Generalization of Yamada-Tanaka theorem

Z

R

sin(u) dµ(u)− Z

R

sin(t) dν(t)

= Z

R

sin(u) d(µ−ν)(u)

≤sup

Z

R^d

h(u) d(µ−ν)(u)

h∈Lips1(R)

=W₁(µ, ν)

≤ |x−y|+W1(µ, ν)

≤ |x−y|+W₂(µ, ν),

where we use Lemma 3.7 to get the nal inequality. This implies that there exists a unique solution to (4.7).

4.4.2 Generalization of Yamada-Tanaka theorem

Next we want to generalize the uniqueness theorem of Yamada and Tanaka, introduced in Theorem 2.28. We consider the case where only the coecientb depends on the distribution variable. In the one dimensional case, this result is proven in [BMM19, Section 3.2]. However, by adapting this proof, we may generalize the theorem even further and consider a specic multidimensional case.

σ_i : [0, T]×R→R and

b_i : [0, T]×R× P₂(R^d)→R

be bounded Borel measurable functions. We dene the coecientsσ: [0, T]×

R^d →R^d and b: [0, T]×R^d× P₂(R^d)→R^d such that

b(t,(x₁, ..., x_d), µ) := (b₁(t, x₁, µ), ..., b_d(t, x_d, µ)) and

σ(t,(x₁, ..., x_d)) :=Diag(σ₁(t, x₁), ..., σ_d(t, x_d)),

where Diag denotes a d×d diagonal matrix, that is Diag(σ1(t, x1), ..., σd(t, xd)) =







σ₁(t, x₁) . . . 0 ... ... ...

0 . . . σ_d(t, x_d)





.

Assume that X = (X¹, ..., X^d) solves

(dX_t=σ(t, X_t) dB_t+b(t, X_t,PXt) dt X₀ =x₀ = (x¹₀, ..., x^d₀).

By denition this is equivalent to the system of one-dimensional MVSDEs where for each i= 1, ..., d, Xⁱ is a solution to

(4.4)

(dX_tⁱ =σ_i(t, X_tⁱ) dB_tⁱ+b_i(t, X_tⁱ,PXt) dt X₀ⁱ =xⁱ₀.

It should be noted that here the coecient for the drift term depends on the law of the whole d-dimensional process, not just its ith component.

In this setting, we may give sucient conditions for the uniqueness of a solution. We assume that for all i = 1, ..., d the following conditions are satised:

(A1) The functionb_i is Lipschitz-continuous with respect to the distribution variable, that is, there exists a constant C > 0such that

|bi(t, x, µ)−bi(t, x, ν)| ≤CW1(µ, ν) for all x∈R, t∈[0, T]and (µ, ν)∈ P₁(R^d)⊗ P₁(R^d)

(A2) There exists a strictly increasing functionρ: [0,∞)→[0,∞)satisfying ρ(0) = 0 and

Z 0

1

ρ²(u)du=∞

for every > 0, and |σ_i(t, x)−σ_i(t, y)| ≤ ρ(|x−y|) for all t ∈ [0, T] and x, y ∈R.

(A3) There exists a strictly increasing concave function κ: [0,∞)→ [0,∞) satisfying κ(0) = 0and

Z 0

1

κ(u)du=∞

for every > 0, and |b_i(t, x, µ)−b_i(t, y, µ)| ≤ κ(|x−y|) for all t ∈ [0, T], x, y ∈R and µ∈ P₂(R^d).

With these conditions we can formulate the following theorem that gives us the uniqueness of a solution.

Theorem 4.8. Under conditions (A1), (A2) and (A3), a solution to (4.4.2) is unique.

Before we can prove this theorem, we need two lemmas. The rst lemma is known as Bihari-LaSalle inequality, which is a non-linear generalization of Gronwall's inequality.

Lemma 4.9 (BihariLaSalle, [Mao07, Section 1.8, theorem 8.2]). Assume constantsT >0andc >0. Letf, u : [0, T]→[0,∞)be continuous functions.

Let κ : [0,∞) → [0,∞) be a continuous and increasing function such that κ(x)>0for allx >0. If the functionuis bounded and satises the following inequality

u(t)≤c+ Z t

0

f(s)κ(u(s)) ds for all t ∈[0, T], then

u(t)≤G⁻¹

G(c) + Z t

0

f(s) ds

for all t ∈[0, T] with

(4.5) G(c) +

Z t 0

f(s) ds∈Dom(G⁻¹), where

G(x) :=

Z x 1

1 κ(u)du, for x >0.

Proof. Let v(t) :=c+Rt

0 f(s)κ(u(s)) ds. We dierentiatev to get v⁰(t) = f(t)κ(u(t)).

By the chain rule we obtain d

dtG(v(t)) =v⁰(t)G⁰(v(t)) = v⁰(t)

κ(v(t)) = f(t)κ(u(t)) κ(v(t)) . Integrating from 0 to t yields

Z t 0

d

dsG(v(s))

ds =G(v(t))−G(v(0))

=G(v(t))−G(c) = Z t

0

f(s)κ(u(s)) κ(v(s)) ds.

Since κ is a strictly increasing function we may apply the estimate κ(u(s))≤κ(v(s)) to see that

G(v(t))−G(c) = Z t

0

f(s)κ(u(s)) κ(v(s)) ds ≤

Z t 0

f(s)κ(v(s)) κ(v(s)) ds=

Z t 0

f(s) ds.

Therefore,

G⁻¹(G(v(t))) =v(t)≤G⁻¹

G(c) + Z t

0

f(s) ds

.

for t ∈ [0, T] where T > 0 is chosen so that it satises the property (4.5).

This completes the proof since u(t)≤v(t)for all t≥0,

With BihariLaSalle inequality we can easily prove Gronwall's inequality.

Lemma 4.10 (Gronwall's inequality, [Mao07, Section 1.8, theorem 8.1]). Let A, B, T ≥0 and let u: [0, T]→[0,∞) be a continuous function satisfying

u(t)≤A+B Z t

0

u(s) ds

for all u∈[0, T]. Then u(t)≤Ae^Bt for all t∈[0, T]. Proof. By choosing κ(x) :=x and f :=B one has

G(x) = Z x

1

1

udu= log(x)−log(1) = log(x) and G⁻¹(x) = e^x. Then, by Lemma 4.9 we have

u(t)≤exp(log(A) + Z t

0

Bds) = Ae^Bt

for t∈[0, T], in the caseA >0. If A= 0, then we use the estimate u(t)≤B

Z t 0

u(s) ds < +B Z t

0

u(s) ds for every >0. Now

u(t)≤exp

log() + Z t

0

Bds

→0 as tends to0.

Now we can prove Theorem 4.8.

Proof of Theorem 4.8. We dene the norm kxk_1,d :=

d

X

i=1

|xi|

in R^d. We see that kxk_2,d :=kxk=

v u u t

d

X

k=1

|x_k|² ≤

d

X

k=1

p|x_k|² =

d

X

k=1

|x_k|=kxk_1,d and by the Cauchy-Schwartz inequality we have that

kxk²_1,d=

d

X

k=1

|x_k|

!2

≤d

d

X

k=1

|x_k|² =kxk²_2,d. Hence

kxk_2,d ≤ kxk_1,d ≤√

dkxk_1,d.

This implies that the norms k·k_1,d and k·k_2,d are equivalent.

We assume two processes X = (X¹, ..., X^d) and Y = (Y¹, ..., Y^d) that solve (4.4.2). Our goal is to show that

EkX_t−Y_tk_1,d = 0

for all t∈[0, T], which implies thatX and Y are indistinguishable.

By assumption (A2) we have that Z

0

1

ρ(u)² du=∞,

for all >0. It follows that for every ξ >0there exists a∈(0,1) such that Z 1

a

1

ρ(u)² du=ξ.

This lets us construct a sequence (a_n)^∞_n=1 of real numbers such that 1> a₁ > a₂ > ... > a_n > an−1 > ... >0

and Z 1

a1

1

ρ(u)²du= 1 and Z an−1

an

1

ρ(u)² du=n for all n≥2. Moreover, we see that a_n→0 asn → ∞.

Next we construct a sequence of functions (ψ_n)^∞_n=1, ψ_n : R → R, such that for every n∈N we have that:

(1) ψ_n is continuous, (2) we have

{x∈R|ψ_n(x)6= 0} ⊆(a_n, an−1), (3) and for all x∈R one has

0≤ψ_n(x)≤ 2 nρ(x)²

and Z an−1

an

ψ_n(u) du= 1.

The idea in this construction is that for alln∈Nwe approximate the function x7→1(an,an−1)(x) 1

nρ(x)²

with a continuous function such that the integral over R is the same. We do not go in details why a function like this exists for every n∈N.

Forn ∈Nwe let

ϕ_n(x) :=

Z |x|

0

Z y 0

ψ_n(u) dudy.

It should be noted that now we think the integral as a Riemann integral to get the required properties. Clearly ϕ_n ∈ C²(R). We see that

ϕ⁰_n(x) = Z x

0

ψn(u) du for x≥0, and ϕ⁰_n(x) = −ϕ⁰_n(−x)for x <0. Therefore

|ϕ⁰_n(x)|= Z |x|

0

ψ_n(u) du≤ Z an−1

an

ψ_n(u) du= 1 for all n≥1. Sincea_n converges to 0as n→ ∞, it follows that

Z y 0

ψ_n(u) du→1

as n tends to ∞ for all y > 0. Therefore the sequence (ϕn)^∞_n=1 converges to the function ϕ(x) :=|x|.

Next we xi= 1, ..., dand consider processesXⁱandYⁱ that are solutions

We apply Theorem 2.24 to the process Z and function ϕ_n to get that ϕ_n(Z_t) = 0 +

It should be noted that since ϕ⁰_n and σ_i are bounded and measurable, the stochastic integral exists and

E

where I₁ is the rst term on the right-hand side and I₂ the second term.

First, we estimateI₂ by

|I₂|= 1

Clearly _n^t →0 as n→ ∞.

We continue by estimating the termI₁. We have shown thatϕ⁰is bounded by 1. Hence We apply triangle inequality to obtain

The Lipschitz property (A1) implies

b_i(t, X_tⁱ,P^Yt)−b_i(t, X_tⁱ,P^Xt)

≤CW₁(P^Yt,P^Xt)

≤CEkXt−Ytk, where Lemma 3.4 implies the latter inequality. Then

CEkX_t−Y_tk ≤CEkX_t−Y_tk_1,d. Finally, assumption (A3) gives us

b_i(t, X_tⁱ,PYt)−b_i(t, Y_tⁱ,PYt)

≤κ

X_tⁱ−Y_tⁱ

. Now we have that

|I₁| ≤E

We recall that |I₂| converges to 0 and the function ϕ_n converges to the function ϕ(x) = |x| asn → ∞, so letting n tend to∞ we get the inequality

for all i= 1, ..., d. Then apply Lemma 4.10 to obtain

f(t)≤A(r) exp(Bt)≤A(r) exp(BT)

By Theorem 2.8 we can take the expectation inside the integral, that is M

We apply Proposition 2.6 to obtain

Since κ is an increasing function, we can continue our estimate

EkX_t−Y_tk_1,d ≤M so by Lemma 4.9 we obtain

EkX_t−Y_tk_1,d ≤G⁻¹

is a bijection. Here we assume that is small enough so that Z 1

1

κ(u)du >(M d)t.

Since G is a strictly increasing function such that G(x) → −∞ as x tends to 0 from the right-hand side, the inverse functionG⁻¹ is also strictly increasing with G⁻¹(x)→0as x→ −∞. It follows that

G⁻¹

− Z 1

1

κ(u)du+ (M d)t

→0 as →0. HenceEkX_t−Y_tk_1,d = 0 for all t ∈[0, T].

This property implies that kX_t(ω)−Y_t(ω)k_1,d = 0 almost everywhere.

By the properties of a norm we have that Xt(ω) =Yt(ω) almost everywhere, that is,

P(X_t=Y_t) = 1

for allt ∈[0, T]. By denitionXandY are modications of each other. Since all the trajectories of processes X and Y are continuous, using Proposition 2.17 we conclude that X and Y are indistinguishable, which completes our proof.

Remark 4.11. It should be noted that Theorem 4.8 only implies uniqueness of the solution, but does not imply the existence. In the case we already know some solution for an MVSDE, we may apply the theorem to conrm that the solution is indeed unique.

Next we give a simple example how one can use Theorem 4.8 to prove the uniqueness of a solution. Since the theorem does not imply the existence, we have to nd some solution rst.

Example 4.12. We consider the following MVSDE:







dX_t= minnp

|X_t|,1o

dB_t+ EX_t 1 + (EX_t)² dt X₀ = 0.

We see that the process X ≡ 0 solves the equation. We apply Theorem 4.8 to prove that this solution is actually the only one. In this example the coecients are the following:

σ(t, x) = minnp

|x|,1o and

b(t, x, µ) =ϕ Z

R

udµ(u)

, where

ϕ(x) := x 1 +x².

We check the conditions (A1), (A2) and (A3) separately, starting from the rst one.

We see that

d

dxϕ(x) = 1−x² (x² + 1)²

is bounded, which implies that ϕ is a Lipschitz-continuous function with some constant L >0. For for all µ, ν ∈ P₁(R) we have that

where the nal equality follows from Theorem 3.5.

Next we verify the condition (A2). We may choose ρ(x) := √

x. It is

for all >0. Using the properties of square root we obtain

|σ(t, x)−σ(t, y)|=

The coecientb depends only on the distribution variable, hence we do not need to check the third condition (A3). Since all the conditions are satised, the uniqueness of the solution follows from Theorem 4.8.

5 Stability and approximation of MVSDEs

In this section we consider various stability and approximation results. In our rst result we introduce an iterative method for the approximation of a solution, and we prove that under certain conditions a sequence of iterated

processes converges to the unique solution of (4.2). In the next three results we consider the stability of the solution. We consider stability from the following points of views:

1. Stability with respect to the initial condition. We prove that if we dene a map that maps the initial value to the solution of (4.2), under certain conditions this map is continuous.

2. Stability with respect to the coecients. One way to approximate the solution is to dene sequences of functions that converge to the coef-cients. We prove that under certain assumptions solutions obtained this way eventually converge to the unique solution of (4.2).

3. Stability with respect to the driving process. So far we have considered MVSDEs with respect to the Brownian motion. In our nal stability result we change this settting. Under sucient conditions we may approximate the driving process with possibly simpler processes, and the solutions obtained in this way converge to the unique solution of (4.2).

5.1 Picard approximation

We start with the Picard approximation, which gives us a method to construct a sequence of processes that eventually converges to the unique solution of an MVSDE. This is a useful method in numeric computations.

Assume a sequence of processes ((X_tⁿ)t∈[0,T])^∞_n=0 such that X⁰ ≡ x₀. For n ≥0dene a process Xⁿ⁺¹ by

(5.1)

(dX_tⁿ⁺¹ =σ(t, X_tⁿ,P^Xtⁿ) dBt+b(t, X_tⁿ,P^Xtⁿ) dt X₀ⁿ⁺¹ =x₀.

With certain conditions, we can prove that this sequence converges in L2(Ω,C([0, T],R^d)) to the unique solution of (4.2). It is shown in A.5 that the space L₂(Ω,C([0, T],R^d)) is a Banach space.

Theorem 5.1 ([BMM19, Theorem 4.1]). Assume that the coecients b and σ satisfy conditions (L1) and (L2). Then the sequence ((X_tⁿ)t∈[0,T])^∞_n=0 con-verges in L₂(Ω,C([0, T],R^d))to a process X = (X_t)t∈[0,T], which is the unique solution to the MVSDE (4.2).

Before we can prove the theorem above, we prove the following lemma, which is an application of Hölder's inequality 2.9.

Lemma 5.2. Assume an integrable and measurable function f : [0, T]→R^d,

Proof. By denition we have that

Proposition 2.9 with exponents p=q= 2 implies

Now we may give a proof for Theorem 5.1.

Proof of Theorem 5.1. Fix n≥1. We use the triangle inequality to obtain X_tⁿ⁺¹−X_tⁿ

Then by the Cauchy-Schwarz inequality we have that

By Lemma 5.2 we have that

Proposition 2.23 gives us the following estimate for the stochastic integral part for some absolute constant C > 0.

Now we have the following inequality

E

Using the Lipschitz property of the coecients b and σ and Lemma 3.4 we

continue to Using similar arguments as earlier and the linear growth condition (L1), we get that

E

where M₂ > 0 is chosen so that the inequality above holds. The choice of M₂ depends on K, C, xand T.

where C := max{M₁, M₂}T.

Letm > n andk =m−n. By triangle inequality we obtain the following estimate

kX^m−Xⁿk_L

2(Ω,C([0,T],R^d)) =

X^n+k−Xⁿ

L2(Ω,C([0,T],R^d))

=

X^n+k−X^n+k−1−(Xⁿ−X^n+k−1)

L2(Ω,C([0,T],R^d))

≤

X^n+k−X^n+k−1

L2(Ω,C([0,T],R^d))

+

X^n+k−1−Xⁿ

L2(Ω,C([0,T],R^d))

≤ C^n+k

(n+k−1)! +

X^n+k−1−Xⁿ

L2(Ω,C([0,T],R^d)). By induction on n we obtain

kX^m−Xⁿk_L

2(Ω,C([0,T],R^d))≤

m−n

X

i=1

Cⁿ⁺ⁱ (n+i−1)!. Letting n and m tend to∞, we conclude that

kX^m−Xⁿk_L

2(Ω,C([0,T],R^d))→0, that is, for any >0we can nd N ∈N such that

kX^m−Xⁿk_L

2(Ω,C([0,T],R^d)) <

for allm, n≥N. Therefore(Xⁿ)^∞_n=1is a Cauchy sequence inL2(Ω,C([0, T],R^d)), which is a complete normed space by corollary A.5. It follows that there ex-ists a unique limit X ∈L₂(Ω,C([0, T],R^d)) such that

kXⁿ−Xk_L

2(Ω,C([0,T],R^d)) →0 as n→ ∞.

Next we need to show thatX is a solution to the MVSDE (4.2). We use

the same estimates as earlier in this proof to see that

as n tends to ∞. It implies that (5.2) X_t =x₀+

Z t 0

b(s, X_s,PXs) ds+ Z t

0

σ(s, X_s,PXs) dB_s

almost surely for any t∈[0, T], and then one gets (5.2) for t∈[0, T]almost surely. Therefore the limit processXis a solution to (4.2). Since assumptions (L1) and (L2) hold for the coecients σ and b, uniqueness follows from Theorem 4.6.

Next we consider a straightforward example of how one can use Picard successive approximation to nd a solution to the given MVSDE.

Example 5.3. Consider the following MVSDE (5.3)

(dXt =λdBt+ min

e^T,|EXt+ 1| dt X₀ = 0,

where λ ∈ R is a given constant. The coecient functions clearly satisfy conditions (L1) and (L2). We use Theorem 5.1 to construct a sequence of processes that converges to a unique solution of (5.3).

Our rst iteration is X_t¹ = 0 +

Z t 0

λdBs+ Z t

0

min

e^T,|E(0) + 1| ds=λBt+t.

We continue by computing the next two iterations, X_t² =λB_t+1

2t²+t and

X_t³ =λB_t+1 6t³+ 1

2t² +t.

By induction we notice that for arbitrary n∈N one has X_tⁿ =λB_t+

n

X

k=1

t^k k!. Letting n tend to ∞ we see that

X_tⁿ →λB_t+

∞

X

k=1

t^k

k! =λB_t+e^t−1 := X_t, which clearly solves (5.3).

This method cannot always be used to approximate a solution for an MVSDE. We return to example 4.4. We recall that it has no solution, so the iterative method for nding a solution should fail.

Example 5.4. We have the following MVSDE (dXt=1Q(EX_t²) dBt

X₀ = 0.

We try to apply Theorem 5.1 to construct a solution.

The rst iteration is X_t¹ =

Z t 0

1Q(E0²) dB_s =B_t. Therefore the second one is

X_t² = Z t

0

1Q(EB_s²) dB_s= Z t

0

1Q(s) dB_s= 0.

Next we notice that the third iteration is the same as the rst one, so we have that

X_tⁿ =

(Bt, if n is odd 0, if n is even

for n ∈N. This sequence does not converge in L₂(Ω,C([0, T],R)).

In document On the uniqueness of a solution and stability of McKean-Vlasov stochastic differential equations (sivua 38-57)