The Heat Equation on Euclidean space

Links: Fourier Transform in Rn, The Heat Equation on the Real line

We consider the time-dependent heat equation in ${\mathbb{R}}^n$ $$\frac{\partial u}{\partial t} = \Delta u$$
with the boundary values $u(x,0)=f(x)\in \mathcal{S}({\mathbb{R}}^n)$. We make a similar procedure as we did for the real line. But now we apply The Fourier transform with respect to the spacial coordinates to the heat equation getting that: $$-4\pi^2 |\omega|^2\hat u(\omega, t) = \frac{\partial \hat u}{\partial t}(\omega,t)$$If we fix $\omega$ this is an ODE in $t$, solving it we get that $$\hat u (\omega, t) = A(\omega) e^{-4\pi^2|\omega|^2 t}$$
We apply the boundary condition, and we see that $A(\omega )=\hat{f}(\omega )$, getting the solution $$\hat u(\omega, t) = \hat f(\omega) e^{-4\pi^2|\omega|^2t}$$
With this in mind we can, calculate the $n$-dimensional heat kernel as, the inverse Fourier transform of $e^{-4{\pi }^2\parallel \omega {\parallel }^{2}t}$, getting that $$\mathcal H_t^{(n)}(x) = \frac1{({4\pi t})^{n/2}} e^{-|x|^2/4t}$$
We see that $$\hat{ \mathcal H}_t^{(n)} (\omega) = e^{-4\pi^2|\omega|^2 t}$$
Th: Given $f\in \mathcal{S}(\mathbb{R})$, let $$u(x, t)= (f*{\mathcal H_t^{(n)}})(x)$$where ${\mathcal{H}}_t$ is the heat kernel. Then:

• The function $u$ is ${\mathcal{C}}^2$ when $x\in {\mathbb{R}}^n$ and $t>0$, and $u$ solves the heat equation
• $u(x,t)\to f(x)$ uniformly in $x$ as $t\to 0$. Hence if we set $u(x,0)=f(x)$, then $u$ is continuous on $\overline{{\mathbb{R}}_{+}^{n+1}}=\{(x,t)\in {\mathbb{R}}^n\times {\mathbb{R}}^{\ge 0}\mid x\in \mathbb{R},t\ge 0\}$
  and

Th: Suppose that $u(x,t)$ satisfies the following conditions:

• $u$ is continuous on $\overline{{\mathbb{R}}_{+}^{n+1}}=\{(x,t)\in {\mathbb{R}}^n\times {\mathbb{R}}^{\ge 0}\mid x\in \mathbb{R},t\ge 0\}$
• $u$ satisfies the heat equation for $t>0$
• $u$ satisfies the boundary condition $u(x,0)=0$
• $u(\cdot ,t)\in \mathcal{S}(\mathbb{R})$ uniformly in $t$
  Then we conclude that $u\equiv 0$