The Heat Equation

The Real Line

Consider an infinite rod with an initial temperature distribution $f(x)$ at time $t=0$

How do we describe the temperature $u(x,t)$ at time $t>0$?

Considerations here motivate the heat equation,
$\frac{\partial u}{\partial t}=\frac{{\partial }^{2}u}{\partial x^2}$
$u(x,0)=f(x)$

We define the heat kernel as ${\mathcal{H}}_t(x)=K_{\delta }(x)$ with $\delta =4\pi t$, so ${\mathcal{H}}_t(x)=\frac{1}{(4\pi t{)}^{1/2}}{e}^{-x^2/4t}$ and ${\hat{\mathcal{H}}}_t(\xi )=e^{-4{\pi }^{2}t{\xi }^2}$

We take the Fourier transform of the heat equation in the $x$ variable to obtain
$\frac{\partial \hat{u}}{\partial t}(\xi ,t)=-4{\pi }^2{\xi }^2\hat{u}(\xi ,t)$

Fixing $\xi$, we get $\hat{u}(\xi ,t)=\hat{f}(\xi )e^{-4{\pi }^2{\xi }^{2}t}$

Theorem: Given $f\in \mathcal{S}(\mathbb{R})$, let $u(x,t)=(f\ast {\mathcal{H}}_t)(x)$,

1. The function $u$ is $C^2$ and $u$ solves the heat equation
2. $u(x,t)\to f(x)$ uniformly in $x$ as $t\to 0$, so setting $u(x,0)=f(x)$ leaves $u$ continuous on the closure of the upper half-plane
3. ${\int }_{-\infty }^{\infty }|u(x,t)-f(x){|}^{2}dx\to 0$ as $t\to 0$

The Fourier transform yields $\hat{u}=\hat{f}\cdot {\hat{\mathcal{H}}}_t$, so $\hat{u}(\xi ,t)=\hat{f}(\xi )e^{-4{\pi }^2{\xi }^{2}t}$
Therefore,
$u(x,t)={\int }_{-\infty }^{\infty }\hat{f}(\xi )e^{-4{\pi }^{2}t{\xi }^2}{e}^{2\pi i\xi x}dx$

We can differentiate under the integral sign indefinitely, verifying our first property

Property two follows from here and property three is also fairly easy to show

We also note that $u(\cdot ,t)$ belongs to $\mathcal{S}(\mathbb{R})$ uniformly in $t$, in the sense that for $T>0$, ${\sup }_{\begin{array}{c}x\in \mathbb{R}\\ 0<t<T\end{array}}|x{|}^k\left|\frac{{\partial }^l}{\partial x^l}u(x,t)\right|<\infty$ for each $k,l>0$

We can use this to help prove uniqueness

Theorem: Suppose $u(x,t)$ is continuous on the closure of the upper half-plane, $u$ satisfies the heat equation, $u(x,0)=0$, and $u(\cdot ,t)\in \mathcal{S}(\mathbb{R})$ uniformly in $t$, then we conclude $u=0$

We define the energy at time $t$ with $E(t)={\int }_{\mathbb{R}}^{}|u(x,t){|}^{2}dx$
$\frac{dE}{dt}={\int }_{\mathbb{R}}^{}[{\partial }_{t}u(x,t)\bar{u}(x,t)+u(x,t){\partial }_t\bar{u}(x,t)]dx$
$={\int }_{\mathbb{R}}^{}[{\partial }_x^{2}u(x,t)\bar{u}(x,t)+u(x,t){\partial }_x^2\bar{u}(x,t)]dx$
$=-{\int }_{\mathbb{R}}^{}[{\partial }_{x}u(x,t){\partial }_x\bar{u}(x,t)+{\partial }_{x}u(x,t){\partial }_x\bar{u}(x,t)]dx$
$=-2{\int }_{\mathbb{R}}^{}|{\partial }_{x}u(x,t){|}^{2}dx\le 0$

Therefore, $E(t)=0$ for all $t$ and $u=0$ follows

${\partial }_x^{2}u(x,t)\bar{u}(x,t)={\partial }_{x}u(x,t){\partial }_x\bar{u}(x,t)$ requires the function to vanish at $\infty$

We can also establish less restrictive uniqueness theorems

Steady-State on the Upper Half-Plane

We are now interested in understanding the steady-state distribution in the upper half-plane

We expand the heat equation to a two dimensional plane with the Laplacian,
$\Delta u=\frac{{\partial }^{2}u}{\partial x^2}+\frac{{\partial }^{2}u}{\partial y^2}=0$ in the upper half-plane ${\mathbb{R}}_{+}^2=\{(x,y):x\in \mathbb{R},y>0\}$, subject to $u(x,0)=f(x)$

To solve this, we use the Poisson kernel, ${\mathcal{P}}_y(x)=\frac{1}{\pi }\frac{y}{x^2+y^2}$

We take the Fourier transform of our heat equation to get $-4{\pi }^2{\xi }^2\hat{u}(\xi ,y)+\frac{{\partial }^2\hat{u}}{\partial y^2}(\xi ,y)=0$

The general solution of this ODE with fixed $\xi$ is $\hat{u}(\xi ,y)=A(\xi )e^{-2\pi |\xi |y}+B(\xi )e^{2\pi |\xi |y}$

We disregard the second term for its exponential increase and set $y=0$ to find $\hat{u}(\xi ,y)=\hat{f}(\xi )e^{-2\pi |\xi |y}$

We prove relevant properties of the Poisson kernel here, showing that the Poisson kernel is a good kernel, and its Fourier transform is $e^{-2\pi |\xi |y}$

Theorem: Given $f\in \mathcal{S}(\mathbb{R})$, let $u(x,y)=(f\ast {\mathcal{P}}_y)(x)$,

1. $u$ is $C^2$ in ${\mathbb{R}}_{+}^2$ and $\Delta u=0$
2. $u(x,y)\to f(x)$ uniformly as $y\to 0$
3. ${\int }_{-\infty }^{\infty }|u(x,y)-f(x){|}^{2}dx\to 0$ as $y\to 0$
4. If $u(x,0)=f(x)$ then $u$ is continous on the closure of the upper half-plane and vanishes at infinity, i.e. $u(x,y)\to 0$ as $|x|+y\to \infty$

The first three are similar to our real line case, the final one uses an estimate $|(f\ast {\mathcal{P}}_y)(x)|\le C\left(\frac{1}{1+x^2}+\frac{y}{x^2+y^2}\right)$

Theorem: Suppose $u$ is continuous on $\overline{{\mathbb{R}}_{+}^2}$, $\Delta u=0$ for $(x,y)\in {\mathbb{R}}_{+}^2$, $u(x,0)=0$, and $u(x,y)$ vanishes at infinity, then $u=0$

This last condition is important to guarantee $u=0$

Lemma (mean-value property): Suppose $\Omega$ is an open set in ${\mathbb{R}}^2$ and let $u$ be a function of class $C^2$ with $\Delta u=0$ in $\Omega$. If the closure of the disc centered in $(x,y)$ of radius $R$ is contained in $\Omega$, then $u(x,y)=\frac{1}{2\pi }{\int }_0^{2\pi }u(x+r\cos \theta ,y+r\sin \theta )d\theta$ for all $0\le r\le R$

I.e., the value of a harmonic function at a point equals its average value around any circle centered at that point

$U(r,\theta )=u(x+r\cos \theta ,y+r\sin \theta )$
$0=\frac{{\partial }^{2}U}{\partial {\theta }^2}+r\frac{\partial }{\partial r}\left(r\frac{\partial U}{\partial r}\right)$ (expression of the Laplacian in polar coordinates)

$F(r)=\frac{1}{2\pi }{\int }_0^{2\pi }U(r,\theta )d\theta dx$
$r\frac{\partial }{\partial r}\left(r\frac{\partial F}{\partial r}\right)=\frac{1}{2\pi }{\int }_0^{2\pi }-\frac{{\partial }^{2}U}{\partial {\theta }^2}(r,\theta )d\theta$

Since $\frac{\partial U}{\partial \theta }$ is periodic, this reduces to $r\frac{\partial }{\partial r}\left(r\frac{\partial F}{\partial r}\right)=0$, so $r\partial F/\partial r$ is a constant

And since this expression is $0$ at $r=0$, $\partial F/\partial r=0$, so $F(r)=u(x,y)$, which is our desired property

Then to prove the uniqueness theorem, we can argue by contradiction

Considering the real and imaginary parts of $u$ separately, assume there is a strictly positive point of $u(x_0,y_0)>0$

Since $u$ vanishes at infinity, we can find a large semi-disc of radius $R$, $D_R^{+}=\{(x,y):x^2+y^2\le R,y\ge 0\}$ outside of which $u(x,y)\le \frac{1}{2}u(x_0,y_0)$

Since $u$ is continuous in $D_R^{+}$, it has a maximum $M$, so there is a point in the region $(x_1,y_1)\in D_R^{+}$ with $u(x_1,y_1)=M$, and therefore outside the semi-plane we have $u(x,y)\le M/2$

So $u(x_1,y_1)=\frac{1}{2\pi }{\int }_0^{2\pi }u(x_1+\rho \cos \theta ,y_1+\rho \sin \theta )d\theta$ for $0<\rho <y_1$

We see by continuity that the region in this integral is all equal to $M$, and taking $\rho \to y_1$ we see this implies $u(x_1,0)=M$, contradicting the initial condition $u(x,0)=0$