Boundary-integral equations and Galerkin BEM — Examples

For $r=\mathrm{\mid }x\mathrm{\mid }>0$, spherical coordinates centered on the axis through $x$ give $S1(x)=\frac{R^2}{2}{\int }_{-1}^1(r^2+R^2-2rRt{)}^{-1\mathrm{/}2}dt=\frac{R}{2r}(r+R-\mathrm{\mid }r-R\mathrm{\mid })\mathrm{.}$ If $r<R$ this equals $R$, and if $r>R$ it equals $R^2\mathrm{/}r$; at $r=R$ the same convergent improper integral gives $R$, and at $r=0$ the constant kernel $1\mathrm{/}(4\pi R)$ times the parametrized area $4\pi R^2$ gives $R$.

Differentiating the two radial formulas gives ${\mathrm{\partial }}_{r}S1=0$ for $r<R$ and ${\mathrm{\partial }}_{r}S1=-R^2\mathrm{/}{r}^2$ for $r>R$. Their boundary limits are $0$ and $-1$, so ${\mathrm{\partial }}_n^{-}S1-{\mathrm{\partial }}_n^{+}S1=1=\lambda$. This direct smooth-sphere calculation checks the kernel normalization and jump orientation independently of the polyhedral theorem.

The single-layer trace and conormal sign are exactly [F2]. For the double layer put $U_R(x)={\int }_{S_R}{G}_3(x,y)d{S}_y$ and $D_{R}1(x)={\int }_{S_R}{\mathrm{\partial }}_{n_y}{G}_3(x,y)d{S}_y$. In sphere parameters $y=R\omega$, differentiating $R^2{G}_3(x,R\omega )$ by [F3] gives ${\mathrm{\partial }}_R{U}_R=D_{R}1+(2\mathrm{/}R)U_R$: the kernel derivative is the outward source-normal derivative, and the second term differentiates area density.

For fixed $\mathrm{\mid }x\mathrm{\mid }<1$ and $R$ sufficiently near one, [F2] gives $U_R(x)=R$, so $D_{R}1=1-2=-1$. For fixed $\mathrm{\mid }x\mathrm{\mid }>1$ it gives $U_R(x)=R^2\mathrm{/}\mathrm{\mid }x\mathrm{\mid }$, so $D_{R}1=2R\mathrm{/}\mathrm{\mid }x\mathrm{\mid }-(2\mathrm{/}R)(R^2\mathrm{/}\mathrm{\mid }x\mathrm{\mid })=0$. At $R=1$ the two potentials are constant on their respective sides; therefore their trace jump is $-1$ and both normal derivatives are zero. This proves both signs without adding a divergence-theorem axiom to the classical example.

The circle $S_1$ is compact. Since $\mathrm{\mid }x-y\mathrm{\mid }\le 2$ on it, every probability energy is at least $-\log 2$. Normalized arclength has finite energy: writing $x=e^{i\theta }$ and $y=e^{i\phi }$ gives $\mathrm{\mid }x-y\mathrm{\mid }=2\mathrm{\mid }\sin ((\theta -\phi )\mathrm{/}2)\mathrm{\mid }$, whose logarithmic singularity is integrable because ${\int }_0^{ϵ}\mathrm{\mid }\log t\mathrm{\mid }dt<\mathrm{\infty }$. Thus $I(S_1)$ is finite and $0<\mathrm{cap}(S_1)<\mathrm{\infty }$. Apply [F1] with $K=S_1$ and $\mathrm{\Gamma }=S_1$ to obtain both scaling identities in the Example. The functional $\lambda \mapsto {\int }_{S_1}\lambda ds$ has one-dimensional range, and its output is a constant function, so the trace correction is rank at most one. Its scalar factor vanishes precisely at $s=1$; otherwise the constant density gives a nonzero correction, proving rank one.

For $\lambda =1\mathrm{/}(2\pi )$ on the unit circle, ${\int }_{S_1}\lambda ds=(1\mathrm{/}(2\pi ))(2\pi )=1$. Substituting $s=e$ into the correction gives $-(\log e)\mathrm{/}(2\pi )=-1\mathrm{/}(2\pi )$. Any zero-total-charge density has no scale correction by the same formula.

In polar coordinates the cross-sectional Laplacian is ${\mathrm{\partial }}_{rr}+r^{-1}{\mathrm{\partial }}_r+r^{-2}{\mathrm{\partial }}_{\theta \theta }$. For $u=r^{\alpha }\sin (\alpha \theta )$, the three terms sum to $[\alpha (\alpha -1)+\alpha -{\alpha }^2]r^{\alpha -2}\sin (\alpha \theta )=0$; the $z$ derivative is zero. Also $u(r,0,z)=u(r,\omega ,z)=0$ because $\alpha \omega =\pi$. Thus it is harmonic with zero side trace in the local geometry [F1].

The squared transverse gradient is $\mathrm{\mid }{\mathrm{\nabla }}_{r,\theta }u{\mathrm{\mid }}^2={\alpha }^2{r}^{2\alpha -2}$, so its local integral is a constant times ${\int }_0^{r_0}{r}^{2\alpha -1}dr<\mathrm{\infty }$ for $\alpha >0$; $u$ itself is also square integrable. On the other hand, $u_{rr}=\alpha (\alpha -1)r^{\alpha -2}\sin (\alpha \theta )$ is nonzero on a set of angles of positive measure, and its squared integral contains $c{\int }_0^{r_0}{r}^{2\alpha -3}dr=\mathrm{\infty }$ because $\alpha <1$. Hence $u\in H^1$ locally but $u\mathrm{\notin }{H}^2$ near the edge.

The eigenvectors $(1,1)$ and $(1,-1)$ of $A$ have eigenvalues $3$ and $1$, respectively. Hence $x^{T}Ax\ge \mathrm{\parallel }x{\mathrm{\parallel }}_2^2$ and $\mathrm{\parallel }A{\mathrm{\parallel }}_2=3$, so $c=1$, $C=3$. The defect matrix is diagonal with entries $\pm 1\mathrm{/}100$, so $\mathrm{\parallel }E{\mathrm{\parallel }}_2=1\mathrm{/}100<1\mathrm{/}50$; alternatively $\mathrm{\parallel }E{\mathrm{\parallel }}_2\le \mathrm{\parallel }E{\mathrm{\parallel }}_F=\sqrt{2}\mathrm{/}100<1\mathrm{/}50$. Thus $\delta =1\mathrm{/}50$ is a certified Euclidean operator defect.

The determinant of $\tilde{A}$ is $(201\cdot 199)\mathrm{/}10000-1=29999\mathrm{/}10000$, and its inverse applied to $f$ gives $\tilde{u}=(19900\mathrm{/}29999,-10000\mathrm{/}29999{)}^T$. The exact inverse of $A$ gives $u=(2\mathrm{/}3,-1\mathrm{/}3{)}^T$. Their difference is $(-298\mathrm{/}89997,-1\mathrm{/}89997{)}^T$, whose squared norm is $88805\mathrm{/}{89997}^2<1\mathrm{/}{49}^2$ because $88805\cdot 2401=213220805<8099460009={89997}^2$.

The discrete space here is the whole two-dimensional space, so the exact Galerkin best-approximation term vanishes; $c=1$, $\delta =1\mathrm{/}50$, $\eta =0$, and $\mathrm{\parallel }{F}_h\mathrm{\parallel }=1$. Applying [F1] gives $\mathrm{\parallel }u-\tilde{u}{\mathrm{\parallel }}_2\le \frac{(1\mathrm{/}50)}{1-1\mathrm{/}50}=\frac{1}{49},$ which agrees with the direct rational check in step 1.2.