| Some effective numerical methods are introduced in this paper to solve two different problems which are defined in the exterior of an open arc.I. A numerical method of a skew-derivative problem in the exterior of an open arc based on Chebyshev polynomialsWe consider a skew-derivative problem in the exterior of an open arc which models Hall effect in two-dimentional space. An electrode is modeled by an open curve Γ∈C2,λ,λ∈(0,1]. We set By z-1,z1we denote the two end points z1:=y(1),z-1:=y(-1) of Γ. Denote the left-hand of Γ by Γ+when s increases, while the opposite side is called Γ-. The normal unit vector which is directed into the side Γ-is denoted by n. The tangent direction τ of Γ is directed into the increasing direction of s. Obviously, whereThe electric-field potential u is the solution of the skew-derivative problem as follows: where β is a constant.Define angular potential where V(x,σ) is defined by the formula It is clear that V(x,σ) is the angle between and x-axis. V(x,σ) is a multi-valued function of x.We fix an arbitary branch of V(x,σ), which is continous with σ along Γ for x∈R2\Γ. From the definition, it is easy to see that an angular potential is a multi-valued function. In order to make it single-valued, it is necessary to require the following additional condition Integrating it by parts, v[μ](x) becomes a double-layer potential as follows: where and Γσ:={y=y(s)=(y1(s),y2(s)),s∈[-1,σ]}.The solution of (1) is constructed in the following form: where C is an arbitrary constant, v is an angular potential, and w is a single-layer potential If μ(σ) in (6) behaves as where μ is a continuous function, the solution u[μ](x) has singularities at the end-points of Γ, i.e., with d=±1and ε∈(-1,0].We put (6) in the boundary condition of (1) and get the following integral equations ([72]), where φ0(x, y) is the angle between the vector xy and the direction of the normal ny. If μ in (6) satisfies (8) and (9), it is the solution of problem (1).In order to prove the existence and uniqueness of solutions for (8) and (9), we introduce two boundary integral operators which are defined by with It is easy to see T o I is the limit of normal derivative of angular potential on Γ. Proposition0.1[1] Under the condition of (9)We rewrite (8) with (9) in the operator formProposition0.2I has a bounded inverse operator.Theorem0.3(11) has a unique solution μ∈H-1/2(Γ) for any f∈H-1/2(Γ).Let Tn(a) and Un(σ) denote the Chebyshev polynomials of first and second kind. We construct the approximation spaces: Let us show the approximation properties over the associated Sobolev space.Proposition0.41. The space PN is a closed subspace of H1/2[-1,1].Proposition0.5[52]1. The space QN is a closed subspace of H-1/2[-1,1].2. Q∞=∪NQn is dense in H-1/2[-l,1]. The idea of constructing the finite dimensional space is to describe μ throughthe truncated expansion:The discrete variational formulation of (11) is to find φN∈Qn satisfying∫-11φN(σ)\y(σ)|dσ=0, andTheorem0.6The variational formulation of (12) has only one solution and the following estimation holds:By some special properties of Chebyshev polynomials, we can simplify the com-putation of the discrete variational formulation. We can rewrite the method as to find cn,n=1,…, N, such that It can be shown that([72]) Rewrite(13) in the formIf|y(s)|≠const, M=N-2.(14) can be computed by The first integral term of (15) can be evaluated by the Gauss-Chebyshev integration of the first kindIf|y(s)|=B, M=N-1, B is a constant.(14) is trival. According to the definition of QN, we have which is obtained by using the orthogonal relation of Chebyshev polynomials Um and the Gauss-Chebyshev integration of the second kind.Assume that the stiffness matrix constructed by the first term in (15) is denoted by I. In the second case, I is a diagonal matrix Consider the computation of the second term of(15). The contribution of stiffness matrix from the second term in (15) is denoted by According to the Gauss-Chebyshev integration of the first kind and the second kind, the entries of matrix A areTherefore, the stiffness matrix for (15) is I+A. The third term can be also evaluated by Finally, we get a linear algebraic system of equations with respect to{c1, c2,…, cN}.Next, we recover single-layer potential w[μ] and angular potential v[μ](x):It is easy to see that we can get approximate electric-field potential u byAlthough angular potential can be also expressed by a double-layer potential, such an expression has stronger singularity near F compared to (2). So (21) is deduced by (2). In order to examining the feasibility of our method, we list some numerical examples under some special cases and compare the numerical solutions with numerical solutions in [73] and analytic solutions.II. A numerical method for scattering from an arcFor a given arc F C R2, we denote the end points of the crack with P and Q, respectively. Denote the left-hand and right-hand side of the crack by Γ+and Γ-. The normal unit vector which is directed into Γ-is written by n+and the opposite direction is written by n-. For the given incident plane wave ui(x)=ui(r,θ)=eikx-d with wave number k∈R+and incident direction d, consider the following scattering problem for total wave v(x)=ui(x)+us(x), In (23),Noting that ui is an entire function, then us satisfies the following system:Define and Let R0be a real number such that the section of Γ in BR0i can approximate as a line segment, where i=P,Q, and R0<3k/π. Let R<min(1,R0).Let us associate with each vertex i of the arc Γ polar coordinates (ri,θi)(such that the lines θi=0coincide with the section of Γ in BR0i).Proposition0.7is a set of orthogonal bases of L2(0,2π).The total wave v(x) in BR0i\Γ has the following expansion from separation of variables, where Jγ(z) is Bessel function of the first kind of order γ. Proposition0.8where M is a constant independent of γ, if γ>0,R0<-π/3k and0<r <R0.Proposition0.9if an,bn≥0.Proposition0.10Under the condition of Proposition0.8, if r <R0, where K is a constant independent of nTheorem0.11If u is the solution of problem (22)-(24); then if (ri,θi)∈BR0i,i=P, Q, Let Under the condition of Proposition0.8, where v≥1, C is a constant independent of v and κ is a positive constant only dependent on R and R0.Let Γ be contained in the interior of the disc BR={x∈R2:|x|<R}. Denote We surround the domain ΩR with a PML layer ΩPML={x∈R2: R <|x|<ρ}. Let andLet α(r)=1+iσ(r) be the model medium property which satisfies σ∈C(R), σ≥0, and σ=0, for r≤R. Denote by r the complex radius which is defined byThe PML solution u in Ωρ=Bρ\Γ is defined as the solution of the following system where A=A(x) is a matrix which satisfies the following equation in polar coordi-nates, The problem (36) is shown as an approximation of problem (26)-(28).Let a:H1(Ωρ)×H1(Ωρ)'C be a sesquilinear form Then the weak formulation for (36) is:Find u E HE1(Ωρ) such that where HE1(Ωρ)={u:u∈H1(Ωρ),u|Γρ=0}, and (·,·)Γ stands for the inner product on L2(Γ).We make the following assumption for some constant σ0>0and some integer m>1.The problem (36) is equivalent to the following system where It is easy to see that f=0in BR. We set U=u-uiφ, where φ∈C∞(Ωρ) satisfies φ=1, on Γρ, and φ=0, in BR. It is easy to see that U is a solution of the following system where Clearly, g=0in ΩR.The variational formulation of the problem (43)-(45) is to find U∈HE1(∮ρ) such thatWe define two operators M: HE1(Ωρ)'H1(Ωρ) and N:HE1(Ωρ)'H1(Ωρ) by Riez representation as follows for all u,v∈HE1(Ωρ). Then the problem (43)-(45) is equivalent to finding U∈HE1(Ωρ) such that Since U=u—uiφ, u satisfies where F is some function in H1(Ωρ).Proposition0.12The problem (40)-(42) has a unique solution u∈H1(Ωρ) for every real k except for a discrete set of values of k.Let ue=u|Ωρe, uj=u|BRj and γj is the trace operator of SRj in H1(Ωρe). Define where Therefore, V C H1(Ωρ). And Obviously, u G V can be decided by u\Ωρe uniquely.Theorem0.13V is a closed subspace of H1(Ωρ). DefineLet (49) multiply with Ⅱ,Thus (49) and (50) have the same unique solution except for a discrete set of k.Let Fh={Kih} be a regular triangulation of the domain Ωρe, where h=maxdiam{Kih}. If Kih∩(BRP∪BRQ)=(?), Kih are triangles. If Kih∩(BRP∪BRQ)≠(?), Kih are curved triangles which have one curve edge aligned with SRQ and SRP. Let Gih be a one-to-one sufficiently smooth mapping, which maps Kih onto the standard triangle Let denote the set of functions u such that u|Khi o (Gih)-1is a polynomial of degree <p.The mesh Fh induces two partitions,where i=P,Q. Denote by the nodal points of FRh,i, which induces two partitions of (0,2π) with θ. Let be the set of functions u such that u|KΔθ,i is a polynomial of degree <p with respect to the polar angle variable. We define an interpolation operator Πθi such that For v∈Hu(0,2π), we have where v>1,0≤t≤1and p≥1. Let where i=P or Q. Define and where Nρ,Jh are the nodal points of Fh in Γρ Define Now we get the following discrete version of (40)-(42): Find uh,N,R∈Xn,Rh,p sucn thatIt is easy to see that XN,Rh,p is ultimately dense in V. So problem (55) has a unique solution expect for a discrete set of values of k. Theorem0.14Let u∈H1(Ωρ) be the solution of problem (40)-(42). Then if u|Ωρe∈Hm(Ωρe) and v>1, we haveTheorem0.15Problem (55) has a unique solution uh,N,r except for a discrete set of values of k, if N and p are large enough, especially, under the condition of Theorem0.14,... |
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