda x)$, $\lambda>1$, we
obtain the blow-up result on the halfline. Multiplying both sides
of \eqref{eKdv-BBM} equation by $\varphi(x)$ and integrating by parts,
we obtain the equality:
\begin{equation}\label{kdf.27}
\frac{d}{dt}\int_0^{\infty} e^{-\lambda x}[\lambda^2-1]u\,dx
= \lambda^3 \int_0^{\infty}e^{-\lambda x}u\,dx
+ \frac{\lambda}{2}\int_0^{\infty} e^{-\lambda x}u^2\,dx.
\end{equation}
Similarly to the bounded domain, forming perfect square and adding
time-inde\-pendent function, we obtain that
\begin{equation}\label{kdf.28}
\frac{d}{dt}\int_0^{+\infty}e^{-\lambda x}[\lambda^2-1](u+\lambda^2)\,dx
= \frac{\lambda}{2}\int_0^{+\infty} (u+ \lambda^2)^2
e^{-\lambda x}\,dx- \frac{\lambda^5}{2}\int_0^{+\infty} e^{-\lambda x}\,dx.
\end{equation}
It is not hard to prove the following chain of inequalities:
\begin{equation}\label{kdf.29}
\begin{aligned}
\Big|\int_0^{+\infty}(u+\lambda^2)e^{-\lambda x}\,dx \Big|^2
&\leqslant \int_0^{+\infty} (u+\lambda^2)^2 e^{-\lambda x}\,dx
 \int_0^{+\infty} e^{-\lambda x}\,dx\\
&= \frac{1}{\lambda}\int_0^{+\infty}(u+\lambda^2)^2 e^{-\lambda x}\,dx.
\end{aligned}
\end{equation}
Thus, we can rewrite \eqref{kdf.28} as
\begin{equation}\label{kdf.30}
\frac{dJ(t)}{dt}\geqslant \frac{\lambda^2 }{2(\lambda^2-1)}J^2
-\frac{\lambda^4}{2(\lambda^2-1)},
\end{equation}
where
\[
J(t)=\int_0^{+\infty}(u+\lambda^2)e^{-\lambda x}\,dx.
\]
After integrating  \eqref{kdf.30}, we can formulate
the following statement.

\begin{theorem} \label{thm9}
Suppose that the initial data satisfies the condition
\begin{equation}\label{kdf.31}
J(0)=\int_0^{+\infty}(u_0(x)+\lambda^2)e^{-\lambda
x}\,dx>1,\quad\lambda>1\,.
\end{equation}
Then the classical solution of  \eqref{eKdv-BBM}, \eqref{bb.18-1}
does not exist globally in time.
Furthermore, the following inequality holds
\begin{equation}\label{kdf.32}
J(t)\geqslant \frac{1+C\exp(\lambda^3t/(\lambda^2-1))}
{1-C\exp(\lambda^3t/(\lambda^2-1))},
\end{equation}
consequently,
\begin{equation}\label{kdf.33}
\lim_{t\to T_b}J(t)=+\infty,\quad
T_b\leqslant -\frac{\lambda^2-1}{\lambda^3}\ln C,\quad
C=\frac{J(0)-\lambda}{J(0)+\lambda}.
\end{equation}
\end{theorem}

\subsection*{Conclusion}
This article shows that the nonlinear capacity method allow us to study
not only standard Kortweg-de Vries equations,
but also other problems of modern mathematical physics:
problems for Benjamin-Bona-Mahony-Burgers,
Rosenau-Burgers and Korteweg-de Vries-Benjamin-Bona-Mahony equations.
Moreover, it is possible to unite the proof of blow-up in problems with
nonclassical boundary conditions and the existence of correct blow-up solutions.

This research was supported by RFBR (grant 08-01-00376-a, 11-01-12018-ofi-m-2011)
and  by grant from the President of the Russian Federation for young
Russian scientists MD-99.2009.1

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