# APPLICATIONS OF THE POINCARÉ INEQUALITY TO EXTENDED KANTOROVICH METHOD

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### APPLICATIONS OF THE POINCARÉ INEQUALITYTO EXTENDED KANTOROVICH METHOD

DER-CHEN CHANG, TRISTAN NGUYEN, GANG WANG, AND NORMAN M. WERELEY

Received 3 February 2005; Revised 2 March 2005; Accepted 18 April 2005

We apply the Poincar´e inequality to study the extended Kantorovich method that was used to construct a closed-form solution for two coupled partial diﬀerential equations with mixed boundary conditions.

Copyright © 2006 Der-Chen Chang et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

1. Introduction

LetΩRnbe a Lipschitz domain inRn. Consider the Dirichlet spaceH01(Ω) which is the collection of all functions in the Sobolev spaceL21(Ω) such that

H01(Ω)=

uL2(Ω) :u|∂Ω=0,uL2+ n k=1

∂u

∂xk

L2

<

. (1.1)

The famous Poincar´e inequality can be stated as follows: foruH01(Ω), then there exists a universal constantCsuch that

Ωu2(x)dxC n k=1

Ω

∂u

∂xk

2dx. (1.2)

One of the applications of this inequality is to solve the modified version of the Dirichlet problem (see, John [5, page 97]): find avH01(Ω) such that

(u,v)=

Ω

n

k=1

∂u

∂xk

∂v

∂xk dx=

Ωu(x)f(x)dx, (1.3)

Hindawi Publishing Corporation Journal of Inequalities and Applications Volume 2006, Article ID 32356, Pages1–21 DOI10.1155/JIA/2006/32356

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where x=(x1,. . .,xn) with a fixed f C( ¯Ω). Then the functionvin (1.3) satisfied the boundary value problem

Δv= −f, inΩ

v=0, on∂Ω. (1.4)

In this paper, we will use the Poincar´e inequality to study the extended Kantorovich method, see . This method has been used extensively in many engineering problems, for example, readers can consult papers [4,7,8,11,12], and the references therein. Let us start with a model problem, see . For a clamped rectangular boxΩ=n

k=1[ak,ak], subjected to a lateral distributed load,ᏼ(x)=ᏼ(x1,. . .,xn), the principle of virtual dis- placements yields

n =1

a

a

η4ΦδΦDx=0, (1.5)

whereΦis the lateral deflection which satisfies the boundary conditions,ηis the flexural rigidity of the box, and

4= n k=1

4

∂x4k+

j=k

2 4

∂x2j∂x2k. (1.6)

Since the domainΩis a rectangular box, it is natural to assume the deflection in the form Φ(x)=Φk1···kn(x)=

n =1

fkx, (1.7)

it follows that when fk2(x2)···fkn(xn) is prescribed a priori, (1.5) can be rewritten as a1

a1

n

=2

a

a

η4Φk1···knfkxdx δ fk1(x1)dx1=0. (1.8)

Equation (1.8) is satisfied when n =2

a

a

η4Φk1···knfkxdx=0. (1.9)

Similarly, whenn=1,=mfk(x) is prescribed a priori, (1.5) can be rewritten as am

am

n

=1,=m

a

a

η4Φk1···knfk

x

dx δ fkm(xm)dxm=0. (1.10)

It is satisfied when

n =1,=m

a

a

η4Φk1···knfkxdx=0. (1.11)

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It is known that (1.9) and (1.11) are called the Galerkin equations of the extended Kan- torovich method. Now we may first choose

f20(x2)···fn0(xn)= n =2

c

x2 a2 1

2

. (1.12)

ThenΦ10···0(x)=f11(x1)f20(x2)···fn0(xn) satisfies the boundary conditions Φ10···0=0, ∂Φ10···0

∂x =0 atx= ±a,x1

a1,a1

, (1.13)

for=2,. . .,n. Now (1.9) becomes n

=2

c a

a

4Φ10···0η

x2 a2 1

2

dx=0, (1.14)

which yields

C4

d4f11

dx4 +C2

d2f11

dx2 +C0f11=B. (1.15)

After solving the above ODE, we can use f11(x1)n=3f0(x) as a priori data and plug it into (1.10) to find f21(x2). Then we obtain the function

Φ110···0(x)= f11

x1

f21

x2

f30

x3

···fn0 xn

. (1.16)

Continue this process until we obtainΦ1···1(x)= f11(x1)f21(x2)···fn1(xn) and there- fore completes the first cycle. Next, we use f21(x2)···fn1(xn) as our priori data and find f12(x1). We continue this process and expect to find a sequence of “approximate solu- tions.” The problem reduces to investigate the convergence of this sequence. Therefore, it is crucial to analyze (1.15). Moreover, from numerical point of view, we know that this sequence converges rapidly (see [1,2]). Hence, it is necessary to give a rigorous mathe- matical proof of this method.

2. A convex linear functional onH02(Ω) Denote

I[φ]=

Ω

|Δφ|22ᏼ(x)φ(x)dx (2.1)

forΩRna bounded Lipschitz domain. Here x=(x1,. . .,xn). As usual, denote

D2φ=

2φ

∂x2

2φ

∂x ∂y

2φ

∂y ∂x

2φ

∂y2

. (2.2)

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ForΩR2, we define the Lagrangian functionLassociated toI[φ] as follows:

L×R×R2×R4−→R,

(x,y;z;X,Y;U,V,S,W)−→(U+V)22ᏼ(x,y)z, (2.3) whereᏼ(x,y) is a fixed function onΩwhich shows up in the integrand ofI[φ]. With the above definitions, we have

Lx,y;φ;φ;D2φ= |Δφ|22ᏼ(x,y)φ(x,y), (2.4) where we have identified

z←→φ(x,y), X←→∂φ

∂x, Y←→∂φ

∂y, U←→2φ

∂x2, V←→2φ

∂y2, S←→ 2φ

∂y ∂x, W←→ 2φ

∂x ∂y.

(2.5)

We also setH02(Ω) to be the class of all square integrable functions such that H02(Ω)=

ψL2(Ω) :

|k|≤2

kψ

∂xk

L2

<,ψ|∂Ω=0, ψ|∂Ω=0

. (2.6)

Fix (x,y)Ω. We know that

L(x,y;z;X,Y;U,V,S,W)=

2ᏼ(x,y) 0 0 2(U+V) 2(U+V) 0 0T. (2.7) Because the convexity of the functionLin the remaining variables, then for all (z;X, Y;U, V,S,W) R×R2×R4, one has

L x,y;z;X, Y;U,V,S,W!L(x,y;z;X,Y;U,V,S,W)2ᏼ(x,y)zz

+ 2(U+V)UU+ VV. (2.8) In particular, one has, withz=φ(x, y),

L x,y;φ;φ;D 2φ!Lx,y;φ;φ;D2φ+ 2Δφφ− ∇φ

2ᏼ(x,y)(φφ). (2.9)

This implies that

Δφ22ᏼ(x,y)φΔφ22ᏼ(x,y)φ+ 2ΔφΔφΔφ2ᏼ(x,y) φφ. (2.10) If instead we fix (x,y;z)Ω×R, then

L x,y;z;X, Y;U, V,S,W!L(x,y;z;X,Y ;U,V,S,W)

+ 2(U+V) UU+ VV. (2.11)

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This implies that

Lx,y;φ;φ;D 2φL(x,y;φ;φ;D2φ+ 2Δφ[φ− ∇φ] (2.12) Therefore,

|Δφ|22ᏼ(x,y)φ≥ |Δφ|22ᏼ(x,y)φ+ 2Δφ[ΔφΔφ]. (2.13) Lemma 2.1. Suppose either

(1)φH02(Ω)C4(Ω) andηCc1(Ω); or (2)φH02(Ω)C3( ¯Ω)C4) andηH02).

LetδI[φ;η] denote the first variation ofIatφin the directionη, that is,

δI[φ;η]=lim

ε0

I[φ+εη]I[φ]

ε . (2.14)

Then

δI[φ;η]=2

Ω

Δ2φᏼ(x,y)η dx d y. (2.15) Proof. We know that

I[φ+εη]I[φ]=

Ω[ΔφΔηη]dx d y+ε2

Ω(Δη)2dx d y. (2.16) Hence,

εI[φ;η]=2

Ω[ΔφΔηη]dx d y. (2.17) If either assumption (1) or (2) holds, we can apply Green’s formula to a Lipschitz domain Ωto obtain

Ω(ΔφΔη)dx d y=

ΩηΔ2φdx d y+

∂Ω

"∂η

∂nΔφη

∂nΔφ#dx d y, (2.18) where∂/∂n is the derivative in the direction normal to∂Ω. Since eitherηCc1(Ω) or ηH02(Ω), the boundary term vanishes, which proves the lemma.

Lemma 2.2. LetφH02(Ω). Then

φH02(Ω)ΔφL2(Ω). (2.19) Proof. The functionφH02(Ω) implies that there exists a sequence{φk} ⊂Cc (Ω) such that limk→∞φk=φinH02-norm. From a well-known result for the Calder ´on-Zygmund operator (see, Stein [10, page 77]), one has

2f

∂xj∂x

LpCΔ fLp, j,=1,. . .,n (2.20)

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for all f Cc2(Rn) and 1< p <. HereCis a constant that depends onnonly. Applying this result to eachφk, we obtain

2φk

∂x2

L2(Ω),2φk

∂x ∂y

L2(Ω),2φk

∂y2

L2(Ω)CΔφk

L2(Ω). (2.21) Taking the limit, we conclude that

2φ

∂x2

L2(Ω), 2φ

∂x ∂y

L2(Ω),2φ

∂y2

L2(Ω)CΔφL2(Ω). (2.22) Applying Poincar´e inequality twice to the functionφH02(Ω), we have

φL2(Ω)C1φL2(Ω)

C2

2φ

∂x2

L2(Ω)+ 2φ

∂x ∂y

L2(Ω)+2φ

∂y2

L2(Ω)

CΔφL2(Ω).

(2.23)

Hence,φL2(Ω)CΔφL2(Ω). The reverse inequality is trivial. The proof of this lemma

is therefore complete.

Lemma 2.3. Let{φk}be a bounded sequence inH02(Ω). Then there existφH02(Ω) and a subsequence{φkj}such that

I[φ]lim infIφkj. (2.24)

Proof. By a weak compactness theorem for reflexive Banach spaces, and hence for Hilbert spaces, there exist a subsequence{φkj}of{φk}andφinH02(Ω) such thatφkjφweakly inH02(Ω). Since

H02(Ω)H01(Ω)⊂⊂L2(Ω), (2.25) by the Sobolev embedding theorem, we have

φkj−→φ inL2(Ω) (2.26)

after passing to yet another subsequence if necessary.

Now fix (x,y,φkj(x,y))R2×Rand apply inequality (2.13), we have

Δφkj22ᏼ(x,y)φkj(x,y)≥ |Δφ|22ᏼ(x,y)φkj(x,y) + 2ΔφΔφkjΔφ. (2.27) This implies that

Iφkj

Ω

|Δφ|22ᏼ(x,y)φkjdx d y+ 2

ΩΔφ·

ΔφkjΔφdx d y. (2.28) ButφkjφinL2(Ω), hence

Ω

|Δφ|22ᏼ(x,y)φkjdx d y−→

Ω

|Δφ|22ᏼ(x,y)φdx d y=I[φ]. (2.29)

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Besidesφkjφweakly inH02(Ω) implies that

ΩΔφ·

ΔφkjΔφdx d y−→0. (2.30) It follows that when taking limit

I[φ]lim inf

j Iφkj

. (2.31)

This completes the proof of the lemma.

Remark 2.4. The above proof uses the convexity ofL(x,y;z;X,Y;U,V,S,W) when (x,y;

z) is fixed. We already remarked at the beginning of this section that when (x,y) is fixed, L(x,y;z;X,Y;U,V,S,W) is convex in the remaining variables, including thez-variable.

That is, we are not required to utilize the full strength of the convexity ofLhere.

3. The extended Kantorovich method

Now, we shift our focus to the extended Kantorovich method for finding an approximate solution to the minimization problem

min

φH02(Ω)I[φ] (3.1)

when Ω=[a,a]×[b,b] is a rectangular region in R2. In the sequel, we will write φ(x,y) (resp.,φk(x,y)) asf(x)g(y) (resp., fk(x)gk(y)) interchangeably as notated in Kerr and Alexander . More specifically, we will study the extended Kantorovich method for the casen=2, which has been used extensively in the analysis of stress on rectangular plates. Equivalently, we will seek for an approximate solution of the above minimization problem in the formφ(x,y)= f(x)g(y) where f H02([a,a]) andgH02([b,b]).

To phrase this diﬀerently, we will search for an approximate solution in the tensor product Hilbert spacesH02([a,a])\$H02([b,b]), and all sequences{φk},{φkj}involved hereinafter reside in this Hilbert space. Without loss of generality, we may assume that Ω=[1, 1]×[1, 1] for all subsequent results remain valid for the general case where Ω=[a,a]×[b,b] by approximate scalings/normalizing of thexandyvariables. As in , we will treat the special caseᏼ(x,y)=γ, that is, we assume that the loadᏼ(x,y) is distributed equally on a given rectangular plate.

To start the extended Kantorovich scheme, we first choose g0(y)H02([1, 1]) Cc (1, 1), and find the minimizerf1(x)H02([1, 1]) of the functional:

If g0

=

Ω

Δf g022γ f(x)g0(y)dx d y

=

Ω

g02(f)2+ 2f fg0g0+f2g022γ f g0

dx d y

= 1

1(f)2dx 1

1g02d y+ 2 1

1

g02d y 1

1(f)2dx +

1

1

g02d y 1

1f2dx1

1g0d y 1

1f dx,

(3.2)

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where the last equality was obtained via the integration by parts of f fandg0g0. Since g0has been chosen a priori; we can rewrite the functionalIas

J[f]=g02

L2

1

1(f)2dx+ 2g02

L2

1

1(f)2dx +g02L2

1

1f2dx1

1g0(y)d y 1

1f dx

(3.3)

for all f H02([1, 1]). Now we may rewrite (3.3) in the following form:

J[f]= 1

1

C1(f)2+C2(f)2+C3f2+C4fdx

1

1K(x,f,f,f)dx

(3.4)

withK:R×R×R×RRgiven by

(x;z;V;W)−→C1W2+C2V2+C3z2+C4z, (3.5) where

C1=g02

L2, C2=g02L2, C3=g02L2, C4= −1

1g0(y)d y. (3.6) As long asg00, as we have implicitly assumed, the Poincar´e inequality implies that

0< C1αC2βC3 (3.7)

for some positive constantsαandβ, independent ofg0. Consequently,K(x;z;V;W) is a strictly convex function in variablez,V,Wwhenxis fixed. In other words,Ksatisfies

K(x;z;V;W) K(x;z;V;W)

∂K

∂z(x;z;V;W)(zz) +∂K

∂V(x;z;V;W)(VV) + ∂K

∂W(x;z;V;W)(WW) (3.8) for all (x;z;V;W) and (x;z;V;W) in R4, and the inequality becomes equality at (x;z;V; W) only ifz=z, orV=V, orW=W.

Proposition 3.1. Letᏸ:R×R×R×RRbe a C function satisfying the following convexity condition:

ᏸ(x;z+z;V+V;W+W)ᏸ(x;z;V;W)

∂ᏸ

∂z(x;z;V;W)z+∂ᏸ

∂V(x;z;V;W)V+ ∂ᏸ

∂W(x;z;V;W)W (3.9)

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for all (x;z;V;W) and (x;z+z;V+V;W+W)R4, with equality at (x;z;V;W) only ifz=0, orV=0, orW=0. Also, let

J[f]= β

αx,f(x),f(x),f(x)dx, f H02(α,β). (3.10) Then

J[f +η]J[f]δJ[f,η], ηCc (α,β) (3.11) and equality holds only ifη0. HereδJ[f,η] is the first variation ofJatf in the directionη.

Proof. Condition (3.9) means that at eachx, ᏸ(x;f+η;f+η;f+η)ᏸ(x;f;f;f)

∂z(x;f;f;f)η(x) +

∂V(x;f;f;f(x) +

∂W(x;f;f;f(x) (3.12) for allηCc (α,β) with equality only ifη(x)=0, orη(x)=0, orη(x)=0. Equivalently, the equality holds in (3.12) atxonly ifη(x)η(x)=0 orη(x)=0. In other words,

η(x) d dx

η2(x)=0. (3.13)

Integrating (3.12) gives J[f+η]J[f]

β

α

"∂ᏸ

∂zη+∂ᏸ

∂Vη+ ∂ᏸ

∂Wη

#

dx=δJ[f,η]. (3.14) Now suppose there existsηCc(α,β) such that (3.14) is an equality. Sinceᏸis a smooth function, this equality forces (3.12) to be a pointwise equality, which implies, in view of (3.13), that

η(x)d dx

η2(x)=0, x. (3.15)

Ifη(x)0, thenη(x)=constant which implies thatη(x)0 (since ηCc (α,β)).

This tells us thatηconstant and conclude thatη0 on the interval (α,β).

Ifη(x)0, setU= {x(α,β) :η(x)=0}. ThenUis a non-empty open set which implies that there existx0Uand some open setᏻx0ofx0contained inU. Thenη(ξ)= 0 for allξx0U. Thus

d dx

η2=0 onᏻx0. (3.16)

Hence,η(ξ)constant onᏻx0. But this creates a contradiction becauseη(ξ)0 onᏻx0. Therefore,

J[f+η]J[f]=δJ[f,η] (3.17) only ifη(x)0, as desired. This completes the proof of the proposition.

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Corollary 3.2. LetJ[f] be as in (3.4). Then f1H02([1, 1]) is the unique minimizer for J[f] if and only if f1solves the following ODE:

g02

L2

d4f

dx4 2g02L2

d2f

dx2 +g02L2f =γ 1

1g0d y. (3.18) Proof. Suppose f1 is the unique minimizer. Then f1 is a local extremum ofJ[f]. This implies thatδJ[f,η]=0 for allηH02([1, 1]). Using the notations in (3.4), we have

0=δJ[f,η]

= 1

1

"∂K

∂zη+∂K

∂Vη+ ∂K

∂Wη

# dx

= 1

1

"∂K

∂z d dx

∂K

∂V

+ d2 dx2

∂K

∂W

# η(x)dx

(3.19)

for allηH02([1, 1]). This implies that

∂K

∂z d dx

∂K

∂V

+ d2 dx2

∂K

∂W

=0, (3.20)

which is the Euler-Lagrange equation (3.18). This also follows fromLemma 2.1directly.

Conversely, assume f1solves (3.18). Then the above argument shows thatδJ[f,η]=0 for allηH02([1, 1]). SinceKsatisfies condition (3.9) inProposition 3.1, we conclude that

Jf1+ηJf1

δJf1, ηCc [1, 1]. (3.21) This tells us thatJ[f1+η]J[f1] for allηCc([1, 1]) andJ[f1+η]> J[f1] ifη0.

Observe thatJ:H01([1, 1])Ras given in (3.4) is a continuous linear functional in theH02-norm. This fact, combined with the density ofCc ([1, 1]) inH02([1, 1]) (in the H02-norm), implies that

Jf1+ηJf1

, ηCc [1, 1]. (3.22) This means that for allϕH02([1, 1]), we haveJ[ϕ]J[f1] and ifϕ f1(almost ev- erywhere), then ϕf10 and hence, J[ϕ]> J[f1]. Thus f1 is the unique minimum

forJ.

Reversing the roles of f andg, that is, fixing f0and findingg1H02to minimizeI[f0g]

overgH02([1, 1]), we obtain the same conclusion by using the same arguments.

Corollary 3.3. Fix f0H02([1, 1]). Theng1H02([1, 1]) is the unique minimizer for J[g]=If0g=f02

L2

1

1(g)2d y+ 2f02L2

1

1(g)2d y +f02L2

1

1g2d yf0

L1

1

1g d y

(3.23)

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if and only ifg1solves the Euler-Lagrange equation f02

L2

d4g

d y42f02L2

d2g

d y2+f02L2g=1

1f0(x)dx. (3.24) Now we search for the solutionf1H02([1, 1]) in (3.18), that is,

g02

L2

d4f

dx4 2g02L2

d2f

dx2 +g02L2f =1

1g0(y)d y. (3.25) Rewrite the above ODE in the following form:

g02

L2

Dg2L2 g02

L2

2

+g2L2 g02

L2

g4L2 g04

L2

f =1

1g0(y)d y, (3.26) whereD=d2/dx2.

Remark 3.4. In general whengH2, that is,gneeds not satisfy the zero boundary con- ditions for function inH02, then the quantity

g02L2

g02

L2

g04L2

g04

L2

(3.27) can take on any values. However, ifgH02andg00, as proved below, this quantity is always positive.

Lemma 3.5. LetΩbe a Lipschitz domain inRn,n1. LetgH02(Ω) be arbitrary. Then

g2L2gL2· ΔgL2, (3.28) and equality holds if and only ifg0.

Proof. Integration by parts yields

g2L2=

Ωg· ∇g dx= −

ΩgΔg dx +

∂Ωg∂g

∂ndσ= −

ΩgΔg dx. (3.29) By the Cauchy-Schwartz inequality, we have

g2L2gL2· ΔgL2, (3.30) and the equality holds if and only if (see Lieb-Loss )

(i)|g(x)| =λ|Δg(x)|almost everywhere for someλ >0, (ii)g(x)Δg(x)=e|g(x)| · |Δg(x)|.

Sincegis real-valued, (i) and (ii) imply

g(x)Δg(x)=λΔg(x)2. (3.31)

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So,gmust satisfy the following PDE:

Δg1

λg=0, (3.32)

wheregH02(Ω). But the only solution to this PDE isg0 (see, Evans [3, pages 300–

302]). This completes the proof of the lemma.

Remark 3.6. Ifn=1, one can solvegλ1g=0 directly without having to appeal to the theory of elliptic PDEs.

Proposition 3.7. The solutions of (3.18) and (3.24) have the same form.

Proof. Using eitherLemma 3.5in casen=1 to the above remark, we see that g2L2

g02

L2

g4L2 g04

L2

>0 ifg00. (3.33)

Hence the characteristic polynomial associated to (3.26) has two pairs of complex con- jugate roots as long asg00. Apply the same arguments to the ODE in (3.24) and the

proposition is proved.

Remark 3.8. The statement inProposition 3.7was claimed in  without verification.

Indeed the authors stated therein that the solutions of (3.18) and (3.24) are of the same form because of the positivity of the coeﬃcients on the left-hand side of (3.18) and (3.24).

As observed inRemark 3.4and proved inProposition 3.7, the positivity requirement is not suﬃcient. The fact that f0,g0H02must be used to conclude this assumption.

4. Explicit solution for (3.26)

We now find the explicit solution for (3.26), and hence for (3.18). Let r=gL2

g0

L2

, t=gL2

g0

L2

, ρ=

% t+r2

2 , κ=

% tr2

2 .

(4.1)

Then from Proposition 3.7and its proof, the 4 roots of the characteristic polynomial associated to ODE (3.26) are

ρ+iκ, ρiκ, ρiκ, ρ+iκ. (4.2)

Thus the homogeneous solution of (3.26) is

fh(x)=c1cosh(ρx) cos(κx) +c2sinh(ρx) cos(κx)

+c3cosh(ρx) sin(κx) +c4sinh(ρx) sin(κx). (4.3)

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