Using the induction hypothesis and the double integrals values given in the pre- vious paragraph we obtain: Rk+1(x, s, y, t) = (p2 - p1). 4. â i=1. Mk,iIi + e2. 4. â.
FUNDAMENTAL SOLUTION OF A PARABOLIC PARTIAL DIFFERENTIAL EQUATION WITH PIECEWISE CONSTANT COEFFICIENTS AND ADMITTING A GENERALIZED DRIFT Mounir ZILI Ecole Pr´eparatoire Aux acad´emies Militaires D.E.U. Rue Mar´echal Tito 4029 Sousse, Tunisie. Abstract: In this article, we construct a fundamental solution of a second-order parabolic partial differential equation, with piecewise constant coefficients and admitting a generalized drift. We present a calcul method, which permits us to give an explicit expression of the solution usable in concrete applications. AMS Subject Classification. 35C10, 35C15, 35D99. Key Words: Fundamental solution, parabolic, piecewise constant coefficients. Introduction In this work, we construct an explicit fundamental solution of the partial differential equation: ∂ (1) (L − )u ≡ 0, ∂t with ∂ ∂ ∂ ∂2 + δ{0} [A (0+ ) + B (0− )], L = a(x) 2 + b(x) ∂x ∂x ∂x ∂x where √ √ √ √ √ p2 [αp1 − (1 − α)p2 ][( p1 + p2 ) − (1 − 2α)( p2 − p1 )] A = (4α − 1)p2 + , (2) √ √ √ √ 2( p1 p2 + α(1 − α)( p2 − p1 )2 ) √ B = (4α − 3)p1 + α∈R
and
√ √ √ √ p1 [αp1 − (1 − α)p2 ][( p1 + p2 ) − (1 − 2α)( p2 − p1 )] , (3) √ √ √ √ 2( p1 p2 + α(1 − α)( p2 − p1 )2 )
∀ x∈R:
a(x) = p1 1R?− (x) + p2 1R+ (x),
b(x) = e1 1R?− (x) + e2 1R+ (x),
p1 , p2 , e1 and e2 are four real constants which satisfy a supplementary condition. δ{0} is the Dirac measure in 0 (for the variable x). Such equations appear notably when we study the particles motion in an heterogenous domain (cf.Mastrangelo [4],[5] and Zili [9]). The case where the coefficients a and b are H¨older continuous, and A = B = 0, has been studied by A.Friedman [1]. When b = 0, the expression of the fundamental solution of such equation is known (cf. Gaveau [3], Mastrangelo [4], [5], Talbi [8] and Zili [9]). In a previous article (cf. Zili [10]), we have constructed explicitly the fundamental 1 solution of the same partial differential equation, but in the special case where α = , 2
√ √ √ √ A = p1 p2 and B = − p1 p2 . The general case is clearly more difficult and more interesting. ∂ using a Zvonkin transformation, A first idea would be to remove the drift b(x) ∂x and to come down to the works [3], [4], [5],[8], and [9]. But this method conduct to an operator which coefficients are not piecewise constant, and it not permit to give an easily calculable solution. Here, the aim is to give an explicit formulation of the fundamental solution usable in concrete applications. Our method is of analytical type, and it is based on a technical and a long calculus. 1. Definitions and notations 1.1 Definition: Let h a real values function, defined and continuous on R. We say that h is of class C 2 in R− [resp. in R+ ], when h is of class C 2 in R?− [resp. in R?+ ] with bounded derivative, such that for all k ≤ 2, 1R?− h(k) [resp. 1R?+ h(k) ] is prolonged continuously on R− [resp. on R+ ]. 1.2 Definition of a fundamental solution: A fundamental solution of the equation (1) in Ω = R × [0, T ] (T > 0), is a function Γ(x, s, y, t) defined for all (y, s) ∈ Ω, (x, t) ∈ Ω, s < t, which satisfies the following conditions: 1. For fixed (y, s) ∈ R? ×[0, T ] it satisfies, as a function of (x, t) ( x ∈ R? , t ∈]s, T ]) the equation (1). 2. For every bounded continuous function h on R, if x ∈ R lim
Z
s→t R
Γ(x, s, y, t)h(y)dy = h(x).
3. For any continuous function h on R of class C 2 in R− and in R+ , for any continuous, compactly supported function g on R: lim
Z
Z
t→0 x∈R
t−1 g(x)[h(y) − h(x)]Γ(x, 0, y, t)dydx =< Lh, g > .
y∈R
2 Zx exp(−t2 )dt and For every x ∈ R, we denote: erf (x) = √ π 0 2 Z +∞ erfc(x) = 1 − erf (x) = √ exp(−t2 )dt. π x 1.3 Definition: For every real α > 0, and every real a, we define the functions and χa by:
ϕαa , ψa , γa ,
ϕαa (t) =
√
0
1 a2 exp(− ) if t > 0 πtα 4t if t ≤ 0.
1 a2 √ [a2 − 2t]exp(− ) if t > 0 4t 16πt5 ψa (t) = 0 if t ≤ 0.
1 a3 a2 [ − 3a]exp(− ) if t > 0 4t γa (t) = 16πt5 2t 0 if t ≤ 0.
√
a erfc( √ ) if t > 0 2 t χa (t) = 0 if t ≤ 0.
1.4 Definition: Denoting a(x) = p1 1R?− (x) + p2 1R+ (x) and b(x) = e1 1R?− (x) + e2 1R+ (x), where e1 , e2 ∈ R and p1 , p2 ∈ R?+ , we introduce the following parametrix functions: For s < t, α ∈ R and (x, y) ∈ R2 , P1α (x, s, y, t) = q
1 4πp1 (t − s)
[exp(−
(x − y)2 ) 4p1 (t − s)
+ sgn(y)(2α − 1)exp(−
P2α (x, s, y, t) = q
1 4πp2 (t − s)
[exp(−
(| x | + | y |)2 )], 4p1 (t − s)
(x − y)2 ) 4p2 (t − s)
+ sgn(y)(2α − 1)exp(−
(| x | + | y |)2 )], 4p2 (t − s)
Z α (x, s, y, t) = P1α (x, s, y, t)1{y 0 and y > 0: e2 2 [1 + (1 − 2α)2 ] √ √ √ ∂ 2 Z2 R2 (x, s, y, t) = 2Sα p1 ( p1 + p2 ) 2 (x, s, y, t) + Z1 (x, s, y, t) ∂x 4p2 −
2α(1 − α)e1 e2 Θ ∂Z2 √ √ (x, s, y, t) √ √ √ Z2 (x, s, y, t) + α(α − 1)( p2 − p1 ) √ √ ( p1 + p2 ) p1 p2 p1 ∂x
+ e2 2 (t − s)
∂ 2 Z 1/2 e2 2 (1 − 2α)(x − y) ∂Z1 (x, s, y, t) + (x, s, y, t). ∂x2 2p2 ∂x
If x < 0 and y > 0: R2 (x, s, y, t) = α [
e2 (p2 − p1 )α ∂Z2 Λα e 1 2 − 2e1 ] (x, s, y, t) − √ Z1 (x, s, y, t) p2 ∂x p1 (p2 − p1 )
2e1 2 αe1 e2 e1 e2 ∂Z2 e2 (p1 − p2 ) ∂ 2 Z2 + [ − ]Z2 (x, s, y, t) + y (x, s, y, t) + y (x, s, y, t) p 2 − p1 p2 p2 ∂x p2 ∂x2 !
e1 Λα ∂Z1 + √ (x, s, y, t) . p2 ∂x
If x > 0 and y < 0: R2 (x, s, y, t) = (1 − α) [
e1 (p1 − p2 )(1 − α) ∂Z2 − 2e2 ] (x, s, y, t) p1 ∂x
2e2 2 (1 − α)e1 e2 Λα e 2 2 Z1 (x, s, y, t) + [ − ]Z2 (x, s, y, t) − √ p2 (p1 − p2 ) p 1 − p2 p1 e1 e2 ∂Z2 e1 (p2 − p1 ) ∂ 2 Z2 e2 Λα ∂Z1 + y (x, s, y, t) + y (x, s, y, t) . (x, s, y, t) + √ 2 p1 ∂x p1 ∂x p1 ∂x !
Proof: We give the proof in the case where x > 0 and y > 0. Z tZ 0
R2 (x, s, y, t) =
−∞
s
R1 (z, s, y, σ)R1 (x, σ, z, t)dzdσ
Z t Z +∞
+
R1 (z, s, y, σ)R1 (x, σ, z, t)dzdσ
0
s
= R21 (x, s, y, t) + R22 (x, s, y, t). Using the lemma 2.1 we obtain: R21 (x, s, y, t) =
Z tZ 0 s
−∞
R1 (z, s, y, σ)R1 (x, σ, z, t)dzdσ 2
= −(p1 − p2 )
+ e2 (p1 − p2 ) + e1 (p2 − p1 ) + e1 e2
s
−∞
Z tZ 0 s
−∞
Z tZ 0
Z tZ 0 s
Z tZ 0
−∞
s
−∞
∂ 2Z α ∂ 2Z α (z, s, y, σ) (x, σ, z, t)dzdσ ∂z 2 ∂x2
∂Z α ∂ 2Z α (z, s, y, σ) (x, σ, z, t)dzdσ ∂z 2 ∂x ∂Z α ∂ 2Z α (z, s, y, σ) (x, σ, z, t)dzdσ ∂z ∂x2
α
∂Z ∂Z α (z, s, y, σ) (x, σ, z, t)dzdσ ∂z ∂x
= I1 + I2 + I3 + I4 . To calculate I1 , we first treat the integral: Iˆ1 (z, t − s) = Iˆ1 (t − s) = =
Z t s
Z t s
∂ 2Z α ∂ 2Z α (z, s, y, σ) (x, σ, z, t)dσ ∂z 2 ∂x2
α (z − y)2 1 (z − y)2 √ [ − ]exp(− ) 2 π 2(p2 (σ − s))5/2 (p2 (σ − s))3/2 4p2 (σ − s)
1 (x − z)2 1 (x − z)2 .(1 − α) √ [ − ]exp(− )dσ. 2 π 2(p1 (t − σ))5/2 (p1 (t − σ))3/2 4p1 (t − σ) y−z x−z Putting v = σ − s, u = t − s, a = √ and b = √ , it may be seen that: p2 p1
Iˆ1 (u) =
α(1 − α) Z u 1 a2 2 √ [a − 2v]exp(− ) p1 3/2 p2 3/2 0 4 πv 5/2 4v
.
1 b2 2 √ )dv [b − 2(u − v)]exp(− 4 π(u − v)5/2 4(u − v)
=
α(1 − α) Z u ψa (u)ψb (u − v)dv p1 3/2 p2 3/2 0
=
α(1 − α) ψa ? ψb (u). p1 3/2 p2 3/2
where ψa ? ψb denotes the convolution product of ψa by ψb . Hence, denoting the Laplace transformation of a function g by L(g), and using the values of L(ψa ) and L(γa ) (cf. Appendix A) we obtain, L(Iˆ1 )(u) =
α(1 − α) L(ψa )L(ψb )(u) p1 3/2 p2 3/2
=
α(1 − α) uexp(−(a + b)u1/2 ) p1 3/2 p2 3/2
=
α(1 − α) L(γa+b )(u). p1 3/2 p2 3/2
Consequently α(1 − α) Iˆ1 (u) = 3/2 3/2 γa+b (u) p1 p2 α(1 − α) 1 q = 3/2 3/2 p1 p2 16π(t − s)5
"
√ √ √ 3 √ − ( p1 + p2 )z + p1 y + p2 x 2(t − s)p1 3/2 p2 3/2
√ √ √ √ # ! √ √ √ √ − ( p1 + p2 )z + p1 y + p2 x (−( p1 + p2 )z + p1 y + p2 x)2 − 3 exp − . √ √ p1 p2 4p1 p2 (t − s)
Therefore
Z 0
I1 = −(p1 − p2 )2 Iˆ1 (z, t − s)dz. −∞ √ √ √ √ −( p1 + p2 )z + p1 y + p2 x q Substituting ξ = , and using the values of the inte2 p1 p2 (t − s) Z
Z
grals u3 exp(−u2 )du and uexp(−u2 )du (cf. Appendix B), we find I1
" √ # ! √ √ √ ( p1 y + p2 x)2 −α(1 − α)(p1 − p2 )2 ( p1 y + p2 x)2 2 − = √ exp − √ √ 4 πp1 p2 ( p1 + p2 ) p1 p2 (t − s)5/2 (t − s)3/2 4p1 p2 (t − s)
2α(1 − α)(p1 − p2 )2 ∂ 2 Z2 (x, s, y, t) = − √ √ . √ p 2 ( p1 + p2 ) ∂x2
Using the same technique we calculate the other double integrals and we obtain : √ √ 2α(1 − α)e2 ( p1 − p2 ) ∂Z2 (x, s, y, t) I2 = . √ p2 ∂x √ √ 2α(1 − α)e1 ( p1 − p2 ) ∂Z2 (x, s, y, t) . I3 = √ p1 ∂x 2α(1 − α)e1 e2 I4 = − √ √ √ Z2 (x, s, y, t). p1 ( p1 + p2 ) R22 (x, s, y, t) = = e22
Z t Z +∞
= e2 2
0
s
Z t Z +∞ s
0
R1 (z, s, y, σ)R1 (x, σ, z, t)dzdσ
∂Z α ∂Z α (z, s, y, σ) (x, σ, z, t)dzdσ ∂z ∂x
[1 + (1 − 2α)2 ] ∂ 2 Z 1/2 (x, s, y, t) Z1 (x, s, y, t) + (t − s) 4p2 ∂x2 !
(1 − 2α)(x − y) ∂Z1 (x, s, y, t) + . 2p2 ∂x We will now generalize the above result, and prove the following theorem: 3.2 Theorem: ∀ k ≥ 2, ∀ x ∈ R? and ∀ y ∈ R? we have ∂ 2 Z2 (x, s, y, t) ∂ 2 Z2 (x, s, y, t) + c (x, y)y. k,2 ∂x2 ∂x2 ∂Z1 (x, s, y, t) ∂Z2 (x, s, y, t) + ck,4 (x, y) + Wk (x, s, y, t) + ck,3 (x, y) ∂x ∂x
Rk (x, s, y, t) = ck,1 (x, y)
where for ν = 1, 2, 3, 4: ck,ν (x, y) = Kk,ν 1R?− (x)1R?− (y) + Lk,ν 1R?− (x)1R?+ (y) + Mk,ν 1R?+ (x)1R?− (y) + Nk,ν 1R?+ (x)1R?+ (y), and for l ≥ 1: √ √ √ K2l,1 = 2 p2 ( p1 + p2 )Sαl , K2l+1,1 = 0; L2l+1,1 = 2(1 − α)(p2 − p1 )Sαl , L2l,1 = 0, √ √ √ M2l+1,1 = 2α(p1 − p2 )Sαl , M2l,1 = 0, N2l,1 = 2 p1 ( p1 + p2 )Sαl , N2l+1,1 = 0; √ √ √ e1 p2 ( p1 + p2 ) l e1 (1 − α)(p2 − p1 ) l−1 K2l+1,2 = Sα , K2l,2 = 0; L2l,2 = Sα , L2l+1,2 = 0, p1 p1 √ √ √ e 2 p1 ( p1 + p2 ) l e2 α(p1 − p2 ) l−1 N2l+1,2 = Sα , N2l,2 = 0; M2l,2 = Sα , M2l+1,2 = 0, p2 p2 √ √ √ √ α(1 − α)( p2 − p1 ) α(1 − α)( p2 − p1 ) l−1 l−1 K2l,3 = θlSα , K2l+1,3 = Sα [(l−1)γα +βα ], √ √ √ p2 p1 p2
√ √ ( p2 + p1 )Λα 1 − √ ) and where γα = Θ( √ 4p1 p2 p1 √ √ √ e1 (p2 − p1 )(1 − α) e2 ( p2 + p1 )Λα 2 p2 e1 + − √ , βα = −2e2 − √ √ p1 2 p 1 p2 p1 e1 (p2 − p1 )(1 − α) Θ ], L2l+1,3 = (1−α)Sαl [l √ +2e2 ], p1 p1 √ √ √ √ α(1 − α)( p1 − p2 ) l−1 α(1 − α)( p1 − p2 ) N2l,3 = θlSαl−1 , N2l+1,3 = Sα [(l−1)γα0 +βα0 ], √ √ √ p2 p1 p1 √ √ ( p2 + p1 )Λα 1 0 where γα = Θ( − √ ) and √ 4p2 p1 p2 √ √ √ e2 (p1 − p2 )α e1 ( p2 + p1 )Λα 2 p1 e2 0 βα = −2e1 − + − √ , √ √ p2 2 p1 p2 p2 L2l,3 = (1−α)Sαl−1 [(l−1)γα −2e2 −
M2l,3 = αSαl−1 [(l − 1)γα0 − 2e1 −
e2 (p1 − p2 )α Θ ], M2l+1,3 = αSαl [l √ + 2e1 ], p2 p2
√ √ e1 ( p2 − p1 )α(1 − α) e2 (1 − α)Λα l−1 K2l+1,4 = Λα Sαl−1 , K2l,4 = 0; L2l,4 = Sα , L2l+1,4 = 0, √ √ √ p1 p2 p1 √ √ e2 ( p1 − p2 )α(1 − α) e1 αΛα Λα Sαl−1 , N2l,4 = 0; M2l,4 = √ Sαl−1 , M2l+1,4 = 0, N2l+1,4 = √ √ p1 p2 p2 Wk satisfies if y < 0: | Wk (x, s, y, t) | ≤ Cte
4 X
! √ √ ( p1 x − p2 y)2 1 |) √ exp − 8p1 p2 (t − s) t−s "
(| Kk,j | + | Lk,j
j=1
1 (x − y)2 + √ exp − 8p1 (t − s) t−s
!#
.
if y > 0: | Wk (x, s, y, t) | ≤ Cte
4 X
! √ √ ( p2 x − p1 y)2 1 |) √ exp − 8p1 p2 (t − s) t−s "
(| Nk,j | + | Mk,j
j=1
1 (x − y)2 + √ exp − 8p2 (t − s) t−s
!#
.
In order to prove this theorem, we shall calculate the following double integrals: 3.3 Preliminary calculus: Using the same technique employed in the calculus of R2 , we obtain: If x > 0 and y > 0: I1 =
Z tZ 0 s
−∞
∂ 2 Z2 ∂ 2Z α (1 − α) ∂ 2 Z2 (z, s, y, σ) (x, σ, z, t)dzdσ = (x, s, y, t). √ √ √ ∂z 2 ∂x2 p2 ( p1 + p2 ) ∂x2
I2 =
Z tZ 0 −∞
s
I3 =
=
∂Z2 ∂ 2Z α ∂Z2 −(1 − α) (z, s, y, σ) (x, s, y, t). (x, σ, z, t)dzdσ = √ √ √ 2 ∂z ∂x p1 ( p1 + p2 ) ∂x
−∞
s
Z tZ 0 −∞
s
−∞
s
I30 =
y
−∞
s
∂ 2 Z2 (1 − α)y ∂Z2 ∂Z α (x, σ, z, t)dzdσ = √ √ (x, s, y, t). (z, s, y, σ) √ 2 ∂z ∂x p2 ( p1 + p2 ) ∂x
Z tZ 0 −∞
s
I40 =
∂Z1 ∂ 2Z α −(1 − α) ∂Z2 (z, s, y, σ) (x, s, y, t). (x, σ, z, t)dzdσ = 2 ∂z ∂x 2p1 ∂x
∂ 2 Z2 (1 − α) ∂Z2 ∂Z α (x, σ, z, t)dzdσ = √ √ (x, s, y, t). (z, s, y, σ) √ 2 ∂z ∂x p2 ( p1 + p2 ) ∂x
Z tZ 0
Z tZ 0
I20 =
∂ 2 Z2 ∂ 2 Z2 ∂ 2Z α (1 − α)y (z, s, y, σ) (x, σ, z, t)dzdσ = (x, s, y, t). √ √ √ ∂z 2 ∂x2 p2 ( p1 + p2 ) ∂x2
Z tZ 0
I4 = I10
y
∂Z2 ∂Z α −(1 − α) (z, s, y, σ) (x, σ, z, t)dzdσ = √ √ √ Z2 (x, s, y, t). ∂z ∂x p 1 ( p1 + p2 )
Z tZ 0 −∞
s
∂Z1 ∂Z α −(1 − α) (z, s, y, σ) (x, σ, z, t)dzdσ = Z2 (x, s, y, t). ∂z ∂x 2p1
∂ 2 Z2 ∂Z α (z, s, y, σ) (x, σ, z, t)dzdσ ∂z 2 ∂x s 0 Λα ∂Z1 1 ∂Z2 = √ (x, s, y, t) − (x, s, y, t). 2 p1 (p2 − p1 ) ∂x p2 − p1 ∂x Z t Z +∞
I”1 =
∂ 2 Z2 ∂Z α (z, s, y, σ) (x, σ, z, t)dzdσ ∂z 2 ∂x s 0 Λα y ∂Z1 y ∂Z2 = √ (x, s, y, t) − (x, s, y, t). 2 p1 (p2 − p1 ) ∂x p2 − p1 ∂x Z t Z +∞
I”2 =
I”3
y
∂Z2 ∂Z α = (z, s, y, σ) (x, σ, z, t)dzdσ ∂z ∂x s 0 Λα 1 = √ Z1 (x, s, y, t) + Z2 (x, s, y, t). 2 p2 (p2 − p1 ) p1 − p2 Z t Z +∞
∂Z1 ∂Z α (z, s, y, σ) (x, σ, z, t)dzdσ ∂z ∂x s 0 x ∂Z1 1−α = − (x, s, y, t) − Z1 (x, s, y, t). 2p2 ∂x 2p2
I”4 =
Z t Z +∞
If x < 0 and y > 0: J1 =
J2 =
Z t Z +∞ s
0
Z t Z +∞
J3 =
s
∂ 2Z α α ∂ 2 Z2 ∂ 2 Z2 (z, s, y, σ) (x, σ, z, t)dzdσ = (x, s, y, t). √ √ √ ∂z 2 ∂x2 p1 ( p1 + p2 ) ∂x2 y
0
Z t Z +∞ s
0
∂ 2 Z2 ∂ 2Z α αy ∂ 2 Z2 (z, s, y, σ) (x, σ, z, t)dzdσ = (x, s, y, t). √ √ √ ∂z 2 ∂x2 p1 ( p1 + p2 ) ∂x2 ∂Z2 ∂ 2Z α −α ∂Z2 (z, s, y, σ) (x, σ, z, t)dzdσ = √ √ (x, s, y, t). √ 2 ∂z ∂x p2 ( p1 + p2 ) ∂x
∂Z1 ∂ 2Z α −α ∂Z2 (z, s, y, σ) (x, s, y, t). (x, σ, z, t)dzdσ = 2 ∂z ∂x 2p2 ∂x s 0 Z t Z +∞ 2 ∂ Z2 α ∂Z2 ∂Z α J10 = (x, σ, z, t)dzdσ = (x, s, y, t). (z, s, y, σ) √ √ √ ∂z 2 ∂x p1 ( p1 + p2 ) ∂x s 0 J4 =
J20 =
Z t Z +∞
Z t Z +∞ 0
s
J30 =
y
∂ 2 Z2 αy ∂Z2 ∂Z α (x, σ, z, t)dzdσ = (x, s, y, t). (z, s, y, σ) √ √ √ ∂z 2 ∂x p1 ( p1 + p2 ) ∂x
Z t Z +∞ s
J40
0
=
∂Z2 ∂Z α −α (z, s, y, σ) (x, σ, z, t)dzdσ = √ √ √ Z2 (x, s, y, t). ∂z ∂x p2 ( p 1 + p2 )
Z t Z +∞ 0
s
∂Z1 ∂Z α −α (z, s, y, σ) (x, σ, z, t)dzdσ = Z2 (x, s, y, t). ∂z ∂x 2p2 ∂ 2 Z2 ∂Z α (x, σ, z, t)dzdσ (z, s, y, σ) ∂z 2 ∂x
Z tZ 0
J”1 =
−∞
s
Λα ∂Z1 1 ∂Z2 (x, s, y, t) − (x, s, y, t). √ 2 p2 (p1 − p2 ) ∂x p1 − p2 ∂x
=
Z tZ 0
J”2 =
y
−∞
s
∂ 2 Z2 ∂Z α (z, s, y, σ) (x, σ, z, t)dzdσ ∂z 2 ∂x
Λα y y ∂Z2 ∂Z1 (x, s, y, t) − (x, s, y, t). √ 2 p2 (p1 − p2 ) ∂x p1 − p2 ∂x
=
∂Z2 ∂Z α (z, s, y, σ) (x, σ, z, t)dzdσ ∂x s −∞ ∂z Λα 1 Z1 (x, s, y, t) − Z2 (x, s, y, t). = √ 2 p1 (p1 − p2 ) p1 − p2
J”3 =
Z tZ 0
J”4 =
Z tZ 0 −∞
s
∂Z1 ∂Z α (z, s, y, σ) (x, σ, z, t)dzdσ ∂z ∂x
x ∂Z1 α (x, s, y, t) − Z1 (x, s, y, t). 2p1 ∂x 2p1 Proof of the theorem: We reason by induction. According to the proposition 3.1, the theorem is satisfied for k = 2. We suppose that it is true for k ≥ 2, and we prove it for k + 1. If x > 0 and y > 0: = −
Rk+1 (x, s, y, t) = +
Z tZ 0 s
−∞
Z tZ 0 s
+
0
Z tZ 0 s
+
Rk (z, s, y, σ)R1 (x, σ, z, t)dzdσ
Z t Z +∞ s
=
−∞
−∞
Rk (z, s, y, σ)(p2 − p1 ) Rk (z, s, y, σ)e2
Z t Z +∞ s
0
Rk (z, s, y, σ)R1 (x, σ, z, t)dzdσ ∂ 2Z α (x, σ, z, t)dzdσ ∂x2
∂Z α (x, σ, z, t)dzdσ ∂x
Rk (z, s, y, σ)e2
∂Z α (x, σ, z, t)dzdσ. ∂x
Using the induction hypothesis and the double integrals values given in the previous paragraph we obtain: Rk+1 (x, s, y, t) = (p2 − p1 )
4 X
Mk,i Ii + e2
i=1
+
Z tZ s
R
4 X
Mk,i Ii0 + e2
i=1
4 X
Nk,i I”i
i=1
Wk (z, s, y, σ)R1 (x, σ, z, t)dzdσ
∂ 2 Z2 (x, s, y, t) ∂ 2 Z2 (x, s, y, t) + N y. k+1,2 ∂x2 ∂x2 ∂Z2 (x, s, y, t) ∂Z1 (x, s, y, t) + Nk+1,3 + Nk+1,4 + Wk+1 (x, s, y, t), ∂x ∂x = Nk+1,1
where Nk+1,1
√ √ (1 − α)( p2 − p1 ) = Mk,1 , √ p2
Nk+1,3
√ √ (1 − α)( p2 − p1 ) e2 (1 − α) (1 − α)(p2 − p1 ) =√ √ Mk,3 − Mk,4 √ √ Mk,1 − p2 ( p 2 + p1 ) p1 2p1
−
Nk+1,2
e 2 Λα Nk+1,4 = √ Nk,1 , 2 p1 (p2 − p1 )
e2 Nk,1 , (p2 − p1 )
Wk+1 (x, s, y, t) = e2
4 X
√ √ (1 − α)( p2 − p1 ) = Mk,2 , √ p2
Mk,i Ii0
+ e2
4 X
Nk,i I”i +
Z tZ
i=2
i=2
s
R
Wk (z, s, y, σ)R1 (x, σ, z, t)dzdσ.
If x < 0 and y > 0: Using the induction hypothesis and reasonnig as in the previous case we get: Rk+1 (x, s, y, t) =
Z t Z +∞ s
0
Rk (z, s, y, σ)(p1 − p2 )
∂ 2Z α (x, σ, z, t)dzdσ ∂x2
∂Z α (x, σ, z, t)dzdσ ∂x s 0 Z tZ 0 ∂Z α + Rk (z, s, y, σ)e1 (x, σ, z, t)dzdσ ∂x s −∞ +
Z t Z +∞
Rk (z, s, y, σ)e1
= (p1 − p2 )
4 X
Nk,i Ji + e1
i=1
+
Z tZ s
R
4 X
Nk,i Ji0 + e1
i=1
4 X
Mk,i J”i
i=1
Wk (z, s, y, σ)R1 (x, σ, z, t)dzdσ
∂ 2 Z2 (x, s, y, t) ∂ 2 Z2 (x, s, y, t) + M y. k+1,2 ∂x2 ∂x2 ∂Z2 (x, s, y, t) ∂Z1 (x, s, y, t) + Mk+1,3 + Mk+1,4 + Wk+1 (x, s, y, t), ∂x ∂x = Mk+1,1
where Mk+1,1
√ √ α( p1 − p2 ) Nk,1 , = √ p1
Mk+1,2
√ √ α( p1 − p2 ) = Nk,2 , √ p1
Mk+1,3 −
√ √ α( p1 − p2 ) α(p1 − p2 ) e1 α Nk,3 − Nk,4 =√ √ √ Nk,1 − √ p 1 ( p2 + p1 ) p2 2p2
e1 Mk,1 , (p1 − p2 )
Wk+1 (x, s, y, t) = e1
4 X
e1 Λα Mk+1,4 = √ Mk,1 , 2 p2 (p1 − p2 ) Nk,i Ji0
+ e1
i=2
4 X
Mk,i J”i +
i=2
Z tZ s
R
Wk (z, s, y, σ)R1 (x, σ, z, t)dzdσ.
And using the induction hypothesis, we obtain the result. 3.4 Corollary: If p1 and p2 satisfy : √ √ α(1 − α)( p2 − p1 )2 |=| Sα |< 1, | √ √ p1 p2 then, for all (x, s, y, t) ∈ (R? × R+ )2 ∩ {0 ≤ s < t ≤ T }, the function Γ is a solution ∂ of the equation (L0 − )u = M u ≡ 0. ∂t Proof: At first we note that for (x, y) ∈ (R? )2 , for each s < t, the function φ defined by +∞ X
φ(x, s, y, t) =
Ri (x, s, y, t),
i=1
satisfies the equation α
φ(x, s, y, t) = M Z (x, s, y, t) +
Z tZ 0
φ(z, s, y, σ)M Z α (x, σ, z, t)dzdσ.
R
We note also that Γ satisfy the following equation: Γ(x, s, y, t) =
+∞ X
Hi (x, s, y, t)
i=1 α
= Z (x, s, y, t) +
Z tZ s
Z α (z, s, y, σ)φ(x, σ, z, t)dzdσ
R
= Z α (x, s, y, t) + Y α (x, s, y, t). Applying the corollary 2.3 with f (x, t) = fs,y (x, t) = φ(x, s, y, t), we get: M Y α (x, s, y, t) = −φ(x, s, y, t) +
Z tZ s
φ(z, s, y, σ)M Z α (x, σ, z, t)dzdσ.
R
Using the equation (5) and the previous equalities we find: M Γ(x, s, y, t) = −φ(x, s, y, t) +
Z tZ s
α
R
+ M Z (x, s, y, t) = 0.
φ(z, s, y, σ)M Z α (x, σ, z, t)dzdσ
(5)
4. Explicit expansion of Hk We start this section by proving the following theorem: 4.1 Theorem: If x ≤ 0 and y < 0: √ √ 2α(1 − α)( p2 − p1 ) (1 − α)2 e1 Z3 (x, s, y, t) H2 (x, s, y, t) = Z (x, s, y, t) + √ 2 p1 2p1 + √
α(α − 1)e2 e1 (y − x) α Z (x, s, y, t). √ √ Z4 (x, s, y, t) + p1 ( p1 + p2 ) 2p1
If x > 0 and y ≥ 0: √ √ 2α(1 − α)( p1 − p2 ) α2 e2 Z2 (x, s, y, t) − Z3 (x, s, y, t) H2 (x, s, y, t) = √ p2 2p2 − √
α(α − 1)e1 e2 (y − x) α Z (x, s, y, t). √ √ Z4 (x, s, y, t) + p2 ( p1 + p2 ) 2p2
If x ≤ 0 and y ≥ 0: αΛα e2 α H2 (x, s, y, t) = √ Z1 (x, s, y, t) − 2αZ2 (x, s, y, t) + yZ2 (x, s, y, t) p2 p2 + (
e2 α2 e1 αΛα e1 α − )Z4 (x, s, y, t) − √ Z3 (x, s, y, t). p2 − p1 2p2 2 p2 (p2 − p1 )
If x > 0 and y < 0: H2 (x, s, y, t) =
+
(1 − α)Λα Z1 (x, s, y, t) − 2(1 − α)Z2 (x, s, y, t) √ p1 e1 (1 − α) e2 (1 − α) e1 (1 − α)2 yZ2 (x, s, y, t) + ( + )Z4 (x, s, y, t) p1 p2 − p1 2p1
e2 (1 − α)Λα + − √ Z3 (x, s, y, t). 2 p1 (p2 − p1 ) Proof: If x > 0 and y ≥ 0; Using the H2 definition we get: H2 (x, s, y, t) = =
Z tZ s
−∞
Z tZ 0 s
+
R1 (z, s, y, σ)Z α (x, σ, z, t)dzdσ
Z tZ 0 s
+
R?
−∞
(p1 − p2 ) e1
Z t Z +∞ s
0
∂ 2Z α (z, s, y, σ)Z α (x, σ, z, t)dzdσ ∂z 2
∂Z α (z, s, y, σ)Z α (x, σ, z, t)dzdσ ∂z
e2
∂Z α (z, s, y, σ)Z α (x, σ, z, t)dzdσ. ∂z
Using the same technique employed in the calculation of R2 and the Laplace trans3/2 formation values of the functions ϕ1/2 and ψa (cf. Appendix A), we obtain the a , aϕa result. Now, we give an explicit expansion of Hk for k ≥ 3. 4.2 Theorem: For each integer k ≥ 3, for each x ∈ R and y ∈ R we have: Hk (x, s, y, t) = dk,1 (x, y)Z1 (x, s, y, t) + dk,2 (x, y)Z2 (x, s, y, t) + dk,3 (x, y)yZ1 (x, s, y, t) + dk,4 (x, y)yZ2 (x, s, y, t) + dk,5 (x, y)xZ1 (x, s, y, t) + dk,6 (x, y)Z3 (x, s, y, t) + dk,7 (x, y)Z4 (x, s, y, t) + Sk (x, s, y, t) where for ν = 1, 2, ..., 7, dk,ν (x, y) = Ak,ν 1R− (x)1R+ (y) + Bk,ν 1R?+ (x)1R?− (y) + Ck,ν 1R?+ (x)1R+ (y) + Dk,ν 1R− (x)1R?− (y), and for l ≥ 1: αΛα A2l,1 = √ Sαl−1 , A2l+1,1 = 0; p2
B2l,1 =
(1 − α)Λα l−1 Sα , B2l+1,1 = 0; √ p1
C2l,1 = 0,
Λα Λα C2l+1,1 = √ √ Sαl ; D2l+1,1 = √ √ S l , D2l,1 = 0; A2l,2 = −2αSαl−1 , ( p2 − p2 ) ( p1 − p2 ) α √ −2 p1 l l−1 l A2l+1,2 = 2αSα ; B2l,2 = −2(1 − α)Sα , B2l+1,2 = 2(1 − α)Sα ; C2l,2 = √ √ Sαl , p1 − p2 √ √ √ 2 p2 −2 p2 2 p1 C2l+1,2 = √ √ Sαl ; D2l,2 = √ √ Sαl , D2l+1,2 = √ √ Sl ; p1 − p2 p1 − p2 p1 − p2 α A2l+1,3 = C2l,3 =
αΛα e2 l−1 S , A2l,3 = 0; 2p2 3/2 α
B2l+1,3 =
(1 − α)Λα e1 l−1 Sα , B2l,3 = 0; 2p1 3/2
e 2 Λα e1 Λα √ √ Sαl−1 , C2l+1,3 = 0; D2l,3 = √ √ S l−1 , D2l+1,3 = 0; 2p2 ( p2 − p2 ) 2p1 ( p1 − p2 ) α
αe2 l−1 (1 − α)e1 l−1 (1 − α)e1 l−1 αe2 l−1 Sα , A2l+1,4 = − Sα ; B2l,4 = Sα , B2l+1,4 = − Sα ; p2 p2 p1 p1 √ √ √ e2 p1 e2 p1 e1 p2 l−1 l C2l,4 = √ √ S , C2l+1,4 = √ √ S ; D2l,4 = √ √ S l−1 , p2 ( p1 − p2 ) α p 2 ( p2 − p1 ) α p1 ( p2 − p1 ) α √ e 1 p2 −αΛα e1 l−1 D2l+1,4 = √ √ Sαl ; A2l+1,5 = √ S , A2l,5 = 0; B2l,5 = 0, p1 ( p 1 − p 2 ) 2p1 p2 α A2l,4 =
B2l+1,5 =
−(1 − α)Λα e2 l−1 e 2 Λα Sα ; C2l,5 = √ √ √ S l−1 , C2l+1,5 = 0; D2l+1,5 = 0, 2p2 p1 2p2 ( p1 − p2 ) α
D2l,5 =
e 1 Λα √ √ S l−1 ; 2p1 ( p2 − p1 ) α
−Ml,3 Λα Ml,4 (1 − α) Al+1,6 = √ + √ √ ; 4 p2 (p2 − p1 ) 4 p2 p1
Ll,4 (1 − α) Nl,3 Λα Nl,4 α Ml,4 (1 − α) Ll,3 Λα − √ √ ; Cl+1,6 = √ − + √ √ ; Bl+1,6 = √ 4 p1 (p1 − p2 ) 4 p2 p1 4 p2 (p1 − p2 ) 4p2 4 p 2 p1 Kl,3 Λα Kl,4 (1 − α) Ll,4 α Dl+1,6 = √ + − √ √ ; 4 p1 (p1 − p2 ) 4p1 4 p 2 p1 Al+1,7 =
Nl,3 α Nl,4 α Ml,3 − √ √ − ; √ 2(p2 − p1 ) 2 p2 ( p1 + p2 ) 4p2
Ll,3 Kl,3 (1 − α) Kl,4 (1 − α) + √ √ + ; √ 2(p2 − p1 ) 2 p1 ( p1 + p2 ) 4p1 √ √ Nl,3 p1 Kl,3 p2 Ml,3 (1 − α) −Ll,3 α = √ √ ; Dl+1,7 = √ √ ; √ + √ √ + √ 2 p2 ( p1 + p2 ) 2 p2 (p2 − p1 ) 2 p1 ( p1 + p2 ) 2 p1 (p2 − p1 ) Bl+1,7 =
Cl+1,7
The rest Sk (x, s, y, t) satisfies : If y < 0 :
| Sk (x, s, y, t) |≤ Cte
7 X
(| Bk,j | + | Dk,j |)
j=1
.
If y > 0 :
h√
√ √ ( p1 x − p2 y)2 1 (x − y)2 i )+ √ ). t − sexp(− exp(− 8p1 p2 (t − s) 8p1 (t − s) t−s
| Sk (x, s, y, t) |≤ Cte
7 X
(| Ak,j | + | Ck,j |)
j=1
√ √ ( p1 y − p2 x)2 (x − y)2 i 1 √ . t − sexp(− exp(− )+ ). 8p1 p2 (t − s) 8p2 (t − s) t−s In order to prove this theorem, we need the following double integral values: h√
4.3 Preliminary calculus: Using the same technique employed in the calculation of R2 , we get: If x > 0 and y ≥ 0: ∂ 2 Z2 (z, s, y, σ)Z α (x, σ, z, t)dzdσ ∂z 2 s 0 Λα 1 Z2 (x, s, y, t). = √ Z1 (x, s, y, t) − 2 p1 (p2 − p1 ) p2 − p1
ζ1 =
Z t Z +∞
∂ 2 Z2 (z, s, y, σ)Z α (x, σ, z, t)dzdσ ∂z 2 s 0 Λα y y = √ Z1 (x, s, y, t) − Z2 (x, s, y, t). 2 p1 (p2 − p1 ) p2 − p1
ζ2 =
Z t Z +∞
y
Z t Z +∞
∂Z2 (z, s, y, σ)Z α (x, σ, z, t)dzdσ ∂z s 0 √ p1 Λα = √ Z3 (x, s, y, t) − √ Z4 (x, s, y, t). 4 p2 (p1 − p2 ) 2 p2 (p1 − p2 )
ζ3 =
ζ4 =
Z t Z +∞ s
0
x α ∂Z1 (z, s, y, σ)Z α (x, σ, z, t)dzdσ = − Z1 (x, s, y, t) − Z3 (x, s, y, t). ∂z 2p2 4p2
ζ10
=
−∞
s
Z tZ 0
ζ20 =
y
−∞
s
ζ30 =
∂ 2 Z2 (1 − α) (z, s, y, σ)Z α (x, σ, z, t)dzdσ = √ √ √ Z2 (x, s, y, t). 2 ∂z p2 ( p1 + p2 )
Z tZ 0
∂Z2 (1 − α) (z, s, y, σ)Z α (x, σ, z, t)dzdσ = √ √ √ Z4 (x, s, y, t). ∂z 2 p 2 ( p1 + p2 )
Z tZ 0 −∞
s
∂ 2 Z2 (1 − α)y (z, s, y, σ)Z α (x, σ, z, t)dzdσ = √ √ √ Z2 (x, s, y, t). 2 ∂z p 2 ( p1 + p2 )
∂Z1 (1 − α) (z, s, y, σ)Z α (x, σ, z, t)dzdσ = √ √ Z4 (x, s, y, t). ∂z 4 p2 p1
Z tZ 0
ζ40 =
−∞
s
If x ≤ 0 and y ≥ 0: χ1 = χ2 = χ3 =
Z t Z +∞ s
0
Z t Z +∞ s
0
∂ 2 Z2 αy y 2 (z, s, y, σ)Z α (x, σ, z, t)dzdσ = √ √ √ Z2 (x, s, y, t). ∂z p1 ( p1 + p2 )
Z t Z +∞ s
0
χ4 =
∂ 2 Z2 α α (z, s, y, σ)Z (x, σ, z, t)dzdσ = √ √ √ Z2 (x, s, y, t). ∂z 2 p 1 ( p1 + p2 )
α ∂Z2 (z, s, y, σ)Z α (x, σ, z, t)dzdσ = √ √ √ Z4 (x, s, y, t). ∂z 2 p2 ( p1 + p2 )
Z t Z +∞ s
0
α ∂Z1 (z, s, y, σ)Z α (x, σ, z, t)dzdσ = Z4 (x, s, y, t). ∂z 4p2
∂ 2 Z2 (z, s, y, σ)Z α (x, σ, z, t)dzdσ s −∞ ∂z 2 Λα 1 = √ Z1 (x, s, y, t) − Z2 (x, s, y, t). 2 p2 (p1 − p2 ) p1 − p2
χ01 =
χ02
Z tZ 0
∂ 2 Z2 (z, s, y, σ)Z α (x, σ, z, t)dzdσ ∂z 2 s −∞ y Λα y = √ Z1 (x, s, y, t) − Z2 (x, s, y, t). 2 p2 (p1 − p2 ) p1 − p2 =
Z tZ 0
y
Z tZ 0
∂Z2 (z, s, y, σ)Z α (x, σ, z, t)dzdσ s −∞ ∂z Λα 1 = √ Z3 (x, s, y, t) − Z4 (x, s, y, t). 4 p2 (p1 − p2 ) 2(p1 − p2 )
χ03 =
χ04
Z tZ 0
∂Z1 (z, s, y, σ)Z α (x, σ, z, t)dzdσ s −∞ ∂z x α = − Z1 (x, s, y, t) − √ √ Z3 (x, s, y, t). 2p1 4 p1 p2 =
If x ≤ 0 and y < 0: ∂ 2 Z2 (z, s, y, σ)Z α (x, σ, z, t)dzdσ s −∞ ∂z 2 Λα 1 = √ Z1 (x, s, y, t) − Z2 (x, s, y, t). 2 p2 (p1 − p2 ) p1 − p 2
ϑ1 =
Z tZ 0
∂ 2 Z2 (z, s, y, σ)Z α (x, σ, z, t)dzdσ ∂z 2 s −∞ y Λα y Z1 (x, s, y, t) − Z2 (x, s, y, t). = √ 2 p2 (p1 − p2 ) p1 − p 2 Z tZ 0
ϑ2 =
y
Z tZ 0
∂Z2 (z, s, y, σ)Z α (x, σ, z, t)dzdσ s −∞ ∂z √ p2 Λα = √ Z3 (x, s, y, t) − √ Z4 (x, s, y, t). 4 p1 (p1 − p2 ) 2 p1 (p1 − p2 )
ϑ3 =
ϑ4 =
Z tZ 0 s
−∞
∂Z1 (z, s, y, σ)Z α (x, σ, z, t)dzdσ ∂z
x (1 − α) Z1 (x, s, y, t) + Z3 (x, s, y, t). 2p1 4p1 ∂ 2 Z2 α α (z, s, y, σ)Z (x, σ, z, t)dzdσ = √ √ √ Z2 (x, s, y, t). ∂z 2 p1 ( p1 + p2 ) = −
Z t Z +∞
ϑ01 =
s
0
Z t Z +∞
ϑ02 =
s
ϑ3 =
0
Z t Z +∞ s
ϑ04
αy ∂ 2 Z2 α (z, s, y, σ)Z (x, σ, z, t)dzdσ = √ √ √ Z2 (x, s, y, t). ∂z 2 p1 ( p1 + p2 )
y
0
=
∂Z2 −α (z, s, y, σ)Z α (x, σ, z, t)dzdσ = √ √ √ Z4 (x, s, y, t). ∂z 2 p1 ( p1 + p2 )
Z t Z +∞ s
0
∂Z1 −α (z, s, y, σ)Z α (x, σ, z, t)dzdσ = √ √ Z4 (x, s, y, t). ∂z 4 p1 p2
If x > 0 and y < 0: Υ1 = Υ2 = Υ3 =
Z tZ 0 s
−∞
Z tZ 0 s
y
−∞
Z tZ 0 s
−∞
Υ4 =
(1 − α) ∂ 2 Z2 (z, s, y, σ)Z α (x, σ, z, t)dzdσ = √ √ √ Z2 (x, s, y, t). 2 ∂z p2 ( p1 + p2 ) ∂ 2 Z2 (1 − α)y (z, s, y, σ)Z α (x, σ, z, t)dzdσ = √ √ √ Z2 (x, s, y, t). 2 ∂z p 2 ( p1 + p2 )
∂Z2 (1 − α) (z, s, y, σ)Z α (x, σ, z, t)dzdσ = √ √ √ Z4 (x, s, y, t). ∂z 2 p 1 ( p1 + p2 )
Z tZ 0 s
Υ01
Υ02
−∞
∂Z1 (1 − α) (z, s, y, σ)Z α (x, σ, z, t)dzdσ = Z4 (x, s, y, t). ∂z 4p1
∂ 2 Z2 (z, s, y, σ)Z α (x, σ, z, t)dzdσ 2 ∂z s 0 Λα 1 = √ Z1 (x, s, y, t) − Z2 (x, s, y, t). 2 p1 (p2 − p1 ) p2 − p1 =
Z t Z +∞
∂ 2 Z2 (z, s, y, σ)Z α (x, σ, z, t)dzdσ ∂z 2 s 0 Λα y y = √ Z1 (x, s, y, t) − Z2 (x, s, y, t). 2 p1 (p2 − p1 ) p2 − p1
=
Z t Z +∞
y
Z t Z +∞
∂Z2 (z, s, y, σ)Z α (x, σ, z, t)dzdσ ∂z s 0 1 Λα Z3 (x, s, y, t) − Z4 (x, s, y, t). = √ 4 p1 (p1 − p2 ) 2(p1 − p2 )
Υ03 =
Υ04
Z t Z +∞
∂Z1 (z, s, y, σ)Z α (x, σ, z, t)dzdσ ∂z s 0 x (1 − α) = − Z1 (x, s, y, t) − √ √ Z3 (x, s, y, t). 2p2 4 p1 p2 =
Proof of the theorem: We present the proof in the case where x ≤ 0 and y ≥ 0. Hk+1 (x, s, y, t) = =
Z t Z +∞ s
0
Z tZ R?
s
Rk (z, s, y, σ)Z α (x, σ, z, t)dzdσ
Rk (z, s, y, σ)Z α (x, σ, z, t)dzdσ +
Z tZ 0 −∞
s
Rk (z, s, y, σ)Z α (x, σ, z, t)dzdσ.
Using the Rk expansion given in the theorem 3.2 and the double integral values given in the paragraph 4.3, we get: Hk+1 (x, s, y, t) =
4 X
4 X
Nk,j χj +
Mk,j χ0j +
Z tZ
j=1
j=1
s
R?
Wk (z, s, y, σ)Z α (x, σ, z, t)dzdσ
= Ak+1,1 Z1 (x, s, y, t) + Ak+1,2 Z2 (x, s, y, t) + Ak+1,3 yZ1 (x, s, y, t) + Ak+1,4 yZ2 (x, s, y, t) + Ak+1,5 xZ1 (x, s, y, t) + Ak+1,6 Z3 (x, s, y, t) + Ak+1,7 Z4 (x, s, y, t) + Sk+1 (x, s, y, t). where Λα Mk,1 αNk,1 Mk,1 Λα Mk,2 Ak+1,1 = √ +√ √ ; Ak+1,2 = ; √ ; Ak+1,3 = √ 2 p2 (p1 − p2 ) p2 − p1 p1 ( p1 + p2 ) 2 p2 (p1 − p2 ) Ak+1,4 =
Mk,2 αNk,2 +√ √ √ ; p2 − p1 p 1 ( p1 + p2 )
Ak+1,5 =
−Mk,4 ; 2p1
Mk,3 αNk,3 αNk,4 − √ √ − ; √ 2(p2 − p1 ) 2 p2 ( p1 + p2 ) 4p2
Ak+1,7 = Z tZ
Sk+1 (x, s, y, t) =
s
Λα Mk,3 (1 − α)Mk,4 Ak+1,6 = √ + √ √ ; 4 p2 (p1 − p2 ) 4 p1 p2
R?
Wk (z, s, y, σ)Z α (x, σ, z, t)dzdσ.
Using the Mk,j and Nk,j values given in the theorem 3.2 and, remarking that: | Sk+1 (x, s, y, t) |≤ cte
4 X
(| Mk,j | + | Nk,j |)
j=1
! √ √ ( p1 z − p2 y)2 1 (z − y)2 √ . exp(− ) + exp(− ) | Z α (x, σ, z, t) | dzdσ 8p1 p2 (σ − s) 8p1 (σ − s) σ−s s R √ √ 7 h√ X ( p1 y − p2 x)2 1 (x − y)2 i ≤ Cte (| Ak,j | + | Ck,j |) t − sexp(− )+ √ ), exp(− 8p1 p2 (t − s) 8p2 (t − s) t−s j=1 Z tZ
we obtain the result 2 Proof of the main theorem: Reminding that: L = [p1 1R?− (x)+p2 1R+ (x)]
∂ ∂ ∂ ∂2 +[e1 1R?− (x)+e2 1R+ (x)] +δ{0} [A (0+ )+B (0− )], 2 ∂x ∂x ∂x ∂x
where √ √ √ √ √ p2 [αp1 − (1 − α)p2 ][( p1 + p2 ) − (1 − 2α)( p2 − p1 )] A = (4α − 1)p2 + , √ √ √ √ 2( p1 p2 + α(1 − α)( p2 − p1 )2 ) √ B = (4α − 3)p1 +
√ √ √ √ p1 [αp1 − (1 − α)p2 ][( p1 + p2 ) − (1 − 2α)( p2 − p1 )] , √ √ √ √ 2( p1 p2 + α(1 − α)( p2 − p1 )2 )
we shall prove that for any continuous function h on R of class C 2 in R− and in R+ , for any continuous, compactly supported function g on R: Ξ = lim
Z
Z
t→0 x∈R
t−1 g(x)[h(y) − h(x)]Γ(x, 0, y, t)dydx =< Lh, g > .
y∈R
Now, Ξ = lim
Z
Z
t→0 x∈R
+ lim
Z
Z
t→0 x∈R
+ lim
t−1 g(x)[h(y) − h(x)]Z α (x, 0, y, t)dydx
y∈R
Z
y∈R
Z
t→0 x∈R
t−1 g(x)[h(y) − h(x)]H2 (x, 0, y, t)dydx t−1 g(x)[h(y) − h(x)]
y∈R
X
Hk (x, 0, y, t)dydx = E + F + G.
k≥3
We start by calculate E. Since h is not of class C 1 in 0, put λ = h0 (0+ ) − h0 (0− )
and
ϕ(x) = h(x) − λx+ .
The function ϕ is continuously differentiable in 0, ϕ0 (0) = h(0− ) and E = lim
Z
Z
t→0 x∈R?
+ lim
Z
Z
t→0 x∈R?
Now, Then
t−1 g(x)[ϕ(y) − ϕ(x)]Z α (x, 0, y, t)dydx
y∈R
y∈R
t−1 g(x)λ[y + − x+ ]Z α (x, 0, y, t)dydx = E1 + E2 .
∀x ∈ R? , ϕ(y) = ϕ(x) + (y − x)ϕ0 (x) +
E1 = lim
Z
t→0 x∈R?
+ lim
Z
t→0 x∈R?
+ lim
Z
t→0 x∈R?
Z
(y − x)2 00 ϕ (x) + o(| y − x |2 ). 2
t−1 g(x)(y − x)ϕ0 (x)Z α (x, 0, y, t)dydx
y∈R
Z y∈R
Z y∈R
g(x) (y − x)2 ϕ00 (x)Z α (x, 0, y, t)dydx 2t g(x) o(| y − x |2 )Z α (x, 0, y, t)dydx = E11 + E12 + E13 . t
And via a simple calculation we get: E11 = [2(p2 − p1 ) + 3(2α − 1)(p1 + p2 )]g(0)ϕ0 (0); E12 =
Z x∈R?
[p1 1R?− (x) + p2 1R?+ (x)]g(x)ϕ00 (x)dx;
E13 = 0.
Now, we calculate E2 . E2 = lim t−1 λg(0) t→0
Z
Z
x∈R?
h Z +∞ Z +∞
−1
= lim t λg(0) t→0
−
0
Z +∞ Z 0
[y + − x+ ]Z α (x, 0, y, t)dydx
y∈R α
[y − x]Z (x, 0, y, t)dydx +
Z 0 Z +∞ −∞
0
yZ α (x, 0, y, t)dydx
0
i
xZ α (x, 0, y, t)dydx .
−∞
0
And via some suitable change of variables we obtain: E2 = [p2 (5α − 1) − (1 − α)p1 ]λg(0), and consequently, Z
E =
x∈R?
[p1 1R?− (x) + p2 1R?+ (x)]g(x)h00 (x)dx
+ [p2 (5α − 1) − (1 − α)p1 ]h0 (0+ ) + [αp2 + (5α − 4)p1 ]h0 (0− )g(0). To calculate F and G, we need the following results: lim
Z +∞ Z +∞
t→0 0
0
Z 0 Z 0
Z +∞ e2 0 2 α e2 ϕ0 (x)g(x)dx. g(x)ϕ (x)(y − x) Z (x, 0, y, t))dydx = 2p2 t 0
Z 0 e1 e1 ϕ0 (x)g(x)dx. g(x)ϕ0 (x)(y − x)2 Z α (x, 0, y, t))dydx = t→0 −∞ −∞ 2p1 t −∞ √ √ √ Z +∞ Z +∞ p1 ( p2 − p1 ) g(x) 0 lim ϕ (x)(y − x)Z2 (x, 0, y, t))dydx = g(0)ϕ0 (0). t→0 0 t 2 0 √ √ √ Z 0 Z 0 p2 ( p 2 − p 1 ) g(x) 0 ϕ (x)(y − x)Z2 (x, 0, y, t))dydx = g(0)ϕ0 (0). lim t→0 −∞ −∞ t 2 √ √ √ Z 0 Z +∞ p1 ( p1 + p 2 ) g(x) 0 lim ϕ (x)(y − x)Z1 (x, 0, y, t))dydx = g(0)ϕ0 (0). t→0 −∞ 0 t 2 √ √ √ Z +∞ Z 0 − p2 ( p1 + p2 ) g(x) 0 lim ϕ (x)(y − x)Z1 (x, 0, y, t))dydx = g(0)ϕ0 (0). t→0 0 t 2 −∞ Z 0 Z +∞ g(x) 0 lim ϕ (x)(y − x)Z2 (x, 0, y, t))dydx = p2 g(0)ϕ0 (0). t→0 −∞ 0 t Z +∞ Z 0 g(x) 0 lim ϕ (x)(y − x)Z2 (x, 0, y, t))dydx = −p1 g(0)ϕ0 (0). t→0 0 t −∞ Z 0 Z +∞ g(x) 0 lim ϕ (x)(y − x)Zj (x, 0, y, t))dydx = 0; j = 3, 4. t→0 −∞ 0 t
lim
lim
Z +∞ Z 0
t→0 0
lim
−∞
Z 0 Z 0
t→0 −∞
lim
g(x) 0 ϕ (x)(y − x)Zj (x, 0, y, t))dydx = 0; t
g(x) 0 ϕ (x)(y − x)Zj (x, 0, y, t))dydx = 0; t
−∞
Z +∞ Z +∞
t→0 0
0
lim
g(x) 0 ϕ (x)(y − x)Zj (x, 0, y, t))dydx = 0; t
Z 0 Z +∞
t→0 −∞
0
j = 3, 4. j = 1, 3, 4. j = 1, 3, 4.
g(x) p2 yZ2 (x, 0, y, t))dydx = g(0). t 2
Z +∞ Z 0
g(x) p2 xZ1 (x, 0, y, t))dydx = g(0). t→0 0 t 2 −∞ √ √ Z 0 Z +∞ p 1 p2 g(x) lim yZ1 (x, 0, y, t))dydx = g(0). t→0 −∞ 0 t 2 lim
Z +∞ Z 0
g(x) p1 xZ2 (x, 0, y, t))dydx = g(0). t 2 −∞ Using the H2 expression given in theorem 4.1 and the same technique used in the calculation of E we obtain: lim
t→0 0
F =
Z x∈R?
[e1 1R?− (x) + e2 1R+ (x)]h0 (x)g(x)dx
1 (1 − α)Λα √ + [p1 (Sα + 1 − α) + αΛα p1 − (α + )p2 ]h0 (0+ ) √ 2 2 p1 1 αΛα √ − [p2 (Sα + α) + (1 − α)Λα p2 − (1 − α + √ )p1 ]h0 (0− ). 2 2 p2 Now, we will calculate G: G = lim
Z
Z
t→0 x∈R?
= lim + lim
+ lim
−∞
X
t−1 g(x)(h(y) − h(x))
0
Hk (x, 0, y, t)dydx
X
Hk (x, 0, y, t)dydx
k≥3
t−1 g(x)(h(y) − h(x))
0
Z +∞ Z 0
t→0 0
Hk (x, 0, y, t)dydx
k≥3
Z +∞ Z +∞
t→0 0
X k≥3
t−1 g(x)(h(y) − h(x))
Z 0 Z +∞
t→0 −∞
+ lim
y∈R
Z 0 Z 0
t→0 −∞
t−1 g(x)(h(y) − h(x))
X
Hk (x, 0, y, t)dydx
k≥3
t−1 g(x)(h(y) − h(x))
−∞
X
Hk (x, 0, y, t)dydx
k≥3
= G1 + G2 + G3 + G4 . Using the Hk expansion given in the theorem 4.2 , we get:
G1 =
X
Dk,1 lim
t→0 −∞
k≥3
+
X
Dk,2 lim
X
Dk,3 lim
+
X
Dk,4 lim
+
X
Dk,5 lim
+
X
Dk,6 lim
+
X
Dk,7 lim
+ lim
Z 0 Z 0
t→0 −∞
−∞
Z 0 Z 0
t→0 −∞
k≥3
−∞
Z 0 Z 0
t→0 −∞
k≥3
−∞
Z 0 Z 0
t→0 −∞
k≥3
−∞
Z 0 Z 0
t→0 −∞
k≥3
−∞
Z 0 Z 0
t→0 −∞
k≥3
−∞
Z 0 Z 0
t→0 −∞
k≥3
+
Z 0 Z 0
−∞
t−1 g(x)(h(y) − h(x))Z1 (x, 0, y, t)dydx t−1 g(x)(h(y) − h(x))Z2 (x, 0, y, t)dydx t−1 g(x)(h(y) − h(x))y.Z1 (x, 0, y, t)dydx t−1 g(x)(h(y) − h(x))y.Z2 (x, 0, y, t)dydx t−1 g(x)(h(y) − h(x))x.Z1 (x, 0, y, t)dydx t−1 g(x)(h(y) − h(x))Z3 (x, 0, y, t)dydx t−1 g(x)(h(y) − h(x))Z4 (x, 0, y, t)dydx
t−1 g(x)(h(y) − h(x))
−∞
X
Sk (x, 0, y, t)dydx.
k≥3
The last six limits are zero. Using the Dk,j expressions (cf. theorem 4.2) we obtain: √ √ X X 2α(1 − α) p1 − p2 ) S α Λα Dk,1 = √ Dk,2 = , , √ √ ( p1 − p2 )(1 − Sα ) p1 k≥3 k≥3 and consequently
G1 = p2 Sα g(0)h0 (0− ).
As above, we calculate the sums: X k≥3
Ck,1
Sα Λα = √ , √ ( p2 − p1 )(1 − Sα )
(1 − α)Sα Λα Bk,1 = √ , p (1 − S ) α k≥3 1 X
X k≥3
Bk,2 = 0,
X
Ck,2
k≥3
X k≥3
and we deduce that: √ √ i αSα Λα p1 h p1 0 − G2 = √ h (0 ) + h0 (0+ ) , 2(1 − Sα ) p2
√ √ 2α(1 − α) p2 − p1 ) = , √ p2
Ak,1 = √
αSα Λα , p2 (1 − Sα )
X
Ak,2 = 0,
k≥3
G3 = −p1 Sα g(0)h0 (0+ ),
√ √ i (1 − α)Sα Λα p2 h 0 − p G4 = − h (0 ) + √ 2 h0 (0+ ) . 2(1 − Sα ) p1
Hence, G =
h
p2 S α +
S α Λα p1 √ i (α √ − (1 − α) p2 ) h0 (0− ) 2(1 − Sα ) p2
−
h
p1 Sα +
S α Λα p2 √ i ((1 − α) √ − α p1 ) h0 (0+ ). 2(1 − Sα ) p1
And consequently we obtain, Ξ = +
Z x∈R?
Z x∈R?
[p1 1R?− (x) + p2 1R?+ (x)]g(x)h00 (x)dx [e1 1R?− (x) + e2 1R+ (x)]h0 (x)g(x)dx
+
h
(4α − 1)p2 +
Λα (αp1 − (1 − α)p2 ) i 0 + h (0 )g(0) √ 2 p1 (1 − Sα )
+
h
(4α − 3)p1 +
Λα (αp1 − (1 − α)p2 ) i 0 − h (0 )g(0). √ 2 p2 (1 − Sα )
Proceeding in the same way, we prove that for every bounded continuous function h on R, if x ∈ R Z lim Γ(x, s, y, t)h(y)dy = h(x), s→t R
which finishs the theorem proof 2 5.Study of some particular cases 5.1 The case where α = If α =
1 2
1 2
:
we obtain the fundamental solution of the equation (L −
L = [p1 1R?− (x)+p2 1R+ (x)] where p1 and p2 satisfy:
∂ )u ≡ 0, with ∂t
∂ √ √ ∂ ∂ ∂2 +[e1 1R?− (x)+e2 1R+ (x)] + p1 p2 δ{0} [ (0+ )− (0− )], 2 ∂x ∂x ∂x ∂x √ √ ( p2 − p1 )2 | |< 1. √ √ 4 p1 p2
Which corresponds to the previous work result of Zili [10]. 5.2 The case where e1 = 0 and e2 = 0 : In this case, the Rk expansion is reduced to: Rk (x, s, y, t) = ck,1 (x, y)
∂ 2 Z2 (x, s, y, t), 1 ∂x2
which allows us to get the following Hk explicit expression: Hk (x, s, y, t) = dk,1 (x, y)Z1 (x, s, y, t) + dk,2 (x, y)Z2 (x, s, y, t).2 1 2
The ck,1 (x, y) expression is given in the theorem 3.2. The dk,j (x, y) expression is given in the theorem 4.2.
And we deduce that:
Γ(x, s, y, t) =
Sα Λα + (1 − 2α)]Z1 (x, s, y, t) [ √ √ ( p1 − p2 )(1 − Sα )
Sα Λα − (1 − 2α)]Z1 (x, s, y, t) [ √ √ ( p2 − p1 )(1 − Sα )
+Z 1/2 (x, s, y, t) √
if x ≤ 0, y < 0
αΛα Z1 (x, s, y, t) p2 (1 − Sα )
if x ≤ 0, y ≥ 0
+Z 1/2 (x, s, y, t)
if x > 0, y ≥ 0
(1 − α)Λα Z1 (x, s, y, t) √ p1 (1 − Sα )
if x > 0, y < 0.
Therefore, 1R? (y) 1 1R (y) − Γ(x, s, y, t) = q + √+ √ p1 p2 4π(t − s) h
. exp(−
(x0 − y 0 )2 (| x0 | + | y 0 |)2 i ) + γα (x, y)exp(− ), 4p1 (t − s) 4p1 (t − s)
√ p where z = √ 1 z + − z − , and p2 0
γα (x, y) =
√ √ −α(1 − α)[p1 − p2 + (1 − 2α)( p2 − p1 )2 ] + (1 − 2α) √ √ √ √ p1 p2 + α(1 − α)( p2 − p1 )2 √ √ √ √ √ α p1 [( p1 + p2 ) − (1 − 2α)( p2 − p1 )] −1 √ √ √ √ p1 p2 + α(1 − α)( p2 − p1 )2 √ √ −α(1 − α)[p2 − p1 − (1 − 2α)( p1 − p2 )2 ] [ − (1 − 2α) √ √ √ √ p1 p2 + α(1 − α)( p1 − p2 )2 √ √ √ √ √ (1 − α) p2 [( p1 + p2 ) − (1 − 2α)( p2 − p1 )] −1 √ √ √ √ p1 p2 + α(1 − α)( p2 − p1 )2
5.3 The case where e1 = 0, e2 = 0, and α =
1 2
if x ≤ 0, y < 0 if x ≤ 0, y ≥ 0 if x > 0, y ≥ 0 if x > 0, y < 0.
:
In this case the expression of Γ is reduced to: 1R? (y) 1 1R (y) − Γ(x, s, y, t) = q + √+ √ p1 p2 4π(t − s) √ √ h ( p1 − p2 ) (x0 − y 0 )2 (| x0 | + | y 0 |)2 i . exp(− )+ √ ), √ sgn(y)exp(− 4p1 (t − s) ( p1 + p2 ) 4p1 (t − s)
where sgn(y) is the sign of y. And we find the results of Mastrangelo [4] (proposition 16 page 43) and Talbi [8] (page 32).
References [1] A. Friedman, Partial differential equation of parabolic type. Prentice Hall Inc 1964. [2] A. Friedman, Partial differential equations. Robert E. Krieger publishing company maladar Florida 1983. [3] B. Gaveau, M. and T. Okada, Second order differential operators and Dirichlet integrals with singular coefficients. Functional calculus of one dimensional operators. Tˆohoku Math. J. t.39(1987) p.465 − 504. [4] M.Mastrangelo, R´esolution stochastique, par des processus de diffusion partiellement r´efl´echis, d’op´erateurs diff´erentiels paraboliques a` coefficients continus par morceaux et admettant un drift g´en´eralis´e. E.D.F. Bulletin de la direction des ´etudes et recherches, s´erie C Math´ematiques, Informatiques N2, 1990, pp.11 − 85. [5] M.Mastrangelo, et T. Talbi, Mouvement browniens asym´etriques modifi´es en dimension finie et op´erateurs diff´erentiels a` coefficients discontinus. Probability mathematical statistics,Vol.11, Fasc.1(1990). [6] A.P. Prudnikov, Integrals and series. V.1:p.707. [7] M.R. Spiegel, Transform´ees de Laplace. S´erie Schaum 1985. [8] M.Talbi, R´esolution stochastique d’´equations aux d´eriv´ees partielles paraboliques a` coefficients discontinus et applications physiques.Th`ese de doctorat de l’universit´e Paris 6, 1987. [9] M.Zili, D´eveloppement asymptotique en temps petits de la solution d’une ´equation aux d´eriv´ees partielles de type parabolique. C.R.Aca.Sci.Paris, t.321, S´erie I, p. 1049 − 1052, 1995. [10] M.Zili, Construction d’une solution fondamentale d’une ´equation aux d´eriv´ees partielles a` coefficients constants par morceaux. Bulletin des Sci. Math. de Paris,123, V.2, p. 115 − 155, 1999. Appendix A 3/2 Laplace transformations of ϕ1/2 a , aϕa , ψa , γa and χc . Denoting L(f ) the Laplace transformation of a function f, we can find the following formulas in Spiegel [7]. 1 1/2 ), ∀a ∈ R? , ∀s > 0, L(ϕ1/2 a )(s) = √ exp(−as s 1/2 L(aϕ3/2 ), a )(s) = 2exp(−as √ L(ψa )(s) = sexp(−as1/2 ),
1 ∀c ∈ R, ∀s > 0, L(χc )(s) = exp(−cs1/2 ), s 1 0 Remarking that γa (t) = (aϕ3/2 a ) (t) we get: 2 1 s 0 3/2 1/2 ∀a ∈ R? , ∀s > 0, L(γa )(s) = L(aϕ3/2 ). a ) (s) = L(aϕa )(s) = sexp(−as 2 2 Appendix B Integrals used in the calculations.
Z
1 erfc(ax)dx = xerfc(ax) − √ exp(−a2 x2 ). a π
x2 1 x erfc(ax) + 2 erf (ax) − √ exp(−a2 x2 ). 2 4a 2a π √ Z +∞ a π u2 exp(−u2 )du = exp(−a2 ) + erfc(a). 2 4 a √ Z b 1 π 2 2 2 2 u exp(−u )du = [aexp(−a ) − bexp(−b )] + [erfc(a) − erfc(b)]. 2 4 a Z +∞ 1 u3 exp(−u2 )du = [a2 + 1]exp(−a2 ).2 2 a Z
xerfc(ax)dx =
