On the Regularization Method for Solving Ill-Posed Problems with Unbounded Operators ()
1. Introduction
Let
be a linear, closed, densely defined unbounded operator, where X and Y are Hilbert spaces. Consider the equation
(1)
Problem-solving solution of Equation (1) is called ill-posed [1] if A is not boundedly invertible. This may happen if the null space
is not trivial, i.e. A is not injective, or if A is injective but
is unbounded, i.e. the range of A,
is not closed [2].
If
, problem-solving stable solution of Equation (1) has been extensively studied in the literature in detail ( [2] [3] [4] [5] [6] and references therein).
If
, the noisy data, are given
(2)
is a stable approximation to the unique minimal norn solution to Equation (1) was constructed by several methods (variational regularization, quasi solution, iterative regularization, ... [2] [3] [4] [5] [6] and references therein).
If A is a linear, closed, densely defined unbounded operator, problem (1) has been some recent research [2] [7] [8] [9] [10] [11], however, there are still many open problems such as parameter choice rules of regularization method with the linear closed, densely defined unbounded operator
.
Our aim is to study problem-solving stable approximation solution of Equation (1) when operator A is a linear, closed, and densely defined from space Hilbert X into space Hilbert Y. We shall present the regularization method for solving the problem (1), we shall present a priori and a posteriori parameter choice rules of regularization; at the same time give an application to the weak derivative operator equation.
The paper structure consists of 3 sections: Section 1 the introduction briefly summarizes the recent research results and come up with the problem that needs to be studied; Section 2 presents some main results; Section 3 presents an application of this method.
2. Some Main Results
Lemma 1. [2] Let
be a linear, closed, densely defined operator, where X and Y are Hilbert spaces, then
1) the operators
and
are densely defined, self-adjoint;
2)
is closed, densely defined and
;
3) the operators
,
are both defined on all of X and are bounded,
. Also,
is self-adjoint;
4) the operator
is bounded and self-adjoint and
is bounded.
Lemma 2. Let
be a linear, closed, densely defined operator, where X and Y are Hilbert spaces. If
,
then y is unique.
Proof. Suppose
, and
satisfy
,
, and
,
then
. Thus
. There exits
such that
imply
. Thus
, it follows that
.
Theorem 1. For any
, the problem
(3)
has a unique solution
, where
is the identity operator on Y.
Proof. Consider the equation
(4)
which is uniquely solvable
(Lemma 1). Let
then
or
We have
(5)
for any
. If
, then
(6)
Thus Equation (6) implies
(7)
and
if and only if
, so
is the unique minimier of
.
Theorem 1 is proved.
Theorem 2. If
,
then
(8)
Proof. It follows from Lemma 2, y is unique. Write Equation (4) as
. Apply
, which is possible because
, we obtain
(9)
Multiply Equation (9) by
, we obtain
or
(10)
Since
this implies
so
Therefore one may assume (taking a subsequence) that
weakly converges to an element z,
, as
.
It follows from Equation (10) that
We shall prove that
.
Let
run through the set such that
is dense in
, where
. Note that
, where
. Because of the formulas
the
is dense in X, and the set
is dense in
.
Multiply the equation
by
and pass to the limit
. We obtain
We have assume
. If
, then
and
, so
.
One may always assume that
because
, where
is the orthogonal projection of
onto
.
Thus, we have
,
. Thus implies
.
For convenience for the reader we prove this claim. Since
, one gets
. The inequality
implies
. Therefore
. This and the weakly converge
imply strong convergence
Theorem 2 is proved.
Theorem 3. If
,
,
and
(11)
then there exists a unique global minimier
to (11) and
, where
and
is properly chosen, in particular
.
Proof. It follows from Lemma 2, y is unique. The existence and uniqueness of the minimizer
of
follows from Theorem 1 and
. We have
By Theorem 2,
, as
.
Let us estimate
By the polar decomposition theorem [12], one has
, where U is a partial isometry, so
. One has,
where the spectral representation for Q was used.
Thus
(12)
For a fixed small
, choose
which minimizes the right side of Equation (12). Then
and
Theorem 3 is proved.
Remark 1. We can also choose
, with any
and
. The constant c can be arbitrary.
We can also choose
by a descrepancy principle. For example, consider the equation for finding
:
We assume that
.
That is the content of the following theorem.
Theorem 4. The equation
(13)
has a unique solution
,
, and if
, then
.
Proof. Let us prove that Equation (13) has a unique root
,
. Indeed, using the spectial theorem [12], one gets
where
is the resolution of the identity of Q.
One has
, and
, where
is the orthoprojector onto the subspace
.
Since
and
, it follows that
, so
. The function
for a fixed
is a continuous strictly increasing function of
on
. Therefore there exists a unique
which solves Equation (13) if
and
. Clearly
, because
and the relation
implies
. The function
is a monotonically growing function of
with
.
Let us prove that
, where
, and
solves Equation (13). By the definition of
, we get
Since
, it follows that
. Thus
, and, as in the proof of Theorem 2, we obtain
and
.
Theorem 4 is proved.
Remark 2. Theorems 1 - 4 are well known in the case of a bounded operator A.
If A is bounded, then a necessary condition for the minimum of the functional
is the equation
(14)
Hence in this case conditions are required
.
If A is unbounded, then f does not necessarily belong to
, so Equation (14) may have no sence. Therefore, some changes in the usual theory are necessary. The changes are given in this paper. We prove, among other things, that for any
, in particular for
, the element
is well defined for any
, provided that A is a closed, linear, densely defined operator in Hilbert space (Theorem 1).
3. Applications
As a simple concrete example of this type of approximation, consider differentiation in
.
We define the operator
as follows
with
.
Then
is dense in H since it contains the complete orthonormal set
.
Clearly, A is a linear operator.
We show that A is a closed operator in Hilbert space H. Indeed, for suppose
and
and
, in each case the convergence being in the
norm. Since
we see that the sequence of constant functions
converges in
and hence the numerical sequence
converges to some real number C.
Now define
by
. Then, for any
, we have by of the Cauchy-Schwarz inequality
and hence
uniformly. Therefore,
and
, verifying that the operator A is closed, linear, densely defined in
.
Let
Then for
and
, we have
Therefore
and
, for
.
On the other hand, if
, let
. Then
for all
. In particular, for
, we find that
.
Now let
Then
and
and hence
. Therefore,
, for all
. But
contains all continuous function and hence
.
We conclude that
According to Theorem 1, for any
, the problem
has a unique solution
, where I is the identity operator on
.
does not necessarily belong to
.
It follows from Theorem 2, that if
,
then
It follows from Theorem 3, that if
,
, and
(15)
then there exists a unique global minimier
to Equation (15) and
, where
and
is properly chosen, in particular
.
It follows from Theorem 4, that the equation
has a unique solution
,
, and if
, then
.