Asymptotic behavior of solution to the heat equation

Question

Given $1>k>0$, I want to show that $\forall t>0$, \begin{equation}%\label{eqn: 1st time derivative} \lim_{|x|\to\infty}\frac{u_x(x,t)}{(u(x,t))^k}=0, \end{equation} where \begin{equation} u(x,t)=\frac{1}{\sqrt{2\,\pi\,t}}\int_{-\infty}^{\infty} g(y)\,e^{-\frac{(y-x)^2}{2\,t}}\,dy \end{equation} is a solution of the heat equation $u_t=\frac{1}{2}u_{xx}$. Here, we assume $g>0$ and $g$ is smooth. My first attempt is to evaluate \begin{equation}%\label{eqn: 1st time derivative} \int_{-\infty}^{\infty}\frac{u_x(x,t)}{(u(x,t))^k}\,dx = \int_{u=0}^{u=0}u(x,t)^{-k}\,du(x,t) =0, \end{equation} using the fact $\lim_{|x|\to\infty}u(x,t)=0$. However, this does not give \begin{equation}%\label{eqn: 1st time derivative} \int_{-\infty}^{\infty}\left|\frac{u_x(x,t)}{(u(x,t))^k}\right|\,dx=0, \end{equation} and we cannot conclude $\lim_{|x|\to\infty}\frac{u_x(x,t)}{(u(x,t))^k}=0$. On the other hand, the initial condition $g(y)=e^{-y^2}$ gives \begin{equation} u(x,t)=\frac{e^{-\frac{x^2}{2 t+1}}}{\sqrt{2 t+1}}, \end{equation} and it is easy to see that $\lim_{|x|\to\infty}\frac{u_x(x,t)}{(u(x,t))^k}=0$, which seems to be true for more general $g(y)$. Any reference, suggestion, idea, or comment is welcome. Thank you!

Answer 1

It seems we need to assume $|g(y)|\le M_1\,e^{ay^2}$ and $|g_y(y)|\le M_2\,e^{by^2}$ for some $a,b,M_1,M2>0$, or assume $|g_y(y)|\le M_3\,|g(y)|$ for some $M_3>0$?

\begin{align}%\nonumber \notag u(x,t)=&\frac{1}{\sqrt{2\,\pi\,t}}\int_{-\infty}^{\infty} g(y)\,e^{-\frac{(y-x)^2}{2\,t}}\,dy && \text{} \\[1ex] \notag u_x(x,t)=&\frac{1}{\sqrt{2\,\pi\,t}}\int_{-\infty}^{\infty} g(y)\,\frac{(y-x)}{t}\,e^{-\frac{(y-x)^2}{2\,t}}\,dy && \text{Leibniz integral rule} \\[1ex] \notag =&\frac{1}{\sqrt{2\,\pi}t^{\frac{3}{2}}}\int_{-\infty}^{\infty} g(y)\,(y-x)\,e^{-\frac{(y-x)^2}{2\,t}}\,dy && \text{simplify} \\[1ex] \notag =&\frac{1}{\sqrt{2\,\pi}t^{\frac{3}{2}}}\int_{-\infty}^{\infty} g(z+x)\,z\,e^{-\frac{z^2}{2\,t}}\,dz && \text{let $z:=y-x$ $\iff$ $y=z+x$} \\[1ex] \notag =&\frac{1}{\sqrt{2\,\pi}t^{\frac{3}{2}}}\int_{-\infty}^{\infty} g(z+x) (-t)\,d\left(e^{-\frac{z^2}{2\,t}}\right) && \text{rewrite} \\[1ex] \notag =&\frac{-1}{\sqrt{2\,\pi\,t}}\int_{-\infty}^{\infty} g(z+x)\,d\left(e^{-\frac{z^2}{2\,t}}\right) && \text{simplify} \\[1ex] \notag =&\frac{-1}{\sqrt{2\,\pi\,t}} \left( g(z+x)\,e^{-\frac{z^2}{2\,t}}\bigg|_{-\infty}^{\infty} - \int_{-\infty}^{\infty} e^{-\frac{z^2}{2\,t}}g_z(z+x)\,dz \right) && \text{IBP} \\[1ex] \notag =&\frac{1}{\sqrt{2\,\pi\,t}} \int_{-\infty}^{\infty} e^{-\frac{z^2}{2\,t}}g_z(z+x)\,dz && \text{assume $g$ is bounded} \\[1ex] \notag =&\frac{1}{\sqrt{2\,\pi\,t}} \int_{-\infty}^{\infty} e^{-\frac{(y-x)^2}{2\,t}}g_y(y)\,dy && \text{$z:=y-x$ $\iff$ $y=z+x$} \end{align}

\begin{align}%\nonumber \notag u(x,t)=&\frac{1}{\sqrt{2\,\pi\,t}}\int_{-\infty}^{\infty} g(y)\,e^{-\frac{(y-x)^2}{2\,t}}\,dy && \text{} \\[1ex] \notag (u(x,t))^k=&\left(\frac{1}{\sqrt{2\,\pi\,t}}\right)^k\left(\int_{-\infty}^{\infty} g(y)\,e^{-\frac{(y-x)^2}{2\,t}}\,dy\right)^k && \text{$k$th power} \end{align}

\begin{align}%\nonumber \notag \frac{u_x(x,t)}{(u(x,t))^k}=& \left(\frac{1}{\sqrt{2\,\pi\,t}}\right)^{1-k} \frac{\int_{-\infty}^{\infty} e^{-\frac{(y-x)^2}{2\,t}}g_y(y)\,dy} {\left(\int_{-\infty}^{\infty} g(y)\,e^{-\frac{(y-x)^2}{2\,t}}\,dy\right)^k} && \text{} \\[1ex] \notag =&\left(\frac{1}{\sqrt{2\,\pi\,t}}\right)^{1-k} \frac{\int_{-\infty}^{\infty} e^{-\frac{(y-x)^2}{2\,t}}g_y(y)\,dy} {\int_{-\infty}^{\infty} g(y)\,e^{-\frac{(y-x)^2}{2\,t}}\,dy} \left(\int_{-\infty}^{\infty} g(y)\,e^{-\frac{(y-x)^2}{2\,t}}\,dy\right)^{1-k} && \text{} \end{align}