Well-posedness of the Dirichlet problem in non-elliptic isotropic elasticity

Z. angew. Math. Phys. 53 (2002) 73–890010-2571/02/010073-17 $ 1.50+0.20/0c© 2002 Birkhauser Verlag, Basel

Zeitschrift fur angewandteMathematik und Physik ZAMP

Well-posedness of the Dirichlet problem in non-elliptic isotropicelasticity

Emil Ernst

Abstract. The well-posedness of the Dirichlet boundary value problem for an isotropic homo-geneous linear elastic material is established for non-elliptic equilibrium systems. Mechanicalproperties of elastic materials with negative shear modulus are also deduced.

Keywords. Negative shear modulus, non-elliptic linear elasticity, well-posed problems.

1. Introduction

When the Lame coefficients of an isotropic homogeneous linear elastic solid satisfyµ > 0 and 2µ + 3λ > 0 , the equilibrium equation is strongly elliptic. Thus, ifΩ , the domain occupied by the body and Γ0 , the surface where the displacementis prescribed, are regular (see Section 2), for every boundary condition u0 ∈H

12 (Γ0)3 , the Dirichlet equilibrium problem

div (2µE(u) + λ tr(E(u))I3) = 0 (1.1)u = u0 on Γ0, (2µE(u) + λ tr(E(u))I3)[n] = 0 on ∂Ω \ Γ0,

where E(u) is the symmetric part of the gradient of u and n is the exteriornormal at ∂Ω , has exactly one H 1(Ω)3 solution.

The H 1 convergence of the solution for fixed Dirichlet data when the coeffi-cients of the equilibrium system move toward the border of the domain in whichthe equilibrium system is elliptical (see [4] for a analysis of the case µ → 0 , or[2] for the study of a class of non-necessarily isotropic systems), suggests that thewell-posedness of the Dirichlet problem could be established also outside the classof elliptic equilibrium systems.

The main result of this paper (Theorem 3) states that the well-posedness sub-sists even in the non-elliptic case µ < 0 , provided that λ > 0 , and that µ/λ isstrictly greater then a strictly negative parameter depending on Ω and Γ0 , buteven the mere existence of the solutions does not longer hold when µ/λ actuallyreaches the value of the geometrical parameter.

For domains with very simple geometry, analytic estimations of the well-posed-

74 E. Ernst ZAMP

ness limit (generally available only through numerical methods) may also be per-formed. The results of the last Section (Propositions 10 and 11), valid for rightcylinders with imposed displacements either on their lateral surface, or on theirterminal ends, allow us to deduce some rather strange mechanical properties ofisotropic elastic materials with negative shear modulus.

Theorem 3 yields that for different domains, the well-posedness of the Dirichletproblem ceases for different values of the Lame’s coefficients. The lost of well-posedness involves thus both the geometrical properties of the domain, and theanalytical properties of the differential operator, character shared with most of theproblems involving stability loss.

2. The main result

In this paper we shall assume that the regularity of the domain occupied by thebody and of the part of the border where the imposed displacement is prescribedis such that classical Sobolev space results hold (for details see [1]). We may take,for instance, open, connected and bounded sets with the cone property and a C1

border, except perhaps on a finite number of compact C1 curves for the domain,and finite reunion of open and connected parts of its border limited by piecewiseC1 curves, for the surface where the displacement is prescribed.

Let us put H for the Hilbert space of all 3 × 3 symmetric L2 matrix fieldsover Ω , endowed with the standard scalar product

〈w1,w2〉 =∫

Ω

tr (w1(x) ·w2T(x)) dx =

∫Ω

tr (w1(x) ·w2(x)) dx. (2.2)

The space H is split both as H1 + H2 , where

H1 =w I3 : w ∈ L2(Ω)

H2 = H⊥

1 =w ∈ H :

∫Ω

tr(w) dx = 0

,

and as U + V , where

U =E(u) : u ∈ H1(Ω)3, u = 0 on Γ0

,

and V = U⊥ . On its turn, V is split as V1 + V2 , where V1 = H1 ∩ V andV2 = V⊥

1 ∩V , as well as W1+W2 , where W1 = E(H1(Ω)3)∩V , and W2 = W⊥1 ∩V

= E(H1(Ω))3)⊥ .Let for every X , closed linear subspace of H , P X be the orthogonal projection

on X ,PX : H 7→ X, s.t. 〈v, u− PX(u)〉 = 0, ∀v ∈ X, u ∈ H;

a measure of the angle between V and H1 is furnished by the scalar

α = supv∈V2

‖PH1(v)‖2‖v‖2 = sup

v∈V2

∫Ω|tr(v)|2

3∫Ω‖v‖2 .

Vol. 53 (2002) Well-posedness of the Dirichlet problem in non-elliptic isotropic elasticity 75

Finally, for every element u0 ∈ H12 (Γ0)3 , we define Uu0 ,

Uu0 =E(u) : u ∈ H1(Ω)3, u = u0 on Γ0

,

which obviously is a linear manifold of H parallel with U .

2.1. Existence and uniqueness of weak solutions of (1.1)

Recall that w ∈ H is a weak solution of equation

div w = 0 on Ω, w[n] = 0 on Γ \ Γ0,

iff it satisfiesw ·E(u) = 0, ∀u ∈ H1(Ω)3, u = 0 on Γ0,

that is iff w ∈ V .Accordingly, u ∈ H1(Ω)3 is a weak solution of (1.1) iff

(3λ + 2µ)PH1(E(u)) + 2µPH2(E(u)) ∈ V, and u = u0 on Γ0.

When µ 6= 0 and 3λ + 2µ 6= 0 , the previous relation may be inverted, and weobtain

E(u) ∈((

13λ + 2µ

PH1 +12µ

PH2

)V

)∩ Uu0 .

As Uu0 is parallel with U , and H = U + V , it follows that the intersection ofUu0 and V has exactly one element, and that

Uu0 = (PV)−1 (Uu0 ∩V) .

Accordingly, u satisfies equation (1.1) iff

E(u) =(

13λ + 2µ

PH1 +12µ

PH2

)(w(u0)),

where w(u0) is a solution of the equation

Uu0 ∩V = Aλ,µ(w) , (2.3)

the linear operator A λ,µ : V 7→ V being defined by

Aλ,µ = PV

(2µ

3λ + 2µPH1 + PH2

).

On the ground of the characterization (2.3) of weak solutions to equation (1.1),we may express the existence and the uniqueness of solutions to (1.1) in terms ofgeometrical properties of the linear operator A λ,µ .

Lemma 1. If µ 6= 0 and 3λ + 2µ 6= 0 , the existence of solutions for equation(1.1) holds iff W1 lays within the image of the operator A λ,µ .

76 E. Ernst ZAMP

Proof of Lemma 1: A solution exist for the equation (1.1) iff equation (2.3) canbe solved for the same limit condition u0 .

Consequently, the assumption of existence of solutions for (1.1) is equivalentwith (Uu0 ∩ V) ⊂ Aλ,µ(V) for every element u0 ∈ H

12 (Γ0)3 , where we deduce

that

W1 =

⋃

u0∈H12 (Γ0)3

Uu0

∩V =

⋃u0∈H

12 (Γ0)3

(Uu0 ∩V) ⊂ Aλ,µ(V),

that is the conclusion of this lemma. ¤

Lemma 2. If µ 6= 0 and 3λ + 2µ 6= 0 , the uniqueness of solutions for equation(1.1) holds iff A λ,µ is one-to-one.

Proof of Lemma 2: As (1.1) is linear, the uniqueness of solutions is established iff(1.1) has exactly one solution for u0 = 0 , that is iff equation

Aλ,µ(w) = 0 (2.4)

has exactly one solution on V . Lemma 2 is proved by noticing that the operatorA λ,µ is linear, and that its kernel coincides with the set of solutions to equation(2.4). ¤

2.2. The proof of Theorem 3

A standard result (see Theorem 2.4 in [5] or Theorem 1.1 in [2]), ensures us thatthe set of all elements of form tr( E(u)) for u ∈ H 1(Ω)3 satisfying u = 0 on Γ0

is, when ∂Ω 6= Γ0 , the entire space L 2(Ω) , and coincides with the hyperplane ofall elements w from L 2(Ω) satisfying

∫Ω

w dx = 0 , when ∂Ω = Γ0 .In both cases, the image of the linear operator obtained by restricting the

orthogonal projection P H1 on U is a closed subspace of H1 . As in the classof linear Hilbert space operators, the closure of the image is a property whichis inherited by the adjoint of the operator, in this case the restriction on V ofthe orthogonal projection P H2 , we deduce that P H2(V) is a closed subspace ofH2 . Equivalently, the angle between V and H1 = H⊥

2 is not zero, that is, asV2 = (V ∩H1)⊥ ∩V ,

1 > α = supv∈V2

‖PH1(v)‖2‖v‖2 .

We are now in position to state the main result of the paper.

Theorem 3. For every Lame coefficients λ > 0 and µ < 0 such that 32 (α−1) <

µλ < 0 , the equation (1.1) is well-posed in the sense of Liapunov, while even themere existence of solutions does not longer hold for µ

λ = 32 (α− 1) .


From 32 (α − 1) ≤ µ

λ it follows that 3λ + 2µ ≥ 3λα , that is, as λ > 0 , that3λ + 2µ > 0 . In proving Theorem 3, we may hence apply Lemmas 1 and 2, whichexpress the existence and the uniqueness of solutions for the equation (1.1) interms of the linear operator A λ,µ . Let us first collect (relations (2.5), (2.8) and(2.9)) some of the properties of this operator.

As for every w ∈ V1 we have P H2(w) = 0 , we deduce that

Aλ,µ(w) =2µ

3λ + 2µw, ∀w ∈ V1. (2.5)

As 〈w1,PH2(w2)〉 = 0 , relation 〈w1,PV(z)〉 = 〈w1, z〉 , valid for every z ∈ H ,yields for z = 2µ

3λ+2µPH1(w2) + PH2(w2)

〈w1,Aλ,µ(w2)〉 = (2.6)⟨w1,

2µ

3λ + 2µPH1(w2) + PH2(w2)

⟩=

2µ

3λ + 2µ〈w1,PH1(w2)〉 .

But w1 ∈ V1 ⊂ H1 , so

〈w1,PH1(w2)〉 = 〈w1,w2〉 = 0, (2.7)

and from relations (2.6) and (2.7) we read

〈w1,Aλ,µ(w2)〉 = 0, ∀w1 ∈ V1, w2 ∈ V2. (2.8)

If we relax the condition w1 ∈ V1 to w ∈ V , we deduce in the same way asfor relation (2.6) that

〈w,Aλ,µ(w)〉 = ‖w‖2(

1− 3λ

3λ + 2µ

‖PH1(w)‖2‖(w)‖2

), ∀w ∈ V. (2.9)

Proof of Theorem 3: Let us first treat the case 32 (α− 1) < µ

λ < 0 .

Lemma 4. For every two Lame coefficients λ > 0 and µ < 0 such that 32 (α −

1) < µλ < 0 , the operator A λ,µ is one-to-one and onto.

Proof of Lemma 4: We prove first that the operator A λ,µ is one-to-one.Let w ∈ V such that A λ,µ(w) = 0 . As V = V1 + V2 , there are w1 ∈ V1

and w2 ∈ V2 such that w = w1 + w2 , so

Aλ,µ(w1) + Aλ,µ(w2) = 0. (2.10)

From 2.5 it follows that A λ,µ(V1) = V1 , while from 2.8 we get A λ,µ(V2) ⊂ V2 ,so relation (2.10) yields

Aλ,µ(w1) = 0, and Aλ,µ(w2) = 0. (2.11)

From relation (2.8) it follows that for every z ∈ V1 , ‖z‖ 6= 0 ,

〈z,Aλ,µ(z)〉 < 0, (2.12)

78 E. Ernst ZAMP

while, as from 32 (α− 1) < µ

λ it follows that

3λ

3λ + 2µ<

1α

,

relation (2.9) implies that for every z ∈ V2 , ‖z‖ 6= 0 ,

〈z,Aλ,µ(z)〉 > 0. (2.13)

Combining (2.11) and (2.12) we get w1 = 0 , while from (2.11) and (2.13) weobtain that w2 = 0 . We have thus proved that w = 0 , and, as w was arbitrarypicked in Ker(A λ,µ ), that the linear operator A λ,µ is one-to-one.

In order to prove that A λ,µ is onto, let us remark that relation 32 (α− 1) < µ

λimplies that 3λα < 3λ + 2µ , and put

ε = 1− 3λ

3λ + 2µα > 0.

Taking into account the definition of α , from relation (2.9) we deduce that therestriction of the operator A λ,µ on V2 is coercive,

〈w,Aλ,µ(w〉 ≥ ε‖w‖2, ∀w ∈ V2,

which implies, since A λ,µ(V2) ⊂ V2 , that A λ,µ(V2) = V2 .Consequently, we have

V = V1 + V2 = Aλ,µ(V1) + Aλ,µ(V2) ⊂ Aλ,µ(V2) ⊂ V,

which concludes the proof of Lemma 4. ¤An important point in the proof of Theorem 3 for the remaining case µ

λ =32 (α− 1) is Lemma 2.15. Let us first clarify the relation between V2 and W2 .

Lemma 5. W2 ⊆ V2 .

Proof of Lemma 5: Let w ∈ V1 = V ∩ H1 . As w ∈ H1 , there is w ∈ L 2(Ω)such that w = w I3 ; but w also belongs to V = U⊥ , so∫

Ω

w (tr(z)) dx = 0, ∀z ∈ U. (2.14)

As we have already noticed, the set of all fields of form tr (z) , for z ∈ Ucoincides with L 2(Ω) when Γ0 6= ∂Ω , and with the hyperplane of all L 2(Ω)elements of zero average, when Γ0 = ∂Ω . Consequently, from relation (2.14)we deduce that, when Γ0 6= ∂Ω , V1 = 0 , and that V1 = IR I3 , in the caseΓ0 = ∂Ω .

The symmetric part of the gradient of the H 1(Ω)3 element uk(x) = kx is k I3

for every k ∈ IR , so in both cases, V1 ⊂ E(u) : u ∈ H1(Ω)3 , which impliesthat V1 ⊂ W1 .

Accordingly, W2 = W⊥1 ∩V ⊂ V⊥

1 ∩V = V2 . ¤


Lemma 6. For every regular domain Ω and every surface Γ0 we have

supw∈W2

‖PH1(w)‖2‖w‖2 =

23

< α. (2.15)

Proof of Lemma 6: Let us first prove that

‖PH1(w)‖2‖w‖2 ≤ 2

3, ∀w ∈ W2, (2.16)

that is the first part of inequality (2.15).Remark that (2.16) is obvious if tr( w) = 0 ; consider thus w ∈ W2 such that

tr (w) 6= 0 . As the domain Ω is bounded, there is R > 0 such that Ω ⊂ CR ,where CR is the cube (−R,+R)3 ⊂ IR3 , and put z for the unique H 1(Ω) solutionof

∆ z = −w

2on CR,

∂

∂nz = 0 on ∂CR, (2.17)

where w is the L 2(CR) element equal with tr (w) on Ω and with zero on CR\Ω .For every x1, x2, x3 ∈ IR , put

y1 = (−1)〈 x12R 〉

(x1 − 2R

⟨ x1

2R

⟩)

y2 = (−1)〈 x22R 〉

(x2 − 2R

⟨ x2

2R

⟩)

y3 = (−1)〈 x32R 〉

(x3 − 2R

⟨ x3

2R

⟩),

where 〈·〉 denotes the closest integer. The measurable function rz ,

rz((x1, x2, x3)) = z((y1, y2, y3)), ∀x ∈ IR3,

is a periodic H 1 function of 4R period, which satisfies on IR3 , thanks to thelimit condition fulfilled by z on ∂CR , the equation ∆ rz = rw , where rw is the4R -periodic L 2 function obtained from w in the same way as rz was obtainedfrom z .

The regularity theorem ([3] Theorem 3, chapter 3.5, part II) implies that rz isa 4R -periodic H 2 element, which means that z , its restriction on CR , belongsto H 2(CR) , and we may put

D(z) = Hess(z)− tr(Hess(z))I3.

As a 4R -periodic H 2 element, rz may be H 2 approximated with 4R -periodictrigonometric polynomials (see [3], Chapter 3.1, Part II). From the definition ofrz we infer that relation

rz((x1, x2, x3)) = rz((2R− x1, x2, x3)) = (2.18)= rz((x1, 2R− x2, x3)) = rz((x1, x2, 2R− x3)),

holds a.e. on IR3 , and the density argument cited above may easily be adaptedin order to prove that rz is the H 2 limit of a sequence (pn)n∈IN∗ of trigonometricpolynomials fulfilling the same symmetry relation (2.18) as rz .

80 E. Ernst ZAMP

As for a C∞ function, relation (2.18) implies that the normal (strong) deriva-tive on ∂CR is zero, standard calculus yields that

∫CR

∂2

∂x2i

pn∂2

∂x2j

pn −(

∂2

∂xi∂xjpn

)2

dx = 0, ∀n ∈ IN, i, j ∈ 1, 2, 3;

passing to the limit in the previous relation yields∫

CR

∂2

∂x2i

z∂2

∂x2j

z −(

∂2

∂xi∂xjz

)2

dx = 0, ∀i, j ∈ 1, 2, 3.

We may now directly verify that∫CR

‖D(z)‖2 dx =12

∫CR

|tr(D(z))|2, (2.19)

or, in an equivalent form, that∫CR

(D(z)− 1

2tr(D(z))I3

)·D(z) dx =

∫CR

Hess(z) ·D(z) dx = 0. (2.20)

Let us now consider w , the 3 × 3 symmetric L 2 -matrix field equal with won Ω and with zero on CR \ Ω . Remark that

tr(D(z)) dx = tr(w) dx. (2.21)

As the restriction on Ω of Hess (z) is in E(H1(Ω)3) , and w is in W2 , wehave ∫

CR

Hess(z) ·w =∫

Ω

Hess(z) ·w = 0. (2.22)

From relations (2.20) and (2.22) we deduce

0 =∫

CR

Hess(z) ·w dx−∫

CR

Hess(z) ·D(z) dx = (2.23)

=∫

CR

(D(z)− 1

2tr(D(z))I3

)· (w −D(z)) dx =

=∫

CR

D(z) · (w −D(z)) dx− 12

∫CR

tr(D(z))I3 · tr(w −D(z))I3 dx.

As tr (w) 6= 0 , it follows that∫CR‖D(z)+t(w−D(z))‖ dx > 0 for any t ∈ IR .

We may accordingly define the function f : IR 7→ IR∗+ ,

f(t) =

∫CR|tr(D(z) + t(w −D(z)))|2 dx∫

CR‖D(z) + t(w −D(z))‖2 dx

,

and remark that

f(1) =

∫CR|tr(w)|2 dx∫

CR‖w‖2 dx

= 3‖PH1(w)‖2‖w‖2 , (2.24)


and that, according to (2.20),

f(0) =

∫CR|tr(D(z))|2 dx∫

CR‖D(z)‖2 dx

= 2. (2.25)

If w = D(z) , (2.16) follows from (2.24) and (2.25).In the case w 6= D(z) , according to (2.21), we have

lim|t|7→∞

f(t) =

∫CR|tr(w −D(z))|2 dx∫

CR‖z−D(z)‖2 dx

= 0. (2.26)

The function f is continuous, non constant (see (2.24) and (2.26)), and itslimits at plus and minus infinity exist and are equal (see (2.26); consequently itachieves its minimum and maximum values on IR , in two of its extremal points.

But

f ′(t) =at2 + bt + c(∫

CR‖D(z) + t(w −D(z))‖2 dx

)2 ,

where

a = 2(∫

CR

|tr(w −D(z))|2 dx

∫CR

D(z) · (w −D(z)) dx−

−∫

CR

‖w −D(z)‖2 dx

∫CR

tr(D(z))tr(w −D(z)) dx

)

b =∫

CR

|tr(w −D(z))|2 dx

∫CR

‖D(z)‖2 dx−∫

CR

‖w−

−D(z)‖2 dx

∫CR

|tr(D(z))|2 dx

c =∫

CR

tr(D(z))tr(w −D(z)) dx

∫CR

‖D(z)‖2 dx−

−∫

CR

|tr(D(z))|2 dx

∫CR

D(z) · (w −D(z)),

so f has at most two extremal points. As from (2.19) and (2.23) it follows thatc = 0 , t = 0 is one of them; relations (2.24) and (2.26) imply that t = 0 can notbe a minimum point, so

f(0) > f(t),∀ t ∈ IR. (2.27)

Inequality (2.16) follows now from (2.24), (2.25) and (2.27).In order to prove the second part

23

< α, (2.28)

of the inequality (6), let us consider the polynomial

φ = (3x1 + 1)(x1 − 1)2(5(x22 + x2

3)− 2)(x22 + x2

3 − 1)2,

82 E. Ernst ZAMP

and the right cylinder

B = (x1, x2, x3) ∈ IR3 : −1 < x1 < 0, x22 + x2

3 < 1;denote by Φ0 and Φ1 its faces laying within the plans x1 = 0 and x1 = −1 ,and put

Φl = (x1, x2, x3) ∈ IR3 : x22 + x2

3 = 1.The polynomial φ was chosen such that

∇φ = 0 on Φ1 ∪ Φl, (2.29)

and that ∫Φ0

∂

∂x1φ dx = 0. (2.30)

Moreover, straightforward calculus yields that∫

B

∑i,j=1,2,3

∂2

∂x2i

φ∂2

∂x2j

φ−(

∂2

∂xi∂xjφ

)2

= (2.31)

=∫

Φ0

∂

∂x1φ

∂2

∂x22 + ∂x3

3

φ dx = −4470

< 0.

From relations (2.29) and (2.30) it follows that the 3×3 symmetric polynomialmatrix field D(φ) = Hess(φ)− tr(Hess(φ))I3 fulfills∫

B

tr(D(φ)) dx = 0, (2.32)

while relation (2.31) implies that∫B|tr(D(φ))|2 dx∫B‖D(φ)‖2 dx

> 2. (2.33)

Put

B(r) = (x1, x2, x3 ∈ IR3 : −1 < x1 < r, x22 + x2

3 < 1, ∀ − 1 < r = ∞.

According to the continuity of the integral with respect to the domain, from (2.33)it follows that there are two positive constants ε and ρ such that, for every regulardomain Ψ such that B(−ρ) ⊆ Ψ ⊆ B(ρ) ,∫

Ψ|tr(wΨ)|2 dx∫Ψ‖wΨ‖2

> 2 + ε, (2.34)

where

wΨ = D(φ)− 13

(∫Ψ

tr(D)(φ)) dx

)I3.

Let us also remark that ∫Ψ

tr(wΨ) = 0. (2.35)


Since D(φ) strongly satisfies

divD(φ) = 0, on IR3, D(φ)[n] = 0 on Φ1 ∪ Φl,

it follows that ∫Ψ

(wΨ) ·E(u) dx = 0, (2.36)

for every u ∈ H1(Ψ)3 such that u = 0 on ∂Ψ \ (Φ1 ∪ Φl) .Let now x0 a regular point of Γ0 , n the outer normal at ∂Ω at x , and Q a

rigid transformation of IR3 which carries the unit vector of the x1 axis, (1, 0, 0) ,in n . For every positive constant r , put

Ωr = (x0 + rQ(B(∞))) ∩ Ω,

andΘr =

1rQ−1 (Ωr − x0) ;

the border of Ω being C1 at x0 , for r sufficiently small, we have B(−ρ) ⊂ Θr ⊂B(ρ) .

For such a value of r , define the 3× 3 symmetric matrix field on Ω

w(x) =

wΘr

(1rQ

−1(x− x0))

x ∈ Ωr

0 x ∈ Ω \ Ωr

Let now u ∈ H1(Ω)3 such that u = 0 on Γ0 ; as

v(x) = u(rQ(x) + x0), ∀x ∈ Θr

is an element from H 1(Θr)3 satisfying v = 0 on ∂Θr \ (Φ1 ∩ Φl) , from relation(2.36) it follows∫

Ω

w ·E(u) dx =∫

Ωr

w ·E(u) dx =∫

Θr

wΘr·E(v) dx = 0,

which means that w belongs to V .From relation (2.35) we get∫

Ω

tr(w) dx =∫

Ωr

tr(w) dx =∫

Θr

tr(wΘr) dx = 0,

which, since V1 ⊆ IRI3 , implies w ∈ V2 .Relation (2.34) implies∫

Ω|w|2∫

Ω‖w‖2 =

∫Ωr|w|2∫

Ωr‖w‖2 =

∫Θr|wΘr

|2∫Θr‖wΘr

‖2 > 2 + ε,

which completely demonstrate relation (2.28), and consequently Lemma 2.15.¤

The proof of Theorem 3 is completed by the following result.

84 E. Ernst ZAMP

Lemma 7. If µλ = 3

2 (1− α) , then W1 does not entirely lay within the image ofA λ,µ .

Let us remark that Lemma 7 is an obvious consequence of the two followingLemmas.


2 (α− 1) , then

Aλ,µ(W2) + W1 = V2.

Proof of Lemma 8: From Lemma 2.15 it follows that

−12

< 1− 1α

=2µ

3λ + 2µ,

so, for every w ∈ V we get

‖Aλ,µ(w)‖2 = (2.37)

=∥∥∥∥PV

(2µ

3λ + 2µPH1(w) + PH2(w)

)∥∥∥∥ ≤≤ 1

2‖PH1(w)‖2 + ‖PH2(w)‖2 ≤ ‖w‖2.

Let us put (see Lemma 2.15)

ε = α− supw∈W2

‖PH1(w)‖2‖w‖2 > 0;

relation (2.9) yields

〈w,Aλ,µ(w)〉 ≥ ε‖w‖2, ∀w ∈ W2. (2.38)

Consider the linear operator B : V 7→ V defined by

B(w) =(

ε

2+

12ε

)PW1(w) + Aλ,µ (PW2(w2)) , ∀w ∈ V.

From relation (2.38) it follows that

〈B(w),w〉 =

=(

ε

2+

12ε

)‖PW1(w)‖2 + 〈PW1(w),Aλ,µ (PW2(w))〉+ ε ‖PW2(w)‖2 ,

and by using (2.37) we get

〈B(w),w〉 ≥≥ ε

2

(‖PW1(w)‖2 + ‖PW2(w)‖2

)+

+(√

ε

2‖PW1(w)‖ − 1√

2ε‖PW2(w)‖

)2

≥ ε

2‖w‖2, ∀w ∈ V.


Accordingly, B( V) = V , and since from the definition of B it follows thatB( V) = A λ,µ(W2) + W1 , Lemma 8 is completely demonstrated. ¤


2 (α− 1) , then the image of A λ,µ is a proper subspace of V .

Proof of Lemma 9: Suppose not. Then, as A λ,µ(V1) = V1 and A λ,µ(V2) ⊂ V2 ,it follows that A λ,µ(V2) = V2 .

Relation µλ = 3

2 (α − 1) may equally be written as 3λ3λ+2µ = 1

α , so, by takinginto account the definition of α , we deduce from relation (2.9) that

〈w,Aλ,µ(w)〉 = ‖w‖2(

1− 1α

‖PH1(w)‖2‖w‖2

)≥ 0, ∀w ∈ V2. (2.39)

Accordingly, the restriction on V2 of A λ,µ is not only a linear and onto oper-ator, but also a positive one, and thus a bicontinuous one. We have consequentlyproved that there is a positive constant ε such that

〈w,Aλ,µ(w)〉 ≥ ε‖w‖2, ∀w ∈ V2. (2.40)

From relations (2.39) and (2.40) it follows that

‖PH1(w)‖2‖w‖2 ≤ (1− ε)α, ∀w ∈ V2,

so

α = supw∈V2

‖PH1(w)‖2‖w‖2 ≤ (1− ε)α < α,

contradiction which closes the proof of Lemma 9. ¤Theorem 3 is a consequence of Lemmas 1, 2, 4 and 7. ¤

3. The right cylinder case

Let S ⊂ IR2 be an open connected C1 plan star-shaped with respect to 0 set ands a positive constant. Put C(s) for the right cylinder of section S and length sdefined by

C(s) = (x1, x2, x3) ∈ IR3 : 0 < x1 < s, (x2, x3) ∈ S;let

Sl(s) = (x1, x2, x3) ∈ IR3 : 0 < x1 < s, (x2, x3) ∈ ∂S,denote its lateral surface, and put

S(r) = (x1, x2, x3) ∈ IR3 : x1 = r, (X1, x2) ∈ S,∀r ∈ IR.

On the IR3 domain C(s) we consider two different Dirichlet problems. In thefirst case (problem L ), the part of the border of C(s) on which the prescribed

86 E. Ernst ZAMP

displacement is imposed is Sl(s) , the lateral surface, while in the second (problemF), this role is played by S(0) ∩ S(s) , that is the reunion of the terminal faces.

As Theorem 3 yields, the well-posedness of the previously defined Dirichletproblems outside the ellipticity area is governed by the value of the geometricalconstant α . The simple geometric form of the domain allows us, in both cases,to obtain some inferior or superior bounds for this constant.

Proposition 10. Problem L : for every 0 < s1 < s2 ,

α(C(s1),Sl(s1)) ≤ α(C(s2),Sl(s2)), (3.41)

andlim

s→∞α(C(s1),Sl(s)) = 1. (3.42)

Proof of Proposition 10: Since Sl(s) 6= ∂C(s) , V1 = 0 for every positive s ;accordingly V2 = V .

Let w ∈ V(C(s1),Sl(s1)) ; for every u ∈ H1(C(s1))3 such that u = 0 onSl(s1) , we have ∫

C(s1)

w ·E(u) dx = 0. (3.43)

Set z for the element of H(C(s2)) equal with w on C(s1) and with zero onC(s2)\C(s1) . As the restriction on C(s1) of every element u ∈ H1(C(s2))3 suchthat u = 0 on Sl(s2) is an element of H 1(C(s1))3 equal with zero on Sl(s1) ,relation (3.43) yields∫

C(s2)

z ·E(u) dx =∫

C(s1)

w ·E(u) dx = 0,

which means that w belongs to V(C(s2),Sl(s2)) , and thus to V2(C(s2),Sl(s2)) .But ∫

C(s1)|tr(w)|2

3∫C(s1)

‖w‖2 =

∫C(s2)

|tr(z)|23

∫C(s2)

‖z‖2 ,

and (3.41) is deduced by taking the supremum over V2(C(s1),Sl(s1)) .In order to infer relation (3.42), let z the polynomial 3× 3 symmetric matrix

field be defined by 2x3

1 − 3x21 −6x1x2(x1 − 1) 0

−6x1x2(x1 − 1) 3x22(2x1 − 1) 0

0 0 0

.

The element z was chosen such that

divz = 0 on IR3

z(x)[(1, 0, 0)] = (0, 0, 0),∀x ∈ S(0), andz(x)[(1, 0, 0)] = (−1, 0, 0),∀x ∈ S(1);


accordingly, for every u from H 1(C(1))3 equal to zero on Sl(1) we have∫C(1)

z ·E(u) dx =∫

S(1)

−u dx. (3.44)

For every s > 4 , put

w((x1, x2, x3)) =

z((x1, x2, x3)) 1 < x1 < 1−I3 1 < x1 < s

2 − 1z(( s

2 − x1, x2, x3)) s2 − 1 < x1 < s

2−z((x1 − s2 , x2, x3)) s

2 < x2 < s2 + 1

I3s2 + 1 < x1 < s− 1

−z((s− x1, x2, x3)) s− 1 < x1 < s

.

Let us remark that, for every 0 < r1 < r2 < s and every u from H 1(C(s))3 ,∫C(r1)\C(r2)

I3 ·E(u) dx =∫

S(r2)

u dx−∫

S(r1)

u dx. (3.45)

From relations (3.44) and (3.45) it follows that∫C(s)

w ·E(u) dx = 0,

for every u ∈ H1(C(s))3 such that u = 0 on Sl(s) , that is w ∈ V(C(s),Sl(s)) ,thus w ∈ V2(C(s),Sl(s)) .

Put M =∫C(1)

‖z‖2 dx and σ for the area of S . From the definition of wwe deduce that

1 ≥ α(C(s), sl(s)) ≥∫C(s)

tr(w) dx

3∫C(s)

‖z2‖2 dx≥ 1− 4M

4M + (s− 4)σ, (3.46)

and relation (3.42) is obtained by passing to the limit in (3.46), closing, in thisway, the proof of Proposition 10. ¤

Proposition 11. Problem F : for every 0 < s1 < s2 ,

α(C(s1),S(0) ∪ S(s1)) ≥ α(C(s2),S(0) ∪ S(s2)). (3.47)

Proof of Proposition 11: For the same reasons as in the proof of Proposition 10,V2(C(s),S(0) ∪ S(s)) = V(C(s),S(0) ∪ S(s)) .

Let w an element of V2(C(s2),S(0) ∪ S(s2)) , and define

z(x) =

w(

s1s2

x)

x ∈ s1s2

C(s2)

0 x ∈ C(s1) \ s1s2

C(s2).

Recall that ∫C(s2)

w ·E(u) dx = 0 (3.48)

88 E. Ernst ZAMP

for every u in H 1(C(s2))3 such that u = 0 on both S(0) and S(s2) . As S isstar-shaped with respect with 0 , on one hand s1C(s2)/s2 ⊂ C(s1) , and on theother v(x) = u(s2x/s1) = 0 on S(0) ∪ S(s2) , so relation (3.48) may be used toobtain∫

C(s1)

zE(u) dx =∫

s1s2

C(ss)

zE(u) dx =(

s2

s1

)−4 ∫C(s2)

wE(v) dx = 0.

Accordingly, z belongs to V(C(s1),S(0)∪S(s1)) ,and thus to V2(C(s1),S(0)∪S(s1)) ; passing to the supremum over w ∈ V2(C(s1),S(0)∪S(s1)) in the equality∫

C(s1)|tr(z)|2 dx

3∫C(s1)

‖z‖2 dx=

∫s1s2

C(s2)|tr(z)|2 dx

3∫

s1s2

C(s2)‖z‖2 dx

=

∫C(s2)

|tr(w)|2 dx

3∫C(s2)

‖w‖2 dx

leads us to the conclusion of Proposition 11. ¤Let us take s0 > 0 and λ > 0, µ < 0 such that both the Dirichlet problems

L and F be well-posed. Theorem 3 and Proposition 11 imply that the problemF is well-posed for every s ≥ s0 , while form Proposition 10 and Theorem 3 wededuce that there is a value, sl > s0 such that problem L is well-posed for everys0 ≤ s < sl , but not for sl .

In other words, if the displacement is imposed on the terminal ends of thecylinder, while the lateral surface is free, the Dirichlet problem is well-posed forcylinders of any length. If the displacement is prescribed on the lateral surface,there is a maximal length up to which the problem is well-posed.

A long object made from an elastic material with negative shear modulus willthus apprehend to be handled on important parts of its border, behavior whichillustrates in a striking way the difference between elastic materials with positiveand negative shear modulus.

References

[1] R. A. Adams, Sobolev Spaces, Academic Press, (1975).[2] D.N. Arnold and R.S. Falk, Well-posedness of the fundamental boundary value problem for

constrained anisotropic elastic materials, Archive for Rational Mechanics and Analysis 98(1987), 143 - 166.

[3] L. Bers, F. John and M. Schechter, Partial Differential Equations. Interscience Pusblishers,New-York - London - Sydney (1964).

[4] E. Ernst, Ellipticity loss in isotropic elasticity J. Elasticity 51 (1998), 203 - 211.[5] R. Temam, Navier-Stokes Equations. North-Holland, Amsterdam (1979).


Emil ErnstLaboratoire de Modelisation en Mecanique et Thermique, L.M.M.TCase 322Faculte des Sciences de Saint-JeromeAvenue Escadille Normandie-NiemenMarseilleFrancee-mail: [email protected]

(Received: November 19, 1999)

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Well-posedness of the Dirichlet problem in non-elliptic isotropic elasticity

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