I find that forFor the stochastic Burgers' equation with linear noise, I can deduce two results. Both of them can be applied to same initial data, but the first result (Proposition 2.1) means the global existence with high probability butand the second (Proposition 3.1) means blow-up happens with probability 1???
- The first onefollowing Result I is obtained by standard $H^s$ energy estimate and some stopping times.
- The second oneResult II is also standard results in studying the shock waves in Buegers' equation.
Only 2 pages, theThe analysis is not difficult, but I can not understand what is wrong??
MY PROBLEM:
Consider the stochastic Burgers' equation
\begin{equation}\label{S-Burgers}
{\rm d}u+uu_x{\rm d}t=b u{\rm d}W,\ b\neq\mathbb{R}.
\end{equation}
Using the following Girsanov type transform
\begin{align}
v=\frac{1}{\beta} u,\ \
\beta(\omega,t)={\rm e}^{b W_{t}-\frac{b^2}{2}t},
\end{align}
then we have
\begin{align}
{\rm d}v=&\frac{1}{\beta} {\rm d}u+u {\rm d}\frac{1}{\beta}+ {\rm d}\frac{1}{\beta} {\rm d}u=-\beta vv_x{\rm d}t,
\end{align}
that is,
$$v_t+\beta vv_x=0.\ \ \ \ (1)$$
Let $\mathbb{T}=\mathbb{R}/2\pi\mathbb{Z}$, $D^s=(1-\partial_{xx}^2)^{s/2}$ and $H^s(\mathbb{T})=\{f:D^s f\in L^2(\mathbb{T})\}$.
Let $s>3/2$ and initial data $u_0\in H^s$ be deterministic. It can be proved that the solution
$$u\in C([0,T_{max});H^s)\bigcap C^1([0,T_{max});H^{s-1}),\ \ \mathbb{P}-a.s.$$
Therefore $v$ also belongs to $C([0,T_{max});H^s)\bigcap C^1([0,T_{max});H^{s-1})$ almost surely.
Global existence with high probability.
The idea of following analysis comes from this paper. To begin with, we apply the operator $D^s$ to (1), multiply both sides of the resulting equation by $D^sv$ and integrate over $\mathbb{T}$ to obtain that for a.e. $\omega\in\Omega$,
\begin{align*}
\frac{1}{2}\frac{\rm d}{{\rm d}t}\|v(t)\|^2_{H^s}
=& -\beta \int_{\mathbb{T}}D^sv\cdot D^s\left[vv_x\right]{\rm d}x\leq C\beta(t)\ \|v\|_{W^{1,\infty}}\|v\|_{H^s}^2.
\end{align*}
From the above estimate, we construct a "damping" term t obtain:
$w={\rm e}^{-b W_{t}}u={\rm e}^{-\frac{b^2}{2} t}v$ satisfies
\begin{align*}
\frac {\rm d}{ {\rm d}t}\|w(t)\|_{H^s}+\frac{b^2}{2}\|w(t)\|_{H^s}
\leq C\alpha(\omega,t) \|w(t)\|_{W^{1,\infty}}\|w(t)\|_{H^s},\ \
\alpha(\omega,t)={\rm e}^{b W_{t'}}.
\end{align*}
For any $R>1$, we fix $R$, let
$
K(b,R)=\frac{b^2}{4CR}
$
and then define
\begin{align}
\tau_{1}(\omega)=\inf\left\{t>0:\alpha(\omega,t) \|w\|_{W^{1,\infty}}
=\|u\|_{W^{1,\infty}}>\frac{b^2}{4C}\right\}.\label{global time tau}
\end{align}
Assume that $\|u_0\|_{H^s}<K(b,R)<\frac{b^2}{4C}$, then
$
\mathbb{P}\{\tau_{1}>0\}=1,
$
and for $t\in[0,\tau_{1})$,
\begin{align*}
\frac {\rm d}{ {\rm d}t}\|w(t)\|_{H^s}+\frac{b^2}{4}\|w(t)\|_{H^s}
\leq 0.
\end{align*}
The above inequality implies that for a.e. $\omega\in\Omega$ and for any $t\in[0,\tau_{1})$,
\begin{align}
\|u(t)\|_{H^s}
\leq& \|u_0\|_{H^s}{\rm e}^{b W_{t}-\frac{b^2}{4} t}.\ \ \ \ (2)
\end{align}
Define the stopping time
$$\tau_2(\omega)
=\inf\left\{t>0:{\rm e}^{b W_{t}-\frac{b^2}{4} t}>R\right\}. $$
Notice that $\mathbb{P}\{\tau_{2}>0\}=1$. From (2), we have
\begin{align}
\|u(t)\|_{H^s}\leq& RK(b,R)=\frac{b^2}{4C},\ \ t\in[0,\tau_{1}\wedge \tau_{2}),\ \ \ \ (3)
\end{align}
which means
$
\mathbb{P}\{\tau_{1}\geq\tau_{2}\}=1.
$
Therefore it follows from (3) that
$$\mathbb{P}\left\{
\|u(t)\|_{H^s}<\frac{b^2}{4C} \ {\rm\ for\ all}\ t>0
\right\}\geq
\mathbb{P}\{\tau_{2}=+\infty\}.$$
It can be estimated (cf. Lemma 9.1 in this paper) that
\begin{equation*}
\mathbb{P}\{\tau_{2}=+\infty\}>1-\left(\frac{1}{R}\right)^{1/2}.
\end{equation*}
In conclusion, we have
Result I: Let $u_0=u_0(x)\in H^s$ be deterministic.
For any $R>1,s>3/2$, if for some $C=C(s)>0$, $\|u_0\|_{H^s}\leq\frac{b^2}{4C R}$, then
$$\mathbb{P} \left\{
\|u(t)\|_{H^s}<\frac{b^2}{4C}
\ {\rm\ for\ all}\ t>0
\right\}
\geq 1-\left(\frac{1}{R}\right)^{1/2}.$$
Blow-up in finite time almost surely. To begin with, we recall
Preliminary result(see this paper): Let $T >0$ and $v\in C^1([0,T); H^2(\mathbb{T}))$. Then given any $t\in[0,T)$, there is at least one point $z(t)$ with \begin{equation*} M(t)\triangleq\min_{x\in\mathbb{T}}[v_x(t,x)]=v_x(t,z(t)). \end{equation*} Moreover, $M(t)$ is almost everywhere differentiable on $(0,T)$ with \begin{equation*} \frac{{\rm d}}{{\rm d}t}M(t)=v_{tx}(t,z(t))\ \ {\rm a.e.\ on}\ (0,T). \end{equation*}
Now we consider $s>4$. Firstly, from (1), we have \begin{equation} v_{tx}+\beta vv_{xx}=-\beta v^2_x,\ \ t\in[0,T_{max}),\ \ \mathbb{P}-a.s.\ \ \ \ (4) \end{equation} Define \begin{equation} M(\omega,t):=\min_{x\in\mathbb{T}}[v_x(\omega,t,x)],\ \ \text{a.e.}\ \omega\in\Omega. \end{equation} Then the preliminary result yields that there is a $z(\omega,t)$ such that $M(\omega,t)=v_x(\omega,t,z(\omega,t))$. Evaluating (4) in $(t,z(t))$ with noticing $v_{xx}(t,z(\omega,t))=0$ and using the above preliminary result yields that for a.e.\ $\omega\in\Omega$, \begin{equation}\label{M equation} \frac{{\rm d}}{{\rm d}t}M(t)=-\beta M^2(t),\ \ {\rm a.e.\ on}\ (0,T_{max}). \end{equation} From the above estimate, it is easy to find that if $M(0)<0$, then $M$ tends to $-\infty$ in finite time almost surely.
In other words, we have:
Result II: Let $u_0=u_0(x)\in H^s$ be deterministic. If $$\min_{x\in\mathbb{T}}\partial_xu_0(x)<0,$$ then the solution $u$ to (1) blows up in finite time almost surely, i.e., $\mathbb{P}\{T_{max}<\infty\}=1.$