An iterative method based on a modified Lagrangian functional for finding a saddle point in the...

23

ISSN 0965-5425, Computational Mathematics and Mathematical Physics, 2006, Vol. 46, No. 1, pp. 23–33. ©

MAIK “Nauka

/Interperiodica” (Russia), 2006.Original Russian Text © G. Woo, R.V. Namm, S.A. Sachkoff, 2006, published in Zhurnal Vychislitel’noi Matematiki i Matematicheskoi Fiziki, 2006, Vol. 46, No. 1, pp.26–36.

An Iterative Method Based on a Modified Lagrangian Functional for Finding a Saddle Point in the Semicoercive

Signorini Problem

G. Woo*, R. V. Namm**, and S. A. Sachkoff**

* Department of Applied Mathematics, Changwon National University, Changwon, 641-773 South Korea** Pacific Ocean State University,

ul. Tikhookeanskaya 136, Khabarovsk, 680035 Russiae-mail: [email protected]

Received March 23, 2005; in final form, August 31, 2005

Abstract

—The iterative Uzawa method with a modified Lagrangian functional is examined in theframework of the Signorini problem.

DOI:

10.1134/S0965542506010052

Keywords:

Signorini problem, Lagrangian functional, saddle point, Gäteaux derivative, Uzawa method.

1. INTRODUCTION

The construction and the analysis of methods for solving elliptic variational inequalities in mechanicsare usually based on the assumption that the functionals to be minimized are strictly coercive (strongly con-vex). However, for a number of practically important variational inequalities, strict coercitivity holds onlyon finite-codimensional subspaces of the original Hilbert space.

A special analysis is required to ensure the numerical stability of methods for the approximate solutionof such semicoercive variational inequalities (see [1–3]).

The use of the classical Lagrangian functional in semicoercive problems does not guarantee the conver-gence of the available methods for finding saddle points (see [4, Theorem 4.1, p. 108, 5, p. 227]. In thispaper, we study the iterative Uzawa method based on a modification of the classical Lagrangian functionalin the framework of the semicoercive Signorini problem. We show that this method converges to a saddlepoint if the appropriate assumptions on the regularity of the solutions to the auxiliary problems are fulfilled.

2. THE UZAWA METHOD WITH A MODIFIED LAGRANGIAN FUNCTIONAL

Consider the following minimization problem (see [6, 7]):

(1)

Here,

Ω ⊂

n

(

n

= 2, 3)

is a bounded domain with a sufficiently smooth boundary

Γ

,

f

∈

L

2

(

Ω

)

is a prescribed

function, and

γ

v

∈

(

Γ

)

is the trace of

v

∈

(

Ω

)

on

Γ

.

Since the functional

J

(

v

)

is not strictly coercive on

Γ

, problem (1) may have no solution. However, if

(2)

then

J

(

v

) +

∞

as

∞

(

v

∈

G

) and, therefore, the problem is solvable (see [6, 7]). Moreover,

in view of (2), the solution is unique. In what follows, we assume that condition (2) is fulfilled.

J v( ) 12--- ∇v 2 Ω fv Ω min,d

Ω∫–d

Ω∫=

v G∈ w W21 Ω( ); γ w 0 almost everywhere in Γ≥∈ .=

W21/2 W2

1

f Ω 0,<d

Ω∫

vW2

1 Ω( )

24

COMPUTATIONAL MATHEMATICS AND MATHEMATICAL PHYSICS

Vol. 46

No. 1

2006

WOO

et al

.

Consider the classical Lagrangian functional

Let

(

L

2

(

Γ

))

+

be the cone of square integrable functions on

Γ

.

Definition 1.

A point

(

v

*,

l

*)

∈

(

Ω

)

×

(

L

2

(

Γ

))

+

is called

a

saddle point of the functional

L

(

v

,

l

)

if

It was shown in monograph [8] that, if the solution

u

*

to the Signorini problem in the coercive case

belongs to the space

(

Ω

)

, then the Lagrangian functional has the unique saddle point

(

u

*,

∂

u

*/

∂

n

)

; i.e.,

v

* =

u

*

almost everywhere in

Ω and l* = ∂u*/∂n almost everywhere in Γ, where n is the unit outer normalto Γ.

An analogous result holds for semicoercive problem (1). The following assertions can be proved oneafter another:

(1) L(u*, l) ≤ L ∀l ∈ (L2(Γ))+.

(2) L ≤ L ∀v ∈ (Ω).

This inequality is valid because –∆u* = f in Ω .(3) L(v, l) has a unique saddle point.

Indeed, assume that, conversely, L(v, l) has the two saddle points (u1, ∂u1/ ) and (u2, ∂u2/ ). It is easyto show that u1 – u2 = const; hence, ∂u1/∂n = ∂u2/∂n. Taking into account the equality L(u1, ∂u1/∂n) = L(u2,∂u1/∂n) and condition (2), we obtain the desired result.

Nevertheless, an application of the Uzawa method to the search for a saddle point of L(v, l) is impossiblein the semicoercive case, because the convergence of this iterative method is ensured by coordinating thestep size for the dual variable l with the positive definiteness constant in the quadratic form of the functionalto be minimized. However, in semicoercive problem (1), the quadratic form

is only positive semidefinite.To overcome this difficulty, consider a modification of the Lagrangian functional.

We define the modified Lagrangian functional M(v, l) on the space (Ω) × L2(Γ) as follows:

Here, r > 0 is a constant, and the symbol w+ denotes max0, w; i.e., (l – rγv)+ = max0, l – rγv.It can be shown that the functional M(v, l) is convex with respect to v for a fixed l and concave with

respect to l for a fixed v. Moreover, M(v, l) is Gäteaux differentiable in both its variables u and l. Further-more, the derivatives ∇vM(v, l) and ∇lM(v, l) satisfy the equalities

L v l,( ) J v( ) lγv Γd∫–12--- ∇v 2 Ω fv Ω lγv Γd

Γ∫–d

Ω∫–d

Ω∫= =

v l,( ) W21 Ω( ) L2× Γ( ).∈∀

W21

L v* l,( ) L v* l*,( ) L v l*,( ) v l,( ) W21 Ω( ) L2 Γ( )( )+.×∈∀≤ ≤

W22

u*∂u*∂n

---------,⎝ ⎠⎛ ⎞

u*∂u*∂n

---------,⎝ ⎠⎛ ⎞ v

∂u*∂n

---------,⎝ ⎠⎛ ⎞ W2

1

∂n ∂n

a v v,( ) ∇v 2 Ωd

Ω∫=

W21

M v l,( ) J v( ) 12r----- l rγv–( )+[ ]2

l2– Γ.d

Γ∫+=

∇vM v l,( ) h,( ) a v h,( ) f h,( )– l rγv–( )+h Γ h W21 Ω( ),∈∀d

Γ∫–=

∇lM v l,( ) θ,( ) 1r--- l rγv–( )+ l–( )θ Γ θ L2 Γ( ),∈∀d

Γ∫=

COMPUTATIONAL MATHEMATICS AND MATHEMATICAL PHYSICS Vol. 46 No. 1 2006

AN ITERATIVE METHOD BASED ON A MODIFIED LAGRANGIAN FUNCTIONAL 25

where

We examine the Uzawa method with the modified Lagrangian functional M(v, l) as applied to problem(1).

Choose an arbitrary initial point l0 ∈ (Γ) that is the trace on Γ of a function u0 ∈ (Ω). The algo-rithm can be described as follows:

(1) at step k (k = 0, 1, …) set

(3)

(2) find lk + 1 by the formula

(4)

Here, r is the step size with respect to the dual variable.Below, we prove that the points uk (k = 1, 2, …) determined by formula (3) exist; moreover, they are

defined uniquely if uk ∈ (Ω).

To investigate method (3), (4), consider the following abstract scheme, which is similar to the schemeproposed for finite-dimensional optimization problems in [9, 10].

Define the following functional on the space (Ω) × L2(Γ) × L2(Γ):

We have

(5)

Thus, M(v, l) = (v, l, m).

Define the functional

It is well known that

Therefore,

where χ(m) = .

a v h,( ) ∇v∇h Ω, f h,( )d

Ω∫ fh Ω.d

Ω∫= =

W21/2 W2

1

uk 1+ M v lk,( );v W2

1 Ω( )∈minarg=

lk 1+ lk rγ uk 1+–( )+.=

W22

W21

K v l m, ,( )J v( ) lm Γ r

2--- m2 Γ, if γv m,≤–d

Γ∫+d

Γ∫+

+∞ otherwise.⎩⎪⎨⎪⎧

=

K v l m, ,( )m

inf J v( ) lmr2---m2+⎝ ⎠

⎛ ⎞ Γd

Γ∫+

⎩ ⎭⎨ ⎬⎧ ⎫

γv– m≤inf J v( ) 1

2r----- l rm+( )2 l2–( ) Γd

Γ∫γv– m≤

inf+= =

= J v( ) 12r----- l rγv–( )+[ ]2

l2– Γd

Γ∫+ M v l,( ).=

Km

inf

M l( ) M v l,( )v

inf K v l m, ,( ).m

infv

inf= =

K v l m, ,( )m

infv

inf K v l m, ,( ).v

infm

inf=

K v l m, ,( )v

infm

inf J v( ) lm Γ r2--- m2 Γd

Γ∫+d

Γ∫+

⎩ ⎭⎨ ⎬⎧ ⎫

vinf

minf=

= J v( ) lm Γ r2--- m2 Γd

Γ∫+d

Γ∫+

γv m≤–inf

⎩ ⎭⎨ ⎬⎧ ⎫

minf χ m( ) lm Γ r

2--- m2 Γd

Γ∫+d

Γ∫+

⎩ ⎭⎨ ⎬⎧ ⎫

,m

inf=

J v( )γv m≤–inf

26


WOO et al.

It is easy to show that χ(m) is a convex function [9, 10] and χ(m) ≠ –∞ for ∀m ∈ L2(Γ) if condition (2)is fulfilled (see [7]).

Thus, the functional (l) has the following two representations:

(6)

(7)

Since

is a strongly convex functional with respect to m ∈ L2(Γ), the infimum in problem (7) is attained at a uniqueelement m(l) for an arbitrary fixed l ∈ L2(Γ).

Theorem 1. Problem (6) is solvable for an arbitrary fixed l ∈ (L2(Γ))+.

Proof. It suffices to show that the functional M(v, l) is coercive with respect to v for l ∈ (L2(Γ))+; i.e.,

We have

For any v ∈ (Ω), divide the boundary Γ into the two parts Γ1 and Γ2 such that γv < 0 on Γ1 and γv ≥0 on Γ2 .

Let v– = min0, v. Then, v = v+ + v– (see [11]). We have

(8)

Here, we used the relations (γv)– = (γ(v+ + v–))– = (γv+ + γv–)– = γv–.

Since a(v, v) = a(v+, v+) + a(v–, v–) (see [11]), we obtain

(9)

M

M l( ) J v( ) 12r----- l rγv–( )+[ ]2

l2– Γd

Γ∫+

⎩ ⎭⎨ ⎬⎧ ⎫

,v

inf=

M l( ) χ m( ) lm Γ r2--- m2 Γd

Γ∫+d

Γ∫+

⎩ ⎭⎨ ⎬⎧ ⎫

.m

inf=

χ m( ) lm Γ r2--- m2 Γd

Γ∫+d

Γ∫+

⎩ ⎭⎨ ⎬⎧ ⎫

M v l,( ) +∞ as vW2

1 Ω( )∞.

M v l,( ) 12--- ∇v 2 Ω fv Ω 1

2r----- l rγv–( )+[ ]2

l2– Γ.d

Γ∫+d

Ω∫–d

Ω∫=

W21

l rγv–( )+[ ]2l2– Γd

Γ∫ l rγv–( )+[ ]2

l2– Γ l rγv–( )+[ ]2l2– Γd

Γ2

∫+d

Γ1

∫=

= l rγv–( )2 l2–[ ] Γ l rγv–( )+[ ]2l2– Γ 2rγv– r2 γv( )2+[ ] Γ l2 Γd

Γ2

∫–d

Γ1

∫≥d

Γ2

∫+d

Γ1

∫

≥ 2r γv( )–– r2 γv( )–[ ]2+ Γ l2 Γd

Γ∫–d

Γ∫ 2rγv –– r2 γv –( )2

+[ ] Γ l2 Γ r2 γv –( )2 Γ l2 Γ.d

Γ∫–d

Γ∫≥d

Γ∫–d

Γ∫=

M v l,( ) 12---a v v,( ) f v,( ) 1

2r----- l rγv–( )+[ ]2

l2– Γd

Γ∫+–=

= 12---a v

+v

+,( ) 12---a v

–v

–,( ) f v+,( )– f v

–,( ) 12r----- l rγv–( )+[ ]2

l2– Γd

Γ∫+–+

≥ 12---a v

+v

+,( ) f v+,( ) 1

2---a v

–v

–,( ) f v–,( ) r

2--- γv –( )2 Γ 1

2r----- l2 Γ.d

Γ∫–d

Γ∫+–+–



Assume that ∞. Since ≤ + , at least one of the norms on the

right-hand side of this inequality tends to infinity. If ∞, then, taking into account (2), we con-

clude that a(v+, v+) – ( f, v+) +∞ (see [12]).

In the case ∞, we take into account the fact that a(v, v) + is equivalent to

(see [7, p. 71]). It follows that

Thus, M(v, l) +∞ as ∞, and problem (6) is solvable for an arbitrary fixed l ∈ (L2(Γ))+.

The theorem is proved.Consider the problem

(10)

It will be called the dual of problem (1).

Assume that the solution u* to the original problem (1) belongs to the space (Ω). Then, as notedabove, the classical Lagrangian functional

has the unique saddle point (u*, ∂u*/∂n); i.e.,

Due to the convexity of the Signorini problem, (u*, ∂u*/∂n) is also a unique saddle point for the modifiedLagrangian functional (see [9]); i.e.,

Moreover, ∂u*/∂n is a unique solution to problem (10).

Let us examine the functional (l) in dual problem (10).

Theorem 2. The functional (l) is Gäteaux differentiable in L2(Γ), and its derivative satisfies the Lip-schitz condition with the constant 1/r; i.e., for any l1, l2 ∈ L2(Γ), it holds that

Proof. For any m ∈ L2(Γ), we have

.

If l1, l2 ∈ L2(Γ); m1 = m(l1); and m2 = m(l2), then

vW2

1 Ω( )v

W21 Ω( )

v+

W21 Ω( ) v

–W2

1 Ω( )

v+

W21 Ω( )

12---

v–

W21 Ω( )

r2--- v

2 ΓdΓ∫

vW2

1 Ω( )2

12---a v

–v

–,( ) f v–,( ) r

2--- γv –( )2 Γ +∞.d

Γ∫+–

vW2

1 Ω( )

M l( ) max, l L2 Γ( ).∈

W22

L v l,( ) 12---a v v,( ) f v,( )– lγv Γd

Γ∫–=

L u* l,( ) L u*∂u*∂n

---------,⎝ ⎠⎛ ⎞ L v

∂u*∂n

---------,⎝ ⎠⎛ ⎞ v l,( ) W2

1 Ω( ) L2 Γ( )( )+× .∈∀≤ ≤

M u* l,( ) M u*∂u*∂n

---------,⎝ ⎠⎛ ⎞ M v

∂u*∂n

---------,⎝ ⎠⎛ ⎞ v l,( ) W2

1 Ω( ) L2 Γ( )( )× .∈∀≤ ≤

M

M

∇M l1( ) ∇M l2( )– L2 Γ( )1r--- l1 l2– L2 Γ( ).≤

χ m l( )( ) lm l( ) Γ r2--- m2 l( ) Γ r

2--- m m l( )– L2 Γ( )

2 χ m( ) lm Γ r2--- m2 Γd

Γ∫+d

Γ∫+≤+d

Γ∫+d

Γ∫+

χ m1( ) l1m1 Γ r2--- m1

2 Γ r2--- m2 m1– L2 Γ( )

2 χ m2( ) l1m2 Γ r2--- m2

2 Γ,d

Γ∫+d

Γ∫+≤+d

Γ∫+d

Γ∫+

χ m2( ) l2m2 Γ r2--- m2

2 Γ r2--- m1 m2– L2 Γ( )

2 χ m1( ) l2m1 Γ r2--- m1

2 Γ.d

Γ∫+d

Γ∫+≤+d

Γ∫+d

Γ∫+

28


WOO et al.

Combining these inequalities, we obtain

(11)

which implies that, for any l1 , l2 ∈ L2(Γ),

(12)

Since (l) is a concave functional and (l) ≠ –∞ ∀l ∈ L2(Γ), the subdifferential ∂( (l)) ≠ ∅ for anyl ∈ L2(Γ) (see [12]). To prove that (l) is differentiable, it suffices to verify that ∂( (l)) consists of a singleelement.

Let l ∈ L2(Γ) be a fixed element and t ∈ ∂( (l)). Then, for any ξ ∈ L2(Γ), it holds that

i.e.,

Hence, for all ξ ∈ L2(Γ) and all β > 0, we have

We set ξ = l + βp in this inequality. Then, taking into account the continuity of m(l) (see (12)) and passingto the limit as β +0, we obtain

which implies that t = m(l). Since m(l) is a single-valued function, we conclude that ∂( (l)) = m(l).

Thus, is Gäteaux differentiable in L2(Γ). Taking into account (12), we obtain the assertion of the the-orem.

Assume that the solution u* to the original problem (1) belongs to the space (Ω). Then, as notedabove, dual problem (10) has the unique solution ∂u*/∂n ∈ L2(Γ). To solve the dual problem, consider thegradient method

(13)

Theorem 3. The mapping P(l) = l + r∇ (l) satisfies the following two conditions:

(1) P = ;

(2) ≤ ∀l ≠ .

Proof. The differentiability of implies that

It follows that P(∂u*/∂n) = ∂u*/∂n.

r m1 m2– L2 Γ( )2 l1 l2–( ) m2 m1–( ) Γ,d

Γ∫≤

m1 m2– L2 Γ( ) r 1– l1 l2– L2 Γ( ).≤

M M MM M

M

M ξ( ) M l( ) t ξ l–( ) Γ;d

Γ∫+≤

χ m ξ( )( ) ξm ξ( ) Γ r2--- m2 ξ( ) Γ χ m l( )( ) lm l( ) Γ r

2--- m2 l( ) Γd

Γ∫+d

Γ∫+≤d

Γ∫+d

Γ∫+

+ t ξ l–( ) Γ χ m ξ( )( ) lm ξ( ) Γ r2--- m2 ξ( ) Γ t ξ l–( ) Γ.d

Γ∫+d

Γ∫+d

Γ∫+≤d

Γ∫

β 1– m ξ( ) t–( ) ξ l–( ) Γ 0.≤d

Γ∫

m l( ) t–[ ] p Γ 0 p L2 Γ( ),∈∀≤d

Γ∫

M

M

W22

lk 1+ lk r∇M lk( ), k+ 0 1 …., ,= =

M

∂u*∂n

---------⎝ ⎠⎛ ⎞ ∂u*

∂n---------

P∂u*∂n

---------⎝ ⎠⎛ ⎞ P l( )–

L2 Γ( )

∂u*∂n

--------- l–L2 Γ( )

∂u*∂n

---------

M

∇M∂u*∂n

---------⎝ ⎠⎛ ⎞ 0.=



Let l ≠ ∂u*/∂n. Since ∇ (∂u*/∂n) = m(∂u*/∂n) and ∇ (l) = m(l), we have

In view of inequality (11), this implies that

The theorem is proved.Let us show that formulas (4) and (13) are equivalent. We have

(14)

It follows from (7) and (14) that

where

Now, (13) implies that

i.e., the definition of lk by (4) corresponds to one step of gradient method (13) as applied to (l).

It follows from Theorem 3 that the sequence lk is bounded in L2(Γ). For any k = 0, 1, …, it holds thatM(uk + 1, lk) ≤ M(0, lk) = 0. Now, the inequality

proved in Theorem 1 for l ∈ (L2(Γ))+ also entails the following assertion.

Theorem 4. The sequence (uk, lk – 1) generated by method (3), (4) is bounded in (Ω) × L2(Γ).

For any k = 0, 1, …, consider the problem

(15)

As shown above, problem (15) is solvable if condition (2) is fulfilled.

M M

P∂u*∂n

---------⎝ ⎠⎛ ⎞ P l( )–

L2 Γ( )

2 ∂u*∂n

--------- r∇M∂u*∂n

---------⎝ ⎠⎛ ⎞ l– r∇M l( )–+

2 ∂u*∂n

--------- l– r m∂u*∂n

---------⎝ ⎠⎛ ⎞ m l( )–⎝ ⎠

⎛ ⎞+L2 Γ( )

2

= =

= ∂u*∂n

--------- l–L2 Γ( )

22r

∂u*∂n

--------- l m∂u*∂n

---------⎝ ⎠⎛ ⎞ m l( )–,–⎝ ⎠

⎛ ⎞L2 Γ( )

r2 m∂u*∂n

---------⎝ ⎠⎛ ⎞ m l( )–

L2 Γ( )

2

.+ +

P∂u*∂n

---------⎝ ⎠⎛ ⎞ P l( )–

L2 Γ( )

2 ∂u*∂n

--------- l–L2 Γ( )

22r2 m

∂u*∂n

---------⎝ ⎠⎛ ⎞ m l( )–

L2 Γ( )

2

r2 m∂u*∂n

---------⎝ ⎠⎛ ⎞ m l( )–

L2 Γ( )

2

+–≤

= ∂u*∂n

--------- l–L2 Γ( )

2r2 m l( ) L2 Γ( )

2 ∂u*∂n

--------- l–L2 Γ( )

2.≤–

M l( ) J v( ) 12r----- l rγv–( )+[ ]2

l2– Γd

Γ∫+

⎩ ⎭⎨ ⎬⎧ ⎫

vinf J v( ) lm

r2---m2+⎝ ⎠

⎛ ⎞ Γd

Γ∫m γv–≥

inf+⎩ ⎭⎨ ⎬⎧ ⎫

vinf= =

= J v( ) 12r----- l rm+( )2 l2–[ ] Γd

Γ∫m γv–≥

inf+⎩ ⎭⎨ ⎬⎧ ⎫

.v

inf

m l( ) maxlr-- γu l( )–,–

⎩ ⎭⎨ ⎬⎧ ⎫

,=

u l( ) J v( ) 12r----- l rγv–( )+[ ]2

l2– Γd

Γ∫+

⎩ ⎭⎨ ⎬⎧ ⎫

.v

minarg=

lk 1+ lk r∇M lk( )+ lk rm lk( )+ lk rmax lk

r---– γu lk( )–,

⎩ ⎭⎨ ⎬⎧ ⎫

+ lk rγu lk( )–( )+lk rγ uk 1+–( )+

;= = = = =

M

M v l,( ) 12---a v

+v

+,( ) f v+,( ) 1

2---a v

–v

–,( ) f v–,( ) v

–( )2 Γ l2 Γd

Γ∫–d

Γ∫+–+–≥

W21

M v lk 1–,( ) min,

v W21 Ω( ).∈

30


WOO et al.

Theorem 5. Let (2) be fulfilled. If a solution to problem (15) belongs to (Ω), then it is unique.

Proof. Assume the contrary; i.e., there are two solutions, and at least one solution, say, u1, belongs to

(Ω). We have

(16)

Since ui (i = 1, 2) satisfy the variational inequalities

(see [12]), relation (16) implies that a(u2 – u1, u2 – u1) = 0; thus, u2 = u1 + c, where c = const. This means that

u2 also belongs to (Ω). For u1 and u2 belonging to (Ω), it is easy to show that ui (i = 1, 2) satisfy theboundary value problem

(17)

Thus,

(18)

Without loss of generality, we can assume that c < 0. Define Γ1 = x ∈ Γ : lk – 1 – rγu1 ≥ 0. Assume thatmesΓ1 > 0; then, lk – 1 – rγ(u1 + c) ≥ 0 on Γ1 , and (18) implies that

which entails the equality c = 0. Therefore, mesΓ1 = 0; i.e.,

Then, it follows from (18) that

This means that lk – 1 – rγu2 ≤ 0 on Γ; i.e., γu2 ≥ lk – 1/r. In this case, we have

It follows that

This contradiction proves the theorem.Theorem 6. Let the points

satisfy the following conditions:

(A) uk ∈ (Ω);

(B) ≤ C, where C > 0 is a constant.

W22

W22

0 M u2 lk 1–,( ) M u1 lk 1–,( ) a u1 u2 u1–,( ) f u2 u1–,( )–=–=

+12---a u2 u1– u2 u1–,( ) 1

2r----- lk 1– rγ u2–( )+[ ]

2lk 1– rγ u1–( )+[ ]

2– Γ.d

Γ∫+

a ui v ui–,( ) f v ui–,( ) 12r----- lk 1– rγ u2–( )+[ ]

2lk 1– rγ u1–( )+[ ]

2– Γ 0 v W2

1 Ω( )∈∀≥d

Γ∫+–

W22 W2

2

∆u– f in Ω,=

∂u/∂n– lk 1– rγu–( )+on Γ.–=

lk 1– rγ u1–( )+lk 1– rγ u1 c+( )–[ ]+

on Γ.=

lk 1– rγ u1– lk 1– rγ u1 rγc,––=

γ u1 lk 1– /r 0 on Γ.≥>

0 lk 1– rγ u1–( )+lk 1– rγ u2–( )+

on Γ.= =

M ui lk 1–,( ) J ui( ) lk 1–( )2 Γ, id

Γ∫– 1 2.,= =

0 M u2 lk 1–,( ) M u1 lk 1–,( )– M u1 c lk 1–,+( ) M u1 lk 1–,( )– c f Ω 0.≠d

Ω∫= = =

uk M v lk 1–,( ) (kv W2

1Ω( )∈

minarg 1 2 …), ,= =

W22

ukW2

2 Ω( )



Then, the sequence (uk, lk – 1) generated by method (3), (4) converges in (Ω) × L2(Γ) to the saddle

point (u*, ∂u*/∂n) for the initial point l0 ∈ (Γ) and any fixed parameter value r > 0.

Proof. It follows from Theorem 4 that (uk, lk – 1) is a bounded sequence in (Ω) × L2(Γ). Assumptions

(A) and (B) ensure that (uk, lk − 1) is compact in (Ω) × L2(Γ). Since ∂uk/∂n = lk (k = 1, 2, …), (uk, lk – 1) is

a compact sequence in (Ω) × L2(Γ). Let ( , ) be a limit point of (uk, lk – 1); i.e.,

= 0 for the appropriate subsequence ( , ) of (uk, lk – 1).

Note that ( , ) satisfies the identity

(19)

We also have for ( , ) the relations

where c1 > 0 is a constant. Thus, passing to the limit in (19) as j ∞, we obtain

(20)

It follows from the conditions for a saddle point (see [8, 13]) that

Set v = uk – u* in the last equality; then,

It is obvious that uk satisfies the relation

Combining the last two equalities, we obtain

W21

W21/2

W21

W21

W21 u l

uk j l

k j 1–,( ) u l,( )– W2

1 Ω( ) L2 Γ( )×j ∞→lim u

k j lk j 1–

uk j l

k j 1–

a uk j v,( ) f v,( )– l

k j 1–rγ u

k j–( )+v Γd

Γ∫– 0 v W2

1 Ω( ).∈∀=

uk j l

k j 1–

lk j 1–

rγ uk j–( )

+l rγ u–( )+

– L2 Γ( ) lk j 1–

l–( ) rγ uk j u–( )+ L2 Γ( )≤

≤ lk j 1–

l– L2 Γ( ) r γ uk j u–( ) L2 Γ( ) l

k j 1–l– L2 Γ( ) c1 u

k j u–( ) W21 Ω( ),+≤+

a u v,( ) f v,( ) l rγ u–( )+γv Γd

Γ∫–– 0 v W2

1 Ω( ).∈∀=

∂u*∂n

--------- ∂u*∂n

--------- rγu*–⎝ ⎠⎛ ⎞ +

,=

a u* v,( ) f v,( )– ∂u*∂n

--------- rγu*–⎝ ⎠⎛ ⎞ +

γv Γd

Γ∫– 0 v W2

1 Ω( ).∈∀=

a u* uk u*–,( ) f uk u*–,( ) ∂u*∂n

--------- rγu*–⎝ ⎠⎛ ⎞ +

γ uk γu*–( ) Γd

Γ∫–– 0.=

a uk u* uk–,( ) f u* uk–,( ) lk 1– rγ uk–( )+ γu* γ uk–( ) Γd

Γ∫–– 0.=

a u* uk uk u*–,–( ) ∂u*∂n

--------- rγu*–⎝ ⎠⎛ ⎞ +

lk 1– rγ uk–( )+– γ uk γu*–( ) Γd

Γ∫– 0,=

0 a uk u*– uk u*–,( )≤ ∂u*∂n

--------- rγu*–⎝ ⎠⎛ ⎞ +

lk 1– rγ uk–( )+– γ uk γu*–( ) Γ,d

Γ∫–=

0 ∂u*∂n

--------- rγu*–⎝ ⎠⎛ ⎞ +

lk 1– rγ uk–( )+–⎝ ⎠

⎛ ⎞ γ uk γu*–( ) Γd

Γ∫≥

32


WOO et al.

(21)

Here, we used the well-known scalar inequality (see [13])

Now,

This implies that

It follows from Theorem 3 that

Therefore,

Then,

Passing to the limit as j ∞, we obtain = . This implies that

Thus, = ( – )+, and equality (19) gives

This is equivalent to the inequality

where L(v, l) is the classical Lagrangian functional. Under the condition = ( – )+, this means that

( , ) = (u*, ∂u*/∂n). Since ( , ) is an arbitrary limit point of (uk, uk − 1), the theorem is proved.Remark. In the coercive case, the regularity of solutions to auxiliary boundary value problems of type

(17) was investigated in [14, Theorem 1.10, p. 43, 15, Theorem 3.2.3.1].

= 1r--- ∂u*

∂n--------- rγu*–⎝ ⎠

⎛ ⎞ +

lk 1– rγ uk–( )+– ∂u*

∂n--------- rγu*– ∂u*

∂n---------–⎝ ⎠

⎛ ⎞ lk 1– rγ uk–( ) lk 1–––⎩ ⎭⎨ ⎬⎧ ⎫

Γd

Γ∫

≥ 1r--- ∂u*

∂n--------- rγu*–⎝ ⎠

⎛ ⎞ +

lk 1– rγ uk–( )+– ∂u*

∂n--------- rγu*–⎝ ⎠

⎛ ⎞ + ∂u*∂n

---------– lk 1– rγ uk–( )+lk 1–––

⎩ ⎭⎨ ⎬⎧ ⎫

Γd

Γ∫

= 1r--- ∂u*

∂n--------- lk–⎝ ⎠

⎛ ⎞ ∂u*∂n

--------- ∂u*∂n

---------–⎝ ⎠⎛ ⎞ lk lk 1––( )–⎝ ⎠

⎛ ⎞ Γd

Γ∫ 1

r--- ∂u*

∂n--------- lk–⎝ ⎠

⎛ ⎞ lk 1– lk–( ) Γ.d

Γ∫=

x+ y+–( ) x y–( ) x+ y+–( )2.≥

lk 1– ∂u*∂n

---------–L2 Γ( )

2lk 1– lk lk ∂u*

∂n---------–+–

L2 Γ( )

2=

= ||lk 1– lk |– L2 Γ( )2

lk ∂u*∂n

---------–L2 Γ( )

22 lk 1– lk–( ) lk ∂u*

∂n---------–⎝ ⎠

⎛ ⎞ Γ.d

Γ∫+ +

lk 1– ∂u*∂n

---------–L2 Γ( )

2lk 1– lk– L2 Γ( )

2lk ∂u*

∂n---------–

L2 Γ( )

2.+≥

0 lk ∂u*∂n

---------–L2 Γ( )

lk 1– ∂u*∂n

---------–L2 Γ( )

.≤ ≤

lk 1– lk– L2 Γ( )k ∞→lim 0.=

lk j l– L2 Γ( ) l

k j lk j 1–

– L2 Γ( ) lk j 1–

l– L2 Γ( ).+≤

lk j

k ∞→lim l

l l rγ u–( )+– L2 Γ( ) l l

k j– L2 Γ( ) lk j l rγ u–( )+

– L2 Γ( )+≤

= l lk j– L2 Γ( ) l

k j 1–rγ u

k j–( )+

l rγ u–( )+– L2 Γ( ) 0, j ∞.+

l l ru

a u v,( ) f v,( )– lv Γd

Γ∫– 0 v W2

1 Ω( ).∈∀=

L u l,( ) L v l,( ) v W21 Ω( ),∈∀≤

l l ru

u l u l



ACKNOWLEDGMENTSThis work was supported by the Russian Foundation for Basic Research, project no. 04-01-97003.

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