Sign change for the Green function and for the first eigenfunction of equations of clamped-plate

type

Hans-Christoph Grunau^{∗}  Guido Sweers^{†}

This article has appeared in Archive Rational Mech. Anal. 150, 179-190 (1999). Copyright: Springer-Verlag.

1 Introduction

By results of Boggio ([2], [3]) it is known that on balls in ℝ^{n}, with n ≥ 1, the biharmonic Dirichlet problem is order preserving, that is, the solution u of

\{ \begin{matrix}Δ^{2}u = f & in Ω, \\ u = \frac{∂}{∂n}u = 0 & on ∂Ω, \end{matrix}  (1)

for Ω = {x ∈ ℝ^{n}; |x| < R} satisfies u ≥ 0 whenever f ≥ 0. One also finds that the corresponding first eigenfunction is strictly positive. This biharmonic Dirichlet problem appears in the linearized clamped plate equation.

For the (second-order) Laplace problem such a positivity preserving property holds for every domain in ℝ^{n}. Although an early conjecture of Boggio and Hadamard ([14], [15]) claimed that such a result would hold for the biharmonic operator on arbitrary nice convex domains, numerous counterexamples have been constructed since then. On many domains both the biharmonic Dirichlet problem is not order preserving nor is the first eigenfunction of fixed sign. See the counterexamples to the Boggio-Hadamard Conjecture of [7], [8], [4], [6], [17] and [23]. Two-dimensional domains which are in an appropriate sense close to a ball however do have the sign preserving property. Such a result was obtained in [11].

The sign-preserving property for (1) is equivalent to having a positive Green function. By an application of Jentzsch’s Theorem ([16]), or the Krein-Rutman Theorem ([19]), it follows that a strictly positive Green function implies that the first eigenvalue of

∗Mathematisches Institut, Universität Bayreuth, D-95440 Bayreuth, Germany,

†Algemene Wiskunde, Faculteit I.T.S., T.U. Delft, Mekelweg 4, 2628 CD Delft, The Netherlands,

\{ \begin{matrix}Δ^{2}φ = λφ & in Ω, \\ φ = \frac{∂}{∂n}φ = 0 & on ∂Ω, \end{matrix}  (2)

is simple and that the corresponding eigenfunction is positive. Hence in two dimensions the eigenfunction also remains positive if the domain is ‘not far’ from a ball. By a direct approach it has even been shown ([12]) that in higher dimensions the sign of the eigenfunction remains fixed under small perturbations of the domain.

A question in this area still left open was mentioned to us by James Serrin [24]. Suppose that one considers a smooth deformation of the domain, say t ↦ Ω_{t} with Ω_{0} = B and B a ball. For such a family of domains we rephrase this question as:

Can it happen that ‘positivity preserving’ and ‘first eigenfunction positive’ fail simultaneously?

Or in other words, if t_{e} is the largest number such that (2) on Ω_{t} has a simple first eigenvalue with positive eigenfunction for all t ∈ [0, t_{e}) and if t_{g} is the largest number such that (1) on Ω_{t} is strongly positivity preserving for all t ∈ [0, t_{g}) , can it happen that t_{e} = t_{g}? By the argument using Jentzsch’s result one finds t_{e} ≥ t_{g}. We will show that for appropriately defined deformations the inequality is strict and hence the answer to the question above is negative: t_{e} > t_{g}.

In order to study this question we include a real parameter λ in (1) and consider for Ω ⊂ ℝ^{n}, a bounded domain with ∂Ω ∈ C^{2m+1}, the problem

\{ \begin{matrix}(−Δ)^{m} u = λu + f & in Ω, \\ (\frac{∂}{∂n})^{k} u = 0 & on ∂Ω & with 0 ≤ k ≤ m − 1, \end{matrix}  (3)

and we allow m ∈ {2, 3, …} . Positivity preserving properties of this system have been studied in [12]. For Ω = B_{1} = {x ∈ ℝ^{n}; |x| < 1} there is λ_{c,m,n} ∈ (−∞, 0) such that (3) is positivity preserving if and only if λ ∈ (λ_{c,m,n}, λ_{1,m,B_{1}}) .

Note that, since the realization in L^{2}(Ω), with D = H^{2m,2}(Ω) ∩ H_{0}^{m,2}(Ω), of the boundary value problem (3) is self-adjoint and has a compact inverse, the spectrum is discrete and consists of real eigenvalues. Since the coefficients involved are constant and the boundary is C^{2m+1}, the eigenfunctions are identical for the realization in L^{2} (Ω) as well as in C \binom{}{Ω̄} . Indeed, the eigenfunctions are in C^{2m} \binom{}{Ω̄} by standard regularity results (see [1]). Note that for each domain Ω the first eigenvalue λ_{1,m,Ω} is well defined by the corresponding Rayleighquotient. Isoperimetric questions for the principal eigenvalue of the biharmonic Dirichlet problem have been studied by Talenti in [25].

2 Varying the parameter in the resolvent

Let us denote by λ_{1,m,Ω} ∈ ℝ^{+} the first eigenvalue and by G_{m,λ,Ω} the Green kernel corresponding to (3). In other words, the solution of (3) for λ < λ_{1,m,Ω} can be written by means of this kernel:

u (x) = (G_{m,λ,Ω}f) (x) := ∫_{Ω} G_{m,λ,Ω} (x, y) f (y) dy.

Since the boundary value problem is self-adjoint with respect to the standard inproduct G_{m,λ,Ω} (x, y) = G_{m,λ,Ω} (y, x) for all x ≠ y ∈ Ω. By [1] it follows that G_{m,λ,Ω} (x, ·) ∈ C^{2m,γ} (Ω̄\ {x}) for all x ∈ Ω, γ ∈ (0, 1) .

Since all eigenvalues are real and satisfy λ ≥ λ_{1,m,Ω} one finds that for all μ ∈ ℝ with λ < μ < λ_{1,m,Ω} the solution of

\{ \begin{matrix}(−Δ)^{m} u = μ u + f & in Ω, \\ (\frac{∂}{∂n})^{k} u = 0 & on ∂Ω & with 0 ≤ k ≤ m − 1, \end{matrix}  (4)

with f ∈ C \binom{}{Ω̄} is well defined by the following Neumann series:

u = ∑_{k=0}^{∞} (G_{m,λ,Ω} (μ − λ) )^{k} G_{m,λ,Ω}f.  (5)

Since G_{m,λ,Ω} is a positive definite operator, the spectral radius of G_{m,λ,Ω} satisfies ν (G_{m,λ,Ω}) = (λ_{1,m,Ω} − λ)^{−1} .

Lemma 2.1 Suppose that Ω and λ < λ_{1,m,Ω} are such that

G_{m,λ,Ω} (x, y) ≥ 0 for all x ≠ y ∈ Ω.

Then, for μ ∈ ℝ with λ < μ < λ_{1,m,Ω} one has

G_{m,μ,Ω} (x, y) > 0 for all x ≠ y ∈ Ω.

A related result in an abstract setting can be found in [5].

Proof. Since the operators G_{m,λ,Ω} are nonnegative, it follows that G_{m,μ,Ω} = ∑_{k=0}^{∞} (μ − λ)^{k} G_{m,λ,Ω}^{k+1} ≥ (μ − λ) G_{m,λ,Ω}^{2}, or in other words:

G_{m,μ,Ω} (x, y) ≥ (μ − λ) ∫_{z∈Ω} G_{m,λ,Ω} (x, z) G_{m,λ,Ω} (z, y) dz.  (6)

Now fix x ≠ y ∈ Ω. Note that G_{m,λ,Ω} (x, ·) cannot be identically zero. Indeed, let φ_{1,1} be the eigenfunction of (3) for m = 1 and use (φ_{1,1})^{m} as a testfunction. Since φ_{1,1} > 0 in Ω and (φ_{1,1})^{m} satisfies the boundary conditions we find that G_{m,λ,Ω} (x, ·) ≡ 0 implies

0 < (φ_{1,1} (x))^{m} = G_{m,λ,Ω} (((−Δ)^{m} − λ) (φ_{1,1})^{m} ) (x) = 0,

a contradiction. Hence there are δ, ε > 0 and z_{0} ∈ Ω such that B_{δ} (z_{0}) ⊂ Ω\ {x} and

G_{m,λ,Ω} (x, z) > ε > 0 for z ∈ B_{δ} (z_{0}) .  (7)

Next we show that G_{m,λ,Ω} (·, y) cannot be identically zero on open sets. Suppose that G_{m,λ,Ω} (·, y) = 0 on some open set U ⊂ Ω. Then the unique continuation principle for elliptic equations (see [20]) applied to ((−Δ)^{m} − λ) G_{m,λ,Ω} (·, y) = 0 on Ω\ {y} , implies that G_{m,λ,Ω} (·, y) ≡ 0 on Ω\ {y} , a contradiction. For U = B_{δ} (z_{0}) it follows that G_{m,λ,Ω} (·, y) ≢ 0.

Combining with (6) and (7), we have, with ε^{∗} = (μ − λ) ε,

G_{m,μ,Ω} (x, y) ≥ ε^{∗} ∫_{z∈B_{δ}(z_{0})} G_{m,λ,Ω} (z, y) dz > 0. □

Lemma 2.2 Suppose that Ω and λ < λ_{1,m,Ω} are such that

G_{m,λ,Ω} (x, y) ≥ 0 for all x ≠ y ∈ Ω.

Then, for μ ∈ ℝ with λ < μ < λ_{1,m,Ω}, one has

(\frac{∂}{∂n_{x}} )^{m} G_{m,μ,Ω} (x, y) > 0  for all x ∈ ∂Ω, y ∈ Ω,  (8)

where n_{x} denotes the inward normal.

Proof. Let y ∈ Ω. By the previous lemma G_{m,μ,Ω} (·, y) > 0 holds in Ω and

since (\frac{∂}{∂n_{x}} )^{k} G_{m,μ,Ω} (·, y) = 0 on ∂Ω for k < m, it follows that

(\frac{∂}{∂n_{x}} )^{m} G_{m,μ,Ω} (·, y) ≥ 0 on ∂Ω.  (9)

Next we fix x ∈ ∂Ω. From the resolvent formula

G_{m,μ,Ω} = G_{m,λ,Ω} (I + (μ − λ) G_{m,μ,Ω})

we find

G_{m,μ,Ω} (x, y) = G_{m,λ,Ω} (x, y) + (μ − λ) ∫_{Ω} G_{m,λ,Ω} (x, z) G_{m,μ,Ω} (z, y) dz.

Using G_{m,μ,Ω} (·, y) ≥ 0 and (9), which also holds for μ = λ, it follows that

(\frac{∂}{∂n_{x}} )^{m} G_{m,μ,Ω} (x, y) = (\frac{∂}{∂n_{x}} )^{m} G_{m,λ,Ω} (x, y) +

(μ − λ) ∫_{Ω} (\frac{∂}{∂n_{x}} )^{m} G_{m,λ,Ω} (x, z) G_{m,μ,Ω} (z, y) dz.

By the Green function estimates of Krasovskiĭ ([18]) the integral is well-defined for quite general 2m^{th}-order elliptic boundary value problems on bounded domains in ℝ^{n} :

|\frac{∂^{|α|+|β|}}{∂x^{α}∂y^{β}} G (x, y)| ≤ c_{G,α,β} \{ \begin{matrix}1 & if 2m − n > |α| + |β| , \\ log (1 + |x − y|^{−1}) & if 2m − n = |α| + |β| , \\ |x − y|^{2m−n−|α|−|β|} & if 2m − n < |α| + |β| . \end{matrix}

Since by the previous lemma G_{m,μ,Ω} (z, y) > 0 for all z ∈ Ω\ {y} , we find that

if (\frac{∂}{∂n_{x}} )^{m} G_{m,λ,Ω} (x, z) > 0 for some z ∈ Ω, then the estimate in (8) holds for

all y ∈ Ω. Note that z ↦ (\frac{∂}{∂n_{x}} )^{m} G_{m,λ,Ω} (x, z) ∈ C (Ω̄\ \{x\}) .

Now suppose that (\frac{∂}{∂n_{x}} )^{m} G_{m,μ,Ω} (x, z) = 0 for all z ∈ Ω. Then for every

f ∈ C \binom{}{Ω̄} the solution u of

\{ \begin{matrix}(−Δ)^{m} u = μ u + f & in Ω, \\ (\frac{∂}{∂n})^{k} u = 0 & on ∂Ω & with 0 ≤ k ≤ m − 1, \end{matrix}  (10)

satisfies (\frac{∂}{∂n})^{m} u (x) = 0. Let d (z, ∂Ω) denote the distance of z to ∂Ω, that is

d (z, ∂Ω) = inf {|z − z′|; z′ ∈ ∂Ω} .

Note that d (·, ∂Ω) is Lipschitz continuous for every Ω. However, since ∂Ω ∈ C^{2m+1}, there exists a neighborhood Γ_{δ} = {z ∈ Ω̄; d (z, ∂Ω) < δ} such that d (·, ∂Ω) ∈ C^{2m+1} (Γ_{δ}), see [9, Lemma 14.16]. Let w ∈ C^{∞} \binom{}{Ω̄} be such that w = 1 on Γ_{δ/2} and w = 0 on Ω̄\Γ_{δ}. Then u_{∗}, defined by u_{∗} (z) = d (z, ∂Ω)^{m} w (z), satisfies (10) for f = ((−Δ)^{m} − μ) u_{∗} and (\frac{∂}{∂n})^{m} u_{∗} (x) > 0, a contradiction. □

Corollary 2.3 Suppose that Ω and λ < λ_{1,m,Ω} are such that

G_{m,λ,Ω} (x, y) ≥ 0 for all x ≠ y ∈ Ω.

Then, after suitable normalization, the first eigenfunction φ_{1,m,Ω} is unique and satisfies for some c_{1} > 0

φ_{1,m,Ω} (x) ≥ c_{1} d (x, ∂Ω)^{m} for all x ∈ Ω.

Remark 1. Note that the boundary conditions and the fact that φ_{1,m,Ω} ∈ C^{2m} \binom{}{Ω̄} imply that there is c_{2} > 0 such that

φ_{1,m,Ω} (x) ≤ c_{2} d (x, ∂Ω)^{m} for all x ∈ Ω.

Proof. [of Corollary 2.3] Since G_{m,μ,Ω} (x, y) > 0 for x ≠ y ∈ Ω and μ ∈ (λ, λ_{1,m,Ω}), one may use Jentzsch’s Theorem (see [21, Theorem 6.6] or [16]) to find that G_{m,μ,Ω} has a unique (normalized) eigenfunction φ_{1} with φ_{1} > 0 a.e. in Ω. The stronger estimate is a consequence of φ_{1,m,Ω} = (λ_{1,m,Ω} − μ) G_{m,μ,Ω}φ_{1,m,Ω} and the estimates for G_{m,μ,Ω} (x, y) in the previous two lemmata. □

Remark 2. For the biharmonic operator in one dimension, Schröder [22, Theorem 6.3] gives conditions where positivity breaks down. We take Ω = (−1, 1) and consider

\{ \begin{matrix}u′′′′ (x) = λu (x) + f (x)  for − 1 < x < 1 \\ u (−1) = u′ (−1) = u (1) = u′ (1) = 0. \end{matrix}  (11)

Using his terminology we set M = {(L_{λ}, B) ; λ ∈ I} with L_{λ}u = u′′′′ − λu and Bu = (u (−1) , u′ (−1) , u (1) , u′ (1) ) . Let I be an interval containing 0. For λ = 0 the boundary value problem (11) is inverse positive. According to [22], inverse-positivity breaks down when:

1. L_{λ}u = 0, Bu = 0 has a nontrivial solution, or

2. L_{λ}u = 0, (u (−1) , u′ (−1) , u′′ (−1) , u (1) ) = 0 has a nontrivial solution.

In our case, his third possibility equals the second by symmetry. The first possibility is reached for λ = λ_{1} > 0, with λ_{1} the first eigenvalue. Since

φ_{1} (x) = cosh ( ^{4} \sqrt{λ_{1}}) cos ( ^{4} \sqrt{λ_{1}}x) − cos ( ^{4} \sqrt{λ_{1}}) cosh ( ^{4} \sqrt{λ_{1}}x)

where λ_{1} is such that φ′_{1} (1) = φ′_{1} (−1) = 0, we find that λ_{1} is the smallest positive zero of

tan ( ^{4} \sqrt{λ_{1}}) = − tanh ( ^{4} \sqrt{λ_{1}}) .

The second possibility is reached for λ = λ_{c} < 0 with

φ_{c} (x) = sin (y) cosh (y) − cos (y) sinh (y) ,

where y = ^{4} \sqrt{\frac{−λ_{c}}{4}} (x + 1) . The number λ_{c} is determined by φ_{c} (1) = 0, that is, λ_{c} is the largest negative number such that

tan (2 ^{4} \sqrt{\frac{−λ_{c}}{4}} ) = tanh (2 ^{4} \sqrt{\frac{−λ_{c}}{4}} ) .

A numerical calculation yields λ_{1} = 31.285243 … and λ_{c} = −59.430266 …. For λ ∈ (λ_{c}, λ_{1}) one concludes that (11) is inverse positive.

In this fourth order example Lemma 2.2 is sharp in the sense that for λ < λ_{c} and |λ − λ_{c}| small there exist y ∈ (−1, 1) (close to 1) such that (\frac{∂}{∂x})^{2} G_{2,λ,(−1,1)} (−1, y) < 0. This can be shown by an explicit calculation of these Green functions.

3 Perturbing the domain

It is not obvious that for small perturbations of the domain the Green function or the eigenfunction remains positive. Such problems have been studied in [11] and [12]. In [11] it is shown that in 2-dimensional domains the Green function for λ = 0, and hence also the first eigenfunction, remains positive whenever the domain Ω is close to the disk in ℝ^{2} in C^{2m+1}-sense. Whether or not the Green function remains positive for small perturbations of Ω ⊂ ℝ^{n} with n > 2 is still an open question. The eigenfunction however is shown to remain positive for small domain perturbations in any dimension ([12]).

For a precise statement we need to define a continuous family of C^{k}-domains. The families of domains that we will consider start with Ω_{0} = B, a ball, and will deform C^{k}-smoothly. We will be interested in the largest t_{e}, t_{g} such that the first eigenfunction, respectively the Green function on Ω_{t}, is strictly positive for all t ∈ [0, t_{e}) , respectively t ∈ [0, t_{g}) .

Definition 3.1 Let k be a positive integer. We say that the collection {Ω_{t}; t ∈ [0, 1]} is a continuous family of C^{k}-domains in ℝ^{n} if there exists a family of functions {h_{t} ∈ C^{k} (Ω̄_{0}; ℝ^{n}); t ∈ [0, 1]} such that:

1. Ω_{0} is bounded and ∂Ω_{0} ∈ C^{k};

2. for every t ∈ [0, 1] the mapping h_{t} : Ω̄_{0} → Ω̄_{t} is a C^{k}-diffeomorphism;

3. for every t ∈ [0, 1] we have lim_{s→t} ∥h_{s} − h_{t}∥_{C^{k}(\overline{Ω}_{0})} = 0.

3.1 Influence on a solution

Let {Ω_{t}; t ∈ [0, 1]} be a continuous family of C^{2m+1}-domains in ℝ^{n} and let {h_{t}; t ∈ [0, 1]} be corresponding mappings mentioned in Definition 3.1. For fixed f ∈ C \binom{}{Ω̄_{0}} we consider

\{ \begin{matrix}(−Δ)^{m} u = f ∘ (h_{t}^{−1}) & in Ω_{t}, \\ (\frac{∂}{∂n})^{k} u = 0 for 0 ≤ k ≤ m − 1 & on ∂Ω_{t}. \end{matrix}  (12)

We will be interested in the case that Ω_{0} = B, the unit ball in ℝ^{n}.

Lemma 3.2 Let {Ω_{t}; t ∈ [0, 1]} be a continuous family of C^{2m+1}-domains in ℝ^{n} and let f ∈ C \binom{}{Ω̄_{0}} . For each t let u_{t} denote the solution of (12). Then

lim_{s→t} ∥u_{s} ∘ h_{s} ∘ (h_{t}^{−1}) − u_{t}∥_{C^{2m}\binom{}{Ω̄_{t}}} = 0.

Proof. For every t ∈ [0, 1] we will denote the unique solution of (12) by u_{t}. Let us define two auxiliary functions on Ω_{t}, namely f_{t} = f ∘ (h_{t}^{−1}) and ũ_{s} = u_{s} ∘ h_{s} ∘ (h_{t}^{−1}) . For simplicity we also write g_{s} = h_{s} ∘ (h_{t}^{−1}) . We will compare u_{t} and ũ_{s}.

First notice that for m even

∫_{Ω_{s}} (Δ^{\frac{m}{2}} u_{s})^{2} dx = ∫_{Ω_{s}} u_{s} f_{s} dx;

this together with Cauchy-Schwarz and the Poincaré-inequality implies that

∥u_{s}∥_{H^{m,2}(Ω_{s})} ≤ c_{R,m,n} ∥f_{s}∥_{L^{2}(Ω_{s})} ,  (13)

where the constant c_{R,m,n} depends only on m, n and the radius R of the smallest ball that contains ⋃_{t∈[0,1]} Ω_{t}. For m odd, the same estimate (13) holds using

∫_{Ω_{s}} |∇ (Δ^{\frac{m}{2}} u_{s})|^{2} dx = ∫_{Ω_{t}} u_{s} f_{s} dx. Since ∥f_{s}∥_{L^{2}(Ω_{s})} is uniformly bounded for

s ∈ [0, 1], so is ∥u_{s}∥_{H^{m,2}(Ω_{s})} and hence also \left∥{ \atop ũ_{s}}\right∥_{H^{m,2}(Ω_{t})} .

For w ∈ C^{2} \binom{}{Ω̄_{s}} we have w ∘ g_{s} ∈ C^{2} \binom{}{Ω̄_{t}} and

(Δw) ∘ g_{s} = A_{s} (w ∘ g_{s})

with

A_{s} = ∑_{k=1}^{n} ∑_{l=1}^{n} (∇ (g_{s}^{−1})_{k} · ∇ (g_{s}^{−1})_{l} ) ∘ g_{s} (·) \frac{∂}{∂x_{k}} \frac{∂}{∂x_{l}} +

∑_{k=1}^{n} (Δ (g_{s}^{−1})_{k} ) ∘ g_{s} (·) \frac{∂}{∂x_{k}} ,

where (g_{s}^{−1})_{k} is the k^{th} component of g_{s}^{−1}. The function ũ_{s} satisfies

\{ \begin{matrix}(−A_{s})^{m} u = f_{t} & in Ω_{t}, \\ (\frac{∂}{∂n})^{k} u = 0 for 0 ≤ k ≤ m − 1 & on ∂Ω_{t}. \end{matrix}  (14)

Since (−A_{s})^{m} is uniformly elliptic on Ω_{t} uniformly in s, there are Λ and b such that (−A_{s})^{m} = ∑_{|α|≤2m} a_{α}^{s} (x) (\frac{∂}{∂x})^{α} satisfies

Λ^{−1} |ξ|^{2m} ≤ ∑_{|α|=2m} a_{α}^{s} (x) ξ^{α} ≤ Λ |ξ|^{2m} and ∥a_{α}^{s}∥_{C^{1}} ≤ b for |α| ≤ 2m.

By [1, Theorem 15.2], respectively [1, Theorem 7.3], we find that for all s ∈ [0, 1] there exist constants C_{Ω_{t},Λ,b,n,2m,p} and C_{Ω_{t},Λ,b,n,2m,γ}, which do not depend on s, such that

\left∥{ \atop ũ_{s}}\right∥_{H^{2m,p}(Ω_{t})} ≤ C_{Ω_{t},Λ,b,n,2m,p} (∥f_{t}∥_{L^{p}(Ω_{t})} + \left∥{ \atop ũ_{s}}\right∥_{L^{p}(Ω_{t})}) and
\left∥{ \atop ũ_{s}}\right∥_{C^{2m+γ} \binom{}{Ω̄_{t}}} ≤ C_{Ω_{t},Λ,b,n,2m,γ} (∥f_{t}∥_{C^{γ} \binom{}{Ω̄_{t}}} + \left∥{ \atop ũ_{s}}\right∥_{C^{0}\binom{}{¯Ω_{t}}}) .

Since we have a uniform bound for \left∥{ \atop ũ_{s}}\right∥_{H^{m,2}(Ω_{t})} we may use Sobolev embeddings (on the fixed domain Ω_{t}) and a bootstrapping argument to find a bound for \left∥{ \atop ũ_{s}}\right∥_{C^{2m+γ} \binom{}{Ω̄}} which is uniform in s ∈ [0, 1] . By the property that h_{s} → h_{t} in

C^{2m+1}, and hence g_{s} → g_{t} in C^{2m+1} as s → t, it follows that a_{α}^{s} → a_{α}^{t} in C^{1} for s → t. Since \left∥{ \atop ũ_{s}}\right∥_{C^{2m+γ} \binom{}{Ω̄}} is uniformly bounded we find that

∥((−Δ)^{m} − (−A_{s})^{m} ) ũ_{s}∥_{C^{γ}} → 0 for s → t.

From (−Δ)^{m} ũ_{s} = f_{t} + ((−Δ)^{m} − (−A_{s})^{m} ) ũ_{s} it follows that

lim_{s→t} ∥ũ_{s} − u_{t}∥_{C^{2m}\binom{}{Ω̄_{t}}} = 0. □

3.2 Dependence on the first eigenfunction

The proof in [12] of the result that the eigenfunction remains positive for Ω close to the ball uses the estimate

c_{1} d (x, ∂B)^{m} ≤ φ_{1,m,Ω} (x) ≤ c_{2} d (x, ∂B)^{m} for x ∈ B.  (15)

Since the proof only uses this estimate, and not the fact that B is a ball, we may refine this proof in order to show the following result.

Theorem 3.3 Let {Ω_{t}; t ∈ [0, 1]} be a continuous family of C^{2m+1}-domains in ℝ^{n} with t_{0} ∈ [0, 1] and Ω_{t_{0}} = Ω. Suppose that λ_{1,m,Ω} is the first eigenvalue of (3), that λ_{1,m,Ω} is simple, and that

0 < \underline{c} = inf_{x∈Ω} \frac{φ_{1,m,Ω} (x)}{d (x, ∂Ω)^{m}} ≤ sup_{x∈Ω} \frac{φ_{1,m,Ω} (x)}{d (x, ∂Ω)^{m}} = c̄ < ∞.  (16)

Then for all ε > 0 there is δ > 0 such that for t ∈ [0, 1] with |t − t_{0}| < δ :

1. the first eigenvalue λ_{1,m,Ω_{t}} is simple and |λ_{1,m,Ω} − λ_{1,m,Ω_{t}}| ≤ ε;

2. the corresponding eigenfunction φ_{1,m,Ω_{t}}, after suitable normalization, satisfies

\underline{c} − ε = inf_{x∈Ω_{t}} \frac{φ_{1,m,Ω_{t}} (x)}{d (x, ∂Ω_{t})^{m}} ≤ sup_{x∈Ω_{t}} \frac{φ_{1,m,Ω_{t}} (x)}{d (x, ∂Ω_{t})^{m}} = c̄ + ε  (17)

Proof. See [12, Lemma 5.1 and Theorem 5.2]. For h_{t} as in Definition 3.1, one finds that there exists δ > 0 such that for |t − t_{0}| < δ

∥φ_{1,m,Ω} (·) − φ_{1,m,Ω_{t}} (h_{t} ∘ h_{t_{0}}^{−1} (·))∥_{C^{2m}\binom{}{Ω̄_{t_{0}}}} ≤ ε.

Since

(\frac{∂}{∂n_{x}} )^{k} φ_{1,m,Ω} = (\frac{∂}{∂n_{x}} )^{k} (φ_{1,m,Ω_{t}} ∘ h_{t} ∘ h_{t_{0}}^{−1}) = 0

for all k ∈ {0, …, m − 1} , it follows that

|φ_{1,m,Ω} (x) − φ_{1,m,Ω_{t}} (h_{t} ∘ h_{t_{0}}^{−1} (x))| ≤ ε d (x, ∂Ω)^{m} ,

implying (17). □

3.3 Conclusion for the sign change

Theorem 3.4 Let {Ω_{t} ⊂ ℝ^{n}; t ∈ [0, 1]} be a continuous family of C^{2m+1}-domains with Ω_{0} = B. Set

t_{g} = sup {t ∈ [0, 1] ; G_{m,0,Ω_{s}} (x, y) > 0 for all x ≠ y ∈ Ω and s ≤ t} .

Then there is ε > 0 such that for all t < t_{g} + ε the first eigenvalue is simple and the corresponding eigenfunction satisfies

φ_{1,m,Ω_{t}} > 0 in Ω_{t}.  (18)

Proof. Suppose that G_{m,0,Ω_{tg}} (x, y) changes sign. Since G_{m,0,Ω_{tg}} is continuous on Ω^{2}\ \{(x, x) ; x ∈ Ω\} there is f ∈ C \binom{}{Ω̄_{t_{g}}} with f ≥ 0 and the corresponding solution u being negative somewhere. Let u_{t} be the solution of (12) with this f on the right hand side. Then we find by Lemma 3.2 that u_{t} ∘h_{t} ∘h_{t_{g}}^{−1} → u_{t_{g}} = u uniformly. Since u_{t} ≥ 0 for all t < t_{g} and u_{t_{g}} ≱ 0 we obtain a contradiction. Hence we have G_{m,0,Ω_{tg}} (x, y) ≥ 0 for all x, y ∈ Ω_{t_{g}}^{2}. Then we find by Corollary 2.3 that φ_{1,m,Ω_{tg}} ≥ c d (x)^{m} for some c > 0. By Theorem 3.3 it follows that there is ε > 0 such that (18) holds for all t ∈ [t_{g}, t_{g} + ε) and hence for all t ∈ [0, t_{g} + ε) . □

We have proven that G_{m,0,Ω_{tg}} ≥ 0 and φ_{1,m,Ω_{tg +ε}} > 0 for small positive ε. In order to answer the question whether or nor the Green function remains positive under small perturbations of the domain, that is t_{g} > 0, one would have to show that G_{m,0,Ω_{t}} ≻ 0 implies G_{m,0,Ω_{t+ε}} ≻ 0 for small positive ε. As mentioned before this question is still open in dimensions n ≥ 3. The symbol ≻ denotes a strict ordering in an appropriate lattice. In [10] one finds ≻ defined as follows for the Green function of (3) with 2m > n : G ≻ 0 if and only if there is c > 0 such that for all x, y ∈ Ω

G (x, y) ≥ c (d (x, ∂Ω) d (y, ∂Ω) )^{m− \frac{1}{2} n} min (1, \frac{d (x, ∂Ω) d (y, ∂Ω)}{|x − y|^{2}})^{\frac{1}{2} n} .

Note that from our proofs one finds that G_{m,λ,Ω_{t}} ≻ 0 implies G_{m,μ,Ω_{t}} ≻ 0 for μ ∈ (λ, λ_{1,m,Ω_{t}}) . Assuming that the Green function remains positive in ≻-sense under domain perturbations for all λ ∈ (λ_{c,t}, λ_{1,t}) , a schematic graph of t and λ for which the Green function is positive should have the appearance that is shown in Fig. 1. In the grey area G_{m,λ,Ω_{t}} > 0 holds true.

Figure 1:

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