Pham Hoang Quan

† -Dang Duc Trong

Alain Pham
email: alain.pham@math.cnrs.fr

Ngoc Dinh

A nonlinearly ill-posed problem of reconstructing the temperature from interior data *

Keywords: Fourier transform, Contraction, ill-posedness AMS Classification 2000: 47J06, 35J60, 42A38, 47H10

We consider the problem of reconstructing, from the interior data u(x, 1), a function u satisfying a nonlinear elliptic equation ∆u = f (x, y, u(x, y)),

x ∈ R, y > 0.

The problem is ill-posed. Using the method of Green function, the method of Fourier transforms and the method of truncated high frequencies, we shall regularize the problem. Error estimate is given.

Introduction

In this paper, we consider the problem of reconstructing the temperature of a body from interior measurements. In fact, in many engineering contexts (see, e.g., [BBC]), we cannot attach a temperature sensor at the surface of a body (e.g., the skin of a missile). Hence, to get the temperature distribution on the surface, we have to use the temperature measured inside the body.

Precisely, we consider a two-dimensional body represented by the half-plane R × R + . Letting u(x, y) be the temperature of the body at (x, y) ∈ R × R + and letting f ≡ f (x, y, u) be a (nonlinear) heat source, we have the following nonlinearly nonhomogeneous equation

∆u = f, x ∈ R, y > 0 (1)
where ∆ = ∂ 2 ∂x 2 + ∂ 2 ∂y 2 . We assume that the temperature on the line y = 1 is known, i. e., u(x, 1) = ϕ(x),

and that u(x, y) → 0 as |x|, y → ∞.

(3)

The problem can be referred as a sideways elliptic problem and the interior measurement ϕ(x) is also called (in geology) the borehole measurement. The problem can be splitted into two problems:

Problem 1: finding the function u satisfying ∆u = f, x ∈ R, y > 1 (4) subject to the conditions (2), (3). Generally, the problem is well-posed. Using the solution of the problem we can calculate the quantity u y (x, 1) = φ(x).

Problem 2: finding a function u(x, y) x ∈ R, 0 < y < 1 satisfying ∆u = f,

x ∈ R, 0 < y < 1 (5) subject to the conditions u(x, 1) = ϕ(x), u y (x, 1) = φ(x).

The latter problem is a Cauchy elliptic problem and, as known, it is severely ill-posed. Hence, a regularization is in order.

The homogeneous problem (f ≡ 0) was studied, by various methods in many papers. Using the mollification method, the homogeneous sideways parabolic problems were considered in [START_REF] Hao | Numerical solution to a sideways parabolic equation[END_REF][START_REF] Hao | On a sideways parabolic equation[END_REF]AS,[START_REF] Levine | Continuous data dependence, regularization and a three lines theorem for the heat equation with data in a space like direction[END_REF][START_REF] Levine | Estimates and regularization for solutions of some illposed problems of elliptic and parabolic type[END_REF][START_REF] Murio | The mollification method and the numerical solution of ill-posed problems[END_REF] and the references therein. Similarly, the number of papers devoted to the Cauchy problem for linear homogeneous elliptic equation are very rich (see, e.g., [HR, T, B, CHWY, KS]).

Although there are many papers on homogeneous cases, we only find a few papers on nonhomogeneous sideways problems (for both parabolic and elliptic equations). Especially, the papers on the nonlinear case are very rare. In [QT], we use the method of integral equations to consider a sideways elliptic equation with a nonlinear heat source. However, we cannot get an effective method to regularize the problem. In the present paper, we shall consider the problem with a nonlinear heat source f ≡ f (x, y, u(x, y)). The remainder of our paper is divided in three sections. In Section 2, using Green functions, we shall transform Problems 1 and 2 into integral equations. In Section 3, we shall prove that Problem 1 has a unique solution. Moreover, we shall give an effective way to approximate the quantity u y (x, 1). In Section 4, using ϕ(x), u y (x, 1), we shall regularize Problem 2. The main result of the section and of the paper is Theorem 3 (in Subsection 4.2). The method of truncated high frequencies (of Fourier images) will be used and the regularized solution can be found as a fixed point of a contraction. In our knowledge, the latter method is new. Error estimates are given.

2 Integral equations of Problems 1 and 2

2.1 Problem 1 Put N(x, y, ξ, η) = - 1 4π ln (x -ξ) 2 + (y -η) 2 (x -ξ) 2 + (y + η -2) 2 . (6) For y > 1, x ∈ R, integrating the identity ∂ ∂ξ (uN ξ -Nu ξ ) + ∂ ∂η (uN η -Nu η ) = -Nf (7)
over the domain (-m, m) × (1, n)\B((x, y), ε), where B((x, y), ε) is the ball with center at (x, y) and radius ε > 0, and letting n → ∞, m → ∞, ε → 0, we get, after some rearrangements,

u(x, y) = Au(x, y), (8)
where

Au(x, y) = h(x, y, ϕ) - +∞ -∞ +∞ 1 N(x, y; ξ, η)f (ξ, η, u(ξ, η))dξdη (9) and h(x, y, ϕ) = +∞ -∞
N η (x, y, ξ, 1)ϕ(ξ)dξ.

Problem 2

We repeat ψ(x) = u y (x, 1) with u being an exact solution of Problem 1. We note that we only consider ϕ as an exact data. The function ψ is a processed data dependent on ϕ. In fact, in Lemma 1 (Section 3), we shall construct a function ψ ε which is an approximation of ψ.

Consider ∆u = f (x, y, u(x, y)) , x ∈ R, y ∈ (0, 1) subject to the boundary conditions below

u(x, 1) = ϕ(x), x ∈ R, (10)
∂u ∂y (x, 1) = ψ(x), x ∈ R.
We assume in addition that the exact solution u satisfying ϕ(ζ)e |ζ| , ψ(ζ)e |ζ| ∈ L 2 (R). We divide this problem into two problems.

Problem 2.1 Consider the problem ∆v = 0, x ∈ R, y ∈ (0, 1), ( 11)

v(x, 1) = ϕ(x), x ∈ R, (12)
∂v ∂y (x, 1) = ψ(x), x ∈ R. ( 13
)
We shall prove that, under appropriate conditions, the problem has a unique solution v 0 approximated by a regularized solution v ε .

Problem 2.2 Let u 0 be an exact solution of (10). If we put w 0 = u 0 -v 0 then w 0 is the solution of the problem ∆w = g(x, y, w), x ∈ R, y ∈ (0, 1) ( 14)

w(x, 1) = 0, x ∈ R, (15)
∂w ∂y (x, 1) = 0, x ∈ R (16)
where g(x, y, w) = f (x, y, w + v 0 ). We shall find a w ε which is an approximation of w 0 and estimate w ε -w 0 2 where . 2 is the norm in L 2 (R × (0, 1)). Let

u ε = v ε + w ε , we shall estimate u ε -u 0 2 . 2.2.1 An integral equation of Problem 2.1 Let Γ(x, y, ξ, η) = - 1 4π ln (x -ξ) 2 + (y -η) 2 and G(x, y, ξ, η) = Γ(x, y, ξ, η) -Γ(x, -y, ξ, η). (17)
For x ∈ R, 0 < y < 1, integrating the identity

∂ ∂ξ (-vG ξ + Gv ξ ) + ∂ ∂η (-vG η + Gv η ) = 0
over the domain (-n, n) × (0, 1)\B((x, y), ε) and letting n → ∞, ε → 0, we get, after some rearrangements

v(x, y) = - +∞ -∞ [ϕ(ξ)G η (x, y, ξ, 1) -G(x, y, ξ, 1)ψ(ξ)] dξ + + +∞ -∞ G η (x, y, ξ, 0)v(ξ, 0)dξ. ( 18
)
Letting y → 1 in (18), we have

1 π +∞ -∞ 1 (x -ξ) 2 + 1 v(ξ, 0)dξ + +∞ -∞ [-ϕ(ξ)G η (x, 1, ξ, 1) + G(x, 1, ξ, 1)ψ(ξ)] dξ = ϕ(x).
It can be rewritten as

F (1) * v (0) (x) = πK (1) (x) + √ π √ 2 ϕ(x) (19)
where

K (y) (x) = - 1 √ 2π +∞ -∞ [-ϕ(ξ)G η (x, y, ξ, 1) + G(x, y, ξ, 1)ψ(ξ)] dξ, F (y) (x) ≡ y x 2 + y 2 , v (y) (x) = v(x, y). ( 20
)
Letting

M (y,1) (x) = 1 -y x 2 + (y -1) 2 - 1 + y x 2 + (y + 1) 2 and L (η,y) (x) ≡ ln x 2 + (y -η) 2 x 2 + (y + η) 2 (0 < y, η < 1, x ∈ R),
we have the Fourier transform of M, L, F as followed

F(y) (ζ) = 1 √ 2π +∞ -∞ F (y) (x)e -ixζ dx = √ π √ 2 e -y|ζ| , L(η,y) (ζ) = √ 2π 1 |ζ| e -(y+η)|ζ| -e -|y-η||ζ| , M(y,1) (ζ) = √ π √ 2 e (y-1)|ζ| -e -(y+1)|ζ| . (21)
From ( 20), ( 21), we get

K (y) (x) = - 1 4π 2ϕ * M (y,1) (x) -ψ * L (1,y) (x) . (22)
From ( 19), (20), we have

v(0) (ζ) = e |ζ| √ 2π K(1) (ζ) + φ(ζ) . ( 23
)
Taking the Fourier transform of (18), we get

v(y) (ζ) = e |ζ| F(y) (ζ) 2 K(1) (ζ) + √ 2 √ π φ(ζ) - √ 2π K(y) (ζ) = 1 2 φ(ζ) e (1-y)|ζ| + e (y-1)|ζ| - 1 2 |ζ| ψ(ζ) e (1-y)|ζ| -e (y-1)|ζ| ≡ ℵ(ζ, y). (24)

An integral equation of Problem 2.2

We recall that w 0 an exact solution of Problem ( 14) -( 16). Let v 0 ∈ L 2 (R × (0, 1)) be the exact solution of ( 11) -( 13) and v ε ∈ L 2 (R × (0, 1)) be a regularized solution.

We write w (y) (x) = w(x, y) and

f (η,w 0 ,v 0 ) (ξ) = f (ξ, η, v 0 (ξ, η) + w 0 (ξ, η)). For x ∈ R, 0 < y < 1, let G be defined in (17), integrating the identity ∂ ∂ξ (-w 0 G ξ + Gw 0ξ ) + ∂ ∂η (-w 0 G η + Gw 0η ) = Gf (η,w 0 ,v 0 ) (ξ) (25)
over the domain (-n, n) × (0, 1)\B((x, y), ε) and letting n → ∞, ε → 0, we get, after some rearrangements,

w 0 (x, y) = +∞ -∞ w 0 (ξ, 0)G η (x, y, ξ, 0)dξ - +∞ -∞ 1 0 G(x, y, ξ, η)f (η,w 0 ,v 0 ) (ξ)dξdη. (26)
From ( 26), we have

w 0 (x, y) = 1 π +∞ -∞ w 0 (ξ, 0) y (x -ξ) 2 + y 2 dξ + 1 4π 1 0 +∞ -∞ ln (x -ξ) 2 + (y -η) 2 (x -ξ) 2 + (y + η) 2 f (η,w 0 ,v 0 ) (ξ)dξdη. ( 27
)
Letting y → 1, we get

1 4 1 0 +∞ -∞ ln (x -ξ) 2 + (1 -η) 2 (x -ξ) 2 + (1 + η) 2 f (η,w 0 ,v 0 ) (ξ)dξdη + +∞ -∞ w 0 (ξ, 0) 1 (x -ξ) 2 + 1 dξ = 0. ( 28
)
From ( 21), we get

F (y) (x) ≡ y x 2 + y 2 , L (η,y) (x) ≡ ln x 2 + (y -η) 2 x 2 + (y + η) 2 (0 < y, η < 1, x ∈ R), F(y) (ζ) = √ π √ 2 e -y|ζ| , L(η,y) (ζ) = √ 2π 1 |ζ| e -(y+η)|ζ| -e -|y-η||ζ| . ( 29
)
We write w 0(y) (x) ≡ w 0 (x, y).

From ( 28), ( 29) can be rewritten as

w 0(0) * F (1) (x) + 1 4 1 0 L (η,1) * f (η,w 0 ,v 0 ) (x)dη = 0.
Taking the Fourier transform, we have

w 0(0) (ζ). F(1) (ζ) + 1 4 1 0 L(η,1) (ζ) f(η,w 0 ,v 0 ) (ζ)dη = 0.
From ( 29), we have

w 0(0) (ζ) = - 1 2 1 0 1 |ζ| e -η|ζ| -e η|ζ| f(η,w 0 ,v 0 ) (ζ)dη. (30)
From ( 26), we have

w 0(y) (x) = √ 2 √ π w 0(0) * F (y) (x) + 1 2 √ 2π 1 0 L (η,y) * f (η,w 0 ,v 0 ) (x)dη. ( 31
)
Taking the Fourier transform of (31), we get

w 0(y) (ζ) = √ 2 √ π w 0(0) (ζ). F(y) (ζ) + 1 2 √ 2π 1 0 L(η,y) (ζ). f(η,w 0 ,v 0 ) (ζ)dη. ( 32
)
From ( 29) and (30), Eq. ( 32) takes the form

w 0(y) (ζ) = 1 2 1 0 1 |ζ| e (η-y)|ζ| -e -|y-η||ζ| f(η,w 0 ,v 0 ) (ζ)dη ( 33
)
for ζ. We have

w 0(y) (ζ) = 1 2 √ 2π 1 0 +∞ -∞ 1 |ζ| e (η-y)|ζ| -e -|y-η||ζ| f (η,w 0 ,v 0 ) (ξ)e -iξζ dξdη. ( 34
)
3 Finding the solution of Problem 1

From ( 8)-( 9), we readily get the following result Theorem 1. (see [QT]) Suppose that for all

(ξ, η, ζ) ∈ R × R + × R f ′ ζ (ξ, η, ζ) ≤ p(ξ, η), (35)
where p(ξ, η)

∈ L 1 (R × (1, +∞)), p ≥ 0 satisfies K ≡ sup (x,y)∈R×(1,+∞) +∞ -∞ +∞ 1 N(x, y; ξ, η)p(ξ, η)dξdη < 1. ( 36
) Put J = u ∈ C(R × (1, +∞)) lim √ x 2 +y 2 →+∞ u(x, y) = 0 . ( 37
)
Then A : J → J is a contraction and hence u is uniquely determined and can be found by successive approximation.

Let ψ(x) = u y (x, 1) with u be an exact solution in the half plane x ∈ R, y > 1. From a measured data ϕ ε of ϕ(x) = u(x, 1), we construct ψ ε being an aproximation of ψ and estimate ψ ε -ψ L 2 (R) by the following lemma Lemma 1 Let u be a solution of (8) as in Theorem 1 and let

ϕ ∈ L 1 (R) ∩ L ∞ (R) satisfy φ(ζ)e |ζ| ∈ L 2 (R).
Suppose that f satisfies the conditions

|f (ξ, η, ζ 1 ) -f (ξ, η, ζ 2 )| ≤ |p(ξ, η)| |ζ 1 -ζ 2 | f or all (ξ, η) ∈ R × R + , ζ 1 , ζ 2 ∈ R where p ∈ L 1 (R × (1, +∞)), K ≡ sup (x,y)∈R×(1,+∞) +∞ -∞ +∞ 1 N(x, y; ξ, η) |p(ξ, η)| dξdη < 1 and L = +∞ -∞   +∞ 1 +∞ -∞ η -1 (x -ξ) 2 + (1 -η) 2 |p(ξ, η)| dξdη   2 dx < +∞. For every 0 < ε < 1, we call ϕ ε ∈ L 2 (R) a measured data such that ϕ ε -ϕ L 2 (R) < ε. From ϕ ε , we can construct a function ψ ε ∈ L 2 (R) such that ψ ε -ψ L 2 (R) < Cε 1/2 where C is independent on ε. Proof Let k(x, y, u) = - +∞ -∞ +∞ 1 N(x, y; ξ, η)f (ξ, η, u(ξ, η))dξdη and h(x, y, ϕ) = +∞ -∞ N η (x, y, ξ, 1)ϕ(ξ)dξ. We have u(x, y) = h(x, y, ϕ) + k(x, y, u). Put ϕ ε (x) = 1 √ 2π |ζ|< 1 ε 1/2 ϕ ε (ζ)e iζx dζ.
We have

ϕ ε (ζ) = ϕ ε (ζ) |ζ| < 1 ε 1/2 0 |ζ| ≥ 1 ε 1/2 (38)
Let u ε be a solution of (8) with ϕ replaced by ϕ ε , i.e.

u ε = h(x, y, ϕ ε ) + k(x, y, u ε ). ( 39
)
We denote

h(x, y, ϕ) = h(x, y), h(x, y, ϕ ε ) = h ε (x, y), k(x, y, u) = k(x, y), k(x, y, u ε ) = k ε (x, y). (40)
Let

ψ ε (x) = h εy (x, 1) + k εy (x, 1). ( 41
)
We have

h(x, y) = +∞ -∞ N η (x, y; ξ, 1)ϕ(ξ)dξ = 1 π +∞ -∞ y -1 (x -ξ) 2 + (y -1) 2 ϕ(ξ)dξ. ( 42
)
If we put F (y) (x) ≡ y x 2 +y 2 , y > 0, then

F(y) (ζ) = 1 √ 2π +∞ -∞ F (y) (x)e -ixζ dx = √ π √ 2 e -y|ζ| .
Taking the Fourier transform of (42), we get

h(ζ, y) = √ 2 √ π F (y-1) * ϕ(ζ) = ϕ(ζ)e -(y-1)|ζ| and h y (ζ, y) = -|ζ| ϕ(ζ)e -(y-1)|ζ| .
Similarly, we have

h ε (ζ, y) = ϕ(ζ)e -(y-1)|ζ| .
We shall find an estimation of h εy (., 1) -h y (., 1) L 2 (R) .

Using the inequality u 4 < e 2u ∀u > 1, we have

h εy (., 1) -h y (., 1) 2 L 2 (R) = +∞ -∞ |ζ| 2 ϕ(ζ) -ϕ ε (ζ) 2 dζ ≤ |ζ|< 1 ε 1/2 |ζ| 2 | ϕ(ζ) -ϕ ε (ζ)| 2 dζ + |ζ|> 1 ε 1/2 e 2|ζ| |ζ| 2 | ϕ(ζ)| 2 dζ. Therefore h εy (., 1) -h y (., 1) 2 L 2 (R) < ε + ε e |ζ| ϕ(ζ) 2 L 2 (R) = C 2 1 ε
where

C 1 = 1 + e |ζ| ϕ(ζ) 2 L 2 (R) . Hence h εy (., 1) -h y (., 1) L 2 (R) < C 1 √ ε. ( 43
)
We have

k(x, y) = - +∞ 1 +∞ -∞ N(x, y; ξ, η)f (ξ, η, u(ξ, η))dξdη.
It follows that

k y (x, 1) = 1 π +∞ 1 +∞ -∞ 1 -η (x -ξ) 2 + (1 -η) 2 f (ξ, η, u(ξ, η))dξdη.
Similarly, we get

k εy (x, 1) = 1 π +∞ 1 +∞ -∞ 1 -η (x -ξ) 2 + (1 -η) 2 f (ξ, η, u ε (ξ, η))dξdη.
We have

k εy (., 1) -k y (., 1) L 2 (R) = = 1 π +∞ -∞   +∞ 1 +∞ -∞ 1 -η (x -ξ) 2 + (1 -η) 2 (f (ξ, η, u(ξ, η)) -f (ξ, η, u ε (ξ, η))dξdη   2 dx ≤ 1 π +∞ -∞   +∞ 1 +∞ -∞ η -1 (x -ξ) 2 + (1 -η) 2 |p(ξ, η)| |u -u ε | dξdη   2 dx ≤ 1 π u -u ε ∞ L (44)
where

L = +∞ -∞   +∞ 1 +∞ -∞ η -1 (x -ξ) 2 + (1 -η) 2 |p(ξ, η)| dξdη   2 dx < +∞. Moreover, we have u -u ε ∞ ≤ sup (x,y)∈R×(1,+∞) +∞ -∞ N η (x, y, ξ, 1) [ϕ(ξ) -ϕ ε (ξ)] dξ - +∞ -∞ +∞ 1 N(x, y, ξ, η) (f (ξ, η, u(ξ, η)) -f (ξ, η, u ε (ξ, η))) dξdη ≤ sup (x,y)∈R×(1+∞)   +∞ -∞ 1 π y -1 (x -ξ) 2 + (y -1) 2 |ϕ(ξ) -ϕ ε (ξ)| dξ + +∞ -∞ +∞ 1 N(x, y, ξ, η) |p(ξ, η)| |u(ξ, η) -u ε (ξ, η)| dξdη   ≤ ϕ -ϕ ε L ∞ (R) + K u -u ε ∞ where K = sup (x,y)∈R×(1+∞) +∞ -∞ +∞ 1 N(x, y, ξ, η) |p(ξ, η)| dξdη ∈ (0, 1). ( 45
) Hence u -u ε ∞ ≤ 1 1 -K ϕ -ϕ ε L ∞ (R) . (46)
We get

ϕ -ϕ ε L ∞ (R) = sup x∈R | ϕ ε (x) -ϕ(x)| = sup x∈R 1 √ 2π |ζ|< 1 ε 1/2 ϕ ε (ζ)e iζx dζ - +∞ -∞ ϕ(ζ)e iζx dζ ≤ 1 √ 2π     |ζ|< 1 ε 1/2 | ϕ ε (ζ) -ϕ(ζ)| dζ + |ζ|> 1 ε 1/2 | ϕ(ζ)| dζ     ≤ 1 √ 2π      1 ε 1/2 -1 ε 1/2 dζ 1 ε 1/2 -1 ε 1/2 | ϕ ε (ζ) -ϕ(ζ)| 2 dζ+ + |ζ|> 1 ε 1/2 e 2|ζ| | ϕ(ζ)| 2 dζ. |ζ|> 1 ε 1/2 e -2|ζ| dζ      ≤ 1 √ 2π √ 2 ε 1/4 ϕ -ϕ ε L 2 (R) + 2 3 e |ζ| ϕ(ζ) L 2 (R) ε 3/4 < C 2 ε 3/4 ( 47
)
where

C 2 = 1 √ 2π √ 2 + 2 3 e |ζ| ϕ(ζ) L 2 (R) .
Inequalities ( 44) -(47) give

k εy (., 1) -k y (., 1) L 2 (R) ≤ L π(1 -K) ϕ -ϕ ε ∞ < LC 2 π(1 -K) ε 3/4 . ( 48
)
In view of ( 38) -( 41), ( 43) and ( 48), we have

ψ ε -ψ L 2 (R) = u εy (., 1) -u y (., 1) L 2 (R)
= h εy (., 1) + k εy (., 1) -h y (., 1) -k y (., 1) L 2 (R)

≤ h εy (., 1) -h y (., 1) L 2 (R) + k εy (., 1) -k y (., 1) L 2 (R)

< LC 2 π(1 -K) ε 3/4 + C 1 ε 1/2 ≤ LC 2 π(1 -K) + C 1 ε 1/2 .
This complete the proof of Lemma 1.

Regularization of problem

We shall first regularize the Problem 2.1 (Subsection 4.1). Using the approximated solution v ε of Problem 2.1, we shall regularize the Problem 2.2 (Subsection 4.2).

Problem 2.1 (Problem (24))

We have the following result Theorem 1

Let ϕ ∈ L 1 (R) ∩ L ∞ (R). Suppose ϕ(ζ)e |ζ| ∈ L 2 (R), ψ(ζ)e |ζ| ∈ L 2 (R). Then Problem (11) -(13) has a unique solution v 0 ∈ L 2 (R × (0, 1)).
Proof From (24) and the inequality e |x| -1 |x| ≤ e |x| we have

|ℵ(ζ, y)| ≤ e |ζ| | ϕ(ζ)| + e |ζ| ψ(ζ) ∈ L 2 (R), for all 0 ≤ y ≤ 1
where ℵ is as in ( 24). Hence Problem ( 11) -(13) has a (unique) solution v 0 ∈ L 2 (R × (0, 1)).

The proof is completed.

Theorem 2 Let assumptions in Theorem 1 hold. For every 0 < ε < e -3 , let

ϕ ε ∈ L 2 (R) be measured data such that ϕ ε -ϕ L 2 (R) < ε.
If, in addition, the assumptions of Lemma 1 hold and we have |ζ|

ϕ(ζ)e |ζ| ∈ L 2 (R), |ζ| ψ(ζ)e |ζ| ∈ L 2 (R).
Then, from ϕ ε , we can construct a regularized solution

v ε such that v 0 -v ε 2 < D ln 1 ε -1 ,
where . 2 is the norm in L 2 (R × (0, 1)) and D is a positive constant independent of ε.

Proof

We recall that ϕ ε is defined in Lemma 1, we get

ϕ ε -ϕ 2 L 2 (R) = |ζ|<ε -1/2 | ϕ ε (ζ) -ϕ(ζ)| 2 dζ + |ζ|>ε -1/2 | ϕ(ζ)| 2 dζ < ε 2 + |ζ|>ε -1/2 | ϕ(ζ)| 2 e 2|ζ| |ζ| 4 dζ < ε 2 + ε 2 ϕ(ζ)e |ζ| 2 L 2 (R) = ε 2 C 2 3 where C 3 = 1 + ϕ(ζ)e |ζ| 2 L 2 (R) . Therefore ϕ ε -ϕ L 2 (R) < C 3 ε. Put φ ε (x) = 1 √ 2π 1 6 ln 1 ε -1 6 ln 1 ε ϕ(ζ)e iζx dζ.
We have

φ ε (ζ) = ϕ ε (ζ) |ζ| < 1 6 ln 1 ε 0 |ζ| ≥ 1 6 ln 1 ε . It follows that φ ε (ζ) -ϕ(ζ) e |ζ| 2 L 2 (R) = 1 6 ln 1 ε -1 6 ln 1 ε ϕ ε (ζ) -ϕ(ζ) 2 e 2|ζ| dζ + |ζ|> 1 6 ln 1 ε | ϕ(ζ)| 2 e 2|ζ| dζ ≤ e 1 3 ln 1 ε ϕ ε -ϕ 2 L 2 (R) + 1 1 6 ln 1 ε 2 ζ ϕ(ζ)e |ζ| 2 L 2 (R) < C 2 3 e 1 3 ln 1 ε ε 2 + 1 1 6 ln 1 ε 2 ζ ϕ(ζ)e |ζ| 2 L 2 (R) < C 2 3 ε 5/3 + 36 ln 2 1 ε ζ ϕ(ζ)e |ζ| 2 L 2 (R) .
Using the inequality ε 5/3 < 1 ln 2 1 ε as ε < e -3 , we get

φ ε (ζ) -ϕ(ζ) e |ζ| 2 L 2 (R) < C 2 4 1 ln 2 1 ε , where C 4 = C 2 3 + 36 ζ ϕ(ζ)e |ζ| 2 L 2 (R) . ( 49
)
Using lemma 1, there exists

ψ ε ∈ L 2 (R) such that ψ ε -ψ L 2 (R) < Cε 1/2 . Put Ψ ε (x) = 1 √ 2π 1 6 ln 1 ε -1 6 ln 1 ε Ψ ε (ζ)e iζx dζ.
We have

Ψ ε (ζ) = ψ ε (ζ) |ζ| < 1 6 ln 1 ε 0 |ζ| ≥ 1 6 ln 1 ε .
Similarly, using the inequality

ε 2/3 < 1 ln 2 1 ε as ε < e -3 , we get Ψ ε (ζ) -ψ(ζ) e |ζ| 2 L 2 (R) < C 2 5 1 ln 2 1 ε , where C 5 = C 2 + 36 ζ ψ(ζ)e |ζ| 2 L 2 (R) . ( 50
)
We put

ℵ ε (ζ, y) = 1 2 φ ε (ζ) e (1-y)|ζ| + e (y-1)|ζ| - 1 2 |ζ| Ψ ε (ζ) e (1-y)|ζ| -e (y-1)|ζ| and v ε (x, y) = 1 √ 2π +∞ -∞ ℵ ε (ζ, y)e iζx dζ. (51)
From ( 24), (51), we get

v ε -v 0 2 = ℵ ε -ℵ 2 ≤ 1 2   ϕ(ζ) -φ ε (ζ) e (1-y)|ζ| + e (y-1)|ζ| 2 + ψ(ζ) -Ψ ε (ζ) |ζ| e (1-y)|ζ| -e (y-1)|ζ| 2   ≤ ( ϕ(ζ) -φ ε (ζ))e |ζ| 2 + ( ψ(ζ) -Ψ ε (ζ))e |ζ| 2 < (C 4 + C 5 ) ln 1 ε -1 . Hence v 0 -v ε 2 < D ln 1 ε -1 .
The proof is completed. 4.2 Problem 2.2 (Problem (34)) Theorem 3 Let assumptions in Theorem 1 and Theorem 2 hold. Let v 0 ∈ L 2 (R × (0, 1)) be an exact solution of (11) -(13) and v ε ∈ L 2 (R × (0, 1)) be the regularized solution of (11) -(13).

Suppose that f satisfies the conditions (27

) and |p(ξ, η)| ≤ k ∀(ξ, η) ∈ R × (0, 1).
Assume in addition that the exact solution u 0 = v 0 +w 0 ∈ L 2 (R×(0, 1)) of ( 10) satisfying

e 3|ζ| f (η,v 0 ,w 0 ) (ζ) ∈ L 2 (R × (0, 1)).
Then there exists a regularized solutions w ε and

u ε = v ε + w ε of (10) such that w ε -w 0 2 < C ln 1 ε -1/2 and u ε -u 0 2 < E ln 1 ε -1/2
where C and E independent of ε. Proof From (33), we have

w 0(y) (ζ) = 1 2 1 0 1 |ζ| e (η-y)|ζ| -e -|y-η||ζ| f (η,w 0 ,v 0 ) (ζ)dη = 1 2 1 y 1 |ζ| e (η-y)|ζ| -e (y-η)|ζ| f (η,w 0 ,v 0 ) (ζ)dη.
We put

T w (α) (x, y) = 1 2 √ 2π 1 y α -α 1 |ζ| e (η-y)|ζ| -e (y-η)|ζ| f (η,w (α) ,vε) (ζ)e iζx dζdη (52)
We shall prove

T m (w (α) (., y)) -T m (w 1(α) (., y)) 2 L 2 (R) ≤ k 2 e 2α m (1 -y) m m! w (α) -w 1(α) 2 (53)
for every α > 0, y ∈ (0, 1), m ≥ 1 and w (α) , w

1(α) ∈ C([0, 1]; L 2 (R)), where | . | denotes sup-norm in C([0, 1]; L 2 (R)).
We shall prove (51) by induction. We define

χ [-α,α] (ζ) = 1 |ζ| ≤ α 0 |ζ| > α .
For m = 1, noting that

T w (α) (ζ, y) = 1 2 χ [-α,α] (ζ) 1 y 1 |ζ| e (η-y)|ζ| -e (y-η)|ζ| f (η,w (α) ,vε) (ζ)dη,
we have

T w (α) (., y) -T w 1 (α) (., y) 2 L 2 (R) = = T w (α) (., y) -T w 1 (α) (., y) 2 L 2 (R) = 1 4 1 y χ [-α,α] (ζ) e (η-y)|ζ| -e (y-η)|ζ| |ζ| f (η,w (α) ,vε) (ζ) -f (η,w 1(α) ,vε) (ζ) dη 2 L 2 (R) ≤ +∞ -∞ 1 y χ [-α,α] (ζ)e (η-y)|ζ| f (η,w (α) ,vε) (ζ) -f (η,w 1(α) ,vε) (ζ) dη 2 dζ ≤ +∞ -∞ 1 y e (η-y)α f (η,w (α) ,vε) (ζ) -f (η,w 1(α) ,vε) (ζ) dη 2 dζ ≤ e 2α (1 -y) +∞ -∞ 1 y f (η,w (α) ,vε) (ζ) -f (η,w 1(α) ,vε) (ζ) 2 dηdζ ≤ k 2 e 2α (1 -y) 1 y +∞ -∞ w (α) (x, η) -w 1(α) (x, η) 2 dxdη ≤ e 2α (1 -y)k 2 1 y w (α) (., η) -w 1(α) (., η) 2 L 2 (R) dη ≤ e 2α (1 -y)k 2 w (α) -w 1(α) 2 .
Therefore (53) holds.

Suppose that (53) holds for m = j. We shall prove that (51) holds for m = j + 1. We have

T j+1 w (α) (., y) -T j+1 w 1(α) (., y) 2 L 2 (R) = = T T j w (α) (., y) -T T j w 1(α) (., y) 2 L 2 (R) ≤ k 2 e 2α (1 -y) 1 y T j w (α) (., η) -T j w 1(α) (., η) 2 L 2 (R) dη ≤ k 2 e 2α (1 -y) 1 y k 2 e 2α j (1 -η) j j! w (α) -w 1(α) 2 dη ≤ k 2 e 2α j+1 (1 -y) j+1 (j + 1)! w (α) -w 1(α) 2 .
Therefore, by the induction principle, (53) holds for every m. From ( 53), we get

T m w (α) -T m w 1(α) 2 ≤ k 2 e 2α m 1 m! w (α) -w 1(α) 2 . Since lim m→∞ (k 2 e 2α )
m 1 m! = 0, there exists a positive integer number m 0 such that

k 2 e 2α m 0 1 m 0 ! < 1.
Hence T m 0 is a contraction in C ([0, 1]; L2 (R)). It follows that the equation T m 0 (w (α) ) = w (α) has a unique solution w (α) ∈ C ([0, 1]; L 2 (R)).

We claim that T (w (α) ) = w (α) . In fact, one has T (T m 0 (w (α) )) = T (w (α) ). Hence T m 0 (T (w (α) )) = T (w (α) ). By the uniqueness of the fixed point of T m 0 , one has T (w (α) ) = w (α) , i.e., the equation T (w (α) ) = w (α) has a unique solution w (α) ∈ C ([0, 1]; L 2 (R)).

We have

4 w 0(y) -w (α) (., y) 2 L 2 (R) = α -α 1 y [e (η-y)|ζ| -e (y-η)|ζ| ] |ζ| f (η,w 0 ,v 0 ) (ζ) -f (η,w (α) ,vε) (ζ) dη 2 dζ + |ζ|>α 1 y [e (η-y)|ζ| -e (y-η)|ζ| ] |ζ| f (η,w 0 ,v 0 ) (ζ)dη ≤ 4 α -α 1 y e (η-y)|ζ| f (η,w 0 ,v 0 ) (ζ) -f (η,w (α) ,vε) (ζ) dη 2 dζ + 4 |ζ|>α 1 y e (η-y)|ζ| f (η,w 0 ,v 0 ) (ζ)dη 2 dζ ≤ 4 α -α 1 y e (η-y)α f (η,w 0 ,v 0 ) (ζ) -f (η,w (α) ,vε) (ζ) dη 2 dζ + 4e -2yα |ζ|>α 1 y e (η-y)|ζ|+yα f (η,w 0 ,v 0 ) (ζ)dη 2 dζ ≤ 4e -2yα    α -α 1 y e 2ηα f (η,w 0 ,v 0 ) (ζ) -f (η,w (α) ,vε) (ζ) 2 dηdζ + |ζ|>α 1 y e (η-y)|ζ|+yα f (η,w 0 ,v 0 ) (ζ)dη 2 dζ      ≤ 4e -2yα    1 y e 2ηα   +∞ -∞ f (η,w 0 ,v 0 ) (ξ) -f (η,w (α) ,vε) (ξ) 2 dξ   dη + |ζ|>α 1 y e |ζ|+α f (η,w 0 ,v 0 ) (ζ) dη 2 dζ      ≤ 4e -2yα    1 y e 2ηα +∞ -∞ k 2 v 0 (ξ, η) -v ε (ξ, η) + w 0 (ξ, η) -w (α) (ξ, η) 2 dξdη + |ζ|>α e 2(|ζ|+α) 1 y f (η,w 0 ,v 0 ) (ζ) 2 dηdζ      ≤ 4e -2yα    2k 2 1 y e 2ηα w 0 (., η) -w (α) (., η) 2 L 2 (R) dη + 2 1 0 +∞ -∞ e 2α k 2 |v 0 (ξ, η) -v ε (ξ, η)| 2 dξdη + |ζ|>α e 2(|ζ|+α) 1 y f (η,w 0 ,v 0 ) (ζ) 2 dηdζ      ≤ 4e -2yα    2k 2 1 y e 2ηα w 0 (., η) -w (α) (., η) 2 L 2 (R) dη + 2k 2 e 2α v 0 -v ε 2 2 + |ζ|>α e 2(|ζ|+α) 1 y f (η,w 0 ,v 0 ) (ζ) 2 dηdζ      .
We put

M 1 = 2k 2 e 2α v 0 -v ε 2 2 and M 2 = |ζ|>α e 2(|ζ|+α) 1 y f (η,w 0 ,v 0 ) (ζ) 2 dηdζ.
Therefore

e 2yα w o(y) -w (α) (., y) 2 L 2 (R) ≤ 2k 2 1 y e 2ηα w o (., η) -w (α) (., η) 2 L 2 (R) dη + M 1 + M 2 .
Using Gronwall's inequality, we have

e 2yα w o (., y) -w (α) (., y) 2 L 2 (R) ≤ (M 1 + M 2 ) e 2k 2 (1-y) , hence w o (., y) -w (α) (., y) 2 L 2 (R) ≤ e -2αy (M 1 + M 2 )e 2k 2 (1-y) . ( 54
)
From Theorem 2, we get

e -2αy M 1 = 2e 2(1-y)α k 2 v 0 -v ε 2 2 < 2D 2 e 2(1-y)α k 2 ln 1 ε -2 . ( 55
)
We have

e -2αy M 2 = e -2αy |ζ|>α e 2(|ζ|+α) 1 y f (η,v 0 ,w 0 ) (ζ) 2 dηdζ 20 ≤ e -2αy |ζ|>α 1 0 e 6|ζ| e 2α f (η,v 0 ,w 0 ) (ζ) 2 dηdζ ≤ e -2α(y+1) e 3|ζ| f (η,v 0 ,w 0 ) (ζ) 2 2 . ( 56
)
From ( 54)-( 56) and choosing α = 1 2(1-y) ln ln 1 ε , we get

w o (., y) -w (α) (., y) 2 L 2 (R) < e 2k 2 2k 2 D 2 ln 1 ε -1 + e 3|ζ| f (η,v 0 ,w 0 ) (ζ) 2 2 ln 1 ε -1 < C 2 ln 1 ε -1
.

Therefore

w o -w (α) 2 < C ln 1 ε -1/2
.

Denoting w ε = w (α) and u ε = v ε + w ε , we get

u 0 -u ε 2 = w 0 + v 0 -v ε -w ε 2 ≤ w 0 -w (α) 2 + v 0 -v ε 2 ≤ C ln 1 ε -1/2 + D ln 1 ε -1 ≤ (C + D) ln 1 ε -1/2
. This completes the proof of Theorem 3.

NUMERICAL RESULTS

Let the problem ∆u = f (x, y, u), x ∈ R, y > 1 u(x, 1) = ϕ(x) (57) with f (x, y, u) = arctan(|u| + x 2 + y 2 ) + g(x, y) g(x, y) = 4Cx (x 2 + 4) 3 + -2C (x 2 + 4) 2 -arctan(

C x 2 + 4 + x 2 + y 2 ), C = 4 √ 2π ϕ(x) = C 4 + x 2
The exact solution of (57) is u(x, y) = C

x 2 + 4 , C = 4/ √ 2π and we solve numerically this problem for x ∈ R, y > 1 by using the iterative sequence defined by whose exact solution is v(x, y) = 1 2 √ 2π 3 -y x 2 + (3 -y) 2 + 1 + y x 2 + (1 + y) 2 , we calculate the regularized solution v ε (x, y) of ( 58) for ε = 10 -2 by the formula (51).

u (n+1) (x, y) = -
Finally for the third problem ∆w = f (x, y, v + w), x ∈ R, 0 < y < 1 w(x, 1) = 0, w y (x, 1) = 0 (59) the regularized solution w ε (x, y) of the problem ( 59) is calculated from its definition (52). So in Fig. 2 we have drawed the regularized solution of the problem 2 i.e.

  Fig.1: approximated solution For the second problem ∆v = 0, x ∈ R, 0 < y < 1 v(x, 1) = ϕ(x), v y (x, 1) = 0 (58)

  Fig.2: regularized solution For comparison Fig.3 gives the exact solution on the interval [-4, 4] × [0, 4].

  Fig.3: exact solution

dζ