Poisson Problems and Interior Harmonic Estimates — Examples

Define $g(e^{it}):=1$ for $0<t<\pi$, $g(e^{it}):=0$ for $\pi \le t\le 2\pi$, and $U_g(r):=(2\pi {)}^{-1}{\int }_0^{2\pi }{P}_r(t)g(e^{it})dt$ for $0\le r<1$, with $P_r$ as in [F1]; this integral is finite because $g$ is bounded and the kernel is continuous on the compact circle. Since $g(e^{it})$ vanishes on $[\pi ,2\pi ]$ and equals one on $(0,\pi )$, $U_g(r)=(2\pi {)}^{-1}{\int }_0^{\pi }{P}_r(t)dt$. Also $g(1)=g(e^{i0})=0$, because $0$ is not an interior point of $(0,\pi )$.

Reflection symmetry. For every $t$ we have $\cos (2\pi -t)=\cos t$, so [F1] gives $P_r(2\pi -t)=P_r(t)$; the substitution $t\mapsto 2\pi -t$ maps $(0,\pi )$ onto $(\pi ,2\pi )$ and preserves the Lebesgue measure. Hence ${\int }_0^{\pi }{P}_r(t)dt={\int }_{\pi }^{2\pi }{P}_r(t)dt$.

Normalization. By [F2], $(2\pi {)}^{-1}{\int }_0^{2\pi }{P}_r(\theta )d\theta =1$; splitting the integral at $\pi$ and using step 2.1, $1=2\cdot (2\pi {)}^{-1}{\int }_0^{\pi }{P}_r(t)dt$. Therefore $U_g(r)=(2\pi {)}^{-1}{\int }_0^{\pi }{P}_r(t)dt=1\mathrm{/}2$ for every $0\le r<1$.

Failure at the jump. The value $U_g(r)=1\mathrm{/}2$ is independent of $r$, so $U_g(r)\to 1\mathrm{/}2$ as $r\uparrow 1$, while the datum assigns $g(1)=0$ at the boundary point $e^{i0}=1$; thus the Poisson integral of a bounded boundary function need not recover the assigned value at a discontinuity, and only continuity of the datum at the point would force it.

Define $u:H\to \mathbb{R}$ by $u(x^{\mathrm{\prime }},t):=t$, i.e. $u(x)=x_n$ with $x_n$ the last coordinate. Its first partial derivatives are ${\mathrm{\partial }}_{i}u\equiv {\delta }_{in}$ and its second partial derivatives all vanish identically, so $\mathrm{\Delta }u={\sum }_i{\mathrm{\partial }}_i{\mathrm{\partial }}_{i}u=0$ on the open half-space $H$; hence $u$ is harmonic by [F1].

The same formula defines a continuous extension of $u$ to the closed half-space $\bar{H}=\{x:x_n\ge 0\}$, and on the boundary plane $\mathrm{\partial }H=\{x_n=0\}$ this extension has the value $0$. The zero function $0$ is harmonic on $H$ with the same zero boundary values.

The two solutions differ and the second is unbounded: $u(e_n)=1\mathrm{\ne }0=0(e_n)$ at the point $e_n=(0,1)\in H$, and along the vertical ray $\{s{e}_n:s>0\}$ the value $u(s{e}_n)=s$ tends to $+\mathrm{\infty }$, so ${\sup }_H\mathrm{\mid }u\mathrm{\mid }=+\mathrm{\infty }$ while ${\sup }_H\mathrm{\mid }0\mathrm{\mid }=0$. Thus $0$ and $u$ are two distinct solutions of the same zero Dirichlet problem on $H$ once boundedness --- or any other growth restriction excluding linear growth --- is dropped.

Steps 1.1, 2.1 and 3.1 exhibit a nonzero unbounded harmonic function with the same continuous zero boundary trace as the zero function on the half-space; uniqueness of the half-space Dirichlet problem therefore requires a growth condition such as boundedness, and this witness is eliminated by it. The computation uses no choice principle.

Put $q(x):={\sum }_{i<n}(x_i-a_i{)}^2=\mathrm{\mid }x-a{\mathrm{\mid }}^2>0$ on $\mathrm{\Omega }$, $h(x):=q(x{)}^{(2-n)\mathrm{/}2}$ and $u(x):=1-R^{n-2}h(x)=1-(R\mathrm{/}\mathrm{\mid }x-a\mathrm{\mid }{)}^{n-2}$. These expressions are continuous on $\{x:\mathrm{\mid }x-a\mathrm{\mid }\ge R\}$.

For $\mathrm{\mid }x-a\mathrm{\mid }>R$ we have $0<R\mathrm{/}\mathrm{\mid }x-a\mathrm{\mid }<1$, so $0<(R\mathrm{/}\mathrm{\mid }x-a\mathrm{\mid }{)}^{n-2}<1$ and $0<u(x)<1$: the function $u$ is bounded on $\mathrm{\Omega }$, while the zero function is bounded as well.

Harmonicity. Write $y_i=x_i-a_i$. Coordinate differentiation using [F1] and [F2] gives ${\mathrm{\partial }}_{i}h=(2-n)y_i{q}^{-n\mathrm{/}2}$ and ${\mathrm{\partial }}_j{\mathrm{\partial }}_{i}h=(2-n){\delta }_{ij}{q}^{-n\mathrm{/}2}+n(n-2)y_i{y}_j{q}^{-(n+2)\mathrm{/}2}$. All these derivatives are continuous because $q>0$, so $h$ and $u$ are $C^2$. Summing the pure second partials yields $\mathrm{\Delta }h=n(2-n)q^{-n\mathrm{/}2}+n(n-2)q{q}^{-(n+2)\mathrm{/}2}=0$. The constant function has zero second partials, hence $\mathrm{\Delta }u=-R^{n-2}\mathrm{\Delta }h=0$; both $u$ and $0$ are harmonic by [F3]. No surface measure or choice assumption is used.

Boundary trace. If $\mathrm{\mid }x-a\mathrm{\mid }=R$ then $R\mathrm{/}\mathrm{\mid }x-a\mathrm{\mid }=1$, so $u(x)=1-1=0$ on $\mathrm{\partial }\mathrm{\Omega }=S_R(a)$; the zero function has the same trace, and $u$ is nonzero on $\mathrm{\Omega }$ by step 2.1.

Far-field behaviour. If $\mathrm{\mid }x-a\mathrm{\mid }\to \mathrm{\infty }$ then $(R\mathrm{/}\mathrm{\mid }x-a\mathrm{\mid }{)}^{n-2}\to 0$ because $n-2>0$, so $u(x)\to 1$, whereas the zero function tends to $0$; in particular $u$ does not satisfy the decay condition $u\to 0$ at infinity.

Steps 2.1, 2.2 and 3.1 exhibit two distinct bounded harmonic functions $0$ and $u$ on the exterior domain $\mathrm{\Omega }$ that agree, with value zero, on $\mathrm{\partial }{B}_R(a)$; step 3.2 shows that they are separated by their far-field behaviour. Hence prescribed boundary data, and boundedness by itself, do not give uniqueness, and a far-field condition such as $u\to 0$ is needed to exclude this witness.

Smoothness. The map $x\mapsto x_1$ is linear, hence smooth, and $\beta$ is smooth by [F2]; the composition $u=\beta \circ (x\mapsto x_1)$ is therefore smooth on ${\mathbb{R}}^n$, with ${\mathrm{\partial }}_{x_2}u(x)={\beta }^{\mathrm{\prime }}(x_1)\cdot 0=0$ and more generally $D^{\alpha }u(x)={\beta }^{(\mathrm{\mid }\alpha \mathrm{\mid })}(x_1)$ if $\alpha ={\alpha }_1{e}_1$ and $D^{\alpha }u(x)=0$ whenever $\alpha$ has a nonzero entry outside the first coordinate.

A first-order PDE. Since $u(x)$ depends on $x$ only through the first coordinate, ${\mathrm{\partial }}_{x_2}u\equiv 0$ on ${\mathbb{R}}^n$; thus $u$ solves the first-order linear equation ${\mathrm{\partial }}_{x_2}u=0$, which is not the Laplace equation.

Vanishing Taylor coefficients. Let $\alpha$ be any multi-index. If ${\alpha }_1=\mathrm{\mid }\alpha \mathrm{\mid }$ then $D^{\alpha }u(0)={\beta }^{(\mathrm{\mid }\alpha \mathrm{\mid })}(0)=0$ by [F2]; otherwise $D^{\alpha }u\equiv 0$ by step 1.1 and again $D^{\alpha }u(0)=0$. Hence every coefficient $c_{\alpha }=D^{\alpha }u(0)\mathrm{/}\alpha !$ of the Taylor expansion of $u$ at the origin vanishes, so the only candidate series is the zero series.

Failure of the representation. For every $\delta >0$ and every $0<x_1<\delta$ we have $u(x_1{e}_1)=\beta (x_1)=\exp (-1\mathrm{/}{x}_1)>0$ by [F1], while the candidate series of step 2.2 sums to $0$; hence no neighbourhood of the origin carries a power-series representation of $u$. By [F3], $u$ is not real analytic at the origin.

Steps 1.1, 2.1 and 3.1 exhibit a $C^{\mathrm{\infty }}$ function that solves a PDE and is not real analytic at a point, so smoothness of a solution does not imply real analyticity for general partial differential equations; the harmonic conclusion of the companion page uses the Laplace equation, not smoothness alone.

Work under [F4], put $\kappa :=R^2-\mathrm{\mid }x-a{\mathrm{\mid }}^2$ (positive because $x$ is interior) and $C:=\{y\in \mathrm{\partial }{B}_R(a):\mathrm{\mid }y-p\mathrm{\mid }<\delta \}$, $D:=\mathrm{\partial }{B}_R(a)\setminus C$. By [F2] the two masses $M_C:={\int }_C{P}_{R,a}(x,\cdot )dS$ and $M_D:={\int }_D{P}_{R,a}(x,\cdot )dS$ are finite and add to ${\int }_{\mathrm{\partial }{B}_R(a)}{P}_{R,a}(x,\cdot )dS=1$ by [F1].

For $y\in D$ one has $\mathrm{\mid }y-p\mathrm{\mid }\ge \delta$, hence $\mathrm{\mid }x-y\mathrm{\mid }\ge \mathrm{\mid }y-p\mathrm{\mid }-\mathrm{\mid }x-p\mathrm{\mid }>\delta \mathrm{/}2$ and $1\mathrm{/}\mathrm{\mid }x-y{\mathrm{\mid }}^n\le 2^n{\delta }^{-n}$; therefore $P_{R,a}(x,y)\le \kappa 2^n\mathrm{/}(R{\omega }_{n-1}{\delta }^n)$ by [F1].

Integrating the bound of step 2.1 over $D$ and using [F2] with $\mathrm{\mid }\mathrm{\partial }{B}_R(a)\mathrm{\mid }={\omega }_{n-1}{R}^{n-1}$ gives $M_D\le (\kappa 2^n\mathrm{/}(R{\omega }_{n-1}{\delta }^n)){\omega }_{n-1}{R}^{n-1}=2^n{R}^{n-2}{\delta }^{-n}\kappa =C_{n,R}{\delta }^{-n}\kappa$.

Hence $M_C=1-M_D\ge 1-C_{n,R}{\delta }^{-n}\kappa$ by step 1.1, and $M_C\le 1$ by the unit-mass identity of [F1]; as $\kappa =R^2-\mathrm{\mid }x-a{\mathrm{\mid }}^2\to 0$ when $x\to p$ (because $\mathrm{\mid }p-a\mathrm{\mid }=R$), the cap mass tends to one and the mass $M_D$ outside the cap tends to zero.

This computation is the mass-split content of the cap/complement estimate [F3] read on constant data: for $g\equiv 1$ one has ${\omega }_{g,p}=0$, $\parallel g{\parallel }_{\mathrm{\infty }}=1$ and $U_g(x)=1$ by [F1], so [F3] reduces to the trivial inequality $0\le 2^{n+1}{R}^{n-2}{\delta }^{-n}\kappa$; the genuine concentration information for kernel mass alone is exactly the bound of step 3.1 and the limit of step 4.1. Both assertions of the statement are therefore proved.

Work under [F4] and put $u(x):=x_j$. By [F2] the function $u$ is smooth with $\mathrm{\Delta }u=0$ on ${\mathbb{R}}^n$, hence on $B_R(a)$, and it lies in $C^2(B_R(a))\cap C(\overline{B_R(a)})$; its restriction to the sphere is $u(y)=y_j=g(y)$.

By [F1] the Poisson integral $U_g$ lies in the same class, is harmonic on $B_R(a)$ and has the same boundary trace $g$. Applying the uniqueness clause of [F1] to the two admissible functions $u$ and $U_g$ gives $U_g(x)=u(x)=x_j$ for every $x\in B_R(a)$.

Evaluating at the centre checks the spherical first moment: step 2.1 gives ${\int }_{\mathrm{\partial }{B}_R(a)}{P}_{R,a}(a,y)y_{j}d{S}_y=a_j$, and by [F3] the kernel there is the constant $1\mathrm{/}\mathrm{\mid }\mathrm{\partial }{B}_R(a)\mathrm{\mid }$, so $\frac{1}{\mathrm{\mid }\mathrm{\partial }{B}_R(a)\mathrm{\mid }}{\int }_{\mathrm{\partial }{B}_R(a)}{y}_{j}d{S}_y=a_j$.

Steps 2.1 and 3.1 prove the displayed identity and its central specialization; the argument uses only the uniqueness clause of the ball Dirichlet theorem together with the elementary harmonicity of $x_j$.

Work under [F4] and define $V(x^{\mathrm{\prime }},t):=\exp (-2\pi \mathrm{\mid }\xi \mathrm{\mid }t)\exp (2\pi i\xi \cdot x^{\mathrm{\prime }})$ on $\bar{H}$. Since $\mathrm{\mid }V(x^{\mathrm{\prime }},t)\mathrm{\mid }=\exp (-2\pi \mathrm{\mid }\xi \mathrm{\mid }t)\le 1$ and $V(x^{\mathrm{\prime }},0)=g(x^{\mathrm{\prime }})$, the function $V$ is bounded and continuous on $\bar{H}$ with trace $g$.

By [F2], $\mathrm{\Delta }V={\mathrm{\Delta }}_{x^{\mathrm{\prime }}}V+{\mathrm{\partial }}_t^{2}V=-4{\pi }^2\mathrm{\mid }\xi {\mathrm{\mid }}^{2}V+4{\pi }^2\mathrm{\mid }\xi {\mathrm{\mid }}^{2}V=0$ on $H$; so $V$ is harmonic (all derivatives exist and are continuous, being those of an exponential).

Applying uniqueness in [F1] to $V$ and to the Poisson integral $U_g$ of the bounded continuous datum $g$ gives $U_g=V$, which is the displayed formula; for $\xi =0$ this reads $U_g\equiv 1$.

Differentiating the formula at $t=0$ gives ${\mathrm{\partial }}_t{U}_g(x^{\mathrm{\prime }},0)=-2\pi \mathrm{\mid }\xi \mathrm{\mid }g(x^{\mathrm{\prime }})$; the outward unit normal of $H$ at the boundary plane is $-e_n$, so the outward normal derivative is $-{\mathrm{\partial }}_t{U}_g(x^{\mathrm{\prime }},0)=+2\pi \mathrm{\mid }\xi \mathrm{\mid }g(x^{\mathrm{\prime }})$. This is the single-mode Dirichlet-to-Neumann computation: the half-space Poisson multiplier $e^{-2\pi \mathrm{\mid }\xi \mathrm{\mid }t}$ differentiates to the boundary multiplier $2\pi \mathrm{\mid }\xi \mathrm{\mid }$ in the outward normal.

The same multiplier is visible in the Fourier description: [F3] says the datum $g$ has Fourier transform ${\delta }_{\xi }$, and the extension multiplies that mode by the factor $e^{-2\pi \mathrm{\mid }\xi \mathrm{\mid }t}$; the constant mode $\xi =0$ is fixed and does not decay.