Boundary-integral equations and Galerkin BEM

Set $h=x-y$ and use translation invariance in [F2] and Tonelli [F3] on the nonnegative integrand. This gives $[f{]}_{1\mathrm{/}2,2}^2={\int }_{{\mathbb{R}}^2}\frac{\mathrm{\parallel }f(\cdot +h)-f(\cdot ){\mathrm{\parallel }}_2^2}{\mathrm{\mid }h{\mathrm{\mid }}^3}dh\mathrm{.}$ For each $h$, Plancherel and the translation formula [F1]–[F2] express the squared norm in the integrand as ${\int }_{{\mathbb{R}}^2}\mathrm{\mid }{e}^{2\pi i\xi \cdot h}-1{\mathrm{\mid }}^2\mathrm{\mid }\widehat{f}(\xi ){\mathrm{\mid }}^{2}d\xi$.

Tonelli [F3] now yields $[f{]}_{1\mathrm{/}2,2}^2={\int }_{{\mathbb{R}}^2}\mathrm{\mid }\widehat{f}(\xi ){\mathrm{\mid }}^{2}J(\xi )d\xi ,J(\xi )={\int }_{{\mathbb{R}}^2}\frac{\mathrm{\mid }{e}^{2\pi i\xi \cdot h}-1{\mathrm{\mid }}^2}{\mathrm{\mid }h{\mathrm{\mid }}^3}dh\mathrm{.}$ If $\xi =0$, then $J(\xi )=0$. If $\xi \mathrm{\ne }0$, rotate $h$ so that $\xi \cdot h=\mathrm{\mid }\xi \mathrm{\mid }{h}_1$, then substitute $z=\mathrm{\mid }\xi \mathrm{\mid }h$ using [F5]; the Jacobian is $\mathrm{\mid }\xi {\mathrm{\mid }}^{-2}$ and $\mathrm{\mid }h{\mathrm{\mid }}^{-3}=\mathrm{\mid }\xi {\mathrm{\mid }}^3\mathrm{\mid }z{\mathrm{\mid }}^{-3}$, so $J(\xi )=c_{\ast }\mathrm{\mid }\xi \mathrm{\mid }$ with $c_{\ast }={\int }_{{\mathbb{R}}^2}\mathrm{\mid }{e}^{2\pi i{z}_1}-1{\mathrm{\mid }}^2\mathrm{\mid }z{\mathrm{\mid }}^{-3}dz$. This constant is finite: for $\mathrm{\mid }z\mathrm{\mid }\le 1$, $\mathrm{\mid }{e}^{2\pi i{z}_1}-1\mathrm{\mid }\le 2\pi \mathrm{\mid }z\mathrm{\mid }$, and for $\mathrm{\mid }z\mathrm{\mid }>1$ it is at most $2$; polar integration gives finite bounds $C{\int }_0^{1}dr$ and $C{\int }_1^{\mathrm{\infty }}{r}^{-2}dr$. It is positive because the numerator is positive on a nonempty open set.

Substitution of $J(\xi )=c_{\ast }\mathrm{\mid }\xi \mathrm{\mid }$ into step 2.1 gives the exact seminorm identity. By [F4], the Bessel norm is the weighted Fourier norm; for every $r\ge 0$, $(1+4{\pi }^2{r}^2{)}^{1\mathrm{/}2}$ is bounded above and below by positive constants times $1+r$. Hence that norm is equivalent to $\mathrm{\parallel }f{\mathrm{\parallel }}_2^2+\int \mathrm{\mid }\xi \mathrm{\mid }\mathrm{\mid }\widehat{f}(\xi ){\mathrm{\mid }}^{2}d\xi$, and step 2.1 proves the asserted equivalence. Equivalent norms have the same Cauchy sequences and completion classes, proving the final claim.

In polar coordinates, ${\int }_{{\mathbb{R}}^2}{P}_t(z)dz=t{\int }_0^{\mathrm{\infty }}r(r^2+t^2{)}^{-3\mathrm{/}2}dr=1$; the same estimate shows $P_t\in L^1$. Direct differentiation of $t(\mathrm{\mid }z{\mathrm{\mid }}^2+t^2{)}^{-3\mathrm{/}2}$ gives $({\mathrm{\Delta }}_z+{\mathrm{\partial }}_t^2)P_t=0$ for $t>0$.

Put $F(t,\xi )=\widehat{P_t}(\xi )$. For $t$ in any compact subinterval of $(0,\mathrm{\infty })$, $P_t$ and its first two $t$-derivatives are dominated by integrable radial functions, so differentiation under the Fourier integral is valid. Fourier transforming the harmonic identity in step 1.1 and using [F1] gives ${\mathrm{\partial }}_t^{2}F-(2\pi \mathrm{\mid }\xi \mathrm{\mid }{)}^{2}F=0$. Also $\mathrm{\mid }F(t,\xi )\mathrm{\mid }\le \mathrm{\parallel }{P}_t{\mathrm{\parallel }}_1=1$. For fixed $\xi$, the change $z=tw$ gives $F(t,\xi )={\int }_{{\mathbb{R}}^2}{P}_1(w)e^{-2\pi itw\cdot \xi }dw$; dominated convergence yields $F(t,\xi )\to 1$ as $t\downarrow 0$. The solutions of this constant-coefficient ODE are linear combinations of $e^{2\pi \mathrm{\mid }\xi \mathrm{\mid }t}$ and $e^{-2\pi \mathrm{\mid }\xi \mathrm{\mid }t}$ when $\xi \mathrm{\ne }0$; boundedness removes the growing term, and the boundary limit fixes the coefficient to one. When $\xi =0$, the bounded affine solutions are constant and the same limit gives one. Hence $F(t,\xi )=e^{-2\pi \mathrm{\mid }\xi \mathrm{\mid }t}$.

Let $\mu$ be any probability measure on $K$ and let $s_{\mathrm{\#}}\mu$ be its pushforward under $y\mapsto sy$. Then $\mathrm{\mid }sx-sy\mathrm{\mid }=s\mathrm{\mid }x-y\mathrm{\mid }$, so its logarithmic energy is ${\int }_{sK}{\int }_{sK}\log \frac{1}{\mathrm{\mid }X-Y\mathrm{\mid }}d(s_{\mathrm{\#}}\mu )(X)d(s_{\mathrm{\#}}\mu )(Y)={\int }_K{\int }_K(\log \frac{1}{\mathrm{\mid }x-y\mathrm{\mid }}-\log s)d\mu (x)d\mu (y)=I(\mu )-\log s,$ using $\mu (K)=1$.

The pushforward map in [F1] is bijective, with inverse dilation by $1\mathrm{/}s$. Taking the infimum over all probability measures therefore gives $I(sK)=I(K)-\log s$. Exponentiating the definition yields $\mathrm{cap}(sK)=e^{-I(sK)}=s{e}^{-I(K)}=s\mathrm{cap}(K)$.

In the curve integral, substitute $Y=sy$. Arclength contributes $d{s}_Y=sd{s}_y$, while ${\lambda }_s(sy)=s^{-1}\lambda (y)$, so their product is $\lambda (y)d{s}_y$. The kernel identity $-\log \mathrm{\mid }sx-sy\mathrm{\mid }=-\log \mathrm{\mid }x-y\mathrm{\mid }-\log s$ then gives $V_{{\mathrm{\Gamma }}_s}{\lambda }_s(sx)=-\frac{1}{2\pi }{\int }_{\mathrm{\Gamma }}(\log \mathrm{\mid }x-y\mathrm{\mid }+\log s)\lambda (y)d{s}_y=V_{\mathrm{\Gamma }}\lambda (x)-\frac{\log s}{2\pi }{\int }_{\mathrm{\Gamma }}\lambda d{s}_y,$ as asserted.

Translation reduces to $y=0$. For $x\mathrm{\ne }0$, direct radial differentiation gives $\mathrm{\Delta }(1\mathrm{/}\mathrm{\mid }x\mathrm{\mid })=0$. Fix $R$ so that $\phi$ vanishes in a neighborhood of $\mathrm{\partial }{B}_R$, and apply Green's identity, obtained by applying [F1] to $G_3\mathrm{\nabla }\phi -\phi \mathrm{\nabla }{G}_3$, on the annulus $B_R\setminus \overline{B_{ϵ}}$. Its inner outward normal is $-e_r$, and ${\mathrm{\partial }}_{-e_r}{G}_3=1\mathrm{/}(4\pi {ϵ}^2)$; therefore ${\int }_{B_R\setminus B_{ϵ}}{G}_3(-\mathrm{\Delta }\phi )dx={\int }_{\mathrm{\partial }{B}_{ϵ}}(\phi {\mathrm{\partial }}_{-e_r}{G}_3-G_3{\mathrm{\partial }}_{-e_r}\phi )dS\mathrm{.}$ The outer flux vanishes by the choice of $R$.

On $\mathrm{\partial }{B}_{ϵ}$, the first inner term is the spherical average of $\phi$ because ${\mathrm{\partial }}_{-e_r}{G}_3=1\mathrm{/}(4\pi {ϵ}^2)$, and so tends to $\phi (0)$ by continuity. The second inner term has absolute value at most $\sup \mathrm{\mid }\mathrm{\nabla }\phi \mathrm{\mid }(4\pi {ϵ}^2)\mathrm{/}(4\pi ϵ)=O(ϵ)$ and tends to zero. The omitted ball integral of $G_3\mathrm{\Delta }\phi$ is also $O({ϵ}^2)$ since $\mathrm{\Delta }\phi$ is bounded and ${\int }_0^{ϵ}{r}^{-1}{r}^{2}dr={ϵ}^2\mathrm{/}2$. Letting $ϵ\downarrow 0$ proves the identity at $y=0$; translating coordinates proves it for every $y$.

For every nonnegative measurable $f$, vertical translation in each fixed $z$ slice and [F2] give $\int f(\mathrm{\Phi }(z,t))dzdt=\int f(z,s)dzds$; the inverse has the same property. The maps are continuous bijections, so Borel measurability is preserved, and this identity also preserves null sets and hence completed Lebesgue measurability. First take smooth compactly supported $u$. Its composition with either Lipschitz shear is locally Lipschitz, so [F1] gives weak derivatives equal almost everywhere to the ordinary chain-rule values. In particular ${\mathrm{\partial }}_t(u\circ \mathrm{\Phi })=({\mathrm{\partial }}_{3}u)\circ \mathrm{\Phi }$ and ${\mathrm{\partial }}_i(u\circ \mathrm{\Phi })=({\mathrm{\partial }}_{i}u)\circ \mathrm{\Phi }+({\mathrm{\partial }}_{3}u)\circ \mathrm{\Phi }{\mathrm{\partial }}_i\psi$ for $i=1,2$. The volume identity and $\mathrm{\mid }\mathrm{\nabla }\psi \mathrm{\mid }\le L$ bound its $H^1$ norm by $C_L\mathrm{\parallel }u{\mathrm{\parallel }}_{H^1}$; the inverse uses $-\psi$. By the whole-space smooth approximation in [F3], both operators extend to all $H^1$ functions, with the correct almost-everywhere compositions identified by the exact $L^2$ volume identity. For a one-sided domain or collar, first use its extension [F3] and localize inside the chart; the same estimates give the asserted local pullbacks independently of the extension, since the compositions agree on that side.

On a flat box $Q^{\mathrm{\prime }}\times (0,a)$, smooth $f$ satisfies $\mathrm{\mid }f(z,0){\mathrm{\mid }}^2\le C_a{\int }_0^a(\mathrm{\mid }f(z,t){\mathrm{\mid }}^2+\mathrm{\mid }{\mathrm{\partial }}_{t}f(z,t){\mathrm{\mid }}^2)dt$ by the fundamental theorem and [F5]. Integrating in $z$ gives a bounded $L^2$ endpoint trace. Approximation by [F3] extends it to $H^1$; the same fundamental-theorem estimate gives convergence of the slice values to that trace in local $L^2$ as $t\downarrow 0$. Define $\tau u$ by applying this trace to the shear pullback in step 1.1. For continuous ambient smooth inputs it is exactly $u(z,\psi (z))$. The bounds make it independent of approximants.

First take an ambient smooth compactly supported $v$ in a chart. Vertical integration gives ${\int }_G{\mathrm{\partial }}_{3}v=-\int v(z,\psi (z))dz$. For $i=1,2$, put $A(z)={\int }_{\psi (z)}^{\mathrm{\infty }}v(z,s)ds$. This is compactly supported and Lipschitz; [F1] and the one-dimensional differentiation of the lower endpoint give ${\mathrm{\partial }}_{i}A={\int }_{\psi (z)}^{\mathrm{\infty }}{\mathrm{\partial }}_{i}v(z,s)ds-v(z,\psi (z)){\mathrm{\partial }}_i\psi$ almost everywhere. Its weak derivative integrates to zero by [F2], so ${\int }_G{\mathrm{\partial }}_{i}v=\int v(z,\psi (z)){\mathrm{\partial }}_i\psi dz$. These identities are exactly ${\int }_G{\mathrm{\partial }}_{j}v=\int v{n}_{j}dS$ with the displayed graph formulas. Apply them to $v=u\eta$ for ambient smooth $u$. Approximate a general one-sided $H^1$ input by restrictions of ambient smooth extensions from [F3]; step 2.1 passes their boundary products in $L^2$ on the bounded graph base and their weak derivatives in $L^2$ on the chart. This proves the stated Green identity. No derivative of the graph normal is taken.

The vector surface measure in step 3.1 is intrinsic: on a chart it represents the distribution $\eta \mapsto {\int }_G\mathrm{\nabla }\eta$, and on an overlap both chart expressions represent the same domain distribution. Locally these are finite Radon vector measures, because they are continuous graph pushforwards of bounded measurable densities times planar Lebesgue measure on compact subcharts. Equality on smooth tests gives equality on continuous compactly supported tests by uniform mollification; [F4] applied to the positive and negative coordinate measures then gives equality of the vector measures. Their variation is exactly the stated scalar graph measure. Indeed write that vector measure as $n\mu$ with $\mathrm{\mid }n\mathrm{\mid }=1$. The triangle inequality gives variation at most $\mu$. For the reverse inequality partition the unit sphere into finitely many measurable sets on which $n$ differs from one fixed unit vector by at most $ϵ$; on each such set projection on that vector bounds its vector-measure norm below by $(1-ϵ)\mu$ of the set. The finite partition definition in [F4], followed by $ϵ\downarrow 0$, gives variation at least $\mu$. Thus both scalar surface measure and its unit normal agree on overlaps.

In a fixed graph chart the graph area density is at most $\sqrt{1+L^2}$, and the base projection of its intersection with a radius-$r$ ball lies in a planar radius-$r$ disk. Thus its area is at most $C_L{r}^2$. A finite compact graph cover gives the same bound for the whole compact boundary, uniformly in the ball center. A point at distance less than $\delta$ from a graph has vertical distance at most $(1+L)\delta$: compare it to a graph point at that distance and use the Lipschitz bound. Choose finitely many smaller graph charts covering the compact boundary with positive margins inside their original charts. For sufficiently small $\delta$, its entire $\delta$ neighborhood is covered by the corresponding enlarged chart cylinders; each has bounded planar base and vertical strip thickness at most $2(1+L)\delta$. Fubini gives volume at most $C\delta$ after the finite sum.

Sum the Green formula in step 3.1 over the two sides and localize by a finite smooth chart partition. The normals are opposite and the trace difference is zero for a matching pair, so its interface terms cancel; derivatives of the partition cancel because their sum is one. Hence its distributional gradient is the piecewise $L^2$ gradient and the glued function is $H^1$ locally. The same argument against the zero function on the other side proves zero extension of a zero-trace field has its zero-extended gradient. In flat coordinates a compactly supported zero extension is $H^1$ and supported in $t\ge 0$; translation to $t\ge ϵ$ converges in $H^1$ by [F3]. Pull back to the graph using step 1.1. This produces $H^1$ approximants supported a positive distance inside the graph side, because vertical separation $ϵ$ gives Euclidean separation at least $ϵ\mathrm{/}\sqrt{1+L^2}$. Mollify in physical coordinates at less than half that distance; [F3] gives interior smooth approximants converging in $H^1$. Cutoffs preserve support margins at chart sides. A finite chart partition and interior cutoffs supply the asserted localized and bounded-domain conclusions.

For a homogeneous polynomial $Q_n$ of degree $n$, direct differentiation and Euler's identity give $\mathrm{\Delta }(\mathrm{\mid }x{\mathrm{\mid }}^{2j}{Q}_n)=2j(2n+2j+1)\mathrm{\mid }x{\mathrm{\mid }}^{2j-2}{Q}_n+\mathrm{\mid }x{\mathrm{\mid }}^{2j}\mathrm{\Delta }{Q}_n$. For homogeneous $P$ of degree $n$, set $H=P+{\sum }_{j=1}^{\lfloor n\mathrm{/}2\rfloor }{a}_j\mathrm{\mid }x{\mathrm{\mid }}^{2j}{\mathrm{\Delta }}^{j}P$, where $a_0=1$ and $a_{j+1}=-a_j\mathrm{/}[2(j+1)(2n-2j-1)]$. Substitution into the displayed identity makes the coefficient of each $\mathrm{\mid }x{\mathrm{\mid }}^{2j}{\mathrm{\Delta }}^{j+1}P$ equal to $a_j+2(j+1)(2n-2j-1)a_{j+1}=0$; the last Laplacian power vanishes by degree. Thus $H$ is homogeneous harmonic and $P-H$ is divisible by $\mathrm{\mid }x{\mathrm{\mid }}^2$. Induction on degree shows that restrictions of polynomials are finite sums of elements of the spaces ${\mathcal{H}}_j$.

The polar-coordinate identity $\mathrm{\Delta }={\mathrm{\partial }}_{rr}+2r^{-1}{\mathrm{\partial }}_r+r^{-2}{\mathrm{\Delta }}_{{\mathbb{S}}^2}$ applied to $r^{\mathrm{\ell }}Y(\omega )$, whose extension is harmonic by [F1], gives ${\mathrm{\Delta }}_{{\mathbb{S}}^2}Y=-\mathrm{\ell }(\mathrm{\ell }+1)Y$. On the sphere, integration by parts gives $⟨v,{\mathrm{\Delta }}_{{\mathbb{S}}^2}w⟩=⟨{\mathrm{\Delta }}_{{\mathbb{S}}^2}v,w⟩$: in $(\theta ,\varphi )$ coordinates integrate the $\varphi$ derivative over its period and the $\theta$ term with weight $\sin \theta$; the pole boundary terms vanish because smooth functions and their spherical gradients are bounded while $\sin \theta \to 0$. Distinct degrees have distinct eigenvalues, so their spaces are orthogonal.

Choose a finite orthonormal basis from [F2] and write $g={\sum }_m{c}_m{Y}_{\mathrm{\ell }m}$. Cauchy–Schwarz gives $\mathrm{\mid }g(\omega ){\mathrm{\mid }}^2\le ({\sum }_m\mathrm{\mid }{c}_m{\mathrm{\mid }}^2)({\sum }_m\mathrm{\mid }{Y}_{\mathrm{\ell }m}(\omega ){\mathrm{\mid }}^2)$. Orthonormality identifies the first sum with $\mathrm{\parallel }g{\mathrm{\parallel }}_2^2$, and [F2] gives the second as $(2\mathrm{\ell }+1)\mathrm{/}(4\pi )$. Taking the supremum proves the evaluation bound.

Real polynomial restrictions contain constants and separate distinct points of ${\mathbb{S}}^2$ by a coordinate function. Compactness and [F3] make them dense in real $C({\mathbb{S}}^2)$; applying this separately to real and imaginary parts gives density in complex $C({\mathbb{S}}^2)$. By step 1.1 this is density of the algebraic sum of the ${\mathcal{H}}_j$. For continuous $f$ and $\epsilon >0$, choose a finite sum $q$ of harmonic modes with $\mathrm{\parallel }f-q{\mathrm{\parallel }}_{\mathrm{\infty }}<\epsilon$. Put $P_{\le N}f={\sum }_{\mathrm{\ell }=0}^N{P}_{\mathrm{\ell }}f$ and $V_N={⨁}_{\mathrm{\ell }=0}^N{\mathcal{H}}_{\mathrm{\ell }}$. Orthogonality from step 1.2 makes $f-P_{\le N}f$ orthogonal to $V_N$, so Pythagoras shows $P_{\le N}f$ is a best $L^2$ approximation in $V_N$. For $N$ at least the largest degree in $q$, $\mathrm{\parallel }f-P_{\le N}f{\mathrm{\parallel }}_2\le \mathrm{\parallel }f-q{\mathrm{\parallel }}_2\le \sqrt{4\pi }\epsilon$. Since $\epsilon$ is arbitrary, the ordered partial sums converge in $L^2$.

For smooth $f$ and $\mathrm{\ell }\ge 1$, self-adjointness just proved gives $P_{\mathrm{\ell }}({\mathrm{\Delta }}_{{\mathbb{S}}^2}^{m}f)=(-\mathrm{\ell }(\mathrm{\ell }+1){)}^m{P}_{\mathrm{\ell }}f$. The finite-dimensional projection is a contraction in $L^2$, hence $\mathrm{\parallel }{P}_{\mathrm{\ell }}f{\mathrm{\parallel }}_2\le [\mathrm{\ell }(\mathrm{\ell }+1){]}^{-m}\mathrm{\parallel }{\mathrm{\Delta }}_{{\mathbb{S}}^2}^{m}f{\mathrm{\parallel }}_2$. Taking $m$ arbitrarily large proves the asserted decay (and degree zero is absorbed into $C_m(f)$).