1
$\begingroup$

I want to show that the heat kernel $\displaystyle\gamma_t(x) := \frac{1}{(4\pi t)^{d/2}} \exp\left( - \frac{|x|^2}{4t} \right), \quad x \in \mathbb{R}^d, \ t > 0$

is a Dirac sequence. $\gamma_t \ge 0$ is obvious and $\int_{\mathbb{R}^d} \gamma_t(x) \ \mathrm{d}x = 1$ I have already proved using Transformation. Now I have to show that $\int_{\mathbb{R}^d - B_{\delta}(0)} \gamma_t(x) \ \mathrm{d}x \to 0$ for $t \to \infty$. I know that $\int_{\mathbb{R}^d - B_{\delta}(0)} \gamma_t(x) \ \mathrm{d}x = \int_{\mathbb{R}^d} \gamma_t(x) \ \mathrm{d}x - \int_{B_{\delta}(0)} \gamma_t(x) \ \mathrm{d}x$. Since $\int_{\mathbb{R}^d} \gamma_t(x) \ \mathrm{d}x = 1$ it is left to prove that $\int_{B_{\delta}(0)} \gamma_t(x) \ \mathrm{d}x \to 1$ for $t \to \infty$. I have tried this: $\int_{B_{\delta}(0)} \frac{1}{(4\pi t)^{d/2}} \exp\left( - \frac{|x|^2}{4t} \right) \ \mathrm{d}x = \frac{1}{\pi^{d/2}} \prod\limits_{i=1}^d \int_{-\delta}^{\delta} \exp(-|z|^2) \ \mathrm{d}z$ with Transformation. But why does this converge to 1 for $t \to \infty$?

$\endgroup$
  • $\begingroup$ Start with $k(t,x) = \int_{\Bbb{R}^d} e^{-4\pi t y^2} e^{2i \pi (x, y)} dy$, it is obviously the fundamental solution of the heat equation because as $t \to 0^+$, $k(t,.)$'s Fourier transform $\to 1$ so $k(t,x) \to \delta(x)$. Ie. the solution with any initial condition $f$ is $F(t,x) = \int_{\Bbb{R}^d} k(t,x-u) f(u)du$. Then using the Fourier transform of the Gaussian $k(t,x)$ has the expression you gave. The Dirac sequence is obtained by periodizing $k(t,x)$ to obtain $\sum_{n\in \Bbb{Z}^d} k(t,x+n)=\sum_me^{-4\pi t |m|^2} e^{2i \pi (x, m)}$ the fundamental solution of the periodic problem $\endgroup$ – reuns Apr 27 at 17:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.