I have read that for the solution $u$ of the heat equation:
$$u_t(t,x) – \frac{1}{2} \nabla^2 u(t,x) =0, \\(t,x) \in (0, \infty)\times\mathbb{R}^n$$
$$u(0,x)=f(x), x \in \mathbb{R}^n$$
with $f \in L^1(\mathbb{R}^n), n \in \mathbb{N}$
When time tends to infinity the u function tends to zero, but I don't know how to prove it, could someone help me?
Best Answer
The unique solution (assuming the usual reasonable growth conditions) is given by convolution with a rescaled version of the heat kernel: $$ u(t,x) = \int_{\mathbb{R}^n} \frac{e^{-|x-y|^2/(2t)}}{(2\pi t)^{n/2}}f(y) dy. $$ Since $f \in L^1(\mathbb{R}^n)$ we can simply bound $$ |u(t,x)| \le \int_{\mathbb{R}^n} \frac{1}{(2\pi t)^{n/2}} |f(y)| dy = \frac{1}{(2\pi t)^{n/2}} \Vert f \Vert_{L^1}, $$ and from this it's clear that $\lim_{t\to\infty} \sup_{x \in \mathbb{R}^n} |u(t,x)| =0$, i.e. the solution decays uniformly to zero as $t \to \infty$.