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ON MIXED DIRICHLET-NEUMANN EIGENVALUES OF TRIANGLES. BARTŁOMIEJSIUDEJA 5 ABSTRACT. WeorderlowestmixedDirichlet-Neumanneigenvaluesofrighttrianglesac- 1 cordingtowhichsidesweapplytheDirichletconditions. ItisgenerallytruethatDirichlet 0 conditiononasupersetleadstolargereigenvalues,butitisnontrivialtocomparee.g. the 2 mixedcaseson triangleswith justone Dirichletside. Asa consequenceof thatorderwe n also classify the lowest Neumann and Dirichlet eigenvaluesof rhombiaccording to their a symmetry/antisymmetrywithrespecttothediagonal. J WealsogiveanorderforthemixedDirichlet-Neumanneigenvaluesonarbitrarytrian- 9 gle,assumingtwoDirichletsides. ThesingleDirichletsidecaseisconjecturedtoalsohave 2 appropriateorder,followingrighttriangularcase. ] P S . h at 1. INTRODUCTION m Laplaceeigenvaluesareofteninterpretedasfrequenciesofvibratingmembranes. Inthis [ context,thenatural(Neumann)boundaryconditioncorrespondstoafreemembrane,while 1 Dirichlet condition indicates a membrane is fixed in place on the boundary. Intuitively, v 8 mixedDirichlet-Neumannconditionsshouldmeanthatthemembraneispartiallyattached, 1 and thelarger theattachedportion,thehigherthefrequencies. 6 Usingvariationalcharacterization ofthefrequencies (see Section 2) onecan easily con- 7 0 clude that increasing the attached portion leads to increased frequencies. In this paper we . 1 investigatea harder, yet stillintuitivelyclear case of imposingDirichlet conditionson var- 0 ioussides oftriangles. ImposingDirichlet conditionon onesidegivessmallereigenvalues 5 1 thanimposingitonthatsideandonemore. However,isittruethatimposingDirichletcon- : dition on shorter side leads to smaller eigenvalue than the Dirichlet condition on a longer v i side? X Note that one can also think about eigenvalues as related to the survival probability of r a theBrownian motionon a triangle, reflecting on the Neumann boundary, and dying on the Dirichlet part. In this context, it is clear that enlarging the Dirichlet part leads to shorter survival time. It is also reasonable, that having Dirichlet condition on one long side gives larger chance of dying, than having shorter Dirichlet side. However, this case is far from obvious to prove, especially that the difference might be very small for nearly equilateral triangles. Let L, M and S denote the lengths of the sides of a triangle T, so that L M S. ≥ ≥ Let thesmallesteigenvaluecorrespondingto Dirichlet conditionappliedto a chosen set of sidesbedenotedbyλset. E.g. λLS wouldcorrespondtoDirichletconditionimposedonthe 1 1 longest and shortest sides. Let also µ and λ denote the smallest nonzero pure Neumann 2 1 and pureDirichleteigenvaluesofthesametriangles. 1 2 Theorem 1.1. Foranyrighttrianglewith smallestanglesatisfyingπ/6 < α < π/4 0 = µ < λS < λM < µ < λL < λMS < λLS < λLM < λ . 1 1 1 2 1 1 1 1 1 When α = π/6 (half-of-equilateral triangle) we have λM = µ , and for α = π/4 (right 1 2 isoscelestriangle)wehaveS = M andλL = µ . Allotherinequalitiesstaysharpinthese 1 2 cases. Furthermoreforarbitrarytriangle min λS,λM,λL < µ λMS < λLS < λLM, { 1 1 1} 2 ≤ 1 1 1 aslongastheappropriatesideshavedifferentlengths. However,itispossiblethatµ > λL 2 1 (foranysmallperturbationoftheequilateraltriangle)orµ < λM (forrighttriangleswith 2 1 α < π/6). Note that, for arbitrary polygonal domains, it is not always the case that a longer restriction leads to a higher eigenvalue (see Remark 3.3). The theorem also asserts that a precise position of the smallest nonzero Neumann eigenvalue in the ordered sequence is an exception, rather than a rule (even among triangles). Nevertheless, we conjecture that mixedeigenvaluesof triangles can be fully ordered. More precisely,we conjecturethat all cases missingintheabovetheoremare stilltrue: Conjecture 1.2. Forarbitrarytriangle λS < λM < λL < λMS, 1 1 1 1 aslong asappropriatesides havedifferentlengths. Even though right triangles are a rather special case, they are of interest in studying other polygonal domains. In particular, recent paper by Nitsch [23] studies regular polygons via eigenvalue perturbations on right triangles. Similar approach is taken in the author’s upcoming collaboration [22]. Finally, right triangles play the main role in the recent progress on the celebrated hot-spots conjecture. Newly discovered approach due to Miyamoto [21, 20] led to new partial results for acute triangles [28] (see also Polymath 7 projectpolymathprojects.org/tag/polymath7/). Theacutecasesrelyon eigenvaluecompar- isonsoftriangles,whichwerefirst considered byMiyamotoon righttriangles. Eigenvalue problems on right triangles were also used to establish symmetry (or antisymmetry) of the eigenfunction for the smallest nonzero Neumann eigenvalue of kites (Miyamoto [20], the author of the present paper [28]) and isosceles triangles [17] (in collaboration with Richard Laugesen). It is almost trivial to conclude that the eigenfunction can be assumed symmetric or antisymmetric with respect to a line of symmetry of a domain. It is however very hard to establish which case actually happens. This problem is also strongly connected to the hot-spots conjecture, given that many known results assume enough symmetry to get a symmetric eigenfunction, e.g. Jerison-Nadirashvili [15] or Ban˜uelos-Burdzy [3]. As a particular case, the latter paper implies that the smallest nonzero Neumann eigenvalue of a narrow rhombus is antisymmetric with respect to the short diagonal. In order to claim the same for all rhombi one needs to look at the very 3 importanthot-spotsresultduetoAtarandBurdzy[1]. TheirCorollary1,partii)canbeap- plied to arbitrary rhombi, but it requires a very sophisticated stochastic analysis argument and asolutionofamorecomplicatedhot-spotsconjecturetoachievethegoal. As a consequence of the ordering of mixed eigenvalues of right triangles we order first four Neumann (and two Dirichlet) eigenvalues of rhombi, depending on their symmetry/antisymmetry. We achieve more than the above mentioned papers, using elementary techniques. Ourresultappliestoallrhombinotnarrowerthanthe“equilateralrhombus”composedof twoequilateraltriangles. Thisparticularcase,aswellasthesquareareinterestingboundary cases duetothepresenceofmultipleeigenvalues. Corollary1.3. Forrhombiwiththesmallestangle2α > π/3 wehave µ , µ , µ andλ aresimple. 2 3 4 2 • µ < λ , 4 1 • theeigenfunctionforµ is antisymmetricwith respectto theshortdiagonal, 2 • theeigenfunctionforµ is antisymmetricwith respectto thelongdiagonal, 3 • theeigenfunctionforµ is doublysymmetric, 4 • theeigenfunctionforλ is antisymmetricwith respect totheshortdiagonal, 2 • Furthermore, if 2α < π/3 then the doubly symmetric mode belongs to µ , and the mode 3 antisymmetricwithrespecttothelongdiagonalcanhavearbitrarilyhighindex(asα 0). → Perhaps the mostinteresting case of our result about rhombi is that thefourth Neumann eigenvalue of nearly square rhombus is smaller than its smallest Dirichlet eigenvalue (and is doubly symmetric). This strengthens classical eigenvalue comparison results: Payne [24], Levine-Weinberger [19], Friedlander [12] and Filonov [11] (on smooth enough domains the third Neumann eigenvalue is smaller than the first Dirichlet eigenvalue, while on convex polygons only the second eigenvalue is guaranteed to be below the Dirichlet case,andthethirdisnotlargerthanit). Thistypeofeigenvaluecomparisonistraditionally used to derive some conclusions about the nodal set of the Neumann eigenfunction, e.g. an eigenfunction for µ cannot have a nodal line that forms a loop. Recent progress on 2 hot-spots conjecture due to Miyamoto [20] and the author [28] relies on such eigenvalue comparisonsandsimilarnodallineconsiderations. Furthermore,author’sforthcomingcol- laboration[22]leveragestheimprovedfourtheigenvaluecomparisononrhombiinstudying regularpolygons. Ourproofsformixedeigenvaluesontrianglesareshortandelementary,yetaverybroad spectrum of techniques is actually needed. Evan though the comparisons look mostly the same, their proofs are strikingly different. Depending on the case, we use: variational techniques with explicitlyor implicitlydefined test functions, polarization (a type of symmetrization) applied to mixed boundary conditions, nodal domain considerations, or an unknowntrial functionmethod(see [17, 18]). 4 2. VARIATIONAL APPROACH AND AUXILIARY RESULTS The mixed Dirchlet-Neumann eigenvalues of the Laplacian on a right triangle T with sidesoflengthL M S can beobtainedbysolving ≥ ≥ ∆u = λDu, onT, u = 0on D L,M,S , ⊂ { } ∂ u = 0on ∂T D. ν \ The Dirichlet condition imposed on D can be any combination of the triangle’s sides, as mentioned in the introduction. For simplicity we denote λ = λLMS (purely Dirichlet eigenvalue)and µ = λ (purely Neumanneigenvalue). ∅ Thesameeigenvaluescan alsobeobtainedby minimizingtheRayleigh quotient u 2 R[u] = T |∇ | . ´ u2 T ´ In particular (1) λD = inf R[u], 1 u H1(T),u=0onD ∈ (2) µ = inf R[u], 2 u H1(T), u=0 ∈ T ´ ForanoverviewofthevariationalapproachwereferthereadertoBandle[2]orBlanchard- Bru¨ning[4]. Foreachkindofmixedboundaryconditionswehaveanorthonormalsequenceofeigen- functionsand 0 < λD < λD λD , 1 2 ≤ 3 ≤ ··· → ∞ as longas D is notempty. When D isempty(purely Neumanncase) wehave 0 = µ < µ µ µ . 1 2 3 4 ≤ ≤ ≤ ··· → ∞ The sharp inequality µ < µ for all nonequilateral triangles was recently proved by the 2 3 author [28]. Similar result λ < λ should hold for purely Dirichlet eigenvalues, but this 2 3 remainsan open problem. Thefact thatλD < λD isaconsequenceofthegeneral smallesteigenvaluesimplicity: 1 2 Lemma2.1. LetΩbeadomainwithDirichletconditiononD = andNeumanncondition 6 ∅ on ∂Ω D. Then 0 < λD < λD and the eigenfunction u belonging to λD can be taken \ 1 2 1 1 nonnegative. Proof. Suppose u is changing sign. Then u is a different minimizer of the Rayleigh 1 1 | | quotient. Any minimizer of the Rayleigh quotient is an eigenfunction (see [2] or a more recent exposition [16, Chapter 9]). But ∆ u = λ u 0, hence minimum principle 1 1 1 | | − | | ≤ ensuresu cannotequalzeroatanyinsidepointofthedomain,givingcontradiction. Hence 1 u has a fixed sign. If there were two eigenfunctions for λD, we could make a linear 1 1 combination that changes sign, which is not possible. Hence the smallest eigenvalue is simple. Finally, λD = 0 would imply that ∂u = 0 a.e., hence the eigenfunction is constant. But itequ1als0 on D,henceu 0. | 1| (cid:3) 1 ≡ 5 If D D then (1) implies that λD1 λD2. Indeed, any test function u that satisfies 1 ⊂ 2 1 ≤ 1 u = 0 on D , can be used in the minimization of λD1. However, the relation between e.g. 2 1 λL and λM isnotclear. 1 1 In the second part of the paper we will consider rhombi R created by reflecting a right triangleT fourtimes. Lemma 2.2. Let u belong to λD(T) or µ (T). Let u¯ be the extension of u to R, that is 1 2 symmetric with respect to the sides of T with Neumann condition and antisymmetric with respecttotheDirichletsides. Thenu¯isaneigenfunctionofR. Furthermore,ifv isanother eigenfunctionofRwiththesamesymmetriesasu¯,thenvbelongstohighereigenvaluethan u¯,or v = Cu¯ forsomeconstantC. Proof. Suppose v is an eigenfunction of R with the same symmetries as u. Its restriction toT satisfiesDirichletandNeumannconditionsonthesamesidesasu. Italsosatisfiesthe eigenvalue equation pointwiseon T. Hence v is an eigenfunction on T. However, λD and 1 µ are simple,hencev = Cu, orv belongstoahighereigenvalueonT. 2 The extension u¯ has the same Rayleigh quotient on R, as on T (due to symmetries). Henceu¯ can beused asa testfunctionforthelowesteigenvalueon R withthesymmetries of u¯. Hence that eigenvalue of R must be smaller or equal to the eigenvalue of u on T. However,it cannotbesmallerby theargumentfrom thepreviousparagraph. (cid:3) In particular,thislemmaimpliesthat λ (R) = λL(T). 1 1 We can also claim that µ (T) equals the smallest Neumann eigenvalue of the rhombus 2 withadoublysymmetriceigenfunction. However,thiseigenvaluewillnotbesecondonR, due to the presence of possibly lower antisymmetricmodes (corresponding to λM(T) and 1 λS(T)). 1 3. INEQUALITIES BETWEEN MIXED EIGENVALUES OF RIGHT TRIANGLES In this section we prove Theorem 1.1. We split the proof into several sections, each treating one or two inequalities. Each section introduces a different technique of proving eigenvaluebounds. Before weproceed wewishtomakeafew remarks. Remark 3.1. All eigenvalues of the right isosceles triangle can be explicitly calculated using eigenfunctionsof the square. ObviouslyS = M in this case, hence someeigenvalue inequalities from Theorem 1.1 become obvious equalities. Furthermore µ = λL, as can 2 1 be seen by taking two orthogonal second Neumann eigenfunctions of the unit square with thediagonalnodallines. Onecorrespondsto µ ontherighttriangle,theotherλL. 2 1 Remark 3.2. Similarly, some of the mixed eigenvalues of the half-of-equilateral triangle can be calculated explicitly using eigenfunctions of the equilateral triangle. In particular λM = µ ,sincethecorrespondingequilateraltrianglehasdoublesecondNeumanneigen- 1 2 value. On the other hand, any mixed case that leads to a mixed case on the equilateral triangle cannot be explicitly calculated. In particular the value of λLS on the half-of- 1 equilateral triangle corresponds to equilateral triangle with Dirichlet condition on two 6 A=(0,h) β 1 α B C=(√1 h2,0) E=(0,0) − D FIGURE 1. Obtuse isosceles triangle O(β) = ABC and acute isosceles triangleA(α) = ADC. sides. The eigenfunctionis not trigonometric(as allother known cases), and to thebestof ourknowledgethereis noclosed formulafortheeigenvalue. Forathoroughoverviewoftheexplicitlycomputablecasesandthegeometricproperties ofeigenfunctionwerefer thereader to[13]and references therein. Remark3.3. Notethatforthetrapeziumwithvertices ( 3,0),(3,0),(3,2)and(0,2)im- − posingtheDirichletconditionontheslopedsideleadstosmallereigenvaluethanimposing iton thetop (numerically). 3.1. For nonisosceles right triangles: λS < λM. Unknown trial function method for 1 1 isoscelestriangles. InthissubsectionO(β)isanobtuseisoscelestrianglewithequalsides of length 1 and aperture angle 2β, with vertices A,B,C equal (0,h), ( √1 h2,0), re- ± − spectively. Let A(α) be an acute isosceles triangle with vertices A,D,C equal (0, h), ± (√1 h2,0)(apertureangle2α). SeeFigure1forbothtriangles. Finallytheirintersection − isaright triangleT = AEC. Notethat handboth anglesare relatedby: h = sinα = cosβ, and β > π/4. Weneed thefollowingthreelemma: Lemma 3.4. Forβ > π/4 π2 µ (O(β)) < 2 4h2(1 h2) − Andtheboundsaturatesfortherightisoscelestriangle(h2 = 1/2orβ = π/4). Proof. Note that h2 < 1/2. Take the second eigenfunction for the right isosceles triangle (0,1),( 1,0)and deform itlinearlyto fit O(β). That istake ± ϕ = sin(πx/2)cos(πy/2) 7 and composewith the linear transformationL(x,y) = (x/√1 h2,y/h). Resulting func- − tioncan beused asa testfunctionforµ (O(β)) 2 (ϕ L) 2 π2 +16h2 8 π2 µ2(O(β)) ≤ ´T(β)|∇ϕ ◦L 2| = 4h2(1 h−2) < 4h2(1 h2) T(β)| ◦ | − − ´ (cid:3) Lemma 3.5. Let ubeanyantisymmetricfunctionon A(α)(so thatu(x, y) = u(x,y)). − − Then π2 u2 > u2. ˆ y 4h2 ˆ A(α) A(α) Proof. Note that for fixed x function u(x, ) is odd, hence it can be used as a test function · forthesecondNeumanneigenvalueonanyverticalintervalcontainedinthetriangleA(α). Weget thelargestinterval[ h,h] whenx = 0. Hence − π2 u2(x,y)dy µ ([ c ,c ]) u2(x,y)dy u2(x,y)dy. ˆ y ≥ 2 − x x ˆ ≥ 4h2 ˆ [ cx,cx] [ cx,cx] [ cx,cx] − − − Integrateoverx toget theresult. (cid:3) Aspecial caseof[17, Corollary 5.5],notingthatα < β,can bestatedas Lemma3.6. Letubetheeigenfunctionbelongingtoµ (A(α)). Thenµ (O(β)) < µ (A(α)) 2 2 2 if u2 (3) A(α) y > tan2(β). ´ u2 A(α) x ´ Supposethecondition(3)isfalse(hencewecannotconcludethatµ (O(β)) < µ (A(α))). 2 2 That is u2 tan2(β) u2 ˆ y ≤ ˆ x A(α) A(α) Then A(α)u2x +u2y 1 A(α)u2y 1 π2 µ (A(α)) = 1+ > = 2 ´ u2 ≥ (cid:18) tan2(β)(cid:19) ´ u2 sin2(β)4h2 A(α) A(α) ´ ´ π2 = > µ (O(β)), 4h2(1 h2) 2 − where the last inequality in the first line follows from Lemma3.5, while the inequality in thesecondlinefollowsfromLemma3.4. Therefore, regardless if we can apply Lemma3.6 or not (condition (3) is true or false), weget µ (O(β)) < µ (A(α)). 2 2 8 Since O(β) is obtuse and isosceles, [17, Theorem 3.2] implies that λS(T) = µ (O(β)). 1 2 The eigenfunction for λM(T) extends to an antisymmetric eigenfunction on A(α), hence 1 µ (A(α)) λM(T). 2 ≤ 1 ThereforeweprovedthatλS < λM foranynonisoscelesrighttriangle. 1 1 3.2. For right triangles: λM < µ if and only if α > π/6. Comparison of Neumann 1 2 eigenfunctionsofanisoscelestriangle. Inthissectionwewillusethenotationintroduced in [17, Section 3]. All isosceles triangles can be splitinto equilateral triangles, subequilat- eraltriangles(withanglebetweenequalsideslessthanπ/3),andsuperequilateral(withthe angleaboveπ/3). Notethatα = π/6meansthatweare workingwithahalfofan equilateraltriangle. The eigenvaluesareexplicitandλM = µ . 1 2 Mirroring a right triangle along middle side M gives a superequilateral triangle if and onlyifα > π/6. Anysuperequilateral trianglehas antisymmetricsecondNeumann eigenfunction [17, Theorem 3.2] and simple second eigenvalue [20] equal λM(T). This proves 1 thatλM(T) < µ (T). 1 2 At the same time any subequilateral triangle has symmetric second eigenfunction [17, Theorem 3.1],withsimpleeigenvalue[20]equalµ (T). HenceλM > µ (T) ifα > π/6. 2 1 2 3.3. For right triangles: µ < λL < λMS. Variational approach and domain mono- 2 1 1 tonicity. Assume that the right triangle T has vertices (0,0), (1,0) and (0,b). We can use four such right triangles to build a rhombus R. Then λL = λ (R), since reflected 1 1 eigenfunction for λL is nonnegative and satisfies Dirichlet boundary condition on R (see 1 Lemma2.2). Hooker and Protter [14] proved the following lower bound for the ground stateofrhombi π2(1+b)2 (4) λL = λ (R) . 1 1 ≥ 4b2 Weneed to provean upperboundforµ that issmallerthan thislowerbound. 2 Considertwoeigenfunctionsoftherightisoscelestrianglewithvertices(0,0),(1,0)and (0,1) ϕ (x,y) = cos(πy) cos(πx), 1 − ϕ (x,y) = cos(πy)cos(πx). 2 Thefirstonebelongstoµ andisantisymmetric,thesecondbelongstoµ andissymmetric. 2 3 In fact all we need is that these integrate to 0 over the right isosceles triangle. Consider a linearcombinationofthelinearly deformedeigenfunctions f(x,y) = ϕ (x,y/b) (1 b)ϕ (x,y/b), 1 2 − − where 0 < b < 1. This function integrates to 0 over the right triangle T, hence it can be used asa testfunctionforµ in(2). As aresultweget thefollowingupperbound 2 3π2((b 1)2 +2)(b2 +1) 64(b 1)2(b+1) (5) µ − − − 2 ≤ 3b2((b 1)2 +4) − Note that when b = 1 bounds (4) and (5) reduce to thesame value. In fact λL = µ in this 1 2 case(right isoscelestriangle). 9 Moreover 3π2((b 1)2 +2)(b2 +1) 64(b 1)2(b+1) π2(1+b)2 − − − = 3b2((b 1)2 +4) − 4b2 − (b 1)2 = − (9π2b2 (256+6π2)b+21π2 256), 12b2((b 1)2 +4) − − − and thequadraticexpressioninbis negativeforb (0,1). Therefore µ < λL. ∈ 2 1 For right triangles, λL is the same as λ for a rhombus built from four triangles, while 1 1 λMS is the same as λ of a kite built from two right triangles. Sharpest angle of the kite is 1 1 the same as the acute angle of the rhombus. We can put the kite inside of the rhombus by puttingthe vertexof the sharpest angleat the vertexof the rhombus. Therefore λL < λMS 1 1 by domain monotonicity(take the eigenfunction of thekite, extend with 0, and use as trial functionontherhombus). 3.4. For arbitrary triangle: min λS,λM,λL < µ λMS. Nodal line consideration { 1 1 1} 2 ≤ 1 and eigenvalue comparisons. To get a lower bound for µ we will define a trial function 2 based on the Neumann eigenfunction, without knowing its exact form, and use it as a trial functionforamixedeigenvalueproblem. Notethattheeigenfunctionofµ hasexactlytwo 2 nodaldomains,byCourant’snodaldomaintheorem([8,Sec. V.5,VI.6])andorthogonality tothefirstconstanteigenfunction. Hencetheclosureofatleastoneofthesenodaldomains must have empty intersection with the interior of one of the sides (nodal line might end in a vertex, but the eigenfunction must have fixed sign on at least one side). Let us call this side D and consider λD. Let u be the eigenfunction of µ restricted to the nodal domain 1 2 not intersecting D. Extend u with 0 to the whole triangle T. We get a valid trial function forλD. Hencemin λS,λM,λL λD < µ . 1 { 1 1 1} ≤ 1 2 Notethat we already provedthat for righttriangles λS < λM, µ < λL, and λM < µ if 1 1 2 1 1 2 andonlyifsmallestangleα > π/6. Henceforrighttrianglestheminimumcanbereplaced by λS, orevenλM ifα > π/6. 1 1 Notealso, thattheresultofthissectiongeneralizes toarbitrary polygons. Lemma3.7. ThesmallestnonzeroNeumanneigenvalueonapolygonwith2n+1or2n+2 sides is bounded below by the minimum of all mixed Dirichlet-Neumann eigenvalues with Dirichletconditionappliedto atleastnconsecutivesides. Furthermore,forarbitrarydomain,theNeumanneigenvalueisboundedbelowbythein- fimumoverallmixedeigenvalueproblemswithhalfoftheboundarylengthhavingDirichlet conditionappliedtoit. As in the previous section, λMS equals λ of a kite built from two triangles. The Neu- 1 1 manneigenfunctionforµ extendedtothekitegivesasymmetriceigenfunctionofthekite. 2 Giventhatµ andµ ofthekitetogethercanhaveatmostoneantisymmetricmode(see[28, 2 3 Lemma 2.1]), we conclude that µ of the triangle is no larger than µ of the kite. Levine 2 3 and Weinberger [19] proved that λ µ for any convex polygon, including kites, giving 1 3 ≥ ustherequired inequality. 3.5. Proofof: λLS < λLM. Symmetrizationofisoscelestriangles. Toprovethisinequal- 1 1 ity we will use a symmetrization technique called the continuous Steiner symmetrization 10 FIGURE 2. Acute isosceles triangle (thick red line) and obtuse isosceles triangle(thinblackline)generatedbythesamerighttriangle(theirintersec- tion). Two cases of continuous Steiner symmetrization based on the shape oftheacuteisoscelestriangle: subequilateralontheleft,superequilateralon theright. introduced by Po´lya and Szego¨ [25, Note B], and studied by Solynin [29, 31] and Brock [5]. Theauthoralready usedthistechniqueforboundingDirichleteigenvaluesoftriangles in [27]. See Section 3.2 in the last reference for detailed explanation. The most impor- tant feature of the transformation is that if one can map one domain to another using that transformation,then thelatterhas smallerDirichleteigenvalue. Note that mirroring a right trianglealong the middle side shows that λLS of the triangle 1 equals λ of an acute isosceles triangle. Similarly, λLM equals λ for an obtuse isosceles 1 1 1 triangle. WeneedtoshowthattheacuteisoscelestrianglehassmallerDirichleteigenvalue. Figure2 shows both isosceles triangles. Position the isosceles triangles as on the figure, and perform the continuous Steiner symmetrization with respect to the line perpendicular tothecommonside. Iftheacuteisoscelestriangleissubequilateral(verticalsideistheshortest,leftpictureon Figure2), then before we fully symmetrizethe obtuse triangle, we will find the acute one. Thearrow on thefigure showshowfar weshouldcontinuouslysymmetrize. Therefore the acuteisoscelestrianglehas smallereigenvalue. Iftheacuteisoscelestriangleissuperequilateral(verticalsideisthelongest,rightpicture on Figure2), then we first reflect the acute triangle across the symmetrization line, then performcontinuousSteinersymmetrization. Again,wegetthattheacuteisoscelestriangle has smallereigenvalue. NotethatthiscaseseemssimilartoSection 3.1. However,wegettousesymmetrization techniqueduetoDirichletboundary,whileintheothersectionwehadtousalesspowerful, butmorebroadlyapplicable,unknowntrialfunctionmethod. 3.6. For arbitrary triangles: λMS < λLS < λLM. Polarization with mixed boundary 1 1 1 conditions. For this inequality we use another symmetrization technique called polarization. It was used by Dubinin [10], Brock and Solynin [6, 30, 7, 31], Draghici [9], and the author [27] to study various aspects of spectral and potential theory of the Laplacian. As with other kinds of symmetrization, if one can map a domain to some other domain, then thelatterhas smallereigenvalue.