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Mathematics LibreTexts

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.

Notations to the maximum principle

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

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

 \(\Box\)

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