On the regularization of the orst kind integral

equation with analytical kernel of logarithmic type

Gottfried Bruckner
Weierstrafl-Institut
f?r
Angewandte Analysis und Stochastik
Mohrenstr. 39
D-10117 Berlin
Germany
bruckner@wias-berlin.de

October 7, 1999

1991 Mathematics Subject Classiocation. 65J20 .

Keywords. regularization by discretzation, selfregularization, projection methods, Tikhonov regularization, severely ill-posed, integral equation of the orst kind, logarithmic convergence rate.

Abstract

We study regularization methods for the integral equation of the orst kind with analytical kernel of logarithmic type. The problem is severely ill-posed. In [1] a logarithmic type convergence rate for the Tikhonov regularized solution was proved. Here we are concerned with numerical aspects of the solution. First we consider the selfregularization of the problem by using projection methods in the sense of [9].Then we will see that the Tikhonov regularization of such methods is in accordance with a discretized version of the Tikhonov regularized solution in [1]. Finally, we describe numerical experiments being in a good agreement with the theoretical results.

1 Introduction

Many inverse problems from applications, such as tomography [8], geophysics [7], nondestructive detection [6], inverse contact problems [3], give rise to integral equations of the orst kind with analytic kernels. In [2], [4], for certain integral equation of the orst kind with analytic kernel, a conditional stability could be proved, provided some a-priori information about the solution is known. Since orst kind integral equations with analytic kernels are severely ill-posed problems, their numerical solution is extremely diOEcult.

Let us consider the integral equation with logarithmic kernel
Z 1

0 log(x ? t)f(t)dt = g(x); x 2 [2; 3]: (1.1)

Since [0; 1] \ [2; 3] = ;, the kernel is analytic with respect to x; t. The integral equation (1.1) is severely ill>=posed in Hadamard's sense. The purpose of this paper is to study fully discretized regularization methods for this problem.

In Section 2 we are engaged with projection methods in the sense of [9]. We describe the methods of least squares, dual least squares and collocation and investigate their properties. In each case the discretized problem is equivalent to a system of linear equations. As the matrix is ill-conditioned in the general case, for its numerical solution a combination with an additional regularization procedure is necessary.

The Section 3 is devoted to a discretized version of the Tikhonov regularized solution in the sense of [1]. This discretized version can be considered as the least squares method combined with additional Tikhonov regularization. In this section the results of [1] are used. The numerical experiments concern the two kinds of regularity assumptions: First, the solution is supposed to be H10 on [0; 1] and second, the solution is supposed to be H1 in a neighborhood of one point. In the orst case the L2-convergence of the approximating sequence to the solution is investigated, while in the second case the pointwise convergence is studied locally. Moreover, near a discontinuity point of the solution the sequence of approximated regularized solutions is growing unboundedly in the L2-sense. We perform three experiments with synthetic data, conorming the theoretical results of [1].

2 Selfregularization by projection methods.

In this paper the problem
Af = g; (2.1)

is considered, where the operator A : L2(0; 1) ! L2(2; 3) is deoned as

(Af)(x) =
Z 1

0 log(x ? t)f(t)dt; x 2 [2; 3]: (2.2)

Let A? : L2(2; 3) ! L2(0; 1) be the adjoint operator,

(A?g)(t) =
Z 3

2 log(x ? t)g(x)dx; t 2 [0; 1]:

Projection methods.

First of all we are concerned with the deonition and properties of abstract projection methods studied in [9].

Let X and Y be Hilbert spaces and A a uniquely invertible operator mapping X into Y
with R(A) = Y . Let Y 0 be the space of linear continuous functionals on Y (Y 0 can be identioed with Y ), let further k ? k and (?; ?) be norm and scalar product in X and Y .

Consider onite dimensional subspaces Xn ae X (trial spaces) and Y 0n ae Y 0 (test spaces) and deone the discretized problem:

Find fn 2 Xn such that
(Afn) = (g) (2.3)

holds for all 2 Y 0n.

We assume that (2.3) is uniquely solvable for any g 2 Y and consider its solution as the approximate solution of the operator equation (2.1). In the case where we have instead of the exact right-hand side g only uncertain data gffl, with kg ? gfflk < ffl, at disposal we denote the solution of (2.3) by fn;ffl.

Now, on the lines of [9] let us deone the linear operators Pn : X ! Xn and Qn : Y ! Xn. Let fn be the solution of (2.3) where g = Af holds. Then

Pnf := fn:

As Pnu = u for u 2 Xn the operator Pn is a projector from X onto Xn: Let fn be the solution of (2.3) where g 2 Y holds. Then

Qng := fn:

It is clear that generally
Pn = QnA (2.4) holds.

Proposition 2.1 If Pnf ! f (n ! 1) for each f 2 X, then kPnk <= c:

If kPnk <= c;
dist(f; Xn) := inf
u2Xn kf ? uk ! 0;
then Pnf ! f:

Proof. The orst assertion follows by the theorem of Banach-Steinhaus, the second one by

kf ? fnk = k(I ? Pn)(f ? u)k <= (1 + kPnk)dist(f; Xn): (2.5)

In the case of unexact data, using (2.5) we have the estimate

kf ? fn;fflk <= (1 + kPnk)dist(f; Xn) + kQnkffl: (2.6)

In the ill-posed case kQnk will grow for growing n. To get a reasonable numerical procedure (2.3) we must

(i) be sure that kPnk is bounded,

(ii) estimate kQnk from above,

(iii) choose n depending on the error level ffl such that kQnkffl decreases for growing n with a rate similar to the rate of the orst summand at the right-hand side of (2.6).

Generalized orthoprojectors.

Let S; Z be n-dimensional subspaces of the Hilbert space X,

S = spanfoe1; ? ? ? ; oeng; Z = spanf?1; ? ? ? ; ?ng:

Let
Z? = f?? 2 X; (??; ?) = 0 8? 2 Zg:

Proposition 2.2 If S \ Z? = f0g; then any f 2 X can be uniquely represented as

f = oe + ??; oe 2 S; ?? 2 Z?: (2.7)

Proof. From
(oe; ?) = (f; ?) 8? 2 Z; (2.8)

oe is uniquely determined: Let be oe = Pn1 yioei: Then the linear system
nX

i=1
yi(oei; ?j) = (f; ?j); j = 1; ? ? ? ; n;

is uniquely solvable as its matrix ((oei; ?j)) is invertible because of the assumption.

Now, let us deone the generalized orthoprojector PZS as

PZS f := oe;

where OE is uniquely determined by (2.7). Clearly, the generalized orthoprojector is a projector. Moreover, in the case S = Z

PS := PSS

is the usual orthoprojector to S, where the assumption S \ S? = f0g is trivially fulolled.

Proposition 2.3 Suppose S\Z? = f0g: The generalized orthoprojector has the following properties:
(PZS f; ?) = (f; ?) 8? 2 Z; (2.9)

PZPZS = PZ ; (2.10)

kPZS k <= kP?1
Z k; (2.11)

kf ? PSfk <= kf ? PZSfk:

where in (2.11) the restriction of PZ to S is considered, being uniquely invertible.

Proof. PZoe = 0 means oe 2 Z?; i.e. oe = 0 if oe 2 S: Therefore, the restriction of PZ to S is uniquely invertible, and (2.11) immediately follows from (2.10). The assertions (2.9) and (2.10) are immediate from (2.7).

Example 2.1 Let be Xn = spanf'1; ? ? ? ; 'ng; Yn = spanf 1; ? ? ? ; ng; with the property Xn \ (A?Yn)? = f0g: Consider the projection method (cf.(2.3)): Find fn 2 Xn such that

(Afn; v) = (g; v) 8v 2 Yn: (2.12)

Then Pn = PZS , where S = Xn; Z = A?Yn:

Proof. Write (2.12) as
(fn; A?v) = (f; A?v) 8v 2 Yn:

The assertion then follows from (2.8).

We are going to study more concrete projection methods, where the assumption of Proposition 2.2 (necessary for the unique solvability) is fulolled.

Method of least squares.

Let be Xn = spanf'1; ? ? ? ; 'ng; Yn = AXn:

Find fn 2 Xn such that
(Afn; Au) = (g; Au) 8u 2 Xn: (2.13)

Let fn = Pni=1 xi'i. Then x = (x1; ? ? ? ; xn) can be calculated from the linear system

nX

i=1
xi(A'i; A'j) = (g; A'j); j = 1; ? ? ? ; n: (2.14)

It is clear that (2.14) is uniquely solvable.

Let Pn be the orthoprojector of Y to Yn and denote the restriction of A to Xn by An. The operators Pn; Qn, deoned above, have the following properties.

Proposition 2.4
AQn = Pn; (2.15)

APn = PnA; (2.16)

kQnk <= kA?1
n k; (2.17) Pn = PZS ; S = Xn; Z = A?AXn; (2.18) kAQng ? gk <= kAu ? gk 8u 2 Xn: (2.19)

Proposition 2.5 If R(A?) = X, and dist(f; Xn) ! 0, then Pnf tends to f in the weak topology.

Proof. Using (2.16), for each y 2 Y we have

(f ? Pnf; A?y) = (Af ? APnf; y) = (Af ? PnAf; y) <= kAf ? PnAfkkyk:

As Pn is the orthoprojector of Y to AXn we obtain
kAf ? PnAfk <= kAf ? APXnfk <= ckf ? PXnfk;
where PXn is the orthoprojector of X to Xn. Since kf ?PXnfk ! 0 the proof is complete.

Method of least error.
This method is also called dual method of least squares. Here we choose

Yn = spanf 1; ? ? ? ; ng; Xn = A?Yn:

Find fn 2 Xn, i.e. fn = A?wn; wn 2 Yn such that

(A?wn; A?v) = (g; v) 8v 2 Yn: (2.20)

Let wn = Pni=1 xi i. Then x = (x1; ? ? ? ; xn) can be calculated from the linear system
nX

i=1
xi(A? i; A? j) = (g; j); j = 1; ? ? ? ; n: (2.21)

Again it is clear that (2.21) is uniquely solvable.

Let Pn be the orthoprojector of X to Xn and f 2 X arbitrary. Then Pn has the following properties:

Proposition 2.6
Pn = Pn; (2.22) kPnf ? fk <= ku ? fk 8u 2 Xn; (2.23) kPnk <= c: (2.24)

Collocation method.
Here we must suppose that X and Y are function spaces. Let
X = L2(0; 1); Y = L2(2; 3):
Suppose further, the data g is taken from C[2; 3]. Given n collocation points
oj 2 (2; 3); j = 1; ? ? ? ; n;
let be

Y 0n := spanfffi1; ? ? ? ; ffing;

where ffij is the point evaluation at oj,

ffijg = g(oj); j = 1; ? ? ? ; n:

Choose
Xn = spanf'1; ? ? ? ; 'ng:
and ond fn 2 Xn such that

(Afn)(oj) = g(oj); j = 1; ? ? ? ; n: (2.25)

Then, setting fn = Pni=1 xi'i, the vector x = (x1; ? ? ? ; xn) is to be calculated from the linear system nX

i=1

xi(A'i)(oj) = g(oj); j = 1; ? ? ? ; n: (2.26)

Proposition 2.7 (i)The system (2.26) is uniquely solvable if the following is true:

If u 2 Xn; (Au)(oj) = 0; j = 1; :::; n then u = 0 identically.

(ii)Let (2.26) be uniquely solvable. For f 2 X we obtain fn = Pnf ! f if the following is true:

If for the sequence ?n 2 X holds (A?n)(oj) = 0; j = 1; ? ? ? ; n then ?n ! 0.

Proof. (i)The matrix ((A'i)(oj)) has full rank if its rows are linearly independent, i.e. Pni=1 >=i(A'i)(oj) = 0; j = 1; :::; n implies >=i = 0; i = 1; :::; n. This means that (Au)(oj) = 0; j = 1; :::; n, where u := Pni=1 >=i'i.
(ii)Take ?n = f ? fn, where Adn(oj) = 0 follows from (2.25).
Now, using the concrete form (2.2) of the operator A we see that

(Af)(oj) = (f; fij); j = 1; :::; n; (2.27)

where for t 2 [0; 1]

fij(t) = log(oj ? t); j = 1; :::; n:

It can easily be proved that fij; j = 1; :::; n; for dioeerent oj are linearly independent. Denote

Bn = spanffi1; :::; fing:

Proposition 2.8 Assume Xn \ B?n = f0g. Then

(i)The matrix of (2.26)
((A'i)(oj)) = (('i; fij)) (2.28)

has full rank.

(ii) There holds
Pn = PBn
Xn; (2.29)

kPnk <= kP?1
Bnk; (2.30) where in (2.30) PBn is the restriction to Xn of the orthoprojector of X to Bn.

Proof. (i) is clear from the assumption. (ii) follows writing (2.26) equivalently as
X xi('i; fij) = (f; fij); j = 1; :::; n;

or (fn; fi) = (f; fi)8fi 2 Bn:

Estimation of kQnk:
In the case of the least squares method let
Xn = spanfe1; :::; eng;
where ei = O[(i?1)=n;i=n] are the characteristic functions. Clearly

(ei; ej) = ffiij=n:

For g 2 Y let Qng = P xiei: Then kQngk = jxj=pn: From (2.14) we obtain

Mx = m; M = ((Aei; Aej)); m = ((g; Aej)):

Then

kQngk = jxj=pn = jM?1mj=pn;

jmj2 = X
j
(g; Aej)2 <= c2kgk2:

We obtain
kQnk <= cjM?1j:

In the case of the dual least squares method (method of least error) let
Yn = spanfe01; :::; e0ng;
where e0i = O[2+(i?1)=n;2+i=n] again are the characteristic functions. Clearly
(e0i; e0j ) = ffiij=n:
For g 2 Y let Qng = P xiA?e0i: Then Qng = A?Rng: From (2.20) we obtain (AQng; Rng) = (g; Rng):
(2.21) gives
Lx = l; L = ((A?e0i; A?e0j)); l = ((g; e0j)):

Then

kQngk2 <= kRngkkgk = kgkjxj=pn <= jL?1jkgk2=pn:

We obtain

kQnk <=

jL?1j

pn

!1=2
:

Finally, in the case of the collocation method let
Xn = spanfe1; :::; eng:
For g 2 C[2; 3] let Qng = P xiei: From (2.26) we obtain
Kx = k; K = (Aei(oj)); k = (g(oj)):
Then
kQngk2 = jxj2=n <= jK?1j2jkj2=n = jK?1j2kgk2;
approximately, as jkj2=n = P g(oj)2=n ss kgk2: We obtain

kQnk <= jK?1j:

3 Tikhonov regularization. A numerical treatment

Here we are engaged with the numerical solution of

Af0 = g;

where A : L2(0; 1) ! L2(2; 3) is deoned in (2.2) and only an approximation gffi of g is given,
kg ? gffikL2(2;3) <= ffi:

We will ond the numerical solution by discretization combined with Tikhonov regularization. To this end we are going to cite some results from [1]. Then we will give an overview over numerical experiments conorming the theoretical results.

Crucial for the numerical approximation is the a priori assumption on the solution f0: Let us start our considerations with the
A priori assumption: f0 2 H10(0; 1):

Let ffi > 0 be oxed and f 2 H10(0; 1): Consider the functional

Fff(f) = kAf ? gffik2L2(2;3) + ffkfk2H10(0;1); (3.1)

where ff > 0: Deone
fi = inf
f2H10 (0;1) Fff(f );
and the regularized solution f ffiff such that

Fff(f ffiff) <= fi + ffi2:

Proposition 3.1 Suppose f0 2 H10(0; 1); ff = ffi2. Then for ffi ! 0 the regularized solution converges to f0 and
kf ffiff ? f0kL2(0;1) <= C1 1
j log 1ffi j ;
where C1 > 0 is a constant which depends on f0.

Computation of a regularized solution.
We assume f0 2 H10 :
Consider in the interval [0,1] the equidistant discretization

ti = i=n; i = 1; ? ? ? ; n ? 1:

Deone
Xn = spanf?i; i = 1; ? ? ? ; n ? 1g;

where ?i is linear and continuous with ?i(tj) = 1 for j = i and = 0 for j 6= i; i = 1; ? ? ? ; n ? 1:

It is known (cf.e.g. [5]), that for ' 2 H10 ; 'n = Pn?1
i=1 '(ti)?i will converge to ' for n ! 1: If ' 2 H1+?
k' ? 'nkH10 <= c ? n??k'kH1+? :

Consider (3.1).
From the identity

Fff ( sf + (1 ? s)g) = sFff(f) + (1 ? s)Fff(g)
?s(1 ? s)fkAf ? Agk2L2 + ffkf ? gk2H10g;

Fff is strongly convex, locally Lipschitz continuous and weakly lower semicontinuous.

There is a unique f ? 2 H10 with

Fff(f ?) = inf
f2H10
Fff(f);

and there is a unique f ?n 2 Xn with

Fff(f ?n) = inf
fn2Xn Fff(fn):
This element f ?n can for n > n0 serve as a regularized solution f ffi; since

Fff(f ?n) ?! Fff(f ?) (n ! 1):

This is clear by going to the limit in

Fff(f ?) <= Fff(f ?n) <= Fff(fn);

where the sequence fn approximates f ?:

Calculation of f ?n: To minimize

min
f2XnfkAf ? gffik2L2 + ff(kfk2L2 + kf 0k2L2)g

let us consider the equivalent problem: Set f = Pn?1
i=1 xi?i; and solve

min
x2IRn?1f< Wx + y; x > + bg;

where x = (xi); W = ((A?i; A?j) + ff((?i; ?j) + (?0i; ?0j)));

y = (?2(A?i; gffi)); b = (gffi; gffi) and < ?; ? > is the scalar product in IRn?1: From the necessary (and suOEcient) condition for a minimum we get

f ?n =

n?1
X

i=1
x0i?i;

where x0 = W?1v; v = ?y=2:
Local regularization of a discontinuous solution

Let x0 2 (0; 1) and consider the neighborhood Or = Or(x0):
For ff; ffi > 0 oxed and f 2 L2(0; 1) \ H1(Or) deone

Gff(f) := kAf ? gffik2L2(2;3) + ff(kfk2L2(0;1) + kfk2H1(Or)): (3.2)

fi1 := inf
f2L2(0;1)\H1(Or) Gff(f):
and the locally regularized solution f ffiff such that

Gff(f ffiff) <= fi1 + ffi2:

Proposition 3.2 Suppose ff = ffi2 and

f0 2 L2(0; 1) \ H1(Or):

Then a locally regularized solution converges for ffi ! 0 to f0 in some neighborhood of x0; and
jf ffiff(x) ? f0(x)j <= C1 1
j log 1ffi jfl ; jx ? x0j <= r1 < r;
where C1 > 0 depends on f0, r and r1.

Now, let us consider discontinuity points of the solution.

Proposition 3.3 Suppose that the exact solution f0 is a piecewise smooth function and x0 is a discontinuity point such that
f0 2 C2((x0 ? ffl; x0), f0 2 C2(x0; x0 + ffl) and f0(x0 + 0) 6= f0(x0 ? 0). Let f ffiff be a locally regularized solution. Then for ff = ffi2

lim
ffi!0 kf ffiffkH1(Or(x0)) = 1:

Proposition 3.4 Let Or be an open subinterval of [0; 1]. There is a discontinuity point of the solution f0 in Or if and only if for a locally regularized solution f ffiff holds for ff = ffi2: kf ffiffkH1(Or) is unbounded for ffi ! 0.

For proofs of Propositions 3.1 to 3.4 we refer to [1].

Our following numerical experiments concern 3 cases:

(i) The reconstruction of f0 2 H10(0; 1), where the approximating sequence converges in the sense of Proposition 3.1.

(ii) The reconstruction of f0 2 L2(0; 1) \ H1(O); where O is an open subinterval of (0; 1) and the approximating sequence converges inside O pointwise in the sense of Proposition 3.2.

(iii) Let O0 be such that f0 has a discontinuity point inside O0. Then the approximating sequence will grow according to Proposition 3.3.

The approximating sequence f(n; ff; ffi) will depend on the discretization number n, the regularization parameter ff and the noise level ffi: It belongs to a onite-dimensional space Un, that is a subspace of H10(0; 1) in the case (i) and of L2(0; 1) \ H1(O) in the cases (ii) and (iii). In the case (iii) the interval O0 is not contained in O.
Now, let n be a oxed natural number and denote

f ffiff = f (n; ff; ffi):

Let O ae [0; 1] be such that its boundary points are points of the equidistant discretization ti = i=n; i = 0; :::; n;
O = (i0=n; i1=n); i0 < i1:

Let [0; 1] = [?i; ?i = [(i ? 1)=n; i=n] and deone as basis functions OEj; j = 1; :::; n + 1; the functions

O?i if ?i \ O = ;; ?i if i=n 2 O; ??i0; ?+i1 ,
and numbering them according to the position of their support in [0,1]. Here O?i is the characteristic function of ?i and

??i (t) =
ae?nt + i + 1 if t 2 ?i+1

0 else ; ?+i (t) =

ae nt ? i + 1 if t 2 ?i

0 else ;

?i(t) =

8<

:

?+i (t) if t 2 ?i
??i (t) if t 2 ?i+1
0 else

are the usual hat-functions. Now deone

Un =
ae spanfOE2; :::; OEng if O = (0; 1)

spanfOE1; :::; OEn+1g if O ae (0; 1) :

The solution f ffiff of the minimum problems

inf
f2UnfkAf ? gffikL2(2;3) + ffkfk2H1g

or
inf

f2UnfkAf ? gffikL2(2;3) + ff

?

kfk2L2(0;1) + kfk2H1(O)
?
g

if f0 2 H10 (0; 1) (case (i)) or f0 2 L2(0; 1) \ H1(O) (case (ii)), respectively, is gained by f ffiff = P xiOEi; where x = (xi) is the solution of the linear system

Wx = u;

W = ((AOEi; AOEj) + fff(OEi; OEj) + (OE0i; OE0j)g) in case (i),
W = ((AOEi; AOEj) + fff(OEi; OEj) + >=ijg) in case (ii), where >=ij = (OEi; OEj) + (OE0i; OE0j) if both OEi; OEj have support in O, and = 0 if not. Moreover,
u = ((AOEi; gffi)):

The scalar products in L2(2; 3) are calculated by using Simpson's rule in an equidistant discretization si; i = 1; :::; m of the interval [2,3]. The data gffi are simulated in the following way.
Let f0 2 L2(0; 1) be given. Deone

gffi(si) = (Af0)(si) + ffi ? z(si); i = 1; :::; m;

where z(si) is a random number,jz(si)j <= 1:

The calculation was performed by using the LAPACK FORTRAN program system.

Let us descibe the numerical experiments. We put always ff = ffi2; n = 50; m = 200:
Experiment 1 (case (i)). Here f0 was taken linear with the properties f0(0:6) = 1; f0(0) = f0(1) = 0:
Experiment 2 (case (ii)). Here f0 = 1 in the interval (0.1,0.6) and f0 = 0 else. We set O = (0:2; 0:4) and calculated at the point t1 = 0:35 2 O:
Experiment 3 (case(iii)). Here we took f0 as in Experiment 2 and O0 = (0:5; 0:7):

The results are given in Table 3.1.

Table 3.1

ffi 10?1 10?2 10?3 10?4 10?5 10?6 10?7
1 kf ffiff ? f0kL2 0.290 0.285 0.283 0.240 0.051 0.049 0.037
2 j(f ffiff ? f0)(t1)j 0.38 0.300 0.28 0.25 0.23 0.019 0.016
3 kf ffiffkH1(O0) 0.43 0.77 0.78 0.74 0.86 2.31 2.29

References

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