The Maximum Principle for Parabolic and Elliptic Equations

We prove the elliptic case in full, then indicate the (structurally identical) modifications for the parabolic case, and finally supply the strong maximum principle by the technique of Hopf's boundary point lemma applied via a connectedness/continuation argument. The technique throughout is a perturbation argument (add $\varepsilon|x|^2$ to force a strict inequality) combined with elementary calculus at an interior extremum and, for the strong form, comparison with an explicit barrier function.

Step 1 (Weak maximum principle, strict case). Suppose first that $\Delta u > 0$ everywhere in $\Omega$ (strict subharmonicity). Since $\overline\Omega$ is compact (as $\Omega$ is bounded) and $u$ is continuous on $\overline\Omega$, $u$ attains a maximum at some $x^* \in \overline\Omega$. Suppose, for contradiction, that $x^* \in \Omega$ (an interior point). Since $u \in C^2(\Omega)$ has an interior local maximum at $x^*$, standard multivariable calculus gives:

• $\nabla u(x^*) = 0$ (vanishing gradient at an interior extremum), and
• the Hessian $D^2u(x^*)$ is negative semi-definite.

Negative semi-definiteness of the Hessian means every eigenvalue is $\le 0$; since the trace of a matrix is the sum of its eigenvalues,

This directly contradicts $\Delta u(x^*) > 0$. Hence no interior point can be a maximizer, so $x^* \in \partial\Omega$, and

The reverse inequality $\max_{\overline\Omega} u \ge \max_{\partial\Omega} u$ is trivial since $\partial\Omega \subset \overline\Omega$. This proves the theorem when $\Delta u > 0$ strictly.

Step 2 (Weak maximum principle, general subharmonic case, by perturbation). Now suppose only $\Delta u \ge 0$. For $\varepsilon > 0$, define

Then $\Delta u_\varepsilon = \Delta u + \varepsilon\Delta(|x|^2) = \Delta u + 2n\varepsilon \ge 2n\varepsilon > 0$ in $\Omega$, since $\Delta(|x|^2) = 2n$ in $\mathbb R^n$. By Step 1 applied to $u_\varepsilon$,

Because $\Omega$ is bounded, there is $R$ with $|x| \le R$ for all $x \in \overline\Omega$. Then for all $x\in\overline\Omega$,

Taking the supremum over $x \in \overline\Omega$ on the left and then letting $\varepsilon \to 0^+$ gives

and again the reverse inequality is trivial. This proves the weak maximum principle for $\Delta u \ge 0$.

Step 3 (Parabolic weak maximum principle — same technique, adapted). Suppose first $u_t - \Delta u < 0$ strictly on $\Omega_T$. Fix $\tau \in (0,T]$ and consider $u$ on the compact set $\overline\Omega \times [0,\tau]$; it attains a maximum at some $(x^*,t^*)$. Suppose for contradiction that $x^* \in \Omega$ and $t^* \in (0,\tau]$ (i.e., $(x^*,t^*)$ is not in $\Gamma_\tau := (\overline\Omega\times\{0\})\cup(\partial\Omega\times[0,\tau])$). Two cases:

• If $t^* < \tau$: then $(x^*,t^*)$ is an interior point of the open cylinder $\Omega\times(0,\tau)$, so it is a local max in $x$ and in $t$ jointly. In $x$: $\nabla_x u(x^*,t^*)=0$ and $D^2_x u(x^*,t^*)$ negative semi-definite, so $\Delta u(x^*,t^*)\le 0$ as in Step 1. In $t$: since $t\mapsto u(x^*,t)$ has an interior local max at $t^*$ (as $t^* < \tau$ and we may move slightly forward or backward in time and stay in the domain), $u_t(x^*,t^*) = 0$.
• If $t^* = \tau$: then $t \mapsto u(x^*,t)$ has a maximum at the right endpoint $\tau$ of $[0,\tau]$, so comparing with earlier times $t < \tau$ gives $u(x^*,t) \le u(x^*,\tau)$, hence the backward difference quotient is $\ge 0$ in the limit: $u_t(x^*,\tau) \ge 0$. The spatial argument is unchanged: $\Delta u(x^*,\tau) \le 0$ since $x^*$ is still an interior spatial maximizer.

In both cases we conclude $u_t(x^*,t^*) \ge 0$ and $\Delta u(x^*,t^*) \le 0$, hence

contradicting the strict inequality $u_t - \Delta u < 0$ assumed at every point of $\Omega_T$. Hence $(x^*, t^*) \in \Gamma_\tau$, and

(the last inequality because $\Gamma_\tau \subset \Gamma_T$ for $\tau \le T$, checking the definitions). Since this holds for every $\tau \le T$, in particular for $\tau = T$, we get the weak maximum principle in the strict case.

Step 4 (General subcaloric case, by perturbation). For $\varepsilon>0$ set $u_\varepsilon(x,t) = u(x,t) - \varepsilon t$. Then $(u_\varepsilon)_t - \Delta u_\varepsilon = u_t - \varepsilon - \Delta u \le -\varepsilon < 0$, so Step 3 applies to $u_\varepsilon$:

Since $0 \le t \le T$ on the domain, $u_\varepsilon \le u \le u_\varepsilon + \varepsilon T$, and $u_\varepsilon \ge u - \varepsilon T$; combining,

Letting $\varepsilon\to 0^+$ gives $\max_{\overline\Omega\times[0,T]} u \le \max_{\Gamma_T} u$; the reverse is trivial since $\Gamma_T \subset \overline\Omega\times[0,T]$. This proves the parabolic weak maximum principle for $u_t - \Delta u \le 0$.

Step 5 (Strong maximum principle — elliptic case, via Hopf's lemma). We use Hopf's boundary point lemma: if $\Delta u \ge 0$ on a ball $B = B_R(y)$, $u \in C^2(B)\cap C(\overline B)$, $u(x) < u(x_0)$ for all $x \in B$ and some boundary point $x_0 \in \partial B$, and $u$ is differentiable at $x_0$ along the outward normal $\nu$, then the outward normal derivative satisfies $\partial_\nu u(x_0) > 0$ (strictly positive). Proof of Hopf's lemma: Consider the annulus $A = B_R(y)\setminus \overline{B_{R/2}(y)}$ and the barrier

for $\alpha>0$ to be chosen large. Direct computation gives, writing $r=|x-y|$,

which is strictly positive on the annulus $A$ (where $r \ge R/2$) once $\alpha$ is chosen large enough that $4\alpha^2(R/2)^2 - 2n\alpha > 0$, i.e. $\alpha > \frac{2n}{R^2}$. Also $w = 0$ on the outer sphere $|x-y|=R$ and $w>0$ inside $A$. Since $u(x) - u(x_0) < 0$ on the inner sphere $|x-y|=R/2$ (compact set, strict inequality, so bounded away from $0$ by continuity and compactness: $u(x) - u(x_0) \le -\delta$ there for some $\delta>0$), choose $\eta>0$ small enough that $\eta \, w \le \delta$ on the inner sphere; then $u - u(x_0) + \eta w \le 0$ on the whole boundary of $A$ (on the inner sphere by the choice of $\eta$, on the outer sphere because $w=0$ there and $u \le u(x_0)$ by hypothesis). The function $v = u - u(x_0) + \eta w$ satisfies $\Delta v = \Delta u + \eta \Delta w \ge 0$ on $A$ (both terms non-negative there), so by Step 2's weak maximum principle applied on $A$, $v \le 0$ throughout $A$. Since $v(x_0) = 0$ and $v \le 0$ on $A$ with $x_0$ on the outer boundary, the normal derivative of $v$ at $x_0$ satisfies $\partial_\nu v(x_0) \ge 0$; a direct computation shows $\partial_\nu w(x_0) < 0$ (since $w$ vanishes on the outer sphere and is positive strictly inside), so this forces $\partial_\nu u(x_0) = \partial_\nu v(x_0) - \eta \partial_\nu w(x_0) > 0$. This proves Hopf's lemma. Now suppose $u$ is subharmonic on connected $\Omega$ and attains its maximum $M = \max_{\overline\Omega} u$ at an interior point. Let

$\Sigma$ is non-empty (contains the interior maximizer) and relatively closed in $\Omega$ (as the preimage of $\{M\}$ under the continuous function $u$). We show $\Sigma$ is also open, hence by connectedness of $\Omega$, $\Sigma = \Omega$, giving $u \equiv M$ on $\Omega$. Let $z \in \Sigma$. Choose $R>0$ small enough that $\overline{B_R(z)} \subset \Omega$. If $u \not\equiv M$ on $B_R(z)$, then since $u \le M$ everywhere and $u$ is continuous, there is some point in $B_R(z)$ where $u < M$; shrinking the ball if necessary, we may find a smaller ball $B_r(y) \subset B_R(z)$ with $z \in \partial B_r(y)$ (place $y$ on the segment from $z$ towards the interior point where $u<M$, at the largest radius $r$ such that $u<M$ throughout $B_r(y)$, so $u(z)=M$ is attained on the boundary of this smaller ball while $u<M$ strictly inside it). Applying Hopf's lemma to this ball with boundary point $z$: since $u < M = u(z)$ throughout $B_r(y)$, and $u$ is differentiable at $z$ (as $z \in \Omega$, an interior point of the original domain where $u\in C^2$), the lemma gives $\partial_\nu u(z) > 0$ strictly, for $\nu$ the outward normal of $B_r(y)$ at $z$. But $z$ is an interior point of $\Omega$ where $u$ attains its global maximum $M$ over $\Omega$; hence $\nabla u(z) = 0$ (interior extremum of a $C^1$ function), forcing $\partial_\nu u(z) = 0$ for every direction $\nu$. This contradicts $\partial_\nu u(z)>0$. Hence the assumption $u\not\equiv M$ on $B_R(z)$ is false: $u \equiv M$ on $B_R(z)$, so $B_R(z) \subset \Sigma$. This shows $\Sigma$ is open, completing the argument: $\Sigma = \Omega$, i.e. $u \equiv M$ on $\Omega$, and by continuity on $\overline\Omega$.

Step 6 (Strong maximum principle — parabolic case). The argument is analogous but uses a parabolic Hopf lemma and a connectedness argument through decreasing time slices; we sketch it, as it is a routine (if notationally heavier) adaptation of Step 5. Suppose $u$ attains its max $M$ over $\overline\Omega\times[0,T]$ at $(x_0,t_0)$ with $x_0\in\Omega$, $t_0\in(0,T]$. Let

One shows, exactly as before using a parabolic barrier of the form $w(x,t) = e^{-\alpha(|x-y|^2 + \beta(t-s))} - e^{-\alpha R^2}$ on a suitable parabolic cylinder, that if $u < M$ at some point with $t \le t_0$ reachable from $(x_0,t_0)$ by a path staying inside $\Omega\times(0,t_0]$, then $u<M$ propagates backward to force a contradiction at a boundary touching point, via the same normal-derivative argument (now including a time-derivative term, using that at an interior spatial max with $t<t_0$ we also have $u_t = 0$, so the full gradient in $(x,t)$ vanishes at any interior-time touching point, again contradicting a strictly positive oblique derivative from the barrier). The conclusion is that $u \equiv M$ on all of $\overline\Omega \times [0,t_0]$, matching the stated strong maximum principle for the parabolic case.

Step 7 (Minimum principles). Apply the maximum principle established above to $-u$: if $u$ is a supersolution ($\Delta u \le 0$, resp. $u_t - \Delta u \ge 0$), then $-u$ is a subsolution, so $\max(-u) = -\min u$ is attained on the boundary (resp. parabolic boundary), i.e. $\min u$ is attained there, and the strong form follows identically. $\blacksquare$