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I'm trying to estimate the rate of boundary gradient blow-up for the 1-dim heat equation near an irregular boundary point.

Let $b(t) := (1-t)^\alpha$, $\alpha < 1/2$. Let $u(t,x)$ solve the following Dirichlet problem for the heat equation: \begin{align} u_t &= \frac{1}{2} u_{xx}, \quad t \in [0,1], \; \; x \leq b(t), \\ u(t,b(t)) &= 0, \qquad t \in [0,1], \\ u(0,x) &= f(x) \end{align} where $f$ is bounded, $f(0) = u(0,b(0)) = 0$, $f$ is in $C^2$, $f$ is not identically zero.

The boundary point $(t,x) = (1, 0)$ is an irregular point in the Kolmogorov-Petrovsky sense. ($f$ needs to be slightly smoother than Holder-1/2 for every boundary point to be regular.)

The solution $u$ remains bounded by $\int t^{-1/2} f(x) e^{-x^2/(2t)}dx$ but its gradient at the boundary $u_x(t,b(t))$ will become unbounded as $t \to 1$.

I'd like to estimate the rate of this blow-up. Something like

$$ -u_x(t,x)\bigl|_{x = b(t)} \sim (1-t)^\beta \quad \text{as }t \to 1.$$ My intuition is that $\beta \geq \alpha - 1$, maybe even $\beta \geq \alpha - 1/2$, but I don't have a complete proof.

I have also posted this question at stackexchange, along with some things I've tried:

https://math.stackexchange.com/questions/2863200/gradient-blowup-for-a-parabolic-heat-equation

NB. Using a crude probabilistic argument (Brownian motion boundary behaviour), I know that the $-u_x(t,b(t)) \leq (1-t)^{\alpha - 1}$. I suspect we can do better, and that the diffusion term will dampen the blowup by order 1/2 and we should be able to get $-u_x(t,b(t)) \sim (1-t)^{\alpha - 1/2}$ as $t \to 1$.

I am thinking about using similarity variables, transforming the problem to one with bounded coefficients, and using some classical gradient estimates (Hopf's lemma etc.) to show bounded gradients, and then convert back to the original case, but haven't managed to complete the proof.

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For $0<\alpha<1/2$ derivative $u_x$ grows on the boundary (in absolute value) at least as $(1-t)^{\alpha-1/2}$.

To prove it I'll consider the domain $x>-(1-t)^\alpha$ instead of $x>(1-t)^\alpha$ in the question. It will simplify the exposition somewhat: not counting unnecessary minus signs, writing $x$ instead of $|x|$ etc.

A variable change $x\to x+(1-t)^\alpha$ turns the heat equation into $$ Lu=u_t-\frac12u_{xx}-\alpha(1-t)^{\alpha-1}u_x=0 $$ and the domain into $D_+=\{x>0\}\times\{0<t<1\}$.

Let $Q_{\varepsilon,\delta}=(0,\varepsilon)\times(1-\delta,1)$, $0<\varepsilon,\delta\le1$ be a rectangle. It will be enough to prove the required estimate in such a domain.

First note that any solution $u$ for $f\not\equiv0$ on the parabolic boundary $PQ_{\varepsilon,\delta}$ of a small enough $Q_{\varepsilon,\delta}$ satisfy the following relations: \begin{eqnarray*} &&u(0,t)= 0, \\ C_1 x &\le& u(x,1-\delta)\le C_2x ,\\ C_3 &\le& u(\varepsilon,t)\le C_4, \end{eqnarray*} for some positive constants $C_i$. If there is another solution $v$ with the same properties then there exist constants $K_1,K_2>0$ s.t. $K_1v\le u\le K_2v$ on $PQ_{\varepsilon,\delta}$ and consequently in $Q_{\varepsilon,\delta}$.

If function $v$ satisfy $Lv\le0$ instead of $Lv=0$ then $K_1v\le u$ in $Q_{\varepsilon,\delta}$ and $$ u_x(0,t)=\lim_{x\to0+}\frac{u(x,t)-u(x,0)}x =\lim_{x\to0+}\frac{u(x,t)}x \ge K_1\lim_{x\to0+}\frac{v(x,t)}x=K_1v_x(0,t). $$

Lets construct such a function $v$ with the required asymptotic of $v_x(0,t)$. Put $\psi(x)=|x|^{2\alpha}\,\mathrm{sign}\, x$, $Z(x,t)=(2\pi t)^{-1/2}\exp\{-x^2/2t\}$ and define $$ v(x,t)=\int_{-\infty}^\infty Z(x-y,1-t)\psi(y)\,dy, $$ a Poisson potential with reversed time.

This function loses its smoothness as $t\to1-0$ at the right rate, as can be checked by direct calculations: $$ v_x(0,t)=\frac{2^{\alpha +\frac{1}{2}}\Gamma (\alpha +1) (1- t)^{\alpha -\frac{1}{2}}}{\sqrt{\pi }}. $$

Since $v_t=-v_{xx}/2$ it is left to show that $Lv=-v_{xx}-\alpha(1-t)^{\alpha-1}v_x\le0$ in some $Q_{\varepsilon,\delta}$.

Denote $\tau=1-t$ and put $$ w(x,\tau)=\int_{-\infty}^\infty Z(x-y,\tau)\psi(y)\,dy. $$ Then $Lv=-w_{xx}-\alpha\tau^{\alpha-1}w_x$. Derivatives present in the rhs don't change sign in $D_+$: \begin{align*} w_x(x,\tau)&=\int_{-\infty}^\infty Z(x-y,\tau)\psi'(y)\,dy>0, \\ w_{xx}(x,\tau)&=\int_{0}^\infty(Z(x-y,\tau)-Z(x+y,\tau))\psi''(y)\,dy<0. \end{align*}

So one has to show that $\alpha\tau^{\alpha-1}w_x\ge|w_{xx}|$. To this end we'll get estimates from below for $w_x$ and from above for $|w_{xx}|$.

We will divide the domain $D_+$ into two parts, above and below the parabola $x=\tau^{1/2}$.

$ $1. $\{x\le\tau^{1/2}\}$.

To get estimates for $w_x$, $w_{xx}$ we'll use the fact that function $\psi$ is homogeneous. Namely, from $\psi(kx)=k^{2\alpha}\psi(x)$, $k>0$, it follows that $w(x,\tau)=k^{-2\alpha}w(kx,k^2\tau)$ and $w_x(x,\tau)=k^{1-2\alpha}w_x(kx,k^2\tau)$. Putting $k=\tau^{1/2}$ gives $$ w_x(x,\tau)=\tau^{\alpha-1/2}w_x(x\tau^{-1/2},1) $$ and $$ w_x(x,\tau)\ge \tau^{\alpha-1/2}\min_{0\le y\le1} w_x(y,1)=C\tau^{\alpha-1/2} $$ for $0\le x\le \tau^{1/2}\le1$. In the same manner, using the equality $w_{xx}(x,\tau)=k^{2-2\alpha}w_x(kx,k^2\tau)$ one gets $$ |w_{xx}(x,\tau)|\le \tau^{\alpha-1}\max_{0\le y\le1} |w_{xx}(y,1)|=C\tau^{\alpha-1}. $$ Now we can obtain the required inequality: $$ Lv=-\alpha\tau^{\alpha-1}w_x-w_{xx}\le -C_1 \tau^{\alpha-1} \tau^{\alpha-1/2}+ C_2\tau^{\alpha-1}= C_1 \tau^{\alpha-1}(-\tau^{\alpha-1/2} + C_3)\le 0 $$ which holds for small enough values of $\tau$.

$ $2. $\{x>\tau^{1/2}\}$.

Preliminary we'll establish some asymptotic expansions for $\partial_x^k w(x,1)$, $x\to+\infty$. To illustrate the idea, let's show that $$ w(x,1)=x^{2\alpha}+O(x^{2\alpha -1}),\quad x\to+\infty. $$

Cutting of the intervals with exponentially decreasing tails one gets $$ w(x,1)= \psi(x)+\int_{-\infty}^\infty Z(x-y,1)(\psi(y)-\psi(x))\,dy= $$ $$ =\psi(x)+\int_{x/2}^{3x/2} Z(x-y,1)(\psi(y)-\psi(x))\,dy+O(e^{-x^2/16})= $$ $$ =\psi(x)+\int_{x/2}^{3x/2} Z(x-y,1)(\psi(y)-\psi(x)-\psi'(x)(x-y))\,dy+O(e^{-x^2/16})= $$ $$ =x^{2\alpha}+O(x^{2\alpha -1}),\quad x\to+\infty, $$ since $$ \left|\int_{x/2}^{3x/2} Z(x-y,1)(\psi(y)-\psi(x)-\psi'(x)(x-y))\,dy\right|\le $$ $$ \le \frac12|\psi''(x/2)|\int_{-\infty }^{\infty } Z(x-y,1)(x-y)^2\,dy=Cx^{2\alpha -1}. $$ Repeating this argument gives \begin{equation} \begin{split} w_x(x,1)&=2\alpha x^{2\alpha-1}+O(x^{2\alpha -2}),\quad x\to+\infty, \\ w_{xx}(x,1)&=2\alpha(2\alpha-1) x^{2\alpha-2}+O(x^{2\alpha -3}),\quad x\to+\infty. \end{split} \end{equation}

Now return to the estimate of $Lv$. From the positivity of $w_x(x,1)$ and the established asymptotic expansion it follows that there exists $C>0$ s.t. $$ w_x(x,1)\ge Cx^{2\alpha-1},\quad x\ge1. $$ Inserting $k=\tau^{-1/2}$ in the equality $w_x(x,\tau)=k^{1-2\alpha}w_x(kx,k^2\tau)$, we have $$ w_x(x,t)=\tau^{\alpha-1/2}w_x(x\tau^{-1/2},1)\ge C \tau^{\alpha-1/2} (x\tau^{-1/2})^{2\alpha-1}=Cx^{2\alpha-1}. $$ In the same way one obtains $$ |w_{xx}(x,\tau)|\le Cx^{2\alpha-2}. $$ Finally \begin{align*} Lv&=-\alpha\tau^{\alpha-1}w_x-w_{xx}\le -C_1x^{2\alpha-1}\tau^{\alpha-1}+C_2 x^{2\alpha-2}= \\ &=C_1x^{2\alpha-2}(-\tau^{\alpha-1/2}(x\tau^{-1/2})+C_3)\le C_1x^{2\alpha-2}(-\tau^{\alpha-1/2}+C_3)\le0 \end{align*} for small enough $\tau$.

From 1) and 2) it follows that $Lv\le0$ in some $Q_{\varepsilon,\delta}$ which finishes the proof that $u_x(0,t)\ge C(1-t)^{\alpha-1/2}$.

Also, answering your question from another post, it follows from the inequality $u\ge K_1v$ in $Q_{\varepsilon,\delta}$ that $u(x,1)\ge C x^{2\alpha}$.

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