Under the given conditions. Prove that $\lim_{t\to\infty} u(x, t) = 0 $, uniformly in x.

Question

For any $(x, t)\in R^n × (0, +∞)$ let $ K(x, t) := (\frac{1}{4πt})^\frac{n}{2} e^-\frac{|x|^2}{4t} $ be fundamental solution of the heat equation (also called the heat kernel) and consider $u(x, t) = \int_{R^n} K(x − y, t)u_0(y) dy $. Suppose $u_0 $ is continuous in $R^n$ and that $u_0(x) $→ 0 uniformly as |x| → +∞. Prove that $\lim_{t\to\infty} u(x, t) = 0 $, uniformly in x. I really need help in this question.

Answer 1

Similarly it can be shown that $\lim_{(x,t) \rightarrow (x_0, 0)} u(x,t)=u_0(x_0)$ $\forall x_0 \in \mathbb{R}^n$, assuming $u_0(x) \in L^\infty(\mathbb{R}^n) \cap C(\mathbb{R}^n)$, note that since $\int_{\mathbb{R}^n} K_t(x) dx = 1$, we have

$\displaystyle u_0(x_0) = \int_{\mathbb{R}^n} K_t(x-y)u_0(x_0) dy$

$\displaystyle u(x,t)-u_0(x_0) = \int_{\mathbb{R}^n} K_t(x-y)(u_0(y)-u_0(x_0))dy = I_1 + I_2$

where $I_1=\int_{B(0,\delta)} ...$ and $I_2=\int_{\mathbb{R}^n \setminus B(0, \delta)}...$ for some $\delta > 0$ to be determined. Since $u_0$ is continuous $\forall \epsilon > 0$ there is $\delta=\delta(\epsilon) > 0$ such that $|u_0(y) - u_0(x_0) | < \epsilon$, then $|I_1| \leq \epsilon$. Since $u_0$ is limited, we have $|u_0| \leq M$. Now, if $x \in B(x_0, \delta /2)$ and $y \in \mathbb{R}^n \setminus B(x_0, \delta)$, we have

$\displaystyle |y-x_0| \leq |y-x|+|x-x_0| \leq |y-x|+\delta /2 \leq |y-x| + |y-x_0| /2$

and $|y-x|\geq |y-x_0|/2 \geq \delta /2$, i.e. $x-y \in \mathbb{R}^n \setminus B(x_0, \delta/2)$. For the properties of the heat kernel is known that

$\displaystyle \lim_{t \rightarrow 0^+} \int_{\mathbb{R}^n \setminus B(0,\delta)} K_t(x) dx =0$ $\forall \delta > 0$

and in this case, we have that $\exists \delta' > 0$ such that

$\displaystyle \int_{\mathbb{R}^n \setminus B(x_0,\delta)} K_t(x-y) dy \leq \epsilon /2M$ $\forall x \in B(x_0, \delta /2)$ and $\forall t \in (0,\delta')$

therefore

$\displaystyle |I_2| \leq \int_{\mathbb{R}^n \setminus B(x_0,\delta)} K_t(x-y) (|u_0(y)|+ |u_0(x_0|) dy \leq \epsilon$

in the end $|u(x,t)-u_0(x_0)| \leq 2 \epsilon$ $\forall x \in B(x_0, \delta /2)$ and $\forall t \in (0,\delta')$.