The biharmonic Steklov boundary value problem: positivity

Figure:The depicted boundary condition for the left endpoint of the beam is clamped whereas for the right endpoint it is hinged. Clamped: u(a) = u0(a)...

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The biharmonic Steklov boundary value problem: positivity preserving and eigenvalues Filippo Gazzola - Politecnico di Milano (Italy)

Minicourse in Mathematical Analysis Padova, June 18-22, 2012 Filippo Gazzola - Politecnico di Milano (Italy)

The biharmonic Steklov problem

OUTLINE 1. The physical model. 2. Boundary conditions. 3. Positivity preserving. 4. Steklov eigenvalues. 5. Minimisation of the least Steklov eigenvalue.

Filippo Gazzola - Politecnico di Milano (Italy)

The biharmonic Steklov problem

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Gazzola • Grunau • Sweers

Lecture Notes in Mathematics

1 LNM 1991

ISSN 0075-8434 ISBN 978-3-642-12244-6

9 783642 122446



springer.com

Polyharmonic Boundary Value Problems

This monograph covers higher order linear and nonlinear elliptic boundary value problems in bounded domains, mainly with the biharmonic or poly-harmonic operator as leading principal part. Underlying models and, in particular, the role of different boundary conditions are explained in detail. As for linear problems, after a brief summary of the existence theory and Lp and Schauder estimates, the focus is on positivity or – since, in contrast to second order equations, a general form of a comparison principle does not exist – on “almost positivity.” The required kernel estimates are also presented in detail. As for nonlinear problems, several techniques well-known from second order equations cannot be utilised and have to be replaced by new and different methods. Subcritical, critical and supercritical nonlinearities are discussed and various existence and nonexistence results are proved. The interplay with the positivity topic from the first part is emphasised and, moreover, a far-reaching Gidas-Ni-Nirenberg-type symmetry result is included. Finally, some recent progress on the Dirichlet problem for Willmore surfaces under symmetry assumptions is discussed.

Lecture Notes in Mathematics

Except for the very last arguments, all the contents of this minicourse and much more can be found in [GGS]

Filippo Gazzola Hans-Christoph Grunau Guido Sweers

Polyharmonic Boundary Value Problems

1991 Positivity Preserving and Nonlinear Higher Order Elliptic Equations in Bounded Domains

123

downloadable for free from my web page. Filippo Gazzola - Politecnico di Milano (Italy)

The biharmonic Steklov problem

THE KIRCHHOFF-LOVE MODEL FOR A THIN PLATE Consider a plate, the vertical projection of which is the planar region Ω ⊂ R2 . A simple model for its elastic energy is Z    2 2 1 − f u dxdy , (∆u) + (1 − σ) u − u u J(u) = xx yy xy 2 Ω

where f = external vertical load, u = vertical deflection. σ is the Poisson ratio: σ =

λ 2(λ+µ)

with the Lam´e constants λ ≥ 0,

µ > 0 that depend on the material, hence 0 ≤ σ < 21 . Usually σ > −1 and some exotic materials have a negative Poisson ratio. For −1 < σ < 1, the quadratic part of the functional is positive.

Filippo Gazzola - Politecnico di Milano (Italy)

The biharmonic Steklov problem

BOUNDARY CONDITIONS: ONE DIMENSIONAL BEAM

Figure: The depicted boundary condition for the left endpoint of the beam is clamped whereas for the right endpoint it is hinged.

Clamped: u(a) = u 0 (a) = 0, also known as homogeneous Dirichlet boundary conditions. Hinged: u(b) = u 00 (b) = 0, also known as homogeneous Navier boundary conditions. This is not the real hinged situation in 2D, due to the boundary curvature!

Filippo Gazzola - Politecnico di Milano (Italy)

The biharmonic Steklov problem

HINGED BOUNDARY CONDITIONS IN 2D For hinged boundary conditions the natural setting is the Hilbert space H 2 ∩ H01 (Ω). Minimising the energy functional leads to the weak Euler-Lagrange equation Z (∆u∆ϕ + (1 − σ) (2uxy ϕxy − uxx ϕyy − uyy ϕxx ) − f ϕ) dxdy = 0 Ω

for all ϕ ∈ H 2 ∩ H01 (Ω). Formally, an integration by parts leads to  Z Z   ∂ 2 ∆u ϕ ds 0 = ∆ u − f ϕ dxdy − Ω ∂Ω ∂ν Z   ∂  2 2 ϕ ds + (1 − σ) ν1 − ν2 uxy − ν1 ν2 (uxx − uyy ) ∂τ ∂Ω Z   ∂ ∆u + (1 − σ) 2ν1 ν2 uxy − ν22 uxx − ν12 uyy + ϕ ds. ∂ν ∂Ω One has u = 0 on ∂Ω and ∆u − (1 − σ) κuν = 0 on ∂Ω. Here κ is the mean curvature of the boundary (κ ≥ 0 for convex boundaries). Filippo Gazzola - Politecnico di Milano (Italy)

The biharmonic Steklov problem

THE STRONG EULER-LAGRANGE EQUATION It reads



∆2 u = f in Ω, u = ∆u − (1 − σ) κuν = 0 on ∂Ω.

In this situation, with an integration by parts, the elastic energy becomes Z Z   1−σ 2 1 κ uν2 dω. J(u) = 2 (∆u) − f u dx − 2 ∂Ω Ω This functional has to be minimised over the space H 2 ∩ H01 (Ω).

Filippo Gazzola - Politecnico di Milano (Italy)

The biharmonic Steklov problem

CLAMPED BOUNDARY CONDITIONS IN 2D For clamped boundary conditions the natural setting is the Hilbert space H02 (Ω). Minimising the energy functional leads to the weak Euler-Lagrange equation Z (∆u∆ϕ + (1 − σ) (2uxy ϕxy − uxx ϕyy − uyy ϕxx ) − f ϕ) dxdy = 0 Ω

for all ϕ ∈ H02 (Ω). Formal integration by parts yield Z Z Z uxy ϕxy dxdy = − ux ϕxyy dxdy = uxx ϕyy dxdy Ω



Z



Z uxy ϕxy dxdy = −



Z uy ϕxxy dxdy =



uyy ϕxx dxdy Ω

so that the weak Euler-Lagrange equation becomes Z  ∆2 u − f ϕ dxdy = 0 ∀ϕ ∈ H02 (Ω). Ω Filippo Gazzola - Politecnico di Milano (Italy)

The biharmonic Steklov problem

THE STRONG EULER-LAGRANGE EQUATION It reads



∆2 u = f in Ω, u = uν = 0 on ∂Ω.

In this situation, the elastic energy becomes Z   2 1 J(u) = (∆u) − f u dx. 2 Ω

This functional has to be minimised over the space H02 (Ω).

Filippo Gazzola - Politecnico di Milano (Italy)

The biharmonic Steklov problem

BOUNDARY CONDITIONS Clamped (Dirichlet): u = uν = 0 on ∂Ω. Hinged (Steklov): u = ∆u − (1 − σ)κuν = 0 on ∂Ω. We may write the fourth order equation as a second order system:   −∆v = f in Ω, −∆u = v in Ω, and v = −(1 − σ)κuν on ∂Ω, u=0 on ∂Ω. Hinged (Navier): u = ∆u = 0 on ∂Ω. We may write the fourth order equation as a second order system:   −∆v = f in Ω, −∆u = v in Ω, and v =0 on ∂Ω, u=0 on ∂Ω.

Filippo Gazzola - Politecnico di Milano (Italy)

The biharmonic Steklov problem

Mixed Navier-Neumann: uν = ∆u = 0 on ∂Ω. We may write the fourth order equation as a second order system:   −∆v = f in Ω, −∆u = v in Ω, and v =0 on ∂Ω, uν = 0 on ∂Ω. R No uniqueness, solvability only under the condition Ω v = 0... too complicated! Neumann-Neumann: uν = (∆u)ν = 0 on ∂Ω. We may write the fourth order equation as a second order system:   −∆v = f in Ω, −∆u = v in Ω, and vν = 0 on ∂Ω, uν = 0 on ∂Ω. R One first needsR Ω f = 0; among solutions v one should choose the one satisfying Ω v = 0; then infinitely many solutions u... too complicated! Filippo Gazzola - Politecnico di Milano (Italy)

The biharmonic Steklov problem

Even worse: ∆u = (∆u)ν = 0 on ∂Ω do not satisfy the complementing condition by Agmon-Douglis-Nirenberg. Well-posedness and elliptic regularity fail! To see this, consider the problem  2 ∆ u=0 ∆u = (∆u)ν = 0

in Ω, on ∂Ω.

Any harmonic function is a solution so that the space of solutions does not have finite dimension. If we take any point x0 ∈ Rn \ Ω, the fundamental solution u0 of −∆ having pole in x0 (namely, u0 (x) = log |x − x0 | if n = 2 and u0 (x) = |x − x0 |2−n if n ≥ 3) is a solution. This shows that it is not possible to obtain uniform a priori bounds in any norm. Indeed, as x0 approaches the boundary ∂Ω it is clear that (for instance!) the H 1 -norm of the solution cannot be bounded uniformly in terms of its L2 -norm. Filippo Gazzola - Politecnico di Milano (Italy)

The biharmonic Steklov problem

Summarising... we consider the following boundary conditions for a bounded domain Ω ⊂ Rn , with n ≥ 2 and a ∈ C 0 (∂Ω): Steklov: u = ∆u − auν = 0 on ∂Ω. Dirichlet: u = uν = 0 on ∂Ω (case a ≡ −∞). Navier: u = ∆u = 0 on ∂Ω (case a ≡ 0). PPP = POSITIVITY PRESERVING PROPERTY Consider the boundary value problem  2 ∆ u=f boundary conditions

in Ω, on ∂Ω.

After defining what is meant by weak solution u ∈ H 2 (Ω) (∩...) we address the following QUESTION: Under which conditions the assumption f ≥ 0 implies that the solution u exists and is positive? Does upwards pushing of a plate yield upwards bending? Remark: Elliptic regularity yields f ∈ L2 (Ω) ⇒ u ∈ H 4 (Ω). Filippo Gazzola - Politecnico di Milano (Italy)

The biharmonic Steklov problem

Before tackling the PPP... DEFINITION A bounded domain Ω ⊂ Rn satisfies a outer ball condition if ∀y ∈ ∂Ω ∃ a ball B ⊂ Rn \ Ω s.t. y ∈ ∂B. It satisfies a uniform outer ball condition if the radius B can be taken independently of y ∈ ∂Ω. In particular, convex domains or domains with smooth boundary are Lipschitz domains which satisfy a uniform outer ball condition. THEOREM 1 Assume that Ω ⊂ Rn is a Lipschitz bounded domain which satisfies a uniform outer ball condition. Then the space H 2 ∩ H01 (Ω) is a Hilbert space when endowed with the scalar product Z (u, v ) 7→ ∆u∆v dx for all u, v ∈ H 2 ∩ H01 (Ω). Ω

This scalar product induces a norm equivalent to k . kH 2 . Filippo Gazzola - Politecnico di Milano (Italy)

The biharmonic Steklov problem

Proof: Under the above assumptions, Adolfsson (Math. Scand. 1992) proved that ∃C > 0 independent of u, such that kukH 2 ≤ C k∆ukL2

for all u ∈ H 2 ∩ H01 (Ω).

For all u ∈ H 2 ∩ H01 (Ω) we also have |D 2 u|2 =

n X

(∂ij u)2 ≥

i,j=1

n X 1 (∂ii u)2 ≥ |∆u|2 n

a.e. in Ω.

i=1

This shows that the two norms are equivalent. 

Filippo Gazzola - Politecnico di Milano (Italy)

The biharmonic Steklov problem

If the domain has a reentrant corner then u 7→ k∆uk2 is not a norm in H 2 ∩ H01 (Ω).  For α ∈ 21 π, π fix the domain  Ωα = (r cos ϕ, r sin ϕ) ∈ R2 ; 0 < r < 1 and |ϕ| < α .  π For ρ = 2α ∈ 12 , 1 the function  vα (r , ϕ) = r −ρ − r ρ cos (ρϕ) satisfies −∆vα = 0 in Ωα , vα = 0 on ∂Ωα \{0} and vα ∈ L2 (Ωα ). Then ∃!bα ∈ H01 (Ωα ) solving −∆bα = vα in Ωα ,

bα = 0 on ∂Ωα .

One has ∆bα 6∈ H01 (Ωα ) and bα 6∈ H 2 (Ωα ).

Filippo Gazzola - Politecnico di Milano (Italy)

The biharmonic Steklov problem

PPP: Navier boundary conditions : u = ∆u = 0 on ∂Ω. The problem may be decomposed into a system.   −∆v = f in Ω, −∆u = v in Ω, and v =0 on ∂Ω, u=0 on ∂Ω. Regardless of the regularity of the boundary ∂Ω, we may apply twice the Lax-Milgram Theorem: ∀f ∈ H −1 (Ω) ∃!v ∈ H01 (Ω)

=⇒

∃!u ∈ H01 (Ω) (∆u ∈ H01 (Ω)).

Applying twice the maximum principle for −∆ we obtain that L2 (Ω) 3 f 0

=⇒ v 0

=⇒ u 0.

The solution u so found is called the system solution. Hence, ∃! system solution and PPP holds. If Ω satisfies a uniform outer ball condition then u has finite energy: u ∈ H 2 (Ω). However,... if Ω has a reentrant corner (nonsmooth) the problem may also admit a finite energy sign changing solution. Filippo Gazzola - Politecnico di Milano (Italy)

The biharmonic Steklov problem

 For α ∈ 12 π, π consider again Ωα . Let f ∈ L2 (Ωα ) and consider the homogeneous Navier problem ∆2 u = f in Ωα ,

u = 0 on ∂Ωα ,

∆u = 0 on ∂Ωα \ {0}.

Let u be the system solution obtained by applying twice the Lax-Milgram Theorem and, ∀c ∈ R let uc = u + cbα . We know that: • ∀c ∈ R we have uc ∈ H01 (Ωα ) and ∆uc ∈ L2 (Ωα ). • ∀c ∈ R uc is a solution and uc ∈ C 0 (Ωα ), ∆uc ∈ C 0 (Ωα \ {0}). • ∆uc ∈ H01 (Ωα ) iff c = 0. • ∀f ∈ L2 (Ωα ) ∃!cα (f ) ∈ R s.t. uc ∈ H 2 ∩ H01 (Ωα ) ⇔ c = cα (f ). Hence, u 6∈ H 2 (Ωα ) whenever cα (f ) 6= 0. See Kondratiev (Trudy Moskovskogo Matematiˇceskogo Obˇsˇcestva, 1967) and Nazarov-Plamenevsky (de Gruyter, 1994). Filippo Gazzola - Politecnico di Milano (Italy)

The biharmonic Steklov problem

Now let f be positive. Nazarov-Sweers (JDE, 2007) show that ucα (f ) ∈ H 2 (Ωα ) but ucα (f ) 0 when α > 43 π and cα (f ) 6= 0. For α ∈ ( 12 π, 43 π) there is only numerical evidence of sign-changing energy solutions:

Figure: The level lines of u and ucα (f ) for f ≥ 0 having a small support near the left top of the domain. Grey region = {x : ucα (f ) (x) < 0}; here, a different scale is used for the level lines.

Filippo Gazzola - Politecnico di Milano (Italy)

The biharmonic Steklov problem

PPP: Dirichlet boundary conditions : u = uν = 0 on ∂Ω. Consider the boundary value problem  2 ∆ u=f in Ω, u = uν = 0 on ∂Ω, where Ω ⊂ Rn is a bounded smooth domain, f a datum in a suitable functional space and u denotes the unknown solution. If Ω is smooth, a unique Green function GΩ exists and Z u(x) = GΩ (x, y )f (y ) dy ∀x ∈ Ω. Ω

PPP ⇐⇒ GΩ ≥ 0.

Filippo Gazzola - Politecnico di Milano (Italy)

The biharmonic Steklov problem

For x, y ∈ B we write q y x 2 2 = |y |x − . [XY ] := |x| |y | − 2x · y + 1 = |x|y − |x| |y | Then  Z

[XY ] |x−y |

GB (x, y ) = kn |x − y |4−n

(σ 2 − 1)σ 1−n dσ > 0 .

1

• T. Boggio, Rend. Circ. Mat. Palermo, 1905 Hence, the implication f 0 =⇒ u 0 is true in balls. A more general formula is available for the polyharmonic Dirichlet Green function.

Filippo Gazzola - Politecnico di Milano (Italy)

The biharmonic Steklov problem

Boggio conjectured that the Green function is always positive (in any domain!). In 1908, Hadamard already knew that this conjecture fails in annuli with small inner radius. He writes that Boggio had mentioned to him that the conjecture was meant for simply connected domains. In the same publication he writes: Malgr´e l’absence de d´emonstration rigoureuse, l’exactitude de cette proposition ne paraˆıt pas douteuse pour les aires convexes. The Boggio-Hadamard conjecture may be formulated as follows: The Green function GΩ for the clamped plate boundary value problem on convex domains is positive. However, this conjecture is wrong.

Filippo Gazzola - Politecnico di Milano (Italy)

The biharmonic Steklov problem

Duffin (J. Math. Phys. 1949, Bull. AMS 1974) showed that the Green function changes sign on a long rectangle.

Garabedian (Pacific J. Math. 1951) showed change of sign of Green’s function in ellipses with ratio of half axes ≈ 1.6. Hedenmalm-Jakobsson-Shimorin (J. Reine Angew. Math. 2002) mention that sign change occurs already in ellipses with ratio of half axes ≈ 1.2. Nakai-Sario (J. Reine Angew. Math. 1977) give a construction how to extend Garabedian’s example also to higher dimensions. Filippo Gazzola - Politecnico di Milano (Italy)

The biharmonic Steklov problem

Sign change is also proved by Coffman-Duffin (Adv. Appl. Math. 1980) in any bounded domain containing a corner, the angle of which is not too large; in particular, squares. Conclusion: neither in arbitrarily smooth uniformly convex nor in rather symmetric domains Green’s function needs to be positive.

Filippo Gazzola - Politecnico di Milano (Italy)

The biharmonic Steklov problem

The Green’s function is positive in suitable perturbations of a planar disc (Grunau-Sweers, Math. Nachr. 1996 & Sassone, Ann. Mat. Pura Appl. 2007). Using the explicit formula from for the lima¸cons de Pascal, Hadamard also claimed to have proven positivity of the Green function GΩ when Ω is such a lima¸con. However, Dall’Acqua-Sweers (Ann. Mat. Pura Appl. 2005) showed that this is not the case.

Figure: Lima¸cons vary from circle to cardioid. The fifth lima¸con from the left is critical for a positive Green function.

Filippo Gazzola - Politecnico di Milano (Italy)

The biharmonic Steklov problem

PPP: Steklov boundary conditions : u = ∆u − auν = 0 on ∂Ω. Let Ω be a bounded domain of Rn (n ≥ 2) with ∂Ω ∈ C 2 and consider the boundary value problem  2 ∆ u=f in Ω, u = ∆u − auν = 0 on ∂Ω, where a ∈ C 0 (∂Ω), f ∈ L2 (Ω). We say that u is a weak solution if u ∈ H 2 ∩ H01 (Ω) and Z Z Z ∆u∆v dx − a uν vν dω = fv dx ∀v ∈ H 2 ∩ H01 (Ω). Ω

∂Ω

Filippo Gazzola - Politecnico di Milano (Italy)



The biharmonic Steklov problem

SUPERHARMONICITY Let Ω ⊂ Rn (n ≥ 2) with ∂Ω ∈ C 2 , H := [H 2 ∩ H01 ] \ H02 (Ω) and Z |∆u|2 Ω Z d1 (Ω) := min . u∈H uν2 ∂Ω −1/2

The minimum is achieved and d1 linear operator

H 2 ∩ H01 (Ω) → L2 (∂Ω)

is the norm of the compact u 7→ uν |∂Ω .

THEOREM 2 Let a ∈ C 0 (∂Ω), f ∈ L2 (Ω), and consider  2 ∆ u=f in Ω, u = ∆u − auν = 0 on ∂Ω. If a < d1 it admits a unique solution u ∈ H 2 ∩ H01 (Ω). If also a ≥ 0 and f 0, then the solution u is strictly superharmonic in Ω. Filippo Gazzola - Politecnico di Milano (Italy)

The biharmonic Steklov problem

Sketch of the proof: On the space H 2 ∩ H01 (Ω) the functional Z Z Z 1 1 |∆u|2 − auν2 − fu I (u) := 2 Ω 2 ∂Ω Ω is strictly convex because a < d1 . The solution is the unique minimiser of I . If a ≥ 0 and f 0, then for all u ∈ H 2 ∩ H01 (Ω) \ {0} the solution w ∈ H 2 ∩ H01 (Ω) to the problem  −∆w = |∆u| in Ω w =0 on ∂Ω satisfies w > 0 in Ω, wν < 0 on ∂Ω, and I (w ) ≤ I (u). 

Filippo Gazzola - Politecnico di Milano (Italy)

The biharmonic Steklov problem

POSITIVITY THEOREM 3 Let a ∈ C 0 (∂Ω), f ∈ L2 (Ω), and consider  2 ∆ u=f in Ω, u = ∆u − auν = 0 on ∂Ω. There exists a number δc := δc (Ω) ∈ [−∞, 0) such that: 1. If a ≥ d1 and if 0  f ∈ L2 (Ω), then 6 ∃ positive solutions. 2. If a = d1 , then ∃ a positive eigenfunction u1 > 0 in Ω for f = 0. Moreover, u1 is unique up to multiples. 3. If a  d1 , then ∀f ∈ L2 (Ω) ∃! solution u. 4. If δc ≤ a  d1 , then 0  f ∈ L2 (Ω) implies u 0 in Ω. 5. If δc < a  d1 , then 0  f ∈ L2 (Ω) implies u ≥ cf d∂Ω > 0 in Ω for some cf > 0. 6. If a < δc , then there are 0  f ∈ L2 (Ω) with 0  u.

Filippo Gazzola - Politecnico di Milano (Italy)

The biharmonic Steklov problem

Flavour of the proof: ∆2 u = f in Ω , u = ∆u − auν = 0 on ∂Ω. EQUIVALENT SYSTEM   −∆v = f in Ω, −∆u = v and v = −auν on ∂Ω, u=0

in Ω, on ∂Ω.

OPERATOR FORMULATION Consider the Green operator G and the Poisson kernel K, formally:  −∆w = f in Ω, w = Gf + Kg ⇐⇒ w =g on ∂Ω. Let (Pw )(x) := −ν · ∇w (x) = −wν (x) for x ∈ ∂Ω. u = Gv = G(Gf + KaPu) = GGf + GKaPu, u = (I − GKaP)−1 GGf .

Filippo Gazzola - Politecnico di Milano (Italy)



The biharmonic Steklov problem

BACK TO THE HINGED PLATE Positivity preserving property for the hinged plate in planar domains. Recall the physical bounds for the Poisson ratio: −1 < σ < 1. COROLLARY 1 Let Ω ⊂ R2 be a bounded convex domain with ∂Ω ∈ C 2 and assume −1 < σ < 1; ∀f ∈ L2 (Ω) ∃!u ∈ H 2 ∩ H01 (Ω) minimiser of the elastic energy functional  Z Z  1−σ |∆u|2 − f u dx − J(u) = κ uν2 dω. 2 2 Ω ∂Ω The minimiser u is the unique weak solution to  2 ∆ u=f in Ω, u = ∆u − (1 − σ) κuν = 0 on ∂Ω. Moreover, f 0 implies that there exists cf > 0 such that u ≥ cf d∂Ω and u is superharmonic in Ω. Filippo Gazzola - Politecnico di Milano (Italy)

The biharmonic Steklov problem

∆2 u = f in Ω, u = ∆u − auν = 0 on ∂Ω. We saw that ∃δc ∈ [−∞, 0) such that: ∃f 0 with u  0

∀f 0 : ∃u and u 0 δc

0

∀f 0 if ∃u then u  0 d1

a −→

QUESTIONS: - What happens if a − d1 changes sign on ∂Ω? - Are there cases where δc = −∞, δc > −∞? For the first question: the function a(x) = n +

2x1 1 + + ε + x1 2 n

satisfies a ∈ C 0 (∂B), a − n changes sign, f 0 implies u 0.

Filippo Gazzola - Politecnico di Milano (Italy)

The biharmonic Steklov problem

THEOREM 4 Let a ∈ C 0 (∂Ω), f ∈ L2 (Ω), and consider  2 ∆ u=f in Ω, u = ∆u − auν = 0 on ∂Ω. If for every m ∈ N and 0  f ∈ L2 (Ω) the solution with a ≡ −m is positive, then for every 0  f ∈ L2 (Ω) the solution u ∈ H02 (Ω) of the Dirichlet problem (u = uν = 0 on ∂Ω) satisfies u 0. This shows that If the Dirichlet problem is not positivity preserving then δc > −∞. The full converse statement is not known: only under additional assumptions on the behaviour of the Green function of the Dirichlet problem.

Filippo Gazzola - Politecnico di Milano (Italy)

The biharmonic Steklov problem

THEOREM 5 If Ω = B, the unit ball in Rn (n ≥ 2), then d1 = n. Moreover, for all 0  f ∈ L2 (B) and all a ∈ C 0 (∂B) such that a  n, there exists c > 0 such that the weak solution u satisfies u ≥ cd∂Ω in B.

Filippo Gazzola - Politecnico di Milano (Italy)

The biharmonic Steklov problem

Sketch of the proof: Assume first that f ∈ Cc∞ (B) so that u ∈ W 2,p (B) for all p > 1. In turn, u ∈ C 1 (B) and hence ∆u = auν ∈ C 0 (∂B). Therefore, u ∈ C ∞ (B) ∩ C 2 (B). Consider the auxiliary function φ ∈ C ∞ (B) ∩ C 0 (B) defined by φ(x) = (|x|2 − 1)∆u(x) − 4x · ∇u(x) − 2(n − 4)u(x). Then φ satisfies the second order Steklov boundary value problem  −∆φ = (1 − |x|2 )f ≥ 0 in B, on ∂B. φν + 12 (n − a)φ = 0 As a  n, by the maximum principle (for this second order problem!) we infer that φ > 0 in B and uν < 0 on ∂B. Therefore, −∆u > 0 in B whenever 0  f ∈ Cc∞ (B). Assume now 0  f ∈ L2 (B) and let u ∈ H 2 ∩ H01 (B) be the unique weak solution. A density argument shows that u 0 in B and therefore δc = −∞. Finally, the lower bound u ≥ cd∂Ω in B follows by comparison with the solutions for smaller a.  Filippo Gazzola - Politecnico di Milano (Italy)

The biharmonic Steklov problem

STEKLOV EIGENVALUES Let Ω ⊂ Rn (n ≥ 2) be a bounded domain with ∂Ω ∈ C 2 , let d ∈ R and consider the boundary eigenvalue problem  2 in Ω  ∆ u=0 u=0 on ∂Ω  ∆u = duν on ∂Ω . EIGENVALUE: A value of d for which the problem admits nontrivial solutions, the corresponding EIGENFUNCTION u ∈ H 2 ∩ H01 (Ω) which satisfies Z Z ∆u∆v dx = d uν vν ds for all v ∈ H 2 ∩ H01 (Ω). Ω

∂Ω

By taking v = u: all the eigenvalues are strictly positive. The least eigenvalue is the threshold for positivity: Z |∆u|2 ZΩ d1 = min , H(Ω) := [H 2 ∩ H01 (Ω)] \ H02 (Ω). u∈H(Ω) 2 uν ∂Ω Filippo Gazzola - Politecnico di Milano (Italy)

The biharmonic Steklov problem

THE SPECTRUM OF THE BIHARMONIC STEKLOV PROBLEM We endow the Hilbert space H 2 ∩ H01 (Ω) with the scalar product Z (u, v ) = ∆u∆v dx . Ω

Consider the space  Z = v ∈ C ∞ (Ω) : ∆2 u = 0, u = 0 on ∂Ω and denote by V the completion of Z with respect to the norm associated to the scalar product (·, ·).

Filippo Gazzola - Politecnico di Milano (Italy)

The biharmonic Steklov problem

THEOREM 6 Assume that Ω ⊂ Rn (n ≥ 2) is an open bounded domain with ∂Ω ∈ C 2 . Then: • there exist infinitely many (countable) eigenvalues; • the set of eigenfunctions is a complete orthonormal system in V ; • the only eigenfunction of fixed sign is the one corresponding to the first eigenvalue; • the space H 2 ∩ H01 (Ω) admits the orthogonal decomposition H 2 ∩ H01 (Ω) = V ⊕ H02 (Ω). • if v ∈ H 2 ∩ H01 (Ω) and if v = v1 + v2 is the decomposition, then v1 ∈ V and v2 ∈ H02 (Ω) solve  2  2  ∆ v1 = 0 in Ω  ∆ v2 = ∆2 v in Ω v1 = 0 on ∂Ω v2 = 0 on ∂Ω and   (v1 )ν = vν on ∂Ω (v2 )ν = 0 on ∂Ω .

Filippo Gazzola - Politecnico di Milano (Italy)

The biharmonic Steklov problem

When Ω = B, much more can be said. Consider the spaces of harmonic homogeneous polynomials: Dk := {P ∈ C ∞ (Rn ); ∆P = 0 in Rn , P is an homogeneous polynomial of degree k − 1}. Denote by µk the dimension of Dk . In particular, we have D1 = span{1} ,

µ1 = 1 ,

D2 = span{xi ; (i = 1, ..., n)} , D3 =

span{xi xj ; x12



xh2 ;

µ2 = n ,

(i, j = 1, ..., n, i 6= j, h = 2, ..., n)} ,

µ3 =

n2 + n − 2 . 2

Filippo Gazzola - Politecnico di Milano (Italy)

The biharmonic Steklov problem

THEOREM 7 If n ≥ 2 and Ω = B, then for all k = 1, 2, 3, ...: (i) the Steklov eigenvalues are dk = n + 2(k − 1); (ii) the multiplicity of dk equals µk ; (iii) for all ψk ∈ Dk , the function φk (x) := (1 − |x|2 )ψk (x) is an eigenfunction corresponding to dk . REMARK If n = 1, the problem u iv = 0

in (−1, 1)

u(±1) = u 00 (−1) + du 0 (−1) = u 00 (1) − du 0 (1) = 0, admits only two eigenvalues, d1 = 1 and d2 = 3, each one of multiplicity 1. The reason of this striking difference is that the “boundary space” has dimension 2, one for each endpoint of the interval (−1, 1). This result is consistent with Theorem 7 since µ1 = µ2 = 1 and µ3 = 0 whenever n = 1.

Filippo Gazzola - Politecnico di Milano (Italy)

The biharmonic Steklov problem

Proof: An eigenfunction u satisfies u ∈ C ∞ (B) and may be written as u(x) = (1 − |x|2 )φ(x) with φ ∈ C ∞ (B) (this is a nontrivial step!). Some computations show that the number d is an eigenvalue with corresponding eigenfunction u if and only if φ is an eigenfunction of the boundary eigenvalue problem  ∆φ = 0 in B φν = aφ on ∂B, where a =

d−n 2 .

The problem reduces to study the eigenvalues of this second order Steklov problem for which a ∈ N. 

Filippo Gazzola - Politecnico di Milano (Italy)

The biharmonic Steklov problem

A PRIORI ESTIMATES FOR HARMONIC FUNCTIONS

Let Ω ⊂ Rn be a bounded domain with smooth boundary. Let g ∈ L2 (∂Ω) and consider the problem  ∆v = 0 in Ω v =g on ∂Ω . Which is the optimal constant δ1 (Ω) for the a priori estimate δ1 (Ω) · kv k2L2 (Ω) ≤ kg k2L2 (∂Ω) ?

Filippo Gazzola - Politecnico di Milano (Italy)

The biharmonic Steklov problem

VARIATIONAL CHARACTERISATION OF δ1 Let H := closure of {v ∈ C 2 (Ω); ∆v = 0 in Ω} with respect to the norm k · kH := k · kL2 (∂Ω) . Let Z δ1 = δ1 (Ω) :=

min h∈H\{0}

h2

Z∂Ω h2

.



THEOREM 8 If Ω ⊂ Rn is open bounded with Lipschitz boundary, then δ1 (Ω) admits a minimiser h ∈ H\{0}. The regularity of the boundary plays a crucial role in the previous statement and in what follows.

Filippo Gazzola - Politecnico di Milano (Italy)

The biharmonic Steklov problem

Proof: Let {hm } ⊂ H\ {0} be a minimising sequence for δ1 (Ω) with khm kL2 (∂Ω) = 1. Hence, ∃h ∈ H s.t. (up to a subsequence) hm * h in L2 (∂Ω). By elliptic estimates (due to Jerison-Kenig), we infer that ∃C > 0 s.t. kw kH 1/2 (Ω) ≤ C kw kL2 (∂Ω)

∀w ∈ H.

Hence, {hm } is bounded in H 1/2 (Ω), hm * h in H 1/2 (Ω) up to a subsequence and, by compact embedding, also hm → h in L2 (Ω). Therefore, since {hm } is a minimising sequence, khm kL2 (∂Ω) = 1 and khm kL2 (Ω) is bounded then δ1 (Ω) > 0, h ∈ H\ {0} and −2 khk−2 L2 (Ω) = lim khm kL2 (Ω) = δ1 (Ω) . m→∞

Moreover, by weak lower semicontinuity of k·kL2 (∂Ω) we also have khk2L2 (∂Ω) ≤ lim inf khm k2L2 (∂Ω) = 1 m→∞

and hence h ∈ H\ {0} satisfies khk2L2 (∂Ω) ≤ δ1 (Ω) khk2L2 (Ω) . This proves that h is a minimiser for δ1 (Ω).  Filippo Gazzola - Politecnico di Milano (Italy)

The biharmonic Steklov problem

THEOREM 9 (FICHERA’S PRINCIPLE OF DUALITY) Assume that ∂Ω is Lipschitz and that Ω satisfies a uniform outer ball condition. Let H := closure of {v ∈ C 2 (Ω); ∆v = 0 in Ω} with respect to the norm k · kH := k · kL2 (∂Ω) . Let H := [H 2 ∩ H01 (Ω)] \ H02 (Ω). Z Z 2 h |∆u|2 ∂Ω Ω Z Z δ1 (Ω) = min = min = d1 (Ω). u∈H h∈H\{0} 2 2 h uν Ω

∂Ω

The already mentioned results by Nazarov-Sweers (JDE, 2007) suggest that this result might become false in domains with a concave corner. Filippo Gazzola - Politecnico di Milano (Italy)

The biharmonic Steklov problem

Proof: (assuming that ∂Ω ∈ C 2 ). We say that δ is an eigenvalue relative to the minimisation problem Z h2 ∂Ω Z min h∈H\{0} h2 Ω

if ∃g ∈ H such that Z Z δ gv dx = Ω

gv ds

∀v ∈ H.

∂Ω

Note that the Euler equation contains no derivatives ! Clearly, δ1 is the least eigenvalue. We show that both δ1 ≥ d1 and δ1 ≤ d1 . Filippo Gazzola - Politecnico di Milano (Italy)

The biharmonic Steklov problem

Let h ∈ H \ {0} be a minimiser for δ1 , then Z Z δ1 hv = hv ∀v ∈ H . Ω

(1)

∂Ω

Let u ∈ H solve ∆u = h in Ω, u = 0 on ∂Ω. By integrating Z Z Z hv = v ∆u = vuν ∀v ∈ H ∩ C 2 (Ω). Ω



∂Ω

By density, this holds ∀v ∈ H. Inserting into (1) gives Z Z δ1 vuν = v ∆u for all v ∈ H . ∂Ω

∂Ω

This yields ∆u = δ1 uν on ∂Ω. Therefore, R R h2 |∆u|2 ∂Ω R R∂Ω δ1 = = = δ12 2 2 h |∆u| Ω Ω In turn, this implies that R |∆u|2 RΩ δ1 = 2 ∂Ω uν

≥ min

Filippo Gazzola - Politecnico di Milano (Italy)

v ∈H

R |∆v |2 RΩ 2 ∂Ω vν

R u2 R ∂Ω ν 2 Ω |∆u|

= d1 .

The biharmonic Steklov problem

.

Let u be a minimiser for d1 , then ∆u = d1 uν on ∂Ω so that ∆u ∈ H 1/2 (∂Ω) ⊂ L2 (∂Ω) and Z Z v ∆u = d1 vuν for all v ∈ H . ∂Ω

∂Ω

Let h := ∆u so that h ∈ L2 (Ω) ∩ L2 (∂Ω). Moreover, ∆h = ∆2 u = 0 (in distributional sense) and hence h ∈ H. Two integrations by parts (and a density argument) yield Z Z Z hv = ∆uv = vuν for all v ∈ H . Ω



∂Ω

Replacing this into (2) gives Z Z hv = d1 hv ∂Ω

for all v ∈ H .



This proves that h is an eigenfunction relative to the harmonic problem with corresponding eigenvalue d1 . Since δ1 is the least eigenvalue, we obtain d1 ≥ δ1 .  Filippo Gazzola - Politecnico di Milano (Italy)

The biharmonic Steklov problem

(2)

As a byproduct of the proof, we see that the minimisers are related by h = ∆φ, up to a multiple.

Z δ1 =

Z

h2

Z∂Ω h2 Ω

=

|∆φ|2

Z∂Ω |∆φ|2 Ω

Filippo Gazzola - Politecnico di Milano (Italy)

=

d12 Z

Z

φ2ν

∂Ω

|∆φ|2



The biharmonic Steklov problem

= d1 .

MINIMISATION OF THE LEAST EIGENVALUE For the second order Steklov problem ∆u = 0

in Ω ,

uν = λu

on ∂Ω ,

the first (nontrivial) eigenvalue satisfies λ1 (Ω) ≤ λ1 (Ω∗ ). • F. Brock, ZAMM 2001 But the fourth order Steklov problem appears completely different! Smith conjectures (and proves!) that for any domain Ω, one has d1 (Ω) ≥ d1 (Ω∗ ). • J. Smith, SIAM J. Numer. Anal. 1968 In particular, for planar domains Ω of measure π (as the unit disk), this means that d1 (Ω) ≥ 2. As noticed by Kuttler and Sigillito, this proof contains a gap. In a “Note added in proof” Smith writes: Although the result is probably true, a correct proof has not yet been found. • J. Smith, SIAM J. Numer. Anal. 1970 Filippo Gazzola - Politecnico di Milano (Italy)

The biharmonic Steklov problem

√ Kuttler also shows that for the square Q√π = (0, π)2 one has d1 (Q√π ) < 1.9889 . • J.R. Kuttler, SIAM J. Numer. Anal. 1972 This estimate may be improved to d1 (Q√π ) < 1.96256. Proof: Let h(x, y ) := x 4 + y 4 − 6x 2 y 2 + 2.69, then ∆h = 0 and Z h2 ∂Q√π

< 1.96256 .

Z h

2



Q√π

• A. Ferrero, F. Gazzola, T. Weth, Analysis 2005 Kuttler suggests a new and weaker conjecture. Let Ω ⊂ Rn be a smooth bounded domain such that |∂Ω| = |∂B|. Then, n = d1 (B) ≤ d1 (Ω). His numerical results on some rectangles support this conjecture. Filippo Gazzola - Politecnico di Milano (Italy)

The biharmonic Steklov problem

THEOREM 10 Let Dε = {x ∈ R2 ; ε < |x| < 1} and let Ωε = Dε × (0, 1)n−2 . Then lim d1 (Ωε ) = 0. ε→0

This Theorem disproves the conjecture by Kuttler: there is no minimiser to d1 among all regions having the same perimeter.

Filippo Gazzola - Politecnico di Milano (Italy)

The biharmonic Steklov problem

1 − |x|2 1 − ε2 − log |x| 4 4 log ε ∀x ∈ Dε . Then wε ∈ H 2 ∩ H01 (Dε ) and Z  |∆wε |2 dx = π 1 − ε2 , Proof: For any ε ∈ (0, 1) let wε (x) =

Ωε

Z ∂Ωε

(wε )2ν



π 1 ds = +o 8 ε log2 ε

Hence,

Z

1 ε log2 ε

→ +∞ as ε → 0.

|∆wε |2 dx

lim d1 (Ωε ) ≤ lim Z Ωε

ε→0



ε→0

∂Ωε

= 0. (wε )2ν ds

For n ≥ 3, let uε (x) =

n Y

! xi (1 − xi ) wε (x1 , x2 )

∀x ∈ Ωε

i=3

and compute as above.  Filippo Gazzola - Politecnico di Milano (Italy)

The biharmonic Steklov problem

BOUNDS IN CONVEX DOMAINS THEOREM 11 Let Ω ⊂ Rn (n ≥ 2) be a bounded convex domain with ∂Ω ∈ C 2 . • For all x ∈ ∂Ω, let κ(x) denote the mean curvature at x and let K := min κ(x) ; x∈∂Ω

then d1 (Ω) ≥ nK with equality if and only if Ω is a ball. • The following isoperimetric bound holds d1 (Ω) ≤

|∂Ω| |Ω|

with equality if and only if Ω is a ball.

Filippo Gazzola - Politecnico di Milano (Italy)

The biharmonic Steklov problem

Proof: For the lower bound, let φ be a first Steklov eigenfunction such that φ > 0 in Ω and φν < 0 on ∂Ω. The boundary condition ∆φ = d1 φν on ∂Ω also reads φνν + (n − 1) κφν = d1 φν on ∂Ω. Hence,  φ2ν ν = 2φνν φν = 2 [d1 − (n − 1) κ] φ2ν n P

so that if we put D 2 φD 2 φ =

(∂ij φ)2 and integrate by parts

i,j=1

Z [d1 − (n −

2

1)κ] φ2ν

Z ds =

∂Ω

Z

2

∆ |∇φ|



Z



= −2

φ∆2 φ dx + 2

=2 Ω



Z



Z

Z φ(∆φ)ν ds + 2

∂Ω

D 2 φD 2 φ dx ≥

2 n

ν

D 2 φD 2 φ dx



Z Ω

|∆φ|2 dx =

2d1 n

Z

φ2ν ds

∂Ω

the latter since φ is the first eigenfunction. Hence, d1 ≥ nK . Filippo Gazzola - Politecnico di Milano (Italy)

ds

D 2 φD 2 φ dx

∇∆φ · ∇φ dx + 2

dx = 2



∂Ω

Z



Z

|∇φ|2

ds =

∂Ω

Z =

(φ2ν )ν

The biharmonic Steklov problem

We prove that equality holds if and only if Ω is a ball. If d1 = nK , then d1 ≤ nκ(x) for x ∈ ∂Ω and since φν < 0 on ∂Ω, from  Z  d1 − κ φ2ν ds ≥ 0, n ∂Ω we infer that κ(x) ≡ dn1 . Hence, Ω is a ball by Alexandrov’s characterisation of spheres (Ann. Mat. Pura Appl. 1962). The upper bound is obtained by taking h ≡ 1 as harmonic test function in Fichera’s characterisation: Z Z 2 d1 |Ω| = d1 1 dx ≤ 12 ds = |∂Ω|. Ω

∂Ω

If equality holds, then the first eigenfunction φ satisfies −∆φ = h = 1

in Ω,

φ = 0, φν = −d1−1

on ∂Ω,

and Ω is a ball by a result of J. Serrin (ARMA, 1971).  Filippo Gazzola - Politecnico di Milano (Italy)

The biharmonic Steklov problem

EXISTENCE OF OPTIMAL CONVEX SHAPES THEOREM 12 Among all convex domains in Rn having the same measure as the unit ball B, there exists an optimal one, minimising d1 . Among all convex domains in Rn having the same perimeter as the unit ball B, there exists an optimal one, minimising d1 . These results should be complemented with the description of the optimal convex shapes. This appears quite challenging since, with the measure constraint, the optimal planar domain is not a disk.

Filippo Gazzola - Politecnico di Milano (Italy)

The biharmonic Steklov problem

Sketch of the proof: STEP 1 The map Ω 7→ d1 (Ω) is continuous with respect to Hausdorff convergence of convex domains. This fact appears nontrivial since there is no monotonicity with respect to inclusions and no obvious extension operator from H 2 ∩ H01 (Ω) to H 2 (Rn ). Upper semicontinuity of the map Ω 7→ δ1 (Ω). Lower semicontinuity of the map Ω 7→ d1 (Ω). STEP 2 A lower bound for d1 on convex domains. By comparison with the solution to the torsion problem, Payne (Indian J. Mech. Math. 1968/69) proved that if ρΩ denotes the minimal distance between parallel planes which define a strip containing Ω then d1 (Ω) ≥ 2ρ−1 Ω . STEP 3 Conclusion. Consider a minimising sequence {Ωm } ⊂ Rn for d1 . By STEP 2 we know that ∃R > 0 such that Ωm ⊂ BR for all m, since otherwise d1 (Ωm ) → +∞. This fact, combined with STEP 1 and with Blaschke selection Theorem, shows that the infimum is achieved.  Filippo Gazzola - Politecnico di Milano (Italy)

The biharmonic Steklov problem

NUMERICAL RESEARCH FOR THE OPTIMAL PLANAR SHAPES We have no theoretical evidence of what the optimal convex shapes could be. Recently, Antunes-Gazzola (ESAIM COCV 2012) performed several numerical experiments. In the plane, we apply the Method of Fundamental Solutions (MFS). The MFS is a meshfree numerical method for which the approximations are linear combinations of fundamental solution associated to the pde, having pole outside Ω.

Filippo Gazzola - Politecnico di Milano (Italy)

The biharmonic Steklov problem

FIXED AREA Let Ωn be a regular n-polygon of measure π having n sides and let D be the unit disk.Then n d1 (Ωn )

3 2.02522

4 1.96179

5 1.95702

6 1.96164

7 1.96733

8 1.97255

9 1.97654

10 1.97974

Remarks. The equilateral triangle Ω3 is the maximiser. We saw before that d1 (Ω4 ) < 1.96256 while d1 (Ω4 ) ≈ 1.96179. The regular pentagon Ω5 is the minimiser. It seems that n 7→ d1 (Ωn ) tends monotonically to 2 = d1 (D) for n ≥ 5 and n → ∞. We tested Reuleaux polygons, irregular polygons up to 8 sides, deformations form regular polygons to the disk, ellipses, stadiums: Ω5 remains a good candidate to be the absolute minimiser.

Filippo Gazzola - Politecnico di Milano (Italy)

The biharmonic Steklov problem

A RELATED PROBLEM? (for a simply supported plate) ∆2 u + cu = f

in Ω,

u = ∆u = 0

on ∂Ω ,

where Ω ⊂ R2 is a bounded domain, c > 0 is the “stiffness of the resistance to deformation” f ∈ L2 (Ω) is the external load, u is the vertical deformation. PPP: under which conditions on c > 0 and Ω the assumption f ≥ 0 implies that the solution u exists and is positive? • P.J. McKenna, W. Walter, ARMA 1987 • B. Kawohl, G. Sweers, Indiana UMJ 2002 ∃ a maximal interval c ∈ (0, c ∗ (Ω)] where ppp holds. Which is the largest c ∗ (Ω) when Ω varies among convex planar domains of given measure? Numerical results show that, among regular polygons, the maximum is attained by the pentagon Ω5 . • R.F. Bass, J. Hor´ak, P.J. McKenna, Proc. AMS 2004 Since d1 is the threshold parameter in order to have the ppp for the Steklov problem, are these results somehow connected? Filippo Gazzola - Politecnico di Milano (Italy)

The biharmonic Steklov problem

FIXED PERIMETER Let Ω]n be a regular n-polygon of perimeter 2π having n sides. Then n d1 (Ω]n )

3 2.60458

4 2.21364

5 2.10443

6 2.05987

7 2.03791

8 2.02586

9 2.01830

10 2.01336

Remarks. It seems that now n 7→ d1 (Ω]n ) tends monotonically to 2 = d1 (D) for n ≥ 3 and n → ∞. The disk D has d1 smaller than any regular polygon. We tested Reuleaux polygons, irregular polygons up to 8 sides, deformations form regular polygons to the disk, ellipses, stadiums: the disk D is a good candidate to be the absolute minimiser.

Filippo Gazzola - Politecnico di Milano (Italy)

The biharmonic Steklov problem

BIBLIOGRAPHY THE KIRCHHOFF-LOVE MODEL FOR A THIN PLATE ¨ • G.R. Kirchhoff, Uber das Gleichgewicht und die Bewegung einer elastischen Scheibe, J. Reine Angew. Math. 1850 • K. Friedrichs, Die Randwert und Eigenwertprobleme aus der Theorie der elastischen Platten, Math. Ann. 1927 • A.E.H. Love, A Treatise on the Mathematical Theory of Elasticity, IV edition, 1927 • R.S. Lakes, Foam structures with a negative Poisson’s ratio, Science, 1987 • P. Villaggio, Mathematical models for elastic structures, 1997 THE STRONG EULER-LAGRANGE EQUATION • W. Stekloff, Sur les probl`emes fondamentaux de la physique ´ math´ematique, Ann. Sci. Ecole Norm. Sup. 1902 • P. Destuynder, M. Salaun, Mathematical Analysis of Thin Plate Models, 1996 Filippo Gazzola - Politecnico di Milano (Italy)

The biharmonic Steklov problem

Theorem 1: see Theorem 2.3.1 p.53 in [GGS] Theorem 2: E. Berchio, F. Gazzola, E. Mitidieri, JDE 2006 Theorems 3-4-5: F. Gazzola, G. Sweers, ARMA 2008 Corollary 1: E. Parini, A. Stylianou, SIAM J. Math. Anal. 2009 Theorems 6-7: A. Ferrero, F. Gazzola, T. Weth, Analysis 2005 Theorem 8-9-10: D. Bucur, A. Ferrero, F. Gazzola, Calc. Var. 2009 Theorem 9: (smooth version) G. Fichera, Atti A.N. Lincei 1955 Theorem 11: L.E. Payne, SIAM J. Math. Anal. 1970 & J.R. Kuttler, SIAM J. Numer. Anal. 1972 & A. Ferrero, F. Gazzola, T. Weth, Analysis 2005 Theorem 12: D. Bucur, F. Gazzola, MJM 2011 & P. Antunes, F. Gazzola, ESAIM COCV 2012

Filippo Gazzola - Politecnico di Milano (Italy)

The biharmonic Steklov problem

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Filippo Gazzola - Politecnico di Milano (Italy)

The biharmonic Steklov problem