$\newcommand{\erf}{\operatorname{erf}}$ $\newcommand{\const}{\mathrm{const}}$

## 1D Heat equation: misc

#### Multidimensional heat equation

In the previous lecture we considered heat equation $$u_t=ku_{xx} \label{eq-1}$$ with $x\in \mathbb{R}$ and $t>0$ and derived formula $$u(x,t)=\int _{-\infty}^\infty G(x,x',t) g(x')\,dx'. \label{eq-2}$$ with $$G(x,x',t)=G_0(x-y,t):=\frac{1}{2\sqrt{k\pi t}}e^{-\frac{(x-x')^2}{4kt}} \label{eq-3}$$ for solution of IVP $u|_{t=0}=g(x)$.

Similar formula was derived for heat equation on half-line or segment with the Dirichlet on Neumann boundary conditions.

Recall that $G(x,y,t)$ quickly decays as $|x-y|\to \infty$ and it tends to $0$ as $t\to +0$ for $x\ne y$, but $\int G(x,y,t)\,dy\to 1$ as $t\to +0$.

Now we claim that for 2D and 3D heat equations \begin{align} &u_t=k\bigl(u_{xx}+u_{yy}\bigr), \label{eq-4}\\ &u_t=k\bigl(u_{xx}+u_{yy}+u_{zz}\bigr), \label{eq-5} \end{align} similar formulae hold: \begin{align} &u=\iint G_2 (x,y;x',y';t) g(x',y')\,dx'dy', \label{eq-6}\\ &u=\iiint G_3 (x,y,z;x',y',z';t) g(x',y',z')\,dx'dy'dz' \label{eq-7} \end{align} with \begin{align} &G_2 (x,y;x',y';t) =&& G_1(x,x',t)G_1(y,y',t), \label{eq-8}\\ &G_3 (x,y,z;x',y',z';t) =&& G_1(x,x',t)G_1(y,y',t)G_1(z,z',t); \label{eq-9} \end{align} in particular for the whole $\mathbb{R}^n$ \begin{align} &G_n (\mathbf{x},\mathbf{x}';t) =&& (2\sqrt{\pi k t})^{-n/2}e^{-\frac{|\mathbf{x}-\mathbf{x}'|}{4kt}}. \label{eq-10} \end{align}

To justify our claim we note that

1. $G_n$ satisfies $n$-dimensional heat equation. Really, consider f.e. $G_2$: \begin{align*} G_{2\,t} (x,y;x',y';t) =& G_{1\,t}(x,x',t)G_1(y,y',t)+G_1(x,x',t)G_{1.t}(y,y',t)=\\ & kG_{1\,xx}(x,x',t)G_1(y,y',t)+kG_1(x,x',t)G_{1\,yy}(y,y',t)=\\ &k \Delta \bigl( G_{1\,t}(x,x',t)G_1(y,y',t) \bigr)= k\Delta G_2(x,y;x',y';t) \end{align*}

2. $G_n (\mathbf{x},\mathbf{x}';t)$ quickly decays as $|\mathbf{x}-\mathbf{x}'|\to \infty$ and it tends to $0$ as $t\to +0$ for $\mathbf{x}\ne \mathbf{x}'$, but

3. $\int G(\mathbf{x},\mathbf{x}',t)\,dy\to 1$ as $t\to +0$;

4. $G(\mathbf{x},\mathbf{x}',t)=G(\mathbf{x}',\mathbf{x},t)$.

The last three properties are due to the similar properties of $G_1$.

Properties (a)-(d) imply integral representation (\ref{eq-6})(or its $n$-dimensional variant).

#### Maximum principle

Consider heat eqution in the domain $\Omega$ like below

We claim that

Proposition 1 (maximum principle). Let $u$ satisfy heat equation in $\Omega$. Then $$\max _\Omega u = \max _\Gamma u. \label{eq-11}$$

Almost correct proof. Let (\ref{eq-11}) be wrong. Then there exist point $P=(\bar{x},\bar{t})\in \Omega\setminus \Gamma$ s.t. $u$ reaches its maximum at $P$. Without any loss of the generality we can assume that $P$ belongs to an upper lid of $\Omega$. Then $$u_t(P )\ge 0 \label{eq-12}$$ (really $u(\bar{x},\bar{t})\ge u(\bar{x},t)$ for all $t: \bar{t}>t>\bar{t}-\epsilon$ and then $\bigl( u(\bar{x},\bar{t})-u(\bar{x},t)\bigr)/(\bar{t}-t)\ge 0$ and as $t\nearrow \bar{t}$) we get (\ref{eq-12}).

Also $u_{xx}( P)\le 0$ (really $u(x,\bar{t})$ reaches maximum as $x=\bar{x}$). This inequality combined with (\ref{eq-12}) almost contradict to heat equation (almost because there could be equalities).

Correct proof. Note first that the above arguments prove (\ref{eq-11}) if $u$ satisfies inequality $u_t-ku_{xx}<0$ because then there will be contradiction.

Note that $v= u-\varepsilon t$ satisfies $v_t-kv_{xx}<0$ for any $\varepsilon >0$ and therefore \begin{equation*} \max _\Omega (u-\varepsilon t) = \max _\Gamma (u-\varepsilon t). \end{equation*} Taking limit as $\varepsilon \to +0$ we get (\ref{eq-11}).

Remark.

1. Sure, the same proof works for multidimensional heat equation.

2. In fact, either in $\Omega \setminus \Gamma$ $u$ is strictly less than $\max _\Gamma u$ or $u=\const$. The proof is a bit more sophisticated.

Corollary 1 (minimum principle). $$\min _\Omega u = \min _\Gamma u. \label{eq-13}$$ Really, $-u$ also satisfies heat equation.

Corollary 2. $u=0$ everywhere on $\Gamma$ $\implies u=0$ everywhere on $\Omega$.

Really, then $\max _\Omega u=\min_\Omega u=0$.

Corollary 3. Let $u,v$ both satisfy heat equation. Then $u=v$ everywhere on $\Gamma$ $\implies u=v$ everywhere on $\Omega$.

Really, then $(u-v)$ satisfies heat equation.

#### Energy method

See Problem 6 of Home Assignment 3. There will be a forum link when solution will be posted.