6.3: Maximum Principle

Let $\Omega\subset \mathbb{R}^n$ be a bounded domain. Set

$$\begin{eqnarray*} D_T&=&\Omega\times(0,T),\ \ T>0,\\ S_T&=&\{(x,t):\ (x,t)\in\Omega\times\{0\}\ \mbox{or}\ (x,t)\in\partial\Omega\times[0,T]\}, \end{eqnarray*}$$

see Figure 6.3.1.

Figure 6.3.1: Notations to the maximum principle

Theorem 6.2

Assume $u\in C(\overline{D_T})$, that $u_t$, $u_{x_ix_k}$ exist and are continuous in $D_T$, and

$$u_t-\triangle u\le 0\ \ \mbox{in}\ D_T.\]

Then

$$\max_{\overline{D_T}} u(x,t)=\max_{S_T} u.\]

Proof

Assume initially $u_t-\triangle u<0$ in $D_T$. Let $\varepsilon>0$ be small and $0<\varepsilon<T$. Since $u\in C(\overline{D_{T-\varepsilon}})$, there is an $(x_0,t_0) \in \overline{D_{T-\varepsilon}}$ such that

$$u(x_0,t_0)=\max_{\overline{D_{T-\varepsilon}}} u(x,t).\]

Case (i)

Let $(x_0,t_0)\in D_{T-\varepsilon}$. Hence, since $D_{T-\varepsilon}$ is open,

$u_t(x_0,t_0)=0$, $u_{x_l}(x_0,t_0)=0$, $l=1,\ldots,n$ and

$$\sum_{l,k=1}^n u_{x_lx_k}(x_0,t_0)\zeta_l\zeta_k\le0\ \ \mbox{for all}\ \zeta\in\mathbb{R}^n.\]

The previous inequality implies that $u_{x_kx_k}(x_0,t_0)\le0$ for each $k$. Thus we arrived at a contradiction to $u_t-\triangle u<0$ in $D_T$.

Case (ii)

Assume $(x_0,t_0)\in\Omega\times\{T-\varepsilon\}$. Then it follows as above $\triangle u\le 0$ in $(x_0,t_0)$, and from $u(x_0,t_0)\ge u(x_0,t)$, $t\le t_0$, one concludes that $u_t(x_0,t_0)\ge0$. We arrived at a contradiction to $u_t-\triangle u<0$ in $D_T$ again.

Summarizing, we have shown that

$$\max_{\overline{D_{T-\varepsilon}}} u(x,t)=\max_{T-\varepsilon} u(x,t).\]

Thus there is an $(x_\varepsilon,t_\varepsilon)\in S_{T-\varepsilon}$ such that

$$u(x_\varepsilon,t_\varepsilon)=\max_{\overline{D_{T-\varepsilon}}} u(x,t).\]

Since $u$ is continuous on $\overline{D}_T$, we have

$$\lim_{\varepsilon\to 0}\max_{\overline{D_{T-\varepsilon}}} u(x,t)=\max_{\overline{D_T}} u(x,t).\]

It follows that there is $(\overline{x},\overline{t})\in S_T$ such that

$$u(\overline{x},\overline{t})=\max_{\overline{D_T}} u(x,t)\]

since $S_{T-\varepsilon}\subset S_T$ and $S_T$ is compact. Thus, theorem is shown under the assumption $u_t-\triangle u<0$ in $D_T$. Now assume $u_t-\triangle u\le 0$ in $D_T$. Set

$$v(x,t):=u(x,t)-kt,\]

where $k$ is a positive constant. Then

$$v_t-\triangle v=u_t-\triangle u-k<0.\]

From above we have

$$\begin{eqnarray*} \max_{\overline{D_T}} u(x,t)&=&\max_{\overline{D_T}} (v(x,t)+kt)\\ &\le&\max_{\overline{D_T}} v(x,t)+kT\\ &=&\max_{S_T} v(x,t)+kT\\ &\le&\max_{S_T}u(x,t)+kT, \end{eqnarray*}$$

Letting $k\to 0$, we obtain

$$\max_{\overline{D_T}} u(x,t)\le\max_{S_T} u(x,t).\]

Since $S_T\subset\overline{D_T}$, the theorem is shown.

$\Box$

If we replace in the above theorem the bounded domain $\Omega$ by $\mathbb{R}^n$, then the result remains true provided we assume an {\it additional} growth assumption for $u$. More precisely, we have the following result which is a corollary of the previous theorem. Set for a fixed $T$, $0<T<\infty$,

$$D_T=\{ (x,t):\ x\in\mathbb{R}^n,\ 0<t<T\}.\]

Proposition 6.2

Assume $u\in C(\overline{D_T})$, that $u_t$, $u_{x_ix_k}$ exist and are continuous in $D_T$,

$$u_t-\triangle u\le 0\ \ \mbox{in}\ D_T,\]

and additionally that $u$ satisfies the growth condition

$$u(x,t)\le Me^{a|x|^2},\]

where $M$ and $a$ are positive constants.

Then

$$\max_{\overline{D_T}} u(x,t)=\max_{S_T} u.\]

It follows immediately the

Corollary

The initial value problem $u_t-\triangle u=0$ in $D_T$, $u(x,0)=f(x)$, $x\in\mathbb{R}^n$, has a unique solution in the class defined by $u\in C(\overline{D_T})$, $u_t$, $u_{x_ix_k}$ exist and are continuous in $D_T$ and $|u(x,t)|\le Me^{a|x|^2}$.

Proof of Proposition 6.2.

See [10], pp. 217. We can assume that $4aT<1$, since the finite interval can be divided into finite many intervals of equal length $\tau$ with
$4a\tau<1$. Then we conclude successively for $k$ that

$$u(x,t)\le\sup_{y\in\mathbb{R}^n}u(y,k\tau)\le\sup_{y\in\mathbb{R}^n}u(y,0)\]

for $k\tau\le t\le (k+1)\tau$, $k=0,\ldots, N-1$, where $N=T/\tau$.

There is an $\epsilon>0$ such that $4a(T+\epsilon)<1$. Consider the comparison function
$$\begin{eqnarray*} v_\mu(x,t):&=&u(x,t)-\mu\left(4\pi(T+\epsilon-t)\right)^{-n/2}e^{|x-y|^2/(4(T+\epsilon-t))}\\ &=&u(x,t)-\mu K(ix,iy,T+\epsilon-t) \end{eqnarray*}$$

for fixed $y\in\mathbb{R}^n$ and for a constant $\mu>0$. Since the heat kernel $K(ix,iy,t)$ satisfies $K_t=\triangle K_x$, we obtain

$$\frac{\partial}{\partial t}v_\mu-\triangle v_\mu=u_t-\triangle u\le0.\]

Set for a constant $\rho>0$

$$D_{T,\rho}=\{(x,t):\ |x-y|<\rho,\ 0<t<T\}.\]

Then we obtain from Theorem 6.2 that

$$v_\mu(y,t)\le\max_{S_{T,\rho}}v_\mu,\]

where $S_{T,\rho}\equiv S_T$ of Theorem 6.2 with $\Omega=B_\rho(y)$, see Figure 6.3.1.

On the bottom of $S_{T,\rho}$ we have, since $\mu K>0$,

$$v_\mu(x,0)\le u(x,0)\le\sup_{z\in\mathbb{R}^n}f(z).\]

On the cylinder part $|x-y|=\rho$, $0\le t\le T$, of $S_{T,\rho}$ it is

$$\begin{eqnarray*} v_\mu(x,t)&\le&Me^{a|x|^2}-\mu\left(4\pi(T+\epsilon-t)\right)^{-n/2}e^{\rho^2/(4(T+\epsilon-t))}\\ &\le&Me^{a(|y|+\rho)^2}-\mu\left(4\pi(T+\epsilon)\right)^{-n/2}e^{\rho^2/(4(T+\epsilon))}\\ &\le&\sup_{z\in\mathbb{R}^n}f(z) \end{eqnarray*}$$
for all $\rho>\rho_0(\mu)$, $\rho_0$ sufficiently large. We recall that $4a(T+\epsilon)<1$.
Summarizing, we have

$$\max_{S_{T,\rho}}v_\mu(x,t)\le\sup_{z\in\mathbb{R}^n}f(z)\]

if $\rho>\rho_0(\mu)$. Thus

$$v_\mu(y,t)\le\max_{S_{T,\rho}}v_\mu(x,t)\le\sup_{z\in\mathbb{R}^n}f(z)\]

if $\rho>\rho_0(\mu)$.

Since

$$v_\mu(y,t)=u(y,t)-\mu\left(4\pi(T+\epsilon-t)\right)^{-n/2}\]

it follows

$$u(y,t)-\mu\left(4\pi(T+\epsilon-t)\right)^{-n/2}\le\sup_{z\in\mathbb{R}^n}f(z).\]

Letting $\mu\to0$, we obtain the assertion of the proposition.

$\Box$

The above maximum principle of Theorem 6.2 holds for a large class of parabolic differential operators, even for degenerate equations.

Set

$$Lu=\sum_{i,j=1}^na^{ij}(x,t)u_{x_ix_j},\]

where $a^{ij}\in C(D_T)$ are real, $a^{ij}=a^{ji}$, and the matrix $(a^{ij})$ is non-negative, that is,

$$\sum_{i,j=1}^na^{ij}(x,t)\zeta_i\zeta_j\ge 0\ \ \mbox{for all}\ \zeta\in\mathbb{R}^n,\]

and $(x,t)\in D_T$.

Theorem 6.3

Assume $u\in C(\overline{D_T})$, that $u_t$, $u_{x_ix_k}$ exist and are continuous in $D_T$, and

$$u_t-L u\le 0\ \ \mbox{in}\ D_T.\]

Then

$$\max_{\overline{D_T}} u(x,t)=\max_{S_T} u.\]

Proof

(i) One proof is a consequence of the following lemma: let $A$, $B$ real, symmetric and non-negative matrices. Non-negative means that all eigenvalues are non-negative. Then trace~$(AB)\equiv\sum_{i,j=1}^na^{ij}b_{ij}\ge0$, see an exercise.

(ii) Another proof is more directly: let $U=(z_1,\ldots,z_n)$, where $z_l$ is an orthonormal system of eigenvectors to the eigenvalues $\lambda_l$ of the matrix $A=(a^{i,j}(x_0,t_0))$. Set $\zeta=U\eta$, $x=U^T(x-x_0)y$ and $v(y)=u(x_0+Uy,t_0)$, then

$$\begin{eqnarray*} 0&\le&\sum_{i,j=1}^na^{ij}(x_0,t_0)\zeta_i\zeta_j=\sum_{i=1}^n\lambda_i\eta_i^2\\ 0&\ge&\sum_{i,j=1}^n u_{x_ix_j}\zeta_i\zeta_j=\sum_{i=1}^n v_{y_iy_i}\eta_i^2. \end{eqnarray*}$$

It follows $\lambda_i\ge0$ and $v_{y_iy_i}\le0$ for all $i$.

Consequently

$$\sum_{i,j=1}^na^{ij}(x_0,t_0)u_{x_ix_j}(x_0,t_0)=\sum_{i=1}^n\lambda_iv_{y_iy_i}\le0.\]

$\Box$

Contributors and Attributions

• Prof. Dr. Erich Miersemann (Universität Leipzig)

• Integrated by Justin Marshall.