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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. enter image description here

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  • $\begingroup$ Er, the inviscid Burgers equation does have blowup solutions (basically because there need not be uniqueness of your quantity $x_0$): en.wikipedia.org/wiki/Burgers%27_equation $\endgroup$
    – Terry Tao
    Apr 6, 2016 at 2:29
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    $\begingroup$ It doesn't directly answer your question, but take a look in www-m6.ma.tum.de/~kuttler/script_reaktdiff.pdf, specifically in 4.1 they discuss a similar problem, only without periodic boundary condition. $\endgroup$
    – Amir Sagiv
    Apr 6, 2016 at 5:55
  • $\begingroup$ $\int_{-\infty}^t e^{-s^2}ds$ does not tend to 0 as $t\to\infty$ $\endgroup$ Jul 7, 2016 at 8:41
  • $\begingroup$ Differentiate the equation with respect to x. It follows that $u_x$ satisfies a maximum principle. From this, you can deduce a bound in the $W^{1,\infty}$ norm, which is more than enough to rule out blowup. $\endgroup$ Jul 7, 2016 at 14:45
  • $\begingroup$ If $g(t)=1$ you do have blow up $\endgroup$ Jul 7, 2016 at 15:10

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At least for initial data verifying $$ \|u_0\|_{L^\infty}\leq 1/3 $$ there is no blow up. The reason is that the $L^\infty$ norm is uniformly bounded:

$$ \|u(t)\|_{L^\infty}\leq \|u_0\|_{L^\infty} $$

That allow us to bound the $H^1$ seminorm:

Multiply by $-u_{xx}$, integrate by parts, use Hölder inequality, then Young inequality to obtain $$ \frac{d}{dt}\|u(t)\|_{\dot{H}^1}^2+\|u_{xx}(s)\|_{L^2}^2 \leq \|u_x\|_{L^4}^4 $$

Now notice that $$ \|u_x(s)\|_{L^4}^4=\int u_x^4dx\leq 3\|u\|_{L^\infty}\|u_{xx}\|_{L^2}\|u_x\|_{L^4}^2, $$ so $$ \frac{d}{dt}\|u(t)\|_{\dot{H}^1}^2+\|u_{xx}(s)\|_{L^2}^2 \leq 9\|u\|_{L^\infty}^2\|u_{xx}\|_{L^2}^2 $$

However, I do think that for initial data with larger $L^\infty$ norm a finite time blow up probably occurs. The reason is that the only quantity that we have uniformly bounded a priori is the $L^\infty$ norm, and that looks somehow critical for this problem

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