The uniqueness of the solution is to be understood as a P -almost sure one: let



u_i(i= 1, 2) be such that u_i(ω^) is a viscosity solution of PDE (4.8) at ω^, for all ω^∈ Ω_i^. Then, by the uniqueness result of viscosity solution of deterministic PDEs (see: Pardoux [16, Theorem 6.14]) we know thatu1(ω^,·) = u2(ω^,·), for all ω^∈ Ω₁^ ∩ Ω₂^. In particular,



u1= u2, P-a.s.

Proof of Theorem 4.9. We adapt the method used in the paper of El Karoui et al. [7] to show that u is a pathwise viscosity subsolution of Eq. (4.8). Recall that the set Ω^ := {ω^∈ Ω^|

I_T^∗(ω^) <+∞} satisfies P^B( Ω^)= 1. We work here on the set Ω^:= {ω^∈ Ω^| u(ω^,·,·) is continuous} which satisfies P^B( Ω^)= 1 in light of Lemma 4.4.

Now, according to the definition of the viscosity solution, for an arbitrarily chosen ω^∈ Ω^, we fix arbitrarily a point (t, x)∈ [0, T ] × R^d and a test function ϕ∈ C_b^∞ such that ϕ(t, x)=

We emphasize that for fixed ω^, this BSDE can be viewed as a classical BSDE with respect to W and we recall that Y_t^t,x_−h(ω^,·) = u(ω^, t− h, X_t^t,x_−h). Now for the fixed ω^∈ Ω^, it holds that (4.13) and Itô’s formula we have

Y_s^t,x

(∇xϕσ )

From the fact that f is Lipschitz and the estimates (4.14) and (4.15), we get

Y_t^t,x

Henceu(ω^, t, x)is a pathwise viscosity subsolution of (4.8). The proof ofubeing a pathwise viscosity supersolution is similar.

The proof of uniqueness becomes clear from Remark 4.10. 2

In analogy to the relation between the solutions (Y, Z) and (Y , Z)of the associated BDSDE and the BSDE, respectively, we shall expect that u(t, x)= Yt^t,x= Y_t^t,x(At)εt, (t, x)∈ [0, T ] × R^d, is a solution of SPDE (4.9). This claim is confirmed by

Proposition 4.11. Suppose that u,u are C^0,2-stochastic fields over Ω^× [0, T ] × R^d such that there exist δ > 0 and a constant Cδ,x>0 (only depending on δ and x) with:

E

w(t, x)^2+δ+

t

0

∇xw(s, x)^2+δ+D_xx² w(s, x)^2+δ ds



 Cδ,x, t∈ [0, T ], for w = u,u. (4.16)

Thenu(t, x) is a classical pathwise solution of Eq.(4.8) if and only if u(t, x) is a classical solution of SPDE (4.9).

Proof. We restrict ourselves to show that u(t, x) solves Eq. (4.9) wheneverusolves (4.8). For this, we proceed in analogy to the proof of Theorem 3.9. Let F be an arbitrary but fixed element ofS_K^ . By using the Girsanov transformation we have

E

u(t, x)F− u(0, x)F

= E

u(A_t, t, x)ε_tF− u(0, x)F

= E

F (T_t)u(t, x)− Fu(0, x)

= E

F (Tt)u(0, x)− Fu(0, x)

+ E

 F (T_t)

t

0

Lu(s, x)+ f

s, x,u(s, x)ε_s(T_s),

∇xu(s, x)σ (x)ε_s(T_s)

ε⁻¹_s (T_s) ds

 .

As in the proof of Theorem 3.9 we use the fact that _dt^dF (T_t)= γt(K^∗KD^BF )(t, T_t)to deduce the following

E

u(t, x)F− u(0, x)F

= E



u(0, x)

t

0

γ_s

K^∗KD^BF (T_s) (s)ds



+ E

t

0

t

s

γr

K^∗KD^BF (Tr)

(r)drLu(s, x)ds



+ E

t

0

F (T_s)

Lu(s, x)+ f

s, x,u(s, x)ε_s(T_s),∇xu(s, x)σ (x)ε_s(T_s)

ε_s⁻¹(T_s) ds



+ E

t

0

t

s

γr

K^∗KD^BF (Tr) (r)drf

s, x,u(s, x)εs(Ts),

∇xu(s, x)σ (x)ε_s(T_s)

ε_s⁻¹(T_s)ds

 .

Thanks to the assumption thatu(t, x)is a pathwise classical solution of Eq. (4.8), we obtain E

u(t, x)F− u(0, x)F

= E

t

0

F (T_s)

Lu(s, x)+ f

s, x,u(s, x)ε_s(T_s),∇xu(s, x)σ (x)ε_s(T_s)

ε_s⁻¹(T_s) ds



+ E

t

0

γ_s

K^∗KD^BF (T_s)

(s)u(s, x)ds



= E

t

0

F

Lu(s, x) + f

s, x, u(s, x),∇xu(s, x)σ (x) ds



+ E

t

0

γ_s

K^∗KD^BF

(s)u(s, x)ds

 .

Consequently,

E

t

0

γ_s

K^∗KD^BF

(s)u(s, x)ds



= E

 F

u(t, x)− u(0, x) −

t

0

Lu(s, x) + f

s, x, u(s, x),∇xu(s, x)σ (x) ds



.

From the integrability condition (4.16) we know that

u(t, x)− u(0, x) −

t

0

Lu(s, x) + f

s, x, u(s, x),∇xu(s, x)σ (x)

ds∈ L²(Ω,F, P ).

Moreover, γ 1_[0,t]u∈ L²([0, T ] × Ω). Indeed,

E

T

0

γ_r1_[0,t](r)u(r, x)²dr



=

T

0

γ_r1_[0,t](r)²E u(r, x)²

dr Cδ,x

T

0

γ_r1_[0,t](r)²dr <∞.

By Definition 3.2 we get

E

 F

t

0

γsu(s, x)dBs



= E

 F

u(t, x)− u(0, x) −

t

0

Lu(s, x) + f

s, x, u(s, x),∇xu(s, x)σ (x) ds



.

It then follows from the arbitrariness of F∈ S_K^ that

u(t, x)= Φ(x) +

t

0

Lu(s, x) + f

s, x, u(s, x),∇xu(s, x)σ (x) ds+

t

0

γ_su(s, x)dBs.

The proof is complete now. 2

Remark 4.12. The regularity ofu in the above proposition is difficult to get under not too re-strictive assumptions (like coefficients of class C_l,b³ , linearity of f in z).

Remark 4.13. Notice that generally speaking, a continuous random field after the Girsanov trans-formation Atis not necessarily continuous in t any more. We give a simple counterexample:

Let 0 < s < T be fixed and

u(t, x)=

(t− s)Bt, 0 t  s;

(t− s) sgn(Bs), s < t T .

It is obvious thatu(t, x)isF_t^B-measurable and continuous in t . But after the Girsanov transfor-mation At, it becomes

u(t, x)= u(A_t, t, x)ε_t =



(t− s)(Bt−_t

0(Kγ 1_[0,t])(r)dr)εt, 0 t  s;

(t− s) sgn(Bs−_s

0(Kγ 1[0,t])(r)dr)εt, s < t T , which is not continuous in t on

 ω^: inf

t∈[s,T ]

B_s−

s

0

(Kγ 1_[0,t])(r)dr



<0 < sup

t∈[s,T ]

B_s−

s

0

(Kγ 1_[0,t])(r)dr



.

However, as we state below, the random field u has a continuous version in our case. To this end we need the following technical result:

Lemma 4.14. Let γ be such that (H1) holds. Then there exist positive constants C and q such that for all r, v, v^∈ [0, T ], v  v^,





T

0

(Kγ 1_[v^,v])(s)(Kγ 1_[0,r])(s)ds



 Cv− v^^q.

Proof. We have





T

0

(Kγ 1_[v^,v])(s)(Kγ 1[0,r])(s)ds





 T

0

(Kγ 1_[v^,v])²(s)ds

1/2 T

0

(Kγ 1_[0,r])²(s)ds

1/2

 C T

0

(Kγ 1_[v^,v])²(s)ds

1/2

,

where the last inequality follows from [11, Lemma 2.3]. Also, by the proof of Lemma 2.3 in [11]

and using the notation α= 1/2 − H , we have

(Kγ 1_[v^,v])(s)= 1_[v^,v](s)

φ_γ(s)+ αs^α Γ (1− α)

T

v

r^−αγr

(r− s)^1+α dr



− 1_[0,v^_](s) αs^α Γ (1− α)

v

v^

r^−αγr

(r− s)^1+αdr

= I1(s)+ I2(s). (4.17)

Now, applying [11, Lemma 2.3] again, we obtain

T

0

I1(s)²ds=

T

0

1_[v,v](s)

1_[0,v](s)



φγ(s)+ αs^α Γ (1− α)

T

v

r^−αγ_r (r− s)^1+α dr

2

ds

v− v^^(p−2)/pφγ1_[0,v]²_Lp  Cv− v^^(p−2)/p. (4.18) On the other hand, for m= 1 + η and qm= m/η, with η small enough, we can write

v

v^

r^−αγr

(r− s)^1+αdr

v− v^^1/qmv v^

r^−mα|γr|^m (r− s)^m(1+α)dr

1/m

= 1

Γ (α)v− v^^1/qmv v^

(_T

r φ_γ(θ )θ^−α(θ− r)^α−1dθ )^m (r− s)^m(1+α) dr

1/m

 Cv− v^^1/qmT v^

φγ(θ )^mθ^−η

θ

v^

(r− s)^−m(1+α)r^−mα+η(θ− r)^m(α−1)dr dθ

1/m

 Cv− v^^1/qm

v^− s_−2η/m

×

T

v^

φ_γ(θ )^mθ^−η

θ

v^

(r− s)^{−mα+η−1}r^−mα+η(θ− r)^mα−η−1dr dθ

1/m

.

Hence, [11, Lemma 2.2] gives

v

v^

r^−αγr

(r− s)^1+α dr

 Cv− v^^1/qm

v^− s_−2η/m

×

T

v^

φγ(θ )^mθ^−ηv^−mα+η

v^− s_−mα+η

(θ− s)⁻¹

θ− v^mα−η

dθ

1/m

 Cv− v^^1/qm

v^− s_−α−η/mT v^

φ_γ(θ )^mθ^−ηv^−mα+η(θ− s)^{−1+mα−η}dθ

1/m

,

which, together with (4.17), implies, for p < 1/α, p^=_{2(1−(mα−η)p)}^pm and 1/p^+ 1/q^= 1,

T

0

I₂(s)²ds Cv− v^^qm2 v^ 0

v^− s2q^(−α−η/m)ds

¹

q

× v^

0

T

v^

φ_γ(θ )^mθ^−η(θ− s)^{−1+mα−η}dθ

^2p_m ds

¹

p

 Cv− v^^2/qm. Thus, we get the wished result. 2

Lemma 4.15. The random field u(t, x):= u(A_t, t, x)ε_t, (t, x)∈ [0, T ] × R^d has a continuous version.

Proof. In the following, for simplicity of notations, we put Θ_r^t,x,v,v=

r, X^t,x_r , Y_r^t,x(Av)εr(TrA_v), Z^t,x_r (Av)εr(TrA_v) . For 0 v^ v  T , we notice that (Y^t,x(A_v))is the solution of the BSDE

Y_s^t,x(Av)= Φ X^t,x₀ 

+

s

0

f

Θ_r^t,x,v,v

ε⁻¹_r (TrAv)dr−

s

0

Z^t,x_r (Av)↓ dWr, (4.19)

while (Y^t,x(A_v))is the solution of the BSDE



Y_s^t,x(A_v)= Φ X^t,x₀ 

+

s

0

f

Θ_r^t,x,v,v

ε_r⁻¹(T_rA_v)dr−

s

0



Z^t,x_r (A_v)↓ dWr.

We set J_r^v = exp{_T

0 (Kγ 1_[0,v])(s)(Kγ 1_[0,r])(s)ds}, and we observe that ε⁻¹_r (T_rA_v) = ε⁻¹_r (T_r)J_r^v. Moreover, Eq. (4.19) can be written as follows:

Y_s^t,x(Av)= Φ X₀^t,x

+

s

0

f

r, X_r^t,x,Y_r^t,x(Av), Z_r^t,x(Av) εr(Tr)

J_r^v₋₁

ε_r⁻¹(Tr)J_r^vdr

−

s

0

Z_r^t,x(Av)↓ dWr, s∈ [0, t],

and a comparison with (4.5) suggests the similarity of arguments which can be applied. So, by a standard BSDE estimate, see, for instance, the proof of Lemma 4.2, we have

E

 sup

0sT

Y_s^t,x(A_v)− Y_s^t,x(A_v)^p

 CE

T

0

f

Θ_r^t,x,v,v

ε⁻¹_r (TrAv)− f

Θ_r^t,x,v,v

ε⁻¹_r (TrA_v)^pdr



 CE

T

0

f

Θ_r^t,x,v,v

ε_r⁻¹(T_rA_v)− ε_r⁻¹(T_rA_v)^p +f

Θ_r^t,x,v,v

− f

Θ_r^t,x,v,v

ε⁻¹_r (T_rA_v)^p dr



. (4.20)

Recalling that ε⁻¹_r (T_rA_v)= ε⁻¹_r (T_r)J_r^vand applying Lemma 4.14, we get

ε⁻¹_r (T_rA_v)− ε⁻¹_r (T_rA_v)^p= εr^−p(T_r)J_r^v− J_r^v^p

 Cεr^−p(T_r)





T

0

(Kγ 1_[0,v])(s)(Kγ 1_[0,r])(s)ds

−

T

0

(Kγ 1_[0,v^_])(s)(Kγ 1[0,r])(s)ds





p

= Cεr^−p(T_r)





T

0

(Kγ 1_[v^,v])(s)(Kγ 1_[0,r])(s)ds





p

 Cεr^−p(T_r)v− v^^pq. (4.21) On the other hand,

f

Θ_r^t,x,v,v

− f

Θ_r^t,x,v,v^pε^−p_r (TrA_v)

 CY_r^t,x(A_v)^p+Z_r^t,x(A_v)^pε_r(T_rA_v)− εr(T_rA_v)^pε^−p_r (T_rA_v)

= CY_r^t,x(A_v)^p+Z_r^t,x(A_v)^p

exp



−

T

0

(Kγ 1_[v^,v])(s)(Kγ 1_[0,r])(s)ds



− 1



p

 CY_r^t,x(A_v)^p+Z_r^t,x(A_v)^pv− v^^pq. (4.22)

Plugging estimates (4.21) and (4.22) into Eq. (4.20), Lemma 4.2 yields that E

 sup

0sT

Y_s^t,x(Av)− Y_s^t,x(A_v)^p

 CE

T

0

f

Θ_r^t,x,v,v

ε⁻¹_r (T_r)^p+Y_r^t,x(A_v)^p+Z_r^t,x(A_v)^p dr

v− v^^pq

 C

1+ |x|^pv− v^^pq.

For the latter inequality we have used the following estimate of Z, the proof of which is post-poned until the end of the current proof.

Lemma 4.16. There exists a constant Cpsuch that E[sup0sT|Z_s^t,x|^p]  Cp(1+ |x|^p).

We continue our proof of Lemma 4.15: According to the proof of Lemma 4.4 we have E

 sup

0sT

Y_s^t,x(A_v)− Y_s^t,x(A_v)^p

= E E

 sup

s∈[0,T ]

Y_s^t,x− Y_s^t,x^pF_T^B

◦ Av



 E

 sup

s∈[0,T ]

Y_s^t,x− Y_s^t,x^2p1/2 E

ε⁻²_v (Tv)1/2

 C

1+ |x|^p+x^^pt− t^^p/2+x− x^^p . Hence, by combining the above estimates we obtain

E

 sup

0sT

Y_s^t,x(A_v)− Y_s^t,x(A_v)^p

 C E

 sup

0sT

Y_s^t,x(A_v)− Y_s^t,x(Av)^p + E

sup

0sT

Y_s^t,x(Av)− Y_s^t,x(Av)^p

 C

1+ |x|^p+x^^pt− t^^p/2+v− v^^pq

+x− x^^p .

Consequently, from the Kolmogorov continuity criterion we know the process{Y_s^t,x(A_v); s, t, v ∈ [0, T ], x ∈ R^d} has an a.s. continuous version. From Lemma 4.14 we have that εt is continuous in t . It then follows that u(t, x)= Y_t^t,x(A_t)ε_t has a version which is jointly continuous in t and x. 2

Proof of Lemma 4.16. For simplicity we suppose all the functions Φ, f, σ, b are smooth. For the proof of the general case, the Lipschitz functions have to be approximated by smooth functions with the same Lipschitz constants. We define (Y^t,x, Z^t,x)to be the solution of the equation

Y^t,x_s = Φ^ X^t,x₀ 

∇xX^t,x₀ +

s

0

f_x^ Ξ_r^

ε⁻¹_r (Tr)∇xX^t,x_r + fy^

Ξ_r^

Y^t,x_r + fz^

Ξ_r^ Z^t,x_r

dr

−

s

0

Z^t,x_r ↓ dWr, (4.23)

where Ξ_r^= (r, X^t,xr , Y_r^t,xε_r(T_r), Z^t,x_r ε_r(T_r))and

∇xX_r^t,x= I −

t

r

∇xX^t,x_s b^ X_s^t,x

ds−

t

r

∇xX^t,x_s σ^ X^t,x_s 

↓ dWs, r∈ [0, t]. (4.24)

With the arguments used in the proof of Lemma 4.2, we get

E

 sup

0sT

Y^t,x_s ^p+ T

0

Z^t,x_s ²ds

p/2

 Cp

1+ |x|^p

. (4.25)

By arguments which by now are standard, it can be seen that the processes X^t,x, Y^t,xand Z^t,x are Malliavin differentiable with respect to W , and thus, for s u  t  T ,

D_u^WY_s^t,x= Φ^ X^t,x₀ 

D_u^WX^t,x₀ −

s

0

D^W_u Z^t,x_r ↓ dWr

+

s

0

f_x^ Ξ_r^

ε⁻¹_r (T_r)D^W_u X_r^t,x+ f_y^ Ξ_r^

D_u^WY_r^t,x+ f_z^ Ξ_r^

D_u^WZ_r^t,x

dr. (4.26)

On the other hand, from Y_s^t,x= Y_θ^t,x+

s

θ

f

r, X^t,x_r , Y_r^t,xεr(Tr), Z^t,x_r εr(Tr)

ε_r⁻¹(Tr)dr

−

s

θ

Z_r^t,x↓ dWr, θ < s u  T ,

we get

D_u^WY_s^t,x= −Z_u^t,x+

u

s

D^W_u Z^t,x_r ↓ dWr

+

u

s

f_x^ Ξ_r^

ε⁻¹_r (T_r)D^W_u X_r^t,x+ fy^

Ξ_r^

D_u^WY_r^t,x+ fz^

Ξ_r^

D_u^WZ_r^t,x dr.

(4.27) From the above three equations (4.23), (4.26) and (4.27) and the relation

D_r^W X_s^t,x

= ∇xX_s^t,x

∇xX^t,x_r ₋₁ σ

X^t,x_r 

(4.28) (see: Pardoux and Peng [18]), we obtain that, for s u  t  T ,

D_u^WY_s^t,x= Y^t,x_s 

∇xX_u^t,x₋₁ σ

X_u^t,x , Z^t,x_s = −Y^t,xs

∇xX_s^t,x₋₁ σ

X_s^t,x .

Finally, from standard estimates for (4.24) and from the estimate (4.25), we get

E

 sup

0sT

Z^t,x_s ^p

 E sup

0sT

Y^t,x_s ^p sup

0sT∇xX^t,x_s ^−p sup

0sT

σ

X^t,x_s ^p

 E sup

0sT

Y^t,x_s ^pq11/q1

E

 sup

0sT∇xX^t,x_s ^−pq21/q2

E

 sup

0sT

σ

X^t,x_s ^pq31/q3

 Cp

1+ |x|^p ,

where 1/q1+ 1/q2+ 1/q3= 1, with q1, q₂, q₃>1. The proof is complete. 2 The above proposition motivates the following definition:

Definition 4.17. A continuous random field u :[0, T ] × R^d× Ω^→ R is a (stochastic) viscosity solution of Eq. (4.9) if and only ifu(t, x)= u(Tt, t, x)ε⁻¹_t (Tt), (t, x)∈ [0, T ] × R^dis a pathwise viscosity solution of Eq. (4.8).

As a consequence of our preceding discussion, we can formulate the following statement:

Theorem 4.18. The continuous stochastic field u(t, x):= u(A_t, t, x)ε_t = Y_t^t,x(A_t)ε_t = Yt^t,x is a stochastic viscosity solution of the semilinear SPDE (4.9). This solution is unique inside the class C_p^Bof continuous stochastic field ˜u : Ω^× [0, T ] × R^d→ R such that

˜u(t, x) Cexp I_T^∗

1+ |x|

, (t, x)∈ [0, T ] × R^d, P -a.s., for some constant C only depending on˜u.

Remark 4.19.

(1) It can be easily checked that ˜u ∈ Cp^B if and only if (˜u(At, t, x)ε_t)∈ Cp^B if and only if (˜u(Tt, t, x)ε⁻¹_t (T_t))∈ Cp^B.

(2) As a consequence of the preceding theorem we have that u(t, x)= Yt^t,xis the unique (in C_p^B) stochastic viscosity solution of SPDE (4.9). This extends the Feynman–Kac formula to SPDEs driven by a fractional Brownian motion.

We conclude the main theorems of Section 4 with the following relation diagram, which shows the mutual relationship between fractional backward SDEs and SPDEs:

u(t, x) is the viscosity solution of (4.8) ←→ u(t, x) is the viscosity solution of (4.9)^GT

 

(Y , Z) is the solution of BSDE(4.5) ←→ (Y, Z) is the solution of BDSDE (4.4)^GT where GT stands for Girsanov transformation.

Finally, in order to illustrate how our method works, we give the example of a linear fractional backward doubly stochastic differential equation.

Example 4.20. Let d= 1 and f (s, x, y, z) = f_s¹x+ f_s²y+ f_s³z, where the coefficients f_s¹, f_s² and f_s³ are bounded and deterministic functions. The associated fractional backward doubly stochastic differential equation is linear and writes:

Y_s^t,x= Φ X₀^t,x

+

s

0

f_r¹X_r^t,x+ fr²Y_r^t,x+ fr³Z_r^t,x dr−

s

0

Z_r^t,x↓ dWr+

s

0

γrY_r^t,xdBr. (4.29) After Girsanov transformation, it becomes

Y_s^t,x= Φ X₀^t,x

+

s

0

f_r¹X_r^t,xε⁻¹_r (T_r)+ fr²Y_r^t,x+ fr³Z_r^t,x dr−

s

0

Z_r^t,x↓ dWr

and has the following solution:

Y_s^t,x= EQ

s

0

f_r¹X^t,x_r ε⁻¹_r (T_r)exp

s

r

f_u²du



dr

+ exp

s

0

f_r²dr

 Φ

X₀^t,xFs,t^W∨ Ft^B

 ,

where EQis the expectation with respect to Q= exp{_t

0f_r³dWr−¹₂_t

0(f_r³)²dr}P. According to Theorem 3.9, the solution of (4.29) is then

Y_s^t,x= EQ

 ε_s

s

0

f_r¹X_r^t,xε_r⁻¹(T_rA_s)exp

s

r

f_u²du



dr

+ εsexp

s

0

f_r²dr

 Φ

X^t,x₀ Fs,t^W∨ F_t^B

 .

Acknowledgements

Part of this work was done while the second author was visiting the Institut Mittag-Leffler.

He thanks it for the hospitality and economical support. The authors also thank Professor Rainer Buckdahn for his valuable advice and discussions.

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