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# 6.3: Maximum Principle

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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 , 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

• Integrated by Justin Marshall.