Pointwise Estimates for Polyharmonic Green's Function

Pointwise Estimates

for Polyharmonic Green’s Function

Submitted to the Department of Mathematics, School of Sciences and Humanities, in partial fulfillment of the requirements for the degree

Master of Science in Applied Mathematics

Author:

Galiya Myrzabayeva

Supervisor:

Durvudkhan Suragan

Spring 2023

Contents

0.1 Abstract . . . 2

1 Introduction 3

1.1 Notations . . . 3 1.2 Short historical and literature review . . . 5 1.3 Structure . . . 11

2 Preliminaries 12

2.1 Useful tools . . . 12 2.2 Known estimates of Green’s function . . . 18

3 Main result 31

3.1 Mazya’s problem in a ball . . . 31 3.2 Conclusion . . . 33

0.1 Abstract

This thesis focuses on the polyharmonic Green’s function, one of fundamental con- cepts in mathematical analysis and partial differential equations. The Green’s func- tion plays a crucial role in solving problems related to the behavior of harmonic functions, and has a wide range of applications in physics and engineering. The in- troduction section starts with an overview of the notation used throughout the thesis.

A brief historical and literature review follows to provide context and a better under- standing of the importance of the topic. The preliminary section then lays out the necessary background knowledge required for further proofs of the main theorems.

The thesis then proceeds into a detailed discussion of known pointwise estimates proofs, which are crucial to understanding the Green’s function’s behavior. The main contribution of this thesis is a partial solution to an open problem from Mazya’s list of ”Seventy Five (thousand) unsolved problems in analysis and partial differential equations” from [27].

Chapter 1 Introduction

1.1 Notations

Let x= (x₁, x₂, x₃, ..., x_n)∈ Rⁿ. As usual, the Laplacian ∆u of a smooth function u is the sum of the second partial derivatives with respect to each component x_i:

∆u=

n

X

i=1

∂²u

∂x²_i.

Our interest is iteration of the above Laplace operator (or the Laplacian), that is, the so-called polyharmonic operator. It is defined inductively by:

(−∆)^mu=−∆((−∆)^m−1u), where m = 1,2, . . . . It is clear to see:

(−∆)^mu= X

l1+...+ln=m

m!

l₁!...l_n!

∂^2mu

∂x^2l₁¹...∂x^2l_nⁿ.

Throughout this thesis B_r⊂Rⁿ denotes a ball with radius r centered at the origin.

Let’s define the distance function between givenxand the boundary∂Ω of a given domain Ω ⊂Rⁿ as the following:

d(x) := dist(x, ∂Ω) = min

y∈∂Ω|x−y|, x∈Ω,

where Ω is a smooth and bounded domain.

The estimates of the Green function will be often presented in the unit ball inRⁿ, for that purpose the following notations will be used:

Forx, y ∈B, where ¯¯ B =B∪∂B, we write [XY] := p

|x|²|y|²−2x·y+ 1 =

|x|y− x

|x|

=

|y|x− y

|y|

. (1.1.1)

[XY] is the distance from |y|x to the projection of y on the unit sphere, which is larger than |x−y|. Indeed,

XY|²− |x−y|² = 1− |x|²

1− |y|²

>0 forx, y ∈B . 1− |x|=d(x) forx∈B.¯

We say f(t) ≃ g(t) if there exists C > 0 such that for all t: _C¹f(t) ≤ g(t) ≤ Cf(t), f, g≥0

We define derivatives ofu as the following:

D^αu= ∂

∂x₁ α1

... ∂

∂x_n αn

u forα ∈Nⁿ with |α| ≤k.

Let Ω ⊂Rⁿ be a smooth bounded set with boundary ∂Ω. Consider the polyhar- monic Dirichlet problem:

(−∆_x)^mG_m,n(x, y) = δ(x−y), x, y ∈Ω, (1.1.2) with the homogeneous Dirichlet boundary conditions:

∂ⁱ

∂nⁱ_xG_m,n|x∈∂Ω = 0, i= 0, ..., m−1, (1.1.3) whereδ is the Dirac delta function. The solutionG_m,n(x, y) of (1.1.2)-(1.1.3) is called the Green function for the polyharmonic Dirichlet problem. For basic properties of δ, please see the next section.

1.2 Short historical and literature review

Higher order linear elliptic equations have a wide range of applications in science, describing models of different phenomena. The first attempt to use polyharmonic operators was made by Jacob Bernoulli II. He tried to model the vibrations of plate by the fourth order operator _∂x^∂44+_∂y^∂44. His attempts were inspired by physicist Chlandni who presented nodal line patterns of vibrating plates around 1800. However, his model didn’t describe vibrations precisely since it was not rotationally symmetric.

Sophie Germain then used ∆² in 1811 to model an elastic plate. The bi-Laplacian operator ∆² appears in problems of linear elasticity (see, e.g. [12]). An example of such problem is the clamped plate problem:

( ∆²u=f in Ω, u= _∂n^∂u = 0 on ∂Ω,

where ^∂u_∂n is the outer unit normal on the boundary ∂Ω of Ω⊂Rⁿ.

George Green was the first to introduce the approach of using Green’s functions to solve many physics problems. Green’s function’s concept can be discussed by considering the dynamics of some particle with zero initial velocity and under the force F(t). First, a short impulse of a force is considered. One chooses an impulse in such a way, so that a unit time momentum is induced at some time t^′. The Green function G(t, t^′) is defined to be a displacement of a particles(t) in some later time t. A force F(t^′) acting for an infinitesimal time ∆t^′ creates an impulse with a magnitude F(t^′)∆t^′. Force that is applied continuously to the particle is considered as an impulse generating force. The motion of particle can be found by summing the effects of impulses in the time interval from t₀ tot. Thus

s(t) = Z t

t0

G(t, t^′)F(t^′)dt^′,

with initial conditions: s(0) = ^ds_dt(0) = 0. Hence, knowing Green’s function, the response of a system to any force can be easily calculated. For Green’s technique’s details we refer to [4].

Originally, Green worked on solving problems in electrostatics in a bounded region.

The Green function G(r, r^′) in that problem is the potential at r which is produced by a point charge atr^′. Analogous to an impulse acting at an instant in time, a point charge acts at a single point in space.

Shortly, the Green function technique is as following:

Let L_x be a linear differential operator, where x = (x₁, x₂, ..., x_n), then Green’s function G(x, y) of a linear operator L_x satisfies the following expression:

LxG(x, y) = δ(x−y),

where δ(x−y) is the Dirac delta function. Let us briefly recall the definition of the Dirac delta function: Dirac delta function (δ), is the generalized function over Rⁿ, whose value is zero everywhere except 0, and integral over a real line is 1. It is also known as the impulse function. Mathematically defined as the following:

δ(x) =

( ∞, if x= 0, 0, if x̸= 0, with R∞

−∞δ(x)dx= 1.

Once we know the Green function G(x, y), we can solve any differential equation containing L_x.

Suppose, we are given a problem:

L_xϕ(x) = f(x).

The solution is :

ϕ(x) = Z

G(x, y)f(y)dy.

Verifying the answer:

L_xϕ(x) = Z

L_xG(x, y)f(y)dy= Z

δ(x−y)f(y)dy =f(x).

In the 19th century, the method of Green’s functions was widely used to solve partial differential equations in thermal, electrical, and mechanical physics, as well

as problems in magnetism. Green’s function is also used in formulating the theory of wave scattering. In the 1950-1960s scientists started to use Green’s function method in many-body interactions theory of a condensed matter physics. Nowadays, physi- cists use Green’s function technique in the areas that were not even known in the times of Green. And it is highly likely that Green’s function will continue to be used in whatever is going to be developed in the future. Green’s function is very pow- erful technique that allows to solve differential equations in different areas, however constructing and solving for Green’s function is not a trivial problem. It is a challeng- ing procedure and requires rigorous mathematics. Going back to the problems with polyharmonic Green’s function (1.1.2), the first explicit solution for polyharmonic problem was calculated by Boggio. Boggio gave a representation of polyharmonic Green’s function in a ball. Boggio’s formula in an integral form is a classical tool to construct an explicit solution to the polyharmonic equation. More precisely, Boggio [3] derived an explicit formula for G−∆^m,B1

( (−∆)^mu=f inB₁,

u= _∂n^∂jju= 0 on∂B₁ with j = 1,2, ..., m−1,

where B₁ is the unit ball centered at the origin, n is the unit normal at the boundary

∂B₁.

The solution of a problem is the following expression:

u(x) = Z

B1

G−∆^m,B1(x, y)f(y)dy.

The representation of Green’s function by Boggio mentioned in [12]:

G_m,n,B1(x, y) =k_m,n|x−y|^2m−n Z

^|x|y−

x

|x|

^/|x−y|

1

(v²−1)^m−1v¹⁻ⁿdv.

The positive constants k_m,n are defined by

k_m,n = 1

nen4^m−1((m−1)!)², e_n= πⁿ Γ(1 +n/2).

The representation of Green’s function above holds true for the domain B₁.

Let us consider the biharmonic case (m= 2) to understand the nature of the problem:

( (−∆)²u=f in B₁, u= ^∂u_∂n = 0 on∂B₁,

Dirichlet boundary value problem with biharmonic operator with n= 2

∆² = ∂⁴

∂x⁴ + 2 ∂⁴

∂x²∂y² + ∂⁴

∂y⁴

appears in many fields of applied mathematics, for example, in fluid mechanics and elasticity theory. In the latter, u(x, y) defines Airy stress function, in the theory of thin plates-displacement of a plate under the external force.

For the case when n= 2 =m, the Green function

G−∆²,B1(x, y) = 1

8π|x−y|² Z

^|x|y−

x

|x|

^/|x−y|

1

v²−1 v dv

is strictly positive [12]. One can show that the polyharmonic Green’s function in a ball is always positive. In these dimensions, it can be seen that the problem models a plate of a circular shape, f is an acting force, and theu is a deflection of a plate.

The positivity of Green’s function is an important question that can help to solve other physical problems. For example, the positivity of Green’s function helps to answer the question presented by Boggio in [3]: ”If a clamped plate of circular shape will be pushed upwards, will the clamped plate bend upwards everywhere, too?”. The answer to the problem above is yes, since Green’s function G_(−∆)²_,B1(x, y) in a given domain and f(x) are positive.

In some general domain Ω

( (−∆)^mu=f in Ω,

∂^j

∂n^ju= 0 on∂Ω with j = 1,2, ..., m−1, (1.2.1) Green’s function is not necessarily positive, it can be sign-changing. There are

many papers that study positivity of Green’s function. For example, in the problem mentioned above, Hadamarad and Boggio [3] conjectured that the Green’s function G−∆²,Ω is positive for convex two-dimensional domains. Garabedian [11] provided a counterexample for smooth convex domains, showing that Green’s function is sign- changing in ellipse. Another counterexample is Duffin’s work [9], where the author shows that the Green’s function changes the sign in a long rectangle. Moreover, the Boggio-Hadamard’s conclusion in not true for dimensions other than two.

For dimensionsm ≥2, n ≥2, Kozlov et al. [24] provided a strictly convex domain where Green’s function is sign-changing.

For the constant right-hand side of (1.2.1), Grunau and Sweers [16] found domains for m = 2 where Green’s function changes sign.

As mentioned above, we cannot claim that Green’s function is positive in domains other than ball. Therefore, much work has been done to identify the domains where the Green’s function’s positivity holds. Grunau and Sweers [14] considered two- dimensional domains which are close to balls. Dal’Acqua and Sweers [8] gave an example of a non-convex domain for m = 2, n = 2. For the case when m = 2 and higher dimensions, Grunau and Roberts [18] proved positivity in the domain close to the ball. The result is true for the dimensions where n ≥2m−1.

Grunau and Sweers in [15] used a method suggested by Nehari [32] for the general case to find the regions where Green’s function is positive. They found a constant δ_m,n which doesn’t depend on the domain Ω such that for every x, y in Ω, and

|x−y|< δ_m,nmax{d(x), d(y)}, where d(x) := dist (x, ∂Ω) =infx∗∈∂Ω|x−x^∗| so that

G−∆^m,Ω(x, y)>0.

The authors proved this for dimensions n > 2m. Kokritz [23] proved it for the case n = 2m . Thus, the negative part is uniformly bounded when x and y are uniformly distanced from the boundary ∂Ω. For example, for the biharmonic case, minimal

distance δ=δ(Ω) can be identified such that for all x, y ∈Ω with x̸=y,

|x−y|< δ implies that G_(−∆)²_,Ω(x, y)>0.

The result of local positivity was obtained by Grunau and Robert [18] for the case when n ≥3 and forn = 2 by Dall’Aaqua et al. in [6].

Eichmann and Sch¨atzle [10] considered the positivity of Green’s function in a clamped plate problem with high tension. The problem is presented by ∆²u−γ∆u=f with clamped boundary conditions. The authors showed that with a given upward pushing force f, there exists such γ₀, so that for all γ ≥γ₀ the bending u is always positive.

Let us give a short overview of what is done for characterization of Green’s function.

In general, for the second-order equations with dimension m = 1, the Green function’s estimates from both sides are known for sufficiently smooth domains. For higher dimensions, Gruter [17] and Widman [31] identified estimates from above.

Zhao [33],[34] found estimates of the Green function from two sides.

Liu and Dai [26] considered a system of polyharmonic equations with Dirichlet boundary conditions in a half-space and proved nonexistence of nonnegative nontrivial classical solutions for the problem.

Ancona [1], Cranston et al. [5], Hueber and Sievking [19] provided estimates for second-order differential equations in sufficiently smooth domains. Estimates in non-smooth domains were presented by Maz’ya and Mayboroda [28].

Karachik [21] provided the Green function for the biharmonic Dirichlet prob- lem and found integral representation of the solution in terms of Green’s function.

Also, Karachik [22] found polynomial solution for Dirichlet problem with polynomial boundary data and right-hand side in unit ball.

Gazzola and Sperone [13] discussed radial properties (symmetry and monotonic- ity) of radial solutions of semilinear higher order elliptic equations. The result yielded in formula for computing partial derivatives of solutions of polyharmonic problems.

Aroua and Bellassoued [2] considered an inverse boundary value problem for poly- harmonic operator, moreover the authors discussed stability estimate for the inverse problem.

1.3 Structure

In Chapter 2, we introduce and discuss several useful tools that are key ingredients in the proofs presented in subsequent chapters. These tools serve as the foundation for the rigorous analysis of the polyharmonic Green’s function and its properties. We discuss then the proofs of the pointwise estimates of Green’s function presented by Gazzola et al. [12], Pulst [30], and the estimates of Green’s function derivatives by Gazzola et al. [12], Krasovskii [25], Dallacqua and Sweers [7]. Moving on to Chapter 3, we proceed into the proof of an inequality from Mazya’s list of ”Seventy Five (thousand) unsolved problems in analysis and partial differential equations” from [27]. Finally, the thesis concludes with a summary of the main findings.

Chapter 2

Preliminaries

2.1 Useful tools

In this section we will provide useful tools that will help to obtain Green’s function’s and its derivatives’ estimates.

The following lemma gives a representation of the Green function:

Lemma 2.1.1. [20] If n >2m, then Green’s function of the polyharmonic Dirichlet problem (1.1.2)- (1.1.3) can be represented in the form

Gm,n,r(x, y) =dm,n

h

X^2m−n−Y_r^2m−n

−

m−1

X

k=1

(−1)^k k!

m− n

2

m− n 2 −1

...

m− n

2 −k+ 1

Y_r^2m−n−2kZ_r^2ki ,

where

d_m,n = 2Γ(n/2−m)Γ(1 +n/2)/(nπ^n/2 ×4^mΓ(n/2)(m−1)!). (2.1.1) Here X(x, y), Yr(x, y) and Zr(X, Y) are the functions such that:

X² =|x−y|², Y_r² = y r

2

x− y

|y|²r²

2

and Z_r =r²

1− y r

2

1− x r

2 , respectively.

Proof. Clearly, for all x, y ∈ B_r the relation X² =Y_r² −Z_r² holds true. Then we can use the relation in the following way:

|x−y|^2m−n= X²m−n/2

= Y_r²m−n/2

1− Z_r² Y²

m−n/2

.

Also, we can represent (1−Z_r²/Y_r⁻²)^m−n/2 by the series. Then |x−y|^2m−n can be expressed as the following:

|x−y|^2m−n= X²m−n/2

= Y_r²m−n/2

1− Z_r² Y_r²

m−n/2

=

= Y_r²m−n/2h 1 +

∞

X

k=1

(−1)^k k

m− n

2 m− n 2 −1

...

m− n

2 −k+ 1 (Z_r²

Y_r²)^ki

=Y_r^2m−n+

m−1

X

k=1

(−1)^k k!

m− n

2 m−n 2 −1

...

m−n

2 −k+ 1

(Y_r²)^m−n/2−k(Z_r²)^k +

∞

X

k=1

(−1)^k k!

m−n

2 m− n 2 −2

...

m− n

2 −k−1

(Y_r²)^m−n/2−k(Z_r²)^k.

The last relation allows us to define Green’s function as follows:

G_m,n,r(x, y) =d_m,nh

X^2m−n−Y_r^2m−n

−

m−1

X

k=1

(−1)^k k!

m− n

2

m− n 2 −1

...

m− n

2 −k+ 1

Y_r^2m−n−2kZ_r^2ki ,

The following lemma shows the positivity of Green’s function in a ball with arbi- trary radius:

Lemma 2.1.2. [15] Suppose n >2m and Br is a ball with radius r. Green’s function can be represented as a function in a ball B1 with radius 1 with re-scaling factor and has the following form:

G_m,n,r(x, y) = r^2m−nG_m,n,B11 rx,1

ry .

Proof. We need to check that r^2m−nG_m,n,B1

1 rx,¹_ry

satisfies equations (1.1.2)- (1.1.3). Let’s first check that r^2m−nG_m,n,B1

1 rx,¹_ry

satisfies equation (1.1.3). Since G_m,n,B1

x, y

is the solution for Dirichlet problem in unit ball, then it satisfies the boundary condition

∂ⁱ

∂nⁱ_xG_m,n = 0 on ∂B₁.

It is clear that for any x∈∂Gm,n,r, ^x_r will be in ∂B1, then _∂n^∂ii

xGm,n,B1

1 rx,¹_ry

is also 0 on ∂B₁. Thus, it satisfies boundary conditions of the problem and

∂ⁱ

∂nⁱ_x

r^2m−nG_1,n,B1(¹_rx,¹_ry)

= 0 on ∂B₁. Now, we need to check that r^2m−nG_m,n,B1

1 rx,¹_ry

satisfies (1.1.2). For arbitrarym, the equation becomes:

(−∆)^mG_m,n(x, y) = δ(x−y) in B₁.

Now, let’s check for r^2m−nG_m,n,B1

1 rx,¹_ry

forx, y ∈B_r. (−∆)^m

r^2m−nG_m,n,B11 rx,1

ry

=r^2m−n 1

r^2m(−∆)^mG_m,n,B11 rx,1

ry

= 1

rⁿ(−∆)^mG_m,n,B11 rx,1

ry . For all (x, y)∈G_m,n,r,

1 rx,¹_ry

∈B₁. Thus, (−∆)^mG_m,n,B1

1 rx,¹_ry

= 0 andr^2m−nG_m,n,B1

1

rx,¹_ry

= 0 for all x̸=y.

For all (x, y) ∈ G_m,n,r, such that x = y,

1 rx,¹_ry

∈ B₁ and ¹_rx = ¹_ry. Thus, (−∆)^mG_m,n,B1

1 rx,¹_ry

=δ(x−y) for any x, y ∈B_r. The statement is proven.

The following proposition presents Green’s second identity for polyharmonic case:

Proposition 2.1.3. [30] Let Ω be a domain for which the divergence theorem holds

and let u, v ∈C^2m( ¯Ω). Then it holds Z

Ω

(−∆)^muv−u(−∆)^mvdx=

m−1

X

ℓ=0

Z

∂Ω

∂

∂n(−∆)^m−1−ℓv(−∆)^ℓu− ∂

∂n(−∆)^ℓu(−∆)^m−1−ℓvdS , where n is the exterior unit normal on ∂Ω.

Proof. For the case whenm = 1:

Z

Ω

(−∆u)vdx = Z

Ω

u(−∆v)dx+ Z

∂Ω

(∂u

∂nv−u∂v

∂n)dS.

Using integration by parts for LHS when m= 2:

Z

Ω

((−∆)²uv −u(−∆)²v)dx=

= Z

Ω

(−∆)u(−∆)v− Z

∂Ω

(∂(−∆)u

∂n v−(−∆)u∂v

∂n)dS

− Z

Ω

(−∆)u(−∆)v+ Z

∂Ω

(∂(−∆)v

∂n u−(−∆)v∂u

∂n)dS Solving for RHS when m= 2:

Z

∂Ω

∂(−∆)v

∂n u− ∂u

∂n(−∆)v+ ∂v

∂n(−∆)u− ∂(−∆)u

∂n vdS

The RHS and LHS are identical. Inserting −∆^lu and −∆^m−1−lv into Green’s second identity will sum to the desired expression.

As a consequence of the previous proposition we have the next corollary:

Corollary 1. [30] Let Ω a domain for which the divergence theorem holds and u, v ∈C^2m( ¯Ω). Let _∂n^∂iiv = 0 on ∂Ω fori= 0, .., m−1. Then fork ∈N the following is true:

1. If m = 2k Pk

l=1

R

∂Ω

∂

∂n(−∆)^l−1u(−∆)^m−lv− _∂n^∂ (−∆)^m−lv(−∆)^l−1udσ +R

Ωv(−∆)^mu−u(−∆)^mvdx = 0. (111) 2. If m = 2k+ 1

Pk l=1

R

∂Ω

∂

∂n(−∆)^l−1u(−∆)^m−lv−_∂n^∂ (−∆)^m−lv(−∆)^l−1udσ

−R

∂Ω

∂

∂n(−∆)^kv(−∆)^kudσ+R

Ωv(−∆)^mu−u(−∆)^mvdx= 0. (111)

The following lemma presents the difference between cases whenxandyare closer to the boundary ∂Ω than to each other and visa versa.

Lemma 2.1.4. [12] Let x, y ∈B.¯ If |x−y| ≥ ¹₂[XY], then

d(x)d(y)≤3|x−y|² (2.1.2)

max{d(x), d(y)} ≤3|x−y| (2.1.3)

If |x−y| ≤ ¹₂[XY], then 3

4|x−y|² ≤ 3

16[XY]≤d(x)d(y) (2.1.4) 1

4d(x)≤d(y)≤4d(x) (2.1.5)

|x−y| ≤3 min{d(x)d(y)} (2.1.6)

[XY]≤5 min{d(x), d(y)}. (2.1.7)

Moreover, for all x, y ∈B¯ we have

dx⩽[XY], d(y)≤[XY], [XY]≈d(x) +d(y) +|x−y|

Proof. In the subsequent chapters, we will use the estimates 2.1.3, 2.1.4. The proof of two estimates provided below.

Let |x−y| ≥ ¹₂[XY].

d(x)d(y) = (1− |x|)(1− |y|)≤(1− |x|²)(1− |y|²) = [XY]²− |x−y|², but since |x−y|² ≥ ¹₄[XY],then

d(x)d(y)≤[XY]²− |x−y|² ≤3|x−y|².

The proof for (2.1.4) is complete. Now, the estimate (2.1.3) follows from the following:

d(x)² ≤d(x)(d(y) +|x−y|)≤3|x−y|²+d(x)|x−y| ≤4|x−y|²+ 1 4d(x)², thus d(x)² ≤ ¹⁶₃|x−y|².

Let|x−y| ≤ ¹₂[XY]. Since (1 +|x|)(1 +|y|)≤4 in unit ball, then (1− |x|)(1− |y|) = d(x)d(y)≥ 1

4(1− |x|)(1− |y|)(1 +|x|)(1 +|y|).

Then

d(x)d(y)≥ 1

4([XY]− |x−y|²)≥ 3

4|x−y|². The proof of (2.1.4) is done.

The following lemma is the consequence of the previous lemma in general domain.

Lemma 2.1.5. [12] Let Ω⊂R^′′ be a bounded domain and let p, q,≥0 be fixed.

For (x, y)∈Ω¯² we have:

min

1, d(y)

|x−y|

≃min

1,d(y)

d(x), d(y)

|x−y|

min

1,d(x)d(y)

|x−y|²

≃min

d(y) d(x)²,d(x)

d(y),d(x)d(y)

|x−y|²

min

1,d(x)^pd(y)^q

|x−y|^p+q

≃min

1, d(x)^p

|x−y|^p, d(y)^q

|x−y|^q,d(x)^pd(y)^q

|x−y|^p+q

min

1,d(x)^pd(y)^p

|x−y|^p+q

≃min

1, d(x)

|x−y|

p

min

1, d(y)

|x−y|

q

and assuming moreover that p+q >0, we also have log

1 + d(x)^pd(y)^p

|x−y|^p+q

≃log

2 + d(y)

|x−y|

min

1,d(x)^pd(y)^q

|x−y|^p+q

2.2 Known estimates of Green’s function

In this section we focus on polyharmonic clamped plate boundary value problem.

( (−∆^m)u=f in Ω,

∂^j

∂n^ju= 0 on ∂Ω with |j| ≤m−1, (2.2.1) Here, Ω ⊂Rⁿis a bounded and smooth domain,f is in suitable functional space. For a smooth domain, Green’s function G−∆^m,Ω exists and the solution of the problem (2.2.1) is of the following form:

u= Z

Ω

G_−∆^m_,Ω(x, y)f(y)dy, x∈Ω.

The following theorem provides two-sided estimate of Green’s function:

Theorem 2. [12]

In B¯×B¯ we have

G_m,n(x, y)≃











|x−y|^2m−nminn

1,^d(x)_|x−y|^md(y)2m^m

o

if n >2m;

log

1 + ^d(x)_|x−y|^md(y)2m^m

if n = 2m;

d(x)^m−n2d(y)^m−n2 minn 1,^d(x)

n 2d(y)ⁿ²

|x−y|ⁿ

o

if n <2m.

Proof. Using the results of Lemma 2.1.4, let’s discuss two cases: |x−y| ≥ ¹₂[XY], and |x−y| ≤ ¹₂[XY].

Case 1: |x−y| ≤ ¹₂[XY].

We are interested in case when n > 2m. For n > 2m it is enough to show that G_m,n(x, y)≃ |x−y|^2m−n, since (2.1.4) applies here.

According to [3], the Green function in a ball has the following form:

G_m,n,B1(x, y) = k_m,n|x−y|^2m−n

Z [XY]/|x−y|

1

(v²−1)^m−1v¹⁻ⁿdv.

Since |x−y| ≤ ¹₂[XY], for any a∈[2,∞) Z a

1

v²−1^m−1

v¹⁻ⁿdv≃ Z a

1

v^2m−n−1dv holds true. According to our assumption, the Green function is then

Gm,n(x, y)≃ |x−y|^2m−n

Z [XY]/|x−y|

1

v²−1m−1

v¹⁻ⁿdv

≃ |x−y|^2m−n

Z [XY]/|x−y|

1

v^2m−n−1dv.

Gm,n(x, y)≃ |x−y|^2m−n

Z [XY]/|x−y|

1

v^2m−n−1dv≃ |x−y|^2m−n.

Case 2: |x−y| ≥ ¹₂[XY].

According to (2.1.3), _|x−y|^d(x) ≤3,_|x−y|^d(y) ≤3. Thus, we have to show that G_m,n(x, y)≃ |x−y|⁻ⁿd(x)^md(y)^m.

Again, using Boggio’s formula [3] with the upper integration bound [XY]/|x−y| is in [1,2], since |x−y| ≥ ¹₂[XY]. On [1,2] we have v⁻ⁿ ≈ 1 and may conclude the following:

G_m,n(x, y)≃ |x−y|^2m−n

Z [XY]/|x−y|

1

v²−1m−1

vdv

≃ |x−y|^2m−n

[XY]²

|x−y|² −1 m

=|x−y|⁻ⁿ [XY]²− |x−y|²m

=|x−y|⁻ⁿ 1− |x|²

1− |y|²m

≃ |x−y|⁻ⁿd(x)^md(y)^m. We completed the proof for n > 2m.

The next theorem presents estimates of Green’s function from two sides:

Theorem 3. [30] Let Ω ⊂ Rⁿ be a bounded and C^2m,γ smooth domain with m ≥ 2 and n ≥ 2. Then there exist constants c₁ ≥ 0, c₂ > 0, c₃ > 0 that depend on the domain and m, such that the following estimate for the polyharmonic Green’s function G(−∆)^m,Ω is true:

c⁻¹₂ H_Ω(x, y)≤G−∆^m,Ω(x, y) +c₁1|x−y|≥c3(x, y)d(x)^md(y)^m ≤c₂H_Ω(x, y), where

H_Ω :=











|x−y|^2m−nmin{1,^d(x)_|x−y|^md(y)2m^m}, if n >2m, log(1 + ^d(x)_|x−y|^md(y)2m^m), if n= 2m

d(x)^m−n/2d(y)^m−n/2min{1,^d(x)_|x−y|^n/2d(y)n^n/2}, if n <2m and indicator function

1{|x−y|≥c₃} =

( 1 if |x−y| ≥c₃ 0 if |x−y|< c₃

Proof. Since we know the estimate of the Green function from Theorem 2, it is necessary to show that on Ω×Ω:

c1d(x)^md(y)^m ≤c2HΩ(x, y)

to complete the proof for the estimate from above. We are interested in case when n > 2m.

Let diam(Ω) =sup{|x−y|:x, y ∈Ω}.

Case 1: d(x)d(y)<|x−y|². We know that |x−y|< diam(Ω), then it follows

|x−y|^2m−n min{1,d(x)^md(y)^m

|x−y|^2m }= d(x)^md(y)^m

|x−y|ⁿ ≥ d(x)^md(y)^m (diam(Ω))ⁿ

Case 2: d(x)d(y)≥ |x−y|².

|x−y|^2m−n min{1,d(x)^md(y)^m

|x−y|^2m }=|x−y|^2m−n ≥(diam(Ω))^2m−n d(x)^md(y)^m (diam(Ω))^2m

≥ d(x)^md(y)^m (diam(Ω))ⁿ

The following theorem is an example of pointwise estimate for biharmonic Green function.

Theorem 4. [12] Let Ω ⊂ Rⁿ be a bounded and smooth domain, then the following estimate for the Green function is true:

G_∆²_,n(x, y)≤C(Ω)







|x−y|⁴⁻ⁿ+max{d(x), d(y)}⁴⁻ⁿ if n >4;

log(1 +|x−y|⁻¹+max{d(x), d(y)}⁻¹) if n = 4;

1 if n = 2,3,

Proof. For brevity, let G(x, y) = G_∆²_,Ω(x, y). The Green function G(x, y) = Fn(|x−y|) +h(x, y), where Fn(x) is the fundamental solution and represented by

Fn(x) =







cn|x|⁴⁻ⁿ if n >4;

−2c₄log|x| if n= 4;

2c₂|x|²log|x| if n= 2,3, where cn is defined by :

c_n =

( ₁

2(n−4)(n−2)ne_n if n /∈ {2,4}

1

8nen if n ∈ {2,4}, and h(x, y) solves the next Dirichlet problem:







∆²_yh(x, y) = 0

h(x, y) = −F_n(|x−y|)

∂

∂nyh(x, y) =−_∂n^∂

yF_n(|x−y|),

Case n >4:

∥h(x, .)∥_C^1,γ_{( ¯Ω)} ≤C(Ω)d(x)^3−n−γ. If d(x)≥d(y), thenh(x, y)≤C(∂Ω)d(x)⁴⁻ⁿ, hence,

|G(x, y)| ≤C(∂Ω)(|x−y|⁴⁻ⁿ+d(x)⁴⁻ⁿ).

For d(y)≥d(x), we use the fact of symmetry of the Green function:

|G(x, y)|=|G(y, x)| ≤C(∂Ω)(|x−y|⁴⁻ⁿ+d(y)⁴⁻ⁿ).

Gazzola et al. also provided estimates for derivatives of the Green function:

Theorem 5. [12] Letα ∈Nⁿ be a multiindex. Then inB¯×B¯ we have

|D^α_xG_m,n(x, y)| ⪯(∗) with (∗) as follows:

1. if |α| ≥2m−n and n odd, or if |α|>2m−n and n even

(∗) =







|x−y|^{2m−n−|α|}minn

1,^{d(x)m−|α||d(y)m}

|x−y|^2m−|α|

o

for|α|< m.

|x−y|^{2m−n−|α|}minn

1,_|x−y|^d(y)mm

o

for|α| ≥m;

2. if |α|= 2m−n and n even

(∗) =





 log

2 + _|x−y|^d(y) minn

1,^d(x)_|x−y|^m−|α|2m−|α|^d(y)m

o

for |α|< m, log

2 + _|x−y|^d(y) minn

1,_|x−y|^d(y)mm

o

for |α| ≥m 3. if |α| ≤2m−n and n odd, or if |α|<2m−n and n even

(∗) =











d(x)^{m−n2−|α|}d(y)^m−n2 minn 1,^d(x)

n2d(y)ⁿ²

|x−y|ⁿ

o

for |α|< m− ⁿ₂, d(y)^{2m−n−|α|}minn

1,^d(x)m−|α|_|x−y|^d(y)n^n−m+|α|

o

for m−ⁿ₂ ≤ |α|< m, d(y)^{2m−n−|α|}minn

1,_|x−y|^{d(y)n−m+|α|}n−m+|α|

o

for |α| ≥m.

Proof. We want to show that:

|D^α_xG_m,n(x, y)| ⪯ |x−y|^{2m−n−|α|}

d(x)

|x−y|

max{m−|α|,0}

d(y)

|x−y|

m

. Recall,

G_m,n,B1(x, y) = k_m,n|x−y|^2m−n

Z [XY]/|x−y|

1

(v²−1)^m−1v¹⁻ⁿdv.

Let’s use transformation s= 1− _v¹2 in the formula above. We have G_m,n(x, y) = k_m,n

2 |x−y|^2m−nf_m,n(A_x,y), where

f_m,n(t) :=Rt

0 s^m−1(1−s)^n/2−m−1ds A_x,y := ^[XY_[XY^]2−|x−y|_]2 ² = (^1−|x|2)(^1−|y|2)

[XY]² ≃ ^d(x)d(y)_|XY]2 .

Now, since Green’s function is the product of ^km,n₂ |x−y|^2m−n and f_m,n(A_x,y), let’s use product rule to find the derivative:

|D^α_xG_m,n(x, y)| ⪯X

β≤α

D_x^α−β

x− y|^2m−n| · |D^β_xf_m,n(A_x,y)|

Pulling out the case when β = 0 and applying chain rule forf_m,n(A_x,y), we get:

|D_x^αG_m,n(x, y)| ⪯ |x−y|^{2m−n−|α|}· |f_m,n(A_x,y)|

+X

β≤α β̸=0

|x−y|2m−n−|α|+|β|·

|β|

X

j=1















f_m,n^(j) (A_x,y)

· X

Pj

i=1β⁽ⁱ⁾=β

|^β(i)|^≥1

j

Y

i=1

D_x^β(i)A_x,y















Using the results (2.1.4), we have

A_x,y ≃ d(x)d(y) [XY]² ≤ 3

4.

Here, i.e for t∈ 0,³₄

, the following holds true:

f_m,n^(j) (t)

⪯t^max{m−j,0}

Since d(x)≤[XY], for every multiindexβ ∈Nⁿ one has D_x^βA_x,y

⪯d(y)[XY]^−1−|β|. Using facts above:

|D^α_xG_m,n(x, y)| ⪯ |x−y|^{2m−n−|α|}d(x)^md(y)^m [XY]^2m

+X

β<a β̸=0

|x−y|2m−n−|α|+|β|·

|β|

X

j=1

(

d(x)d(y) [XY]²

max{m−j,0}

· d(y)^j [XY]^j+|β|}

)

⪯X

β≤α

|x−y|^{2m−n−|α|}

|x−y|

|XY|

|β|

d(x) [XY]

max{m−|β|,0}

d(y) [XY |

m

⪯ |x−y|^{2m−n−|α|}

d(x) [XY]

max{m−|α|,0}

d(y) [XY]

m

The estimate is proven.

Krasovskii provided estimates in a general contest in [25]. The next theorem provides estimates without boundary terms.

Theorem 6. [12] Let Ω be a bounded C^2m,γ-smooth domain in Rⁿ with n ≥ 2 and G be a Green’s function in Ω for the Dirichlet boundary value problem. Then there exists a constant C =C(Ω), such that for all α, β ∈Nⁿ0 with |α|+|β| ≤2m : - If |α|+|β|+n >2m:

D_x^αD_y^βG(x, y)

≤C(Ω)|x−y|2m−n−|α|−|β|

for all x, y ∈Ω - If |α|+|β|+n = 2m and n is even:

D^α_xD^β_yG(x, y)

≤C(Ω) log 1 +|x−y|⁻¹

for all x, y ∈Ω.

- If |α|+|β|+n = 2m and n is odd, or if|α|+|β|+n <2m : D_x^αD_y^βC(x, y)

≤C(Ω) for all x, y ∈Ω.

Proof. We are interested in case whenn >2m.

In Theorem 2 we showed that

|G(x, y)| ≤C|x−y|^2m−n.

Let x ∈ Ω, y ∈ Ω\{x} and R = ^|x−y|₄ . First, let’s consider the case when |α| = 0.

Then in B_R(y)⊂B_2R(y) the following can be derived:

D^β_yG(x,·)

L^∞(BR(y)∩Ω) ≤ C

|x−y|^|β|∥G(x, .)∥_L^∞_(B2R(y)∩Ω)

≤ C

|x−y|^|β|∥|x− ·|^2m−n∥_L^∞_(B2R(y)∩Ω)

≤C|x−y|^{2m−n−|β|}

where we used for z ∈B2R(y) that

|x−z| ≥ |x−y| − |y−z|> 1

2|x−y|.

Since the Green function is symmetric, the same can be proved for|β|= 0 and|α|>0 in a similar way. Moreover, since y 7→ D^α_xG(x, y) solves the homogeneous Dirichlet boundary value problem, we can proceed as before for the mixed derivatives.

Next, let us present the following estimates proved by Maz’ya and Mayboroda:

Theorem 7. [28] For every x, y in a bounded domain Ω, there exist constants C, C^′ that depend only on m, n, such that the following estimates hold:

- if n ∈ {3,2m+ 1} is odd ∇ⁱ_x∇^j_yG(x, y)

≤Cmin (

1,

d(x)

|x−y|

m−ⁿ₂+¹₂−i

,

d(y)

|x−y|

m−ⁿ₂+¹₂−j)

× 1

|x−y|^n−2m+i+j,

whenever 0≤i, j ≤m−ⁿ₂ +¹₂ are such that i+j ≥2m−n, and

| ∇ⁱ₂∇^j_yG(x, y)≤Cmin (

1,

d(x)

|x−y|

m−⁵₂+¹₂−i

,

d(y)

|x−y|

m−⁴₂+¹₂−j)

×

× 1

|x−y|^n−2m+i+j min

|x−y|

d(x) ,|x−y|

d(y) ,1

^n−2m+i+j

if 0≤i, j ≤m−ⁿ₂ +¹₂ are such that i+j ≤2m−n.

-Ifn ∈[2,2m]∩N is even, then ∇ⁱ_x∇^j_yG(x, y)

≤Cmin

1,

d(x)

|x−y|

m−ⁿ

2−i

,

d(y)

|x−y|

m−ⁿ

2−j

×

×_|x−y|n−2m+i+j¹ minn_|x−y|

d(x),^|x−y|_d(y) ,1on−2m+i+j

×

×log

1 + min{d(x),d(y)}

|x−y|

for all 0≤i, j ≤m−ⁿ₂.

Dall’Acqua and Sweers [7] presented the next estimates:

Theorem 8. Let G_m(x, y) be the Green function. Let k ∈ Nⁿ. The following estimates hold for every x, y ∈Q:

(1) For |k|⩾m :

(a) if 2m−n− |k|<0, then

|D_x^kG_m(x, y)| ⪯ |x−y|^2m−n−|k|min

1, d(y)

|x−y|

m

,

(b) if 2m−n− |k|= 0, then D_x^kG_m(x, y)

⪯log

1 + d(y)^m

|x−y|^m

∼log

2 + d(y)

|x−y|

min

1, d(y)

|x−y|

m

, (c) if 2m−n− |k|>0, then

|D_x^kG_m(x, y)| ⪯d(y)^2m−n−|k|min

1, d(y)

|x−y|

n+|k|−m

, (2) For |k|< m :

(a) if 2m−n− |k|<0, then D^k_xG_m(x, y)

⪯ |x−y|^2m−n−|k|min

1, d(x)

|x−y|

m−|k|

min

1, d(y)

|x−y|

m

, (b) if 2m−n− |k|= 0, then

D^k_xG_m(x, y) ⪯log

1 + d(y)^md(x)^m−|k|

|x−y|^2m−|k|

∼log

2 + d(y)

|x−y|

min

1, d(y)

|x−y|

m

min

1, d(x)

|x−y|

m−|k|

(c) if 2m−n− |k|>0, and moreover (i) m− ¹₂n ⩽|k|, then

|D_x^kGm(x, y)⪯d(y)^2m−n−|k|min

1, d(x)

|x−y|

m−|k|

×min

1, d(y)

|x−y|

n−m+|k|

, (ii) |k|< m− ¹₂n, then

|D_x^kG_m(x, y)⪯d(y)^m−n/2d(x)^{m−n/2−|k|}min

1,d(x)d(y)

|x−y|^{2 n}₂

Proof. We refer to [8] and provide sketch of the proof.

Dall’Acqua and Sweers started from estimates of Krasovskii [25] for higher-order derivatives of Green’s function. To obtain lower-ordered estimates for derivatives, authors integrated higher-ordered estimates along some path γ_x^y. For α, β ∈ Nⁿ, and

˜

x∈∂Ω they found:

D^α_xD^β_yG_k(x, y) =D^α_xD^β_yG_m(˜x, y) + Z

γ^yx

∇_zD_z^αD_y^βG_m(z, y)dz.

If |α| ≤m−1, then

|D^α_xD^β_yG_k(x, y)|= Z l

0

|∇_xD_x^αD_y^βG_m(˜γ_x^y(s), y)|ds If |β| ≤m−1,integrating with respect to y:

|D^α_xD^β_yG_k(x, y)|= Z l

0

|∇_yD_y^βD^α_xG_m(˜γ_x^y(s), y)|ds After that the authors considered two cases when r ≥m and r < m.

Case r≥m: According to [25],|D_x^αD_y^βG_k(x, y)| ≤ |x−y|^m−n−r with β =m−1.

Case 1 (a) of the Theorem was proven then by applying mtimes the estimate proven by the authors earlier :

|H(x, y)| ⪯ |x−y|^−k+1min{1, dx

|x−y|}^ν1+1min{1, dy

|x−y|}^ν2, where ν1, ν2, k∈N.

Case 1 (b) was proven by applying the same estimate as in 1(a), m−1 times. Case 1(c) was proven by first applying estimate as above n+r−m−1 times. And after that the authors suggested to use the following estimate:

|H(x, y)| ⪯log

2 + ^d(x)d(y)_|x−y|2

minn

1,_|x−y|^d(x)oν1+1

minn

1,_|x−y|^d(y) oν2

, once they found

|D^β_y^˜D_x^kG_m(x, y)| ⪯log(1 + d(y)^n+r−m

|x−y|^n+r−m),

where ˜β ∈Nⁿ,β˜≤β and |β|˜ = 2m−n−r. Then, the estimate

|H(x, y)| ⪯d(x) minn

1,_|x−y|^d(x)oν1

minn

1,_|x−y|^d(y)oν2

was applied once and

|H(x, y)| ⪯ d(x)^α1+1d(y)^α2min n

1,_|x−y|^d(x) oν1+1

min n

1,_|x−y|^d(y) oν2

was applied 2m−n − r−1 times for someα1, α2 ∈N.

Case 2: r < m, |α|=m−r,|β|=m. Again, starting from Krasovskii’s estimate for

|D_y^βD^α_xD^k_xG_m(x, y)|, the authors integrated the estimate m times with respect to y and m−r times with respect to x.

Case 2(a) was proven by applying the estimate

|H(x, y)| ⪯ |x−y|^−k+1min{1, dx

|x−y|}^ν1+1min{1, dy

|x−y|}^ν2 m times with respect toy and m−r times with respect tox.

Case 2(b) was proven by applying the same estimate as in case 2(a) m times with respect toyandm−r−1 times with respect toxand applying the estimate|H(x, y)| ⪯ log

2 + ^d(x)d(y)_|x−y|2

minn

1,_|x−y|^d(x)oν1+1

minn

1,_|x−y|^d(y)oν2

with respect to x.

Case 2(c), for m−r ≤ n −1 the statement was proven by applying the following estimates:

|H(x, y)| ⪯ |x−y|^−k+1min{1, dx

|x−y|}^ν1+1min{1, dy

|x−y|}^ν2 n−1 times,

|H(x, y)| ⪯log

2 + d(x)d(y)

|x−y|²

min

1, d(x)

|x−y|

ν1+1

min

1, d(y)

|x−y|

ν2

one time,

|H(x, y)| ⪯d(x) min

1, d(x)

|x−y|

ν1

min

1, d(y)

|x−y|

ν2

and

|H(x, y)| ⪯d(x)^α1+1d(y)^α2min

1, d(x)

|x−y|

ν1+1

min

1, d(y)

|x−y|

ν2

2m−n−r−1 times.

Case 2(c), when m−r > n−1, was proven by applying the following estimates:

|H(x, y)| ⪯ |x−y|^−k+1min{1, dx

|x−y|}^ν1+1min{1, dy

|x−y|}^ν2 n−1 times,

|H(x, y)| ⪯log

2 + d(x)d(y)

|x−y|²

min

1, d(x)

|x−y|

ν1+1

min

1, d(y)

|x−y|

ν2

one time, and

|H(x, y)| ⪯d(x)^α1+1d(y)^α2min

1, d(x)

|x−y|

ν1+1

min

1, d(y)

|x−y|

ν2

m times with respect toy and m−r−n with respect to x.

Chapter 3 Main result

3.1 Mazya’s problem in a ball

The next problem is taken from collection of open problems published in the article

”Seventy Five (Thousand) Unsolved Problems in Analysis and Partial Differential Equations” by Vladimir Maz’ya. The statement of the problem is the following:

Let Ω ⊂Rⁿ be an arbitrary domain. Green’s function is a solution of

(−∆_x)^mG_m,n(x, y) = δ(x−y), x, y ∈Ω, (3.1.1) with the homogeneous Dirichlet boundary conditions:

∂ⁱ

∂nⁱ_xG_m,n|x∈∂Ω = 0, i= 0, ..., m−1, (3.1.2) where δ is the Dirac delta function.

Letn >2m. Prove or disprove that

|G_m,n,r(x, y)| ≤ d(m, n)

|x−y|^n−2m, x, y ∈Ω, x̸=y, where d(m, n) does not depend on Ω.