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???

  1. The following Result I is obtained by standard $H^s$ energy estimate and some stopping times.
  2. 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!!!!


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

  • $\begingroup$ What is the source of the paper you have included pictures of? Google doesn't find a paper with that title. It is not much better than useless to include pictures but no link to a version with actual textual data. If this is your work, please provide an online copy, or else type the relevant information here into the question. $\endgroup$
    – David Roberts
    Aug 3, 2019 at 5:25
  • $\begingroup$ Thanks. That is my manuscript just for stating my problem. If you think it is not useful, sorry. I have changed it. $\endgroup$ Aug 3, 2019 at 13:39
  • $\begingroup$ Thanks for editing it in. I know it's a lot of work, but giving it as text means that the question is more accessible, and amenable to analysis by a computer looking at the whole MO corpus etc; the semantic context is extractable now by means other than the eyeballs of mathematicians. $\endgroup$
    – David Roberts
    Aug 3, 2019 at 14:30
  • $\begingroup$ Yes you are right. In the beginning I was afraid that I cannot express my problem clearly so I wrote a manuscript. Thanks a lot! $\endgroup$ Aug 3, 2019 at 15:11

1 Answer 1


It is not true that the bound $dM/dt = -\beta M^2$ implies that $M$ blows up almost surely. For example, with $b = 2$, there is a non-zero probability that $\beta < ce^{-t}$ for all $t>0$, for any given $c > 1$. Therefore, if $M(0) > -1/c$, no blow-up occurs.

  • $\begingroup$ Thanks a lot!! You make it clear. But $dM/dt=-\beta M^2$ implies $-1/M(0)>\int_0^t\beta(t') dt'$ almost surely. Taking expectation yields $-1/M(0)>\int_0^t\mathbb{E}\beta dt'=t$. What's wrong here? $\endgroup$ Aug 3, 2019 at 17:50
  • $\begingroup$ I see, $-1/M(0)>t$ for all $t\in[0,T_{max})$ but not for $t=T_{max}$. When $t=T_{max}$, which is also a random variable, then $\mathbb{E}\int_0^{T_{max}}\beta dt'\neq\int_0^{T_{max}}\mathbb{E}\beta dt'=T_{max}$. So we can not say $T_{max}<\infty$ almost surely. $\endgroup$ Aug 3, 2019 at 17:56

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