Maximum Principles Harnack and Liouville in Rn — Examples

Every second derivative of $u$ is zero, so $\mathrm{\Delta }u=0$. In the open ball, $\mathrm{\mid }{x}_1\mathrm{\mid }\le \mathrm{\mid }x\mathrm{\mid }<1$. On its closure, equality $x_1=1$ forces $x=e_1$, and $x_1=-1$ forces $x=-e_1$. Thus the explicit extrema have exactly the boundary location allowed by the weak principle.

The ball itself is an interior tangent ball at $e_1$, with outward direction $e_1$. The quotient $(u(e_1)-u(e_1-t{e}_1))\mathrm{/}t$ equals $1$ for $0<t<2$. It has a strictly positive limit, agreeing with Hopf since $u<1$ throughout the interior.

The ball mean property and continuous mean-value theorem give smoothness. Fix $\mathrm{\mid }x\mathrm{\mid }<s<R$. The smooth trace on $\mathrm{\partial }{B}_s$ and harmonic replacement represent $u$ by the kernel there; at the center the same formula gives ${\int }_{\mathrm{\partial }{B}_s}u={\omega }_{n-1}{s}^{n-1}u(0)$.

For $\mathrm{\mid }y\mathrm{\mid }=s$, the inequalities $s-\mathrm{\mid }x\mathrm{\mid }\le \mathrm{\mid }x-y\mathrm{\mid }\le s+\mathrm{\mid }x\mathrm{\mid }$ bound the positive kernel above and below. Integrating against the nonnegative trace gives $\frac{1-q}{(1+q{)}^{n-1}}u(0)\le u(x)\le \frac{1+q}{(1-q{)}^{n-1}}u(0)$, where $q=\mathrm{\mid }x\mathrm{\mid }\mathrm{/}s<1$.

Let $s\uparrow R$ so $q\to t<1$. This proves the two bounds. If $u(0)=0$, ball Harnack already gives $u\equiv 0$ on the ball, and both displayed inequalities are equalities. At $x=0$, both factors equal one.

$\mathrm{\Delta }u=-2n<0$, so $u$ is superharmonic, with the required regularity.

Its unique maximum on the closure is $u(0)=0$, while it is $-1$ at every boundary point. Thus a superharmonic maximum can be strictly interior.

The half-space is open, connected and unbounded, with nonempty boundary $\{x_n=0\}$. The affine function is smooth on all of ${\mathbb{R}}^n$, has Laplacian zero, and vanishes on that boundary.

For every $t>0$, $u(t{e}_n)=t>0$, and these values tend to infinity as $t\to \mathrm{\infty }$. Hence the proposed interior bound fails. The bounded-open-set hypothesis of the weak maximum principle does not hold for this example.

For a direct real differentiation check, use pairs written as complex numbers only for algebra: $z=x+iy$, $L=\log r+i\theta$, and $F=z\mathrm{/}L=u+iv$ with $v=(y\log r-x\theta )\mathrm{/}((\log r{)}^2+{\theta }^2)$. The derivatives $(\log r{)}_x=x\mathrm{/}{r}^2$, $(\log r{)}_y=y\mathrm{/}{r}^2$, ${\theta }_x=-y\mathrm{/}{r}^2$, ${\theta }_y=x\mathrm{/}{r}^2$ give $L_x=1\mathrm{/}z$, $L_y=i\mathrm{/}z$. The quotient rule gives $F_x=(L-1)\mathrm{/}{L}^2$, $F_y=i(L-1)\mathrm{/}{L}^2$. Equating real and imaginary parts yields $u_x=v_y$, $u_y=-v_x$. All functions are smooth for $x>0$, $r<1\mathrm{/}2$, so commuting mixed real derivatives gives $\mathrm{\Delta }u=v_{yx}-v_{xy}=0$. No theorem about holomorphic functions is used.

In polar coordinates the sign condition is $\theta \tan \theta <-\log r$ for $\mathrm{\mid }\theta \mathrm{\mid }<\pi \mathrm{/}2$. The function $\theta \tan \theta$ is even and strictly increasing from zero to infinity as $\theta$ runs from zero to $\pi \mathrm{/}2$. Thus at each radius the permitted angles form an interval containing zero. Moving the angle to zero at fixed radius, and then moving along the positive axis, joins any two points by a path in $\mathrm{\Omega }$. In particular the open nonempty set is connected.

The bound $\mathrm{\mid }u\mathrm{\mid }\le \mathrm{\mid }F\mathrm{\mid }=r\mathrm{/}\sqrt{(\log r{)}^2+{\theta }^2}\le r\mathrm{/}\mathrm{\mid }\log r\mathrm{\mid }$ gives the continuous value zero at the origin. Away from the origin the formula is continuous on the closure; its other points cannot lie on $x=0$, since there the limiting numerator is $\mathrm{\mid }y\mathrm{\mid }\pi \mathrm{/}2>0$ for $y\mathrm{\ne }0$. Thus the full extension is continuous, and closure values are at most zero. Along the positive axis $u(t,0)=t\mathrm{/}\log t<0$, so $(u(0,0)-u(t,0))\mathrm{/}t=-1\mathrm{/}\log t\to 0$.

Any ball in $\mathrm{\Omega }\subset \{x>0\}$ tangent at zero must have center $(b,0)$ and radius $b>0$: containment in the half-plane forces its center’s first coordinate to be at least its radius, while passage through zero makes that radius equal to the center’s Euclidean norm. For small $y>0$, the point $(y^2\mathrm{/}b,y)$ is in this ball because its squared distance to $(b,0)$ is $b^2-y^2+y^4\mathrm{/}{b}^2<b^2$. But at these points $(x\log r+y\theta )\mathrm{/}y\to \pi \mathrm{/}2>0$, since $x\log r\mathrm{/}y=(y\mathrm{/}b)\log r\to 0$ and $\theta \to \pi \mathrm{/}2$. For sufficiently small $y$ they have $r<1\mathrm{/}2$ and $u>0$, so are not in $\mathrm{\Omega }$. This rules out every such ball.

The local zero-level boundary has angles $\pm \beta (r)$ determined by $\beta \tan \beta =-\log r$, so $\beta \to \pi \mathrm{/}2$ and $x\mathrm{/}\mathrm{\mid }y\mathrm{\mid }=\beta \mathrm{/}(-\log r)\to 0$. Its tangent at zero is therefore the vertical line, with outward side $-e_1$ because $\mathrm{\Omega }$ lies to the right. Step 2.1 computes the accessible derivative in precisely that direction as zero, despite the strict interior inequality. The interior sphere hypothesis in Hopf is the missing one.

All second derivatives vanish, so $u$ is entire harmonic, and $u(e_1)=1\mathrm{\ne }0=u(0)$.

As $t\to \mathrm{\infty }$, $u(t{e}_1)=t\to \mathrm{\infty }$ and $u(-t{e}_1)=-t\to -\mathrm{\infty }$. Thus neither global one-sided boundedness hypothesis of Liouville is present, and constancy fails.

For $r=\mathrm{\mid }x\mathrm{\mid }>0$, differentiating $f(r)$ gives $\mathrm{\Delta }f(r)=f\mathrm{\prime }\mathrm{\prime }(r)+(n-1)f\mathrm{\prime }(r)\mathrm{/}r$. For $n=2$, $f\mathrm{\prime }=1\mathrm{/}r$ and $f\mathrm{\prime }\mathrm{\prime }=-1\mathrm{/}{r}^2$ cancel. For $n\ge 3$, $f\mathrm{\prime }=(2-n)r^{1-n}$ and $f\mathrm{\prime }\mathrm{\prime }=(2-n)(1-n)r^{-n}$ also cancel. Thus both profiles are harmonic away from zero.

The logarithm tends to $-\mathrm{\infty }$ as $r\downarrow 0$, and the negative power tends to $+\mathrm{\infty }$. Continuity at zero would require a finite limit, so no harmonic extension exists. These profiles violate boundedness near the point. Their signed ratios to the critical profiles in the stronger little-o removability condition are $-1$ for $n=2$ and $1$ for $n\ge 3$, so in both cases the absolute ratio is one rather than tending to zero.

The two open balls are disjoint, since their centers are distance three apart. Every point has a neighborhood on which $u$ is constant; consequently $u$ is smooth with $\mathrm{\Delta }u=0$ everywhere in $\mathrm{\Omega }$.

Every point of the first ball attains the global maximum $1$, whereas every point of the second has value $0$. Thus $u$ is not constant on $\mathrm{\Omega }$, which is disconnected. This is precisely the hypothesis absent from the strong theorem.

The function is smooth with zero Laplacian and zero boundary values. At each $p\in \mathrm{\partial }{B}_1$, the ball itself is an interior tangent ball and its outward direction is $p$.

For $0<t<2$, $(u(p)-u(p-tp))\mathrm{/}t=0$, so the outward derivative is zero. The strict condition $u(x)<u(p)$ inside the domain, required by Hopf, fails; all its other stated conditions are satisfied.

Differentiation gives ${\mathrm{\partial }}_{i}q=4\mathrm{\mid }x{\mathrm{\mid }}^2{x}_i$ and ${\mathrm{\partial }}_{ii}q=8x_i^2+4\mathrm{\mid }x{\mathrm{\mid }}^2$, so $\mathrm{\Delta }q=4(n+2)\mathrm{\mid }x{\mathrm{\mid }}^2\ge 0$. It is subharmonic, with value one on the unit sphere.

The constant $h=1$ is harmonic, and $\mathrm{\Delta }q\ge \mathrm{\Delta }h$ with equal boundary values. Comparison gives $q\le h$ on the closed ball, also directly visible from $\mathrm{\mid }x\mathrm{\mid }\le 1$.

For $s$, the only nonzero pure second derivatives are ${\mathrm{\partial }}_{11}s=2$ and ${\mathrm{\partial }}_{22}s=-2$, so $\mathrm{\Delta }s=0$ and $\mathrm{\nabla }s(0)=0$. But $s(t{e}_1)=t^2$ and $s(t{e}_2)=-t^2$ for $0<\mathrm{\mid }t\mathrm{\mid }<1$, proving that zero is neither a local maximum nor a local minimum. There is no interior global extremum to which the strong theorem would apply.