# Boundary gradient blowup of a parabolic PDE with an unbounded coefficient

I am exploring the boundary gradient blowup of following parabolic PDE on the semi-infinite strip $(-\infty,0]\times[0,1]$:

Let $0< \alpha < 1/2$.

$\partial_t u(x,t) = -\alpha(1-t)^{\alpha - 1} \partial_x u(x,t) + \frac{1}{2} \partial_{xx} u(x,t), \quad (x,t) \in (-\infty,0] \times [0,1]$

with the initial condition $u(x,0)$ a non-negative continuous $L^2$ function with $u(x,0) \to 0$ as $x \nearrow 0$ and $x \searrow -\infty$, (for example, $u(x,0) = |x| e^{-x^2}$),

and boundary conditions $u(0,t) = u(-\infty,t) = u(-\infty,t) = 0$ for $t \in [0,1]$.

I suspect that the solution is bounded but the gradient $\partial_x u(x,t)$ becomes unbounded as $(x,t) \to (0,1)$. In fact, I think

$u(x,1) \sim |x|^{2 \alpha}$ as $x \nearrow 0$,

and

$u(1,t) \sim (1-t)^\alpha$ as $t \nearrow 1$.

PDEs are not my main area of study and I'm not familiar with the literature. I'd appreciate some pointers and references.

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I've retagged. Both (parabolic) and (gradient) are too general (applicable to too many things). And the (blowup) tag doesn't mean what you think it means (it means the blowup construction in algebraic geometry), not singularity formation for differential equations. – Willie Wong Apr 12 '13 at 8:23
Also, what values are $\alpha$ allowed to take? – Willie Wong Apr 12 '13 at 8:26
Thanks Willie. Good points. In my problem $\alpha < 1/2$. For $\alpha \geq 1/2$, the diffusion term dominates the $\partial_x$ term and no singularity is formed. Also it is known that the solution $u$ itself remains bounded. – zab Apr 12 '13 at 9:08
Another question: you cannot expect blowup for all non-negative continuous $L^2$ data: $u\equiv 0$ is a solution. Are you claiming that for all non-zero data one should expect blow up? Is this backed by numerical studies or perhaps by some sort of heuristics? (Yes, in general it is a good practice to share your "good reason to suspect" if you are asking a question.) – Willie Wong Apr 12 '13 at 9:25
Lastly, do you mean data is only in $L^2$ with continuity? You do not assume anything on the first derivative level for the data? – Willie Wong Apr 12 '13 at 9:26

Thanks Willie. I have read the article. Unfortunately it only deals with cases when my $\alpha \geq 1/2$ so that no blow-up occurs. – zab Apr 12 '13 at 12:15