Let $u(x,t)$ be a solution to the following parabolic PDE:
With $\alpha \in (0,1)$, \begin{align} \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,1] \times [0,1] \\ u(x,0) &= f(x),\\ u(1,t) &= u(-\infty,t) = 0 \quad \text{(boundary condition)} \end{align} Here $f \in C^2$, $\; f \geq 0$, $\; \int f^2(x)dx < \infty$, $\; f(1) = f(-\infty) = 0$.
I'm especially interested in the case $\alpha < 1/2$, but $\alpha \in [1/2,1)$ is also valuable.
Note that the coefficient of $\partial_x u$, i.e $\alpha (1-t)^{\alpha -1}$ becomes unbounded at $t = 1$.
I'm trying to show that $$-\partial_xu(x,t)|_{x=1} \geq \kappa (1-t)^{\alpha-1} \quad \text{ as } \quad t \to 1, \quad (\kappa > 0 \text{ constant})$$ i.e. the right side boundary $(x = 1)$ gradient blows up at the rate at least $(1 - t)^{\alpha - 1}$ as $t \to 1$.
We can try the transformation $(z,r) = (\mu x, \mu^2 t)$ for some $\mu > 0$, and define $\xi(x,t) = u(z,r)$ for $(x,t) \in (-\infty, 1/\mu] \times [0,1/\mu^2].$ Then we see that \begin{align} \xi_x(x,t) &= \mu u_z(z,r), \\ \xi_{xx}(x,t) &= \mu^2 u_{zz}(z,r), \\ \xi_t(x,t) &= \mu^2 u_r(z,r), \end{align} so that $\xi$ solves the Dirichlet problem
\begin{align} \partial_t \xi(x,t) &= \alpha \mu (1-\mu^2 t)^{\alpha - 1} \partial_x \xi(x,t) + \frac{1}{2} \partial_{xx} \xi(x,t), \quad (x,t) \in (-\infty,1/\mu] \times [0,1/\mu^2] \\ \xi(x,0) &= f(\mu x),\\ \xi(1/\mu,t) &= \xi(-\infty,t) = 0 \quad \text{for }t \in[0,1/\mu^2] \end{align} Next, for each fixed $T < 1$, we restrict the solution of the above problem to $t \in [0,T/\mu^2]$ and set the scaling $$\mu = \mu_T = (1 - \mu^2 t)^{1 - \alpha}|_{t = T/\mu^2} = (1 - T)^{1 - \alpha}.$$ This means that on each strip $(x,t) \in (-\infty,1/\mu_T)\times [0,T/\mu_T^2]$ and with the scaling $\mu = \mu_T$, the coefficients of the equation are (uniformly) bounded, and it has a solution in the classical sense.
Also, for each $T$, by Hopf's boundary gradient lemma, $\partial_x \xi(x, T/\mu_T^2 )|_{x = 1/\mu_T} = C_T > 0$ (boundary gradient is strictly positive).
This means that $$\partial_x u(1,T) = C_T / \mu_T = C_T(1-T)^{\alpha - 1}$$.
This would be complete the proof, but unfortunately, as $T \to 1$, the right lateral boundary $x = 1/\mu_T$ becomes unbounded and Hopf lemma fails. In other words, the asymptoic factor $C_T$ is not uniformly boubded from below. In fact, I suspect $C_T \to 0$ as $T \to 1$, and the argument is not enough to complete the proof. I need to show that $C_T > C > 0, \forall T$ (probably not true), or revise the estimate by showing that $C_T \searrow 0$ slowly (logarithmic etc), or use a completely different tactic.
NB. I already have an asymptotic upper bound on $-\partial_x u(x,t)|_{x = 1}$: $$-\partial_x u(x,t)|_{x = 1} \leq (1 - t)^{1 - \alpha} \text{ as } t \to 1,$$ $\alpha \in (0,1)$.
Any ideas? I'll start a bounty when I'm allowed.