Existence, uniqueness and continuity

S_q is a Banach space as well. The proof is a standard argument and we skip it.

Throughout this section, we make the following assumptions, which are adapted from [41, Hypothesis 6.2].

Assumption 4.11: There exists q≥2 such that (i) u0 ∈ Sq,

(ii) f and g are predictable and satisfy the following Lipschitz and linear growth conditions: There exists L > 0 and a predictable process M : Ω×[0, T] →R such that kMk_q,T := sup_s∈[0,T]kM(s)k_L^q_(Ω) < ∞ and for all (w, t, x, y) ∈ Ω× [0, T]×R²

|f(ω, t, x)−f(ω, t, y)|+|g(ω, t, x)−g(ω, t, y)| ≤L|x−y|,

|f(ω, t, x)|+|g(ω, t, x)| ≤M(w, t) +L|x|.

Predictability of a process f : Ω×[0, T]×R can be defined in the same way as in Section2.3 (see also [69]).

We are now able to prove stochastic continuity properties ofv₁ and v₂, which are defined for (t, x)∈[0, T]×[0,1] and v₀ ∈ S_q,T by

v1(t, x) :=

Z t 0

Z 1 0

p^b_t−s(x, y)g(s, v0(s, y))ξ(s, y)dµ(y)ds, (58) v₂(t, x) :=

Z t 0

Z 1 0

p^b_t−s(x, y)f(s, v₀(s, y))dµ(y)ds. (59) Proposition 4.12: Let q ≥ 2 be fixed. Then, there exists a constant C₃ > 0 such that for v₀ ∈ S_q,T v₁ and v₂ are well-defined and we have for all s, t ∈ [0, T], x, y ∈[0,1], i∈ {1,2}

E[|v_i(t, x)−v_i(t, y)|^q]≤C₃ 1 +kv₀k^q_q,T

|x−y|^q2, E[|v_i(s, x)−v_i(t, x)|^q]≤C₃ 1 +kv₀k^q_q,T

|s−t|^q(¹2−^γδ₂ ). Remark 4.13: Note thatγδ < 1is equivalent to

i=1,...,Nmax

logµ_i

log(µ_ir_i) <1, which is satisfied as we assume0< µ_i, r_i <1.

Proof. We adapt ideas from [41, Proposition 6.7]. First, we consider v₁. For x ∈ [0,1], (t, y) → p^b_t(x, y) is continuous on (0, T]×[0,1] and g and v0 are predictable.

The integrand is thus predictable. By Hypothesis 4.11, for q ≥ 2, (t, x) ∈ [0, T]× [0,1],

E[|g(t, v₀(t, x))|^q]≤2^q−1(E[|M(t)|^q] +L^qE[|v₀(t, x)|^q]). (60) Hence, by Lemmas 4.4(iii)and 4.5,

E Z t

0

Z 1 0

p^b_t−s(x, y)²g(s, v₀(s, y))²dµ(y)ds

≤ 2 sup

s∈[0,T]

kM(s)k²_L2(Ω)+ 2L²kv₀k²_q,T

!Z t 0

Z 1 0

p^b_t−s(x, y)²dµ(y)ds

= 2 sup

s∈[0,T]

kM(s)k²_L2(Ω)+ 2L²kv₀k²_q,T

!Z t 0

p^b_2s(x, x)ds

≤ 2 sup

s∈[0,T]

kM(s)k²_L2(Ω)+ 2L²C₀(2T)kv₀k²_q,T

!Z t 0

(2s)^−γδds, which is finite due to γδ <1. Hence, v1(t, x) is well-defined.

In order to prove the spatial estimate for v₁, let (t, x, y)∈[0, T]×[0,1]² be fixed.

There exists a constantC(q) such that E[|v₁(t, x)−v₁(t, y)|^q]

=E

Z t 0

Z 1 0

p^b_t−s(x, z)−p^b_t−s(y, z)

g(s, v₀(s, z))ξ(s, z)dµ(z)ds

q

≤C(q)E

"

Z t 0

Z 1 0

p^b_t−s(x, z)−p^b_t−s(y, z)2

g(s, v₀(s, z))²dµ(z)ds

q 2#

(61)

≤C(q)

Z t 0

Z 1 0

p^b_t−s(x, z)−p^b_t−s(y, z)q

E[|g(s, v₀(s, z))|^q]

2

q dµ(z)ds

q 2

(62)

=C(q)

Z t 0

Z 1 0

p^b_t−s(x, z)−p^b_t−s(y, z)2

(E[|g(s, v₀(s, z))|^q])^2q dµ(z)ds

q 2

≤2^q−1C(q) kMk^q_q,T +L^qkv₀k^q_q,T

Z t 0

Z 1 0

p^b_t−s(x, z)

−p^b_t−s(y, z)2

dµ(z)ds

q 2

,

(63)

where we have used the Burkholder-Davis-Gundy inequality (see [49, Theorem B.1])

in (61), Minkowski’s integral inequality in (62) and relation (60) in (63). We proceed by dealing with the integral term in (63), whereby we first treat the case x, y ∈K.

Applying Lemma 4.3, Lemma 4.6, Lemma 4.7 and the Lipschitz continuity of the resolvent density (see (55)),

Z t 0

Z 1 0

p^b_t−s(x, z)−p^b_t−s(y, z)2

dµ(z)ds

= lim

n→∞

Z t 0

Z 1 0

p^b_t−s(·, z), f_n^x−f_n^y2

µdµ(z)ds

≤ e^2t 2 lim

n→∞

Z 1 0

Z 1 0

ρ^b₁(z₁, z₂)(f_n^x(z₁)−f_n^y(z₁))(f_n^x(z₂)−f_n^y(z₂))dµ(z₁)dµ(z₂)

= e^2t 2

ρ^b₁(x, x)−2ρ^b₁(x, y) +ρ^b₁(y, y)

≤L1e^2t|x−y|.

For i ∈ N and x ∈ (a_i, b_i), recall that we evaluate the D_µ²-representative of p^b_t(·, y) for(t, y)∈(0, T]×[0,1]. We thus have

p^b_t(x, y) = p^b_t(a_i, y) +

x−a_i b_i−a_i

p^b_t(b_i, y)−p^b_t(a_i, y)

, y∈[0,1]

and consequently,

v₁(t, x) = Z t

0

Z 1 0

p^b_t−s(x, y)g(s, v₀(s, y))ξ(s, y)dµ(y)ds

=v₁(t, a_i) +

x−a_i b_i−a_i

(v₁(t, b_i)−v₁(t, a_i)) almost surely. Consequently, for allx, y ∈[0,1]

Z t 0

Z 1 0

p^b_t−s(x, z)−p^b_t−s(y, z)2

dµ(z)ds

q 2

≤(L₁)^q2 e^qt|x−y|^2q. We conclude

E[|v1(t, x)−v1(t, y)|^q]≤2^q−1e^qtC(q) (L1)^q2 kMk^q_q,T +L^qkv0k^q_q,T

|x−y|^q2. This proves the spatial estimate.

We turn to the temporal estimate. Lets, t∈[0, T]withs < tandx∈[0,1]. Then, by using the Burkholder-Davis-Gundy inequality, Minkowski’s integral inequality and inequality (60) in the same way as before, we get

E[|v1(t, x)−v1(s, x)|^q]

≤C(q)

Z t 0

Z 1 0

p^b_t−u(x, y)−p^b_s−u(x, y)1[0,s](u)2

E[|g(s, v₀(s, y))|^q]

2

q dµ(y)du

q 2

≤2^q−1C(q) kMk^q_q,T +L^qkv₀k^q_q,T

Z t 0

Z 1 0

p^b_t−u(x, y)

−p^b_s−u(x, y)1[0,s](u)2

dµ(y)du

q 2

. We split the latter integral in the time intervals [0, s]and (s, t] and get for the first part by Proposition4.5(iii)

Z s 0

Z 1 0

p^b_t−u(x, y)−p^b_s−u(x, y)2

dµ(y)du

= Z s

0

Z 1 0

p^b_u(x, y)−p^b_u+t−s(x, y)2

dµ(y)du

= Z s

0

p^b_2u(x, x)−2p^b_2u+t−s(x, x) +p^b_2(u+t−s)(x, x)du

≤2^−γδC₂(2T) Z s

0

u^−γδ−2

u+t−s 2

−γδ

+ (u+t−s)^−γδdu (64)

≤2^−γδ+1C₂(2T) Z s

0

u^−γδ−

u+ t−s 2

−γδ

du

= 2^−γδ+1

1−γδC₂(2T) s^1−γδ− s

2 + t 2

1−γδ

+

t−s 2

1−γδ!

≤ C2(2T)

1−γδ(t−s)^1−γδ.

For the second part by Proposition 4.5(i) Z t

s

Z 1 0

p^b_t−u(x, y)²dµ(y)du = Z t−s

0

p^b_2u(x, x)du

≤2^−γδC0(2T) Z t−s

0

u^−γδdu

≤ 2^−γδ

1−γδC0(2T)(t−s)^1−γδ.

The statements about v₂ can be shown in the same way using Jensen’s inequality instead of the Burkholder-Davis-Gundy inequality.

Corollary 4.14: Let q ≥2 and v₀ ∈ S_q,T. Then, v₁ and v₂ defined as in (58)-(59) are elements of S_q,T.

Proof. By setting s = 0 in Proposition 4.12 we obtain kv_ik_q,T < ∞, i = 1,2. We need to show that v₁ is predictable. Let n∈N, (t, x)∈[0, T]×[0,1]and

v₁ⁿ(t, x) :=

2ⁿ−1

X

i,j=0

v1

i 2ⁿT, j

2ⁿ

1(2ⁱnT,ⁱ⁺¹_2nT](t)1(2^jn,^j+1_2n](x).

Evidently,kv₁ⁿk_q,T <∞. To prove thatv₁ⁿis predictable, we show thatv₁ⁿis theSq,T -limit of a sequence of simple processes. To this end, letN ≥1,(t, x)∈[0, T]×[0,1]

and

v₁^n,N(t, x) := ((−N)∨v₁ⁿ(t, x))∧N.

This defines a simple process since (−N)∨v_{1 2}ⁱnT,₂^jn

∧N isFiT

2n-measurable and bounded for0≤i, j ≤2ⁿ⁻¹. Further, it converges inS_q,T tov₁ⁿ asN → ∞. Indeed, let X be a set, f : X → R. We define f⁺(x) := f(x)∨0 and f⁻(x) := f(x)∧0, x∈X. With that,

lim

N→∞

v₁ⁿ(t, x)−v^n,N₁ (t, x) q,T

≤ lim

N→∞

2ⁿ−1

X

i,j=0

v⁺₁ i

2ⁿT, j 2ⁿ

−

v₁⁺ i

2ⁿT, j 2ⁿ

∧N

q,T

+

v₁⁻ i

2ⁿT, j 2ⁿ

−

(−N)∨v₁⁻ i

2ⁿT, j 2ⁿ

q,T

= lim

N→∞

2ⁿ−1

X

i,j=0

v₁⁺ i

2ⁿT, j 2ⁿ

−

v⁺₁ i

2ⁿT, j 2ⁿ

∧N

L^q(Ω)

+

v₁⁻ i

2ⁿT, j 2ⁿ

−

(−N)∨v₁⁻ i

2ⁿT, j 2ⁿ

L^q(Ω)

= 0,

where the last equation follows by the monotone convergence theorem. We conclude

that v₁ⁿ is predictable for n ∈N. By Proposition4.12, kv₁−v₁ⁿk_q,T ≤ sup

|s−t|<^T_n

sup

|x−y|<¹

n

kv₁(s, x)−v₁(t, y)k_Lq(Ω)

≤ sup

|s−t|<^T_n

sup

x∈[0,1]

kv1(s, x)−v1(t, x)k_Lq(Ω)

+ sup

t∈[0,T]

sup

|x−y|<_n¹

kv₁(t, x)−v₁(t, y)k_Lq(Ω)

≤ C₃ 1 +kv₀k^q_q,T¹_q T

n ¹₂−^γδ

2

+ 1

n ¹₂!

→0, n→ ∞.

Hence, v1 is predictable. The predictability of v2 follows in the same way.

We can now follow the methods in [41, Theorem 6.9] to establish existence and uniqueness.

Theorem 4.15: Assume Condition 4.11 with q ≥ 2. Then, SPDE (51) has a unique mild solution in S_q,T.

Proof. Uniqueness: Letu,eu∈ S_q,T be mild solutions to (51). Thenv := u−eu∈ S_2,T. By setting G(t) := sup_x∈[0,1]E[v²(t, x)], t ∈ [0, T], we calculate for (t, x) ∈ [0, T]× [0,1]

E

v(t, x)²

≤2TE Z t

0

Z 1 0

p^b_t−s(x, y)2

(f(s, u(s, y))−f(s,eu(s, y)))²dµ(y)ds

+ 2E Z t

0

Z 1 0

p^b_t−s(x, y)2

(g(s, u(s, y))−g(s,eu(s, y)))²dµ(y)ds

(65)

≤2(T + 1)L²E Z t

0

Z 1 0

v²(s, y) p^b_t−s(x, y)2

dµ(y)ds

≤2(T + 1)L² Z t

0

G(s) Z 1

0

p^b_t−s(x, y)2

dµ(y)ds

= 2(T + 1)L² Z t

0

G(s)p^b_2(t−s)(x, x)ds

≤2^1−γδ(T + 1)C₀(2T)L² Z t

0

G(s)(t−s)^−γδds, (66)

where we have used Walsh’s isometry (see Section 2.3) and Hölder’s inequality in

(65) and Proposition 4.5(i) in (66). For t∈[0, T], it follows G(t)≤2^1−γδ(T + 1)C₀(2T)L²

Z t 0

G(s)(t−s)^−γδds.

Settingh_n =G,n ∈Nin [69, Lemma 3.3] givesG(t) = 0 fort∈[0, T]. We conclude u(t, x) =u(t, x)e almost surely for every (t, x)∈[0, T]×[0,1].

Existence: As usual, we apply Picard iteration to find a solution. For that, let u₁ = 0∈ S_q,T and for n ≥1, (t, x)∈[0, T]×[0,1],

un+1(t, x) = Z 1

0

p^b_t(x, y)u0(y)dµ(y) + Z t

0

Z 1 0

p^b_t−s(x, y)f(s, un(s, y))dµ(y)ds +

Z t 0

Z 1 0

p^b_t−s(x, y)g(s, u_n(s, y))ξ(s, y)dµ(y)ds.

(67)

Let n ≥ 1, assume that un ∈ Sq,T and define un+1 as in (67). The last two terms on the right-hand side are elements ofS_q,T by Corollary4.14. The first term is pre-dictable because it isF₀-measurable and thus adapted and almost surely continuous due to the dominated convergence theorem and Proposition4.5(i). Furthermore, by Minkowski’s integral inequality and Lemma 4.4(vi)

E

Z 1 0

p^b_t(x, y)u₀(y)dµ(y)

q

≤

Z 1 0

p^b_t(x, y)E[|u₀(y)|^q]^1q dµ(y)

q

≤ ku₀k^q_q

Z 1 0

p^b_t(x, y)dµ(y)

q

≤ ku₀k^q_q. It followsu_n+1 ∈ S_q,T.

We prove that(u_n)n∈N is a Cauchy sequence in S_q,T. For n ∈N, let v_n:= u_n+1− u_n ∈ S_q,T. By Hölder’s and the Burkholder-Davis-Gundy inequality, for (t, x) ∈ [0, T]×[0,1]

E[|v_n+1(t, x)|^q]

≤2^q−1T^q2E

"

Z t 0

Z 1 0

p^b_t−s(x, y)²(f(s, u_n+1(s, y))−f(s, u_n(s, y)))²dµ(y)ds

q 2#

+ 2^q−1C(q)E

"

Z t 0

Z 1 0

p^b_t−s(x, y)²(g(s, u_n+1(s, y))−g(s, u_n(s, y)))²dµ(y)ds

q 2#

and using the Lipschitz property off and g and Minkowski’s integral inequality, E[|v_n+1(t, x)|^q]

≤2^q−1

T^q2 +C(q) L^qE

"

Z t 0

Z 1 0

p^b_t−s(x, y)²v_n²(s, y)dµ(y)ds

q 2#

≤2^q−1

T^q2 +C(q) L^q

Z t 0

Z 1 0

p^b_t−s(x, y)²(E[|v_n(s, y)|^q])^2q dµ(y)ds

q 2

.

Let n ∈ N and Hn(s) := sup_y∈[0,1](E[|vn(s, y)|^q])^2q, s ∈ [0, T]. Then, there ex-ists a constant ec_n such that |H_n(t)| ≤ ec_n for each t ∈ [0, T]. By setting C :=

2^q−1 T^q2 +C(q) L^q2_q

we get for (t, x)∈[0, T]×[0,1] and n∈N (E[|v_n+1(t, x)|^q])^2q ≤C

Z t 0

H_n(s)p^b_2(t−s)(x, x)ds

≤2^−γδC0(2T)C Z t

0

Hn(s)(t−s)^−γδds, where we have used Proposition4.5(i) and thus

H_n+1(t)≤2^−γδC₀(2T)C Z t

0

H_n(s)(t−s)^−γδds.

By [69, Lemma 3.3], there exists a constant C⁰ >0 and an integer k ≥1 such that forn, m≥1, t∈[0, T]

H_n+mk(t)≤ C^0m (m−1)!

Z t 0

H_n(s)(t−s)ds.

Therefore, P

m≥1

pH_n+mk(t) converges uniformly for t ∈ [0, T], which is straight forward to check by the ratio test using that

q_H

n+(m+1)k(t) Hn+mk(t) ≤q

C⁰

m forn, m≥1. We conclude

sup

t∈[0,T]

pH_n(t)→0, n→ ∞,

which implies the same for kv_nk_q,T. Hence, (u_n)_n≥1 is a Cauchy sequence in S_q,T with limit, say u. To show that u satisfies (57), let (t, x) ∈ [0, T]×[0,1] be fixed and take the limit in L^q(Ω) forn → ∞on both sides of (67). We get u(t, x) on the left-hand side. For the right-hand side, note that there is a constant C⁰⁰ ≥ 0 such

that E

Z t 0

Z 1 0

p^b_t−s(x, y) (f(s, u(s, y))−f(s, u_n(s, y)))ξ(s, y)dµ(y)ds

q

+E

Z t 0

Z 1 0

p^b_t−s(x, y) (g(s, u(s, y))−g(s, u_n(s, y)))dµ(y)ds

q

≤C⁰⁰ Z t

0

Z 1 0

p^b_t−s(x, y)²(E[|u(s, y)−u_n(s, y)|^q])^2q dµ(y)ds

q 2

,

which goes to zero asntends to infinity with the same argumentation as before.

Before stating the main result, we defineu^sto ∈ Sq,T by u^sto(t, x) :=

Z t 0

Z 1 0

p^b_t−s(x, y)f(s, u(s, y))dµ(y)ds +

Z t 0

Z 1 0

p^b_t−s(x, y)g(s, u(s, y))ξ(s, y)dµ(y)ds almost surely for(t, x)∈[0, T]×[0,1]. That is,

u^sto(t, x) = u(t, x)− Z 1

0

p^b_t(x, y)u₀(y)dµ(y)

almost surely for(t, x)∈[0, T]×[0,1]

Theorem 4.16: Assume Condition 4.11 with q ≥ 2. Then, there exists a version of u^sto, denoted by u˜^sto, such that the following holds:

(i) Ifq >2andt∈[0, T], u˜^sto(t,·)is essentially ¹₂−¹_q-Hölder continuous on[0,1].

(ii) If q > ¹₂ − ^γδ₂ ⁻¹

and x ∈ [0,1], u˜^sto(·, x) is essentially ¹₂ − ^γδ₂ − ¹_q-Hölder continuous on [0, T].

(iii) If q > 2 ¹₂ − ^γδ₂ ⁻¹

, u˜^sto is essentially 1

2 − ^γδ₂ − ²_q

-Hölder continuous on [0, T]×[0,1].

Proof. The continuity properties in part (i) and (ii) of a version of u^sto are direct consequences of Proposition 4.12 by setting v₀ := u and Kolmogorov’s continuity theorem (see e.g. [52, Proposition 21.6]).

Now, let (s, t, x, y) ∈ [0, T]² ×[0,1]² and without loss of generality, we assume

|s−t| ≤1. Then, E

u^sto(s, x)−u^sto(t, y)

q

≤2^q−1 E

u^sto(s, x)−u^sto(t, x)

q +E

u^sto(t, x)−u^sto(t, y)

q

≤2^qC₃ 1 +kuk^q_q,T

|x−y|^q2 +|s−t|^q(¹₂^−γδ₂ )

≤2^qC₃ 1 +kuk^q_q,T

maxn

|x−y|^q2,|s−t|^q(¹2−^γδ₂ )o

≤2^qC₃ 1 +kuk^q_q,T

max{|x−y|,|s−t|}^q(¹₂^−γδ₂ ).

The result follows by Kolmogorov’s continuity theorem in two dimensions (see e.g.

[52, Remark 21.7]).

Remark 4.17: We consideru^sto because the regularity ofuis, in general, restricted by the regularity of u−u^sto. However, we can formulate the following: There is a version ofu−u^sto such that this version is for fixedt ∈(0, T] ¹₂-Hölder continuous on [0,1]and for fixedx∈[0,1]and T₁ >0 ¹₂-Hölder continuous on [T₁, T]. This can be checked in the same way as [41, Lemma 6.10] using Proposition 4.5. Consequently, there exists a versionu˜ of u such that

(i) If q >2 and t∈(0, T], u(t,˜ ·) is essentially ¹₂ − ¹_q-Hölder continuous on [0,1].

(ii) If q > ¹₂ − ^γδ₂ ⁻¹

and x∈[0,1], u(·, x)˜ is essentially ¹₂−^γδ₂ −¹_q-Hölder contin-uous on [T1, T].

Example 4.18: (i) If u₀, f and g satisfy Assumption 4.11 for some q ≥ 2 and, in addition, are uniformly bounded, q can be chosen arbitrarily large such that we obtain ¹₂ for the ess. spatial and ¹₂ − ^γδ₂ for the ess. temporal Hölder exponent. If, moreover, the measure µ is the natural measure on K, then

1 2 − γδ

2 = 1 2 − 1

2 max

1≤i≤N

log r_i^dH log

r_i^dH+1 = 1

2 − d_H

2d_H + 2 = 1 2d_H + 2. The left-hand side of Figure 6visualizes that for 0< d_H ≤1.

For d_H = 1 and the natural measure on K, we meet the well-known Hölder continuity properties for the stochastic heat equation defined by the standard Laplacian. Further, in case of ∆^b_µ being a Laplacian on a p.c.f. self-similar set with 1 ≤ dH < 2, Hambly and Yang [41, Theorem 6.14]) established the

Figure 6: Hölder exponent graphs for the stochastic heat equation

essential spatial Hölder exponent ¹₂ and the essential temporal Hölder exponent

1

2dH+2. Therefore, our results can be understood as an extension of their results to the case of d_H <1.

(ii) If µis not the natural measure on K, then it can be easily checked that 1

2− γδ

2 < 1 2d_H + 2.

As an example, consider the weighted IFS given byS1, S2 : [0,1]→R,S1(x) =

x

2, S₂(x) = ^x₂ +¹₂, x∈[0,1]and weights µ₁, µ₂ ∈(0,1). Then, 1

2− γδ 2 = 1

2− 1 2max

i=1,2

logµ_i log(µ_ir_i) = 1

2 − 1 2

logµ_min log ^µmin₂ ,

which goes to zero as µmin → 0. The right-hand side of Figure 6 indicates that.

Im Dokument Stochastic partial differential equations on Cantor-like sets (Seite 68-79)