We know that a solution to the Cauchy problem on $\mathbb{R}$ : $u_{xx}=u_t$ with condition $u|_{t=0}=\varphi(x)$ is of the form
$$u(x,t)=\dfrac{1}{2\sqrt{\pi t}}\int_{-\infty}^{\infty}\exp\left({\dfrac{-|x-y|^2}{4t}}\right)\varphi(y)dy$$
I want to know what happen as $t\to\infty$ in the case that $\varphi(x)$ is a bounded continuous function that satisfies boundary condition at infinity
$\lim_{x\to\infty}\varphi(x)=A$ and $\lim_{x\to-\infty}\varphi(x)=B$
Any hint on how to use these assumptions on $\varphi$ to simplify $u(x,t)$ ?
The limit should be $(A+B)/2$.
Given $\epsilon > 0$, there is $N$ such that $|\varphi(x) - A| < \epsilon$ for $x \ge N$ and $|\varphi(x) - B| < \epsilon$ for $x \le -N$. The integral over $(-\infty, -N]$ is then within $\epsilon$ of $$\dfrac {B}{2\sqrt{\pi t}} \int_{-\infty}^{-N} \exp\left(\frac{-|x-y|^2}{4t}\right)\; dy $$ which goes to $B/2$ as $t \to \infty$, and similarly for the integral over $[N,\infty)$. The integral over $[-N,N]$ goes to $0$.