For 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 means the global existence with high probability and the second means blow-up happens with probability 1???

- The following
**Result I**is obtained by standard $H^s$ energy estimate and some stopping times. **Result II**is also standard results in studying the shock waves in Buegers' equation.

**The analysis is not difficult, but I can not understand what is wrong??**

Which one is right, or both are correct but the underlying probability space is changed? Actually, in first one, the solution is viewed as an $H^s$ element, in the second one, the solution is viewed as a random field???

I have been confused for many days. Thanks in advance!!!!

**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., $\mathbb{P}\{T_{max}<\infty\}=1.$