The Heat Equation on the Real line

Subjects: Fourier Analysis, Partial Differential Equations
Links: Fourier Transform in R, The Heat Equation, Convolution for Rapidly Decreasing Functions

Consider an infinite rod, which we model by the real line, and suppose that we are ficen an initial temperature distribution $f(x)$ on the rod at time $t=0$, We wish now to determine the temperature $u(x,t)$ at the point $x$ and time $t>0$. When $u$ is properly normalised, it solves the following pde: $$ \frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x^2}$$called the heat equation. The initial condition we impose is $u(x,0)=f(x)$.

Just as in the case of circle the solution is given in terms of a convolution. Indeed, define the heat kernel of the line by $$\mathcal H_t(x) = K_\delta(x), \qquad \text{with } \delta = 4\pi t$$so that $$\mathcal H_t(x) = \frac1{(4\pi t)^{1/2}}e^{-x^2/4t}\qquad \text{and} \quad \hat{\mathcal H_t}(\omega) = e^{-4\pi^2t\omega^2}$$
We see that as $t\to 0$, then $\delta \to 0$, then we see that ${\mathcal{H}}_t(x)$ is a good kernel as $t\to 0$.

If we take the Fourier transform of the heat equation in the $x$ variable, leads to $$\frac{\partial \hat u}{\partial t}(\omega, t) = -4\pi^2\omega^2\hat u(\omega, t)$$If we fix $\omega$, this is an ode in the variable $t$, so there exists $A(\omega )$ so that $$\hat u(\omega, t) = A(\omega) e^{-4\pi^2 \omega^2 t}$$we mat also take the Fourier transform of the initial condition and obtain $\hat{u}(\omega ,0)=\hat{f}(\omega )$, hence $A(\omega )=\hat{f}(\omega )$, then the solution is $$\hat u(\omega, t) = \hat f(\omega) e^{-4\pi^2\omega^2 t} = \hat f(\omega) \hat{\mathcal H_t}(\omega)$$
Th: Given $f\in \mathcal{S}(\mathbb{R})$, let $$u(x, t)= (f*{\mathcal H_t})(x)$$where ${\mathcal{H}}_t$ is the heat kernel. Then:

• The function $u$ is ${\mathcal{C}}^2$ when $x\in \mathbb{R}$ 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 the closure of the upper half-plane $\overline{{\mathbb{R}}_{+}^2}=\{(x,t)\mid x\in \mathbb{R},t\ge 0\}$
  and

Cor: $u(\cdot ,t)$ belongs to $\mathcal{S}(\mathbb{R})$ uniformly in $t$, in the sense that for any $T>0$, $$\sup_{
\substack{x\in \Bbb R\ 0< t< T}}|x|^k \left| \frac{\partial^n}{\partial x^n} u(x, t)\right| < \infty \qquad \forall k, n \ge 0$$
Th: Suppose that $u(x,t)$ satisfies the following conditions:

• $u$ is continuous on the closure of the upper half-plane
• $u$ 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 $u\equiv 0$

The idea of the proof is to define the energy at time $t$ of the solution $u(x,t)$ by $$E(t) = \int_\Bbb R |u(x,t)|^2, dx$$$E(t)\ge 0$, we proved that $E^{\prime }(t)\le 0$, and since $E(0)=0$, thus it must be that $E\equiv 0$, thus also $u$

The idea that seems interesting is that the initial condition must also have "finite" energy to have a Fourier transform, and seems like a prerequisite to be solvable.

Th (Tychonoff): Supose $u(x,t)$ satisfies the conditions:

• $u(x,t)$ solves the heat equation for all $x\in \mathbb{R}$ and $t>0$
• $u(x,t)$ is continuous for all $x\in \mathbb{R}$ and $0\le t\le c$
• $u(x,0)=0$
• $|u(x,t)|\le M{e}^{a{x}^2}$ for some $M,a$ and all $x\in \mathbb{R}$, $0\le t\le c$
  Then $u$ is identically equal to $0$