A formula for the first positive eigenvalue of a one-dimensional elliptic transmission problem is derived. The eigenvalue problem arises during  the spectral analysis of a Laplacian on the disc with angular transmission conditions. The formula for the first eigenvalue provides an explicit link to the transmission conditions in the problem

Zerulla, Konstantin

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A FORMULA FOR THE FIRST POSITIVE EIGENVALUE OF AONE-DIMENSIONAL TRANSMISSION PROBLEMKonstantin ZerullaDepartment of Mathematics, Karlsruhe Institute of TechnologyEnglerstr. 276131 Karlsruhe, GermanyAbstract. A formula for the first positive eigenvalue of a one-dimensionalelliptic transmission problem is derived. The eigenvalue problem arises duringthe spectral analysis of a Laplacian on the disc with angular transmissionconditions. The formula for the first eigenvalue provides an explicit link to thetransmission conditions in the problem.In this note, we determine the first eigenvalue of the one-dimensional system(ψ(i))′′(ϕ) = −κ2ψ(i)(ϕ) for ϕ ∈ Ii, i ∈ {1, . . . , 4},ψ(1)(0) = ψ(4)(2π), ε(1)(ψ(1))′(0) = ε(4)(ψ(4))′(2π),ψ(1)(π2 ) = ψ(2)(π2 ), (ψ(1))′(π2 ) = (ψ(2))′(π2 ),ψ(2)(π) = ψ(3)(π), (ψ(2))′(π) = (ψ(3))′(π),ψ(3)( 32π) = ψ(4)( 32π), ε(1)(ψ(3))′( 32π) = ε(4)(ψ(4))′( 32π),(1)whereI1 := (0, π2 ), I2 := (π2 , π), I3 := (π,32π), I4 := (32π, 2π),and ε is a piecewise constant function on the partition I1 ∪ · · · ∪ I4. By f (i) wedenote the restriction of a function f ∈ L2(0, 2π) to the interval Ii, i ∈ {1, . . . , 4}.We furthermore assume the relationε(1) = ε(2) = ε(3) 6= ε(4). (2)System (1) naturally arises during the analysis of the Laplacian∆u := 1ε̃ div(ε̃∇u),u ∈ D(∆) := {v ∈ H10 (D) | div(ε̃∇u) ∈ L2(D)}on the unit disc D with transmission conditions, see [2, 3, 5, 1, 4] for instance. Hereε̃(r, ϕ) := ε(ϕ) for r ∈ [0, 1] and ϕ ∈ [0, 2π). In fact, (1) corresponds to the angularpart of the eigenvalue problem for the Laplacian on the disc. Note that the finalresult of this note in Lemma 2 is essential for the explicit nature of the regularityresults in [5].In the next two lemmas, we calculate the first nonzero eigenvalue of (1). Due totechnical issues, we distinguish between two cases for the ratio ε(4)/ε(1).Lemma 1. Letε(4)ε(1)6= 1− 2cos( 157 π) sin(67π)cos( 127 π) sin(97π).Funded by the Deutsche Forschungsgemeinschaft (DFG, German Research Foundation) –Project-ID 258734477 – SFB 1173.12 KONSTANTIN ZERULLAThen the first nonzero eigenvalue κ21 of (1) satisfies the identity(ε(4) − ε(1))2ε(4)ε(1)= − 4 sin2(κ1π)sin(κ12 π) sin(32κ1π). (3)It has a one-dimensional eigenspace, and the next eigenvalue is greater than one.Proof. 1) We first assume that λ ∈ (0, 1) is an eigenvalue of (1) with an associatedeigenfunction ψ 6= 0. The first line of (1) then implies the representationψ(i)(ϕ) = a(i) cos(√λϕ) + b(i) sin(√λϕ), a(i), b(i) ∈ R, ϕ ∈ Ii (4)for i ∈ {1, . . . , 4}. The third and fourth line of (1) lead to the relations a(1) =a(2) = a(3) and b(1) = b(2) = b(3). The second and fifth lines of (1) further result inthe formulasa(1) = a(4) cos(√λ2π) + b(4) sin(√λ2π), (5)b(1) = ε(4)ε(1)(− a(4) sin(√λ2π) + b(4) cos(√λ2π)),a(4) cos(√λ 32π) + b(4) sin(√λ 32π) = a(1) cos(√λ 32π) + b(1) sin(√λ 32π)= a(4) cos(√λ2π) cos(√λ 32π) + b(4) sin(√λ2π) cos(√λ 32π) (6)− ε(4)ε(1)a(4) sin(√λ2π) sin(√λ 32π)+ ε(4)ε(1)b(4) cos(√λ2π) sin(√λ 32π).Reformulating the last identity, the equationa(4)(cos(√λ 32π)− cos(√λ2π) cos(√λ 32π) +ε(4)ε(1)sin(√λ2π) sin(√λ 32π))= b(4)(sin(√λ2π) cos(√λ 32π)− sin(√λ 32π) +ε(4)ε(1)cos(√λ2π) sin(√λ 32π))=: b(4)A1(λ) (7)is derived. Relating the derivative condition in the fifth line of (1) to (5), weconclude the formulasa(4)(− ε(4) sin(√λ 32π) + ε(1) cos(√λ2π) sin(√λ 32π) + ε(4) sin(√λ2π) cos(√λ 32π))= b(4)(− ε(4) cos(√λ 32π)− ε(1) sin(√λ2π) sin(√λ 32π) + ε(4) cos(√λ2π) cos(√λ 32π))=: b(4)A2(λ). (8)2) We next show by contradiction that a(4) is nonzero. Assume hence thata(4) = 0. Equations (5) and (7) then imply that b(4) is nonzero and that A1(λ)vanishes. In the following, the numbersω := ε(4)/ε(1) − 1, ξ := ε(4) − ε(1)are employed. Manipulating (7) by means of trigonometric identities, we infer therelations0 = A1(λ) = sin(√λ2π) cos(√λ 32π)− sin(√λ 32π) + (ε(4)ε(1)− 1) cos(√λ2π) sin(√λ 32π)+ cos(√λ2π) sin(√λ 32π)= ω cos(√λ2π) sin(√λ 32π) + 2 cos(√λ 52π) sin(√λπ).Since the two summands in the last line have no common zeros on (0, 1), the lastline gives rise to the formulaω = ω(λ) = −2 cos(√λ 52π) sin(√λπ)cos(√λ2π) sin(√λ 32π). (9)FORMULA FOR THE FIRST EIGENVALUE OF A TRANSMISSION PROBLEM 3From (8) we further deduce that the expression A2(λ) vanishes. Manipulatingthe defining relation for A2(λ) by means of trigonometric identities, the equations0 = − ε(4)ε(1)cos(√λ 32π)− sin(√λ2π) sin(√λ 32π) +ε(4)ε(1)cos(√λ2π) cos(√λ 32π)= − ε(4)ε(1)cos(√λ 32π)− sin(√λ2π) sin(√λ 32π) + ω cos(√λ2π) cos(√λ 32π)+ cos(√λ2π) cos(√λ 32π)= − ε(4)ε(1)cos(√λ 32π) + cos(√λ 72π) + ω cos(√λ2π) cos(√λ 32π)= −ω cos(√λ 32π)− 2 sin(√λ 52π) sin(√λπ) + ω cos(√λ2π) cos(√λ 32π) (10)follow. The right hand side is next multiplied with the factor cos(√λ2π) sin(√λ 32π).Using also (9), we then infer the formulas0 = −ω cos(√λ 32π) cos(√λ2π) sin(√λ 32π)− 2 sin(√λ 52π) sin(√λπ) cos(√λ2π) sin(√λ 32π)+ ω cos2(√λ2π) cos(√λ 32π) sin(√λ 32π)= 2 cos(√λ 52π) sin(√λπ) cos(√λ 32π)− 2 sin(√λ 52π) sin(√λπ) cos(√λ2π) sin(√λ 32π)− 2 cos(√λ 52π) sin(√λπ) cos(√λ2π) cos(√λ 32π). (11)We next divide (11) by sin(√λπ) 6= 0, and we use besides the angle sum formulafor cosine the trigonometric relationscos(√λ 52π) cos(√λ 32π) =12 (cos(√λ4π) + cos(√λπ)),cos(√λ2π) cos(√λπ) = 12 (cos(√λ3π) + cos(√λπ)).In this way, we arrive at the equations0 = cos(√λ 52π) cos(√λ 32π)− sin(√λ 52π) cos(√λ2π) sin(√λ 32π)− cos(√λ 52π) cos(√λ2π) cos(√λ 32π)= cos(√λ 52π) cos(√λ 32π)− cos(√λ2π) cos(√λπ)= 12 (cos(√λ4π)− cos(√λ3π)) = − sin(√λ 72π) sin(√λ 12π).As sin(√λ 12π) 6= 0, we conclude that λ is an element of the set {449 ,1649 ,3649}. Pluggingthese values for λ into the formula (9) for ω(λ), we obtain, however, thatω( 449 ) < ω(1649 ) < −1(thus contradicting (2)), while ω(36/49) is excluded in the assumption. We concludethat a(4) is different from zero.3) Taking the results of part 2) into account, we can assume that a(4) = 1. Inthe following, we distinguish the cases of A2(λ) being zero and nonzero, see (8).3.i) Suppose A2(λ) = 0, and proceed similar to part 2). The formula for A2(λ)in (8) is divided by ε(4), and the numberω0 := 1 + ε(1)/ε(4)is introduced. By means of trigonometric identities, the equations0 = − cos(√λ 32π)−ε(1)ε(4)sin(√λ2π) sin(√λ 32π) + cos(√λ2π) cos(√λ 32π)= − cos(√λ 32π)− ω0 sin(√λ2π) sin(√λ 32π) + cos(√λ 12π)= 2 sin(√λπ) sin(√λ 12π)− ω0 sin(√λ2π) sin(√λ 32π) (12)4 KONSTANTIN ZERULLAare then obtained. Since the summands on the right hand side of (12) have nocommon zero on (0, 1), the formulaω0 =2 sin(√λπ) sin(√λ 12π)sin(√λ2π) sin(√λ 32π)(13)follows. Treating the left hand side of (8) in the same way, we further arrive at0 = −2 cos(√λπ) sin(√λ 12π) + ω0 cos(√λ2π) sin(√λ 32π).Multiplying by sin(√λ2π) and inserting (13), we deduce0 = −2 sin(√λ2π) cos(√λπ) sin(√λ 12π) + 2 cos(√λ2π) sin(√λπ) sin(√λ 12π).Dividing by sin(√λ 12π) and using the sum formula for sine, we arrive at the identity0 = sin(√λπ). Since λ is assumed to belong to (0, 1), this is a contradiction.3.ii) In consideration of the results in 3.i), we infer that A2(λ) has to be nonzero.Dividing in (8) by A2(λ), and using trigonometric identities as well as the numberξ = ε(4) − ε(1), we then obtain the equationsb(4) =−ε(4) sin(√λ 32π) + ε(1) cos(√λ2π) sin(√λ 32π) + ε(4) sin(√λ2π) cos(√λ 32π)−ε(4) cos(√λ 32π)− ε(1) sin(√λ2π) sin(√λ 32π) + ε(4) cos(√λ2π) cos(√λ 32π)=−ε(4) sin(√λ 32π) + ε(4) sin(√λ 72π)− ξ cos(√λ2π) sin(√λ 32π)−ε(4) cos(√λ 32π) + ε(4) cos(√λ 72π) + ξ sin(√λ2π) sin(√λ 32π)= −2ε(4) cos(√λ 52π) sin(√λπ)− ξ cos(√λ2π) sin(√λ 32π)2ε(4) sin(√λ 52π) sin(√λπ)− ξ sin(√λ2π) sin(√λ 32π). (14)We next reformulate (7) algebraically with the number ω = ε(4)/ε(1)−1 and therelation a(4) = 1. We derive the identities0 = cos(√λ 32π)− cos(√λ2π) cos(√λ 32π) +ε(4)ε(1)sin(√λ2π) sin(√λ 32π)− b(4)(sin(√λ2π) cos(√λ 32π)− sin(√λ 32π) +ε(4)ε(1)cos(√λ2π) sin(√λ 32π))= cos(√λ 32π)− cos(√λ 72π) + ω sin(√λ2π) sin(√λ 32π)− b(4)(− sin(√λ 32π) + sin(√λ 72π) + ω cos(√λ2π) sin(√λ 32π))= 2 sin(√λ 52π) sin(√λπ) + ω sin(√λ2π) sin(√λ 32π)− b(4)(2 cos(√λ 52π) sin(√λπ) + ω cos(√λ2π) sin(√λ 32π)). (15)The representation (14) for b(4) is next inserted into the right hand side of (15),and all arising expressions are multiplied with the denominator in (14). In this way,we deduce the equations0 = 4ε(4) sin2(√λ 52π) sin2(√λπ) + 2ωε(4) sin(√λ2π) sin(√λ 32π) sin(√λ 52π) sin(√λπ)− 2ξ sin(√λ 52π) sin(√λπ) sin(√λ2π) sin(√λ 32π)− ωξ sin2(√λ2π) sin2(√λ 32π)+ 4ε(4) cos2(√λ 52π) sin2(√λπ)− 2ξ cos(√λ 52π) sin(√λπ) cos(√λ2π) sin(√λ 32π)+ 2ωε(4) cos(√λ2π) sin(√λ 32π) cos(√λ 52π) sin(√λπ)− ωξ cos2(√λ2π) sin2(√λ 32π) (16)= 4ε(4) sin2(√λπ) + 2(ωε(4) − ξ) cos(√λ 12π) sin(√λπ) sin(√λ 32π)− ωξ sin2(√λ 32π). (17)To further simplify the expressions on the right hand side, we use the formulascos(√λ 12π) sin(√λπ) = 12 (sin(√λ 12π) + sin(√λ 32π)),FORMULA FOR THE FIRST EIGENVALUE OF A TRANSMISSION PROBLEM 5ωε(4) − ξ = (ε(4))2ε(1)− 2ε(4) + ε(1) = (ε(4)−ε(1))2ε(1)= ( ε(4)ε(1)− 1)(ε(4) − ε(1))= ωξ.Inserting these relations in (17), we arrive at the identities0 = 4ε(4) sin2(√λπ) + 2ωξ cos(√λ 12π) sin(√λπ) sin(√λ 32π)− ωξ sin2(√λ 32π)= 4ε(4) sin2(√λπ) + ωξ sin(√λ 12π) sin(√λ 32π).As the two summands on the right hand side have no common zeros on (0, 1), weconclude the representation(ε(4) − ε(1))2ε(1)= ωξ = − 4ε(4) sin2(√λπ)sin(√λ 12π) sin(√λ 32π). (18)Note that λ is uniquely determined by (18). Altogether, there is at most one eigen-value of (1) in (0, 1). The associated eigenspace is furthermore one-dimensional.4) Let λ ∈ (4/9, 1) satisfy (18). It remains to show that λ is indeed an eigenvalueof (1). Let ψ be given by (4) with numbers a(i), b(i) that have to be determined.4.i) We first consider the case A2(λ) = 0. Due to the choice of λ, (16) is stillvalid. Rewriting (16) in product formula, we then deduce the identities0 =(2 sin(√λ 52π) sin(√λπ) + ω sin(√λ2π) sin(√λ 32π))·(2ε(4) sin(√λ 52π) sin(√λπ)− ξ sin(√λ2π) sin(√λ 32π))+(2 cos(√λ 52π) sin(√λπ) + ω cos(√λ2π) sin(√λ 32π))·(2ε(4) cos(√λ 52π) sin(√λπ)− ξ cos(√λ2π) sin(√λ 32π))=(2 cos(√λ 52π) sin(√λπ) + ω cos(√λ2π) sin(√λ 32π))·(2ε(4) cos(√λ 52π) sin(√λπ)− ξ cos(√λ2π) sin(√λ 32π))= A1(λ) ·(2ε(4) cos(√λ 52π) sin(√λπ)− ξ cos(√λ2π) sin(√λ 32π)). (19)In case A1(λ) = 0, the arguments in part 2) lead to a contradiction. Hence, thesecond factor on the right hand side of (19) is zero. The trigonometric manipulationsin (14) now lead to0 = −ε(4) sin(√λ 32π) + ε(1) cos(√λ2π) sin(√λ 32π) + ε(4) sin(√λ2π) cos(√λ 32π).This means that (8) is true. We now define a(4) := 1 andb(4) :=cos(√λ 32π)− cos(√λ2π) cos(√λ 32π) +ε(4)ε(1)sin(√λ2π) sin(√λ 32π)A1(λ).Consequently, (7) is valid. Choosing a(1) = a(2) = a(3) and b(1) = b(2) = b(3) by(5), ψ is an eigenfunction of (1) with eigenvalue λ.4.ii) We finally treat the case A2(λ) 6= 0. We then define b(4) according to (14),set a(4) = 1, and define a(1) = a(2) = a(3) as well as b(1) = b(2) = b(3) by (5). Thefunction ψ is chosen as in (4).Altogether, it only remains to validate the required transmission conditions forψ. By definition of b(4), formula (8) is satisfied. Due to the choice of λ, identity(17) is also true. Dividing the right hand side of (17) by A2(λ), we then concludethat (15) holds. This finally means that also the first transmission condition (7) isfulfilled. Based on Lemma 1, we can now conclude formula (3) without an additionalrestriction on the ratio ε(4)/ε(1).6 KONSTANTIN ZERULLALemma 2. Let ε satisfy (2). Then the statements of Lemma 1 on the first positiveeigenvalue κ21 of (1) are valid.Proof. 1) In view of Lemma 1, it remains to treat the caseε(4)ε(1)= 1− 2cos( 157 π) sin(67π)cos( 127 π) sin(97π).Assume first that λ ∈ (0, 1) is an eigenvalue of (1) with associated eigenfunctionψ 6= 0. The reasoning in part 1) of the proof for Lemma 1 applies also here, whencewe employ the constructions and formulas from there.2) We first show by contradiction that the second factor on the left hand side of(8) is not zero. So assume it is zero. We then arrive at the formulaε(4)ε(1)= −cos(√λ2π) sin(√λ 32π)sin(√λ2π) cos(√λ 32π)− sin(√λ 32π). (20)With (8) we further conclude that b(4) = 0 or A2(λ) = 0. We distinguish betweenthese cases.2.i) Let first b(4) = 0. From (7) we then conclude the relationε(4)ε(1)=cos(√λ 32π)(cos(√λ2π)− 1)sin(√λ2π) sin(√λ 32π). (21)Studying the signs of the right hand side expressions in (20) and (21), we infer thatλ ∈ (4/9, 1). Then the right hand side of (20) is, however, smaller than 1, and weobtain a contradiction to (2).2.ii) Let A2(λ) = 0. Then (10) is again valid and leads to the relationω =2 sin(√λ 52π) sin(√λπ)(cos(√λ2π)− 1) cos(√λ 32π)(22)for the number ω = ε(4)/ε(1)−1. In consideration of the signs of the expressions onthe right hand side of (20) and (22), we deduce λ ∈ (16/25, 1). Again the right handside of (20) is now smaller than 1, and we arrive at a contradiction. Altogether,the second factor on the left hand side of (8) is not zero.3) We next show by contradiction that the expression A2(λ) from (8) is zero. So,assume A2(λ) 6= 0. Relation (8) then shows that a(4) and b(4) are not zero, whencewe can choose a(4) = 1 without loss of generality. The arguments in part 3.ii) of theproof for Lemma 1 are then also valid in the current setting, and lead to relation(18). The assumption on ε(4)/ε(1) then implies that λ = 3649 , and that relation (9) isvalid. Reversing the reasoning in part 2) of the proof for Lemma 1, we additionallyinfer that formula (11) holds true. Dividing (11) by cos(√λ2π) sin(√λ 32π), we thenconclude that A2(λ) = 0, see (10). This leads to a contradiction.4) By part 3), A2(λ) = 0 and a(4) = 0. The reasoning in part 2) of the prooffor Lemma 1 then implies that λ = 36/49. Taking the assumption on ε(4)/ε(1) intoaccount, formula (3) is valid for κ1 =√λ.5) It remains to show that λ = 36/49 is indeed an eigenvalue of (1). The abovereasoning in this proof already implies that the associated eigenspace is then one-dimensional. We choose a(4) = 0, b(4) = 1, and define a(1) = a(2) = a(3) as well asb(1) = b(2) = b(3) by (5). A desired eigenfunction ψ of (1) is defined via (4).It then remains to show that ψ satisfies the transmission condition in the fifthline of (1). This is equivalent to the validity of (7) and (8). Due to the assumptionon ε(4)/ε(1), relation (9) is valid, implying that A1(λ) = 0. As a(4) = 0, we inferthat (7) is true. As λ = 36/49, we obtain that also A2(λ) = 0 by repeating theFORMULA FOR THE FIRST EIGENVALUE OF A TRANSMISSION PROBLEM 7arguments in part 2) of the proof for Lemma 1 in reverse order. Thus, also (8) isfulfilled. References[1] M. Costabel and M. Dauge: Singularities of electromagnetic fields in polyhedral domains,Arch. Ration. Mech. Anal. 151 (3) (2000), 221–276.[2] R.B. Kellogg: Singularities in interface problems, in: Numerical solution of partial differentialequations-II. SYNSPADE 1970 (ed. B. Hubbard), Academic Press, New York (1971), 351–400.[3] R.B. Kellogg: On the Poisson equation with intersecting interfaces, Appl. Anal. 4 (1974/75),101–129.[4] H. Triebel: Higher analysis, Johann Ambrosius Barth, Leipzig, 1992.[5] K. Zerulla: Analysis of a dimension splitting scheme for Maxwell equations with low regularityin heterogeneous media, J. Evol. Equ. 22, 90 (2022).Email address: konstantin.zerulla@kit.edu

A formula for the first positive eigenvalue of a one-dimensional transmission problem

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A formula for the first positive eigenvalue of a one-dimensional transmission problem

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