1. Introduction
Consider the following Hamilton-Jacobi-Bellman (HJB) equation:
(1.1)
where
is a bounded domain in
are elliptic operators of second order. Equation (1.1) is arising in stochastic control problems. See [2] and the references therein.
Equation (1.1) can be discretized by finite difference method or finite element method. See [1] [3] and the references therein. Then we obtain the following discrete HJB equation:
(1.2)
where
. Equation (1.2) is a system of nonsmooth nonlinear equations. Many numerical algorithms for solving (1.2) have been proposed. See [4] -[12] and the references therein.
[1] has given two iterative algorithms for solving (1.2). At each iteration, a linear complementarity subproblem or a linear equation system subproblem is solved. See also [4] .
Scheme I.
Step 1: Given
for some
we find
such that

Step 2: Let
For
we find
such that

Step 3: If
then the output is
otherwise
and it goes to Step 2.
Assume
Let
(1.3)
That is: the lth row of matrix
is the lth row of matrix
; the lth component of vector
is the lth component of vector
. Now we formulate Scheme II of Lions and Mercier in the notation above.
Scheme II.
Step 1:
for some
we find
such that
(1.4)
Step 2: For
we find
such that
(1.5)
Step 3: Compute
as the solution of
(1.6)
Step 4: If
then the output is
, otherwise
and it goes to Step 2.
In the last decade many numerical schemes have been given for solving (1.2). But the above schemes are still playing a very important role. See [4] -[6] and the references therein.
In this paper we propose, based on Scheme II above, a relaxation scheme with a parameter
, which for
is just Scheme II. In our numerical example, the new scheme with
is faster than Scheme II
. The monotone convergence of the new scheme has been proved.
2. New Scheme and Convergence
We propose a new scheme which is an extension of Scheme II.
New Scheme II.
Step 1: Given
for some
find
such that
(2.1)
Step 2: For
find
such that
(2.2)
Step 3: Compute
as the solution of
(2.3)
Step 4: Compute
(2.4)
Step 5: If
then output
otherwise
and go to Step 2.
In [13] we proposed the following conditions for (1.2).
Condition
All the matrices
are M-matrices.
In [13] we have proved the following theorem.
Theorem 2.1 If Condition
holds then (1.2) has a unique solution.
We have the following convergence theorem.
Theorem 2.2 Assume that Condition
holds, and that
are produced by New Scheme II. Then
is monotonely decreasing and convergent to the solution of (1.2).
Proof Since all
are M-matrices,
in New Scheme II are well defined.
First, we prove
is decreasing monotonically, i.e.,
(2.5)
By (2.3) we have
(2.6)
which combining with (2.1) and (2.2) yields
(2.7)
Since
are M-matrices, (2.7) means
(2.8)
By (2.4) we obtain
(2.9)
By
, (2.8) and (2.9) we know
(2.10)
and
(2.11)
which and (2.10) implies
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Similarly, by (2.3) we derive
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which combining with (2.2) and (2.6) implies
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Hence we have
(2.12)
By (2.4), we have
(2.13)
By (2.12), (2.13) and
, we know
(2.14)
which combining with
and (2.11) we derive
(2.15)
By (2.11), (2.12) and (2.13) ,we get
![]()
which combining with (2.15) implies
![]()
It is easy to derive by induction that
(2.16)
and
(2.17)
It follows that (2.5) holds.
It follows from (2.2) and (2.3) that
(2.18)
Since the set
is a finite set there exist positive integers
and
with
such that
![]()
Therefore, we have
![]()
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Then by (2.2) we obtain
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which and (2.17) results in
(2.19)
From (2.4), (2.16) and (2.19) we have
(2.20)
It follows from (2.18), (2.19) and (2.20) that
![]()
which means
is a solution of (1.2). The existence of solution has been proved.
Finally, we prove the uniqueness of solution. Assume
and
are solutions of (1.2), i.e.,
(2.21)
(2.22)
It is easy to see from (2.21) and (2.22) that there exist
and
such that
(2.23)
(2.24)
(2.25)
(2.26)
(2.23) and (2.26) implie
. But (2.24) and (2.25) implies
. Hence
. The proof is complete. ,
3. Numerical Example
We use example 2 in [4] , i.e., ![]()
(3.1)
where ![]()
![]()
![]()
![]()
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The discretization of the above second order derivatives are:
![]()
![]()
where
denote the forward and backward difference respectively in
and
,
,
. We use New Scheme II to solve the discrete problem. Take
,
and 1.1, 1.3, 1.5, 1.8, 1.9 respectively.
Table 1 and Table 2 show the ∞-norm of the residual
when iteration terminates.
We see that
for
and
is big for
.
Table 3 shows the relation between iteration number
and relaxation number
. Table 4 and Table 5 show the value of
at
for
and
respectively.
We can see from Table 3 that the algorithm for
is faster than that for
. Table 4 and Table 5 display the monotonicity of the algorithm.
Funding
This work was supported by Educational Commission of Guangdong Province, China (No. 2012LYM-0066) and the National Social Science Foundation of China (No. 14CJL016).