On the principal eigenvalue of a Robin problem with a large parameter 5 0 0 Michael Levitin 2 Department of Mathematics, Heriot-Watt University n a Riccarton, Edinburgh EH14 4AS, U. K. J 7 email [email protected] 1 Leonid Parnovski ] P Department of Mathematics, University College London S . Gower Street, London WC1E 6BT, U. K. h t email [email protected] a m [ Dedicated to Viktor Borisovich Lidski˘ı on the occasion of his 80th birthday 2 v 17 January 2005 9 7 1 3 Abstract 0 4 We study the asymptotic behaviour of the principal eigenvalue of a 0 / Robin (or generalised Neumann) problem with a large parameter in the h boundary condition for the Laplacian in a piecewise smooth domain. We t a show that the leading asymptotic term depends only on the singularities m of the boundary of the domain, and give either explicit expressions or : v two-sided estimates for this term in a variety of situations. i X r a 1 Introduction Let Ω be an open bounded set in Rm (m 1) with piecewise smooth, but not ≥ necessarily connected, boundary Γ := ∂Ω. We investigate the spectral boundary value problem ∆u = λu in Ω, (1.1) − ∂u γGu = 0 on Γ. (1.2) ∂n − 1 M Levitin and L Parnovski Eigenvalue of the Robin problem In (1.1), (1.2), ∂ denotes the outward unit normal derivative, λ is the spec- ∂n tral parameter, γ is a positive parameter (which we later on assume to be large), and G : Γ R is a given continuous function. We will always assume that → supG(y) > 0. (1.3) y∈Γ We treat the problem (1.1), (1.2) in the variational sense, associating it with the Rayleigh quotient v 2dx γ G v 2ds |∇ | − | | Z Z (v;γ,G) := Ω Γ , v H1(Ω), v 0. (1.4) J ∈ 6≡ v 2dx | | Z Ω For every fixed γ, the problem (1.1), (1.2) has a discrete spectrum of eigen- values accumulating to + . By ∞ Λ(Ω;γ,G) := inf (v;γ,G) (1.5) v∈H1(Ω), v6≡0J we denote the bottom of the spectrum of (1.1), (1.2). Our aim is to study the asymptotic behaviour of Λ(Ω;γ,G) as γ + and → ∞ its dependence upon the singularities of the boundary Γ. The problem (1.1)–(1.2) naturally arises in the study of reaction-diffusion equation where a distributed absorption competes with a boundary source, see [2, 3] for details. Remark 1.1. Sometimes, we shall also consider (1.1)–(1.2) for an unbounded domain Ω. In this case, we can no longer guarantee either the discreteness of the spectrum of (1.1)–(1.2), or its semi-boundedness below. We shall still use, however, the notation (1.5), allowing, in principle, for Λ(Ω;γ,G) to be equal to . −∞ 2 Basic properties of the principal eigenvalue We shall mostly concentrate our attention on the case of constant boundary weight G 1; in this case, we shall denote for brevity ≡ (v;γ) := (v;γ,1), Λ(Ω;γ) := Λ(Ω;γ,1). J J See Remark 3.3 for the discussion of the case of an arbitrary smooth G 1. 6≡ We start with citing the following simple result of [3]: Page 2 M Levitin and L Parnovski Eigenvalue of the Robin problem Lemma 2.1. For any bounded and sufficiently smooth Ω Rm, Λ(Ω;γ) is a ⊂ real analytic concave decreasing function of γ 0, Λ = 0, and ≥ |γ=0 d Γ m−1 Λ(Ω;γ) = | | . dγ − Ω (cid:12)γ=0 | |m (cid:12) (cid:12) The problem (1.1)–(1.2) with G 1 admits a solution by separation of (cid:12) ≡ variables in several simple cases. Example 2.2. For a ball B (0,1) = x < 1 Rm, Λ = Λ(B (0,1);γ) is m m {| | } ⊂ given implicitly by √ Λtanh√ Λ = γ, m = 1, − − I (√ Λ) √ Λ m/2 − = γ, m 2, − I (√ Λ) ≥ m/2−1 − where I denotes a modified Bessel function. This implies that for any ball B(a,R) := x : x a < R Rm, { | − | } ⊂ Λ(B(a,R);γ) = γ2 +O(γ2), γ + − → ∞ (independently of the dimension m and radius R); it may be shown that the same asymptotics holds for an annulus A (R ,R) = x (R ,R) . m 1 1 {| | ∈ } Example 2.3. ForaparallelepipedP(l ,...,l ) := x < l : j = 1,...,m 1 m j j {| | } ⊂ Rm we get m µ2 j Λ(P(l ,...,l );γ) = , 1 m − l2 j=1 j X where µ > 0 solves a transcendental equation j µ tanhµ = γl . j j j Thus we obtain Λ(P(l ,...,l );γ) = mγ2 +O(γ2), γ + . 1 m − → ∞ Example 2.4. Let Ω = (0,+ ), and Γ = 0 . It is easy to see that the bot- ∞ { } tom of the spectrum is an eigenvalue Λ((0,+ );γ) = γ2, the corresponding ∞ − eigenfunction being exp( γx). Thus we arrive at a useful (and well-known) − inequality ∞ ∞ v′(x) 2dx γ(v(0))2 γ2 v(x) 2dx, (2.1) | | − ≥ − | | Z Z 0 0 valid for all v H1((0,+ )). ∈ ∞ Page 3 M Levitin and L Parnovski Eigenvalue of the Robin problem A slightly more complicated example is that of a planar angle U := z = α { x+iy C : argz < α of size 2α. ∈ | | } Example 2.5. Let Ω = U with α < π/2. Again the spectrum is not purely α discrete; moreover, the separation of variable does not produce a complete set of generalised eigenfunctions. However, one can find an eigenfunction u (x,y) = 0 exp( γx/sinα) and compute an eigenvalue λ = γ2sin−2α explicitly. Thus − − Λ(U ;γ) γ2sin−2α. We shall now prove that this eigenvalue is in fact the α ≤ − bottom of the spectrum. Lemma 2.6. If α < π/2, Λ(U ;γ) = γ2sin−2α. (2.2) α − Proof. It is sufficient to show that for all v H1(U ), we have α ∈ v 2dz γ v 2ds γ2(sin−2α) v 2dz. (2.3) |∇ | − | | ≥ − | | Z Z Z Uα ∂Uα Uα As ds = dy/sinα, the left-hand side of (2.3) is bounded below by 2 ∂v γ dy dx v 2 . ∂x − sinα| | Z Z (cid:12) (cid:12) ! (cid:12) (cid:12) (cid:12) (cid:12) For each y the integrand is not(cid:12)sma(cid:12)ller than γ2(sin−2α) v 2dx by (2.1). − | | Integrating over y gives (2.3). R Example 2.7. Let us now consider the case of an angle U with α [π/2,π). α ∈ Lemma 2.8. If α π/2, ≥ Λ(U ;γ) = γ2. (2.4) α − Proof. To prove an estimate above, we for simplicity consider a rotated angle U := z = x + iy C : 0 < argz < 2α . In order to get an upper bound α { ∈ } Λ(U ;γ) γ2, we construct a test function in the following manner. Let α ≤ − ψe(s) be a smooth nonnegative function such that ψ(s) = 1 for s < 1/2, and | | ψ(se) = 0 for s > 1 Set now | | 1, if s < τ 1, | | − χ (s) = ψ( s (τ 1)), if τ 1 s < τ , τ  | |− − − ≤ | | 0, otherwise   Page 4 M Levitin and L Parnovski Eigenvalue of the Robin problem (a parameter τ is assumed to be greater than 1). Consider the function v (x,y) = e−γyχ (xγ τ). τ τ − Then one can easily compute that ∞ χ′(s) 2ds (v ;γ) = γ2 1+ −∞| τ | τ ∞ J − χ (s) 2ds R−∞| τ | ! R 1 ψ′(s) 2ds = γ2 1+ −1| | , − 1 ψR(s) 2ds+2(τ 1)! −1| | − R and therefore (v ;γ) γ2 as τ . Thus, Λ(U ;γ) = Λ(U ;γ) γ2. τ α α J → − → ∞ ≤ − To finish the proof, we need only to show that for v H1(U ), α ∈ e v 2dz γ v 2ds γ2 v 2dz. (2.5) |∇ | − | | ≥ − | | Z Z Z Uα ∂Uα Uα Denote V = z : α π/2 < argz < α U . The estimate (2.5) will α α { − | | } ⊂ obviously be proved if we establish v 2dz γ v 2ds γ2 v 2dz. |∇ | − | | ≥ − | | Z Z Z Vα ∂Uα Vα ButthiscanbedoneasintheproofofLemma2.6, byintegratingfirstalong∂U , α and then using one-dimensional inequalities (2.1) in the direction orthogonal to ∂U . α We now consider a generalization of two previous examples to the multi- dimensional case. Example 2.9. LetK Rm = x : x/ x M beaconewiththe cross-section ⊂ { | | ∈ } M Sm−1. Any homothety f : x ax (x Rm, a > 0) maps K onto itself. ⊂ 7→ ∈ Then, as easily seen by a change of variables w = γ−1x, Λ(K;γ) = γ2Λ(K;1). (2.6) In particular, if K contains a half-space, then, repeating the argument of Lemma 2.8 with minor adjustments, one can show that Λ(K,1) = 1 and so − Λ(K;γ) = γ2. (2.7) − Page 5 M Levitin and L Parnovski Eigenvalue of the Robin problem All the above examples suggest that in general one can expect Λ(Ω;γ) = C γ2 +O(γ2), γ + . (2.8) Ω − → ∞ Some partial progress towards establishing (2.8) was already achieved in [3]. In particular, the following Theorems were proved. Theorem 2.10. Let Ω Rm be a domain with piecewise smooth boundary Γ. ⊂ Then Λ(Ω;γ,1) limsup 1. γ2 ≤ − γ→+∞ Theorem 2.11. Let Ω Rm be a domain with smooth boundary ∂Ω. Then ⊂ Λ(Ω;γ) = γ2(1+o(1)), γ + . − → ∞ Remark 2.12. The actual statements in [3] are slightly weaker than the versions above, but the proofs can be easily modified. Note that the proof of Theo- rem 2.10 can be done by constructing a test function very similar to the one used in the proof of Lemma 2.8. The situation, however, becomes more intriguing even in dimension two, if Γ is not smooth. Suppose that Ω R2 is a planar domain with n corner points ⊂ y ,...,y on its boundary Γ. The following conjecture was made in [3]: 1 n Conjecture2.13. LetΩ R2 beaplanardomainwithncornerpointsy ,...,y 1 n ⊂ on its boundary Γ and let α , j = 1,...,n denote the inner half-angles of the j boundary at the points y . Assume that 0 < α < π. Then (2.8) holds with j j 2 C = max sin−2(α ) . Ω j j=1,...,n (cid:8) (cid:9) This conjecture was proved in [3] only in the model case when Ω is a triangle. As we shall see later on, formula (2.8) does not, in general, hold if we allow Γ to have zero angles (i.e., outward pointing cusps, see Example 3.4). We shall thus restrict ourselves to the case when Ω is piecewise smooth ina suitable sense, see below for the precise definition. Under this assumption, we first of all prove that the asymptotic formula (2.8) holds. Moreover, we compute C explicitly in Ω the planar case, thus proving Conjecture 2.13. In the case of dimension m 3, ≥ we give some upper and lower bounds on C , which, in some special cases, Ω amount to a complete answer. Page 6 M Levitin and L Parnovski Eigenvalue of the Robin problem 3 Reduction to the boundary We shall only consider the case when Ω is piecewise smooth in the following sense: for each point y Γ there exists an infinite “model” cone K such that y ∈ for a small enough ball B(y,r) of radius r centred at y there exists an infinitely smooth diffeomorphism f : K B(0,r) Ω B(y,r) with f (0) = y and the y y y ∩ → ∩ derivative of f at 0 being the identity matrix (we shall write in this case that y Ω K near a point y Γ). For example, if y is a regular point of Γ, then K y y ∼ ∈ is a half-space. We require additionally that Ω satisfies the uniform interior cone condition [1], i.e. there exists a fixed cone K with non-empty interior such that each K y contains a cone congruent to K. (See Example 3.4 for a discussion of a case where this condition fails.) Definition 3.1. Let Ω K near a point y Γ. We denote C := Λ(K ;1). y y y ∼ ∈ − Our main result indicates that the asymptotic behaviour of Λ(Ω;γ,1) is in a sense “localised” on the boundary. Theorem 3.2. Let Ω be piecewise smooth in the above sense and satisfy the uniform interior cone condition. Then Λ(Ω;γ) = γ2supC +o(γ2), γ + . (3.1) y − → ∞ y∈Γ Remark 3.3. This result can be easily generalised for the case of our original setting of a non-constant boundary weight G(y) satisfying (1.3): Λ(Ω;γ,G) = γ2 sup G(y)2C +o(γ2), γ + . y − { } → ∞ y∈Γ G(y)>0 Example 3.4. Formula (2.8) does not, in general, hold if Γ is allowed to have outward pointing cusps. In particular, for a planar domain Υ = (x,y) R2 : x > 0, y < xp , p > 1 p { ∈ | | } one can show that γ2/(2−p) for 1 < p < 2, Λ(Υ ;γ) const p ≤ − (γN with any N > 0 for p 2, ≥ by choosing the test function v = exp( γxqp) with q = 2 p for 1 < p < 2 p − − and q = 2 for p 2. p ≥ Page 7 M Levitin and L Parnovski Eigenvalue of the Robin problem InordertoprovidetheexplicitasymptoticformulaforΛ(Ω;γ)inthepiecewise smooth case it remains to obtain the information on the dependence of the constants C upon the local geometry of Γ at y. y Itiseasy to do this, firstly, in thecase ofa regular boundary inanydimension, and, secondly, in the two-dimensional case, where the necessary information is already contained in Lemmas 2.6 and 2.8. Theorem 3.5. Let Γ be smooth at y. Then C = 1. y Moreover, C = 1 whenever there exists an (m 1)-dimensional hyperplane y − H passing through y such that for small r, B(y,r) H Ω. y y ∩ ⊂ Theorem 3.6. Let Ω R2 and let y Γ be such that Ω U near y. Then α ⊂ ∈ ∼ 1, if α π/2; C = ≥ y (sin−2α, if α π/2. ≤ Theorems 3.2, 3.5, and 3.6 prove the validity of Conjecture 2.13. In more general cases, we are only able to provide the two-sided estimates on C , and obtain the precise formulae only under rather restrictive additional y assumptions. These results are collected in Section 5. 4 Proof of Theorem 3.2 We proceed via a sequence of auxiliary Definitions and Lemmas. Definition 4.1. Let K Rm be a cone with cross-section M Sm−1, and let ⊂ ⊂ r > 0. By K = K (K) we denote the family of “truncated” cones K such r r r,R that K = x Rm : θ := x/ x M Sm−1, x < rR(θ) , r,R { ∈ | | ∈ ⊂ | | } where R : M [1,m] is a piecewise smooth function. Thus, for any K K r,R r → ∈ we have K B(0,r) K K B(0,mr). r,R ∩ ⊂ ⊂ ∩ Let K K , and let ♯ be an index assuming values D or N (which r,R r ∈ in turn stand for Dirichlet or Neumann boundary conditions). By Λ♯(K ;γ) r,R we denote the bottom of the spectrum of the boundary value problem (1.1) considered in K with boundary conditions (1.2) on ∂K ∂K = x r,R r,R ∩ { ∈ ∂K : x/ x ∂M and with the boundary condition defined by ♯ on the rest r,R | | ∈ } of the boundary x : x/ x M , x = R(θ) (this boundary value problem is { | | ∈ | | } of course considered in the variational sense). Page 8 M Levitin and L Parnovski Eigenvalue of the Robin problem It is important to note that a simple change of variables as in Example 2.9 leads to the re-scaling relations Λ♯(K ;γ) = γ2Λ♯(K ;1). (4.1) r,R rγ,R These formulae show that the bottoms of the spectra Λ♯(K ;γ) are determined r,R (modulo a multiplication by γ2) by a single parameter µ := rγ via Λ♯(K ;1). µ,R It is therefore the latter which we proceed to study. The first Lemma gives a relation between the bottoms of the spectra for an infinite cone K and its finite “cut-offs”. Lemma 4.2. Let K K (K) and let µ = rγ. Then, as µ , r,R r ∈ → ∞ Λ♯(K ;γ) r,R = Λ(K;1)+o(1). γ2 Proof of Lemma 4.2. By (4.1), we need to prove that lim Λ♯(K ;1) = Λ(K;1). µ,R µ→∞ This can be done by considering a function v H1(K) and comparing the ∈ Rayleigh quotients J(v;1) with “truncated” quotients J(vψ( /µ);1), where ψ · is the same as in the proof of Lemma 2.8. An easy but somewhat tedious computation which we omit shows that as µ + , we have J(vψ( /µ);1) → ∞ · → J(v;1), which finishes the proof. Let y Γ, and let K be a cone with cross-section M such that Ω K y y ∈ ∼ near y. Let r > 0 and K K (K ). We define Ω := f (K ), and y,r,R r y y,r,R y y,r,R ∈ introduce the numbers Λ♯(Ω ;γ) similarly to Λ♯(K ;γ). y,r,R y,r,R Lemma 4.3. Let µ > 0 be fixed. Then Λ♯(K ;µ/r) y,r,R lim = 1 (4.2) r→+0 Λ♯(Ωy,r,R;µ/r) uniformly over y Γ. ∈ Proof of Lemma 4.3. Let us denote by Ω an image of Ω under the ho- y,r,R y,r,R mothety h : z y+r−1(z y). Conditions imposed on the mapping f imply y,r y 7→ − that as r +0, Ω K in the efollowing sense. The volume element of y,r,R y,1,R → → Ω at a point (h f )(xr) tends to the volume element of K at point y,r,R y,r y y,1,R ◦ x, and the analoegous statement holds for the area element of the boundary. Seince µ is fixed, this implies that the bottoms of the spectra Λ♯(Ω ;µ) (with y,r,R boundary conditions as described above) converge to Λ♯(K ;µ) as r +0. y,1,R → Now the same re-scaling arguments as before imply (4.2). A simpele compactness argument shows that this convergence is uniform in y Γ. ∈ Page 9 M Levitin and L Parnovski Eigenvalue of the Robin problem Remark 4.4. It is easy to see that the estimates of Lemmas 4.2 and 4.3 are uniform in R if we assume that all first partial derivatives of R are bounded by a given constant. We can now conclude the proof of Theorem 3.2 itself. First of all, given an arbitrary positive ǫ and y Γ, we use Lemma 4.2 to find a positive µ(y) such ∈ that ǫ γ−2Λ♯(K ;γ)+C < , (4.3) y,r,R y | | 2 whenever γ r−1µ(y). It is easy to see that µ(y) can be chosen to be con- ≥ tinuous on each smooth component of the boundary. Therefore, there exists µ = supµ(y). Let us fix this value of µ for the rest of the proof. y∈Γ Formula (3.1) splits into two asymptotic inequalities. The inequality e e Λ(Ω;γ) γ2supC +ǫγ2, γ + y ≤ − → ∞ y∈Γ follows immediately from formula (4.3), Lemma 4.3 (with ♯ = D and µ = µ) and the obvious inequality e Λ(Ω;γ) ΛD(Ω ;γ). y,r,R ≤ In order to prove the opposite inequality Λ(Ω;γ) γ2supC +ǫγ2, γ + , (4.4) y ≥ − → ∞ y∈Γ we consider a partition Ω = N Q by disjoint sets Q satisfying the following ℓ=0 ℓ ℓ properties: Q ⋐ Ω (i.e. Q Γ = ), and for each ℓ 1, Q = Ω = 0 0 ℓ y,r,R f (K ) with some r > 0,Fy∩ Γ, a∅nd K K (K ),≥such that Ω K y r,R r,R r y y ∈ ∈ ∼ near y. Such a partition can be constructed for each sufficiently small r > 0 by considering, for example, a partition of Rm into cubes of size r, and including into Q all the cubes which lie strictly inside Ω. Note that Γ = N (Γ Q ). 0 ℓ=1 ∩ ℓ Now we use the following inequality: assuming that J(v;γ) is negative for S Page 10