How to deal integral at infinity?

Question

Let $u\in C^{2}(\mathbb R^n \times (0, \infty))$ satisfies the heat equation $\partial_t u - \Delta u = 0 \ (t>0)$ and $u(x, 0)=0.$ Assume that for every $\epsilon>0$ there exists $C>0$ such that

$$|u(x,t)| \leq C e^{\epsilon |x|^2} , \ |\nabla_x u(x,t)| \leq C e^{\epsilon |x|^2}.$$ We recall the heat kernel: $K(x,t)= (4\pi t)^{-n/2} e^{-|x|^2/4t} \ (t>0)$.

$$I_r=\sum_{j=1}^n \int_a^b \int_{|x|=r}[u(x,t) \partial_j K(x,t) -K(x,t) \partial_ju(x,t)]\frac{x_j}{r} d\sigma(x) dt$$ where $d\sigma $ is a surface measure on $\{x\in \mathbb R^n: |x|=r\},$ $0<a<t<b.$

Question:How to show that $I_r\to 0$ as $r\to \infty$?

Motivation: This idea has been used [Folland's Intro. to PDE book, p.145] in the proof that heat equation has unique solution under some conditions.

Answer 1

Sketch. We use the following crude bound,$$\int_{t=a}^b \int_{|x|=r}\left|(u\partial_jK - K\partial_j u)\frac{x_j}{r} \right| \, d\sigma \ dt ≤ \int_{t=a}^b \int_{|x|=r}|u\partial_jK| +| K\partial_j u| \, d\sigma \ dt $$

We compute $\partial_j K = C'(t)x_j e^{-|x|^2/4t} $ for some constant $C'(t)$ depending on $t$ only, bounded on the range $t\in[a,b]$ if $a>0$. By choosing $\epsilon $ smaller and $C$ bigger if necessary, since $t\in[a,b]$, both $K$ and $\partial_jK$ satisfy "opposite" estimates to the given ones for $u$, $$ |K|+|\partial_j K| ≤ Ce^{-2\epsilon |x|^2}$$ Hence, $$ \int_{|x|=r}|u\partial_jK| +| K\partial_j u| \, d\sigma ≤ 2C^2 \int_{|x|=1} e^{-\epsilon r^2} r^{n-1}d\sigma_{S^{n-1}}(x) = \tilde C e^{-\epsilon r^2} r^{n-1}$$

The fact that $a>0$ means that the integral in $t$ only makes things messier to write down but qualitatively the same. Hence, $I_r\to 0$ as $r\to\infty$.

(I should note that I have been sloppy, the $\epsilon$ here is not quantified over, rather it is a special $\epsilon=\epsilon(a)$ that has already been fixed by the choice of $a$, that I did not compute.)

Answer 2

You can still integrate it just remove $r$ out the integral since it is treated as a constant. We have that: \begin{align*} I_r &= \sum_{j = 1}^n \int_a^b [u(x,t)\partial_jK(x,t) - K(x,t) \partial_ju(x,t)]\frac{x_j}{r}d\sigma(x)dt \\ I_r &= \sum_{j = 1}^n \frac{1}{r} \int_a^b [u(x,t)\partial_jK(x,t) - K(x,t) \partial_ju(x,t)]x_jd\sigma(x)dt \end{align*} We know that $\lim_{r \rightarrow \infty} \frac{1}{r} \rightarrow 0$. Thus by the zero product property we clearly see that $I_r \rightarrow 0$ as $r \rightarrow \infty$. Also, the sum of zeros is 0. May need to argue or prove and show some basic Riemann Integral theorems here if you're in advanced calculus/real analysis but yeah.