My original problem is to see if the following pde develops blow-ups in $(-L,L)$

$$u_{t}=u_{xx}+g(t)(u_{x})^{2}$$

for periodic boundary $u_{0}(-L)=u_{0}(L)$, where $0<g(t)<1$; specifically $g(t)=\Phi(\frac{\beta-t}{\beta})=\int_{-\infty}^{t}e^{-s^{2}}ds$ for some $\beta>0$.

Do you think it blows-up? Any suggestions on books or papers that treat a similar pde?

For $g(t)\equiv 1$, this is called the Potential Burgers equation as can be seen by differentiating and setting $v=u_{x}$.

**Attempt**

So to start I am checking $u_{t}=(u_{x})^{2}$ for blow-ups. We differentiate it and set $v=u_{x}$ to obtain

$$v_{t}-2vv_{x}=0,$$

and by MOD $v(x,t)=v_{0}(x+t2v_{0}(x_{0}))$. So there is a blow-up if there exist $x_{1}<x_{2}$ with

$$\frac{1}{-2v_{0}(x_{1})}>\frac{1}{-2v_{0}(x_{2})}.$$

Assuming $v_{0}>0$ we obtain $v_{0}(x_{1})>v_{0}(x_{2})$, which can happen for periodic IC as in our case (eg. exp(cos(x))).

Next we check $u_{t}=g(t)(u_{x})^{2}\Rightarrow v_{t}-2g(t)vv_{x}=0$. By MOC we get curve $x(s)=-2v_{0}(x_{0})\int^{s}_{0} g(r)dr+x_{0}$ and thus

$$v(x,t)=v_{0}(x+2v_{0}(x_{0})\int^{t}_{0} g(r)dr).$$

The derivative of the curve is $\frac{dx}{dt}=-2v_{0}(x_{0})g(t)$, so it's not clear from here. Two curves are equal if $$v_{0}(x_{1})-v_{0}(x_{2})=\frac{x_{2}-x_{1}}{\int^{t_{int}}_{0} g(r)dr},$$

for some $x_{i}$ and time $t_{int}$. Now I am trying to use Implicit FT to prove existence of such $x_{i},t_{i}$.

But even if it does it is not clear that the original PDE blows-up For example, Burgers ($u_{t}=u_{xx}-u_{x}u$) doesn't blow-up but the semilinear heat equation ($u_{t}=u_{xx}+u^{2}$) does.

So next I will try to mimic techniques from Burgers and Semilinear heat equation (Evans chapter 9) to find blow-ups.

Interestingly, by assuming $g(t)\equiv 1$ we obtain that $u=Log(\phi)$, where $\phi$ satisfies the heat eqn $\phi_{t}=\phi_{xx}$. Since $\phi$ can be negative, we consider the complex logarithm. This can present blow-ups, so it is natural to guess that the PDE above will also. However, it is not totally clear because as $t\to \infty$ the $g(t)\to 0$. Very cool stuff.

From numerical analysis (pseudo-spectral method with IC 8*exp(cos(x/128)) a blowup develops close to time 0.17.

doeshave blowup solutions (basically because there need not be uniqueness of your quantity $x_0$): en.wikipedia.org/wiki/Burgers%27_equation $\endgroup$1more comment