Smoothing estimate for heat equation with compactly supported initial data

Let us consider \begin{equation} \partial_t u- u_{xx}=0\quad \text{in } (0,\infty)\times \mathbb{R},\quad u=u_0\quad \text{on } \{t=0\}\times \mathbb{R}. \end{equation} It is well-known that if $u_0 \in L^2(\mathbb{R})$, then the solution $u$ is instantaneously smooth for $t>0$. However, the solution does not have good norm control in derivatives. However, if we assume that the initial data has a compact support, say $\mathrm{supp}\, u_0 \subset (-1/4,1/4)$, then we can show that $$ u\in L^{\infty}(0,\infty;L^{2}((-2,2)^c). $$

To show this, we first recall the global estimate for the heat equation: \begin{equation} \int_{\mathbb{R}}|u(t,x)|^2\,{dx}+2\int_0^t \int_{\mathbb{R}} |\partial_x u(t,x)|^2 \,dx=\int_{\mathbb{R}} |u_0(x)|^2 \,dx. \end{equation}

Choose a cut-off function $\chi$ so that $\chi=1$ in $(-1,1)$ and $\chi=0$ outside $(-2,2)$. For $h\neq 0$, define $u_h(t,x) = (u(t,x+h)-u(t,x))/h$. Then $u_h$ satisfies $$ \partial_t (u_h)-(u_h)_{xx}=0\quad \text{in } (0,\infty)\times \mathbb{R},\quad u_h=(u_0)_h\quad \text{on } \{t=0\}\times \mathbb{R}. $$

Choose $\chi$ so that $0\leq \chi\leq 1$, $\chi=0$ in $(-1,1)$, and $\chi=1$ on $(-2,2)^c$. Multiplying $u_h \chi$ and taking integration by part, we get \begin{equation} \begin{aligned} &\frac{1}{2}\int_{\mathbb{R}} |u_h(t)|^2 \chi \,dx + \int_0^t \int_{\mathbb{R}} |\partial_x (u_h)|^2 \chi \,dx ds \\ &=\frac{1}{2}\int_0^t \int_{\mathbb{R}} |u_h|^2 \chi'' \,dxds+\frac{1}{2}\int_{\mathbb{R}} |u_h(0)|^2 \chi\,dx. \end{aligned} \end{equation} For sufficiently small $h$, we note that $u_h(0,x)\chi(x)=0$ since $\mathrm{supp}\, u_0 \subset (-1,1)$ and $\chi=0$ on $(-1,1)$. Hence it follows that $$ \frac{1}{2}\int_{\mathbb{R}} |u_h(0)|^2 \chi\,dx=0.$$ Hence by a method of finite difference, we get $$ \int_{\mathbb{R}} |u_x|^2 \chi \,dx+\int_0^t \int_{\mathbb{R}} |\partial_{xx} u|^2 \chi \,dxds \leq C \int_0^t \int_{\mathbb{R}} |u_x|^2|\chi''| \,d{x}ds. $$ Moreover, it follows from the energy estimate that $$ \int_{\mathbb{R}} |u_x(t)|^2 \chi \,dx+\int_0^t \int_{\mathbb{R}} |\partial_{xx} u|^2 \chi \,dxds \leq C \int_{\mathbb{R}} |u_0|^2 \,d{x}. $$ Hence it follows that $u \in L^\infty((0,\infty);H^1((-2,2)^c))$. Then by induction, we can further show that $u\in L^\infty ((0,\infty);H^k((-3,3)^c))$ for any $k$.

In fact, we can obtain better smoothing estimate. It belongs to Gevrey $1/s$-class $G^s$ for $0<s<1$ outside $(-1/2,1/2)$. Choose the following parameters $$ x_1 = \frac{3}{8},\quad x_{n+1}=x_n+\frac{c_\sigma}{n^{1+\sigma}},\quad y_n = x_n + \frac{c_\sigma}{100 n^{1+\sigma}},$$ where the constant $c_\sigma$ is chosen so that $c_\sigma \sum_{n=1}^\infty \frac{1}{n^{1+\sigma}}<1/8$. Define $\chi_n$ so that $\chi_n(y)=0$ for $-x_n<y<x_n$, and $\chi_n(y)=1$ for $\{-\infty<y<-y_n\}\cup \{y_n<y<\infty\}$ for $n\geq 1$. Then note that $$ \bigcap _{n=1}^\infty \{x_n=1\}\supset \left(-\infty,-\frac{1}{2}\right)\cup\left(\frac{1}{2},\infty\right),$$ $$ \mathrm{supp}\, (\nabla^k \chi_{n+1})\subset \{\chi_n=1\},$$ and $$ |\partial_x^j \chi_n|\leq C n^{j(1+\sigma)} \chi_{n-1}.$$ Following a similar argument, we get $$ \frac{d}{dt}\int_{\mathbb{R}} |\partial_x^{n+1} u|^2 \chi_{n+1}^2 dy + 2\int_{\mathbb{R}} |\partial_x^{n+2} u|^2 \chi_{n+1}^2 dy=\int_{\mathbb{R}} |\partial_x^{n+1} u|^2 \partial_x^2 (\chi_{n+1}^2)\,dx.$$ We will show that $$ \sum_{n=0}^\infty \left(\frac{1}{n!}\right)^{2/s} \Vert{\partial_x^n u}\Vert_{L^2}^2<\infty.$$

Note that $$ \partial_y^2 (\chi_{n+1}^2)\leq C (n+1)^{2(1+\sigma)}\chi_n^2.$$ So it follows that \begin{equation} \begin{aligned} &\int_{\mathbb{R}}|\partial_x^{n+1} u(t)|^2 \chi_{n+1}^2 d{x}+\int_0^t \int_{\mathbb{R}}|\partial_x^{n+2}u|^2 \chi_{n+1}^2 dxds\\ &\leq C(n+1)^{2(1+\sigma)}\int_0^t \int_{\mathbb{R}}|\partial_x^{n+1}u|^2 \chi_n^2 dxds. \end{aligned} \end{equation} Since $\sigma<1/s-1$, it follows that \begin{equation} \begin{aligned} &\left(\frac{1}{(n+1)!}\right)^{2/s}\int_{\mathbb{R}}|\partial_x^{n+1} u(t)|^2 \chi_{n+1}^2 d{x}+\left(\frac{1}{(n+1)!}\right)^{2/s}\int_0^t \int_{\mathbb{R}}|\partial_x^{n+2}u|^2 \chi_{n+1}^2 dxds\\ &\leq C\left(\frac{1}{n!}\right)^{2/s}\int_0^t \int_{\mathbb{R}}|\partial_x^{n+1}u|^2 \chi_n^2 dxds. \end{aligned} \end{equation}

If we set \begin{equation} \begin{aligned} a_n &=\int_{\mathbb{R}}|\partial_x^{n} u(t)|^2 \chi_{n}^2 d{x},\\ b_n &=\int_0^t\int_{\mathbb{R}}|\partial_x^{n+1} u|^2 \chi_{n+1}^2 d{x}ds, \end{aligned} \end{equation} then by induction, we have $$ b_n \leq C^n (n!)^{2(1+\sigma)}\Vert b_0\Vert_{L^2} $$ and so $$ \sum_{n=0}^\infty \left(\frac{1}{(n+1)!}\right)^{2/s}a_{n+1} \leq \Vert{b_0}\Vert_{L^2} \sum_{n=0}^\infty C^n \left(\frac{1}{n!} \right)^{2/s-2(1+\sigma)} .$$ Since $2/s-2(1+\sigma)>0$, the series converges. This implies that $$ \sum_{n=0}^\infty \left(\frac{1}{n!}\right)^{2/s}\int_{(-1/2,1/2)^c} |\partial_x^n u(t)|^2 dx<\infty.$$ Hence $u$ belongs to $L^\infty(0,\infty;G^{s}((-1/2,1/2)^c))$.