$u(x,t)=\left(f*\mathcal{H}_{t}^{(n)}\right)(x)$ is solution to the n-dimensional heat equation

Question

Consider the time-dependent heat equation in $\mathbb{R^{n}}$:

$$\displaystyle\frac{\partial u}{\partial t}=\frac{\partial^{2}u}{\partial x_{1}^{2}}+\cdots+\frac{\partial^{2}u}{\partial x_{n}^{2}},\quad\textrm{where}\;t>0 $$ with boundary values $u(x,0)=f(x)\in \mathcal{S}(\mathbb{R}^{n})$ (i.e. $f$ belongs to Schwartz space of $\mathbb{R}^{n})$.

Define the n-dimensional heat kernel by

$$\displaystyle\mathcal{H}_{t}^{(n)}(x):=\frac{1}{(4\pi t)^{n/2}}e^{|x|^2/4t}=\int_{\mathbb{R}^{n}}e^{-4\pi^{2}t|\xi|^{2}}e^{2\pi i x\cdot \xi}d\xi, $$ where $|\cdot|$ is the euclidian norm in $\mathbb{R}^{n}$.

I need to prove three statements about $u(x,t)=\left(f*\mathcal{H}_{t}^{(n)}\right)(x)$.

(1) $u$ is indefinitely differentiable when $x\in\mathbb{R}^{n}$ and $t>0.$

(2) $u$ solves the heat equation above

(3) $u$ is continuous up to the boundary $t=0$ with $u(x,0)=f(x).$

---

By definition,

$$u(x,t)=\underbrace{\int_{-\infty}^{\infty}\cdots\int_{-\infty}^{\infty}}_{n\;\textrm{times}}\frac{f(y)}{(4\pi t)^{n/2}}e^{-\frac{\left[(x_{1}-y_{1})^{2}+\cdots+(x_{n}-y_{n})^{2}\right]^{1/2}}{4t}}dx_{1}\dots dx_{n}$$

That's all I got. I really want some tips (I don't want the solution ifself)

Answer 1

1. This is an easy consequence of the fact that $f'*g = f*g'$ for any $f$ and $g$, as well as the fact that the heat kernel is $C^{\infty}$ for $t>0$.

2. This can be shown with the above fact also, namely we have that $$(\partial_t-\Delta)u = f*((\partial_t-\Delta\mathcal){H}) = f*0=0$$ since the heat kernel is a solution to the homogenous heat equation.

3. This is pretty tough to prove and the difficulty varies depending on space $f$ lies in. I will try to sketch a proof. You are trying to show that $\lim_{t\to0}u(x,t) = f(x)$. The way this typically goes is you divide up the region into an $\varepsilon$-ball centered at $x$ and its compliment then show that as $t\to 0$, the integral of the outer part goes to $0$ and the interior approaches $f(x)$. For the outer part, use the fact that the integral converges (i.e. the heat kernel decreases rapidly) and the fact that $f$ is rapidly decreasing (Schwartz). For the inner part, you can use some the continuity of $f$ to relate the value of $f$ in this region to $f(x)$. You will use the fact that the heat kernel integrates to 1 at this point somewhere.