Section 6.2 The weak maximum principle
In this section we prove weak maximum principles for linear parabolic operators, analogous to Theorem 3.2. The role of \(\Omega\) is now played by the parabolic interior \(D\) (as opposed to \(Q\)), while the role of \(\partial\Omega\) is played by the parabolic boundary \(\Sigma\text{.}\)
Theorem 6.3. Weak maximum principle.
Suppose that \(\Omega\) is bounded, \(L\) is uniformly parabolic, and that \(u \in C^2(D) \cap C^0(\overline{D})\) satisfies \(Lu \ge 0\) in \(D\text{.}\) If any of the conditions
- \(\displaystyle c \equiv 0\)
- \(c \le 0\) and \(\displaystyle\max_{\overline D} u \ge 0\)
- \(\displaystyle \max_{\overline D} u = 0\)
hold, then \(\displaystyle\max_{\overline D} u = \max_{\Sigma} u
\text{.}\)
Proof.
First case: \(c \equiv 0\).
As in the proof of Theorem 3.2, we first modify \(u\) to obtain a nearby function \(v\) which satisfies the strict inequality \(Lv \gt 0\text{.}\) Consider
\begin{gather*}
v = u + \varepsilon e^{-t},
\end{gather*}
where \(\varepsilon \gt 0\) is a constant which we will eventually send to \(0\text{.}\) Since \(c \equiv 0\text{,}\) we calculate
\begin{gather}
Lv = Lu + \varepsilon e^{-t} \ge \varepsilon e^{-t} \gt 0
\qquad \ina D.\tag{6.3}
\end{gather}
Suppose for the sake of contradiction that \(v\) achieves \(\max_{\overline D}
v\) at some point \((\bar x,\bar t) \in Q\text{.}\) Then we have in particular that \(x=\bar x\) is a local maximum of the function \(x \mapsto v(x,\bar
t)\text{,}\) and so by Proposition 2.34 we have that \(\nabla v(\bar x,\bar t)
= 0\) and that \(D^2 v(\bar x,\bar t)\) is negative semi-definite. Thus uniform parabolicity and Lemma 2.31 imply that \(a_{ij}(\bar
x,\bar t) \partial_{ij} v(\bar x,\bar t) \le 0\text{.}\) Similarly, \(t=\bar t\) is a local maximum of the function \(t \mapsto v(\bar x,t)\text{,}\) and so \(\partial_t v(\bar x,
\bar t) =0\text{.}\) Putting this all together we find
\begin{gather*}
Lv(\bar x, \bar t)
= a_{ij}(\bar x,\bar t) \partial_{ij} v(\bar x, \bar t)
\le 0,
\end{gather*}
which contradicts (6.3).
Next suppose that \(v\) achieves \(\max_{\overline D}v\) at some point \((\bar
x,T) \in \Omega \times \{T\}\text{.}\) Note that \((\bar x,T) \in D\text{,}\) and so from \(u \in
C^2(D)\) we know that \(\partial_i u, \partial_{ij} u, \partial_t u\) are all defined and continuous there. Arguing as above we still find that \(\nabla v(\bar x,T)
= 0\) and \(a_{ij}(\bar x,T) \partial_{ij} v(\bar x,T) \le 0\text{,}\) but now instead of \(\partial_t v(\bar x,T) = 0\) we only get the inequality \(\partial_t
v(\bar x,T) \ge 0\text{.}\) Still, we obtain
\begin{gather*}
Lv(\bar x, T)
= -\partial_t v(\bar x,T)+a_{ij}(\bar x,T) \partial_{ij} v(\bar x, T)
\le 0,
\end{gather*}
which again contradicts (6.3).
Since \(v\) cannot achieve \(\max_{\overline D} v\) in \(Q\) or on \(\Omega \times
\{T\}\text{,}\) the only option left is that \(\max_{\overline D} v = \max_\Sigma v\text{.}\) As in the proof of Theorem 3.2, the result now follows by sending \(\varepsilon \to 0\text{.}\)
Second case: \(c \le 0\) and \(\max_D u \ge 0\).
Third case: \(\max_D u = 0\).
We use exactly the same trick as in the proof of the third case of Theorem 3.2.
Corollary 6.4. Comparison principle.
Suppose that \(\Omega\) is bounded and \(L\) is uniformly parabolic. If \(u,v
\in C^2(D) \cap C^0(\overline D)\) satisfy \(Lu \le Lv\) in \(D\) and \(u \ge v\) on \(\Sigma\text{,}\) then \(u \ge v\) in \(D\text{.}\)
Proof.
Exercise 6.2.3. Note that, unlike in the elliptic case, we do not need to make an assumption about the sign of \(c\text{.}\)
Corollary 6.5. Uniqueness.
Let \(\Omega\) be bounded and let \(L\) be uniformly parabolic. If \(u,v \in
C^2(D) \cap C^0(\overline D)\) satisfy \(Lu=Lv\) in \(D\) and \(u=v\) on \(\Sigma\text{,}\) then \(u\equiv v\text{.}\)
Proof.
Remark 6.6.
As in the elliptic case, we can obtain corresponding minimum principles by considering \(-u\text{.}\)
Exercises Exercises
1. (PS10) Weak maximum principle for \(c\le0\).
Prove the second case of Theorem 6.3 as follows.
(a)
Show that, for \(k \gt 0\) sufficiently large, \(Le^{-kt} \gt 0\) in \(D\text{.}\)
Solution.
We have
\begin{equation*}
Le^{-kt} = (k+c)e^{-kt} \gt 0
\end{equation*}
provided \(k \gt \sup_D (-c)\text{.}\)
(b)
With \(k \gt 0\) chosen in the previous part, let \(\varepsilon \gt 0\) and consider the function \(v=u+\varepsilon e^{-kt}\text{.}\) Argue that \(\max_{\overline D} v \gt
0\text{.}\)
Solution.
This follows at once from the fact that \(v \gt u\) in \(\overline
D\text{.}\)
(c)
By emulating the proof of Theorem 6.3, show that \(v\) cannot achieve \(\max_{\overline D} v\) in \(Q\) or on \(\Omega \times
\{T\}\text{.}\)
Solution.
If \(v\) were to achieve this positive maximum at a point \((\bar
x,\bar t) \in Q \cup (\Omega \times \{T\})\text{,}\) then arguing exactly as in the proof of Theorem 3.2 we get
\begin{equation*}
Lv(\bar x,\bar t) \le c(\bar x,\bar t) \max_{\overline D} v \le 0,
\end{equation*}
where in the last step we have used that \(c \le 0\text{.}\) But this contradicts Part a.
(d)
Conclude by sending \(\varepsilon \to 0\text{.}\)
Solution.
Thus, exactly as in the proof of Theorem 3.2, we have \(\max_{\overline D} v = \max_\Sigma v\text{,}\) and so sending \(\varepsilon \to 0\) we get
\begin{equation*}
\max_{\overline D} u = \max_\Sigma u
\end{equation*}
as desired.
2. Asymptotics for the heat equation.
Let \(\Omega = B_1(0) \subset \R^N\text{,}\) and suppose that
\begin{equation*}
u \in C^2(\Omega \times
(0,\infty)) \cap C^0(\overline\Omega \times [0,\infty))
\end{equation*}
satisfies
\begin{equation*}
\left\{
\begin{alignedat}{2}
L u \amp= 0 \amp\qquad\amp \ina \Omega \times (0,\infty),\\
u \amp= 0 \amp\qquad\amp \ona \partial\Omega \times (0,\infty),\\
\abs u \amp \le 1 \amp\qquad\amp \ona \Omega \times \{0\},
\end{alignedat}
\right.
\end{equation*}
where \(L = -\partial_t + \Delta\) is the heat operator.
(a)
Show that the comparison function
\begin{equation*}
v(x,t) = (2-\abs x^2) e^{-Nt}
\end{equation*}
satisfies \(Lv \le 0\) in \(\Omega \times (0,\infty)\text{.}\)
Hint.
Be careful with all of the minus signs when calculating \(\partial_t v\text{!}\)
Solution.
We calculate
\begin{align*}
Lv
\amp = -\partial_t v + \Delta v \\
\amp = N(2-\abs x^2) e^{-Nt} - 2N e^{-Nt} \\
\amp = -N\abs x^2 e^{-Nt}\\
\amp \le 0\text{.}
\end{align*}
(b)
For any \(T \gt 0\text{,}\) apply Corollary 6.4 on \(\Omega \times
(0,T]\) to find that \(u \le v\) on \(\Omega \times (0,T]\text{.}\)
Solution.
First we observe that \(Lv \le 0 = Lu\) in \(\Omega \times (0,T]\text{.}\) Since \(\abs x = 1\) on \(\partial \Omega \times [0,T]\text{,}\) we also have \(v=(2-1)e^{-Nt} \ge 0 = u\) on this set. Finally, on \(\Omega \times \{0\}\) we have
\begin{equation*}
v = 2-\abs x^2 \ge 2-1 = 1 \ge u\text{.}
\end{equation*}
Applying Corollary 6.4, we conclude that \(u \le v\) in \(\overline \Omega \times [0,T]\) as desired.
(c)
Conclude that \(u(x,t) \to 0\) as \(t \to\infty\text{,}\) uniformly in \(x\text{.}\)
Solution.
Since \(T\) in the previous part was arbitrary, we have \(u \le v\) on \(\Omega \times (0,\infty)\text{.}\) Arguing similarly with \(-u\) we find \(-u \le
v\) and hence \(\abs u \le v\) on \(\Omega \times (0,\infty)\text{.}\) Since
\begin{equation*}
v(x,t) \le 2e^{-Nt} \to 0
\end{equation*}
as \(t \to \infty\text{,}\) uniformly for \(x \in \Omega\text{,}\) we conclude that \(u(x,t) \to
0\) uniformly as desired.
Comment.
In this context of this problem and this chapter, this is not a particularly important point, but the above convergence is uniform because our estimate \(\abs{u(x,t)-0} \le 2e^{-Nt}\) is independent of \(x\) (‘uniform in \(x\)’). Thus
\begin{equation*}
\lim_{t \to \infty} \sup_{x \in \Omega} \abs{u(x,t)-0} = 0\text{.}
\end{equation*}
3. (PS10) Comparison principle.
Prove Corollary 6.4. First reduce to the case where \(c \le 0\) by considering the parabolic inequalities satisfied by \(\bar u = e^{\gamma t}
u\) and \(\bar v = e^{\gamma t} v\) for an appropriate constant \(\gamma\text{.}\)
Hint.
Show that change of variable \(\bar u = e^{\gamma t} u\) and \(\bar v
= e^{\gamma t}v\) transforms the inequality \(L u \le Lv\) into an inequality \(\bar L \bar u \le \bar L \bar v\) where \(\bar L\) is a uniformly parabolic operator. Then choose \(\gamma\) appropriately so that the zeroth order coefficient \(\bar c\) of \(\bar L\) is \(\le 0\text{.}\)
Solution.
Following the hint, we first reduce to the case where \(c \le 0\) by making the change of variable \(\bar u = e^{\gamma t} u\) and \(\bar v = e^{\gamma t}
v\) where \(\gamma \in \R\) is a constant to be determined. The product rule gives
\begin{align*}
L u \amp= L (e^{-\gamma t}\bar u) \\
\amp= e^{-\gamma t} L\bar u + \gamma e^{-\gamma t} \bar u\\
\amp= e^{-\gamma t} (L\bar u + \gamma \bar u) \\
\amp= e^{-\gamma t} \bar L \bar u,
\end{align*}
where \(\bar L\) is the operator
\begin{gather*}
\bar L = -\partial_t + a_{ij} \partial_{ij} + b_i \partial_i + (c+\gamma),
\end{gather*}
and similarly for \(v\) and \(\bar v\text{.}\) Since the factor \(e^{\gamma t} \gt 0\text{,}\) our assumptions on \(u\) and \(v\) are equivalent to \(\bar L \bar
u \le \bar L \bar v\) in \(D\) and \(\bar u \ge \bar v\) on \(\Sigma\text{,}\) while the desired conclusion is similarly equivalent to \(\bar u \ge \bar v\) in \(D\text{.}\) Moreover, since \(c\) is bounded, we can pick \(\gamma\) large and negative enough that the zeroth order coefficient of \(\bar L\) is \(\bar c = c+\gamma \le 0\text{.}\)
Thus we can assume without loss of generality that \(c \le 0\text{.}\) Defining \(w = v-u\) we have \(Lw \ge 0\) in \(D\) and \(w \le 0\) on the parabolic boundary \(\Sigma\text{.}\) We now conclude as in the proof of Proposition 3.4: if \(M = \max_{\overline D} w \le 0\) then we are already done, while if \(M \gt 0\) then we can apply (the second case of) Theorem 6.3 to get a contradiction.
