In this paper, the dynamical behaviors for a stochastic SIRS epidemic model with nonlinear incidence and vaccination are investigated. In the models, the disease transmission coefficient and the removal rates are all affected by noise. Some new basic properties of the models are found. Applying these properties, we establish a series of new threshold conditions on the stochastically exponential extinction, stochastic persistence, and permanence in the mean of the disease with probability one for the models. Furthermore, we obtain a sufficient condition on the existence of unique stationary distribution for the model. Finally, a series of numerical examples are introduced to illustrate our main theoretical results and some conjectures are further proposed. As is well known, transmissions of many infectious diseases are inevitably affected by environment white noise, which is an important component in realism, because it can provide some additional degrees of realism compared to their deterministic counterparts. Therefore, in recent years, stochastic differential equation (SDE) has been used widely by many researchers to model the dynamics of spread of infectious disease (see –5 and thβSI is frequently used (see Λ/(dS(dR + ε) + pdR). Applying the Lyapunov function method and the theory of persistence for dynamical systems, we can prove that, when R0 < 1, model  and, when R0 > 1, model  and disease I is permanent.It is well known that the basic reproduction number for model  is defin1, model  has a gl1, model  has a unβ, dS, dI, and dR in model (i dt (0 ≤ i ≤ 3) may be approximated by a normal distribution with zero mean and variance σi2dt (0 ≤ i ≤ 3), respectively. That is, i dt may correlate with each other, we represent them by l-dimensional Brownian motion B(t) = (B1(t),…, Bl(t)) as follows: σij are all nonnegative real numbers. Therefore, model  = S, g(I) = I, and p = q = 0 has been investigated by Yang and Mao in  by 〈h(t)〉 = (1/t)∫0th(s)ds.For any function H1) and (H2) hold. Let (S(t), I(t), R(t)) be any positive solution of model (φ(t) is defined for all t ≥ 0 satisfying limt→∞φ(t) = 0.Assume that  is defined byt→∞φ(t) = 0  a.s.Taking integration from 0 to or model , we get , we havedR+εSt−S0ion from  with whiH1) and (H2) hold and σj1 = σj2 = σj3 = 0  (1 ≤ j ≤ l). Then, for any solution (S(t), I(t), R(t)) of system (S(0), I(0), R(0)) ∈ R+3, one has μ = min{dS, dI, dR}.Assume that  = S(t) + I(t) + R(t), from model  and (H2) hold, σj1 = σj2 = σj3 = 0  (1 ≤ j ≤ l), dS = dR, and dI = dS + α with constant α ≥ 0. Then, for any solution (S(t), I(t), R(t)) of model (S(0), I(0), R(0)) ∈ R+3, one has Assume that (of model  with (S(N(0) = S(0) + I(0) + R(0). From the third equation of model (y(t) = ∫0tedS + ε)s(S(s)ds; then y(t), we obtain Since of model  we have e obtain yt=e−ptΛ1erefore, St=Λ1−qdStituting –23). Th. Th(24)ddS ≠ dR in model , σr(x) = (σr1(x), σr2(x),…, σrn(x)), and Br(t)  (1 ≤ r ≤ m) are standard Brownian motions defined on the above probability space. The diffusion matrix is defined byV(x), we define Consider the following D in Rn with a regular  boundary such that(i)i = 1,2,…, n and positive constant η > 0 such that aii(x) ≥ η for all x ∈ D;there exist some (ii)V(x) : Dc → R such that V(x) is second-order continuously differentiable function and that, for some θ > 0, LV(x)≤−θ for all x ∈ Dc, where Dc = Rn∖D.there exists a nonnegative function Assume that there is a bounded open subset π. That is, if function f is integrable with respect to the measure π, then for all x0 ∈ RnThen  has a unTo study the permanence in mean with probability one of model  we need Y ∈ C and Z ∈ C satisfy limt→∞(Z(t)/t) = 0  a.s. If there is T > 0 such thatt ≥ T, thenAssume that functions For the convenience of following statements, we denote H1) and (H2) hold. If one of the following conditions holds: (a)σ02f(S0)g′(0) ≤ β + ∑j=1lσj0σj2,(b)σ0 > 0  and  (β + ∑j=1lσj0σj2)2/2σ02 − (dI + γ + (1/2)σ22) < 0,then, for any initial value (S(0), I(0), R(0)) ∈ R+3, one has I is stochastically extinct exponentially with probability one. Moreover, Assume that  and Applying Itô's formula to model  leads toAssume that condition (b) holds. Sincehen from (43)lnItϵ > 0 such that β + ∑j=1lσj0σj2 ≥ g′(0)f(ϵ)σ02. We compute thatAssume that condition (a) holds. Choose constant σ02 = 0, which implies σj0 = 0  (1 ≤ j ≤ l), we have from , from (H2), we can obtain f′(ξ)(S − S0) ≤ f′(S0)(S − S0). Hence, we haveWhen ave from (47)fxβfSrding to , 40), a, aσ02 = , we havefS≤fS0+f′ce, from  and Lemmσ02 ≠ 0, from (F(u) is a monotone increasing for u ∈  and monotone decreasing for u ∈ /dS(dS + ε + p).Firstly, when U(I) = ∫I(0)I(t)(1/g(I))dI; similarly to above proof of S2 = S02 + 2S0(S − S0)+(S − S0)2, we further havea + b)2 ≤ 2(a2 + b2), it follows that H(t), we easily have limt→∞〈H2(t)〉 = 0. By (S(t) + I(t) + R(t) ≤ Λ/dS  a.s. for all t ≥ 0. Hence,t→∞Φ(t) = 0. Therefore, taking t → ∞ in (Let tituting  and 21)U(I) = ∫Ither haveUIt≥βSt−dpression  of H(t), = 0. By , without0It.From , 89), a, aU(I) =lly have G2t≤M0It.t.From (91t∫0tIrdrIt.From (1t∫0tIrdrt → ∞ in  holds. TA(x) = h(x)hT(x), where x = ,aii(x)   the diagonal elements of matrix A(x). We have aii(x) = ∑j=1lhij2(x).In this section, we discuss the stationary distribution of model  by usingH1) holds, f(S) = S, and there is a constant ρ > 0 such that aii(x) > ρ, for any x ∈ R+3 and i = 1,2, 3, γ > p, dI > dS, and γ(dS + dR) > p(dI + dR). If R0 > 1 and S∗, I∗, R∗) is the unique endemic equilibrium of model (Assume that (> 1 and 9dS+pγdS+dg(I), the Lyapunov function structured in the following is different from that given in [We here use the Lyapunov function method to prove this theorem. The proof given here is similar to Theorem 5.1 in . But, dugiven in .V(x) and compact set K ⊂ R+3 such that LV(x)≤−C for some C > 0 and x ∈ R+3/K.By x =  ∈ R+3. Define the function LV1(x), we have Denote LV2(x), we have Define the function LV3(x), we get Define the functionLV4(x), we get Define the functionR+3. Hence, we can easily obtain that there exists a constant C > 0 and a compact set K of R+3 such that, for any x ∈ R+3/K, Define the Lyapunov function for model  as follohen from , 101), , 106)Vxhen from , and  = I/(1 + ωI2), where ω is a positive constant. It is easy to verify that assumption (H1) holds. By Milstein's higher-order method [ξji   are N-distributed independent Gaussian random variables and Δt > 0 is time increment.Throughout the following numerical simulations, we choose r method , 30, we f(S) = S/(1 + 0.2S), Λ = 1.85, q = 0.52, β = 0.52, p = 0.24, ε = 0.2, γ = 0.3, ω = 2, dS = 0.4, dI = 0.21, dR = 0.3, σ01 = 0.15, σ02 = 0.99, σ11 = 0.23, σ12 = 0.17, σ21 = 0.14, σ22 = 0.72, σ31 = 0.47, and σ32 = 0.93. By computing, we obtain σ02f(S0)g′(0)−(β + ∑j=12σj0σj2) = 0.3442 > 0, and (β + ∑j=12σj0σj2)2/2σ02 − (dI + γ + (1/2)σ22) = 0.005 > 0. This shows that conditions (a) and (b) of I(t) of model  holds. The disease I(t) in model (Assume (in model  is stochf(S) = S/(1 + 1.5S), Λ = 3, q = 0.2, β = 2.1, p = 0.3, ε = 0.8, γ = 0.1, ω = 2, dS = 0.5, dI = 0.8, dR = 0.4, σ01 = 0.8, σ02 = 1.2, σ11 = 0.3, σ12 = 0.75, σ21 = 0.45, σ22 = 0.8, σ31 = 0.8, and σ32 = 0.3. By computing, we obtain I(t) of model  holds. The disease I(t) in model (Assume (in model  is stochf(S) = S/(1 + 0.1S), Λ = 1.2, q = 0.5, β = 1.5, p = 0.9, ε = 1.1, γ = 0.9, ω = 2, dS = 0.6, dI = 0.35, dR = 0.4, σ01 = 0.4, σ02 = 0.2, σ11 = 0.1, σ12 = 0.45, σ21 = 0.2, σ22 = 0.1, σ31 = 0.2, and σ32 = 0.3. By computing, we obtain I(t) of model (I(t).In model , we takeH1) holds. The disease I(t) in model (Assume (in model  is stochf(S) = S, Λ = 0.67, q = 0.02, β = 1.7, p = 0.05, ε = 3, γ = 0.99, ω = 4, dS = 0.29, dI = 0.53, dR = 0.39, σ01 = 0.025, σ02 = 0.02, σ11 = 0.0121, σ12 = 0.01, σ21 = 0, σ22 = 0, σ31 = 0.02, and σ32 = 0.01. By computing, we obtain that the basic reproduction number for deterministic model  = . Furthermore, we can verify that there is a constant ρ > 0 such that aii(x) > ρ for any x ∈ R+3  , dI  −  dS = 0.24 > 0, γ  −  p = 0.94 > 0, γ(dS + dR) − p(dI + dR) = 0.6272 > 0, and S∗ = 1.4230, I∗ = 0.3845, and R∗ = 0.1372.In model , we takef(S) = S/(1 + 0.4S), Λ = 2.5, q = 0.5, β = 1.4, p = 0.7, ε = 0.9, γ = 0.51, ω = 1.89, dS = 0.7, dI = 0.45, dR = 0.58, σ01 = 0.4, σ02 = 0.2, σ11 = 0.21, σ12 = 0.1, σ21 = 0.1, σ22 = 0.24, σ31 = 0.2, and σ32 = 0.1. By computing, we obtain that the basic reproduction number for deterministic model  = . Furthermore, we can verify that there is not a constant ρ > 0 such that aii(x) > ρ for any x ∈ R+3 and i = 1,2, 3, dI − dS = −0.25 < 0, γ − p = −0.19 < 0, γ(dS + dR) − p(dI + dR) = −0.0682 < 0 and In model , we takeH1) holds. Model  ≡ S and (1) σj0 = 0  (1 ≤ j ≤ l) or (2) σj1 = σj2 = σj3 = 0  (1 ≤ j ≤ l). However, for the general model (f(S) ≠ S and ≠  (1 ≤ j ≤ l), whether we also can establish similar results still is an interesting open problem.The stochastic persistence and permanence in the mean of the disease for model  are estaal model , particuβf(S(t)) and f2(S(t))g′(I(t)). If we may get v1 and v2 are two positive constants; then the following perfect result may be established.In fact, under the above case, from the proofs of Theorems H1) holds. If I in model (Assume that (in model  is stochf(S) is a nonlinear function. The best perfect result on the stationary distribution is to prove that model (Another important open problem is about the existence of stationary distribution of model , that isat model  possesseHowever, the numerical examples given in


Scientific Reports 10.1038/s41598-017-00198-5, published online 13 March 2017Correction to: This Article contains an error in the Methods section under subheading ‘Data analysis’.E[λij|Xij]) = ln(workerij) + β0 + bi0 + β1new criteriai + β2working hourij + β3salaryij + β4unemployment rateij + β5yeari”“ln(should read:E[Yij|Xij]) = ln(workerij) + β0 + b0i + β1new criteriai + β2working hourij + β3salaryij + β4unemployment rateij + β5yeari”“ln(


Scientific Reports6: Article number: 3827610.1038/srep38276; published online: 12062016; updated: 01252017This Article contains a typographical error. In the Results section,εo(497 nm) of 50294 ± 259 M−1 cm−1 for Ca2+-saturated GCaMP6f and 1564 ± 1140 M−1 cm−1 for Ca2+-saturated GCaMP6fu, brightness values of 29673 and 7197 M−1 cm−1 were obtained for GCaMP6f and GCaMP6fu, respectively.’‘With measured molar extinction coefficients should read:εo(497 nm) of 50294 ± 259 M−1 cm−1 for Ca2+-saturated GCaMP6f and 15646 ± 1140 M−1 cm−1 for Ca2+-saturated GCaMP6fu, brightness values of 29673 and 7197 M−1 cm−1 were obtained for GCaMP6f and GCaMP6fu, respectively’.‘With measured molar extinction coefficients


There are a number of errors in Table 1. The “Application” value for “P38MAPK” in line 4 should read “WB”. The “Dilution” value for “MAPKAPK2” in line 13 should read “1:1000(WB)”. The “Dilution” value for “HSP27” in line 15 should read “1:1000”. Please view the correct In Fig 5, Fig 5C shows the incorrect image under the “AF 1.0 μM” header for “0h”. Please view the correct


PP2A is composed of a scaffolding subunit (A), a catalytic subunit (C) and a regulatory subunit (B) that is classified into four families including B, B′, B′′ and B′′′/striatin. Here, we found that a distinct PP2A complex regulates NF‐κB signalling by dephosphorylation of IKKβ, IκBα and RelA/p65. The PP2A core enzyme AC dimer and the holoenzyme AB′′′C trimer dephosphorylate IKKβ, IκBα and RelA, whereas the ABC trimer dephosphorylates IκBα but not IKKβ and RelA in cells. In contrast, AB′C and AB′′C trimers have little effect on dephosphorylation of these signalling proteins. These results suggest that different forms of PP2A regulate NF‐κB pathway signalling through multiple steps each in a different manner, thereby finely tuning NF‐κB‐ and IKKβ‐mediated cellular responses. IκB, inhibitor of kappa BIKKβ, inhibitor of kappa B kinase βIKKβKN, kinase‐negative IKKβ mutantNF‐κB, nuclear factor kappa BPPPs, protein phosphatasesNuclear factor kappa B (NF‐κB) is a critical transcription factor that regulates many cellular and organismal processes including immune and inflammatory responses, cellular growth and cell survival. The NF‐κB signalling pathway is regulated by the phosphorylation of several proteins including inhibitor of kappa B (IκB) kinase β (IKKβ), an inhibitory protein IκBα and an essential subunit of NF‐κB RelA/p65 Aberrant activation of NF‐κB is linked to various diseases such as inflammatory disorders and cancer. Thus, there are numerous regulatory mechanisms at multiple levels to ensure the tight control of NF‐κB activity Among these phosphatases, PP2A is the most abundant, constituting approximately 1% of total cellular proteins A previous RNAi screen revealed that PP2A plays an important role in the regulation of NF‐κB signalling and identified the core enzyme subunits, Aα, Aβ, Cα and Cβ, as negative regulators of the NF‐κB signalling pathway in mouse astrocytes m L‐glutamine at 37 °C.cDNA‐encoding Aα, Aβ, Cα, Cβ, Bα, Bβ, Bγ, Bδ, B′α, B′β, B′γ, B′δ, B′ε, B′′β, Strn and Strn3 subunits of PP2A were amplified from a human cDNA library by PCR. The cDNAs were inserted into pRK‐HA and pRK‐Flag expression vectors. Expression plasmids encoding IKKβ, kinase‐negative IKKβ mutant (IKKβKN) and IκBα have been reported previously −1 TNF‐α for 18 h. After lysing cells with buffer from the assay system, luciferase activity was analysed following the manufacturer's instructions.Nuclear factor kappa B activity was estimated by Dual‐Luciferase Reporter Assay System (Promega) using pNF‐κB and pRL‐TK Luciferase reporter plasmids. After transfection of 0.01 μg NF‐κB reporter plasmids with or without 0.01 μg expression plasmids of various PP2A subunits, IKKβ, IκBα and/or RelA into GP2‐293 cells in collagen‐coated 96‐well dishes, cells were incubated in the presence or absence of 50 ng·mLm Tris/Cl, pH 7.5, 150 mm NaCl, 1 mm EGTA, 10 mm MgCl2, 60 mm β‐glycerophosphate, 1 mm Na3VO4, 1 mm 4‐amidino phenyl methyl sulfonyl fluoride, 50 KIU·mL−1 aprotinin, 20 μg·mL−1 pepstatin, 20 μg·mL−1 leupeptin, 2 mm DTT and 1% Triton X‐100. After centrifugation at 16 000 × g for 20 min at 4 °C, the supernatants were used as cell lysates. The cell lysates were subjected to SDS/PAGE and then gel‐separated proteins were transferred to PVDF membranes (Millipore) and subjected to immunoblotting using the SuperSignal West Pico Chemiluminescence System (Pierce). Antibodies used were as follows: anti‐Flag (M2) , anti‐HA , anti‐IκBα , anti‐RelA/p65 , anti‐phospho‐IκBα , anti‐phospho‐RelA/p65(S468) , anti‐phospho‐RelA/p65(S536) , anti‐PP2A , anti‐PP2A , anti‐IKKβ  and anti‐phospho‐IKKα/β .GP2‐293 cells in collagen‐coated 12‐well dishes were transfected with 0.2 μg expression plasmids of various PP2A subunits, IKKβ, IκBα and/or RelA. After 24‐h transfection, cells were washed with PBS and solubilized with buffer A consisting of 20 mvia anti‐Flag (M2) agarose beads (Sigma). IKKβ proteins were eluted in a buffer containing 20 mm Tris/Cl, pH 7.5, 150 mm NaCl, 2 mm DTT, 1% Triton X‐100 and 3 × Flag peptides (Sigma) from anti‐Flag‐agarose beads with ultra‐free Centrifugal Filter Units (Millipore). For the dephosphorylation assay of IκBα and RelA, cells were transfected with plasmids encoding Flag‐RelA, HA‐IκBαRR and IKKβEE, and then phosphorylated IκBαRR and RelA protein complexes were purified from cell lysates via anti‐Flag (M2) agarose beads. PP2A proteins were purified from GP2‐293 cells following transfection with plasmids encoding the Flag‐tagged Aα subunit and HA‐tagged Cα subunit, together with or without HA‐tagged B subunits including Bα, Bβ, Bγ, Bδ, B′α, B′β, B′γ, B′δ, B′ε, B′′β, Strn and Strn3. PP2A core enzyme composed of Aα and Cα, and PP2A holoenzyme composed of Aα, Cα and various B subunits were purified from cell lysates via anti‐Flag (M2) agarose beads with ultra‐free Centrifugal Filter Units in the previously described buffer. Phosphorylated IKKβ proteins (0.001 μg) or protein complexes (0.001 μg) of phosphorylated IκBαRR and RelA were incubated with purified PP2A (0.001 μg) for 30 min at 37 °C, and then reaction mixtures were subjected to SDS/PAGE. Gel‐separated proteins were transferred to PVDF membranes and subjected to immunoblotting using anti‐phospho‐IKKα/β, anti‐phospho‐IκBα and anti‐phospho‐RelA antibodies, respectively.For the dephosphorylation assay of IKKβ, GP2‐293 cells in 10‐cm dishes were transfected with expression plasmids encoding Flag‐IKKβ, and then phosphorylated IKKβ was recovered from cell lysates PP2A is composed of three subunits scaffolding (A), regulatory (B) and catalytic (C) subunit Fig. A. Cells To investigate the involvement of the PP2A in regulation of NF‐κB signalling, we constructed expression plasmids encoding each subunit of the B, B′′, B′′ and B′′′/Strn family (Fig. We next investigated the effects of distinct B subunits on phosphorylation of the IKKβ activation loop. Cells were transfected with expression plasmids encoding IKKβ and plasmids encoding Aα, Cα and each B subunit, and phosphorylation of the activation loop was investigated by immunoblotting. Expression of the AC core enzyme markedly suppressed IKKβ phosphorylation in a phosphatase activity‐dependent manner Fig. A. The ACTo reveal the effects of each B subunit on the NF‐κB/IκBα complex, we transfected cells with expression plasmids encoding IκBαRR, RelA and IKKβEE, together with plasmids encoding Aα, Cα and each B subunit. Then, the effects of PP2A on IKKβEE‐mediated phosphorylation of IκBαRR and RelA were investigated by immunoblotting. Expression of the AC core enzyme suppressed phosphorylation of IκBαRR and RelA in a phosphatase activity‐dependent manner Fig. C. AC, ABThen, we investigated the effects of distinct B subunits on phosphorylation of endogenous NF‐κB signalling proteins in TNF‐α‐stimulated cells. Cells were transfected with expression plasmids of PP2A, and then stimulated with TNF‐α. AC and AB′′′C suppressed TNF‐α‐induced phosphorylation of IKKβ and RelA, whereas ABC, AB′C and AB′′C had little effect Fig. E. AC, ABin vitro. Expression plasmids encoding Flag‐tagged Aα, HA‐tagged Cα and HA‐tagged B subunits were transfected in cells, and PP2A complexes of various isoforms of B subunit family were purified from cells by using Flag‐agarose beads (Fig. in vitro, AC and AB′′′C preferentially dephosphorylate IKKβ and RelA, and AC, ABC and AB′′′C preferentially dephosphorylate IκBα in vivo, suggesting that intracellular mechanisms may relate to the preference of AC, ABC and AB′′′C to dephosphorylate IKKβ, IκBα and RelA.Dephosphorylation of IKKβ, IκBαRR and RelA was analysed by using purified PP2A enzyme ads Fig. A. Phosphads Fig. B. Followads Fig. C. Then, ads Fig. D. Followads Fig. E. These in vitro and in vitro, we investigated intracellular mechanisms of the preference of AC, ABC, AB′′C and AB′′′C to dephosphorylate NF‐κB signalling proteins. The effects of B subunits on the interaction between PP2A and IKKβ were analysed by immunoprecipitation assay. Cells were transfected with plasmids encoding Flag‐tagged CαD85N, HA‐tagged Aα, HA‐tagged B subunits and HA‐tagged IKKβ. PP2A complexes were immunoprecipitated by using Flag‐agarose beads from cells, and the association of IKKβ proteins to the PP2A complexes was analysed by immunoblotting (Fig. To resolve the discrepancies in the dephosphorylation of IKKβ and RelA between in vitro and in vivoin vivoMany studies have revealed important and complicated roles of PP2A in IKKβ regulation. For example, it has been reported that PP2A suppresses IKKβ activity NF‐κB signalling is not uniformly regulated by PP2A but rather is subjected to specific regulation by distinct PP2A complexes at multiple steps. In contrast to AC and AB′′′C, which dephosphorylate IKKβ, IκBα and RelA, ABC dephosphorylates IκBα but not IKKβ and RelA in cells. These results indicate that ABC suppresses NF‐κB activity without inhibition of IKKβ. A recent study revealed that IKKβ not only activates NF‐κB through the phosphorylation of IκBα but also regulates many cellular functions by phosphorylating various proteins in an NF‐κB‐independent manner via A subunits of PP2A in the complexes. STRIPAK complexes have a critical role in protein dephosphorylation and act as important regulators of multiple vital signalling pathways, including the Hippo pathway, mitogen‐activated protein kinases and cytoskeleton remodelling. Recent studies suggest that the dysregulation of STRIPAK complexes correlates with human diseases including cancer Among the four classes of B subunit families, the B′′′/Strn family proteins only facilitate dephosphorylation of three proteins including IKKβ, IκBα and RelA. Members of the B′′′/Strn family are evolutionarily conserved and have critical roles in biological processes such as development and cell growth PP2A is a confirmed tumour suppressor protein that is genetically altered or functionally inactivated in many cancers HK and YT designed and performed the experiments, interpreted the study, and wrote the paper. KO, MK, YN, ET and TH performed the experiments.


Scientific Reports6: Article number: 38356; 10.1038/srep38356 published online: 12062016; updated: 12222017.This Article contains errors in the legend of Figure 3.c = μ|a〉 + ν|b〉ψshould read:c = μ|0〉 + ν|1〉ϕIn addition,a = μ|a〉 + ν|b〉ϕshould read:a = μ|a〉 + ν|b〉ψ


AbstractLeptomias Faust, 1886 occurring in the Sichuan Province of China, including the description of a new species, Leptomiasverticalis Ren, Zhang & Song, sp. n. from Jiulong County, Southwest Sichuan. New locality data and remarks for the other eleven species, a key to and distribution map of all twelve Sichuan species are provided. Leptomiaschenae Alonso-Zarazaga & Ren is transferred to Geotragus Schoenherr, 1845, where its valid name is G.granulatus , comb. n. in application of Art. 59.4. Structural details of Leptomiasverticalis and Geotragusgranulatus are illustrated.An account is given of the twelve species of  Leptomias Faust, 1886 is a diverse genus of flightless weevils  in the subfamily Entiminae, with a centre of distribution in China, India, Nepal, Afghanistan and Myanmar. It differs from related genera by having (i) the metanepisternum completely separated from the metaventrite and (ii) the upper edge of the scrobes directed towards the lower margin of the eye  does not belong in Leptomias but instead to the genus Geotragus Schoenherr, 1845. With the transfer of this species out of Leptomias effected and the new species described in this paper, the number of current species of Leptomias remains at 159 and that of species recorded from China at 89. The new species and the new combination are here documented and illustrated, along with a key to the species present in Sichuan. Moreover, species documented in Sichuan are usually described in Chinese and it might be difficult for most people to obtain information of their distribution.Ninety per cent of the Institute of Zoology, Chinese Academy of Sciences, Beijing, China (IZCAS); Forschungsmuseum Alexander Koenig, Bonn, Germany; Natural History Museum, London, UK; Naturkundemuseum, Berlin, Germany; Senckenberg Naturforschendes Museum, Frankfurt am Main, Germany; Senckenberg Naturhistorische Sammlungen, Dresden, Germany. The types of the new species are deposited in IZCAS.All specimens, including types, examined for this study are located in the following collections: Specimens were dissected after soaking them in soapy water overnight, for cleaning and softening, and the dissected parts were placed in a cold 10 % NaOH solution for 20 hours to macerate the soft tissues. After dissection, all parts were photographed and stored in glycerine in microvials pinned beneath the specimen from which they were dissected.The morphological terminology used in this study mainly follows Ren (2013). Measurements were made using an ocular micrometre as follows: standard length – in dorsal view from anterior margin of thorax to apex of elytra along midline; pronotal length – in dorsal view from anterior margin to base along midline; pronotal width – in dorsal view across widest part; elytral length – in dorsal view along suture of elytra from base to apex; elytral width – in dorsal view across widest part; rostral length – in lateral view in a straight line from apex to anterior margin of eye; rostral width – in dorsal view across base of rostrum. Measurements are made in millimetres.All observations and dissections were performed using a Nikon SMZ1500 stereo microscope. The habitus photographs were taken with a MP-E 65 macro lens mounted on a CANON EOS700D digital camera. Other photographs were taken with a CCD Qimagine MicroPublisher 5.0 RTV camera mounted on a Zeiss SteREO Discovery V.12 microscope. Extended-focus images were generated with Auto-Montage Pro 5.03.0061 and edited with Adobe Photoshop CS 14.0 if required.verbatim, with pinyin romanisation and comments in square brackets if labels are in Chinese; labels are separated by semicolons and lines by slashes.Label data are given Taxon classificationAnimaliaColeopteraCurculionidaeRen, Zhang & Songsp. n.http://zoobank.org/02A25235-8C44-4C3D-9270-AAD7BB9E9EC6L.ochrolineatus Chen, 1987 but differs by the following characters: elytra in lateral view abruptly sloping posteriorly, dorsal edge of slope of declivity straight, almost parallel to anterior margin; elytra in dorsal view at apical 1/3 with symmetrical crescent-shaped dark brown patches.This new species resembles Measurements (mm): Standard length: 8.00; pronotal length: 3.00; pronotal width: 3.00; elytral length: 5.00; elytral width: 3.20; rostral length: 1.30; rostral width: 1.20.Holotype, male. Habitus and colour : Standard length: 8.20; pronotal length: 3.20; pronotal width: 3.20; elytral length: 5.00; elytral width: 3.30; rostral length: 1.35; rostral width: 1.20; inner margin of protibiae with eleven blunt teeth (apex worn out), mesotibiae with ten small sharp teeth.Female paratypes. Measurements (in mm): Standard length: 10.40–11.30; pronotal length: 3.00–3.20; pronotal width: 3.10–3.20; elytral length: 6.00–6.70; elytral width: 3.70–3.90; rostral length: 1.29–1.40; rostral width: 1.28–1.32. Pronotum with anterior and posterior margins not truncate, slightly curved; greatest width just behind midpoint. Elytra much longer and wider than in male; in lateral view with posterior declivity straight and overhanging elytral apex; ventrite 5 : 四川九龙县南 [Sìchuān Jǐulóngxiàn nán] J08 / 2120m 核桃林 [Hétáolín] 杯诱 [bēiyòu] / 2001.VII.9–12 于晓东 [Yú Xiǎodōng] / 中科院动物所 ; : IOZ(E) 1965001. Paratypes : 1 ♂: : same data as holotype except IOZ(E) 1965002; 2 ♀: : same data as holotype except IOZ(E) 1965003, IOZ(E) 1965004; 2 ♀: : same data as holotype except J09 / 2200m 青冈灌丛 [Qīnggāng Guàncóng] and IOZ(E) 1965005, IOZ(E) 1965006.The specific epithet refers to the straight declivity of the elytra in lateral view.Sichuan (China).Taxon classificationAnimaliaColeopteraCurculionidaeChao, 1981Leptomiaselongitus Chao, 1981. Insects of Xizang 1: 543, pl. III–19.Holotype, ♂: (white): 西藏  芒康  / 3800 公尺  / 中国科学院 ; (white): 1976.VI.9 [handwritten] / 采集者: 韩寅恒 ; : HOLOTYPE; : IOZ(E) 906241. Paratypes: 1 ♀: same data as holotype except ALLOTYPE printed on sea-green paper and IOZ(E) 906242. 1 ♀: same data as holotype except PARATYPE printed on yellow paper and IOZ(E) 906243. 4 ♀, 2 ♂: same data as holotype except locality 芒康盐井 , 2700 公尺 , collecting date 1976.VI.3, PARATYPE printed on yellow paper and IOZ(E) 906244–IOZ(E) 906249. 1 ♀: (white): 西藏芒康什草  / 19 [printed] 76 [handwritten] 年  7 [handwritten] 月  15 [handwritten] 日  / 采集者  2600 米  / 中国科学院 ; : PARATYPE; : IOZ(E) 906250. 2♀: ditto, IOZ(E) 906251, IOZ(E) 906252.1 ♂: : 3 / 得荣鱼根 [Déróng Yúgēn] / 高山新 [Gāoshānxīn] / 3700 公尺 [Gōngchǐ], 何多吉 [Hé Duōjí] 80.6.8; : IOZ(E) 906253. 1♀: ditto, IOZ(E) 906254.Leptomiaselongitus is known from the province of Sichuan (Derong) and Xizang (Mangkang). It is narrowly distributed in the southwest of Sichuan : 58.PageBreak者: 陈元清  and IOZ(E) 906860. 1 ♀: (white): 四川宝兴  / 硗碛  2200–2700m [printed] / 1963.VI. [printed] 25 [handwritten] / 天津自然博物馆 ; (white): 采集者: 熊江 ; : IOZ(E) 906836. 2 ♀: ditto, IOZ(E) 906837, IOZ(E) 906830.1 ♂: (white): 四川  康定瓦新  / 沟  1450m [handwritten] / 中国科学院 ; (white): 1983.VI–22 [handwritten] / 采集者 王書永; : IOZ(E) 906879. 1 ♂: (white): 四川  康定  / 2600m [handwritten] / 中国科学院 ; (white): 1983.V–30 [handwritten] / 采集者 王書永 ; : IOZ(E) 906854. 1 ♂: same data as 906854 except collecting date 1983.VI–25, 采集Leptomiasfoveicollis is widely distributed in Sichuan , Liaoning (Changtu) and Heilongjiang (Haerbing). From southwest to northeast of China, this species has a wide distribution range. Leptomiasfoveicollis is widely distributed in the central-western region of Sichuan : 409, fig. 7.Holotype, ♂: (white): 四川  马尔康  / 2500m [handwritten] / 中国科学院 ; (white): 1983.VIII.17 [handwritten] / 采集者 王書永 ; : HOLOTYPE; : IOZ(E) 905588.Leptomiasglobosus is an endemic species of China which recorded from Maerkang, central region of Sichuan  Insects of the Hengduan Mountains Region 2: 843, fig. 10.PageBreak905925. Paratypes: 1 ♀: same data as holotype except ALLOTYPE printed on sea-green paper and IOZ(E) 905926. 5 ♂, 3 ♀: same data as holotype except PARATYPE printed on yellow paper and IOZ(E) 905927–905934. 1 ♂: (white): 四川  泸定磨西海  / 螺沟 1550m  / 中国科学院 ; (white): 1982.IX.16 [handwritten] / 采集者 王書永 ; : PARATYPE; : IOZ(E) 905935. 1 ♀: ditto, IOZ(E) 905936. 1 ♀: same data as holotype except PARATYPE printed on yellow paper, 1500m, 采集者: 张学忠  and IOZ(E) 905937. 1 ♂: ditto, IOZ(E) 905938.Holotype, ♂: (white): 四川  泸定磨西  / 1650m [handwritten] / 中国科学院 ; (white): 1983.VI.20 [handwritten] / 采集者: 陈元清 ; : HOLOTYPE; : IOZ(E) PageBreakLeptomiasmoxiensis is also an endemic species of China. It is recorded only from Luding, the central region of Sichuan : 397–398, fig. 2.Holotype, ♂: (white): 西藏芒康  / 中国科学院 ; (white): 1977.9.15 [handwritten] / 采集者  李继均 ; : HOLOTYPE; : IOZ(E) 906290. Paratypes: 1 ♀: same data as holotype except PARATYPE printed on yellow paper and IOZ(E) 906291.1 ♂: (white): 四川  乡城  / 2900m [handwritten] / 中国科学院 ; (white): 1983.VI.28 [handwritten] / 采集者: 张学忠 ; : IOZ(E) 906309. 2 ♂: ditto, IOZ(E) 906310, IOZ(E) 906311. 1 ♀: same data as 906309 except collecting date 1982.VI.17, 采集者: 王書永  and IOZ(E) 906293. 2 ♀: ditto, IOZ(E) 906294, IOZ(E) 906296.Leptomiasnubilus is recorded from Sichuan (Xiangcheng) and Xizang (Mangkang). Xiangcheng is located in the southwest region of Sichuan : 406–407, fig. 4.PageBreak905521. 8 ♀, 4 ♂: same data as holotype except PARATYPE printed on yellow paper and IOZ(E) 905522–905528, IOZ(E) 905531, IOZ(E) 905537, IOZ(E) 905540–905542. 3 ♀, 6 ♂: same data as holotype except PARATYPE printed on yellow paper, collecting date 1983.VII.4 and IOZ(E) 905529, IOZ(E) 905530, IOZ(E) 905532–905536, IOZ(E) 905538, IOZ(E) 905539.Holotype, ♂: (white): 四川  德格  / 3200m [handwritten] / 中国科学院 ; (white): 1983.VII.6 [handwritten] / 采集者: 陈元清 ; : HOLOTYPE; : IOZ(E) 905520. Paratypes: 1 ♀: same data as holotype except ALLOTYPE printed on sea-green paper and IOZ(E) Leptomiasochrolineatus is endemic to China and collected only from Sichuan (Dege). It is narrowly distributed in the northwest region of Sichuan : 404–405, fig. 1.PageBreakprinted on yellow paper, collecting date 1981.VII.28 and IOZ(E) 905450, IOZ(E) 905456, IOZ(E) 905459, IOZ(E) 905460, IOZ(E) 905464, IOZ(E) 905475, IOZ(E) 905479, IOZ(E) 905484. 1 ♂, 2 ♀: same data as holotype except PARATYPE printed on yellow paper, collecting date 1981.VII.27 and IOZ(E) 905451, IOZ(E) 905469, IOZ(E) 905470. 1 ♂, 1♀: same data as holotype except PARATYPE printed on yellow paper, collecting date 1981.VII.27, IOZ(E) 905453, IOZ(E) 905455 and with  采集者 王書永 . 1 ♂, 1 ♀: same data as 905455 except collecting date 1981.VII.28 and IOZ(E) 905457, IOZ(E) 905480. 1 ♂, 1 ♀: same data as holotype except PARATYPE printed on yellow paper, collecting date 1981.VII.27, IOZ(E) 905476, IOZ(E) 905487 and with  采集者 张学忠 . 1 ♂, 1 ♀: same data as holotype except PARATYPE printed on yellow paper, IOZ(E) 905486, IOZ(E) 905490 and with  采集者 王書永 . 1 ♀: (white): 云南  维西白济汎  / 1780m [handwritten] / 中国科学院 ; (white): 1981.VII.12 [handwritten] / 采集者:  廖素柏  / : PARATYPE; : IOZ(E) 905504. 1 ♀: (white): 云南  维西白济汎  / 1780m [handwritten] / 中国科学院 ; (white): 1981.VII.10 [handwritten] / 19号  / 采集者 王書永  / : PARATYPE; : IOZ(E) 905505. 1 ♀: (white): 四川  德格  / 3200m [handwritten] / 中国科学院 ; 1983.VII.4 [handwritten] / 采集者: 陈元清 ; : PARATYPE; : IOZ(E) 905506.Holotype, ♂: (white): 云南  维西攀天  / 阁  2500m [handwritten] / 中国科学院 ; (white): 1981.VII.24 [handwritten] / 采集者:  廖素柏  / : HOLOTYPE; : IOZ(E) 905433. Paratypes: 1 ♀: same data as holotype except ALLOTYPE printed on sea-green paper and IOZ(E) 905434. 14 ♂, 5 ♀: same data as holotype except PARATYPE printed on yellow paper and IOZ(E) 905435–905437, IOZ(E) 905448, IOZ(E) 905473, IOZ(E) 905482, IOZ(E) 905483, IOZ(E) 905485, IOZ(E) 905488, IOZ(E) 905489, IOZ(E) 905491–905493, IOZ(E) 905495, IOZ(E) 905497–905500, IOZ(E) 905503. 7 ♂, 4 ♀: same data as holotype except PARATYPE printed on yellow paper,  采集者 张学忠  and IOZ(E) 905438, IOZ(E) 905442, IOZ(E) 905447, IOZ(E) 905452, IOZ(E) 905472, IOZ(E) 905477, IOZ(E) 905478, IOZ(E) 905481, IOZ(E) 905494, IOZ(E) 905501, IOZ(E) 905502. 12 ♂, 5 ♀: same data as holotype except PARATYPE printed on yellow paper, collecting date 1981.VII.26, IOZ(E) 905439–905441, IOZ(E) 905443, IOZ(E) 905446, IOZ(E) 905449, IOZ(E) 905454, IOZ(E) 905458, IOZ(E) 905462, IOZ(E) 905463, IOZ(E) 905465–905468, IOZ(E) 905471, IOZ(E) 905474, IOZ(E) 905496 and with 寄主: 黑桃 . 1 ♂, 2 ♀: same data as holotype except PARATYPE printed on yellow paper, collecting date 1981.VII.26, IOZ(E) 905444, IOZ(E) 905445, IOZ(E) 905461 and with  采集者 王書永 . 4 ♂, 4 ♀: same data as holotype except PARATYPE Leptomiassublongicollis is recorded from Sichuan (Dege)  Fig.  and YunnTaxon classificationAnimaliaColeopteraCurculionidaeHeteromiasthibetanus Faust, 1888. Stett. Entomol. Zeit, 49(7–9): 285–286.Leptomiasthibetanus (Faust): 1 ♂: Thibet / Oayrollv ; Coll J. Faust / Ankauf 1900 ; Type ; Staatl. Museum für / Tierkunde Dresden .PageBreak院 ; 1983.VI.10 [handwritten] / 采集者: 陈元清 ; : IOZ(E) 906817. 1 ♀: (white): 四川  红原  / 3500m [handwritten] / 中国科学院 ; 1983.VIII.27 [handwritten] / 采集者:  牛春来 ; : IOZ(E) 906819. 1 ♂: (white): 四川  贡嘎山  / 燕子沟 2500m  / 中国科学院 ; 1983.VI.8 [handwritten] / 采集者: 陈元清 ; : IOZ(E) 906823. 1 ♂: (white): 四川  泸定新兴  / 1900m [handwritten] / 中国科学院 ; 1983.VI.12 [handwritten] / 采集者: 陈元清 ; : IOZ(E) 906825.1 ♀: (white): 四川  贡嘎山  / 燕子沟 2500m  / 中国科学Leptomiasthibetanus is known from Sichuan  and Xizang. It is widely distributed from the central to northwest region of Sichuan : 405, fig. 2.Holotype, ♂: (white): 四川  德格  / 3200m [handwritten] / 中国科学院 ; 1983.VII.4 [handwritten] / 采集者: 陈元清 ; : HOLOTYPE; : IOZ(E) 905414. Paratypes: 1 ♀: same data as holotype except ALLOTYPE printed on sea-green paper and IOZ(E) 905415. 6 ♂, 3 ♀: same data as holotype except PARATYPE printed on yellow paper and IOZ(E) 905416–905424.Leptomiasvarians is an endemic species of China, collecting from Sichuan (Dege)  Fig. . L.variPageBreakTaxon classificationAnimaliaColeopteraCurculionidaeChen, 1992Leptomiaswenchuanensis Chen, 1992. In Chen S (Ed) Insects of the Hengduan Mountains Region 2: 843, fig. 11.PageBreakprinted] 柴怀成 ; (white): 四川  汶川  / 映秀  / 900m [handwritten] / 中国科学院 ; : PARATYPE; : IOZ(E) 905922. 1 ♀: (white): 四川  马尔康  / 2900m [handwritten] / 中国科学院 ; (white): 1983.VIII.18 [handwritten] / 采集者:  张学忠 ; : PARATYPE; : IOZ(E) 905923. 1 ♀: (white): 四川  马尔康  / 梦笔山  / 4000m [handwritten] / 中国科学院 ; (white): 1983.VIII.19 [handwritten] / 采集者: 王書永 ; : PARATYPE; : IOZ(E) 905924.Holotype, ♂: (white): 四川  汶川卧龙  / 1920m [handwritten] / 中国科学院 ; (white): 1983.VII.24 [handwritten] / 采集者 王書永 ; : HOLOTYPE; : IOZ(E) 905907. Paratypes: 1 ♀: (white): 四川  汶川  / 卧龙  1920m [handwritten] / 中国科学院 ; (white): 1983.VII.24 [handwritten] / 采集者 王書永 ; : ALLOTYPE; : IOZ(E) 905908. 2 ♂: same data as allotype except PARATYPE printed on yellow paper and IOZ(E) 905909, IOZ(E) 905910. 3 ♀: same data as holotype except PARATYPE printed on yellow paper, collecting date 1983.VII.25, 1780m and IOZ(E) 905411, IOZ(E) 905413, IOZ(E) 905414. 1 ♂: (white): 四川  汶川  / 卧龙  1780m [handwritten] / 中国科学院 ; (white): 1983.VII.25 [handwritten] / 采集者 王書永 ; : PARATYPE; : IOZ(E) 905912. 1 ♀: ditto, IOZ(E) 905415. 1 ♂: same data as 905912 except 1800m, 采集者 张学忠  and IOZ(E) 905416. 1 ♀: (white): 四川  卧龙  / 2700m [handwritten] / 中国科学院 ; (white): 1983.VIII.9 [handwritten] / 采集者  牛春来 ; : PARATYPE; : IOZ(E) 905917. 1 ♀: (white): 四川  汶川  / 1300m [handwritten] / 中国科学院 ; (white): 1983.IX.13 [handwritten] / 采集者: 张学忠 ; : PARATYPE; : IOZ(E) 905918. 1 ♀: (white): 四川  汶川  / 三圣沟 2500m  / 中国科学院 ; (white): 1983.VIII.6 [handwritten] / 采集者: 柴怀成 ; : PARATYPE; : IOZ(E) 905919. 1 ♀: (white): 四川  汶川木江坪 [Wènchuān Mùjiāngpíng handwritten] / 1200m [handwritten] / 中国科学院 ; (white): 1983.VIII.8 [handwritten] / 采集者: 柴怀成 ; : PARATYPE; : IOZ(E) 905920. 1 ♀: (white): 四川  汶川映 秀 [Wènchuān Yìngxìu handwritten] / 900–1000m [handwritten] / 中国科学院 ; (white): 1983.VIII.1 [handwritten] / 采集者: 张学忠 ; : PARATYPE; : IOZ(E) 905921. 1 ♀: (white): 1983.VIII.3 [handwritten] / 采集者  / 4200m [handwritten] / 中国科学院 ; (white): 1983.VII.13 [handwritten] / 采集者:  牛春来 ; : PARATYPE; : IOZ(E) 905958.Holotype, ♂: (white): 四川  康定  / 4200m [handwritten] / 中国科学院 ; (white): 1983.VII.13 [handwritten] / 采集者: 陈元清 ; : HOLOTYPE; : IOZ(E) 905953. Paratypes: 1 ♀: same data as holotype except ALLOTYPE printed on sea-green paper and IOZ(E) 905954. 2 ♂, 2 ♀: same data as holotype except PARATYPE printed on yellow paper and IOZ(E) 905955, IOZ(E) 905957, IOZ(E) 905959, IOZ(E) 905961. 1 ♂: (white): 四川  康定折多  / 山 垭口 4200m  / 中国科学院 ; (white): 1983.VII.13 [handwritten] / 采集者 王書永 ; : PARATYPE; : IOZ(E) 905956. 1 ♂: ditto, IOZ(E) 905960. 1 ♂: (white): 四川  康定  / 3100m [handwritten] / 中国科学院 ; (white): 1983.VI.24 [handwritten] / 采集者  陈元清 ; : PARATYPE; : IOZ(E) 905962. 1 ♀: ditto, IOZ(E) 905963. 1 ♀: (white): 四川  Leptomiaszheduoshanensis is another endemic species of China which recorded from Sichuan (Kangding). It is narrowly distributed in the central region of Sichuan comb. n.Leptomiasgranulatus Chao, 1980. Entomotaxonomia 2(1): 29.LeptomiaschenaeColeoptera 8: 89, 396 (replacement name for secondary homonymy). Alonso-Zarazaga & Ren, 2013. Catalogue of Palaearctic Geotragus is G.granulatus, not G.chenae, because of Art. 59.4 of the Code: “59.4. Reinstatement of junior secondary homonyms rejected after 1960. A species-group name rejected after 1960 on grounds of secondary homonymy is to be reinstated as valid by an author who considers that the two species-group taxa in question are not congeneric, unless it is invalid for some other reason.”The correct name for this species under PageBreakshort and stout, exceeding anterior margin of eyes but not surpassing middle of eyes. Funicles with desmomere 1 elongate clavate, apical stout, 1.70× longer than desmomere 2, distinctly wider than 2. Prementum with four setae. Prothorax transverse, sides evenly rounded, broadest behind middle, pronotum with extremely shallow, fine, incomplete, median longitudinal groove. Elytral interstriae slightly elevated, unequal in width, without tubercles. Proventriculus : 四川康定 [Sìchuān Kāngdìng] / 2400–2700 公尺  / 中国科学院 [Zhōngguó Kēxuéyuàn]; (white): 1963.VII.28 [handwritten] / 采集者 张学忠 ; : HOLOTYPE; : IOZ(E) 906893.PageBreakPageBreakPageBreak29.68912 / E 102.06859 / 中国科学院 [Zhōngguó Kēxuéyuàn]; : IOZ(E) 1506001–1506006, IOZ(E) 1506022, IOZ(E) 1506032, IOZ(E) 1506033, IOZ(E) 1506052–1506057, IOZ(E) 1506063, IOZ(E) 1506064. 1 ♂, 2 ♀: same data as 1506064 except 和平组 [Hépíngzǔ], 1845m, 2011.VI.15, N 29.64937, E 102.09164, IOZ(E) 1506027, IOZ(E) 1506031 and IOZ(E) 1506062. 3 ♂, 5 ♀: same data as 1506064 except 火草坪 [Huǒcǎopíng], 2116m, 2011.VI.30, N 29.512675, E 102.133512, IOZ(E) 1506035, IOZ(E) 1506040–1506045 and IOZ(E) 1506061. 1 ♂: (white): 四川  泸定磨西  / 1650m [handwritten] / 中国科学院 ; (white): 1983.VI.20 [handwritten] / 采集者: 陈元清 ; : IOZ(E) 907104. 1 ♂: same data as 907104 except 采集者 王書永  and IOZ(E) 907106. 1 ♀: same data as 907104 except 1500m, 采集者: [Cǎijízhě] 张学忠  and IOZ(E) 907009. 1 ♀: (white): same data as 907104 except collecting date 1983.VI.17, and IOZ(E) 907010. 6 ♂, 7 ♀: same data as 907104 except 四川泸定 , 1800m, 1983.VI.14 and IOZ(E) 906944–906956.8 ♂, 9 ♀: : 四川甘孜州泸定县 [Sìchuān Gānzī Zhōu Lúdì Xiàn] / 折田坝 [Zhétiánbà] 2110m / 2011.VII.03 / 中国科学院 [Zhōngguó Kēxuéyuàn]; : leg. 张华康 [Zhāng Huákāng] / N Geotragusgranulatus mainly occurs northeast and east of Gongga Mountain, which is the highest mountain in Sichuan province, China.Leptomias occurring in Sichuan Province, accounting for 14 % of the species presently known from China. Seven of them appear to be endemic to an area that stretches from Dege County to Wenchuan County (Fig. Leptomias diversity in China. The species occur in Sichuan at elevations between 900 and 4200 m, in a geographical rectangle delimited by 31°48.600'N 98°34.120'W and 31°27.600'N 103°36.600'W. Geotragus is recorded for the first time from Sichuan, which also presents a new northern-most record for the genus. Geotragus currently comprises 13 species, six of which occur in China.There are 12 species of nty Fig. . Seven e


This significantly improves the previously best upper bound of 1.6667n by Fomin et al. [STOC 2016] and 1.6740n by Gaspers and Mnich . Our new upper bound almost matches the best‐known lower bound of 21n/7≈1.5448n, due to Gaspers and Mnich. Our proof is algorithmic, and shows that all minimal FVS of tournaments can be enumerated in time O(1.5949n).We study feedback vertex sets (FVS) in tournaments, which are orientations of complete graphs. As our main result, we show that any tournament on  This problem belongs to Karp's original list of 21 NP‐hard problems The minimum fvs problem remains NP‐hard even in tournaments T is a digraph with exactly one arc between any two of its vertices. Various approaches have been suggested to solve the minimum fvs problem on tournaments, including approximation algorithms M(T) of minimal FVS in T. Therefore, using this approach, the complexity of the minimum fvs problem in tournaments is within a polynomial factor of the maximum of M(T) over all n‐vertex tournaments, which we denote by M(n).The M(n) was Moon 1.4757n≤M(n)≤1.7170n. This was improved by Gaspers and Mnich 1.5448n≤M(n)≤1.6740n. Very recently, an improvement on the upper bound was made by Fomin et al. M(n)≤1.6667n. The problem of exactly determining M(n) was explicitly posed by Woeginger The first one to provide nontrivial bounds on 1.1M(n). Our main combinatorial result is as follows:Theorem 1n has at most M(n)≤1.5949n minimal FVS.Any tournament of order In this article, we make significant progress on establishing better bounds for M(n) is attained by regular tournaments. For regular tournaments, we show an upper bound on M(n) that matches the lower bound:Theorem 2n has at most 21n/7 minimal FVS, and this is sharp: some regular tournament of order n has exactly 21n/7 minimal FVS.Any regular tournament of order We also consider regular tournaments , because the best‐known lower bound on M(n):The following Table n of nodes in the input tournament T. Starting with T, they consider a vertex v with maximum out‐degree Δ, and depending on the value of Δ and neighbors of v, they construct subtournaments by deleting distinct vertices, such that each maximal transitive vertex set of T is contained in at least one subtournament. Applying the induction hypothesis to the subtournaments then implies their upper bound.Our proof of Theorem M for the maximum number of maximal transitive vertex sets in a tournament of order n containing a fixed set of k vertices, and we will show that M≤1.5949n−k for all 0≤k≤n. A similar approach has been used by Gupta et al. r‐regular induced subgraphs in undirected graphs.Here, we use a refined technique, that yields upper bounds on the number of inclusion‐maximal vertex sets with certain properties. Namely, in addition to deleting vertices to generate subtournaments, we also keep fixed vertex sets. Within these subtournaments we only consider maximal transitive vertex sets that contain all the fixed vertices. We introduce a new function n can be listed in time O(1.5949n). Second, using an algorithm by Gaspers and Mnich Corollary 1T of order n, all its minimal FVS can be listed in time M(T)·nO(1)=O(1.5949n) with polynomial delay and in polynomial space.Given any tournament Our combinatorial result has algorithmic consequences. First, our proof of Theorem v with in‐degree 0 in a subtournament induced by a maximal transitive vertex set. Brandt et al. NP‐complete, a feasible approach to determine the Banks set is to enumerate all minimal FVS. For this purpose, Brandt et al. Enumerating the minimal FVS in tournaments has several interesting applications. For example, Banks 2tournamentT= is a directed graph with exactly one edge between each pair of vertices. We denote the set of all tournaments with n vertices by Tn. A feedback vertex set (FVS) of T is a set F⊆V(T) such that T−F is free of (directed) cycles, where T−F is the induced subgraph of T after removing all vertices in F. An FVS is minimal if none of its proper subsets is an FVS.A M(T) the number of minimal FVS in a tournament T, and definen.Denote by T= be a tournament. For a set V′⊆V, let T[V′] be the subtournament of T induced by V′. For each v∈V, let N−(v)={u∈V|∈A} and let N+(v)={u∈V|∈A}. We write v→u if u∈N+(v) and call v a predecessor of u and u a successor of v. For each v∈V, its in‐degree is d−(v)=|N−(v)| and its out‐degree is d+(v)=|N+(v)|; call Tregular if all its vertices have the same out‐degree. Let Δ+(T) denote the maximum out‐degree over all vertices of T. Further, T is strong if there is a directed path from v to u for each pair of vertices v,u∈V; let Tn★ denote the set of strong tournaments of order n. Note that any tournament can uniquely be decomposed into strong subtournaments S1,⋯,Sr such that v→u for all v∈V(Si), u∈V(Sj) for all i<j.Observation 1T, we obtain M(T)=M(S1)·⋯·M(Sr).For any tournament Let M(n) from above by βn for some β by considering strong tournaments of every order n.Therefore, we can bound Lemma 1In a tournament, any vertex contained in a cycle is contained in a directed triangle.Our proofs will use the following well‐known observation about cycles in tournaments:v1,⋯,vℓ be a shortest cycle containing v1 with ℓ>3, vi→vi+1 for all i∈{1,⋯,ℓ−1} and vℓ→v1. Depending on the orientation of the arc between v1 and v3, either v1,v2,v3 form a triangle or v1,v3,v4,⋯,vℓ is a shorter cycle containing v1. ▪Let Henceforth, throughout the article by “triangle” we always mean “directed triangle.”transitive if its induced subtournament is acyclic. Thus, a vertex set is a maximal transitive vertex set if and only if its complement is a minimal FVS. Instead of counting minimal FVS, we count maximal transitive vertex sets. The next property of maximal transitive vertex sets was already used by Moon Lemma 2T, M(T)≤∑v∈V(T)M(d+(v)).For any tournament We call a vertex set W of T has a vertex v with in‐degree 0 in T[W]. Hence, W is also a maximal transitive vertex set in T[N+(v)∪{v}]; this yields the bound. ▪Any maximal transitive vertex set M(T) in terms of a recurrence relation, in particular in combination with the next lemma that extends Lemma Lemma 3n∈N and let T∈Tn★. Then either T is regular, or for any d∈N at most 2d vertices in T have out‐degree at least n−d−1.Let Lemma V∼ be the set of vertices in T with out‐degree at least n−d−1. Then any vertex in V∼ has in‐degree at most d. Hence,Let V∼≠∅, for otherwise the statement of the lemma holds. We distinguish two cases.We may suppose that V∼≠V(T). Then, since T is strong and V∼≠∅, there is some arc from V(T)∖V∼ to V∼. There are |V∼|2 arcs between vertices in V∼. Therefore, ∑v∈V∼|N−(v)|≥|V∼|2+1. Combining this inequality with d∈N yields |V∼|≤2d.Consider first the case that V∼=V(T). We may suppose that T is not regular, for otherwise the statement of the lemma holds. Note that not every vertex of V∼=V(T) can have in‐degree exactly d, since T is not regular. Hence, some vertex in V∼ has in‐degree at most d−1. Consequently,|V∼|2 arcs between vertices in V∼. Thus, ∑v∈V∼|N−(v)|≥|V∼|2. Combining these two inequalities and solving for d∈N yields |V∼|≤2d. ▪Second, consider the case that d vertices of out‐degree at least n−d−1, as witnessed for instance by the triangle and d=1.We remark that a regular tournament may have more than 23M(n) of minimal FVS in any tournament of order n is bounded from above by 1.5949n. For this purpose, for a tournament T and V′⊆V(T) let M be the number of maximal transitive vertex sets in T that contain all vertices in V′. Also, letM(n)=M.In this section, we show that the maximum number Example.To clarify the definition, we compute M. Precisely, we show that M=2. There are two nonisomorphic tournaments for n=3:T1 is acyclic and thus has only a single maximal transitive vertex set, V(T1). Thus, M=1 for all v∈V(T1). The tournament T2 has three maximal transitive vertex sets, each consisting of exactly two vertices. Thus, each vertex of T2 is contained in exactly two maximal transitive vertex sets. This yields M=2 for all v∈V(T2). Summarizing, we get M=2.The tournament β=1.5949. We will show that M≤βn−k for all n∈N and k∈{0,⋯,n}. To this end, ideally we would like to prove the following statements:(I)It holdsM≤βn−kfor alln≥k>0.(II)It holdsM≤βn.Unfortunately, we are unable to do prove these directly. The reason is that our proof of Statement (I) for a fixed pair  with n≥k>0depends on the validity of Statement (II) for values n∼<n. Vice versa, our proof of the validity of Statement (II) for fixed n∈N depends on the validity of Statement (I).Henceforth, fix Lemma 4n∈N. If M(n∼)≤βn∼ and M≤βn∼−k∼ holds for all 0<k∼≤n∼<n, then M≤βn−k for 0<k≤n.Let We will therefore establish the following two lemmas:Lemma 5n∈N. If M(n∼)≤βn∼, M≤βn∼−k∼ and M≤βn−k∼ for all 0<k∼≤n∼<n, then M(n)≤βn.Let The proof of Lemma The proof of Lemma We are ready to prove Theorem Proof of Lemma 1n∈N, it holds M(n)≤1.5949n. Clearly, M(1)≤1≤1.5949 and M≤1≤1.59491−k for all k∈{0,1}. This yields our induction hypothesis. Lemma M(n) for all n∈N. ▪We show that for all 4 be a minimum counterexample, that is, T is a tournament and V′⊆V(T) such that |V(T)|−|V′| is minimum and M>β|V(T)|−|V′|. Throughout this section, write n=|V(T)| and k=|V′|>0.In this section, we prove Lemma M≤βn−k for each of them; this yields the desired contradiction (and hence the truth of the statement of the lemma). In each case, we will use the minimality of  to bound M from above.Case 1:V′ form a triangle.Three vertices in M=0≤βn−k.Then, as no transitive vertex set contains all of these three vertices, Case 2:V′ form a triangle with some vertex v∈V(T)∖V′.Two vertices in V′ does not contain v. Hence,Any transitive vertex set that contains all vertices in Case 3:v∈V′ that is not contained in any cycle of T.There is a vertex W⊇V′ is a maximal transitive vertex set of T if and only if W∖{v}⊇V′∖{v} is a maximal transitive vertex set of T−v. This yieldsThen, a set We will distinguish several cases and show that Remark 1M(n∼)<βn∼ for n∼<n. The reason is that possibly V′∖{v}=∅, in which case k−1=0 and we need that M≤βn−1.We remark that it is this case where we rely on the validity of Lemma  to which Cases 1–3 do not apply.Observation 2, then (i) any vertex of V′ is contained in at least one triangle ≤β0n−k for β0=1.6181 (under the conditions imposed by the lemma). By  Observation v∈V′ that forms a triangle with two vertices w1,w2∉V′. Any maximal transitive vertex set W⊇V′ (and thus containing v) cannot contain both w1 and w2. Therefore, w1∈W implies w2∉W and we getβ0n for β0=1.6181.We remark that with Case 1–3 we can already show a bound of β0=1.6181 to β=1.5949.The subsequent cases allow us to improve Case 4:w∉V′ that is contained in two distinct triangles, both of which contain a vertex from V′ (possibly shared by both triangles). Then we are in one of two cases, where vertices in V′ are circled:There is a vertex , be distinct triangles containing w, such that v1,v2∈V′ where possibly v1=v2. Let W be a maximal transitive vertex set of T containing V′. Then either w∉W or w∈W. Clearly, if w∈W then u1,u2∉W. We therefore haveβn−k, since β≥1.4656.Case 5:v∈V′ and w1,w2∈V(T)∖V′ that form a triangle, such that w1 also belongs to triangles , for some u1,u2,u3∈V(T)∖{v,w2}.There are vertices Let u1,u2,u3∈V(T)∖V′, as otherwise Case 2 or Case 4 would apply. Any transitive vertex set W⊇V′ either contains w1 or not. If w1∈W then w2∉W. Moreover, w1∈W implies that either u2∉W, or u2∈W but u1,u3∉W. Thus,βn−k, since β≥1.5702.Then we can assume that . Then some vertex v0∈V′ forms a triangle with some w1,w2∈V(T)∖V′, as Cases 1–3 do not apply. For i=1,2, let Δi be the set of triangles ti= that are disjoint from w3−i and for which T is acyclic for all v′∈V′. Consequently, all triangles in Δ1∪Δ2 are disjoint from V′, as Case 4 does not apply. Further, all triangles in Δi are pairwise edge‐disjoint (as Case 5 does not apply), and therefore intersect only in wi.Henceforth, we assume that Cases 1–5 do not apply to M, we again distinguish the maximal transitive vertex sets that contain w1 or w2, from those that do not contain either of them. Let W be a maximal transitive vertex set of T containing V′.To prove an upper bound on w1,w2∉W. Then, T[W∪{wi}] contains a cycle for i=1,2, by maximality of W. Thus, by Lemma t= for some z1,z2∈W. We have that t∈Δi, since z1,z2 do not form a triangle with any v′∈V′ as z1,z2∈W. Thus, those W with w1,w2∉W can be partitioned into |Δi| classes, where the r‐th class contains the sets W that contain the two vertices of the r‐th triangle in Δi.First consider that Δ1∪Δ2. Consider two triangles tir=,tis=∈Δi:To use this argument effectively, we need some further observations about the relation among triangles in wi are pairwise edge‐disjoint (as Case 5 does not apply), the edge between uir and vis has to be directed from vis to uir; else, wi,uir,vis would form a triangle that is not edge‐disjoint from the triangle wi,uir,vir. Likewise, the edge between uis and vir has to be directed from vir to uis. Ignoring symmetries obtained by swapping the roles of tir and tis, there are only two possibilities how the two remaining edges  can be oriented:Since all triangles that contain Case A, and to the situation in the right figure as Case B. Note that in Case A,  and  form triangles; while in Case B, triangles are formed by  and .Observation 3uis,vis∈W implies that uir,vir∉W. In Case B, uir,vir∈W implies that vis∉W; and uis,vis∈W implies that uir∉W.In Case A, We refer to the situation in the left figure as tir=∈Δi let Vtir be the set of vertices that are excluded from those W with uir,vir∈W due to Observation Thus, for each 1 and Δ2 are vertex‐disjoint. Therefore, for each tir∈Δi, every vertex in Vtir is not contained in any triangle of Δ3−i. This implies that for any pair of triangles t1∈Δ1,t2∈Δ2 the sets Vt1,Vt2 are disjoint. Altogether, this means that we can bound the number of maximal transitive vertex sets W⊇V′ not containing w1,w2 from above byIn Lemma i∈{1,2}. Let ti1,⋯,ti|Δi| be an ordering of the triangles in Δi such that |Vtir|≤|Vtis| for 1≤r<s≤|Δi|. Then for any pair r,s∈{1,⋯,Δi} with r≠s, Observation r<s, since β≥1, we getr<s,|Vtir|=2(r−1) for all r=1,⋯,|Δi|. We obtain(β2β2−1)·(β2β2−1)=β4(β2−1)2.Thus, our goal is now to bound (⋆). Fix 1 is disjoint from every triangle in Δ2.Lemma 6v0,w1,w2,Δ1, and Δ2 be defined as before. Then any triangle in Δ1 is vertex‐disjoint from every triangle in Δ2.Let Let us now prove that indeed any triangle in ΔV′ is a transitive set, as Case 1 does not apply. Thus, the vertices in V′ admit a topological order such that vx′→vy′ for all vx′,vy′∈V′ with x>y. Second, for each vertex z∈V(T)∖V′ the set V′∪{z} is a transitive set, as Case 2 does not apply. Therefore, the vertices of V(T)∖V′ can be partitioned into layers Z1,⋯,Zℓ such that for each z∈Zr, z→vs′ if and only if s<r.First note that i=1,2, the vertices of any triangle ∈Δi all belong to the same layer. This implies in particular that for i=1,2, all vertices in triangles of Δi belong to the same layer. Since w1 and w2 are in different layers , this shows that any triangle in Δ1 is vertex‐disjoint from any triangle in Δ2.We claim that for i∈{1,2} and let ∈Δi be a triangle with wi→uir,uir→vir,vir→wi. Suppose that uir∈Zu,vir∈Zv,wi∈Zw for some u,v,w∈{1,⋯,l}. So we must show that u=v=w to prove the claim.To show the claim, let u<w then wi,uir,vu′ form a triangle, contradicting that Case 4 does not apply. If v>w then wi,vir,vv′ form a triangle, again contradicting that Case 4 does not apply. Hence, v≤w≤u holds. If v<u then uir,vir,vv′ form a triangle, contradicting the definition of Δi. So indeed u=v=w, and the claim holds. ▪If W⊇V′ that contain exactly one of w1,w2 . Overall, if Cases 1–5 do not apply, with the obtained bound on (⋆), by βn−k, since β≥1.5703.To complete the proof of Lemma This completes the proof of Lemma 5M(n)≤1.5949n for all n∈N. For sake of contradiction, suppose that the statement of the lemma is not true. Let T∈Tn be a counterexample with minimum number of vertices. By Observation T is strong. In particular, any vertex of T has out‐degree at most n−2.We will show by induction that T satisfies the following stronger restriction:T is regular, or for any d∈N at most 2d vertices have out‐degree at least n−d−1.First suppose that T is regular, then any vertex has out‐degree exactly (n−1)/2. Using that M(T)≤∑v∈V(T)M(d+(v)) by Lemma n≥11 we obtain 21n/7>n·1.5949n−12, and so any regular tournament with at least 11 vertices has at most 21n/7 maximal transitive vertex sets. For n≤9, the inequality M(n)≤21n/7 was shown explicitly by Gaspers and Mnich If d∈N, at most 2d vertices have out‐degree at least n−d−1. Let v1,⋯,vn be a labeling of the vertices of T such that d+(v1)≥d+(v2)≥⋯≥d+(vn). Then Lemma d∈N,d=i−12 implies i>2d, and we obtainSo we may assume that for any βn since β≥1.5462. This completes the analysis of tournaments T satisfying It follows thatM(T) in each of the four cases Δ+(T)=n−i for i=2,3,4,5. In each case, we bound M(T) via M(n∼) for values n∼<n and the result of Lemma branching vectorb=, where t is the number of branches we consider of how a particular maximal transitive vertex set W of T can look like. Each number bi,i=1,⋯,t is a positive integer that is the sum of Δn and Δk, where Δn is the number of vertices by which the tournament order decreases in that branch and Δk is the number of vertices by which the parameter kincreases. It might be that one of Δn and Δk is equal to zero in some branch, but the sum bi=Δn+Δk is always positive. We will show that M(T)≤∑i=1tβn−bi≤βn in each case.We will now work toward removing assumption M(T), we classify all maximal transitive vertex sets of T. Let W be a maximal transitive vertex set of T. We branch on carefully selected vertices whether they belong to W or not. These choices either exclude certain other vertices from W (by acyclicity of W) and thus yield Δn>0; they force certain other vertices to be included into W, based on the following observation:Observation 4W be a maximal transitive vertex set of T, and let v be a vertex of T. If v∉W then at least one predecessor of v in T belongs to W. Equivalently, if no predecessor of v belongs to W, then v∈W.Let To bound of v. These choices and their implications will be depicted by case trees that also show the pair  for each branch.Case 1:Δ+(T)=n−2v★ be a vertex with maximum out‐degree and unique predecessor b. We distinguish the following subcases.Case 1.1:d+(b)=n−2b∼ be the only predecessor of b. Let W be a maximal transitive vertex set of T.Let b∉W, we have that v★∈W by applying Observation v=v★. We further have that b∼∈W by applying Observation v=b.When W can be categorized as follows, with branching pairs :This shows that b=, which solves to 1.4656.This yield a branching vector Case 1.2:d+(b)≤n−3W be a maximal transitive vertex set of T. As in Case 1.1, at least one of v★,b belongs to W, by Observation v★,b∈W then no predecessor of b belongs to W, as T[W] is acyclic. Since b has at least two predecessors, these observations yield:Let Let This yields a branching vector of , which solves to 1.5538.β≥1.5538.This completes the analysis of Case 1, where we used that Case 2:Δ+(T)=n−3V★ be the set of vertices with maximum out‐degree.Let Lemma 7v★∈V★ has two predecessors of out‐degree exactly n−3 or two predecessors of out‐degree at most n−4, or V★ induces a triangle.Either some vertex We will apply this observation with various choices for V★ contains at most four vertices.By Lemma V★ contains a single vertex v★ then its two predecessors are outside V★, and thus have out‐degree at most n−4.If V★ consists of two vertices v1★,v2★ then assume v1★→v2★. Thus, the two predecessors of v1★ are outside V★, and thus have out‐degree at most n−4.If V★ consists of three vertices v1★,v2★,v3★ then we may assume that V★ does not induce a triangle . Thus, we can assume that vi★→vj★ for 1≤i<j≤3. Thus, the two predecessors of v1★ are outside V★, and thus have out‐degree at most n−4.If V★ consists of four vertices. Then the subtournament T[V★] has exactly six arcs. As each vertex in V★ has two incoming arcs, this implies that for some vertex in V★ its two predecessors are also in V★ and therefore have out‐degree n−3. ▪Finally, assume that v★∈V★ has two predecessors with out‐degree exactly n−3 (Case 2.1),some v★∈V★ has two predecessors with out‐degree at most n−4 (Cases 2.2–2.7),or some T has exactly three vertices with out‐degree n−3 that form a triangle (Case 2.8).or Therefore, we can distinguish whetherN−(v★)={b1,b2} and b1→b2.Case 2.1:d+(b1)=n−3 and d+(b2)=n−3b1 and b2 have at most one common predecessor. Let N−(b1)={u1,u2}.Case 2.1.1:N−(b1)∩N−(b2)=∅.Then N−(b2)={b1,u3} for some u3∉N−(b1). Note that if v★∉W,b2∈W, then u3∈W.  Then we branch as follows:Let Case 2.1.2N−(b1)∩N−(b2)={u2}.This yields a branching vector of  that solves to 1.5748.In Cases 2.1–2.7, let Then we branch as follows:Case 2.2:d+(b1)≤n−4, d+(b2)≤n−4 and b1,b2 share no common predecessor.b1 forms a triangle with b1,b2 and therefore any maximal transitive vertex set containing both b1,b2 cannot contain any predecessor of b1. Then we branch as follows:Any predecessor of This yields a branching vector of  that solves to 1.5923.Case 2.3:d+(b1)≤n−4, d+(b2)=n−4 and one predecessor of b2 is a predecessor of b1N−(b1)={w1,...,wl} with l≥3 and let N−(b2)={b1,u,w1}, so that w1 is the unique predecessor common to b1,b2.Case 2.3.1:wi→u for some i∈2,...,l.Then {u,b2,wi} induces a triangle.Let Then we branch as follows:Case 2.3.2:u→wi for all i∈2,…,l{u,wi,b1} induces a triangle for all i∈2,…,l.Then Then we branch as follows:This yields a branching vector of  that solves to 1.5780.This yields a branching vector of  that solves to 1.5772.Case 2.4:d+(b1)=n−4, d+(b2)=n−4 and all predecessors of b2 other than b1 are predecessors of b1u be the only predecessor of b1 that is not a predecessor of b2. Then we branch as follows:Let This yields a branching vector of  that solves to 1.5912.Case 2.5:d+(b1)≤n−5,d+(b2)=n−4 and all predecessors of b2 are predecessors of b1Then we branch as follows:This yields a branching vector of  that solves to 1.5820.Case 2.6:d+(b1)=n−4 and d+(b2)≤n−5b1,b2 have no common predecessor in Case 2.2. Hence, b1,b2 have either one, two, or three common predecessors.Case 2.6.1:b1 and b2 have exactly one common predecessor.Then we branch as follows:We already considered that Case 2.6.2:b1 and b2 share exactly two predecessors.sb2<n−5, or b2 has exactly one predecessor that is not a predecessor of b1. Let u1 be the only predecessor of b1 that is not a predecessor of b2. Note that any maximal transitive vertex set that contains n,b2 but not b1 has to contain u1.Then either For sb2<n−5 we branch as follows:This yields a branching vector of  that solves to 1.5923.This yields a branching vector of  that solves to 1.5817.sb2=n−5 and u2 being the only predecessor of b2 that is not a predecessor of b1, we branch as follows:For Case 2.6.3:b1 are predecessors of b2.All predecessors of v★,b2 also contains b1. Then we branch as follows:Then any maximal transitive vertex set containing This yields a branching vector of  that solves to 1.5861.This yields a branching vector of  that solves to 1.5702.Case 2.7:d+(b1)≤n−5 and d+(b2)≤n−5Then we branch as follows:This yields a branching vector of  that solves to 1.5912.Case 2.8:n−3 and they form a triangle.There are exactly three vertices with score This yields a branching vector of  that solves to 1.4736.v1,v2,v3 and assume v1→v2, v2→v3, and v3→v1.Call these vertices v1,v2,v3 have distinct predecessors, and let bi be the predecessor of vi.Suppose that vi,vi+1,bi form a triangle and that bi+1 is the only predecessor of vi+1 other than vi. Therefore, we branch as follows:Note that This yields a branching vector of  that solves to 1.5903.v1,v2 have a common predecessor b and b3 is the predecessor of b3.Now suppose that v3 is the only vertex where the orientations of the arcs incident to v1 differ from those incident to v2, any maximal transitive vertex set that does not contain v3 either contains both v1,v2 or none of them. We branch as follows:Since This yields a branching vector of  that solves to 1.5702.b, any maximal transitive vertex set that contains some vertex of v1,v2,v3 contains at exactly two of them. We branch as follows:If all three vertices have a common predecessor Case 3:Δ+(T)=n−4V★ be the set of vertices with out‐degree n−4. Since T is strong, some vertex v★∈V★ has a predecessor in V(T)∖V★. Let b1,b2,b3 be the predecessors of v★. For i=1,2,3 let s∼i=|N−(bi)∖{b1,b2,b3}|; assume, without loss of generality, that s∼1≥s∼2≥s∼3.Let v★ we have |N−(b1)|+|N−(b2)|+|N−(b3)|≥10 and hence s∼1+s∼2+s∼3≥7. Note that |N−(bi)|≥3 for i=1,2,3 and |N−(bj)|≥4 for some j∈{1,2,3}. Therefore s∼1≥3, s∼2≥2, s∼3≥1.By the choice of s∼1,s∼2,s∼3.Case 3.1:s∼3=1If s∼3=1, then either s∼2≥3 or s∼1≥4. Let {u}=N−(b3)∖{b1,b2,b3}. Then we branch as follows:We will estimate bounds for different values of Case 3.2:s∼1≥4 or s∼2≥3, s∼3≥2We branch as follows:This yields a branching vector of  that solves to 1.5812.This yields a branching vector of  that solves to 1.5949.s∼2≥s∼3, we can now assume that s∼1=3,s∼2=2,s∼3=2.Case 3.3:s∼1=3,s∼2=2,s∼3=2 and all predecessors of b2 except the bi are predecessors of b3We branch as follows:Since Case 3.4:s∼1=3,s∼2=2,s∼3=2, and b2,b3 have exactly one common predecessor u1 in V(T)∖{b1,b2,b3}W be a maximal transitive vertex set, u2≠u1 be the other predecessor of b2 and u3≠u1 be the one of b3. If v★∈W but b2,b3∉W, then either u1∈W or u2,u3∈W, as otherwise b2 or b3 could be added to W. Therefore, we branch as follows:Let This yields a branching vector of  that solves to 1.5516.This yields a branching vector of  that solves to 1.5793.b2,b3 have no common predecessor except possibly b1.Case 3.5:s∼1=3,s∼2=2,s∼3=2, b2,b3 have no common predecessor in V(T)∖{b1,b2,b3} and two predecessors of b1 in V(T)∖{b1,b2,b3} are predecessors of bi for some i=2,3b3=bi and let u1 be the only predecessor of b1 that is not a predecessor of b3. Since s∼1=3, there is at least one predecessor of b2 that is not a predecessor of b1. We branch as follows:Assume, without loss of generality, that Henceforth, we can assume that Case 3.6:s∼1=3,s∼2=2,s∼3=2, b2,b3 have no common predecessor in V(T)∖{b1} and both b2 and b3 share at most one predecessor of b1.b1 shares no predecessor in V(T)∖{b1,b2,b3} with b2 and none with b3, orb1 shares one predecessor with each of b2 and b3, orb1 shares a predecessor with one of b2,b3 but none with the other; without loss of generality, let b1 share a predecessor with b3.There are three cases to consider:b1,b2, and b3 are forming a triangle.Case 3.6.1:b1 shares one predecessor in V(T)∖{b1,b2,b3} with b2 and one with b3.b1,b2, and b3 do not form a triangle.w1 the common predecessor of b1 and b2 and let w2 be the common predecessor of b1 and b3. For i=1,2,3 let ui be the predecessor of bi that is neither w1 nor w2. We branch as follows:Let This yields a branching vector of  that solves to 1.5828.Case 3.6.2:b1 shares one predecessor in V(T)∖{b1,b2,b3} with b2 and one with b3.b1,b2, and b3 form a triangle.u1,u2,u3,w1,w2 be as in Case 3.6.1. Then we branch as follows:Let This yields a branching vector of  that solves to 1.5805.Case 3.6.3:b1 shares no predecessor in V(T)∖{b1,b2,b3} with b2 and one with b3.b1,b2, and b3 do not form a triangle.u3 the only predecessor of b3 that is not a predecessor of b1. Then we branch as follows:Let This yields a branching vector of  that solves to 1.5912.Case 3.6.4:b1 shares no predecessor in V(T)∖{b1,b2,b3} with b2 and one with b3.b1,b2, and b3 form a triangle.u3 be as defined in Case 3.6.3. Then we branch as follows:Let  This yields a branching vector of  that solves to 1.5889.Case 3.6.5:b1 shares no predecessor in V(T)∖{b1,b2,b3} with b2 and none with b3.b1,b2, and b3 do not form a triangle.s∼2=s∼3=2 but |N−(bi)|≤n−4 for i=2,3, there is a vertex bj≠bi, j=1,2,3 such that bj→bi. As the vertices b1,b2, and b3 form a transitive set, we have b1→b2 and b1→b3. Let u1,u2,u3 the predecessors of b1. Note that b2,b3,v★ and each ui for i=1,2,3 form a cycle with b1. Therefore, if W contains some ui, it can not contain b2,b3, or v★. If it contains b2,b3, or v★, it can not contain any ui. Thus, we branch as follows:Since  This yields a branching vector of  that solves to 1.5437.Case 3.6.6:b1 shares no predecessor in V(T)∖{b1,b2,b3} with b2 and none with b3.We will split each of the cases into two subcases where we consider if This yields a branching vector of  that solves to 1.5904.This yields a branching vector of  that solves to 1.5397.b1,b2, and b3 form a triangle. Assume without loss of generality that b1→b2,b2→b3 and b3→b1. Therefore all predecessors of b1 form a triangle with b1 and b2 and all predecessors of b3 form a triangle with b3 and b1. We branch as follows:                          This yields a branching vector of  that solves to 1.5800.Case 4:Δ+(T)=n−5Let V★ be the set of vertices with maximum out‐degree; Since T is strong, there is a vertex v★ that has a predecessor in V(T)∖V★.This completes the analysis of Case 3.N−(v★)={b1,b2,b3,b4}. By the choice of v★, the vertices b1,b2,b3,b4 together have at least 17 incoming arcs. Exactly 6 of these arcs are incident to two of b1,b2,b3,b4. The remaining 11 arcs are incoming from vertices in V(T)∖{b1,b2,b3,b4}. For i=1,⋯,4 let s∼i be the number of arcs incoming to bi from V(T)∖{b1,b2,b3,b4}, i.e. s∼i=|N−(bi)∖{b1,b2,b3,b4}|. Assume, without loss of generality, that s∼1≥s∼2≥s∼3≥s∼4.Observation 5s∼4≥1 and s∼3≥2. If s∼2=2, then s∼3=s∼4=2.It holds, Let |N−(bi)|≥4 for all i=1,2,3,4, we get |N−(bi)∖{b1,b2,b3,b4}|≥1. There is at most one vertex with with three incoming arcs from b1,b2,b3,b4. This yields s∼3≥2.Since s∼2=s∼3=2 and s∼4=1. If s∼4=1, then b1→b4, b2→b4, and b3→b4. Since s∼3=2, b1→b3, and b2→b3. But then we get s∼2≥3 that is a contradiction. ▪Assume that ; see Table M(T) in terms of pairs  for each of these six vectors in Table This leaves exactly six feasible vectors s∼1,s∼,s∼3,s∼4;s∼1,s∼,s∼3,s∼4;W that contains v★,bi cannot contain any predecessor of bi in V(T)∖{b1,b2,b3,b4,v★}.To this end, note that any maximal transitive vertex set βn since β≥1.5612.This yields for four of the casesM(T) is bounded byβn since β≥1.5691.For the remaining two cases, This last case completes the proof of Lemma 6M(n) of minimal FVS in n‐vertex tournaments, to 1.5448n≤M(n)≤1.5949n. It remains to determine the growth of M(n) exactly—Gaspers and Mnich M(n)≤21n/7≈1.5448n for all n∈N, and we repose this conjecture here.In this article, we narrowed the gap between the lower and upper bounds for the maximum number cn for some constant c<2, on the number of minimal FVS in general directed graphs. As far as we know, currently only a bound of 2n/n is known, implied by Sperner's Lemma.In a different direction, it would be interesting to prove nontrivial upper bounds of the form


E×B “polarization” drift.The relativistic trajectory of a charged particle driven by the Lorentz force is different from the classical one, by velocity-dependent relativistic acceleration term. Here we show that the evolution of optical polarization states near the polarization singularity can be described in analogy to the relativistic dynamics of charged particles. A phase transition in parity-time symmetric potentials is then interpreted in terms of the competition between electric and magnetic ‘pseudo’-fields applied to polarization states. Based on this Lorentz pseudo-force representation, we reveal that zero Lorentz pseudo-force is the origin of recently reported strong polarization convergence to the singular state at the exceptional point. We also demonstrate the deterministic design of achiral and directional eigenstates at the exceptional point, allowing an anomalous linear polarizer which operates orthogonal to forward and backward waves. Our results linking parity-time symmetry and relativistic electrodynamics show that previous PT-symmetric potentials for the polarization singularity with a chiral eigenstate are the subset of optical potentials for the  V(x) = V*(−x), providing physical observables even for non-equilibrium systems. The existence of real observables in PT-symmetric complex potentials has opened the field of non-Hermitian quantum mechanics6616n(x) = n*(−x), the physics of PT-symmetric potentials has been applied to the exotic control of light flows1791320222325272526293031With the universal existence of open systems1346891011121314Ẽ = mc2/[1 − (v/c)2]1/2, shows that observers in different frames will see different values of total energy for moving charged particles of velocity v. We note that this non-equilibrium condition results in non-Hermitian form of Hamiltonians, the necessary condition for the achievement of PT symmetry. In the context of the multidisciplinary realization of PT symmetry891011Meanwhile, it is known that relativistic electrodynamicsE×B drift  of charged particlesE×B “polarization” drift of light. The phase transition in PT-symmetric potentials6161525263738Inspired by the polarization equation of motion from the Schrödinger-like form of Maxwell’s equationsz-axis of the electrically anisotropic material, with the unity permeability (μ = 1). For the later discussion, we express the arbitrary permittivity tensor in the x-y plane, in terms of Pauli matrices26σ1–3 as ε = (ε2σ1 + ε3σ2 + ε1σ3), orConsider the planewave propagating along the εo(z) and ε1−3(z) have slowly-varying complex values, σ1 = , σ2 = , and σ3 = . By applying the spin (x ± iy)/21/2 basesdψe/dz = Hs·ψe with the temporal-like z-axisψe = T is the electric field amplitude of (±) optical spin waves ((+) for right-circular polarization (RCP) (x + iy)/21/2, and (−) for left-circular polarization (LCP) (x − iy)/21/2), and Hs is the traceless Hamiltonian expressed with Pauli matrices36Hs = (ε1σ1 + ε2σ2 + ε3σ3)/(iλ) for λ = 2εo1/2/k0 and the free-space wavenumber k0. If we assign the symmetry axis between x and y axes, the condition of PT-symmetric potentials15εo, ε2, and ε3, and imaginary-valued ε1. Note that imaginary-valued ε1 corresponds to the linear dichroism42ε2 represents the birefringence. ε3 represents the magneto-optical change of plasma permittivity induced by an external static magnetic fieldε3 = 0.where Sj = ψe†·σj·ψe . The change of the SOP can then be expressed in view of light-matter interactionsdψe/dz = Hs·ψe and its conjugate form dψe†/dz = ψe†·Hs†, which leads to the Lorentz pseudo-force equation of motion for the SOPIn this representation, the SOP of light is described by the Stokes parametersSn = T/S0 is the pseudo-velocity of the ‘hypothetical’ charged particle corresponding to the SOP of light, and E = 2·ImT/λ and B = −2·ReT/λ are the electric and magnetic pseudo-field in relation to imaginary- and real-parts of the permittivity, respectively. Note that 20∂tβ = E + β×B −(β·E)β. In this representation of optical polarization states, the acceleration of optical SOP comes from the Lorentz pseudo-force, F ~ dSn/dz Sn corresponds to the Joule effectβ·E)β in the relativistic equation of motion.where ~ dSn/dz . The firε1, real ε2, and zero ε3). While ε2 of the birefringence derives the circulating acceleration on the Poincaré sphere , PT-symmetric potentials naturally satisfy the ideal E×B drift34/λ, Sn·E  providesE×B drift34Δεeig for the Hamiltonian equation dψe/dz = Hs·ψe, as a function of the imaginary potential . First, before the EP where eigenvalues are real and non-degenerate (εi < εc), the magnetic pseudo-field is larger than the electric pseudo-field, resulting in the counter-directive acceleration of SOP (lower panels in d point in εi = εc), the equal magnitude of E = 2·εi·e1/λ and B = −2εc·e2/λ fields derives the suppression of total Lorentz pseudo-force on the southern Poincaré sphere, especially with the zero net force at the south pole Sn = 2·εi·e1/λ + e3 × 2εc·e2/λ = 0, εi > εc), the strong electric pseudo-field dominates the motion equation of the SOP, with the co-directive force (lower panels in B-dominant (before the EP), (ii) B = E (at the EP) and (iii) E-dominant regime (after the EP). It is worth mentioning that the stable point with the stationary polarization can also be obtained at the north pole by changing the sign of εc (converting the fast and slow axes for the birefringence) or εi (converting the gain and loss axes for the linear dichroism), allowing perfect RCP chirality. In terms of this Lorentz pseudo-force representation of SOP, we also note that PT-symmetric potentials15ε2 and imaginary-valued ε1 are the special case of the E×B drift with specific field vectors E = 2·εi·e1/λ and B = −2εc·e2/λ, implying the existence of unconventional polarization singularity at other SOPs  which will be discussed later.Based on the Lorentz pseudo-force equation of 616generate , εi < εcE×B drift (B| > |E|), the SOP for each spin simply rotates around the B field following the Sn×B of E×B drift34E×B axis (S3 axis) is thus reproduced by the slow evolution of SOPs near the Sn = −e3 on the Poincaré sphere , the (+) spin state converges to the (−) spin when the state approaches the south pole through the gyration by the magnetic pseudo-field  and linear dichroism42Im[εx] ≠ Im[εy] for different x- and y-dissipation). The spin-based Hamiltonian equation dψe/dz = Hs·ψe, for slowly-varying εx = εo + Δε(z) and εy = εo − Δε(z) and constant χ = χo with real-valued εo and χo, derives the Hamiltonian Hs for general chiral materials, in the form ofExtending the special case of the k2 = ω2·(μ0εo + χo2) (see χ(z)). The Pauli expressionHs = a1σ1 + a2σ2 + a3σ3 has the coefficients of a1 = −(ik/2)·(Δε/εo), a2 = −(ωχ/2)·(Δε/εo), and a3 = iωχ. The electric and magnetic pseudo-fields for and ρ = Δε/εo.with the degree of electrical anisotropy E and B driving the SOP are strongly dependent on the type of the anisotropy: birefringence  or linear dichroism (imaginary ρ) both satisfying the condition of the E×B drift (E⊥B). For the case of birefringence with E = −ωχoρ·e2 and B = −kρ·e1 + 2ωχo·e3, the pseudo-field satisfies |E| < |B| in most cases and the condition of |E| ≥ |B| enforces εo < −4χo2/(μ0·ρ2) and thus prohibits the existence of propagating waves at the EP. It is interesting to note that this restriction proves the necessity of complex potentials for obtaining the singularity; therefore we employ linear dichroism for achieving the EP for the propagating wave, by fulfilling the condition of |B| = |E|.ρ = i·ρi) chiral materials, which derive pseudo-fields of E = kρi·e1 and B = −ωχoρi·e2 + 2ωχo·e3 with the EP condition of ρi2 = 4χo2/(μ0εo) for |B| = |E|. The PT-symmetry-like phase transition , the linear-polarizing functionality in the structure of This directionality with achiral designer eigenstate at the EP allows the implementation of unconventional polarizers based on the polarization convergence at the EP. E×B drift toward a single direction for every initial velocity vectors.In summary, we found the link between the seemingly unrelated fields of PT symmetry optics and relativistic electrodynamics. This reinterpretation of PT symmetry brings insight to the singularity in polarization space, broadening the class of parity-time symmetric Hamiltonians in vector wave equations47485153How to cite this article: Yu, S. et al. Acceleration toward polarization singularity inspired by relativistic E×B drift. Sci. Rep.6, 37754; doi: 10.1038/srep37754 (2016).Publisher’s note: Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.


Scientific Reports6: Article number: 37077; 10.1038/srep37077 published online: 11282016 updated: 01162017.This Article contains a typographical error. In the Results section, under the subheading “SiC Sensor”“These intermediate coupled pair states can be described by the singlet-triplet basis: the symmetric triplet states, T+ = |↑↑〉, T0 = (|↑↓〉 + |↓↑〉)/, and T− = |↓↓〉, each having spin angular momentum S = 1 with ms = +1, 0, −1 respectively or the anti-symmetric singlet state, S0 = (|↑↓〉 + |↓↑〉)/ which has spin angular momentum S = 0 with ms = 0”.should read:“These intermediate coupled pair states can be described by the singlet-triplet basis: the symmetric triplet states, T+ = |↑↑〉, T0 = (|↑↓〉 + |↓↑〉)/, and T− = |↓↓〉, each having spin angular momentum S = 1 with ms = +1, 0, −1 respectively or the anti-symmetric singlet state, S0 = (|↑↓〉 − |↓↑〉)/ which has spin angular momentum S = 0 with ms = 0”.


There are two errors in the “miR-200/182 expression was induction in HSV-1 positive areas of the brain” section of the Results. “Fig B in S1 File” should be listed as “Fig D in S1 File” and “Fig C in S1 File” should be listed as “Fig E in S1 File”.in situ hybridization in brain tissue” section of the Results. In the first paragraph, “Fig D in S1 File” should be listed as “Fig B in S1 File”. In the second paragraph, “Fig E in S1 File” should be listed as “Fig C in S1 File”.There are two errors in the “Induction of miR-200/182 expression visualized by Panel C in


H-parallel case with a fixed incident wavelength. We denote η > 0 as the typical size of the complex structure and obtain the effective equations by letting η → 0. For metallic permittivities with negative real part, plasmonic waves can be excited on the surfaces of the slits. For the waves to be in resonance with the height of the metallic layer, the corresponding results can be perfect transmission through the layer.We investigate the transmission properties of a metallic layer with narrow slits. We consider (time-harmonic) Maxwell's equations in the  In , the reldone. In  where thdone. In –13.Two-scale convergence has proven to be a very efficient tool in homogenization theory while dealing with the problems where the underlying medium is heterogenous. The concept of two-scale convergence is first introduced by Nguetseng in   Two-scalγ, γ)× and therefore by defining a suitable test function, they have shown that the first component of j0 vanishes and they obtain j0 =   for  x ∈ R  and  y ∈ Y∖Σ. This is clearly not the case in this paper as the metallic subpart Σ is chosen to be sinusoidal along y2-axis due to the geometry of the metallic structure given by j0 and we end up having a different j0 compared to that in  and 1 ≤ r, s ≤ ∞ be such that 1/r + 1/s = 1. Assume that Ξ ∈ {Ω, Ση} and l ∈ ℕ0; then as usual Lr(Ξ) and Hl,r(Ξ) denote the Lebesgue and Sobolev spaces with their usual norms and they are denoted by ‖·‖r and ‖·‖l,r. For the sake of clarity if ϕ ∈ Lr(Ξ), thenϕ ∈ Hl,r(Ξ), thenα =  ∈ ℕn is a multi-index, |α | = α1 + α2 + ⋯+αn, and Dα = ∂α||/∂x1α1∂x2α2 ⋯ ∂xnαn. Similarly, θ,r, and θ are the Hölder, real, and complex interpolation spaces, respectively, endowed with their standard norms; for definition confer ,,j0 and Lloc2(Ω) which implies that up to a subsequence By , it follThe Field outside of R. For uη → ∇U uniformly on compact subsets of The Field in the Metal Part of R. We note that |aη| ≤ Cη2 in Ση and |aη| ≤ C in R∖Ση; therefore (jη‖L2(Ση)2 ≤ Cη2 and ‖jη‖L2(R∖Ση)2 ≤ C. This implies R × Σ. Moreover, j0 = j0(x) = ∇U(x) for a.e.  ∈ R × (Y − Σ).herefore  gives ‖jDivergence of j0. Due to boundedness assumption on uη, by (jη‖L2(Ω) ≤ k2  supη>0‖uη‖L2(Ω) < ∞, ∀η > 0. For Θ ∈ C0∞(Ω) and ψ ∈ Cper∞(Y), we test (x)ψ(x/η) which givesC0∞(Ω) is arbitrary, ∫Y∇yψ(y) · j0  dy = 0 for a.e. x ∈ Ω which implies ∇y · j0 = 0 for y ∈ Y in distributional sense. This shows that y; that is, x only. uη, by  we have  we test  by Θ(x)ψU as shown in [ψ ∈ L2:∇·ψ = 0, ψ is Y-periodic, and ψ = 0  in  Σ}. We choose a test function φ≔Θ(x)ψ(x/η), where Θ ∈ C0∞ and ψ ∈ Ξ. We use φ≔∇·(Θ(x)ψ(x/η)) = ψ(x/η)·∇Θ(x) as ∇y · ψ(y) = 0. Thenψ(y)|Σ = 0 and is nonvanishing in Y∖Σ and by  = U(x), all these lead toj0 = ∇U(x) for j0 = 0 for  ∈ R × Σ.Next we determine the relation between shown in . We defiΣ and by  it impliProof of (i). To conclude this part, the arguments rely on that of [x ∈ R. To show this, we define a function Yj0dy = ∫Yj(x)dy = j(x) and y · j0 = 0 and ∇y · j1 = 0, and (iii) ψ0(y) = 0, which shows that  that of . We consies that  holds goy=0 from , we have=0 from . To verify the claim, let us choose C0∞(Ω). Note that U ∈ L2(Ω) and we find also that x ∈ R and x ∈ Ω∖R; that is, a(x) = α for x ∈ R and a(x) = 1 for x ∈ Ω∖R.hen from , we haveΩ′ with R ⊂ Ω′ ⊂ ⊂Ω, L2(Ω′), then, up to a subsequence, L2(Ω′).The proof of U and U; see U and j→;Case 1. Let ϕ ∈ C0∞(Ω) from ((Ω) from  we have from ((Ω) from  we have(The combination of  and 55)55) givesx ∈ R. Due to their rectangular metallic gratings inside R, the component along x1 direction vanishes; that is, the first component of α∂x22U = −k2μeffU.Here we can compare our upscaled equation with the limit problem obtained in , especiaThe proof is devided into three steps which are demonstrated below.(i) Uniqueness of the Limit Problem. With a fixed incident field ui we will show that the limit problem  = 0  for  x ∈ ℝ  and  y ∈ Y. The main ingredient for this uniqueness result is Rellich's first lemma and the fact that aeff is real and μeff has positive imaginary part. In fact aeff is identity and μeff is 1 outside of R. Let us denote the surface of a sphere Br(0) of radius r by Sr  (≔∂Br(0)), where r is chosen so large such that R ⊂ Br(0). Let r0 be such r; then by (Br0(0). Since  > 0, thenu = 0 in R. Since r0 is chosen arbitrarly, for every r from   compare .(ii) Convergence to the Limit Problem Assuming an Lloc2-Bound. Let the radius r0 > 0 be such that Ω≔Br0(0). We begin with the assumption thatη → 0, we obtain that U(x) = u(x)χ(x)|Ω∖R + Nw−1u(x)χ(x)|R solves (tion thattη≔∫Ωuη21R solves .uη and ∇uη are uniformly convergent on every compact subset of ℝ2∖R. Let us choose r < r0 such that R ⊂ ⊂Br(0)⊂⊂Ω. By [uηs = uη − ui coincides on ℝ2∖Br(0) with its Helmholtz representation through values and derivatives of uη − ui on ∂Br(0). By the similar representation formula, using the values and derivatives of U − ui on ∂Br(0), we can extend U into all of ℝ2 to a solution of the Helmholtz equation ∇2U = −k2U outside of R. Thus this construction of U shows that U − ui satisfies the Sommerfeld radiation condition. The uniform convergence of uη → U and ∇uη → ∇U on ∂Br(0) implies the uniform convergence of uη and its derivatives on all compact subsets of exterior of R. Finally by uniqueness of the limit from part (i), η → 0 for the whole sequence. This shows that the Sommerfeld radiation condition holds for r = |x | → ∞. which establishes (We only need to verify the radiation condition . By Lemmablishes .(iii) Boundedness of tη. In the previous step the limit problem is obtained assuming |Ω∖R + Nw−1vχ(x)|R is the unique solution of .assuming  is true.assuming  holds trution of  and satiR, the gradient estimate . Then by Rellich compactness lemma ∫Ω∖R|vη|2 → 0 as η → 0. For inside of R, we use the estimate  = 1. Thus tη has to be bounded.For outside of estimate  for vη estimate  on uη aBy Theorems R be ℝ ×  for h > 0. We assume planar front of waves that reaches the metallic slab (−h < x2 < 0) from above (x2 > 0). The incoming waves would be partially reflected and partially transmitted through the metallic structure. Before we proceed any further we define the following parameters:M = 1amplitude of the incident wave, where  θ≔ incident angle, where θ ∈  T≔ complex amplitude and phase shift, where T ∈ ℂ  Ai, Bi≔ complex amplitudes in the structure, where Ai, Bi ∈ ℂ, i = 1,2 R≔ complex amplitude of the reflected wave, where R ∈ ℂLet the rectangle U of (τ≔√(Nw/α). We are yet to determine the coefficients Ai, Bi, R, and T by using the interface (x2 = 0 and x2 = −h) and transmission conditions (see (We write the solution U of  as(67)whons (see ).The Transfer Matrix M. We will calculate a transfer matrix M which basically gives a transformation relation between the solutions on the upper boundary x2 = 0 and the lower boundary x2 = −h. To be precise, we define a map M : ℂ2 → ℂ2  asx2 = 0 and x2 = −h which we choose as  and , respectively. In short the matrix M maps the vector M can be expressed as a ℂ2  ×  2 matrix. Now we determine the transfer matrix M where the two columns are obtained by M · t and M · t.Columns of M. To obtain the first column of M, we study a solution U of the effective system such that U|x2=0+ = 1 and U in the interval  is given by |x2=0+ = U|x2=0− = 1, α∂x1U|x2=0+ = α∂x1U|x2=0− = 0, and α∂x2U|x2=0+ = α∂x2U|x2=0− = 0. With the help of these conditions, we obtain A2 = B2 = 0 and (1/2)A1B1cos(τkx1) = 1 which gives U = cos(τkx2). With the help of similar transmission condition we obtaingiven by . By tranU|x2=−h− = U|x2=−h+ = cos(τkh), ∂x2U|x2=−h− = α∂x2U|x2=−h+ = ατksin(τkh), and ∂x1U|x2=−h− = α∂x1U|x2=−h+ = 0. This gives first column of M as (cos(τkh), ατksin(τkh))t. A similar computation by taking M · t in account will yield the second column of M as (−(ατk)−1sin(τkh), cos(τkh))t. Thus the required transfer matrix is given byτ≔√(Nw/α) and α = 1 − (3/2)γ.The Transmission Coefficient. After having the matrix M in hand, our next step is to calculate the transfer coefficient T. With the help of matrix M, we map the values x2 = 0+ to the values x2 = −h−; that is, (−1 + R))eikx1sin(θ) will get mapped to )eikx1sin(θ). In other words,T, we eliminate the unknown R. Now we follow a simple elimination technique shown in [v ∈ ℂ2 and w ∈ ℂ2 byv⊥ will result in the elimination of R from , where Nw is defined by the help of an eigenvalue problem in the metallic part Σ and we also notice that T depends on the wave number k by the relation Nw = Nw(k). For a rather simple Nw, the graph of |T|2 against the wave number k is shown in figure  4 in [By , we havere  4 in .εr < 0; then Nw ∈ . Also α = 1 − (3/2)γ ∈ , where 0 < γ < (1/2). This implies that the term ((ατ/cos(θ)) + (cos(θ)/ατ)) in (T| ≤ 1 and we get |T| = 1⇔cos(τkh) = 1. This corresponds to a resonance of the plasmon waves in the metallic structure ) with height h.Let us focus again on the case of a material that permits perfect plasmon waves, that is, of a lossless material with negative permittivity, /ατ)) in  is great M of (τkh) = 1, sin(τkh) = 0 and we get the transfer matrix M = I, the Identity matrix, corresponding to perfect transmission.We see that this effect can also be deduced from the transfer matrix M of , since f


Scientific Reports6: Article number: 3852910.1038/srep38529; published online: 12062016; updated: 05172017In Figure 1, the latitudes ‘80.5 N’ and ‘81.0 N’ were incorrectly given as ‘81.0 N’ and ‘81.5 N’ respectively. In addition, the scale between 0 km and 40 km was incorrectly given as between 0 km and 80 km. The correct Figure 1 appears below as