3.1. 1D Heat equation

$\renewcommand{\Re}{\operatorname{Re}}$ $\renewcommand{\Im}{\operatorname{Im}}$ $\newcommand{\erf}{\operatorname{erf}}$ $\newcommand{\dag}{\dagger}$ $\newcommand{\const}{\mathrm{const}}$ $\newcommand{\arcsinh}{\operatorname{arcsinh}}$

# Chapter 3. Heat equation in 1D

In this Chapter we consider 1-dimensional heat equation (also known as diffusion equation). Instead of more standard Fourier transform method (which we will postpone a bit) we will use the method of self-similar solutions.

## 3.1. 1D Heat equation

### Introduction

Heat equation which is in its simplest form \begin{equation} u_t = ku_{xx} \label{eq-3.1.1} \end{equation} is another classical equation of mathematical physics and it is very different from wave equation. This equation describes also a diffusion, so we sometimes will refer to it as diffusion equation. See Subsection 1.4.2.

It also describes random walks (see Project "Random walks").

### Self-similar solutions

We want to solve IVP for equation (\ref{eq-3.1.1}) with $t>0$, $-\infty< x < \infty$. Let us plug \begin{equation} u_{\alpha,\beta,\gamma}(x,t):=\gamma u(\alpha x, \beta t). \label{eq-3.1.2} \end{equation}

Proposition 1. If $u$ satisfy equation (\ref{eq-3.1.1}) then $u_{\alpha,\beta,\gamma}$ also satisfies this equation provided $\beta =\alpha^2$.

Proof is just by calculation. Note that $\beta=\alpha^2$ because one derivative with respect to $t$ is "worth" of two derivatives with respect to $x$.

We impose another assumption:

Condition 1. Total heat energy \begin{equation} I(t):=\int_{-\infty}^\infty u(x,t)\, dx \label{eq-3.1.3} \end{equation} is finite and does not depend on $t$.

The second part is due to the first one. Really (not rigorous) integrating (\ref{eq-3.1.1}) by $x$ from $-\infty$ to $+\infty$ and assuming that $u_x (\pm \infty)=0$ we see that $\partial_t I(t)=0$.

Note that $\int_{-\infty}^\infty u_{\alpha,\beta,\gamma} \,dx = \gamma |\alpha|^{-1} \int_{-\infty}^\infty u\,dx$ and to have them equal we should take $\gamma=|\alpha|$ (actually we restrict ourselves by $\alpha>0$). So (\ref{eq-3.1.2}) becomes \begin{equation} u_{\alpha}(x,t)=\alpha u(\alpha x, \alpha^2 t). \label{eq-3.1.4} \end{equation} This is transformation of similarity. Now we are looking for a self-similar solution of (\ref{eq-3.1.1}) i.e. solution such that $u_{\alpha}(x,t)=u(x,t)$ for all $\alpha>0, x, t>0$. So we want \begin{equation} u(x,t)=\alpha u(\alpha x, \alpha^2 t)\qquad \forall \alpha>0, t>0, x. \label{eq-3.1.5} \end{equation} We want to get rid off one of variables; so taking $\alpha = t^{-\frac{1}{2}}$ we get \begin{equation} u(x,t)=t^{-\frac{1}{2}} u(t^{-\frac{1}{2}} x, 1) = t^{-\frac{1}{2}} \phi (t^{-\frac{1}{2}} x) \label{eq-3.1.6} \end{equation} with $\phi (\xi ):= u(\xi, 1)$. Equality (\ref{eq-3.1.6}) is equivalent to (\ref{eq-3.1.5}).

Now we need to plug $u(x,t)=t^{-\frac{1}{2}} \phi (t^{-\frac{1}{2}} x)$ into equation (\ref{eq-3.1.1}). Note that \begin{multline*} u_t = -\frac{1}{2} t^{-\frac{3}{2}}\phi (t^{-\frac{1}{2}}x) + t^{-\frac{1}{2}}\phi '(t^{-\frac{1}{2}}x)\times \bigl(-\frac{1}{2}t^{-\frac{3}{2}}x\bigr)= \\ -\frac{1}{2} t^{-\frac{3}{2}} \bigl( \phi (t^{-\frac{1}{2}}x) + t^{-\frac{1}{2}}x \phi' (t^{-\frac{1}{2}}x)\bigr) \end{multline*} and \begin{equation*} u_x = t^{-1}\phi ' (t^{-\frac{1}{2}}x), \qquad u_{xx} = t^{-\frac{3}{2}}\phi '' (t^{-\frac{1}{2}}x). \end{equation*}

Multiplying by $t^{\frac{3}{2}}$ and plugging $t^{-\frac{1}{2}}x=\xi$ we arrive to \begin{equation} -\frac{1}{2} \bigl( \phi (\xi) + \xi \phi '(\xi )\bigr)= k\phi '' (\xi). \label{eq-3.1.7} \end{equation} Good news: it is ODE. Really good news: $\phi (\xi) + \xi \phi '(\xi )= \bigl( \xi\phi(\xi)\bigr)'$. Then integrating we get \begin{equation} -\frac{1}{2} \xi \phi (\xi)= k\phi ' (\xi). \label{eq-3.1.8} \end{equation}

Remark 1. Sure there should be $+C$ but we are looking for a solution fast decaying with its derivatives at $\infty$ and it implies that $C=0$.

Separating in (\ref{eq-3.1.8}) variables and integrating we get \begin{equation*} \frac{d\phi}{\phi}= -\frac{1}{2k}\xi d\xi \implies \log \phi = -\frac{1}{4k}\xi^2+\log c\implies \phi(\xi)= ce^{-\frac{1}{4k}\xi^2} \end{equation*} and plugging into (\ref{eq-3.1.6}) we arrive to \begin{equation} u(x,t)= \frac{1}{2\sqrt {\pi kt}} e^{-\frac{x^2}{4kt}}. \label{eq-3.1.9} \end{equation}

Remark 2. We took $c=\frac{1}{2\sqrt {\pi k}}$ to satisfy $I(t)=1$. Really, \begin{equation*} I(t)=c\int_{-\infty}^{+\infty} t^{-\frac{1}{2}} e^{-\frac{x^2}{4kt}}\,dx = c\sqrt{4k}\int_{-\infty}^{+\infty} e^{-z^2}\,dz= 2c\sqrt{k\pi} \end{equation*} where we changed variable $x=z/\sqrt{2kt}$ and used the equality \begin{equation} J= \int_{-\infty}^{+\infty} e^{-x^2}\,dx=\sqrt{\pi}. \label{eq-3.1.10} \end{equation} To prove (\ref{eq-3.1.10}) just note that \begin{equation*} J^2= \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty}\ e^{-x^2}\times e^{-y^2}\,dxdy= \int_0^{2\pi} d\theta \int_0^\infty e^{-r^2}rdr =\pi \end{equation*} where we used polar coordinates; since $J>0$ we get (\ref{eq-3.1.10}).

Remark 3. Solution which we got is a very important one. However we have a problem understanding what is $u|_{t=0^+}$ as $u(x,t)\to 0$ as $t\to 0^+$ and $x\ne 0$ but $u(x,t)\to \infty$ as $t\to 0^+$ and $x= 0$ and $\int_{-\infty}^\infty u(x,t)\,dx=1$.

In fact $u|_{t=0^+}=\delta(x)$ which is Dirac $\delta$-function which actually is not an ordinary function but a distribution (see Section 11.1). Distributions play important role in the modern Analysis and applications, in particular, to physics.

To work around this problem we consider \begin{equation} U(x,t)=\int_{-\infty}^x u(x,t)\,dx. \label{eq-3.1.11} \end{equation}

We claim that

Proposition 2.

1. $U(x,t)$ also satisfies equation (\ref{eq-3.1.1}).
2. U(x,0^+)=\theta(x)=\left\{\begin{aligned} &0 &&x<0,\\ &1 && x>0 \end{aligned}\right.\quad is a Heaviside step function.

Proof. Plugging $u=U_x$ into (\ref{eq-3.1.1}) we see that $(U_t-kU_{xx})_x=0$ and then $(U_t-kU_{xx})=\Phi(t)$. However one can see easily that as $x\to -\infty\$ $U$ is fast decaying with all its derivatives and therefore $\Phi(t)=0$ and  is proven.

Note that \begin{equation} U(x,t)=\frac{1}{\sqrt{4\pi }} \int_{-\infty}^{\frac{x}{\sqrt{4kt}}} e^{- z^2}\,dz=: \frac{1}{2}+\frac{1}{2}\erf \bigl(\frac{x}{\sqrt{4kt}}\bigr) \label{eq-3.1.12} \end{equation} with \begin{equation} \erf(z):= \frac{2}{\sqrt{\pi}}\int_0^z e^{-z^2}\,dz \label{Erf}\tag{erf} \end{equation} and that an upper limit in integral tends to $\mp \infty$ as $t\to 0^+$ and $x\lessgtr 0$. Then since an integrand is very fast decaying at $\mp \infty$ we using (\ref{eq-3.1.10}) arrive to .

Remark 4.

1. One can construct $U(x,t)$ as a self-similar solution albeit with $\gamma=1$.
2. We can avoid analysis of $U(x,t)$ completely just by noting that $u(x,t)$ is a $\delta$-sequence as $t\to0^+$: $u(x,t)\to 0$ for all $x\ne 0$ but $\int_{-\infty}^\infty u(x,t)\,dx=1$.

Consider now a smooth function $g(x)$, $g(-\infty)=0$ and note that \begin{equation} g(x)=\int _{-\infty}^\infty \theta (x-y) g'(y)\,dy. \label{eq-3.1.13} \end{equation} Really, the r.h.e. is $\int_{-\infty}^x g'(y)\,dy=g(x)-g(-\infty)$.

Also note that $U(x-y,t)$ solves the IVP with initial condition $U(x-y,0^+)=\theta (x-y)$. Therefore $u(x,t)=\int _{-\infty}^\infty U (x-y,t) g'(y)\,dy$ solves the IVP with initial condition $u(x,0^+)=g(y)$.

Integrating by parts with respect to $y$ we arrive to \begin{gather} u(x,t)= \int _{-\infty}^\infty G(x-y,t) g(y)\,dy \label{eq-3.1.14} \end{gather} with \begin{gather} G(x-y,t):= \frac{1}{2\sqrt{k\pi t}} e^{-\frac{(x-y)^2}{4kt}} . \label{eq-3.1.15} \end{gather}

So we have proven:

Theorem 3. Formulae (\ref{eq-3.1.14})--(\ref{eq-3.1.15}) give us a solution of \begin{align} &u_t = ku_{xx}\qquad &&-\infty <x< \infty, t>0, \label{eq-3.1.16}\\ &u|_{t=0}=g(x). \label{eq-3.1.17} \end{align}

Remark 5. We will recreate the same formulae in Section 5.3 using Fourier transform.

Example 1. Find the solution $u(x,t)$ to \begin{align*} &u_t=4u_{xx} && -\infty<x<\infty, \ t>0,\\[2pt] &u|_{t=0}=\left\{\begin{aligned} &1-|x| &&|x|<1,\\ &0 &&|x|\ge 1, \end{aligned}\right. \\ &\max |u|<\infty. \end{align*}

Solution. \begin{align*} u(x,t)=&\frac{1}{\sqrt{16\pi t}} \int_{-\infty}^\infty e^{-\frac{1}{16t} (x-y)^2}g(y)\,dy\\ =& \frac{1}{\sqrt{16\pi t}} \Bigl(\int_{-1}^0 e^{-\frac{1}{16t} (x-y)^2}(1+y)\,dy + \int_0^1 e^{-\frac{1}{16t} (x-y)^2}(1-y)\,dy \Bigr). \end{align*}

Plugging $y= x+4z\sqrt{t}$ and changing limits \begin{multline*} u(x,t)=\frac{1}{\sqrt{\pi }} \Bigl(\int_{-(x+1)/4\sqrt{t}}^{-x/4\sqrt{t}} e^{-z^2} \bigl(1+ x+4z\sqrt{t}\bigr)\,dz \\ + \int_{-x/4\sqrt{t}}^{-(x-1)/4\sqrt{t}} e^{-z^2}\bigl(1- x-4z\sqrt{t}\bigr)\,dz \Bigr) \end{multline*} we evaluate integrals of the terms without factor $z$ as $\erf$ functions and terms with factor $z$ we simply integrate resulting in

\begin{multline*} u(x,t)=\frac{1}{2} (1+x) \Bigl(\erf(\frac{x+1}{4\sqrt{t}})-\erf(\frac{x}{4\sqrt{t}})\Bigr)\\ + \frac{1}{2} (1-x) \Bigl(\erf(\frac{x}{4\sqrt{t}})-\erf(\frac{x-1}{4\sqrt{t}})\Bigr) \ +\frac{2\sqrt{t}}{\sqrt{\pi}}\Bigl( e^{-z^2}\Bigl|_{z=x/4\sqrt{t}} ^{z=(x+1)/4\sqrt{t}}+e^{-z^2}\Bigl|_{z=x/4\sqrt{t}} ^{z=(x-1)/4\sqrt{t}}\Bigr). \end{multline*}

### Inhomogeneous Right-hand Expression

Consider instead problem with the right-hand expression \begin{align} &u_t = ku_{xx}+f(x,t) \qquad &&-\infty < x\ < \infty,\ t>0, \label{eq-3.1.18}\\ &u|_{t=0}=g(x). \tag{17} \end{align}

By Duhamel principle the contribution of this right-hand expression is \begin{equation} \int_0^t \int G(x-y,t-\tau) f(y,\tau)\,dyd\tau \label{eq-3.1.19} \end{equation} Indeed, Duhamel integral (2.5.8) remains except now auxiliary function $U(x,t;\tau)$ satisfies heat equation (\ref{eq-3.1.6}) with initial condition $U_t|_{t=\tau}=f(x,\tau)$. One can prove it exactly as in Subsection 2.5.2. Therefore we arrive to

Theorem 4. Solution of problem (\ref{eq-3.1.18})-(\ref{eq-3.1.17}) is given by \begin{equation} u(x,t)= \int_0^t \int_{-\infty}^\infty G(x-y,t-\tau) f(y,\tau)\,dyd\tau+ \int _{-\infty}^\infty G(x-y,t) g(y)\,dy. \label{eq-3.1.20} \end{equation}

Theorem 5. Let $u$ satisfy (\ref{eq-3.1.18}) in domain $\Omega\subset \mathbb{R}_t\times \mathbb{R}_x$ with $f\in C^\infty(\Omega)$. Then $u\in C^\infty(\Omega)$.

We will prove it later.

Remark 6.

1. So far we have not discussed the uniqueness of the solution. In fact, as formulated, the solution is not unique. But do not worry: "extra solutions'' are so irregular (fast growing and oscillating) at infinity that they have no "physical sense". We discuss it later.
2. IVP in the direction of negative time is ill-posed. Indeed, if $f\in C^\infty$ it follows from Theorem 5 that solution cannot exist for $t\in (-\epsilon, 0]$ unless $g \in C^\infty$ (which is only necessary, but not sufficient).
3. Domain of dependence for $(x,t)$ is ${(x',t')\colon -\infty<x'<\infty,\ 0<t'<t}$ and the propagation speed is infinite! No surprise: heat equation is not relativistic!

### References  -

$\erf(x)$ and its derivative $\erf'(x)$ original and scaled.