$\newcommand{\erf}{\operatorname{erf}}$

1D Heat equation

Introduction

Heat equation which is in its simplest form $$u_t = ku_{xx} \label{eq-1}$$ 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.

Self-similar solutions

We want to solve IVP for (\ref{eq-1}) with $t>0$, $-\infty<x <\infty$. Let us plug $$u_{\alpha,\beta,\gamma}(x,t)=\gamma u(\alpha x, \beta t). \label{eq-2}$$

Proposition 1. If $u$ satisfy (\ref{eq-1}) then $u_{\alpha,\beta,\gamma}$ also satisfies (\ref{eq-1}) 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 $$I(t):=\int_{-\infty}^\infty u(x,t)\, dx \label{eq-3}$$ is finite and does not depend on $t$.

The second part is due to the first one. Really (not rigorous) integrating (\ref{eq-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-2}) becomes $$u_{\alpha}(x,t)=\alpha u(\alpha x, \alpha^2 t). \label{eq-4}$$ This is transformation of similarity. Now we are looking for a self-similar solution of (\ref{eq-1}) i.e. solution such that $u_{\alpha}(x,t)=u(x,t)$ for all $\alpha>0, x, t>0$. So we want $$u(x,t)=\alpha u(\alpha x, \alpha^2 t)\qquad \forall \alpha>0, t>0, x. \label{eq-5}$$ We want to get rid off one of variables; so taking $\alpha = t^{-\frac{1}{2}}$ we get $$u(x,t)=t^{-\frac{1}{2}} u(t^{-\frac{1}{2}} x, 1) = t^{-\frac{1}{2}} \phi (t^{-\frac{1}{2}} x) \label{eq-6}$$ with $\phi (\xi ):= u(\xi, 1)$. (\ref{eq-6}) is equivalent tp (\ref{eq-5}).

Now we need to plug it into (\ref{eq-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*} and after multiplication by $t^{\frac{3}{2}}$ and plugging $t^{-\frac{1}{2}}x=\xi$ we arrive to $$-\frac{1}{2} \bigl( \phi (\xi) + \xi \phi '(\xi )\bigr)= k\phi '' (\xi). \label{eq-7}$$ Good news: it is ODE. Really good news: $\phi (\xi) + \xi \phi '(\xi )= \bigl( \xi\phi(\xi)\bigr)'$. Then integrating we get $$-\frac{1}{2} \xi \phi (\xi)= k\phi ' (\xi). \label{eq-8}$$ Remark. 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-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-6}) we arrive to $$u(x,t)= \frac{1}{2\sqrt \pi kt} e^{-\frac{x^2}{4kt}}. \label{eq-9}$$

Remark. 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{2k}\int_{-\infty}^{+\infty} e^{-\frac{1}{2}z^2}\,dz= 2c\sqrt{k\pi} \end{equation*} where we changed variable $x=z/\sqrt{2kt}$ and used the equality $$J= \int_{-\infty}^{+\infty} e^{-\frac{1}{2}x^2}\,dx=\sqrt{2\pi}. \label{eq-10}$$ To prove (\ref{eq-10}) just note that \begin{equation*} J^2= \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty}
e^{-\frac{1}{2}x^2}\times e^{-\frac{1}{2}y^2}\,dxdy= \int_0^{2\pi} d\theta \int_0^\infty e^{-\frac{1}{2}r^2}rdr =2\pi \end{equation*} where we used polar coordinates; since $J>0$ we get (\ref{eq-10}).

Remark. 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.

To work around this problem we consider $$U(x,t)=\int_{-\infty}^x u(x,t)\,dx. \label{eq-11}$$

We claim that

Proposition 2.

1. $U(x,t)$ also satisfies (\ref{eq-1}).
2. U(x,0)=\theta(x)=\left\{\begin{aligned} &0 \qquad &&x<0,\\ &1 && x>0. \end{aligned}\right.

Proof. Plugging $u=U_x$ into (\ref{eq-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 (a) is proven.

Note that $$U(x,t)=\frac{1}{\sqrt{2\pi }} \int_{-\infty}^{\frac{x}{\sqrt{2kt}}} e^{-\frac{1}{2}z^2}\,dz=: \frac{1}{2}+\frac{1}{2}\erf \bigl(\frac{x}{\sqrt{2kt}}\bigr) \label{eq-12}$$ with $$\erf(z):= \sqrt{\frac{2}{\pi}}\int_0^z e^{-z^2/2}\,dz \label{Erf}\tag{erf}$$ 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-10}) arrive to (b).

Remark. One can construct $U(x,t)$ as a self-similar solution albeit with $\gamma=1$.

Consider now a smooth function $g(x)$, $g(-\infty)=0$ and note that $$g(x)=\int _{-\infty}^\infty \theta (x-y) g'(y)\,dy. \label{eq-13}$$ 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 $u(x,t)=\int _{-\infty}^\infty U_x (x-y,t) g(y)\,dy$ and finally to $$u(x,t)=\frac{1}{2\sqrt{k\pi t}}\int _{-\infty}^\infty e^{-\frac{(x-y)^2}{4kt}} g(y)\,dy. \label{eq-14}$$

So we have proven:

Proposition 3. Fiormula (\ref{eq-14}) gives us a solution of \begin{align*} &u_t = ku_{xx}\qquad &&-\infty <x< \infty, t>0,\\ &u|_{t=0}=g(x). \end{align*}