Weighted asymptotic Korn and interpolation Korn inequalities with singular weights

# Weighted asymptotic Korn and interpolation Korn inequalities with singular weights

Davit Harutyunyan111University of California Santa Barbara, harutyunyan@ucsb.edu and Hayk Mikayelyan222University of Nottingham Ningbo China, Hayk.Mikayelyan@nottingham.edu.cn
###### Abstract

In this work we derive asymptotically sharp weighted Korn and Korn-like interpolation (or first and a half) inequalities in thin domains with singular weights. The constants (Korn’s constant) in the inequalities depend on the domain thickness according to a power rule where and are constants independent of and the displacement field. The sharpness of the estimates is understood in the sense that the asymptotics is optimal as The choice of the weights is motivated by several factors, in particular a spacial case occurs when making Cartesian to polar change of variables in two dimensions.

Keywords  Korn inequality; Weighted Korn inequality; thin domains
Mathematics Subject Classification  00A69, 35J65, 74B05, 74B20, 74K25

## 1 Introduction

Since the pioneering work of Korn [References,References], Korn and Korn-like inequalities, such as geometric rigidity estimates [References,References] as well as extensions [References] have been known to play a central role in the theories of linear [References,References,References,References,References] and nonlinear [References,References] elasticity. Korn’s first inequality has been introduced by Korn [References,References] to prove the coercivity of the linear elastic energy, and it asserts the following: Given and a closed subspace of vector fields that has a trivial intersection with the subspace of rigid body motions, i.e., there exists a constant depending only on and such that for any vector field the inequality holds:

 C∥∇u∥2L2(Ω)≤∥e(u)∥2L2(Ω). (1.1)

Here, is the symmetric part of the gradient, i.e., strain in linear elasticity. Korn’s second inequality reads as follows: Given there exists a constant depending only on such that for any vector field the inequality holds:

 C∥∇u∥2L2(Ω)≤∥e(u)∥2L2(Ω)+∥u∥2L2(Ω). (1.2)

It has been known that in thin structure, such as rod plate and shell theories, the dependence of the constant in Korn’s inequalities on the geometric parameters of the domain becomes crucial, e.g., [References,References,References,References,References]. Especially it is important to know how the optimal constant scales with the thickness of the thin structure as goes to zero. For plates the constant has been proved to scale like by Friesecke, James and Müller [References] even in the geometric rigidity estimate, which is the nonlinear analog of Korn’s first inequality. When the shell has a nonzero principal curvature, then the scaling is no longer optimal, and new exponents satisfying occur as shown in [References,References,References]. We will call such inequalities sharp. The recent survey book chapter by Stefan Müller [References] gives a complete picture on the above issues and applications as well as the open problems in the field. Sharp Kotn’s inequalities for thin structures, such as rods, plates, shells and combinations of those, have been recently studied by several authors and groups. We refer to the works as well as the above mentioned ones and the references therein for more detailed information. [References,References,References,References,References,References,References,References]. In the present work we deal with weighted Korn and Korn-like inequalities, on which there is relatively less information in the literature [References,References,References,References,References]. The recent work of Lopez Garcia [References] goes further and establishes a generalization of Korn inequalities in the case when the domain is not necessarily thin and thus one is not interested in sharp estimates. We refer to the book of Acosta and Duran [References] for a more detailed discussion of the subject and possible applications. Another motivation of ours of considering weighted Korn and Korn-like inequalities is the following: When dealing with radially symmetric structures, it is convenient to make a Cartesian to polar change of variables, where a weight occurs in the norms, which vanishes at the origin and thus becomes singular. The Korn and similar inequalities under consideration become weighted ones with the above weight, which do not follow from the non-weighted analogues due to the singularity of the weight. The case of two spatial dimensions and is partially studied in [References], and applied to prove optimal Korn inequalities for washers. Another aspect is that the classical Kotn’s first inequality requires a least one condition on the displacement, such as a boundary or a normalization condition, whereas the analogous geometric rigidity estimate does not. Therefore, in order to avoid the imposed boundary conditions, one may be able to apply a localization argument in some parts of the domain, by considering the analogous weighted version of the inequality under consideration. Of course the last is a delicate question and is task for out future studies.

## 2 Main results

We assume in the sequel that and are open bounded connected Lipschitz domains. Let the constant matrix be positive definite with eigenvalues between the positive constants i.e.,

 λ|ξ|2≤ξAξT≤Λ|ξ|2 for allξ∈Rn. (2.1)

Set next the elliptic operator

 L(u)=div(A∇u),for allu∈H1(Ω). (2.2)

The following gradient separation estimate for solutions of elliptic equations is one of the main results of the paper. It has been shown that this kind of estimates derive Korn’s first and Korn interpolation inequalities with the same weight in two space dimensions [References,References,References].

###### Theorem 2.1.

Let let be an open bounded connected Lipschitz domain with the following properties: there exists a partition of the boundary and a number such that

• is a dimensional simplex.

• is star shaped with respect to i.e., for any points and the ray starting from and going through meets the boundary of second time at such that the segment is the only common part of and

Assume further and denote Let the matrix and the operator be as in (2.1) and (2.2). Assume and the exponents and the coefficients satisfy the conditions and for . Denote and for If the function solves the equation in and satisfies the boundary conditions on and the exponent satisfies the bound

 λ>4nΛβ(1−2β)2, (2.3)

then there exists a computable constant depending only on the quantities and such that

 ∥w∇u∥2L2(Ω)≤C(∥wu∥L2(Ω)⋅∥wux1∥L2(Ω)h+∥wux1∥2L2(Ω)), (2.4)

whenever for . Moreover, if the operator is the Laplacian and i.e., and for some then the estimate (2.4) holds true for all values of which means that imposing the condition (2.3) is not necessary.

Next theorem is the analogous Korn’s interpolation inequality in two space dimensions.

###### Theorem 2.2.

For set Let and let the exponents and the function be as in Theorem 2.1. Assume the displacement satisfies the boundary condition for all in the sense of traces. Then the weighted Korn interpolation inequality holds:

 ∥w∇U∥2L2(R)≤C(∥wu∥L2(Ω)⋅∥we(U)∥L2(Ω)h+∥we(U)∥2L2(Ω)). (2.5)

Finally, we emphasize the sharpness of the estimates (2.4) and (2.5) in terms of the asymptotics of as

###### Theorem 2.3.

For the case the estimate (2.4) is sharp in terms of the asymptotics of as Also, the estimate (2.5) is sharp as well in terms of the asymptotics of as

## 3 Preliminaries

###### Lemma 3.1.

Assume is an open bounded set with Lipschitz boundary and assume Denote by the distance function from the part of the boundary i.e., for Assume the weight function is such that

 w(x)≥0,|δ(x)∇w(x)|≤K⋅w(x)for allx∈Ω, (3.1)

for some If the function satisfies the boundary condition for in the sense of traces, then there exists a constant depending only on the quantities and , such that the estimate holds:

 ∥wδ∇u∥2L2(Ω)≤C(∥wu∥2L2(Ω)+∥wδ2L(u)∥2L2(Ω)). (3.2)
###### Proof.

Let us mention that in the proof the constant depends only on the quantities and Following [References], we evaluate for any using the boundary conditions on

 I3=∫Ωw2δ2uL(u)dx=∫Ωw2δ2u[div(A∇u)]dx=I+I1+I2, (3.3)

where

 I=−∫Ωw2δ2∇u⋅(A∇u)dx, (3.4)
 I1=−2∫Ωw2δu∇δ⋅(A∇u)dx,I2=−2∫Ωwδ2u∇w⋅(A∇u)dx. (3.5)

Observe that as is a distance function, thus it is Lipschitz and weeakly differentiable a.e. with Consequently we have by the geometric-arithmetic mean inequality in the form that

 |I1|≤ϵ∫Ωw2δ2|∇u|2dx+Cϵ∫Ωw2u2dx, (3.6)

where is a number yet to be chosen. We have similarly that

 |I2|≤ϵ∫Ωw2δ2|∇u|2dx+Cϵ∫Ωδ2|∇w|2u2dx,

thus owing to the bound (3.1) we get

 |I2|≤ϵ∫Ωw2δ2|∇u|2dx+CK2ϵ∫Ωw2u2dx. (3.7)

Finally we have for by the Schwartz inequality, that

 |I3|≤12∫Ωw2u2dx+12∫Ωw2δ4L(u)2dx. (3.8)

By the positive definiteness condition (2.1) of we get the lower bound

 I=∫Ωw2δ2∇u⋅(A∇u)dx≥λ∫Ωw2δ2|∇u|2, (3.9)

thus combining the estimates (3.3)-(3.9) and choosing we obtain (3.2).

Next we give some useful examples of weights satisfying the hypothesis (3.1). The statement is formulated in the below lemma.

###### Lemma 3.2.

Exponents of distance functions such as and fulfill the condition (3.1) for any point with If functions satisfy (3.1) with the same constant , then any linear combination with positive coefficients satisfies (3.1) with the same constant Also, the product weight satisfies (3.1) with a constant

###### Proof.

The proof is elementary. We have for the case that thus For the case we have thanks to the inequality that

 |δ(x)∇w(x)|=|α|(dist(x,x0))α−1δ(x)|∇dist(x,x0)|≤|α|(dist(x,x0))α=|α|w(x).

The proofs of the two remaining statements being trivial are skipped. ∎

## 4 Proofs of the main resuts

###### proof of Theorem 2.1.

We assume fist that and are general and the condition (2.3) is satisfied. Due to the convenience of the reader we divide the proof is into some steps.
Step 1. It is sufficient to prove Theorem 2.1 for a single summand where
We aim to verify that if (2.4) is valid for the weights , with the same constant then it is valid for the sum with the new constant We have by the Cauchy-Schwartz inequality that

 ∥w∇u∥2L2(Ω) =∫Ωw2|∇u|2dx (4.1) =∫Ω(k∑i=1wi)2|∇u|2dx ≤k∫Ωk∑i=1w2i|∇u|2dx =kk∑i=1∥wi∇u∥2L2(Ω).

In the other hand we have the obvious inequality

 ∥wu∥L2(Ω)⋅∥wux1∥L2(Ω)h +∥wux1∥2L2(Ω) ≥∥wiu∥L2(Ω)⋅∥wiux1∥L2(Ω)h+∥wiux1∥2L2(Ω),

thus combining it with (4.1), the validity of the statement follows. Thus without loss of generality, we will assume that in what follows in the proof, the weight is a single summand, i.e., where We also set
Step 2. Assume the total number of dimensional subfaces of is , for and denote them by Then there exists open connected disjoint subsets of for and with the following properties:

• The function satisfies if for and

•  |ω∖(∪dl=0∪Nls=1ωls)|=0,

where stands for the dimensional Lebesgue measure.

• The estimate holds

 n∑i,j=1aij∂2δγ∂xi∂xj≤nΛγδγ−2, (4.2)

for any and

Observe that upon rotation and translation of the coordinate system, the distances function from an dimensional hyperplane in is given by

 δ(x)=(x21+x22+…+x2n−i)1/2.

Thus the equality of the distance of the points from two and dimensional hyperplanes gives a hypersurface of dimensional measure zero. Next, it is clear that the distance function from the simplex is locally the distance function from one of the subfaces of i.e., the domain can be partitioned into open domains and modulo zero measure, such that in each of the sets the distance is the distance from the subface of the simplex Assume now and are fixed. Given any point we have We aim next to prove a Hardy-like estimate for the weighted norms of and that holds in To that end we can assume without loss of generality that (upon a translation of the coordinates). Assume that is dimensional, then we can rotate the coordinate system to put in the hyperplane thus we get

 δ(x)=(m∑k=1y2k)1/2, (4.3)

where and is a rotation. We can calculate for any

 ∂δγ∂xi=γδγ−2m∑k=1bkiyk,

and

 ∂2δγ∂xi∂xj=γ(γ−2)δγ−4m∑k=1bkiykm∑l=1bljyl+γδγ−2m∑k=1bkibkj. (4.4)

Consequently

 n∑i,j=1aij∂2δγ∂xi∂xj=γ(γ−2)δγ−4n∑i,j=1aijm∑k=1bkiykm∑l=1bljyl+γδγ−2n∑i,j=1aijm∑k=1bkibkj. (4.5)

Using the fact that the matrix is positive definite and we get that the first summand in (4.5) is nonpositive, thus we obtain the estimate

 n∑i,j=1aij∂2δγ∂xi∂xj≤γδγ−2n∑i,j=1aijm∑k=1bkibkj (4.6)

The condition (2.1) and the orthogonality of gives the bound

 n∑i,j=1aijbkibkj≤Λn∑i=1b2ki≤Λ,for all1≤k≤m,

thus we derive from (4.6) the bound

 n∑i,j=1aij∂2δγ∂xi∂xj≤mΛγδγ−2≤nΛγδγ−2, (4.7)

which is (4.2), and Step 2 is done.
Step 3. For any and the Hardy-like estimate holds:

 ∫ωlsδγ−2u2≤4(γ−1)2∫ωlsδγ|∇u|2. (4.8)

Observe that if , is absolutely continuous with then we have integrating by parts,

 ∫a0tγ−2f2(t)dt =1γ−1∫a0f2(t)dtγ−1 =21−γ∫a0tγ−1f(t)f′(t)dt,

thus we have by the Schwartz inequality,

 ∫a0tγ−2f2(t)dt≤21−γ(∫a0tγ−2f2(t)dt)1/2(∫a0tγf′2(t)dt)1/2,

consequently we get

 ∫a0tγ−2f2(t)dt≤4(1−γ)2∫a0tγf′2(t)dt. (4.9)

Next we fix a point It is clear that there exists a cone with an apex at such that for each point one has Moreover, if such that and Take now any pooint , then by the assumption, the ray starting at and going through meets the boundary of second time at It is clear that one has and as well. Denote which is clearly absolutely continuous. An application of (4.9) to gives

 ∫10tγ−2u2(x+t(z−x))dt≤4(1−γ)2∫10tγ|z−x|2|∇u(x+t(z−x))|2dt. (4.10)

Denoting for and noting that we obtain from the following segmental integral estimate

 ∫zxδγ−2u2(yt)dt≤4(1−γ)2∫zxδγ|∇u(yt)|2dt. (4.11)

By integrating (4.11) over all directions from to in and then integrating the obtained estimtes over all we arrive at (4.8). Before starting the last step of the proof, first observe that by Step 2, the estimate (4.2) holds a.e. in , and second, by summing up the bounds (4.8) over the indeces and we get the analogous estimate for

 ∫ωδγ−2u2≤4(γ−1)2∫ωδγ|∇u|2. (4.12)

Step 4. In the last step we conclude the proof the Theorem 2.1. For any denote and Recalling that we have by integration by parts and using the condition (2.1), that

 λ∫Ωt δγ|∇u|2dx≤∫Ωtδγn∑i,j=1aijuxiuxjdx (4.13) =−∫Ωtδγun∑i=1n∑j=2aijuxixjdx−∫Ωtun∑i=1n∑j=2aijuxi∂∂xj(δγ)dx−∫Ωtδγun∑j=2a1jux1xjdx −∫Ωtun∑j=2a1jux1∂∂xj(δγ)dx−∫Ωta11uδγux1x1dx+a11∫ω[(uδγux1)|x1=h/2+tx1=h/2−t]dx′ =−∫ΩtuδγL(u)−2a12∫Ωtδγ−1uux1dx−∫Ωtun∑i,j=2aijuxi∂∂xj(δγ)dx +a11∫ω[(uδγux1)|x1=h/2+tx1=h/2−t]dx′ =−2a12γ∫Ωtδγ−1uux1dx−∫Ωtun∑i,j=2aijuxi∂∂xj(δγ)dx+a11∫ω[(uδγux1)|x1=h/2+tx1=h/2−t]dx′.

For the second summand we have integrating by parts and using the estmates (4.2) and (4.12), that

 ∫Ωtun∑i,j=2aijuxi∂∂xj(δγ)dx =−12∫Ωtu2n∑i,j=2aij∂2(δγ)∂xi∂xjdx (4.14) ≥−nΛγ2∫Ωtδγ−2u2dx ≥−2nΛγ(1−γ)2∫Ωtδγ|∇u|2dx.

For the first summand we have by the Schwartz inequality and by (4.12),

 ∣∣∣2a12γ∫Ωtδγ−1uux1dx∣∣∣ ≤2|a12|γ(∫Ωtδγ−2u2dx)1/2(∫Ωtδγ|ux1|2dx)1/2 (4.15) ≤4|a12|γ1−γ(∫Ωtδγ|∇u|2dx)1/2(∫Ωtδγ|ux1|2dx)1/2.

Combining (4.13), (4.14) and (4.15) we establish the bound

 (λ−2nΛγ(1−γ)2)∫Ωtδγ|∇u|2dx ≤Λ∫Ωt[(uδγux1)|x2=h/2+tx2=h/2−t]dx′ (4.16) +4Λγ1−γ(∫Ωtδγ|∇u|2dx)1/2(∫Ωtδγ|ux1|2dx)1/2

Next we integrate (4.11) in over the interval and utilize the Schwartz and the Cauchy-Schwartz inequalities to get under the condition (2.3) the bound

 ∫h/20dt∫Ωtδγ|∇u|2dx≤C(∫Ωδγu2dx)1/2(∫Ωδγ|ux1|2dx)1/2+Ch∫Ωδγ|ux1|2dx. (4.17)

Observe, that the function increases in thus we get from (4.17) the estimate

 ∫Ωh/4δγ|∇u|2dx≤Ch(∫Ωδγu2dx)1/2(∫Ωδγ|ux1|2dx)1/2+C∫Ωδγ|ux1|2dx. (4.18)

Having (4.18) in hand, it remains to estimate the quantity Next we recall the following Hardy-like estimate established by Kondratiev and Oleinik [References,References,References].

###### Lemma 4.1.

Assume and is absolutely countinuous. Then there holds:

 ∫a/20f2(t)dt≤4∫aa/2f2(t)dt+4∫a0t2|f′(t)|2dt.

We fix any index any point and apply Lemma 4.1 to the function over the interval with the endpoints and We have that

 ∫h/40δγ|uxk(t,x′)|2dt≤4∫h/2h/4δγ|uxk(t,x′)|2dt+4∫h/20δγt2|uxkx1(t,x′)|2dt.

Summing the above inequalities over and addind the missing summand on the left hand side we arrive at

 ∫h/40δγ|∇u(t,x′)|2dt≤4∫h/2h/4δγ|∇u(t,x′)|2dt+∫h/40δγ|ux1(t,x′)|2dt+4∫h/20δγt2|∇ux1(t,x′)|2dt.

Upon integrating the last estimate in over we obtain

 ∫Ω′h/4δγ|∇u|2≤4∫Ωh/4δγ|∇u|2+∫Ω′h/4δγ|ux1|2+4∫Ω′h/2δγx21|∇ux1|2. (4.19)

From the boundary conditions on we have that on