solutions of some elliptic equations associated with a piecewise ...

SOLUTIONS OF SOME ELLIPTIC EQUATIONS ASSOCIATED WITH A PIECEWISE CONTINUOUS PROCESS B. IFTIMIE, I. MOLNAR and C. VÂRSAN

We provide a quasimartingale representation and a stochastic rule of dierentiation for test functions belonging to some class of polynomials, which depend on the solution of a stochastic dynamic system dened via a piecewise continuous function λ(t). This is achieved by applying the classical Itô's dierentiation rule on each continuity interval. The quasimartingale representation on each continuity interval is obtained by solving some possibly degenerate elliptic equations. AMS 2000 Subject Classication: Key words:

35B40, 35J70, 60G44, 91B28.

stochastic rule of dierentiation, piecewise continuous process, degenerate elliptic equation. 1. INTRODUCTION

The

evolution

of

a

multi-dimensional

piecewise

continuous

process

{(z(t), µ(t)); t > 0} is analyzed in connection with the stochastic rule of derivation by using polynomial solutions for some (possibly degenerate) elliptic equations. In our setting, the process

{z(t) ∈ Rn ; t > 0}

is generated as a solution

of a stochastic dierential system depending on a piecewise constant process

{µ(t) ∈ S ⊆ Rd ; t > 0}.

ϕ ∈ Pp (z; µ) consists of z = (z1 , . . . , zn ), d whose coecients are continuous bounded functions of µ ∈ R . It agrees with d the elliptic operators Lµ depending on the parameter µ ∈ R which determine the drift part when a pondered functional {exp(γt)ϕ(z(t); µ(t)); t > 0} is The set of test functions

all polynomial functions of degree

p

in the state variables

considered and the decomposition formula

t

Z exp(γt)ϕ(z(t); µ(t)) = ψt (z(·), µ(·)) +

exp(γs)[γϕ+Lµ (ϕ)](z(s); µ(s))ds+ 0

+ Mt (z(·), µ(·)), is valid. Here,

t > 0}

Lµ : Pp (z; µ) → Pp (z; µ)

t>0

{Mt (z(·), µ(·)); {ψt (z(·); µ(·)); t > 0} is a piecewise

is an elliptic operator,

is a continuous martingale and

constant process.

REV. ROUMAINE MATH. PURES APPL.,

53

(2008), 4, 323338

324

B. Iftimie, I. Molnar and C. Vârsan

2

The above decomposition formula is meaningfull when describing the asymptotic behaviour of the augmented process provided the nontrivial solution

ϕγ ∈ P2 (z; µ)

{exp(γt)kz(t)k2 | t > 0},

of the elliptic equation

γϕγ (z; µ) + kyk2 + Lµ (ϕγ )(z, µ) = 0,

∀z ∈ Rn , µ ∈ Rd ,

satises the nonsingularity condition

δ(γ)kzk2 6 ϕγ (z, µ) 6 for some positive constants

1 kzk2 C(γ)

δ(γ), C(γ).

2. DEFINITIONS, SETTING OF THE PROBLEM AND AUXILIARY RESULTS

m-dimensional Wiener process over a complete {Ω, F, {Ft }t>0 , P}. On the same probability space we have piecewise constant adapted processes {µ(t); t > 0} and {y(t); t > 0} of dimension d, respectively n. Set λ(t) = (y(t), µ(t)), t > 0, and Let

W (t)

be a standard

probability space

λ(t, ω) = λk (ω) = (yk (ω), µk (ω)),

t ∈ [tk (ω), tk+1 (ω)),

{tk ; k > 0} is an increasing sequence of positive adapted random varit0 = 0, tk → ∞, a.s., as k → ∞ and (yk , µk ), k > 0, are multidimensional Ftk -measurable random variables. We assume that the process W (t) and the sequence {(tk , λk ); k > 1} are independent. n We are next given a nite set of vector elds gi (z; µ) ∈ R , linear in z ,

where

ables with

dened as

i = 0, 1, . . . , m, µ ∈ Rd , z ∈ Rn ,

gi (z; µ) = ai (µ) + Ai (µ)z,

(1)

where the

(n × n)-matrices Ai (µ)

and the

n-dimensional

vectors

ai (µ)

are

assumed to be continuous and bounded. For each

x ∈ Rn

dene a piecewise continuous process

{z(t, x); t > 0}

as

the unique solution of the dynamical system

 m X    gj (z(t); µ(t))dWj (t),   dz(t) = g0 (z(t); µ(t)) +

(2)

 z(tk ) = z_ (tk ) + y(tk ),     z(0) = x,

where

t ∈ [tk , tk+1 ),

j=1

for any

k > 1,

z_ (tk ) = lim z(t). t↑tk

By denition, the unique solution of system (2) is a piecewise continuous and

{Ft }-adapted process {z(t, x); t > 0} such that, z(tk , x) − z_ (tk , x) = y(tk ) occurs.

jump

at each jump time tk , the

3

Solutions of some elliptic equations

Denition

325

Pp (z; µ) the set consisting of the polynomial (z1 , z2 , . . . , zn ) = z , whose coecients d functions of µ ∈ R ; Pp (z) ⊆ Pp (z; µ) consists of

1. Denote by

functions of degree

p

in the variables

are continuous bounded

polynomials with constant coecients. Consider the piecewise continuous scalar process

Vzγ (t, x) = exp(γt) ϕ(z(t, x); µ(t)) +

(3)

Z

t

exp(γs) f (z(s, x))ds,

t > 0,

0 where

γ

is a constant,

f ∈ Pp (z)

and

ϕ ∈ Pp (z; µ).

We provide a stochastic rule of dierentiation involving the piecewise continuous process

{Vzγ (t, x); t > 0} which contains both a continuous component

and a piecewise constant process dened as

Dzγ (t, x) =

(4)

 P  exp(γtk )ψk (z(·, x)), 0,

 where

t > t1 ,

0 1.

Lemma 1. Let f ∈ Pp (z), ϕ ∈ Pp (z; µ) dene the piecewise continuous process {Vzγ (t, x); t > 0} as in (3), corresponding to the unique solution {z(t, x); t > 0} of system (2). Then Vzγ (t, x) satises the integral equation

γ (5) Vz (t, x)

Z = ϕ(z(0, x); µ(0)) +

t

exp(γs) [γϕ + f + L(ϕ)](z(s, x); µ(s))ds 0

+

m Z X j=1

t

exp(γs) h∇z ϕ(z(s, x); µ(s)), gj (z(s, x); µ(s))idWj (s) + Dzγ (t, x),

0

for t > 0 and x ∈ Rn , where the piecewise constant process {Dzγ (t, x); t > 0} is dened by (4) and the elliptic operator L : Pp (z; µ) → Pp (z; µ) is given by m

(6)

L(ϕ)(z; µ) = h∇z ϕ(z; µ), g0 (z; µ)i +

1X h∇z ∇z ϕ(z; µ)gj (z; µ), gj (z; µ)i. 2 j=1

Proof.

The stochastic rule of dierentiation from (5) can be obtained

by applying the standard rule of stochastic dierentiation associated with the continuous process (7)

{Vzγ (t, x); t ∈ [tk , tk+1 )} for each k .

In this respect, rewrite

Vzγ (t, x) = Vzγ (tk , x) + Uk (t, z(t, x)) − Uk (tk , z(tk , x)) Z

t

+

exp(γs) f (z(s, x))ds, tk

326 for


t ∈ [tk , tk+1 ),

where

Uk (t, z) = exp(γt)ϕ(z; µ(tk )).

4 According to the sto-

chastic rule of dierentiation, we get

Z

t

Uk (t, z(t, x)) − Uk (tk , z(tk , x)) =

exp(γs) [γϕ + L(ϕ)](z(s, x); µ(tk ))ds tk

+

m Z X j=1

for every

t

exp(γs) h∇z ϕ(z(s, x); µ(tk )), gj (z(s, x); µ(tk ))idWj (s),

tk

t ∈ [tk , tk+1 ).

Using also (7), it follows that

Vzγ (t, x) = Vzγ (tk , x) +

(8)

Z

t

exp(γs) [γϕ + f + L(ϕ)](z(s, x); µk )ds tk

+

m Z X

exp(γs) h∇z ϕ(z(s, x); µk ), gj (z(s, x); µk )idWj (s),

tk

j=1 for

t

t ∈ [tk , tk+1 )

µk = µ(tk ) and Z tk γ Vz (tk , x) = Uk (tk , z(tk , x)) + exp(γs) f (z(s, x))ds.

(9)

where, by denition,

0 On the other hand, we notice that

Uk (tk , z(tk , x)) = Uk−1 (tk , z_ (tk , x)) + exp(γtk ) ψk (z(·, x)) for

k > 1. Uk−1 (tk , z_ (tk , x))

is computed as

Uk−1 (tk , z_ (tk , x)) = Uk−1 (tk−1 , z(tk−1 , x)) Z tk exp(γs)[γϕ + L(ϕ)](z(s, x); µk−1 )ds + tk−1

+

m Z tk X j=1

exp(γs)h∇z ϕ(z(s, x); µk−1 ), gj (z(s, x); µk−1 )idWj (s).

tk−1

Inserting now the last two formulas into (9), we get

γ γ (10) Vz (tk , x) = Vz (tk−1 , x)

Z

tk

+

exp(γs)[γϕ + f + L(ϕ)](z(s, x); µk−1 )ds tk−1

+

m Z X j=1

tk

exp(γs)h∇z ϕ(z(s, x); µk−1 ), gj (z(s, x); µk−1 )idWj (s)

tk−1

+ exp(γtk )ψk (z(·, x)). Now, iterating the last formula, we can rewrite equation (8) as it appears in (5).

5


327

Problem A. For a xed f ∈ Pp (z), nd a constant γ 6= 0 and ϕγ ∈ Pp (z; µ), such that the piecewise continuous process

{Vzγ (t, x); t > 0}

dened by (3) is

a quasi-martingale, i.e.,

Vzγ (t, x) = ϕγ (z(0, x); µ(0)) + Dzγ (t, x)

(11)

+

m Z X

t

exp(γs)h∇z ϕγ (z(s, x); µ(s)), gj (z(s, x); µ(s))idWj (s).

0

j=1

Problem

B. For a xed

f ∈ Pp (z), nd a constant γ 6= 0 and ϕγ ∈ Pp (z; µ)

solution of the elliptic equation

γϕγ (z; µ) + f (z) + L(ϕγ )(z; µ) = 0

(12)

where the linear operator

Remark

for any

L : Pp (z; µ) → Pp (z; µ)

1. Notice that each solution

z ∈ Rn , µ ∈ Rd ,

is dened by (6).

(γ, ϕγ )

of Problem A. In this respect, the scalar process

of Problem B is a solution

{Vzγ (t, x); t > 0}

dened

by (3) is a quasi-martingale expressed as in (11), provided the stochastic rule of dierentiation from Lemma 1 is used for the given solution of Problem B. Lemma 2.

be given as in

Let f ∈ P2 (z) and the vector elds gi (z; µ), i = 0, 1, . . . , m, . Let γ < 0 be such that

(1)

m 1X 2 |γ| > 4 kg0 k + kgj k , 2

(13)

j=1

where kgj k = sup (kaj (µ)k + kAj (µ)k). Then there exists ϕγ ∈ P2 (z; µ) satisµ∈Rd

fying the elliptic equation (12). In addition, ϕγ is the sum of the convergent series ∞ 1 X k L|γ| (f )(z; µ) , ϕγ (z; µ) = |γ|

(14)

(z, µ) ∈ Rn × Rd ,

k=0

where L|γ| = Proof.

1 |γ| L.

Lk|γ| (f )(z; µ) ∈ P2 (z; µ), k > 1, Write each ϕ ∈ P2 (z; µ) as

Estimates of the polynomials

be obtained by an induction argument.

can

ϕ(z; µ) = C(µ) + hh0 (µ), zi + hH(µ)z, zi, C(µ) ∈ R, hi (µ) ∈ Rn , i = 0, 1, . . . , n are n P continuous bounded functions with respect to µ. Denote kϕk = kCk+ khi k, where

H(µ) = (h1 (µ), . . . , hn (µ)

and

i=0

328


where

6

kCk = sup |C(µ)| and khi k = sup khi (µ)k, using khi (µ)k = µ∈Rd

i = 0, 1, . . . , n.

µ∈Rd

n P j=1

|hji (µ)|,

It is easily seen that

|ϕ(z; µ)| 6 kϕk(1 + kzk2 ),

∀(z, µ) ∈ Rn × Rd ,

and an elementary computation shows that

 n 1X 4  2 2   kg0 k + kgj k kϕk, (a) kL|γ| (ϕ)k 6   |γ| 2 j=1 n  4 1X  2 2  kgj k (1 + kzk2 )   (b) |L|γ| (ϕ)(z; µ)| 6 kϕk |γ| kg0 k + 2

(15)

j=1

for any

(z, µ).

Using an induction argument, we can easily prove that

  k n X 4 1 k kgj k2  kf k (1 + kzk2 ) kg0 k2 + L|γ| (f )(z; µ) 6  |γ| 2

(16)

j=1

for any

(z, µ) ∈ Rn × Rd

k > 0. Lk+1 |γ| : P2 (z; µ) → P2 (z; µ) can k Lk+1 k > 0, |γ| (f )(z, µ) = L|γ| L|γ| (f ) (z, µ), and

In this respect, recall that

be written as

Lk|γ| (f ) ∈ P2 (z; µ) and   k n X

k

L (f ) 6  4 kg0 k2 + 1 kgj k2  kf k |γ| |γ| 2

and assume that

j=1

ϕ = Lk|γ| (f ), we get n X

k+1 2

L (f ) = L|γ| (ϕ) 6 4 kg0 k2 + 1 kgj k kϕk |γ| |γ| 2 j=1   k+1 n X 4 1 kgj k2  kf k = ρk+1 kf k 6 kg0 k2 + |γ| 2

for some xed

k > 1.

Then, using estimates (15) for

j=1

and

k+1 L (f )(z, µ) 6 ρk+1 kf k(1 + kzk2 ) |γ|

for any

(z, µ),

where

ρ :=

4 |γ|

h

kg0 k2 +

1 2

n P j=1

kgj k2

i

.

7


329

This shows that estimate (16) is satised for any that

ρ 0

and assuming

(see (13)), we easily get that the series in (14) is convergent. In

addition, using (16) we obtain that

1 |ϕγ (z, µ)| 6 |γ|

X ∞

ρ

k

kf k(1 + kzk2 ),

k=0

where

∞

1 X k 1 1 ρ = = |γ| |γ| 1 − ρ k=0

By denition,

ϕγ ∈ P2 (z; µ)

1 "

n 1 P |γ| − 4 kg0 k2 + kgj k2 2 j=1

and using (14) we get that

ϕγ

# > 0.

satises the elliptic

equation (12), as the following simple computation shows

" L(ϕγ )(z; µ) = L|γ|

∞ X

# Lk|γ| (f )(z, µ)

k=0 The proof is complete.

Remark

=

∞ X

Lk|γ| (f )(z, µ) = −γϕγ (z; µ) − f (z).

k=1

2. The functional

ϕγ ∈ P2 (z; µ)

constructed in Lemma 2 is

meaningful for the asymptotic analysis of the piecewise continuous process

{Vzγ (t, x); t > 0}

provided

ϕγ (z; µ)

is a quadratic positive form satisfying

δ(γ)kzk2 6 ϕγ (z; µ) 6 for any

µ ∈ Rd .

Here,

δ(γ)

and

C(γ)

1 kzk2 , C(γ)

are positive constants.

This can be achieved by imposing the additional conditions on the vector elds (17)

gi (z; µ), i = 0, 1, . . . , m given in (1) (a) gi (z; µ) = Ai (µ)z, i = 0, 1, . . . , m, (b) Aj (µ)Ak (µ) = Ak (µ)Aj (µ) for any j, k ∈ {0, 1, . . . , m}, µ ∈ Rd ,

where recall that

Ai (µ)

are continuous and bounded symmetric matrices.

Assume that the vector elds gi (z; µ) given by sumptions (17). Let f (z) = kzk2 and let the constant γ full Lemma 3.

(1)

satisfy as-

m 1X γ < 0, |γ| > 4 kg0 k + kgj k2 . 2

(18)

j=1

Then there exists ϕγ ∈ P2 (z; µ) satisfying the equations (19)

  (a)

γϕγ (z; µ) + kzk2 + L(ϕγ )(z; µ) = 0, ∀(z, µ) ∈ Rn × Rd , 1 kzk2 ,  (b) ϕγ (z; µ) = hRγ (µ)z, zi, δ(γ)kzk2 6 ϕγ (z; µ) 6 C(γ)

330


8

where m

L(ϕγ )(z; µ) = h∇z ϕ(z, µ), A0 (µ)zi +

1X h∇z ∇z ϕ(z; µ) Aj (µ)z, Aj (µ)zi, 2 j=1

−1 C(γ), δ(γ) are positive constants, Rγ (µ) = |γ|In − B(µ) and B(µ) = m P A2j (µ). 2A0 (µ) + j=1

Proof.

The assumptions of Lemma 2 are satised. Let

ϕγ ∈ P2 (z; µ)

be

the solution of the elliptic equation (12) given by the series

"∞ # 1 X k L|γ| (f )(z, µ) , ϕγ (z; µ) = |γ|

(z, µ) ∈ Rn × Rd ,

k=0

where

f (z) =

kzk2 and the linear operator

L|γ| =

1 |γ| L satises

m 1X 1 h∇z ϕ(z, µ), A0 (µ)zi+ h∇z ∇z ϕ(z; µ) Aj (µ)z, Aj (µ)zi . L|γ| (ϕ)(z, µ) = |γ| 2 j=1

By denition,

1 L|γ| (f )(z, µ) = |γ|

* 2A0 (µ) +

m X j=1

+ 1 A2j (µ) z, z = hB(µ)z, zi. |γ|

It is obvious that the symmetric matrix

i = 0, 1, . . . , m.

B(µ)

Therefore,

Lk|γ| (f )(z, µ) = Here, the matrix

B(µ)

*

B(µ) |γ|

z, z

m X

kAj (µ)k2 6 2kg0 k +

j=1

µ ∈ Rd .

Dene

+

k

.

satises

kB(µ)k 6 2kA0 (µ)k + for any

commutes with each

m X j=1

kBk = sup kB(µ)k.

Using (18), we get

µ∈Rd

ρ=

(20)

ν 1 kBk 6 < |γ| |γ| 2

and the matrix series

∞

(21)

1 X Rγ (µ) = |γ| k=0

kgj k2 = ν,

B(µ) |γ|

k ,

µ ∈ Rd ,

Ai (µ),

9


|γ| > 2ν .

is convergent if

Hence

1 ϕγ (z; µ) = |γ| provided

|γ| > 2ν .

331

*∞ X B(µ) k k=0

|γ|

+ = hRγ (µ)z, zi,

z, z

Rγ (µ)

Recall that the symmetric matrix

fulls

1 B(µ) −1 Rγ (µ) = In − . |γ| |γ|

(22)

Conclusion (19) is a direct application of the following argument.

T (µ)y

be an orthogonal mapping, i.e.

T ∗ (µ) = [T (µ)]−1 ,

Let

z =

such that

T −1 (µ)B(µ)T (µ) = D(µ) = diag(γ1 (µ), . . . , γn (µ)) ϕγ (T (µ)y; µ) = hT −1 (µ)Rγ (µ)T (µ)y, yi as *∞ + 1 X D(µ) k y, y = hQγ (µ)y, yi, y ∈ Rn , ϕγ (T (µ)y; µ) = |γ| |γ| k=0 −1 , . . . , (|γ| − γ (µ))−1 . Using B(µ)e (µ) = where Qγ (µ) = diag (|γ| − γ1 (µ)) n k γk (µ)ek (µ), k = 1, . . . , n, where T (µ) = col[e1 (µ), . . . , en (µ)], it is easily seen that kB(µ)k > max |γk (µ)| = |b γ (µ)| and |b γ (µ)| 6 kBk for any µ ∈ Rd . The and write

16k6n

hypothesis

|γ| > 2ν

(see (18)) implies that the matrix

uniformly with respect to

µ,

and

δ(γ)kyk2 6 ϕγ (T (µ)y; µ) 6 with some positive constants

kyk2

=

Qγ (µ) is positive dened,

δ(γ), C(γ).

1 kyk2 , C(γ)

Take now

y = T −1 (µ)z .

Using

kzk2 , we obtain 1 kzk2 > ϕγ (z; µ) = hRγ z, zi > δ(γ)kzk2 , C(γ)

Remark

∀z ∈ Rn .

3. The stochastic rule of dierentiation from Lemma 1 and the

result stated in Lemma 3 are relevant for the piecewise continuous process

{z(t, x); t > 0},

dened as the unique solution of system (2). In this respect,

we use the decomposition

z(t, x) = zc (t, x) + y(t),

(23)

where the continuous component

t > 0,

{zc (t, x); t > 0}

is the unique solution of the

linear stochastic system

(24)

 m X   dzc (t) = A0 (µ(t))zc (t)dt + A0 (µ(t))zc (t)dWj (t), t > 0, j=1

 

zc (0) = x.

332


The stochastic rule of dierentiation written for

zc (t, x)

10 reads as

Vzγc (t, x) = ϕγ (x; µ(0)) + Dzγc (t, x)

(25)

Z +

t

exp(γs)[γϕγ + f + L(ϕγ )](zc (s, x); µ(s))ds 0

+

m Z X

t

exp(γs)h∇z ϕγ (zc (s, x); µ(s)), Aj (s, x); µ(s))zc (s, x)idWj (s),

0

j=1 where

Dzγc (t, x) =

X

exp(γtk ) ψk (zc (·, x))

0 0, x ∈ Rn ,

Lemma 4.

k > 1.

for f (z) = kzk2 and ϕγ (z; µ) = hRγ z, zi. Then the piecewise continuous process Uγ (t, x) satises the integral equation Uγ (t, x) = ϕγ (zc (0, x); µ(0)) + Dzγc (t, x)

(26)

Z th

+

i − C(γ)Uγ (s, x) + exp(γs)βγ (zc (s, x); µ(s)) ds

0

+

m Z X j=1

t

exp(γs)h∇z ϕγ (zc (s, x); µ(s)), Aj (µ(s))zc (s, x)idWj (s),

0

where  2 n d   βγ (z; µ) = C(γ)ϕγ (z; µ) − kzk 6 0, ∀(z, µ) ∈ R × R , X Dγ (t, x) = exp(γtk ) ϕγ (zc (tk , x); µ(tk )) − ϕγ (zc (tk , x); µ(tk−1 )) .   zc 0 0}.

If we apply it for

we get equation (25). Using the elliptic equation

γϕγ (z; µ) + kzk2 + L(ϕγ )(z; µ) = 0,

∀(z, µ) ∈ Rn × Rd

and the integral equation (25), we obtain that the piecewise continuous process

{Uγ (t, x); t > 0} (27)

satises

Uγ (t, x) = ϕγ (x; µ(0)) +

Dzγc (t, x)

Z − 0

t

exp(γs)kzc (s, x)k2 ds

11


+

m Z X j=1

333

t

exp(γs)h∇z ϕγ (zc (s, x); µ(s)), Aj (µ(s))zc (s, x)idWj (s).

0

Conclusion (19) of Lemma 3 shows that the equation

−kzk2 = −C(γ)ϕγ (z; µ) + βγ (z; µ), is satised, where

βγ (z; µ) 6 0

for any

(z, µ) ∈ Rn × Rd .

Rewrite now (27) using the last formula. We thus get conclusion (26).

Remark

4. Assuming that the set

Rγ (µ(t)) = Rγ (µ(0)),

for any

S ⊆ Rd

is chosen such that

t = tk , k > 0 (see µ(tk ) ∈ S),

γ the piecewise constant process {Dzc (t, x);

t > 0}

equation (26) can be written accordingly.

vanishes and the integral

In addition, the same equation

reduces to

t

Z Uγ (t, x) = ϕγ (x; µ(0)) −

(28)

C(γ)Uγ (s, x)ds 0

+2

m Z X j=1

t

Z Uγ (s, x)θj (s)dWj (s) +

0

t

exp(γs)βγ (zc (s, x); µ(s))ds, 0

θj (µ) : S → R, j = 1, . . . , m, b θj (t) = µj (µ(t)).

where

are given such that

Aj (µ) = θj (µ)In

and

If it is the case, then dene an exponential martingale

 (29)

ξ(t) = exp 2

m Z X j=1

t

0

Z θbj (s)dWj (s) −

t

 (θbj (s))2 ds  ,

t > 0,

0

which satises the equation

(30)

 m X   dξ(t) = 2ξ(t) θbj (t)dWj (t), j=1   ξ(0) = 1.

t > 0;

Lemma 5. Let hypotheses (17), (18) and (19) hold. In addition, assume that Aj (µ) = θj (µ)In , j = 1, . . . , m, and that the matrix B(µ) = 2A0 (µ) +

m P

j=1

A2j (µ) is constant. Denote hγ (t, x) = exp(γt)kzc (t, x)k2 , t > 0. Then

(31)

0 6 Uγ (t, x) 6 ϕγ (x; µ(0)) exp(−C(γ)t)ξ(t),

(32)

0 6 hγ (t, x) 6

for any t > 0, x ∈ Rn .

ϕγ (x; µ(0)) exp(−C(γ)t)ξ(t), δ(γ)

334


Proof.

12

By hypothesis, the conclusions of Lemma 4 hold. Using Remark 2

Φγ (t) = exp(−C(γ)t)ξ(t), Uγ (t, x) of equation (26) as Z t −1 Uγ (t, x) = Φγ (t) ϕγ (x; µ(0))+ Φγ (s) exp(γs) βγ (zc (s, x); µ(s))ds .

we rewrite the integral equation (26) as in (28). Set

t > 0. (33)

Represent the solution

0

βγ (z; µ) 6 0 for 1 hγ (t, x) 6 δ(γ) Uγ (t, x).

(z, µ) ∈ Rn × S .

Recall that

any

that

We easily get (32).

Remark

Thus, (31) follows. Notice

5. A relaxation of the assumptions made consists of admitting a

{q1 , . . . , qd } ⊆ Rn , Bi (µ) such that

common eigen-subspace generated by the constant vectors

d 6 n.

In this respect, real symmetric

∗

Q Ai (µ)z = Bi (µ)y,

(d × d)-matrices

i = 0, 1, . . . , m, µ ∈ S, z ∈ Rn ,

Q = [q1 , . . . , qd ], y = col(q1∗ z, . . . , qd∗ z). gi (z, µ) = ai (µ) + Ai (µ)z will be replaced by

will exist, where elds

hi (y, µ) = bi (µ) + Bi (µ)y,

The original vector

bi (µ) = Q∗ ai (µ),

bi (µ) ∈ Rd and the (d × d)-matrices Bi (µ) are continuous and bounded. n For each x ∈ R , dene a piecewise continuous process {b z (t, x); t > 0}

where

as the unique solution of the dynamic system

(34)

 m X   db z (t) = h0 (b z (t); µ(t))dt + hj (b z (t); µ(t))dWj (t), t ∈ [tk , tk+1 ), j=1   zb(tk ) = zb_ (tk ) + Q∗ µ(tk ) for any k > 1, zb(0) = Q∗ [x + µ(0)].

Remark

6. The analysis contained in Lemmas 15 has an adequate ver-

sion for the piecewise continuous process

{b z (t, x); t > 0}

and, in particular,

the decomposition

can be used.

Here, the

zb(t, x) = zbc (t, x) + Q∗ µ(t) continuous argument z bc (t, x)

stands for the unique

solution of the linear stochastic system

(35)

 m X   db zc (t) = h0 (b zc (t); µ(t))dt + hj (b zc (t); µ(t))dWj (t), j=1   zbc (0) = Q∗ x.

t > 0,

Problem C. Prove that under suitable conditions imposed on the matrices Bi (µ), we get the asymptotic behaviour of the continuous process {˜ z (t, x); t > 0}, satisfying (a) exp(γt)Ek˜ z (t, x)k2 → 0 as t → ∞ for each x ∈ Rn provided γ < 0 and m P kBj k2 , where hi (y; µ) = Bi (µ)y and kBi k = sup kBi (µ)k. |γ| > kB0 k + 21 j=1

µ∈S

13

Solutions of some elliptic equations (b) For any

γ

as above,

335

lim exp(γt)k˜ z (t, x)k2 = 0 P-a.s.

t↑∞

3. SOLUTIONS OF PROBLEM C Assume that there exist matrices

Bi (µ)

q 1 , q 2 , . . . , q d ∈ Rn , d 6 n,

and symmetric

Q∗ gi (z; µ) = Bi (µ)y,

i ∈ {0, 1, . . . , m}, µ ∈ S, z ∈ Rn ,

gi (z; µ) = ai (µ) + Ai (µ)z , Q = [q1 , q2 , . . . , qd ] and qd∗ z). Associate the linear stochastic system  m X

where

  z˜(t) = B0 (µ(t))˜ z (t)t +

(36)

d × d-

such that

y=

Bj (µ(t))˜ z (t)dWj (t),

∗

∗

col(q1 z, q2 z, . . . ,

t > 0,

j=1

 

z˜(0) = Q∗ x,

x ∈ Rn .

Moreover, assume the matrices

Bi (µ)

are bounded, continuous and mutually

commuting, i.e.,

Bi (µ)Bj (µ) = Bj (µ)Bi (µ) for any µ ∈ S ; m P (b) B(µ) = 2B0 (µ) + Bj2 (µ) is a constant (a)

matrix.

j=1

be the unique h Let z˜(t) P i solution of system (36) and γ < 0 m 1 2 such that |γ| > 4 kB0 k + 2 j=1 kBj k . Then (37) lim exp(γt) E k˜ z (t, x)k2 = 0, for any x ∈ Rn . Theorem 1.

t↑∞

Proof.

ϕγ ∈ P2 (z; µ) be the positive dened quadratic functional ϕγ (y; µ) = hRγ (µ)y, yi such that the conclusions in Lemma 3 hold. Here, z ∈ Rn is replaced by y ∈ Rd and the equations mentioned are replaced by  2 n  (a) γϕγ (y; µ) + kyk + L(ϕγ )(y; µ) = 0, ∀(y, µ) ∈ R × S, (38) 1  (b) δ(γ)kyk2 6 ϕγ (y; µ) 6 kyk2 , C(γ) Let

where

m

L(ϕγ )(y; µ) = h∇y ϕ(y, µ), B0 (µ)yi +

1X h∇y ∇y ϕ(y; µ) Bj (µ)y, Bj (µ)yi, 2 j=1

C(γ), δ(γ)Pare positive constants, Rγ (µ) = [|γ|Id − B(µ)]−1 and B(µ) = 2 2B0 (µ) + m j=1 Bj (µ). bγ (t, x) = exp(γt)ϕγ (˜ Set U z (t, x); µ(t)), t > 0, x ∈ Rn . Then the piecewise bγ (t, x); t > 0} fulls all the necessary conditions needed continuous process {U

336


to apply Lemma 4.

Ubγ (t, x)

As a consequence,

14

is a solution of the integral

equation

b yγ (t, x) Ubγ (t, x) = ϕγ (˜ z (0, x); µ(0)) + D

(39)

+ +

Z th 0 t

m Z X j=1

i − C(γ)Ubγ (s, x) + exp(γs)βγ (˜ z (s, x); µ(s)) ds

exp(γs)h∇y ϕγ (˜ z (s, x); µ(s)), Bj (µ(s))˜ z (s, x)idWj (s),

0

where

 2 n   βγ (y; µ) = C(γ)ϕγ (y; µ) − kyk 6 0, ∀(y, µ) ∈ R × S, X b γ (t, x) = D exp(γtk ) [ϕγ (˜ z (tk , x); µ(tk )) − ϕγ (˜ z (tk , x); µ(tk−1 ))] .   y 00 , P}. Taking v(t, x) = EUbγ (t, x), we get

space

now the expectation in equation (39) and

setting

Z

t

[−C(γ) v(s, x) + exp(γs) β(s, x)] ds,

v(t, x) = v(0, x) + D(t, x) +

(40)

0 where

D(t, x) =

X

exp(γtk )E {h[Rγ (µ(tk )) − Rγ (µ(tk−1 ))] z˜(tk , x), z˜(tk , x)i}

0 0, x ∈ Rn .

k

In addition, since

, B(µ) is a constant matrix,

µ ∈ S and γ < 0. Therefore, {D(t, x); t > 0} vanishes and by a standard

is also constant for any

the piecewise constant component

integral representation formula we have

(41)

t

Z

0 6 v(t, x) = exp[−C(γ)t] v(0, x) +

exp(C(γ)s) exp(γs)β(s, x)ds

0

6 exp(−C(γ)t) v(0, x)

for any

Notice that

exp(γt)Ek˜ z (t, x)k2 6

t > 0, x ∈ Rn .

1 v(t, x). δ(γ)

15


337

Using (41) we get (37).

Let the assumptions of Theorem 1 hold. Moreover, let Bj (µ) = θj (µ)Id , j = 1, . . . , m, and assume µj (µ) continuous and bounded. Dene the exponential martingale Theorem 2.

 ξ(t) = exp 2

(42)

m Z X

t

θbj (s)dWj (s) −

0

j=1

t

Z

 (θbj (s))2 ds  ,

t > 0,

0

where θbj (t) = θj (µ(t)). Then

2

exp(γt)k˜ z (t, x)k 6 exp[−C(γ)t] ξ(t)

Proof.

.

Ubγ (t, x) dened in Theorem 1, satises the integral b yγ (t, x) = 0 matrix Rγ (µ) is constant, we get that D

By hypothesis,

equation (27). Since the for any

k˜ z (0, x)k2 δ(γ)

t > 0, x ∈ Rn .

Hence, (27) becomes

Ubγ (t, x) = ϕγ (˜ z (0, x); µ(0))

(43)

+

Z th

i − C(γ)Ubγ (s, x) + exp(γs)βγ (˜ z (s, x); µ(s)) ds

0

+2

m Z X j=1

t

Ubγ (s, x) θbj (s)dWj (s).

0

A standard integral representation formula allows us to write (44)

0 6 Ubγ (t) 6 exp[−C(γ)t] ξ(t) [ϕγ (˜ z (0, x), µ(0))] ,

∀ t > 0.

Notice that

exp(γt)k˜ z (t, x)k2 6

1 b Uγ (t, x), δ(γ)

and estimate (44) leads us to the conclusion.

Acknowledgement.

t > 0, x ∈ Rn

The authors were supported by Contract 2-CEx06-11-18/2006.

REFERENCES [1] R. Cont and P. Tankov, Financial Modelling with Jump Processes. Chapman and Hall, 2004. [2] I. Karatzas and S. Shreve, Brownian Motion and Stochastic Calculus, 2nd Edition. Springer Verlag, 1991.

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16

[3] D. Lamberton and B. Lapeyre, Introduction to Stochastic Calculus Applied to Finance. Chapman and Hall, 1996. [4] M. Musiela and M. Rutkovski, Martingale Methods in Financial Modelling. Springer Verlag, 1998. [5] B. Oksendal, Stochastic Dierential Equations. An Introduction with Applications, 5th Edition. Springer, 2000. Received 12 November 2007

B. Iftimie Academy of Economic Studies Department of Mathematics Piaµa Roman , 6 010374 Bucharest, Romania

and I. Molnar, C. Vârsan Romanian Academy S. Stoilow Institute of Mathematics P.O. Box 1-764 014700 Bucharest, Romania