Z. angew. Math. Phys. 53 (2002) 1075–1098 0044-2275/02/061075-24 c 2002 Birkh¨auser Verlag, Basel Zeitschrift f¨ ur angewandte Mathematik und Physik ZAMP Elastic cavitation is incompatible with the well-posedness of the radially symmetric equilibrium problem on spherical shells Emil Ernst Abstract. Isotropic compressible elastic materials such that every spherical annulus admits exactly one weak radially symmetric solution every time when both its inner and outer surfaces are subjected to either hydrostatic pressures, or pure dilatations, do not allow cavitation. Mathematics Subject Classification (2000). 74B20. Keywords. Cavitation, well-posed radially symmetric equilibrium problem. 1. Introduction In this article we prove (Theorem 2.1) that if for an isotropic elastic material the equilibrium problem obtained by imposing on both the inner and the outer borders of any spherical annulus either hydrostatic pressures or pure dilatations has always exactly one radially symmetric solution, then the elastic material can not undergo cavitation in the sense of [1]. Accordingly, cavitation is disallowed solely by the fact that the elastic mate- rial exhibits a regular behavior with respect to the radially symmetric equilibrium problem. We may thus see Theorem 2.1 as another possible illustration of the re- mark (see [2]) that ”the constitutive character inducing radial cavitation of elastic balls is carried by the liquid-like portion, if any, of their stored-energy density.” Let us remark that, for strongly elliptic hyper-elastic materials, this result may be obtained as a consequence of the proof of the main result of [1]. Indeed, under the hypothesis of [1], no equilibrium displacement can exist for a spherical shell subjected to zero pressure on the inside and to a pressure greater then the pressure at which the cavitation takes place, on the outside. The main interest of Theorem 2.1 is that it proves that the connection between ill-posedness of the radially symmetric equilibrium problem and cavitation, far from being a consequence of the strong ellipticity of the elastic energy, as it might appear from [1], is a general feature, valid for every elastic material. However,
Zeitschrift fur angewandteMathematik und Physik ZAMP
Elastic cavitation is incompatible with the well-posednessof the radially symmetric equilibrium problem on sphericalshells
Emil Ernst
Abstract. Isotropic compressible elastic materials such that every spherical annulus admitsexactly one weak radially symmetric solution every time when both its inner and outer surfacesare subjected to either hydrostatic pressures, or pure dilatations, do not allow cavitation.
In this article we prove (Theorem 2.1) that if for an isotropic elastic material theequilibrium problem obtained by imposing on both the inner and the outer bordersof any spherical annulus either hydrostatic pressures or pure dilatations has alwaysexactly one radially symmetric solution, then the elastic material can not undergocavitation in the sense of [1].
Accordingly, cavitation is disallowed solely by the fact that the elastic mate-rial exhibits a regular behavior with respect to the radially symmetric equilibriumproblem. We may thus see Theorem 2.1 as another possible illustration of the re-mark (see [2]) that ”the constitutive character inducing radial cavitation of elasticballs is carried by the liquid-like portion, if any, of their stored-energy density.”
Let us remark that, for strongly elliptic hyper-elastic materials, this result maybe obtained as a consequence of the proof of the main result of [1]. Indeed, underthe hypothesis of [1], no equilibrium displacement can exist for a spherical shellsubjected to zero pressure on the inside and to a pressure greater then the pressureat which the cavitation takes place, on the outside.
The main interest of Theorem 2.1 is that it proves that the connection betweenill-posedness of the radially symmetric equilibrium problem and cavitation, farfrom being a consequence of the strong ellipticity of the elastic energy, as it mightappear from [1], is a general feature, valid for every elastic material. However,
1076 E. Ernst ZAMP
verifying the existence and the uniqueness of the radially symmetric solution for agiven elastic material might be rather difficult, so the main result of this article cannot be directly used as a condition disallowing cavitation, like the ones presentedin [4], [3] or [7].
Section 2 is concerned with the statement of the radially symmetric boundary-value problem, in both its weak and strong forms.
Theorem 2.1 refers to the class of elastic isotropic materials exhibiting exactlyone radially symmetric solution for every radially symmetric problem on any spher-ical shell. It is thus necessary, in order to prove that our result is not meaningless,to show that this class is sufficiently large. In Section 3 we construct, for everytwo positive constants λ and µ , an isotropic elastic material belonging to thisclass, such that λ and µ are the Lame coefficients of its linearized part.
The main result is proved in Section 4. All the technical results used withoutproof in Section 4 are proved in Section 5.
2. Statement of the problem
Let us consider an elastic material occupying the reference configuration
B(X1, X2) = {X : X1 < |X| < X2} , 0 ≤ X1 < X2,
where |X | denotes the norm of the 3-vector X . A radially symmetric dis-placement on the spherical annulus B(X1, X2) is a field of 3-vectors of formu ( X ) = f(|X |) · X /|X | , where f : [X1, X2] → R+ is a bicontinuous map.
If f is of class C1 , the associated finite strain at X ∈ B(X1, X2) is
F(X) = f ′(|X|)(
X
|X| ⊗X
|X|)
+f(|X|)|X|
(I3 − X
|X| ⊗X
|X|)
.
We assume that the reference configuration is stress-free, that the material isisotropic and that the stress - strain relation T = φ( F ) is of class C1 . As thestrain F is symmetric, the Cauchy strain raised at x = u ( X ) by the strainF has the form
Accordingly, N and S are the values of the normal and of the shear stress. Asg1, g2, g3 are symmetric, it follows that N(a, a) = S(a, a) , for every a > 0 , and,as the reference configuration is stress - free, it follows that N(1, 1) = S(1, 1) = 0 .Finally, as the stress - strain relation is of class C1 , we deduce that N,S ∈C1(R∗+ × R∗+, R) .
If f is of class C2 , at x = u ( X ) , the divergence of the tensor field T ( x )takes the form
divxT(x) =
2
N
(f ′(|X|), f(|X|)
|X|)− S
(f ′(|X|), f(|X|)
|X|)
f(|X|)
+f ′′(|X|)f ′(|X|) N,1
(f ′(|X|), f(|X|)
|X|)
+f ′(|X|)− f(|X|)
|X||X|f ′(|X|) N,2
(f ′(|X|), f(|X|)
|X|)
X
|X| .
A bicontinuous C2 function f gives thus rise to a equilibrium displacement ifand only if
f ′′(|X|)N,1
(f ′(|X|), f(|X|)
|X|)
f ′(|X|) =
f(|X|)|X| − f ′(|X|)|X|f ′(|X|) N,2
(f ′(|X|), f(|X |)
|X |
)
+2S
(f ′(|X|), f(|X|)
|X|)−N
(f ′(|X|), f(|X|)
|X|)
f(|X|) .
(2.1)
The weak sense (in accord with [1]) of the previous equation may be defined inthis case as follows.
Definition 2.1. A radially symmetric weak equilibrium on the spherical annulusB(X1, X2) is a 3 - vector field of form u ( X ) = f(|X |) · X /|X | such that
1078 E. Ernst ZAMP
(i)function f belongs to W1,1(X1, X2) and f ′ ≥ 0 a.e. on (X1, X2) (hereafternotation f will always design the only continuous and increasing representa-tive of this W1,1(X1, X2) element, while by f ′ it will be designed one of therepresentatives of its L1(X1, X2) derivative),
(ii)for almost every (with respect to the Lesbegue measure) values X3, X4 from(X1, X2) , relation
N
(f ′(|X4|), f(|X4|)
|X4|)−N
(f ′(|X3|), f(|X3|)
|X3|)
= (2.2)
∫ X4
X3
2f ′(|X|)f(|X|)
(S
(f ′(|X|), f(|X|)
|X|)−N
(f ′(|X|), f(|X|)
|X|))
dX,
have sense and holds.
Thus, for every f , solution of (2.2), the application X 7→ N(f ′(X), f(X)/X)belongs to W1,1(X1, X2) , and let Nf its continuous representative on [X1, X2] ,that is the unique continuous function on [X1, X2] such that relation
Nf (X) = N
(f ′(X),
f(X)X
)(2.3)
holds almost everywhere (with respect to the Lesbegue measure) on (X1, X2) .When completed with the boundary conditions of place f(X1) = d1 and
f(X2) = d2 , 0 ≤ d1 < d2 , we will refer to equation (2.2) as to the boundaryproblem of place (d1, d2) , while when completed with the boundary conditions ofpressure Nf (X1) = p1 and Nf (X2) = p2 , p1, p2 ∈ R , as to the boundary problemof pressure (p1, p2) .
Let us also recall (according to [1]) the definition of cavitation solutions.
Definition 2.2. When X1 = 0 , every solution of equation (2.2) with f(0) > 0and Nf (0) = 0 will be called cavitation equilibrium or cavitation solution.
We are now in position to state the main result of our paper.
Theorem 2.1. If for every 0 < X1 < X2 < +∞ , every boundary problem of place(d1, d2) , 0 < d1 < d2 , as well as every boundary problem of pressure (p1, p2) ,p1, p2 ∈ R , has exactly one W1,1(X1, X2) solution, then equation (2.2) does notadmit any cavitation equilibrium.
so λ and µ are the Lame coefficients of the material. Since N,1 (a, b) > 0 forevery a, b > 0 , standard regularity results imply that every weak solution ofequation (2.1) is also a strong solution, so all what we have to prove is that theboundary problems of place and pressure are well-posed for (2.1). From (3.4) itfollows that
2a(b− a)N,1 (a, b) = (b− a)bN,2 (a, b) + 2a(S(a, b)−N(a, b)),
so equation (2.1) takes the form
Xf ′′(X) = 2f ′(X)f(X)
(f(X)−Xf ′(X)).
It is easy to see that any solution of (2.2) must have the form f(X) =3√
β1 X3 + β2 , where β1 > 0 and β2 > −β1 X32 . It is now obvious that function
f(X) = 3
√d32 − d3
1
X32 −X3
1
X3 +X3
2d31 −X3
1d32
X32 −X3
1
,
is the unique solution of the boundary problem of place (d1, d2) .In order to construct the solution of the boundary problem of pressure, let
us notice that, for every two positive constants α1 α2 , the application g :(−α2,+∞) 7→ R defined by
g(x) =(
α1 + x
α1− α1
α1 + x
)−
(α2 + x
α2− α2
α2 + x
),
is one-to-one and onto. For α1 = X32 and α2 = X3
1 , we thus set P for the uniquevalue from the interval (−X3
1 ,+∞) which fulfills
g(P) =32µ(p1 − p2),
and put
β1 = e1
λ+2µ
(P+ 2µ
3
(X3
1+P
X31− X3
1X3
1+P
)), and β2 = Pβ1.
>
1080 E. Ernst ZAMP
It is now easy to prove that the function f(X) = 3√
β1 X3 + β2 is the uniquesolution of the boundary problem of pressure (p1, p2) . Thus, relation (3.4) definesan isotropic elastic material of Lame coefficients µ and λ for which the radiallysymmetric problem is well-posed.
4. The proof of the main result
The proof of Theorem 2.1 relies on the fact that several non trivial properties ofthe stress - strain function and of the equilibrium displacements are implied solelyby the fact that the radially symmetric problem is well-posed.
4.1. Technical results
Under the hypothesis of well-posedness for the radially symmetric problem, weare able to prove the following weaker versions of the pressure - compression andtension - extension inequalities.
Lemma 4.1. If the radially symmetric problem is well-posed, then(i)the application s 7→ N(s, s) is strictly monotone,(ii)the application s 7→ N(s, a) is either increasing for all 0 < a , or decreasing
for all 0 < a .
Put m = infa>0 N(a, a) , and M = supa>0 N(a, a) . As N(1, 1) = 0 , we havem < 0 < M . The equilibrium displacements are characterized by the followinginequalities.
Lemma 4.2. If the radially symmetric problem is well-posed, and if f is a solu-tion of (2.2) on the interval (X1, X2) , 0 < X1 < X2 , then(i)the map X 7→ f(X)/X is monotone,(ii)the map X 7→ Nf (X) is monotone every time when Nf (Xi) ∈ (m,M) , i =
1, 2 ,(iii)if M ≤ Nf (X1) and M ≤ Nf (X2) , then M ≤ Nf (X) for every X in
(X1, X2) ,(iv)if the map X 7→ Nf (X) is not constant, then the same holds for the map
X 7→ f(X)/X ; if, in addition, Nf (Xi) ∈ (m,M) , i = 1, 2 , then the mapX 7→ f(X)/X is constant if and only if the map X 7→ Nf (X) is constant.
We recall the following standard Sobolev space results.
Lemma 4.3. For every f ∈ W1,1(X1, X2) , 0 < X1 < X2 , if the applicationx 7→ f(X)/X is increasing, then f ′(X) ≥ f(X)/X a.e. on (X1, X2) , while ifx 7→ f(X)/X is decreasing, then f ′(X) ≤ f(X)/X a.e. on (X1, X2) .
Lemma 4.4. Let f1 and f2 two solutions of (2.2) on (X1, X2) and on (X2, X3) .Then(i)for every 0 < α , the map X 7→ f1(αX)/α is a solution of (2.2) on (X1/α,X2/α) ,(ii)if f1(X2) = f2(X2) and Nf1(X2) = Nf2(X2) , then the map g , g(X) = f1(X)
if X ∈ [X1, X2] and g(X) = f2(X) if X ∈ (X2, X3] , is a solution of (2.2)on (X1, X3) .
Let f1 and f2 two solutions of (2.2) on (X1, X2) , 0 < X1 < X2 , and(Y1, Y2) , 0 < Y1 < Y2 , such that f1(Xi)/Xi = f2(Yi)/Yi = ai for i = 1, 2 . Thefollowing result clarifies the relations between two arbitrary equilibrium displace-ments.
Lemma 4.5. If the radially symmetric problem is well-posed, then(i) (Nf1(X1)−Nf2(Y1))(Nf1(X2)−Nf2(Y2)) ≥ 0 ,(ii)relation (Nf1(X1)−Nf1(X2))(Nf2(Y1)−Nf2(Y2)) > 0 holds every time when
a1 6= a2 and Nfi(Xj) ∈ (m,M) for i, j = 1, 2 .
4.2. Proof of Theorem 2.1
If a cavitation solution exists for the isotropic elastic material defined by the stress-strain relation T = φ( F ) , the same displacement will be a cavitation equilibriumfor the isotropic elastic material defined by relation T = −φ( F ) . Moreover, ifthe radially symmetric problem is well-defined for the first material, the sameholds for the latter. As Lemma 4.1 - (i) states that application s 7→ N(s, s) iseither strictly increasing, or strictly decreasing, it is sufficient to prove Theorem2.1 for the case when the application s 7→ N(s, s) is strictly increasing.
We consequently suppose that the map s 7→ N(s, s) is strictly increasing,that the radially symmetric problem is well-posed, and that the r -radius balladmits a cavitation equilibrium of form u ( X ) = f(|X |)( X /|X |) , wheref ∈ W1,1(0, r) , and f(0) > 0 .
Outline of the proof. Under the above hypothesis, we prove:Step 1: that there is X0 , 0 < X0 ≤ r such that, for every two values p1 ,
m < p1 ≤ m/2 and p2 , N(2f(X0)/X0, 2f(X0)/X0) ≤ p2 < M , both mapsX 7→ f1(X)/X and X 7→ f2(X)/X are increasing, where f1 and f2 are the so-lutions of the boundary problems of pressure (p1, p2) and (p2, p1) on the sphericalannulus of inner radius 1 and outer radius 2 ,
Step 2: that inequality g(1) ≤ f2(1) holds for every two values p1 from the in-terval (m,m/2] and p2 from the interval [N(2f(X0)/X0, 2f(X0)/X0),M) , whereg is the solution on (1, 2) of the boundary problem of pressure (m/2, N(2f(X0)/X0,2f(X0)/X0)) , and
Step 3: that every time when p1 < N(g(1), g(1)) , inequality g(1) ≤ f2(1)leads us to a contradiction.
1082 E. Ernst ZAMP
Step 1: maps X 7→ f1(X)/X and X 7→ f2(X)/X are increasing. The mapNf is continuous and Nf (0) = 0 , so i(X) = inf0≤Y≤X Nf (Y ) and s(X) =sup0≤Y≤X Nf (Y ) are continuous applications and i(0) = s(0) = 0 . On the otherhand, f(X)/X goes to infinity when X goes to 0 , so N(f(X)/X, f(X)/X)converges to M when X converges to 0 .
Accordingly, there is 0 < X0 ≤ r such that i(X0), s(X0) ∈ [m/4, N(f(X0)/X0,f(X0)/X0)−M/4] , that is
m/4 ≤ Nf (X) ≤ N
(f(X0)
X0,f(X0)
X0
)−M/4 ∀0 < X ≤ X0. (4.5)
The map X 7→ f(X)/X is monotone on every interval (α, r) , 0 < α < r ,(apply Lemma 4.2 - (i) to the restriction of f on (α, r) ), and f(X)/X goes toinfinity when X goes to 0 . Accordingly, the map X 7→ f(X)/X is decreasingon (0, r] , and, as the map s 7→ N(s, s) is strictly increasing, we deduce that
N
(f(X0)
X0,f(X0)
X0
)≤ N
(f(X)
X,f(X)
X
)∀X ∈ (0, X0). (4.6)
From relations (2.3), (4.5) and (4.6), we deduce that
N
(f′(X),
f(X)X
)= Nf (X) < N
(f(X0)
X0,f(X0)
X0
)≤ N
(f(X)
X,f(X)
X
)(4.7)
for almost all X from (0, X0) .Applying Lemma 4.3 to the restriction of f on (α, r) for every 0 < α < r we
deduce that f′(X) ≤ f(X)/X almost everywhere on (0, r) . Accordingly, the set
where either (4.7) fails, or f(X)/X < f′(X) , is negligible, and, as the measure of
the interval (0, X0) is not zero, there is at least one value X ∈ (0, X0) such thatf′ (
X) ≤ f
(X
)/X and N
(f′ (
X), f
(X
)/X
)< N
(f
(X
)/X, f
(X
)/X
).
Consequently, the map s 7→ N(s, f
(X
)/X
)can not be decreasing, and from
Lemma 4.1 - (ii) we deduce that the maps s 7→ N(s, a) are increasing for all thepositive values of a .
We are now in position to prove the following result.
Lemma 4.6. The maps X 7→ f1(X)/X and X 7→ f2(X)/X are increasing forevery p1 from (m,m/2] and p2 from [N(2f(X0)/X0, 2f(X0)/X0),M) , wheref1 and f2 are the solutions on the interval (1, 2) of the boundary problems ofpressure (p1, p2) and (p2, p1) .
Proof of Lemma 4.6. Suppose that X 7→ f1(X)/X is decreasing; let us first provethat this assumption implies that 2f(X0)/X0 ≤ f1(2)/2 .
Indeed, if we assume the opposite, that is that f1(2)/2 < 2f(X0)/X0 , then we
deduce that there is 1 <∪X < 2 and ε > 0 such that f1(X)/X ≤ 2f(X0)/X0− ε
As the map X 7→ f1(X)/X is supposed decreasing, from Lemma 4.3 we obtainf ′1(X) ≤ f1(X)/X almost everywhere on (1, 2) , and, as the maps s 7→ N(s, a)
are increasing for any 0 < a , for almost every X from (∪X, 2) we have
Nf1(X) = N
(f ′1(X),
f1(X)X
)≤ N
(f1(X)
X,f1(X)
X
)
≤ N
(2f(X0)
X0, 2
f(X0)X0
)− ε ≤ p2 − ε.
The map Nf1 is continuous, so “almost every” may be dropped from theprevious inequality to obtain Nf1(X) ≤ p2 − ε for every X from the interval
[∪X, 2] . Thus p2 = Nf1(2) ≤ p2−ε , contradiction which proves that 2f(X0)/X0 ≤f1(2)/2 .
As Nf1(1) = p1 < p2 = Nf1(2) , the application Nf1 is not constant on (1, 2) .According to Lemma 4.2 - (iv), neither is the decreasing map X 7→ f1(X)/X , sof1(2)/2 < f1(1) .
Combining the two previous inequalities, we get
f(X0)X0
≤ f1(2)2
<f1(1)
1. (4.8)
The map X 7→ f(X)/X is continuous, and f(X)/X goes to infinity when Xgoes to 0 , so from (4.8) we deduce that there are Y1 , Y2 , 0 < Y1 < Y2 ≤ X0 ,such that f(Y1)/Y1 = f1(1)/1 and f(Y2)/Y2 = f1(2)/2 .
Set X = Y1 in relation (4.5) to deduce that Nf1(X1) − m/4 = p1 − m/4 ≤m/4 ≤ Nf (Y1) , and X = Y2 to obtain Nf (Y2) ≤ N(f(X0)/X0, f(X0)/X0) −M/4 ≤ Nf1(X2)−M/4 . Accordingly Nf1(1)−Nf (Y1) ≤ m/4 < 0 and Nf1(2)−Nf (Y2) ≥ M/4 > 0 , which contradicts Lemma 4.5, when applies for f1 andthe restriction to (Y1, Y2) of f . Thus, the assumption that X 7→ f1(X)/X isdecreasing is false, and Lemma 4.6 is proved for f1 .
Suppose that the map X 7→ f2(X)/X is decreasing. In the same way as forproving that 2f(X0)/X0 ≤ f1(2)/2 , we show that 2f(X0)/X0 ≤ f2(1)/1 . Asf2(1) < f2(2) , we have
f(X0)X0
≤ f2(1)2
<f2(2)
2. (4.9)
Since Nf2(2) = p1 < p2 = Nf2(1) , the application Nf2 is not constant on(1, 2) , and from Lemma 4.2 - (iv) we deduce that neither is the map X 7→f2(X)/X . Accordingly, from (4.9) we deduce that
f(X0)x0
≤ f2(2)2
<f2(1)
1.
In the same way as relation (4.8), the previous relation leads us to a contra-diction, completing thus the proof of Lemma 4.6.
1084 E. Ernst ZAMP
Step 2: g(1) ≤ f2(1) for every m < p1 ≤ m/2 and N(2f(X0)/X0, 2f(X0)/X0)≤ p2 < M .
Let us suppose that there are p1 , m < p1 ≤ m/2 , and p2 , N(2f(X0)/X0,2f(X0)/X0) ≤ p2 < M , such that f2(1) < g(1) , and set a1 = f2(1) and a2 =g(2)/2 . According to Lemma 4.6, the map X 7→ g(X)/X is increasing, so f2(1) <g(1)/1 ≤ g(2)/2 , that is a1 < a2 .
For every a ∈ [a1, a2] , put d(a) = sup0≤x≤1{x : N(a + x, a) < M} . As forevery 0 < a , N(a, a) < M and the map s → N(s, a) is increasing and continuous,we deduce that d(a) > 0 for every a ∈ [a1, a2] , and as N ∈ C0(R+ × R+, R) ,it follows that d : [a1, a2] 7→ [0, 1] is continuous. Moreover, as the set [a1, a2] iscompact, there is 0 < d such that d ≤ d(a) for every a ∈ [a1, a2] .
Set h : (1, 2e(a2−a1)/d) 7→ R∗+ the solution of (2.2) with the boundary condi-tions h(1) = a1 and h
(2e(a2−a1)/d)
)= a22e(a2−a1)/d .
As the map X 7→ h(X)/X is increasing, Lemma 4.3 implies that h(X)/X ≤h′(X) a . e . on (1, 2e(a2−a1)/d) , which, taking into account that maps s 7→N(s, a) are increasing for every 0 < a , implies that m < N(h(X)/X, h(X)/X) ≤N(h′(X), h(X)/X) for almost every X from (1, 2e(a2−a1)/d) . As Nh(X) iscontinuous and almost everywhere equal to N(h′(X), h(X)/X) , we deduce thatm < Nh(X) for every X from (1, 2e(a2−a1)/d) .
Let us prove that at least one of the inequalities Nh(1) < M andNh
(2e(a2−a1)/d)
)< M holds. Indeed, let us suppose that M ≤ Nh(1) and M ≤
Nh
(2e(a2−a1)/d)
). Then Lemma 4.2 - (iv) implies M ≤ Nh(X) for every X from
(1, 2e(a2−a1)/d) . As Nh(X) is almost everywhere equal to N(h′(X), h(X)/X) ,from the definition of d it follows that h′(X) ≥ d+h(X)/X for almost every X in(1, 2e(a2−a1)/d) . The map j = ln(X)−h(X)/(dX) belongs to W1,1(1, 2e(a2−a1)/d) ,and its derivative is
j′(X) =h(X)
X + d− h′(X)dX
.
Consequently, j′ ≤ 0 a . e . on (1, 2e(a2−a1)/d) , so j(2e(a2−a1)/d) ≤ j(1) that is
ln 2− a1
d= ln(2e(a2−a1)/d)− h(2e(a2−a1)/d)
2e(a2−a1)/d
= j(2e(a2−a1)/d) ≤ j(1) = ln(1)− h(1)d
= −a1
d,
contradiction which proves that at least one of the inequalities Nh(1) < M andNh
(2e(a2−a1)/d)
)< M holds.
Let us consider the case Nh(1) < M . On (1, 2) , the map Nf2 is not constant,so (according to Lemma 4.2 - (iv)) neither is the application X 7→ f2(X)/X . Con-sequently, a1 = f2(1) < f2(2)/2 . On the other side, either M ≤ Nh(2e(a2−a1)/d) ,and thus Nh(1) < Nh(2e(a2−a1)/d) , either Nh(2e(a2−a1)/d) < M , and then wemay apply Lemma 4.2 - (iv). In both cases, the map Nh is non constant and con-
By the definition of∩X , the map Nh is not constant on (1,
∩X) , so (as a
consequence of Lemma 4.2 - (iv)) the same holds for the map X 7→ h(X)/X ;
accordingly, a1 = h(1) < h(∩X))/
∩X) = a3 .
The map X 7→ f2(X)/X is continuous, and f2(1) = a1 < a3 ≤ f2(2)/2 , so
there is◦X in the interval (1, 2) such that f2(
◦X)/
◦X = a3 . On (1,
◦X)) , the map
X 7→ f2(X)/X is not constant, so from Lemma 4.2 - (iv) we deduce that Nf2 is
decreasing and non constant. Consequently, Nf2(◦X) < Nf2(1) , and Lemma 4.5
implies that Nh(∩X) < Nh(1) .
Accordingly, Nh is decreasing, so Nh(X) ≤ Nh(1) < M for every X in(1, 2e(a2−a1)/d) . Let g(1) < a4 < a2 ; as, on one hand, X 7→ g(X)/X iscontinuous and g(1) < a2 < g(2)/2 , and, on the other, h(1) < a2 < a2 =
h(2e(a2−a1)/d)/2e(a2−a1)/d and X 7→ h(X)/X is continuous, there are4X in
(1, 2) and¤X in (1, 2e(a2−a1)/d) such that g(
4X)/
4X = h(
¤X)/
¤X = a4 . On (
4X, 2)
the map X 7→ g(X)/X is non constant, so (see Lemma 4.2 - (iv)), Ng is in-
creasing and non constant. Consequently, Ng(4X) < Ng(2) , and from Lemma 4.5
it follows that Nh(¤X) < Nh(2e(a2−a1)/d) , which contradicts the fact that Nh is
decreasing.As a similar contradiction is obtained in the case Nh(2e(a2−a1)/d) < M in
almost the same way as in the case Nh(1) < M , we can conclude that g(1) ≤ f2(1)for every p1 such that m < p1 ≤ m/2 and every p2 fulfilling N(2f(X0)/X0,2f(X0)/X0) ≤ p2 < M .
Step 3: conclusion of the proof. We have proved that, for every p1 , m <p1 ≤ m/2 , and every p2 , N(2f(X0)/X0, 2f(X0)/X0) ≤ p2 < M , the mapsX 7→ g(X)/X and X 7→ f2(X)/X are increasing, and that g(1) ≤ f2(1) .
As m < N(g(1), g(1)) , it is always possible to pick p1 such that m < p1 <N(g(1), g(1)) . The map X 7→ f2(X)/X is increasing, thus f2(X)/X ≤ f ′2(X)a . e . on (1, 2) As the applications s 7→ N(s, a) are increasing for every 0 < a ,for almost every X from (1, 2) , we have
N
(f2(X)
X,f2(X)
X
)≤ N
(f ′2(X),
f2(X)X
)= Nf2(X).
Since the maps x 7→ N(f2(X)/X, f2(X)/X) and Nf2 are continuous, we maydrop “almost every” from the previous inequality, and deduce that N(f2(X)/X,f2(X)/X) ≤ Nf2(X) for every X from (1, 2) .
Setting X = 2 in the previous inequality implies N(f2(2)/2, f2(2)/2) ≤Nf2(2) = p1 , and as p1 was picked such that p1 < N(g(1), g(1)) , we deducethat N(f2(2)/2, f2(2)/2) < N(g(1), g(1)) .
1086 E. Ernst ZAMP
On the other hand, the map s 7→ N(s, s) is strictly increasing and g(1) <f2(1) < f2(2)/2 , so N(g(1), g(1)) < N(f2(2)/2, f2(2)/2) , contradiction whichallows us to conclude.
5. Proofs of technical results
This section is concerned with some of the properties of radially symmetric equi-libriums and of stress - strain functions for isotropic elastic materials for whichthe radially symmetric problem is well-posed.
5.1. General properties of equilibriums: proof of Lemma 4.2.
Proof of Lemma 4.2 - (i): Let f be a solution of (2.2) on the interval (X1, X2) ,0 < X1 < X2 , and suppose that the application X 7→ f(X)/X is not monotone.It is obvious that a map ζ : [z1, z2] 7→ R is not monotone if and only if there arethree values z1 , z2 and z3 from the interval [z1, z2] , such that z1 < z2 < z3 ,while ζ(z2) does not belong to the closed interval bounded by ζ(z1) and ζ(z3) ,that is (ζ(z1)− ζ(z2))(ζ(z2)− ζ(z3)) < 0 .
Applying this result to the map X 7→ f(X)/X , we deduce that there are Z1 ,Z2 and Z3 in the interval [X1, X2] such that Z1 < Z2 < Z3 and (f(Z1)/Z1 −f(Z2)/Z2)(f(Z2)/Z2 − f(Z3)/Z3) < 0 , and set
a =f(Z2)
Z2−min
((f(Z2)
Z2− f(Z1)
Z1
),
(f(Z2)
Z2− f(Z3)
Z3
)).
Obviously, a is either f(Z1)/Z1 , or f(Z3)/Z3) , and thus belongs to boththe closed intervals bounded by f(Z1)/Z1 and f(Z2)/Z2 , and by f(Z3)/Z3) andf(Z2)/Z2 . As the map X 7→ f(X)/X is continuous, there are Y1 , Z1 ≤ Y1 ≤ Z2 ,and Y2 , Z2 ≤ Y2 ≤ Z3 , such that f(Y1)/Y1 = f(Y2)/Y2 = a .
Since f(Z2) 6= aZ2 , we have Y1 < Z2 < Y2 , and the boundary problem ofplace (aY1, aY2) has two different solutions: the restriction on (Y1, Y2) of f , onone side, and the map X 7→ aX on the other, contradiction which implies thatthe map X 7→ f(X)/X is monotone.
Proof of Lemma 4.2 - (ii): The proof in this case reposes on the following technicalresult.
Lemma 5.1. Let Y1 and Y2 two values from the interval [X1, X2] , Y1 < Y2 ,such that Nf (Yi) ∈ (m,M) , i = 1, 2 . Then, for every X ∈ [Y1, Y2] , Nf (X)belongs to the closed interval bounded by Nf (Y1) and Nf (Y2) , that is (Nf (Y1)−Nf (X))(Nf (X)−Nf (Y2)) ≥ 0 .
Proof of Lemma 5.1. Suppose that there is an element X in the interval (Y1, Y2)such that (Nf (Y1) − Nf (X))(Nf (X) − Nf (Y2)) < 0 , and put p = Nf (X) −min
((Nf (X)−Nf (Y1)), (Nf (X)−Nf (Y2))
). Then p is either Nf (Y1) or Nf (Y2) ,
and it belongs to both intervals bounded by Nf (Y1) and Nf (X) , and by Nf (X)and Nf (Y2) .
The map Nf is continuous, thus there are X1 , Y1 ≤ X1 ≤ X , and X2 ,X ≤ X2 ≤ Y2 , such that Nf (X1) = Nf (X2) = p . As p 6= Nf (X) , it follows thatX1 < X < X2 , so the boundary problem of pressure (p, p) has, on (X1,X2) ,two solutions: the restriction on (X1,X2) of f , and the map g(X) = aX , wherep = N(a, a) ( a exists as m < p < M ).
Accordingly, f = g , so Nf = Ng . Obviously, Ng(X) = p , for every X from(X1,X2) , while p 6= Nf (X) , contradiction which proves Lemma 5.1.
In order to prove Lemma 4.2 - (ii), let us apply the previous Lemma for Y1 = X1
and Y2 = X2 , and deduce that Nf (X) ∈ (m,M) for every X from (X1, X2) .Accordingly, Lemma 5.1 may be applied for every two elements Y1 , Y2 from theinterval (X1, X2) , and we obtain that
relation which implies that the map Nf is monotone.
Proof of Lemma 4.2 - (iii): Suppose that there is an element X from the interval(X1, X2) such that Nf (X) < M , and set
p = max(
m + M2
,Nf (X) + M
2
).
Obviously, p ∈ (m,M) , so there is 0 < a such that p = N(a, a) . From the def-inition of p if follows also that p ∈ (Nf (X), Nf (X1)) and p ∈ (Nf (X), Nf (X2)) ,and, as Nf is continuous, there are Y1 in (X1,X) and Y2 in (X,X2) such thatNf (Y1) = Nf (Y2) = p . On (Y1, Y2) , the restriction of f and g(X) = aX aresolutions to the same boundary problem of pressure (p, p) , thus coincides. Ac-cordingly, p = Ng(X) = Nf (X) 6= p , contradiction which allows us to conclude.
Proof of Lemma 4.2 - (iv): Let f such that f(X)/X is constant, and put a forthe value of f(X)/X . Accordingly, f ′(X) = a for almost every X in (X1, X2) ,so Nf (X) = N(f ′(X), f(X)/X) = N(a, a) almost everywhere on (X1, X2) . Themap Nf is continuous, so “almost everywhere” may be dropped from the previousequality, that is Nf (X) = N(a, a) for every X from (X1, X2) .
Consider now f such that Nf (X) = p for every X from (X1, X2) , andp ∈ (m,M) . Let 0 < a such that N(a, a) = p . The map X 7→ aX and fare two solutions of the boundary problem of pressure (p, p) on (X1, X2) , thuscoincide. Accordingly, f(X)/X = a for every X from (X1, X2) .
Proof of Lemma 4.1 - (i): Let a1 , a2 such that N(a1, a1) = N(a2, a2) . On everyinterval (X1, X2) , the maps X 7→ a1 X and X 7→ a2 X are solutions to the sameboundary problem of pressure, thus coincides.
Accordingly, a1 = a2 every time when N(a1, a1) = N(a2, a2) , that is theapplication a 7→ N(a, a) is one-to-one. As a 7→ N(a, a) is also continuous, itfollows that a 7→ N(a, a) is a strictly monotone continuous map.
Proof of Lemma 4.1 - (ii): Fix 0 < a . Let us prove the following partial versionof Lemma 4.1 - (ii).
Lemma 5.2. The application s 7→ N(s, a) is monotone.
Proof of Lemma 5.2: Suppose that s 7→ N(s, a) is not a monotone map, thatis that there are s1 , s2 and s3 , 0 < s1 < s2 < s3 , such that (N(s1, a) −N(s2, a)) (N(s2, a) − N(s3, a)) < 0 . Put p = N(s2, a) − min((N(s2, a) −N(s1, a)), (N(s2, a) − N(s3, a))) . The Sard’s Lemma yields that the Lebesguemeasure of the set S = {N(x, a) : x ∈ R∗+, N,1 (x, a) = 0} (the image of the sin-gular set of the map s 7→ N(s, a) ) is zero. Accordingly, as p 6= N(s2, a) , there isa value p in the open interval bounded by p and N(s2, a) such that p 6= N(a, a)and p /∈ S .
The map N(s, a) is continuous, and p belongs to both the open intervalsbounded by N(s2, a) and N(s1, a) , and by N(s2, a) and N(s3, a) , so thereare t1 in (s1, s2) , and t2 in (s2, s3) such that N(t1, a) = N(t2, a) = p . Sincep /∈ S , N,1 (t1, a) 6= 0 and N,1 (t2, a) 6= 0 . We may thus apply the general O.D.E.existence theorem to the differential equation (2.2) (the strong version of (2.1)),and deduce that there are 0 < η and h1, h2 ∈ C1((1− η, 1 + η), R) , two solutionsof (2.2) such that h1(1) = h2(1) = a , h′1(1) = t1 , and h′2(1) = t2 .
For i = 1, 2 , set ei(X) = hi(X)/X ; obviously, e′i(1) = ti − a . Since p 6=N(a, a) , we deduce that ti 6= a , so e′i(1) 6= 0 , i = 1, 2 . Accordingly, maps e1
and e2 are invertible in a vicinity of 0 : there are α > 0 and gi : [a−α, a+α] 7→(1− η, 1 + η) such that ei(gi(t)) = t , for all t ∈ [a− α, a + α] and i = 1, 2 .
As g′1(a) = 1/(t1 − a) and g′2(a) = 1/(t2 − a) , it follows that |g′1(a)− g′2(a)| ,|g1(a)| and |g2(a)| are three strictly positive numbers. The maps t 7→ (g′1(t) −g′1(t)) and t 7→ (g′2(t) − g′2(t)) are continuous, and their common value at t = ais 0 ; accordingly, there is 0 < β ≤ α such that
|g′i(t)− g′i(a)| ≤ 14|g′1(a)− g′2(a)| and |g′i(t)− g′i(a)| ≤ 1
2|g′i(a)|,
∀t ∈ [a− β, a + β] and i = 1, 2.(5.10)
Function of the disposition of t1 and t2 with respect to a , we distinguish
three cases: t1 < t2 < a , t1 < a < t2 and a < t1 < t2 .Case t1 < t2 < a : In this case, g′i(a) < 0 , i = 1, 2 , and the second part of
relation (5.10) yields g′i(t) < 0 , i = 1, 2 , for every t from [a− β, a + β] . On theother hand, g′2(a) < g′1(a) , and from the first part of relation (5.10) we obtaing′2(t) < g′1(s) for every s, t from [a− β, a + β] . As g1(a) = g2(a) = 1 , from theprevious inequalities we deduce that
Relation (5.11) yields ω(0) < 0 < ω(β) ; as ω is continuous, there is a in(0, β) such that ω(a) = 0 , that is
g2(a− a)g1(a + β − a)
=g1(a− a)
g2(a + β − a),
and put d for the common value of the two terms of the above equality.From Lemma 4.4 it follows that both maps
ζ1(X) =
h1 (g1(a + β − a)X)g1(a + β − a) , X ∈
[1, 1
g1(a + β − a)
]
h2 (g1(a + β − a)X)g1(a + β − a) , X ∈
[1
g1(a + β − a) , d = g2(a− a)g1(a + β − a)
]and
ζ2(X) =
h2 (g2(a + β − a)X)g2(a + β − a) , X ∈
[1, 1
g2(a + β − a)
]
h1 (g2(a + β − a)X)g2(a + β − a) , X ∈
[1
g2(a + β − a) , d = g1(a− a)g2(a + β − a)
]are, on the interval (1, d) , solutions of the boundary problem of place (1, d(a−a)) .
Accordingly, ζ1 = ζ2 . As ζ ′1(1) = h′1(g1(α + β − a) = 1/g′1(α + β − a) andζ ′2(1) = h′1(g2(α + β − a) = 1/g′2(α + β − a) , and as g′2(t) < g′1(s) for every s, tfrom [a− β, a + β] , it follows that ζ ′1(1) < ζ ′2(1) , contradiction which closes theanalysis of the case t1 < t2 < a .
Case t1 < a < t2 : Relation (5.10) implies in this case that g′1(t) < 0 < g′2(s) ,for every s, t in [a− β, a + β] . Consequently, g1(a + β) < 1 < g2(a + β) , so (seeLemma 4.4)
ζ(X) =
h1 (g1(a + β)X)g1(a + β) , X ∈
[1, 1
g1(a + β)
]
h2 (g1(a + β)X)g1(a + β) , X ∈
[1
g1(a + β) ,g2(a + β)g1(a + β)
]
1090 E. Ernst ZAMP
and g(X) = (a + β)X are, on the interval (1, g2(a + β)/g1(a + β)) solutions ofthe boundary problem of place (1, (a + β)g2(a + β)/g1(a + β)) , thus ζ = g . Asζ ′(1) = h′1(g1(a + β)) = 1/g′1(a + β) < 0 , and 0 < a + β = g′(1) , it follows thatζ ′(1) < g′(1) , contradiction which closes the analysis of the case t1 < a < t2 .
With minor changes, the analyse of the case t1 < t2 < a may be transposedfor the third case, a < t1 < t2 , so any of the three possible cases leads us to acontradiction. Our initial assumption is thus false, and Lemma 5.2 is established.
Let us return to the proof of Lemma 4.1 - (ii).Put I for the set of values 0 < a < +∞ for which s 7→ N(s, a) is non constant
and increasing, and D for the set of the values a for which s 7→ N(s, a) is nonconstant and decreasing. As N(·, ·) is continuous, it is obvious that both I andD are open sets.
Let us suppose that Lemma 4.1 - (ii) does not hold, that is both I and Dare non-void. Accordingly, as (0,∞) is a connected set, it can not be covered bythe reunion of I and D , two disjoint open sets, so there is a point a0 , 0 < a0 ,such that a0 /∈ I and a0 /∈ D . Accordingly, the monotone (see Lemma 5.2) maps 7→ N(s, a0) can not be at the same time not constant and monotone, thus isconstant.
Set f1 and f2 for the solutions of the boundary problem of place (2a1, 3a1)and (a1, 6a1) on the interval (1, 3) . Lemma 4.4 implies that the maps
ζ(X) ={
f1(X) , X ∈ [1, 3]3f2(X/3) , X ∈ [3, 9]
and g(X) = 2a1 X are, on the interval (1, 9) , solutions of the same boundaryproblem of place, thus ζ = g . As ζ(3) = 3a1 and g(3) = 6a1 , it follows thatζ(3) < g(3) , contradiction which proves that our assumption was false, that isthat Lemma 4.1 - (ii) holds.
5.3. Relations between equilibria: proof of Lemma 4.5.
Proof of Lemma 4.5 - (i): If a1 = a2 , then f1 and f2 are solutions on (X1, X2)and (Y1, Y2) of the boundary problems of place (aX1, aX2) and (aY1, a Y2) ,where a is the common value of a1 and a2 . Thus, f1(X) = aX for every X in(X1, X2) and f2(X) = aX for every X in (Y1, Y2) , and Nf1(X1) = Nf1(X2) =Nf2(Y1) = Nf2(Y2) = N(a, a) . Accordingly, Lemma 4.5 - (i) is verified in the casea1 = a2 .
Let us consider the case a1 < a2 (the case a2 < a1 is analogous). Suppose thatLemma 4.5 - (i) does not hold. Accordingly, there are f1 and f2 , solutions of (2.1)on (X1, X2) , 0 < X1 < X2 , and (Y1, Y2) , 0 < Y1 < Y2 such that f1(Xi)/Xi =f2(Yi)/Yi = ai , i = 1, 2 , and (Nf1(X1)−Nf2(Y1))(Nf1(X2)−Nf2(Y2)) < 0 .
Let hi(X) = fi(X)/X , i = 1, 2 . As h1(X1) = a1 < a2 = h1(X2) and
h2(Y1) = a1 < a2 = h2(Y2) , Lemma 4.2 - (i) implies that both the maps g1 andg2 are increasing. A standard real analysis result yields that there are g1 : [0, 1] 7→[X1, X2] and g2 : [0, 1] 7→ [Y1, Y2] , two continuous and increasing maps such thatg1(0) = X1 , g1(1) = X2 , g2(0) = Y1 , g2(1) = Y2 , and h1(g1(t)) = h2(g2(t)) forevery t ∈ [0, 1] .
Maps N1 : [0, 1] 7→ R N2 : [0, 1] 7→ R , defined as N1(t) = Nf1(g1(t))and N2(t) = Nf2(g2(t)) are continuous. We have N1(0) = Nf1(X1) , N1(1) =Nf1(X2) , N2(0) = Nf2(Y1) and N2(1) = Nf2(Y2) , so (N1(0)−N2(0))(N1(1)−N2(1)) < 0 . Accordingly, M = {t ∈ (0, 1) : N1(t) = N2(t)} is a non void, closedand bounded set, and let t = min M . Clearly,
Nf1(g1(t)) 6= Nf2(g2(t)) ∀t ∈ (0, t). (5.12)
Set z(t) = g1(t)/g2(t) . We will distinguish two cases.Case when z(t) does not belong to the open interval bounded by z(0) and
z(1) : In this case, the value
z = z(t)− 12
(min(z(t)− z(0), z(t)− z(1))
)belongs to both the closed intervals bounded by z(t) and z(0) , and by z(t) andz(1) . The map z is continuous, so there are t1 in [0, t] and t2 in [t, 1] such thatz(t1) = z(t2) = z . Moreover, as z = z(t) if and only if either z = z(t) = z(0) , orz = z(t) = z(1) , we can always pick t1 and t2 such that t1 < t < t2 .
Consequently, g1(t1)/g2(t1) = g1(t2)/g2(t2) , so g1(t2)/g1(t1) = g2(t2)/g2(t1) ,and put g for the common value of g1(t2)/g1(t1) and g2(t2)/g2(t1) . For everyX from (1, g) , set
ζ1(X) =f1(g1(t1)X)
g1(t1)and ζ2(X) =
f2(g2(t1)X)g2(t1)
.
From Lemma 4.4 - (i) it follows that ζ1 and ζ2 are solutions of (2.2). Moreover,ζ1(1) = ζ2(1) = t1g1(t1) and ζ1(g) = ζ2(g) = t2g1(t2) , so ζ1 and ζ2 are twosolutions of the same boundary problem of place, and thus ζ1 = ζ2 .
For any X from (1, g) and any i = 1, 2 , ζ ′i(X) = f ′i(gi(t1)X) and ζi(X)/X =fi(gi(t1)X)/(gi(t1)X) , and thus
Nζi(X) = N
(ζ ′i(X),
ζi(X)X
))
= N
(f ′i(gi(t1)X),
fi(gi(t1)X)gi(t1)X
)= Nfi
(gi(t1)X)
(5.13)for almost everywhere (1, g) , and i = 1, 2 . As Nζi
and Nfiare continuous
maps, “almost everywhere” may be dropped from (5.13), to deduce that Nζi(X) =
Nfi(gi(t1)X) for every X in (1, g) . For X = 1 we get Nζ1(1) = Nf1(g1(t1))
and Nζ2(1) = Nf2(g2(t1)) ; as t1 < t , relation (5.12) implies that Nf1(g1(t1)) 6=Nf2(g2(t1)) , and thus Nζ1(1) 6= Nζ2(1) , contradiction.
Case when z(t) belongs to the open interval bounded by z(0) and z(1) :As the maps s 7→ z(t)s and s 7→ z(t)/s are continuous, and as their limit
when s goes to 1 is z(t) , there is 0 < ε < 1 such that both z(t)ε and z(t)/ε
1092 E. Ernst ZAMP
are in the open interval bounded by z(0) and z(1) .As z(t)ε < z(t) < z(t)/ε , one of the values z(t)ε and z(t)/ε belongs to the
open interval bounded by z(0) and z(t) , while the other is in the open intervalbounded by z(t) and z(1) . We deduce that there are t1 , 0 < t1 < t and t2 ,t < t2 < 1 , such that {z(t1), z(t2)} = {z(t)ε, z(t)/ε} , and thus z(t1)z(t2) = z(t)2 .
Consequently,g2(t2)g1(t)g1(t1)g2(t)
=g1(t2)g2(t)g2(t1)g1(t)
,
and put g for the common value of the two terms of the previous equality.The applications ζ1, ζ2 : [1, g] 7→ R ,
ζ1 =
f1(g1(t1)X)g1(t1)
, 1 ≤ X ≤ g1(t)g1(t1)
f2
(g2(t)g1(t1)
g1(t)X
)g2(t)g1(t1)
g1(t)
,g1(t)g1(t1)
≤ X ≤ g2(t2)g1(t)g1(t1)g2(t)
= g
and
ζ2 =
f2(g2(t1)X)g2(t1)
, 1 ≤ X ≤ g2(t)g2(t1)
f1
(g1(t)g2(t1)
g2(t)X
)g1(t)g2(t1)
g2(t)
,g2(t)g2(t1)
≤ X ≤ g1(t2)g2(t)g2(t1)g1(t)
= g
are (see Lemma 4.4) solutions of (2.2), and ζ1(1) = ζ2(1) = t1 , while ζ1(g) =ζ2(g) = gt2 . Accordingly, ζ1 and ζ2 are solutions to the same problem of place(t1, gt2) on the interval (1, g) . In the same way as for the previous case, we provethat Nζi
(X) = Nfi(gi(t1)X) for every X in (1, gi(t)/gi(t1)) , i = 1, 2 , and from
relation (5.12) we deduce that Nζ1(1) = Nf1(g1(t1)) 6= Nf2(g2(t1)) = Nζ2(1) .Thus ζ1 and ζ2 are two different solutions for the same boundary problem ofplace, contradiction.
Proof of Lemma 4.5 - (ii): The two cases a1 < a2 and a2 < a1 being similar,we may consider that a1 < a2 . For every X , 1 < X , put fX for the solution on(1, X) of the boundary problem of place (a1, a2X) , and set hX(X) = fX(X)/X .
Lemma 4.2 - (i) states that hX is monotone for every X . When the continuousmap hX is strictly monotone, its inverse, h−1
X, is also a bicontinuous monotone
map, and thus admits a L1 derivative. Moreover, the chain rule holds, and then
for almost every a from (a1, a2) .Accordingly, every time when hX is strictly monotone, we have
ln(
X1
X2
)=
∫ hfX
(X2)
hfX
(X1)
1f ′
X(h−1
X(a))− a
da, (5.14)
for every X1 , X2 , 1 ≤ X1 < X2 ≤ X .Let X1 and X2 , 1 < X1 < X2 , such that both hX1
and hX2are strictly in-
creasing. From Lemma 4.5 - (i) it follows that either NfX1(h−1
X1(a)) ≤ NfX2
(h−1
X2(a)) ,
for every a in (a1, a2) , or NfX2(h−1
X2(a)) ≤ NfX1
(h−1
X1(a)) for every a in (a1, a2) .
As for every 1 < X we have
NfX(h−1
X(a)) = N(f ′
X(h−1
X(a)), a) a.e. on (a1, a2),
and as the map s 7→ N(s, a) is either increasing for every 0 < a , or decreasing forevery 0 < a (Lemma 4.1 - (ii)), it follows that either f ′
X1(h−1
X1(a)) ≤ f ′
X2(h−1
X2(a))
for almost every a in (a1, a2) , or f ′X2
(h−1
X2(a)) ≤ f ′
X1(h−1
X1(a)) for almost every
a in (a1, a2) .Applying relation (5.14) to 1 and X1 , and to 1 and X2 , we deduce that
∫ a2
a1
1f ′
X1(h−1
X1(a))− a
da = ln(X1) < ln(X2) =∫ a2
a1
1f ′
X2(h−1
X2(a))− a
da.
It is therefore impossible to have f ′X1
(h−1
X1(a)) ≤ f ′
X2(h−1
X2(a)) for almost every
a in (a1, a2) ; accordingly, every time when X1 < X2 and both maps hX1and
hX2are strictly monotone, then
f ′X2
(h−1
X2(a)) ≤ f ′
X1(h−1
X1(a)) a.e. on (a1, a2). (5.15)
It is thus important to determine the values of X leading to strictly monotonemaps hX .
Lemma 5.3. Either hX is strictly monotone for every X , 1 < X , or there isX∗ , 1 < X∗ , such that hX is strictly monotone for every X from the interval(1, X∗] , and NfX∗ (1) = NfX
(1) and NfX∗ (X∗) = NfX(X) for every X∗ ≤ X .
Proof of Lemma 5.3: If the map hX is not strictly monotone for every X∗ , thereis X1 , 1 < X1 and a∗ , a1 ≤ a∗ ≤ a2 such that h−1
X1(a∗) = [Y1, Y2] , where
1 < Y1 < Y2 < X1 and h−1
Xis the (possible multi-valued) inverse of the map
hX .
1094 E. Ernst ZAMP
Put X∗ = Y1X1/Y2 ; for every X∗ ≤ X , the map
ζX(X) =
fX1(X), 1 ≤ X ≤ Y1
a∗ X, Y1 ≤ X ≤ XX1
Y2
fX1
(X1X
X)
X1X
, XX1
Y2 ≤ X ≤ X
is (see Lemma 4.4) a solution on (1, X) of the boundary problem of place (a1, a2X) ,so fX = ζX .
It is now immediate that NfX(X) = NfX1
(X) for every X , 1 ≤ X ≤ Y1 ,equality which implies that NfX
(1) = NfX1(1) , and that NfX
(X) = NfX1(X1X/X) ,
for every XY2/X1 ≤ X ≤ X , relation implying that NfX(X) = NfX1
(X1) .Accordingly, we have proved that NfX∗ (1) = NfX
(1) and NfX∗ (X∗) = NfX(X)
for every X∗ ≤ X . The proof of Lemma 5.3 is completed by showing that, forevery 1 < X ≤ X∗ , the map hX1
is strictly monotone.Suppose that there is X2 , 1 < X2 ≤ X∗ , and b∗ , a1 ≤ b∗ ≤ a2 such that
h−1
X2(b∗) = [Z1,Z2] for some values Z1 and Z2 , 1 < Z1 < Z2 < X2 .The applications
ζ1(X) =
fX2(X), 1 ≤ X ≤ Z1
b∗ X, Z1 ≤ X ≤ X∗
X2Z2
fX2
(X2X∗ X
)X2X∗
, X∗
X2Z2 ≤ X ≤ X∗
and fX∗ are (see Lemma 4.4), solutions on (1, X∗) of the same problem of place,so ζ1 = fX∗ .
Put h−11 for the inverse of the map h1(X) = ζ1(X)/X . As h−1
1 (b∗) =[Z1, X
∗Z2/X2] , and as ζ1 = fX∗ , we deduce that h−1X∗(b∗) = [Z1, X
∗Z2/X2] .Since ζX∗ = fX∗ , from the definition of ζX∗ it follows that h−1
X∗(b∗) = h−1
X1(b∗) .
Accordingly,
h−1
X1(b∗) =
[Z1,
X∗Z2
X2
]. (5.16)
From Lemma 4.4 it follows that the maps
ζ2(X) =
fX2(X), 1 ≤ X ≤ Z1
b∗ X, Z1 ≤ X ≤ X1
X2Z2
fX2
(X2X1
X)
X2X∗
, X1
X2Z2 ≤ X ≤ X1
and fX1are solutions on (1,X1) of the same boundary problem of place. If we
put h−12 for the inverse of h2(X) = ζ2(X)/X , we have h−1
From (5.16) and (5.17) we obtain Y1X1/Y2 = X∗ = X1 , contradiction whichproves Lemma 5.3.
Lemma 5.3 has the following consequence.
Lemma 5.4. The maps X 7→ NfX(1) and X 7→ NfX
(X) are either both in-creasing, or both decreasing.
Proof of Lemma 5.4: Let X1 and X2 , 1 < X1 < X2 such that both hX1
and hX2are strictly monotone maps, and let a in [a1, a2] . From relation (5.15)
it follows that, if the maps s 7→ N(s, a) are increasing for every 0 < a , thenNfX2
(h−1
X2(a)) ≤ NfX1
(h−1
X1(a)) , while if s 7→ N(s, a) is decreasing for every
0 < a , then NfX1(h−1
X1(a)) ≤ NfX2
(h−1
X2(a)) .
Applying the previous inequalities for a = a1 and a = a2 we deduce that,NfX2
(1) ≤ NfX1(1) and NfX2
(X2) ≤ NfX1(X1) , if the map s 7→ N(s, a) is
increasing for all 0 < a , and NfX1(1) ≤ NfX2
(1) and NfX1(X1) ≤ NfX2
(X2) ,when s 7→ N(s, a) is a decreasing map for every 0 < a .
Thus, if maps hX are strictly monotone for every 1 < X , then the maps X 7→NfX
(1) and X 7→ NfX(X) are either both decreasing, or both increasing. If hX
are not strictly monotone maps for every 1 < X (see Lemma 5.3), the maps X 7→NfX
(1) and X 7→ NfX(X) are either both decreasing on (1, X∗] , and constant
on [X∗,∞) , or both increasing on (1, X∗] , and constant on [X∗,∞) .Let us now return to the proof of Lemma 4.5 - (ii). Put I = {X : 1 <
X,NfX(1) < NfX
(X)} , and D = {X : 1 < X,NfX(X) < NfX
(1)} . Suppose thatLemma 4.5 - (ii) does not hold, that is that there are X1 in I and X2 in Dsuch that NfXi
(1), NfXi(Xi) ∈ (m,M) for i = 1, 2 .
Lemma 5.4 implies that, for every X from the closed interval bounded byX1 and X2 , NfX
(1) belongs to the closed interval bounded by NfX1(1) and
NfX2(1) , while NfX
(X) belongs to the closed interval bounded by NfX1(X1)
and NfX2(X2) . Consequently, both NfX
(1) and NfX(X) belongs to (m,M) ,
and from Lemma 4.2 - (iv) it follows that NfX(1) 6= NfX
(X) .Thus, every element X from the closed interval bounded by X1 and X2
belongs either to I or to D . As any real interval is a connected set, there is atleast a common cluster point of I and D , say X0 , in the closed interval boundedby X1 and X2 .
Let us first remark that hX0is strictly monotone. Indeed, Lemma 5.3 implies
that either the map hX is strictly monotone for every 1 < X , case where isnothing to prove, or that there is X∗ , 1 < X∗ , such that hX is strictly monotonefor 1 < X ≤ X∗ , and NfX∗ (1) = NfX
(1) and NfX∗ (X∗) = NfX(X) for every
X∗ ≤ X .In the second case, the set [X∗,∞) is either completely included in I , or
completely included in D , and its interior, (X∗,∞) is thus included either in the
1096 E. Ernst ZAMP
interior of I , or in the interior of D . As a common cluster point of two disjointsets can not belong to the interior of any of them, it follows that X0 /∈ (X∗,∞) .Accordingly, X0 ≤ X∗ , and Lemma 5.3 yields that the map hX0
is strictlymonotone.
The point X0 is a cluster point for both I and D ; consequently, there is asequence (Xn)n∈N∗ , Xn 7→ X0 , such that {Xn : n ∈ N∗} is a subset of I , ifX0 is in D , and a subset of D , if X0 belongs to I .
Lemma 5.3 states that either the map hX is strictly monotone for every X ,and in this case the map hXn
is strictly monotone for every n ∈ N , or that thereis X∗ such that hX is strictly monotone when X ≤ X∗ .
In this case, we have already proved that X0 ≤ X∗ . If X0 < X∗ , we deduce,as Xn 7→ X0 , that for large enough n , Xn < X∗ , thus that the map hXn
isstrictly monotone. If X0 = X∗ , supposing that hXn
≥ X∗ for some n impliesthat X0 and Xn are either both in I , or both in D , contradiction. Accordingly,Xn < X∗ for every n ∈ N , so once again hXn
is strictly monotone.We have thus proved that for every common cluster point X0 of I and D ,
the map hX0is strictly monotone, and, at least for large enough n , the same
holds for hXn.
For every i ∈ N , put mi = min1≤X≤X0
(NfXi
(X))
and
Mi = max1≤X≤X0
(NfXi
(X))
.
When X0 belongs to I and Xi to D for every i ∈ N∗ , it follows thatm0 = NfX0
(1) , M0 = NfX0(X0) , mi = NfXi
(Xi) and Mi = NfXi(1) . When
X0 belongs to D , and, for every i ∈ N∗ , Xi belongs to I , we deduce thatm0 = NfX0
(X0) , M0 = NfX0(1) , mi = NfXi
(1) and Mi = NfXi(Xi) .
In both cases, Lemma 4.5 - (i) applied to fx0 and fXiimplies that (m0 −
Mi)(M0−mi) ≥ 0 for every i ∈ N∗ . Accordingly, either Mi ≤ m0 , or M0 ≤ mi ,that is either NfXi
(X) ≤ m0 for every 1 ≤ X ≤ Xi , or M0 ≤ NfXi(X) for
every 1 ≤ X ≤ Xi .The map NfXO
is continuous, so there are Z1 , Z2 , 1 < Z1 < Z2 < X0
such that NfX0(Z1) = (2NfX0
(1) + NfX0(X0))/3 and NfX0
(Z2) = (NfX0(1) +
2NfX0(X0))/3 . Set b1 = hX0
(Z1) and b2 = hX0(Z2) .
As the map NfX0is monotone, we deduce that
2m0 + M0
3≤ NfX0
(h−1
X0(a)
)= N
(f ′
X0
(h−1
X0(a)
), a
)≤ m0 + 2M0
3a.e. on [b1, b2].
(5.18)Let a , a value from (b1, b2) for which the previous relation holds. As the
map s 7→ N(s, a) is continuous and monotone, there are four values d1 , d2 , d3
and d4 , d1 < d2 < f ′X0
(h−1
X0(a)
)< d3 < d4 , such that N(d1, a) and N(d4, a)
belong to the interval (m0,M0) , and that N(d2, a) and N(d3, a) lay outside the
interval ((2m0 + M0)/3, (m0 + 2M0)/3) .The map N(·, ·) is continuous, so there are c1 , c2 , such that b1 ≤ c1 < c2 ≤
b2 and N(d1, a), N(d4, a) ∈ (m0,M0) and N(d2, a), N(d3, a) /∈ ((2m0 + M0)/3,(m0 +2M0)/3) for every a in the interval (c1, c2) . From relation (5.18) it followsthat N
(f ′
X0
(h−1
X0(a)
), a
)belongs to the closed interval bounded by N(d2, a)
and N(d3, a) for almost every a from (c1, c2) . Accordingly,
d2 ≤ f ′X0
(h−1
X0(a)
)≤ d3 a.e. on (c1, c2).
Let i ∈ N∗ large enough to ensure that hXiis strictly monotone, and suppose
that Xi < X0 .Relation (5.15) implies that
f ′X0
(h−1
X0(a)) ≤ f ′
Xi(h−1
Xi(a)) a.e. on [a1, a2];
from the two previous inequalities it follows that, for almost every a from (c1, c2)we have d1 < d2 ≤ f ′
Xi(h−1
Xi(a)) .
As, for almost every a from (a1, a2) , N(f ′Xi
(h−1
Xi(a)), a) lays outside the
interval m0,M0 , and, for every a from (c1, c2) , N(d1, a) and N(d4, a) laywithin the same interval, it follows that f ′
Xi(h−1
Xi(a)) does not belong to (d1, d4)
for almost every a from (c1, c2) .Accordingly, if hXi
is strictly monotone, and Xi < X0 , then d4 ≤ f ′Xi
(h−1
Xi(a))
almost everywhere on (c1, c2) . Thus
f ′X0
(h−1
X0(a)
)≤ d3 < d4 ≤ f ′
Xi(h−1
Xi(a)) a.e. on (c1, c2). (5.19)
Let Y1 = h−1
X0(c1) , Y1i = h−1
Xi(c1) , Y2 = h−1
X0(c2) , and Y2i = h−1
Xi(c2) .
From relations (5.19) and (5.14) applied for fX0between Y1 and Y2 , and for
fXibetween Y1i and Y2i , we deduce that
Y2
Y1≤ Y2i
Y1i
(d3 − c2)(d4 − c1)(d3 − c1)(d4 − c2)
. (5.20)
Relations (5.15) and (5.14) applied for fX0between 1 and Y1 and for fXi
between 1 and Y1i , and for fX0between Y2 and X0 , and for fXi
between Y2i
and Xi , imply that
Y1 ≤ Y1i andX0
Y2≤ Xi
Y2i. (5.21)
From (5.20) and (5.21) we obtain that, for every i ∈ N∗ such that hXiis
strictly monotone and Xi ≤ X0 we have
Xi ≤ X0(d3 − c2)(d4 − c1)(d3 − c1)(d4 − c2)
. (5.22)
1098 E. Ernst ZAMP
The same reasoning may be applied for the case when hXiis strictly monotone,
while X0 ≤ Xi , to obtain
X0 ≤ Xi(d1 − c2)(d2 − c1)(d1 − c1)(d2 − c2)
. (5.23)
From relations(5.22) and (5.23) it follows that, for every i ∈ N∗ such thathXi
is strictly monotone, there is 1 < γ such that X0 ≤ γXi every time whenX0 < Xi , and Xi ≤ γX0 , every time when Xi < X0 , inequalities contradictingthe fact that Xi 7→ X0 6= 0 .
References
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[2] M. R. Lancia, P. Podio-Guidugli, G. Vergaracaffarelli, Gleanings of radial cavitation, J.Elasticity 44 (1996), 183–192.
[3] V. Sverak, Regularity properties of deformations with finite energy, Arch. Rational Mech.Anal. 100 (1988), 105–127.
[4] R. Knops, C. A. Stuart, Quasiconvexity and uniqueness of equilibrium solutions in nonlinearelasticity, Arch. Rational Mech. Anal. 86 (1984), 233–249.
[5] J. Sivaloganathan, On the stability of cavitating equilibria, Quart. Appl. Math. 53 (1995),301–313.
[6] J. G. Murphy, S. Biwa, Nonmonotonic cavity growth in finite, compressible elasticity, Inter-nat. J. Solids Structures 34 (1997), 3859–3872.
[7] R. D. James, S.Spector, The formation of filamentary voids in solids, J. Mech. Phys. Solids39 (1991), 783–813.
Emil ErnstLaboratoire de Modelisationen Mecanique et Thermodynamique (LMMT)Casse 322Faculte de Sciences et Techniques de Saint JeromeAvenue Escadrille Normandie-Niemen13397 Marseille Cedex 20e-mail: [email protected]
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