Stability of Stochastic Differential Equations › ma › ws2010 › doc › mao_notes.pdf · Stability of Stochastic Differential Equations Part 1: Introduction Xuerong Mao FRSE

Lyapunov stability theory for ODEsStability of SDEs

Stability of Stochastic Differential EquationsPart 1: Introduction

Xuerong Mao FRSE

Department of Mathematics and StatisticsUniversity of Strathclyde

Glasgow, G1 1XH

December 2010

Xuerong Mao FRSE Stability of SDE


Outline

1 Lyapunov stability theory for ODEsConcept of stabilityThe Lyapunov method

2 Stability of SDEsSDEsDefinition of stochastic stabilityDiffusion operator



Outline





Concept of stabilityThe Lyapunov method

Outline






In 1892, A.M. Lyapunov introduced the concept of stabilityof a dynamic system.Roughly speaking, the stability means insensitivity of thestate of the system to small changes in the initial state orthe parameters of the system.For a stable system, the trajectories which are “close" toeach other at a specific instant should therefore remainclose to each other at all subsequent instants.












Consider a d-dimensional ordinary differential equation (ODE)

dx(t)dt

= f (x(t), t) on t ≥ 0,

where f = (f1, · · · , fd)T : Rd × R+ → Rd . Assume that for everyinitial value x(0) = x0 ∈ Rd , there exists a unique globalsolution which is denoted by x(t ; x0). Assume furthermore that

f (0, t) = 0 for all t ≥ 0.

So the ODE has the solution x(t) ≡ 0 corresponding to theinitial value x(0) = 0. This solution is called the trivial solutionor equilibrium position.




DefinitionThe trivial solution is said to be stable if, for every ε > 0, thereexists a δ = δ(ε) > 0 such that

|x(t ; x0)| < ε for all t ≥ 0.

whenever |x0| < δ. Otherwise, it is said to be unstable.

The trivial solution is said to be asymptotically stable if it isstable and if there moreover exists a δ0 > 0 such that

limt→∞

x(t ; x0) = 0

whenever |x0| < δ0.




If the ODE can be solved explicitly, it would be rather easyto determine whether the trivial solution is stable or not.However, the ODE can only be solved explicitly in somespecial cases.Fortunately, Lyapunov in 1892 developed a method fordetermining stability without solving the equation. Thismethod is now known as the method of Lyapunov functionsor the Lyapunov method.




If the ODE can be solved explicitly, it would be rather easyto determine whether the trivial solution is stable or not.However, the ODE can only be solved explicitly in somespecial cases.Fortunately, Lyapunov in 1892 developed a method fordetermining stability without solving the equation. Thismethod is now known as the method of Lyapunov functionsor the Lyapunov method.




Outline






Notation

Let K denote the family of all continuous nondecreasingfunctions µ : R+ → R+ such that µ(0) = 0 and µ(r) > 0 if r > 0.

Let K∞ denote the family of all functions µ ∈ K such thatlimr→∞ µ(r) =∞.

For h > 0, let Sh = x ∈ Rd : |x | < h.

Let C1,1(Sh × R+; R+) denote the family of all continuousfunctions V (x , t) from Sh×R+ to R+ with continuous first partialderivatives with respect to every component of x and to t .




Basic ideas of the Lyapunov method

Let x(t) be a solution of the ODE and V ∈ C1,1(Sh × R+; R+).Then v(t) = V (x(t), t) represents a function of t with thederivative

dv(t)dt

= V (x(t), t),

where V (x , t) = Vt(x , t) + (Vx1(x , t), · · · ,Vxd (x , t))f (x , t).

If dv(t)/dt ≤ 0, then v(t) will not increase so the “distance”of x(t) from the equilibrium point measured by V (x(t), t)does not increase.If dv(t)/dt < 0, then v(t) will decrease to zero so thedistance will decrease to zero, that is x(t)→ 0.




Basic ideas of the Lyapunov method

Let x(t) be a solution of the ODE and V ∈ C1,1(Sh × R+; R+).Then v(t) = V (x(t), t) represents a function of t with thederivative

dv(t)dt

= V (x(t), t),

where V (x , t) = Vt(x , t) + (Vx1(x , t), · · · ,Vxd (x , t))f (x , t).

If dv(t)/dt ≤ 0, then v(t) will not increase so the “distance”of x(t) from the equilibrium point measured by V (x(t), t)does not increase.If dv(t)/dt < 0, then v(t) will decrease to zero so thedistance will decrease to zero, that is x(t)→ 0.




Theorem

Assume that there exist V ∈ C1,1(Sh ×R+; R+) and µ ∈ K suchthat

V (0, t) = 0, µ(|x |) ≤ V (x , t)

andV (x , t) ≤ 0

for all (x , t) ∈ Sh × R+. Then the trivial solution of the ODE isstable.




Theorem

Assume that there exist V ∈ C1,1(Sh × R+; R+) andµ1, µ2, µ3 ∈ K such that

µ1(|x |) ≤ V (x , t) ≤ µ2(|x |)

andV (x , t) ≤ −µ3(|x |)

for all (x , t) ∈ Sh × R+. Then the trivial solution of the ODE isasymptotically stable.



SDEsDefinition of stochastic stabilityDiffusion operator

Outline






Consider a d-dimensional stochastic differential equation (SDE)

dx(t) = f (x(t), t)dt + g(x(t), t)dB(t) on t ≥ 0,

where f : Rd × R+ → Rd and g : Rd × R+ → Rd×m, andB(t) = (B1(t), · · · ,Bm(t))T is an m-dimensional Brownianmotion.As a standing hypothesis in this course, we assume that both fand g obey the local Lipschitz condition and the linear growthcondition.Hence, for any given initial value x(0) = x0 ∈ Rd , the SDE hasa unique global solution denoted by x(t ; x0). Assumefurthermore that

f (0, t) = 0 and g(0, t) = 0 for all t ≥ 0.

Hence the SDE admits the trivial solution x(t ; 0) ≡ 0.




When we try to carry over the principles of the Lyapunovstability theory to to SDEs, we face the following problems:

What is a suitable definition of stochastic stability?With what should the derivative dv(t)/dt or V (x , t) bereplaced?What conditions should a stochastic Lyapunov functionsatisfy?














Outline






It turns out that there are various different types of stochasticstability. In this course, we will only concentrate on

stability in probability;pth moment exponential stability;almost sure exponential stability.














DefinitionThe trivial solution of the SDE is said to be stochastically stableor stable in probability if for every pair of ε ∈ (0,1) and r > 0,there exists a δ = δ(ε, r) > 0 such that

P|x(t ; x0)| < r for all t ≥ 0 ≥ 1− ε

whenever |x0| < δ. Otherwise, it is said to be stochasticallyunstable.




DefinitionThe trivial solution is said to be stochastically asymptoticallystable if it is stochastically stable and, moreover, for everyε ∈ (0,1), there exists a δ0 = δ0(ε) > 0 such that

P limt→∞

x(t ; x0) = 0 ≥ 1− ε

whenever |x0| < δ0.




DefinitionThe trivial solution is said to be stochastically asymptoticallystable in the large if it is stochastically stable and, moreover, forall x0 ∈ Rd

P limt→∞

x(t ; x0) = 0 = 1.




DefinitionThe trivial solution is said to be almost surely exponentiallystable if for all x0 ∈ Rd

lim supt→∞

1t

log(|x(t ; x0)|) < 0 a.s.

It is said to be pth moment exponentially stable if for all x0 ∈ Rd

lim supt→∞

1t

log(E |x(t ; x0)|p) < 0.




Outline






To figure out with what the derivative dv(t)/dt or V (x , t) shouldbe replaced, we naturally consider the Itô differential of theprocess V (x(t), t), where x(t) is a solution of the SDE and V isa Lyapunov function.

According to the Itô formula, we of course requireV ∈ C2,1(Sh × R+; R+), which denotes the family of allnonnegative functions V (x , t) defined on Sh × R+ such thatthey are continuously twice differentiable in x and once in t .




By the Itô formula, we have

dV (x(t), t) = LV (x(t), t)dt + Vx(x(t), t)g(x(t), t)dB(t),

where

LV (x , t) = Vt(x , t)+Vx(x , t)f (x , t)+12

trace[gT (x , t)Vxx(x , t)g(x , t)

],

in which Vx = (Vx1 , · · · ,Vxd ) and Vxx = (Vxi xj )d×d .




We shall see that V (x , t) will be replaced by the diffusionoperator LV (x , t) in the study of stochastic stability. Forexample, the inequality V (x , t) ≤ 0 will sometimes be replacedby LV (x , t) ≤ 0 to get the stochastic stability. However, it is notnecessary to require LV (x , t) ≤ 0 to get other stabilities e.g.almost sure exponential stability.

The study of stochastic stability is therefore much richer thanthe classical stability of ODEs. Let us begin to explore thiswonderful world of stochastic stability.


TheoryExamples

Stability of Stochastic Differential EquationsPart 2: Stability in Probability

Xuerong Mao FRSE


Glasgow, G1 1XH

December 2010


TheoryExamples

Outline

1 TheoryStochastic stabilityStochastic asymptotic stabilityStochastic asymptotic stability in the large

2 ExamplesScale SDEsMulti-dimensional SDEs


TheoryExamples

Outline




TheoryExamples

Stochastic stabilityStochastic asymptotic stabilityStochastic asymptotic stability in the large

In this part, we shall see how the classical Lyapunov method isdeveloped to study stochastic stability in such a similar way thatthe results in this part are natural generalizations of theLyapunov stability theory for ODEs. Of course, such resultsmay not be surprising but we will see some surprising results inthe next part.


TheoryExamples


Outline




TheoryExamples


Theorem

Assume that there exist V ∈ C2,1(Sh ×R+; R+) and µ ∈ K suchthat

V (0, t) = 0, µ(|x |) ≤ V (x , t)

andLV (x , t) ≤ 0

for all (x , t) ∈ Sh × R+. Then the trivial solution of the SDE isstochastically stable.


TheoryExamples


Proof. Let ε ∈ (0,1) and r ∈ (0,h) be arbitrary. Clearly, we canfind a δ = δ(ε, r) ∈ (0, r) such that

1ε

supx∈Sδ

V (x ,0) ≤ µ(r).

Now fix any x0 ∈ Sδ and write x(t ; x0) = x(t) simply. Define

τ = inft ≥ 0 : x(t) 6∈ Sr.

(Throughout this course we set inf ∅ =∞.) By Itô’s formula, forany t ≥ 0,

V (x(τ ∧ t), τ ∧ t) = V (x0,0) +

∫ τ∧t

0LV (x(s), s)ds

+

∫ τ∧t

0Vx(x(s), s)g(x(s), s)dB(s).


TheoryExamples


Taking the expectation on both sides, we obtain

EV (x(τ ∧ t), τ ∧ t) = V (x0,0)+E∫ τ∧t

0LV (x(s), s)ds ≤ V (x0,0).

Noting that |x(τ ∧ t)| = |x(τ)| = r if τ ≤ t , we get

EV (x(τ ∧ t), τ ∧ t) ≥ E[Iτ≤tV (x(τ), τ)

]≥ µ(r)Pτ ≤ t.

(Throughout this course IA denotes the indicator function of setA.) We therefore obtain Pτ ≤ t ≤ ε. Letting t →∞ we getPτ <∞ ≤ ε, that is

P|x(t)| < r for all t ≥ 0 ≥ 1− ε

as required.


TheoryExamples


Outline




TheoryExamples


Theorem

Assume that there exist V ∈ C2,1(Sh × R+; R+) andµ1, µ2, µ3 ∈ K such that

µ1(|x |) ≤ V (x , t) ≤ µ2(|x |)

andLV (x , t) ≤ −µ3(|x |)

for all (x , t) ∈ Sh × R+. Then the trivial solution of the SDE isstochastically asymptotically stable.


TheoryExamples


Proof. We know from the previous theorem that the trivialsolution is stochastically stable. So we only need to show thatfor any ε ∈ (0,1), there is a δ0 = δ0(ε) > 0 such that

P limt→∞

x(t ; x0) = 0 ≥ 1− ε

whenever |x0| < δ0, or for any β ∈ (0,h/2),

Plim supt→∞

|x(t ; x0)| ≤ β ≥ 1− ε.

By the previous theorem, we can find a δ0 = δ0(ε) ∈ (0,h/2)such that

P|x(t ; x0)| < h/2 ≥ 1− ε

2. (1.1)

whenever x0 ∈ Sδ0 .


TheoryExamples


Moreover, in the same way as the previous theorem wasproved, we can show that for any β ∈ (0,h/2), there is aα ∈ (0, β) such that

P|x(t ; x0)| < β for all t ≥ s ≥ 1− ε

2(1.2)

whenever |x(s; x0)| ≤ α and s ≥ 0. Now fix any x0 ∈ Sδ andwrite x(t ; x0) = x(t) simply. Define

τα = inft ≥ 0 : |x(t)| ≤ α

andτh = inft ≥ 0 : |x(t)| ≥ h/2.


TheoryExamples


By Itô’s formula and the conditions, we can show that

0 ≤ V (x0,0) + E∫ τα∧τh∧t

0LV (x(s), s)ds

≤ V (x0,0)− µ3(α)E(τα ∧ τh ∧ t).

Consequently

tµ3(α)Pτα ∧ τh ≥ t ≤ E(τα ∧ τh ∧ t) ≤ V (x0,0).

This implies immediately that

Pτα ∧ τh <∞ = 1.

But, by (1.1), Pτh <∞ ≤ ε/2. Hence


TheoryExamples


1 = Pτα∧τh <∞ ≤ Pτα <∞+Pτh <∞ ≤ Pτα <∞+ε

2,

which yieldsPτα <∞ ≥ 1− ε

2.

We now compute, using (1.2),

Plim supt→∞

|x(t)| ≤ β

≥ Pτα <∞ and |x(t)| ≤ β for all t ≥ τα= Pτα <∞P|x(t)| ≤ β for all t ≥ τα |τα <∞ ≥ Pτα <∞(1− ε/2) ≥ (1− ε/2)2 ≥ 1− ε

as required.


TheoryExamples


Outline




TheoryExamples


Theorem

Assume that there exist V ∈ C2,1(Rd × R+; R+) andµ1, µ2 ∈ K∞ and µ3 ∈ K such that

µ1(|x |) ≤ V (x , t) ≤ µ2(|x |)

andLV (x , t) ≤ −µ3(|x |)

for all (x , t) ∈ Rd × R+. Then the trivial solution of the SDE isstochastically asymptotically stable in the large.


TheoryExamples


Proof. Clearly, we only need to show

P limt→∞

x(t ; x0) = 0 = 1

for all x0 ∈ Rd , or for any pair of ε ∈ (0,1) and β > 0,

Plim supt→∞

|x(t ; x0)| ≤ β ≥ 1− ε.

To show this, let us fix any x0 and write x(t ; x0) = x(t) again.Let h sufficiently large for h/2 > |x0| and

µ1(h/2) ≥ 2V (x0,0)

ε.

As in the previous proof, define the stopping time

τh = inft ≥ 0 : |x(t)| ≥ h/2.


TheoryExamples


By Itô’s formula, we can show that for any t ≥ 0,

EV (x(τh ∧ t), τh ∧ t) ≤ V (x0,0).

ButEV (x(τh ∧ t), τh ∧ t) ≥ µ1(h/2)Pτh ≤ t.

Hence Pτh ≤ t ≤ ε2 . Letting t →∞ gives Pτh <∞ ≤ ε/2.

That isP|x(t)| < h/2 for all t ≥ 0 ≥ 1− ε

2,

which is the same as (1.1). From here, we can show in thesame way as in the previous proof that

Plim supt→∞

|x(t)| ≤ β ≥ 1− ε

as desired.


TheoryExamples

Scale SDEsMulti-dimensional SDEs

Outline




TheoryExamples


Consider a scale SDE

dx(t) = f (x(t), t)dt + g(x(t), t)dB(t) on t ≥ 0

with initial value x(0) = x0 ∈ R. Assume that f : R × R+ → Rand g : R × R+ → Rm have the expansions

f (x , t) = a(t)x+o(|x |), g(x , t) = (b1(t)x , · · · ,bm(t)x)T +o(|x |).

in a neighbourhood of x = 0 uniformly with respect to t ≥ 0,where a(t), bi(t) are all bounded Borel-measurable real-valuedfunctions. We impose a condition that there is a pair of positiveconstants θ and K such that

−K ≤∫ t

0

(a(s)− 1

2

m∑i=1

b2i (s) + θ

)ds ≤ K for all t ≥ 0.


TheoryExamples


Let0 < ε <

θ

supt≥0∑m

i=1 b2i (t)

and define the Lyapunov function

V (x , t) = |x |ε exp[− ε

∫ t

0

(a(s)− 1

2

m∑i=1

b2i (s) + θ

)ds].

Then, by the condition,

|x |εe−εK ≤ V (x , t) ≤ |x |εeεK .


TheoryExamples


Moreover, compute

LV (x , t) = ε|x |ε exp[−ε∫ t

0

(a(s)− 1

2

m∑i=1

b2i (s) + θ

)ds]

×( ε

2

m∑i=1

b2i (t)− θ

)+ o(|x |ε)

≤ −12εθe−εK |x |ε + o(|x |ε).

We hence see that LV (x , t) is negative-definite in a sufficientlysmall neighbourhood of x = 0 for t ≥ 0. We can thereforeconclude that the trivial solution of the scale SDE isstochastically asymptotically stable.


TheoryExamples


Outline




TheoryExamples


Assume that the coefficients f and g of the underlying SDEhave the expansions

f (x , t) = F (t)x+o(|x |), g(x , t) = (G1(t)x , · · · ,Gm(t)x)+o(|x |)

in a neighbourhood of x = 0 uniformly with respect to t ≥ 0,where F (t), Gi(t) are all bounded Borel-measurabled × d-matrix-valued functions. Assume that there is asymmetric positive-definite matrix Q such that

λmax

(QF (t) + F T (t)Q +

m∑i=1

GTi (t)QGi(t)

)≤ −λ < 0

for all t ≥ 0, where (and throughout this course) λmax(A)denotes the largest eigenvalue of matrix A.


TheoryExamples


Now, define the Lyapunov function V (x , t) = xT Qx . Clearly,

λmin(Q)|x |2 ≤ V (x , t) ≤ λmax(Q)|x |2.

Moreover,

LV (x , t) = xT(

QF (t) + F T (t)Q +m∑

i=1

GTi (t)QGi(t)

)x + o(|x |2)

≤ −λ|x |2 + o(|x |2).

Hence LV (x , t) is negative-definite in a sufficiently smallneighbourhood of x = 0 for t ≥ 0. We therefore conclude thatthe trivial solution of the SDE is stochastically asymptoticallystable.


TheoryExamples

Stability of Stochastic Differential EquationsPart 3: Almost Sure Exponential Stability

Xuerong Mao FRSE


Glasgow, G1 1XH

December 2010


TheoryExamples

Outline

1 TheoryAlmost sure exponential stabilityAlmost sure exponential instability

2 ExamplesLinear SDEsNonlinear case


TheoryExamples

Outline




TheoryExamples

Almost sure exponential stabilityAlmost sure exponential instability

In this part, we shall develop the classical Lyapunov method tostudy the almost sure exponential stability. In contrast to theclassical Lyapunov stability theory, we will no longer requireLV (x , t) be negative-definite but we still obtain the almost sureexponential stability making full use of the diffusion (noise)terms. It is this new feature that makes the stochastic stabilitymore interesting and more useful as well.


TheoryExamples


To establish the theory on the almost sure exponential stability,we need prepare an important lemma. Recall that we assume,throughout this course, that both coefficients f and g obey thelocal Lipschitz condition and the linear growth condition and,moreover, f (0, t) ≡ 0, g(0, t) ≡ 0. Under these standinghypotheses, we have the following useful lemma.

Lemma

For all x0 6= 0 in Rd

Px(t ; x0) 6= 0 for all t ≥ 0 = 1.

That is, almost all the sample path of any solution starting froma non-zero state will never reach the origin with probability 1.


TheoryExamples


Proof. If the lemma were false, there would exist some x0 6= 0such that Pτ <∞ > 0, where

τ = inft ≥ 0 : x(t) = 0

in which we write x(t ; x0) = x(t) simply. So we can find a pair ofconstants T > 0 and θ > 1 sufficiently large for P(B) > 0,where

B = τ ≤ T and |x(t)| ≤ θ − 1 for all 0 ≤ t ≤ τ.

But, by the standing hypotheses, there exists a positiveconstant Kθ such that

|f (x , t)| ∨ |g(x , t)| ≤ Kθ|x | for all |x | ≤ θ, 0 ≤ t ≤ T .


TheoryExamples


Let V (x , t) = |x |−1. Then, for 0 < |x | ≤ θ and 0 ≤ t ≤ T ,

LV (x , t) = −|x |−3xT f (x , t)

+12

(−|x |−3|g(x , t)|2 + 3|x |−5|xT g(x , t)|2

)≤ |x |−2|f (x , t)|+ |x |−3|g(x , t)|2

≤ Kθ|x |−1 + K 2θ |x |−1

= Kθ(1 + Kθ)V (x , t).

Now, for any ε ∈ (0, |x0|), define the stopping time

τε = inft ≥ 0 : |x(t)| 6∈ (ε, θ).

By Itô’s formula,


TheoryExamples


E[e−Kθ(1+Kθ)(τε∧T )V (x(τε ∧ T ), τε ∧ T )

]− V (x0,0)

= E∫ τε∧T

0e−Kθ(1+Kθ)s

[−(Kθ(1 + Kθ))V (x(s), s) + LV (x(s), s)

]ds

≤ 0.

Note that for ω ∈ B, τε ≤ T and |x(τε)| = ε. The aboveinequality therefore implies that

E[e−Kθ(1+Kθ)T ε−1IB

]≤ |x0|−1.

Hence P(B) ≤ ε|x0|−1eKθ(1+Kθ)T . Letting ε→ 0 yields thatP(B) = 0, but this contradicts the definition of B. The proof iscomplete.


TheoryExamples


We will also need the well-known exponential martingaleinequality which we state here as a lemma.

Lemma

Let g = (g1, · · · ,gm) ∈ L2(R+; R1×m), and let T , α, β be anypositive numbers. Then

P

sup0≤t≤T

[∫ t

0g(s)dB(s)− α

2

∫ t

0|g(s)|2ds

]> β

≤ e−αβ.


TheoryExamples


Outline




TheoryExamples


Theorem

Assume that there exists a function V ∈ C2,1(Rd × R+; R+),and constants p > 0, c1 > 0, c2 ∈ R, c3 ≥ 0, such that for allx 6= 0 and t ≥ 0,

c1|x |p ≤ V (x , t),LV (x , t) ≤ c2V (x , t),

|Vx (x , t)g(x , t)|2 ≥ c3V 2(x , t).

Thenlim sup

t→∞

1t

log |x(t ; x0)| ≤ −c3 − 2c2

2pa.s. (1.1)

for all x0 ∈ Rd . In particular, if c3 > 2c2, then the trivial solutionof the SDE is almost surely exponentially stable.


TheoryExamples


Proof. Clearly, the assertion holds for x0 = 0 since x(t ; 0) ≡ 0.Fix any x0 6= 0 and write x(t ; x0) = x(t). By the lemma, x(t) 6= 0for all t ≥ 0 almost surely. Thus, one can apply Itô’s formulaand the condition to show that, for t ≥ 0,

log V (x(t), t) ≤ log V (x0,0) + c2t + M(t)

− 12

∫ t

0

|Vx (x(s), s)g(x(s), s)|2

V 2(x(s), s)ds, (1.2)

where

M(t) =

∫ t

0

Vx (x(s), s)g(x(s), s)

V (x(s), s)dB(s)

is a continuous martingale with initial value M(0) = 0. Assignε ∈ (0,1) arbitrarily and let n = 1,2, · · · . By the exponentialmartingale inequality,


TheoryExamples


P

sup0≤t≤n

[M(t)− ε

2

∫ t

0

|Vx (x(s), s)g(x(s), s)|2

V 2(x(s), s)ds]>

2ε

log n≤ 1

n2 .

Applying the Borel–Cantelli lemma we see that for almost allω ∈ Ω, there is an integer n0 = n0(ω) such that if n ≥ n0,

M(t) ≤ 2ε

log n +ε

2

∫ t

0

|Vx (x(s), s)g(x(s), s)|2

V 2(x(s), s)ds

holds for all 0 ≤ t ≤ n. Substituting this into (1.2) and thenusing the condition we obtain that

log V (x(t), t) ≤ log V (x0,0)− 12

[(1− ε)c3 − 2c2]t +2ε

log n

for all 0 ≤ t ≤ n, n ≥ n0 almost surely.


TheoryExamples


Consequently, for almost all ω ∈ Ω, if n − 1 ≤ t ≤ n and n ≥ n0,

1t

log V (x(t), t) ≤ −12

[(1− ε)c3 − 2c2] +log V (x0,0) + 2

ε log nn − 1

.

This implies

lim supt→∞

1t

log V (x(t), t) ≤ −12

[(1− ε)c3 − 2c2] a.s.

Hence

lim supt→∞

1t

log |x(t)| ≤ −(1− ε)c3 − 2c2

2pa.s.

and the required assertion follows since ε > 0 is arbitrary.


TheoryExamples


Outline




TheoryExamples


Theorem

Assume that there exists a function V ∈ C2,1(Rd × R+; R+),and constants p > 0, c1 > 0, c2 ∈ R, c3 > 0, such that for allx 6= 0 and t ≥ 0,

c1|x |p ≥ V (x , t) > 0,LV (x , t) ≥ c2V (x , t),

|Vx (x , t)g(x , t)|2 ≤ c3V 2(x , t).

Thenlim inft→∞

1t

log |x(t ; x0)| ≥ 2c2 − c3

2pa.s.

for all x0 6= 0 in Rd . In particular, if 2c2 > c3, then almost all thesample paths of |x(t ; x0)| will tend to infinity exponentially.


TheoryExamples


Proof. Fix any x0 6= 0 and write x(t ; x0) = x(t). By Itô’s formulaand the conditions, we can easily show that for t ≥ 0,

log V (x(t), t) ≥ log V (x0,0) +12

(2c2 − c3)t + M(t), (1.3)

where

M(t) =

∫ t

0

Vx (x(s), s)g(x(s), s)

V (x(s), s)dB(s)

is a continuous martingale with the quadratic variation

〈M(t),M(t)〉 =

∫ t

0

|Vx (x(s), s)g(x(s), s)|2

V 2(x(s), s)ds ≤ c3t .


TheoryExamples


By the strong law of large numbers for the martingales,

limt→∞

M(t)t

= 0 a.s.

It therefore follows from (1.3) that

lim inft→∞

1t

log V (x(t), t) ≥ 12

(2c2 − c3) a.s.

which implies the required assertion immediately by using thecondition.


TheoryExamples

Linear SDEsNonlinear case

Outline




TheoryExamples


Consider the scalar linear SDE

dx(t) = ax(t) +m∑

i=1

bix(t)dBi(t) on t ≥ 0.

It is known that it has the explicit solution

x(t) = x0 exp(

[a− 0.5m∑

i=1

b2i ]t +

m∑i=1

biBi(t)).

This implies that, for x0 6= 0,

limt→∞

1t

log |x(t)| = a− 0.5m∑

i=1

b2i a.s.


TheoryExamples


Let us now apply the stability theorem to obtain the sameconclusion. Let V (x , t) = x2. Then

LV (x , t) =(

2a +m∑

i=1

b2i

)|x |2

and, writing g(x , t) = (b1x , · · · ,bmx),

|Vx (x , t)g(x , t)|2 = 4m∑

i=1

b2i |x |4.

In other words, the conditions in the Theorems holds with

p = 2, c1 = 1, c2 = 2a +m∑

i=1

b2i , c3 = 4

m∑i=1

b2i .


TheoryExamples


We hence have

lim supt→∞

1t

log |x(t)| ≤ a− 12

m∑i=1

b2i a.s.

and

lim inft→∞

1t

log |x(t)| ≥ a− 12

m∑i=1

b2i a.s.

Combining these gives what we want.


TheoryExamples


Consider, for example,

dx(t) = x(t)dt + 2x(t)dB(t)

with initial value x(0) = x0 6 0, where B(t) is a one-dimensionalBrownian motion. The theory above shows that the solution ofthis linear sde obeys

lim inft→∞

1t

log |x(t)| = −1 a.s.

The following simulation shows a typical sample path of thesolution with initial value x(0) = 10.


TheoryExamples


0 2 4 6 8 10

010

2030

4050

60

t

x(t)


TheoryExamples


Outline




TheoryExamples


Consider the two-dimensional SDE

dx(t) = f (x(t))dt + Gx(t)dB(t) on t ≥ 0

with initial value x(0) = x0 ∈ R2 and x0 6= 0, where B(t) is aone-dimensional Brownian motion,

f (x) =

(x2 cos x12x1 sin x2

), G =

(3 −0.3−0.3 3

)


TheoryExamples


Let V (x , t) = |x |2. It is easy to verify that

4.29|x |2 ≤ LV (x , t) = 2x1x2 cos x1+4x1x2 sin x2+|Gx |2 ≤ 13.89|x |2

and

29.16|x |2 ≤ |Vx (x , t)Gx |2 = |2xT Gx |2 ≤ 43.56|x |4.

Applying the Theorems we then have

−8.745 ≤ lim inft→∞

1t

log |x(t ; x0)| ≤ lim supt→∞

1t

log |x(t ; x0)| ≤ −0.345

almost surely. The following figure is a compute simulation.


TheoryExamples


0 2 4 6 8 10

01

23

4

t

X1(t)

0 2 4 6 8 100

24

6t

X2(t)

x1(0) = x2(0) = 1.


Moment verse Almost Sure Exponential StabilityCriteria

A Case Study

Stability of Stochastic Differential EquationsPart 4: Moment Exponential Stability

Xuerong Mao FRSE


Glasgow, G1 1XH

December 2010



A Case Study

Outline

1 Moment verse Almost Sure Exponential Stability

2 CriteriaNonlinear caseLinear case

3 A Case Study



A Case Study

Outline



3 A Case Study



A Case Study

Outline



3 A Case Study



A Case Study

Generally speaking, the pth moment exponential stability andthe almost sure exponential stability do not imply each otherand additional conditions are required in order to deduce onefrom the other. The following theorem gives the conditionsunder which the pth moment exponential stability implies thealmost sure exponential stability. However we still do not knowunder what conditions the almost sure exponential stabilityimplies the pth moment exponential stability.



A Case Study

TheoremAssume that there is a positive constant K such that

xT f (x , t) ∨ |g(x , t)|2 ≤ K |x |2 for all (x , t) ∈ Rd × R+.

Then the pth moment exponential stability of the trivial solutionof the SDE implies the almost sure exponential stability.



A Case Study

To prove this theorem we need the Burkholder–Davis–Gundyinequality which we cite as a lemma.

Lemma

Let g ∈ L2(R+; Rd×m). Define, for t ≥ 0,

x(t) =

∫ t

0g(s)dB(s) and A(t) =

∫ t

0|g(s)|2ds.

Then for every p > 0, there exist universal positive constantscp,Cp (depending only on p), such that

cpE |A(t)|p2 ≤ E

(sup

0≤s≤t|x(s)|p

)≤ CpE |A(t)|

p2 .



A Case Study

In particular, one may take

cp = (p/2)p, Cp = (32/p)p/2 if 0 < p < 2;

cp = 1, Cp = 4 if p = 2;

cp = (2p)−p/2, Cp =[pp+1/2(p − 1)p−1]p/2 if p > 2.



A Case Study

Proof of the theorem. Fix any x0 6= 0 in Rd and writex(t ; x0) = x(t) simply. By the definition of the pth momentexponential stability, there is a pair of positive constants and Csuch that

E |x(t)|p ≤ Ce−λt on t ≥ 0.

Let n = 1,2, · · · . By Itô’s formula and the condition, one canshow that for n − 1 ≤ t ≤ n,

|x(t)|p ≤ |x(n − 1)|p + c1

∫ t

n−1|x(s)|pds

+

∫ t

n−1p|x(s)|p−2xT (s)g(x(s), s)dB(s),

where c1 = pK + p(1 + |p − 2|)K/2. Hence



A Case Study

E(

supn−1≤t≤n

|x(t)|p)≤ E |x(n − 1)|p + c1

∫ n

n−1E |x(s)|pds

+E(

supn−1≤t≤n

∫ t

n−1p|x(s)|p−2xT (s)g(x(s), s)dB(s)

).

On the other hand, by the well-knownBurkholder–Davis–Gundy inequality we compute



A Case Study

E(

supn−1≤t≤n

∫ t

n−1p|x(s)|p−2xT (s)g(x(s), s)dB(s)

)

≤ 4√

2E(∫ n

n−1p2|x(s)|2(p−2)|xT (s)g(x(s), s)|2ds

) 12

≤ 4√

2E(

supn−1≤s≤n

|x(s)|p∫ n

n−1p2K |x(s)|pds

) 12

≤ 12

E(

supn−1≤s≤n

|x(s)|p)

+ 16p2K∫ n

n−1E |x(s)|pds.

Substituting this into the previous inequality yields



A Case Study

E(

supn−1≤t≤n

|x(t)|p)≤ 2E |x(n − 1)|p + c2

∫ n

n−1E |x(s)|pds,

where c2 = 2c1 + 32p2K . By the property of the pth momentexponential stability, we then have

E(

supn−1≤t≤n

|x(t)|p)≤ c3e−λ(n−1),

where c3 = C(2 + c2). Now, let ε ∈ (0, λ) be arbitrary. Then

P

supn−1≤t≤n

|x(t)|p > e−(λ−ε)(n−1)≤ c3e−ε(n−1).



A Case Study

In view of the Borel–Cantelli lemma we see that for almost allω ∈ Ω,

supn−1≤t≤n

|x(t)|p ≤ e−(λ−ε)(n−1)

holds for all but finitely many n. Hence, there exists ann0 = n0(ω), for all ω ∈ Ω excluding a P-null set, for which theinequality above holds whenever n ≥ n0. Consequently, foralmost all ω ∈ Ω,

1t

log |x(t)| =1pt

log(|x(t)|p) ≤ −(λ− ε)(n − 1)

pn

if n − 1 ≤ t ≤ n, n ≥ n0.



A Case Study

Hencelim sup

t→∞

1t

log |x(t)| ≤ −(λ− ε)

pa.s.

Since ε > 0 is arbitrary, we must have

lim supt→∞

1t

log |x(t)| ≤ −λp

a.s.

By definition, the trivial solution of the SDE is almost surelyexponentially stable.



A Case Study

Nonlinear caseLinear case

Outline



3 A Case Study



A Case Study


Theorem

Assume that there is a function V (∈ C2,1(Rd × R+; R+), andpositive constants c1–c3, such that

c1|x |p ≤ V (x , t) ≤ c2|x |p and LV (x , t) ≤ −c3V (x , t)

for all (x , t) ∈ Rd × R+. Then

E |x(t ; x0)|p ≤ c2

c1|x0|pe−c3t on t ≥ 0

for all x0 ∈ Rd .



A Case Study


Proof. Fix any x0 ∈ Rd and write x(t ; x0) = x(t). For eachn ≥ |x0|, define the stopping time

τn = inft ≥ 0 : |x(t)| ≥ n.

Obviously, τn →∞ as n→∞ almost surely. By Itô’s formula,we can derive that for t ≥ 0,

E[ec3(t∧τn)V (x(t ∧ τn), t ∧ τn)

]− V (x0,0)

= E∫ t∧τn

0ec3s[c3V (x(s), s) + LV (x(s), s)

]ds ≤ 0



A Case Study


Hence

c1E[ec3(t∧τn)E |x(t ∧ τn)|p

]≤ V (x0,0) ≤ c2|x0|p.

Letting n→∞ yields that

c1ec3tE |x(t)|p ≤ c2|x0|p

which implies the desired assertion.



A Case Study


TheoremAssume that there exists a symmetric positive-definite d × dmatrix Q, and constants α1 ∈ R, 0 ≤ α2 < α3, such that for all(x , t) ∈ Rd × R+,

xT Qf (x , t) +12

trace[gT (x , t)Qg(x , t)] ≤ α1xT Qx

andα2xT Qx ≤ |xT Qg(x , t)| ≤ α3xT Qx .

(i) If α1 < 0, then the trivial solution of the SDE is pth momentexponentially stable provided p < 2 + 2|α1|/α2

3.(ii) If 0 ≤ α1 < α2

2, then the trivial solution of equation (1.2) ispth moment exponentially stable provided p < 2− 2α1/α

22.



A Case Study


Proof. Let V (x , t) = (xT Qx)p2 . Then

λp2min(Q)|x |p ≤ V (x , t) ≤ λ

p2max(Q)|x |p.

It is also easy to verify that

LV (x , t) = p(xT Qx)p2−1(

xT Qf (x , t) +12

trace[gT (x , t)Qg(x , t)])

+ p(p

2− 1)

(xT Qx)p2−2|xT Qg(x , t)|2.



A Case Study


(i) Assume that α1 < 0 and p < 2 + 2|α1|/α23. Without loss of

generality, we can let p ≥ 2. Then

LV (x , t) ≤ −p[|α1| −

(p2− 1)α2

3

]V (x , t).

(ii) Assume that 0 ≤ α1 < α22 and p < 2− 2α1/α

22. Then

LV (x , t) ≤ −p[(p

2− 1)α2

2 − α1

]V (x , t).

So in both cases the stability assertion follows from theprevious theorem.



A Case Study


Outline



3 A Case Study



A Case Study


Consider a d-dimensional linear SDE

dx(t) = Fx(t)dt +m∑

i=1

Gix(t)dBi(t),

where F , Gi ∈ Rd×d . This is of course a special case of theunderlying SDE where

f (x , t) = Fx , g(x , t) = (G1x , · · · ,Gmx).



A Case Study


CorollaryAssume that there exists a symmetric positive-definite d × dmatrix Q such that the following LMI holds:

QF + F T Q +m∑

i=1

GTi QGi < 0.

Then the trivial solution of the linear SDE is mean-squareexponentially stable as well as almost surely exponentiallystable.



A Case Study


Proof. Let V (x , t) = xT Qx . Then

λmin(Q)|x |2 ≤ V (x , t) ≤ λmax(Q)|x |2.

Moreover

LV (x , t) = xT Qx ≤ λmax(Q)|x |2.

where Q = QF + F T Q +∑m

i=1 GTi QGi . By the condition,

λmax(Q) < 0. Hence

LV (x , t) ≤ −|λmax(Q)|λmax(Q)

V (x , t).

The assertions follow therefore from the theory establishedabove.



A Case Study


In the case when

QF + F T Q +m∑

i=1

GTi QGi

is not negative-definite, the following result is useful.



A Case Study


CorollaryAssume that there exists a symmetric positive-definite d × dmatrix Q, and nonnegative constants β and βi (1 ≤ i ≤ m),such that β <

∑mi=1 βi ,

QF + F T Q +m∑

i=1

GTi QGi − βQ ≤ 0,

and, moreover, for each i = 1, · · · ,m,

either QGi + GTi Q −

√2βiQ ≥ 0 or QGi + GT

i Q +√

2βiQ ≤ 0.

If 0 < p < 2− 2β/(∑m

i=1 βi), then the trivial solution of thelinear SDE is pth moment exponentially stable, whence it isalso almost surely exponentially stable.



A Case Study


Proof. We will use the 2nd theorem established above to showthis corollary. We first have that

xT Qf (x , t) +12

trace[gT (x , t)Qg(x , t)]

= 0.5xT(

QF + F T Q +m∑

i=1

GTi QGi

)x ≤ 0.5βxT Qx .

We also observe from the condition that for each i ,

|xT QGix |2 = 0.25|xT (QGi + GTi Q)x |2 ≥ 0.5βi(xT Qx)2.

Hence



A Case Study


|xT Qg(x , t)| =

√√√√ m∑i=1

|xT QGix |2 ≥

√√√√0.5m∑

i=1

βi xT Qx .

Applying the theorem with

α1 = 0.5β, α2 =

√√√√0.5m∑

i=1

βi ,

we can therefore conclude that the trivial solution of the linearSDE is pth moment exponentially stable if0 < p < 2− 2β/(

∑mi=1 βi). This implies that the trivial solution

of the linear SDE is also almost surely exponentially stable.



A Case Study


As an even more special case, let us consider the scalar linearSDE

dx(t) = ax(t)dt +m∑

i=1

bix(t)dBi(t),

where a, bi are all real numbers. Using the corollaries above,we can conclude:



A Case Study


If 2a +∑m

i=1 b2i < 0, then the trivial solution of this scalar

linear SDE is mean-square exponentially stable as well asalmost surely exponentially stable.If 0 ≤ 2a +

∑mi=1 b2

i < 2∑m

i=1 b2i , then the trivial solution of

this scalar linear SDE is pth moment exponentially stableprovided

0 < p < 1− 2a∑mi=1 b2

i,

whence it is also almost surely exponentially stable.



A Case Study


If 2a +∑m

i=1 b2i < 0, then the trivial solution of this scalar

linear SDE is mean-square exponentially stable as well asalmost surely exponentially stable.If 0 ≤ 2a +

∑mi=1 b2

i < 2∑m

i=1 b2i , then the trivial solution of

this scalar linear SDE is pth moment exponentially stableprovided

0 < p < 1− 2a∑mi=1 b2

i,

whence it is also almost surely exponentially stable.



A Case Study

This example is from the satellite dynamics. Sagirow in 1970derived the equation

y(t) + β(1 + αB(t))y(t) + (1 + αB(t))y(t)− γ sin(2y(t)) = 0

in the study of the influence of a rapidly fluctuating density ofthe atmosphere of the earth on the motion of a satellite in acircular orbit. Here B(t) is a scalar white noise, α is a constantrepresenting the intensity of the disturbance, and β, γ are twopositive constants.



A Case Study

Introducing x = (x1, x2)T = (y , y)T , we can write this equationas the two-dimensional SDE

dx1(t) = x2(t)dt ,dx2(t) = [−x1(t) + γ sin(2x1(t))− βx2(t)]dt

−α[x1(t) + βx2(t)]dB(t).

For the Lyapunov function, we try an expression consisting of aquadratic form and integral of the nonlinear component:

V (x , t) = ax21 + bx1x2 + x2

2 + c∫ x1

0sin(2y)dy

= ax21 + bx1x2 + x2

2 + c sin2 x1.



A Case Study

This yields

LV (x , t) = −(b − α2)x21 + bγx1 sin(2x1)− (2β − b − α2β2)x2

2

+ (2a− bβ − 2 + 2α2β)x1x2 + (c + 2γ)x2 sin(2x1).

Setting 2a− bβ − 2 + 2α2β = 0 and c + 2γ = 0 we obtain

V (x , t) =12

(bβ + 2− 2α2β)x21 + bx1x2 + x2

2 − 2γ sin2 x1

and

LV (x , t) = −(b − α2)x21 + bγx1 sin(2x1)− (2β − b − α2β2)x2

2 .



A Case Study

Note that

V (x , t) ≥ 12

(bβ + 2− 2α2β − 4γ)x21 + bx1x2 + x2

2 .

So V (x , t) ≥ ε|x |2 for some ε > 0 if

2(bβ + 2− 2α2β − 4γ) ≥ b2

or equivalently

β −√β2 + 4− 8γ − 4α2β < b < β +

√β2 + 4− 8γ − 4α2β.

Note also that

LV (x , t) ≤ −(b − α2 − 2bγ)x21 − (2β − b − α2β2)x2

2 .



A Case Study

So LV (x , t) ≤ −ε|x |2 for some ε > 0 provided bothb − α2 − 2bγ > 0 and 2β − b − α2β2 > 0, that is

2γ < 1 and α2/(1− 2γ) < b < 2β − α2β2.

We can therefore conclude that if γ < 1/2 and

maxα2/(1− 2γ), β −

√β2 + 4− 8γ − 4α2β

< min

2β − α2β2, β +

√β2 + 4− 8γ − 4α2β

then the trivial solution of the SDE is exponentially stable inmean square.


Motivating Examples and HistoryStochastic Stabilization

Stochastic Destabilization

Stability of Stochastic Differential EquationsPart 5: Stochastic Stabilization and Destabilization

Xuerong Mao FRSE


Glasgow, G1 1XH

December 2010




Outline

1 Motivating Examples and HistoryDestabilizationStabilizationA brief history

2 Stochastic StabilizationTheoryExamples and Simulations

3 Stochastic DestabilizationTheoryCase study and simulations




Outline







Outline







DestabilizationStabilizationA brief history

Outline








It is not surprising that noise can destabilize a stable system.





Consider a 2-dimensional ODE

y(t) = −y(t) on t ≥ 0, y(0) = y0 ∈ R2.

This is an exponentially stable system. Perturb it by noise andassume the stochastically perturbed system is described by anSDE

dx(t) = −x(t)dt + Gx(t)dB(t) on t ≥ 0, x(0) = y0 ∈ R2,

where B(t) is a scalar Brownian motion and

G =

(0 −22 0

)





The SDE has the explicit solution

x(t) = exp[(−I − 0.5G2)t + GB(t)]x(0) = exp[It + GB(t)]x(0),

where I is the 2× 2 identity matrix. Consequently

limt→∞

1t

log(|x(t)|) = 1 a.s.

That is, the stochastically perturbed system has becomeunstable with probability one.





0.0 0.5 1.0 1.5 2.0

−4−2

02

46

8

t

x1(t)

or y

1(t)

x1(t)y1(t)

0.0 0.5 1.0 1.5 2.0

05

10

t

x2(t)

or y

2(t)

x2(t)y2(t)





Do you believe that noise can also stabilize an unstablesystem?





Outline








Consider the scalar ODE

y(t) = y(t) on t ≥ 0, y(0) = y0 ∈ R.

The solution isy(t) = y(0)et .

So |y(t)| → ∞ if y(0) 6= 0. That is, the ODE is an exponentiallyunstable system. Perturb it by noise and assume thestochastically perturbed system is described by an SDE

dx(t) = x(t)dt + σx(t)dB(t) on t ≥ 0, x(0) = y0 ∈ R,





The SDE has the explicit solution

x(t) = x(0) exp[(1− 0.5σ2)t + σB(t)].

Consequentlyx(t)→ 0 a.s. if σ >

√2.

That is, the stochastically perturbed system has become stablewith probability one.





0 2 4 6 8 10

01

23

45

67

t

x(t) o

r y(t)

x(t)y(t)

σ = 2





Of course, if the noise is not strong enough, it will not be able tostabilize the system.





0 2 4 6 8 10

010

020

030

040

050

0

t

x(t) o

r y(t)

x(t)y(t)

σ = 0.5





0 2 4 6 8 10

01

23

45

67

t

x(t) o

r y(t)

x(t)y(t)

σ =√

2





Outline








Has’minskii (1969): The pioneering work where two whitenoise sources were used to stabilize a particular system.Arnold, Crauel & Wihstutz (1983) and Arnold (1990): Anylinear system x(t) = Ax(t) with trace(A) < 0 can bestabilized by one real noise source.Scheutzow (1993): Stochastic stabilization for two specialnonlinear systems.Mao (1994): The general theory on stochastic stabilizationfor nonlinear SDEs.Mao (1996): Design a stochastic control that canself-stabilize the underlying system.Caraballo, Liu and Mao (2001): Stochastic stabilization forpartial differential equations (PDEs).Appleby and Mao (2004): Stochastic stabilization forfunctional differential equations.


































TheoryExamples and Simulations

Outline








Suppose that the given system is described by a nonlinear ODE

y(t) = f (y(t), t) on t ≥ 0, y(0) = x0 ∈ Rd .

Here f : Rd × R+ → Rd is a locally Lipschitz continuousfunction and particularly, for some K > 0,

|f (x , t)| ≤ K |x | for all (x , t) ∈ Rd × R+.





We now use the m-dimensional Brownian motionB(t) = (B1(t), · · · ,Bm(t))T as the source of noise to perturb thegiven system. For simplicity, suppose the stochasticperturbation is of a linear form, that is the stochasticallyperturbed system is described by the semi-linear Itô equation

dx(t) = f (x(t), t)dt+m∑

i=1

Gix(t)dBi(t) on t ≥ 0, x(0) = x0 ∈ Rd ,

where Gi , 1 ≤ i ≤ m, are all d × d matrices.





TheoremAssume that there are two constants λ > 0 and ρ ≥ 0 such that

m∑i=1

|Gix |2 ≤ λ|x |2 andm∑

i=1

|xT Gix |2 ≥ ρ|x |4

for all x ∈ Rd . Then

lim supt→∞

1t

log |x(t)| ≤ −(ρ− K − λ

2

)a.s.

for all x0 ∈ Rd . In particular, if ρ > K + 12λ, then the

stochastically perturbed system is almost surely exponentiallystable.





Outline








ExampleLet Gi = σi I for 1 ≤ i ≤ m, where I is the d × d identity matrixand σi a constant. Then the SDE becomes

dx(t) = f (x(t), t)dt +m∑

i=1

σix(t)dBi(t).

Moreover,

m∑i=1

|Gix |2 =m∑

i=1

σ2i |x |2 and

m∑i=1

|xT Gix |2 =m∑

i=1

σ2i |x |4.

The solution of the SDE has the property

lim supt→∞

1t

log |x(t)| ≤ −(1

2

m∑i=1

σ2i − K

)a.s.





Therefore, The SDE is almost surely exponentially stableprovided

12

m∑i=1

σ2i > K .

An even simpler case is that when σi = 0 for 2 ≤ i ≤ m, i.e. theSDE

dx(t) = f (x(t), t)dt + σ1x(t)dB1(t).

This SDE is almost surely exponentially stable provided12σ

21 > K . These show that if we add a strong enough stochastic

perturbation to the given ODE, then the system is stabilized.





The Theorem above ensures that there are many choices forthe matrices Gi in order to stabilize a given system and ofcourse the above choices are just the simplest ones. Forillustration, we give one more example here.

For each i , choose a positive-definite matrix Di such that

xT Dix ≥√

32||Di || |x |2.

Obviously, there are many such matrices. Let σ be a constantand Gi = σDi . Then





m∑i=1

|Gix |2 ≤ σ2m∑

i=1

||Di ||2|x |2

andm∑

i=1

|xT Gix |2 ≥3σ2

4

m∑i=1

||Di ||2|x |3.

By the Theorem, the solution of the SDE satisfies

lim supt→∞

1t

log |x(t)| ≤ −(σ2

4

m∑i=1

||Di ||2 − K)

a.s.

The SDE is therefore almost surely exponentially stable if

σ2 >4K∑m

i=1 ||Di ||2.





TheoremAny nonlinear system y(t) = f (y(t), t) can be stabilized byBrownian motions provided the following condition is fulfilled


Moreover, one can even use only a scalar Brownian motion tostabilize the system.





ExampleGiven an unstable 2-dimensional ODE

y(t) = f (y(t), t),

where

f (y , t) =

(y1 cos(t) + y2 sin(y1)y2 sin(t) + y1 cos(y2)

).

It is easy to see

|f (y , t)| ≤ 2|y | ∀(y , t) ∈ R2 × R+.





Perturbing this ODE by a scalar Brownian motion results in anSDE

dx(t) = f (x(t), t)dt + σ1x(t)dB1(t).

The Theorem above shows that this SDE is almost surelyexponentially stable provided

σ1 > 2.





0 1 2 3 4 5

01

23

45

6

t

x1(t)

or y1

(t)

x1(t)y1(t)

0 1 2 3 4 50

12

34

t

x2(t)

or y2

(t)

x2(t)y2(t)

σ1 = 3





0 1 2 3 4 5

01

23

45

67

t

x1(t)

or y1

(t)

x1(t)y1(t)

0 1 2 3 4 50

12

34

t

x2(t)

or y2

(t)

x2(t)y2(t)

σ1 = 2.5





0 1 2 3 4 5

02

46

8

t

x1(t)

or y1

(t)

x1(t)y1(t)

0 1 2 3 4 50

24

68

t

x2(t)

or y2

(t)

x2(t)y2(t)

σ1 = 2





0 1 2 3 4 5

1.01.5

2.02.5

3.03.5

t

x1(t)

or y1

(t)

x1(t)y1(t)

0 1 2 3 4 51

23

4

t

x2(t)

or y2

(t)

x2(t)y2(t)

σ1 = 1.5




TheoryCase study and simulations

Outline








TheoremAssume that there are two positive constants λ and ρ such that

m∑i=1

|Gix |2 ≥ λ|x |2 andm∑

i=1

|xT Gix |2 ≤ ρ|x |4

for all x ∈ Rd . Then

lim inft→∞

1t

log |x(t)| ≥(λ

2− K − ρ

)a.s.

for all x0 6= 0. In particular, if λ > 2(K + ρ), then the SDE isalmost surely exponentially unstable.





Can we find the matrices Gi as described in the theorem abovein order to destabilize the given ODE?





Outline








Case 1. d ≥ 3

Choose the dimension of the Brownian motion m = d . Let σ bea constant. For each i = 1,2, · · · ,d − 1, define the d × d matrixGi = (g i

uv ) by g iuv = σ if u = i and v = i + 1 or otherwise

g iuv = 0. Moreover, define Gd = (gd

uv ) by gduv = σ if u = d and

v = 1 or otherwise gduv = 0. Then SDE becomes

dx(t) = f (x(t), t)dt + σ

x2(t)dB1(t)

...xd(t)dBd−1(t)x1(t)dBd(t)

.





Compute thatm∑

i=1

|Gix |2 =m∑

i=1

(σxi+1)2 = σ2|x |2

andm∑

i=1

|xT Gix |2 = σ2m∑

i=1

x2i x2

i+1,

where we use xd+1 = x1. Notingm∑

i=1

x2i x2

i+1 ≤12

m∑i=1

(x4i + x4

i+1) =m∑

i=1

x4i ,

we have

3m∑

i=1

x2i x2

i+1 ≤ 2m∑

i=1

x2i x2

i+1 +m∑

i=1

x4i ≤ |x |4.





Thereforem∑

i=1

|xT Gix |2 ≤σ2

3|x |4.

By the theorem, the solution of the SDE has the property that

lim inft→∞

1t

log |x(t)| ≥(σ2

2− K − σ2

3

)=σ2

6− K a.s.

for any x0 6= 0. If σ2 > 6K , then the SDE will be almost surelyexponentially unstable.





Example

Given a stable 3-dimensional ODE

y(t) = f (y(t), t),

where

f (y , t) =

−2y1 + sin(y2)−2y2 + sin(y3)−2y3 + sin(y1)

.It is easy to see

|f (y , t)| ≤ 3|y | ∀(y , t) ∈ R3 × R+.





Perturbing this ODE by a 3-dimensional Brownian motionresults in an SDE


x2(t)dB1(t)x3(t)dB2(t)x1(t)dB3(t)

.This SDE is almost surely exponentially unstable provided

σ >√

18.





0.0 0.2 0.4

−10

12

t

x1(t)y1(t)

0.0 0.2 0.4

−10

12

3

t

x2(t)y2(t)

0.0 0.2 0.4

−3−2

−10

12

3

t

x3(t)y3(t)

σ = 5





0 2 4 6 8 10

0e+0

01e

+13

2e+1

33e

+13

4e+1

3

t

x1(t)y1(t)

0 2 4 6 8 10

0e+0

01e

+13

2e+1

33e

+13

t

x2(t)y2(t)

0 2 4 6 8 10

−1e+

130e

+00

1e+1

32e

+13

3e+1

3

t

x3(t)y3(t)

σ = 4





0 1 2 3 4 5

0.00.5

1.01.5

t

x1(t)y1(t)

0 1 2 3 4 5

−0.5

0.00.5

1.0

t

x2(t)y2(t)

0 1 2 3 4 5

−0.5

0.00.5

1.0

t

x3(t)y3(t)

σ = 2





Case 2. d is an even number

Let d = 2k(k ≥ 1) and σ be a constant. Define

G1 =

0 σ−σ 0

0

. . .

00 σ−σ 0

but set Gi = 0 for 2 ≤ i ≤ m. So the SDE becomes


x2(t)−x1(t)

...x2k (t)−x2k−1(t)

dB1(t).





In this case we have

m∑i=1

|Gix |2 = σ2|x |2 andm∑

i=1

|xT Gix |2 = 0

Hence, the solution of SDE obeys

lim inft→∞

1t

log |x(t)| ≥ σ2

2− K a.s.

for any x0 6= 0. If σ2 > 2K , then the SDE will be almost surelyexponentially unstable.





Example

Given a stable 4-dimensional ODE

y(t) = f (y(t), t),

where

f (y , t) =

−2y1 + sin(y2)−2y2 + sin(y3)−2y3 + sin(y4)−2y4 + sin(y1)

.It is easy to see

|f (y , t)| ≤ 3|y | ∀(y , t) ∈ R4 × R+.





Perturbing this ODE by a scale Brownian motion results in anSDE


x2(t)−x1(t)x3(t)−x4(t)

dB1(t).

This SDE is almost surely exponentially unstable provided

σ >√

6.





0.0 0.5 1.0 1.5 2.0

−6−2

24

6

t

x1(t)y1(t)

0.0 0.5 1.0 1.5 2.0

−22

46

8

t

x2(t)y2(t)

0.0 0.5 1.0 1.5 2.0

−6−2

24

6

t

x3(t)y3(t)

0.0 0.5 1.0 1.5 2.0

−22

46

8

t

x4(t)y4(t)

σ = 2.5





0.0 0.5 1.0 1.5 2.0

−8−4

02

4

t

x1(t)y1(t)

0.0 0.5 1.0 1.5 2.0

−40

24

6

t

x2(t)y2(t)

0.0 0.5 1.0 1.5 2.0

−8−4

02

4

t

x3(t)y3(t)

0.0 0.5 1.0 1.5 2.0

−40

24

6

t

x4(t)y4(t)

σ =√

6





0 1 2 3 4 5

−0.2

0.20.6

1.0

t

x1(t)y1(t)

0 1 2 3 4 5

0.00.4

0.8

t

x2(t)y2(t)

0 1 2 3 4 5

−0.2

0.20.6

1.0

t

x3(t)y3(t)

0 1 2 3 4 5

0.00.4

0.8

t

x4(t)y4(t)

σ = 1





Case 3. d = 1

Consider a stable scale linear ODE

y(t) = −ay(t) (a > 0),

and its perturbed linear SDE

dx(t) = −ax(t) +m∑

i=1

bix(t)dBi(t).

It is known that

limt→∞

1t

log |x(t)| = −a− 12

m∑i=1

b2i < 0 a.s.

That is, the perturbed system remains stable.

Noise does not destabilize the given system in this case.Xuerong Mao FRSE Stability of SDE




TheoremAny d-dimensional nonlinear system y(t) = f (y(t), t) can bedestabilized by Brownian motions provided the dimension ofthe state d ≥ 2 and



The LaSalle-Type TheoremsMean-Square Exponential Stability of Numerical Methods

Almost Sure Exponential Stability of Numerical Methods

Stability of Stochastic Differential EquationsPart 6: New Developments

Xuerong Mao FRSE


Glasgow, G1 1XH

December 2010




Outline

1 The LaSalle-Type TheoremsOriginal theoremImproved resultsExamples

2 Mean-Square Exponential Stability of Numerical MethodsStability of numerical methodsThe Euler-Maruyama methodStability of the EM method

3 Almost Sure Exponential Stability of Numerical MethodsLinear scalar SDEsMulti-Dimensional SDEsThe Backward Euler-Maruyama (BEM) Method




Outline







Outline







Original theoremImproved resultsExamples

Outline








The Lyapunov method has been developed and applied bymany authors during the past century. One of the importantdevelopments in this direction is the LaSalle theorem forlocating limit sets of nonautonomous ODE established by

LaSalle, J.P., Stability theory of ordinary differentialequations, J. Differential Equations 4 (1968), 57–65.

The first LaSalle-type theorem for SDEs established by

Mao, X., Stochastic versions of the LaSalle theorem, J.Differential Equations 153 (1999), 175–195.

is stated below:





Theorem

Assume that there are functions V ∈ C2,1(Rn × R+; R+),γ ∈ L1(R+; R+) and w ∈ C(Rn; R+) such that

lim|x |→∞

inf0≤t<∞

V (x , t) =∞

andLV (x , t) ≤ γ(t)− w(x), (x , t) ∈ Rn × R+.

Moreover, for each initial value x0 ∈ Rn there is a p > 2 suchthat

sup0≤t<∞

E |x(t ; x0)|p <∞.

Then, for every x0 ∈ Rn, limt→∞ V (x(t ; x0), t) exists and is finitealmost surely, and moreover, limt→∞w(x(t ; x0)) = 0 a.s.





Outline








This LaSalle-type theorem for SDEs has been developedsignificantly for the past 10 years. In this course we willhighlight a couple of developments.

Although the boundedness of the pth moment of the solution

sup0≤t<∞

E |x(t ; x0)|p <∞

has its own right, it is somehow too restrictive. Can we removethis condition?

The answer is yes. To state the improved LaSalle-typetheorem, let us introduce a few more notations.





L1(R+;R+): the family of all continuous functions γ : R+ → R+

such that∫∞

0 γ(t)dt <∞.

d(x ,A) = infy∈A |x − y | for x ∈ Rn and set A ⊂ Rn.

If µ ∈ K, its inverse function is denoted by µ−1 with domain[0, µ(∞)).

If w ∈ C(Rd ; R+), then Ker(w) = x ∈ Rd : w(x) = 0.





Theorem

Assume that there are functions V ∈ C2,1(Rn × R+; R+),γ ∈ L1(R+; R+) and w ∈ C(Rn; R+) such that

lim|x |→∞

inf0≤t<∞

V (x , t) =∞ and LV (x , t) ≤ γ(t)− w(x)

for (x , t) ∈ Rn × R+. Then, for every initial value x0 ∈ Rd , thesolution x(t ; x0) = x(t) of the SDE has the following properties:∫∞

0 Ew(x(t))dt <∞.∫∞0 w(x(t))dt <∞ a.s.

lim supt→∞ V (x(t), t) <∞ a.s.Ker(w) 6= ∅ and limt→∞ d(x(t),Ker(w)) = 0 a.s.





The proof of the theorem is very technical so is omitted in thiscourse.

To see the powerfulness of this theorem, let us demonstratethat many classical stability results follow from this theorem. Infact, under the conditions of the theorem, we have:





If w(x) = 0 iff x = 0, then the SDE is almost surelyasymptotically stable in the sense that limt→∞ x(t) = 0 a.s.If V (x , t) ≥ eλt |x |p on (x , t) ∈ Rn × R+ for some λ > 0 andp > 0, then the SDE is almost surely exponentially stable.If V (x , t) ≥ (1 + t)λ|x |p on (x , t) ∈ Rn × R+ for some λ > 0and p > 0, then the SDE is almost surely polynomiallystable in the sense that

lim supt→∞

log(|x(t)|)log t

≤ −λp

a.s.





If there is a positive constant p and a convex functionµ ∈ K∞ such that

lim suph→0+

µ−1(h)

h<∞ (1.1)

andw(x) ≥ µ(|x |p) ∀x ∈ Rn, (1.2)

then ∫ ∞0

E|x(t)|pdt <∞; (1.3)

and ∫ ∞0|x(t)|pdt <∞ a.s. (1.4)





All items except the last one are obvious. To show the last item,we observe that there is a constant C > 0 such that∫ t

0Eµ(|x(s)|p)ds ≤ C, ∀t ≥ 0.

Since µ convex, we may apply the Jensen inequality to obtain

tµ(1

t

∫ t

0E |x(s)|pds

)≤ C, ∀t ≥ 0.

This implies∫ t

0E |x(s)|pds ≤ tµ−1(C/t) = C

µ−1(C/t)C/t

, ∀t ≥ 0.





Letting t →∞ yiels ∫ t

0E |x(s)|pds <∞.

By the Fubini theorem, we also have

E∫ t

0|x(s)|pds <∞.

Hence ∫ t

0|x(s)|pds <∞ a.s.





This improved LaSalle-type theorem can also be used tohandle the problem of partially asymptotic stability. Let1 ≤ n ≤ n and 1 ≤ i1 < i2 < · · · < in ≤ n be all integers. Letx = (xi1 , xi2 , · · · , xin ) be the partial coordinates of x , which can

be regarded as in Rn with the norm |x | =√

x2i1

+ · · ·+ x2in

.





Under the assumptions of the theorem, we have:

If w(x) = 0 iff x = 0, then limt→∞ x(t) = 0 a.s.If there is a positive constant p and a convex functionµ ∈ K∞ such that

lim suph→0+

µ−1(h)

h<∞

andw(x) ≥ µ(|x |p) ∀x ∈ Rn,

then∫ ∞0

E|x(t)|pdt <∞ and∫ ∞

0|x(t)|pdt <∞ a.s.





In many practical situations, we require the solutions of theSDEs to stay in some regime of the state space Rd . Forexample, the SDEs used in finance or population systemsrequire the solutions remain in the positive conex ∈ Rd : xi > 0, 1 ≤ i ≤ d.

To develop the LaSalle-type theorems to cope with thesecases, let us recall the definition of an invariant set with respectto the solutions of the SDE.





Definition

An open subset G of Rd is said to be invariant with respect tothe solutions of the SDE if

Px(t ; x0) ∈ G for all t ≥ 0 = 1 for every x0 ∈ G,

that is, the solutions starting in G will remain in G.

Under our standing hypotheses, we know that Rd − 0 is aninvariant set of the underlying SDE. As another example,consider the one-dimensional equation

dx(t) = − sin(x(t))dt + sin(x(t))dB(t)

where B(t) is a scalar Brownian motion. The open interval(0, π) is an invariant set of this equation.





Theorem

Let G be an invariant set and G be its closure. Assume thatthere are functions V ∈ C2,1(G × R+; R+), γ ∈ L1(R+; R+) andw ∈ C(G; R+) such that

LV (x , t) ≤ γ(t)− w(x), (x , t) ∈ G × R+.

If G is bounded; or otherwise if

limx∈G,|x |→∞

inf0≤t<∞

V (x , t) =∞,

then, for every initial value x0 ∈ G, the solution x(t ; x0) = x(t) ofthe SDE has the following properties:





Theorem ∫ ∞0

Ew(x(t))dt <∞,∫ ∞0

w(x(t))dt <∞ a.s.

lim supt→∞

V (x(t), t) <∞ a.s.

KerG(w) := x ∈ G : w(x) = 0 6= ∅,

limt→∞

d(x(t),KerG(w)) = 0 a.s.





Outline








In the following examples we will let B(t) be a scalar Brownianmotion.





Example 1Let α and β be bounded functions and consider a scalar SDE

dx(t) = α(t)x(t)dt + β(t)x(t)dB(t), t ≥ 0.

It is known that G = R − 0 is an invariant set. Assume thatthere is a δ ∈ (0,1) such that

ε := inf0≤t<∞

(1− δ2

β2(t)− α(t))> 0.

Let V (x , t) = |x |δ for x 6= 0 and t ≥ 0. Then

LV (x , t) = −δ(1− δ

2β2(t)− α(t)

)|x |δ ≤ −δε|x |δ.

We can therefore conclude that limt→∞ x(t) = 0 a.s.





Example 2Consider

dx(t) = − sin(x(t))dt + sin(x(t))dB(t), t ≥ 0.

It is known that G = (0, π) is an invariant set. Let V (x , t) = |x |2for x ∈ (0, π) and t ≥ 0. Then

LV (x , t) = − sin(x)[2x − sin(x)] ≤ 0.

Hence, for any x0 ∈ (0, π), almost every sample path of x(t ; x0)will tend to either 0 or π.





Example 3Let α be a function on R+ such that 0 < δ1 ≤ α(t) ≤ δ2 <∞.Consider a stochastic oscillator

y(t) +[α(t) +

√α(t)B(t)

]y(t) + y(t) = 0, t ≥ 0.

Introducing a new variable x = (x1, x2)T = (y , y)T , thisoscillator can be written as an Itô equation

dx(t) =

[x2(t)

−x1(t)− α(t)x2(t)

]dt +

[0

−√α(t)x2(t)

]dB(t).

Let 2 < p < 3 and define V (x , t) = |x |p for (x , t) ∈ R2 × R+.Then





LV (x , t) ≤ p|x |p−2x1x2 + p|x |p−2x2(−x1 − α(t)x2)

+p(p − 1)

2α(t)|x |p−2x2

2

= −p(3− p)

2α(t)|x |p−2x2

2

≤ −p(3− p)

2δ1|x |p−2x2

2 .

We can therefore conclude that we see that

limt→∞

x2(t ; x0)→ 0 a.s.

and limt→∞ x1(t ; x0) exists and is finite almost surely.




Stability of numerical methodsThe Euler-Maruyama methodStability of the EM method

Outline








Two key questions

Q1. Given a stable SDE, for what choices of stepsize does thenumerical method reproduce the stability property of thetest equation?

Q2. Given that the numerical solution to an SDE is stable for asufficiently small stepsize, can we conclude confidentlythat the SDE is stable?





Mitsui, Saito et al. (93, 94, 95, 96), Higham (00) discussedQ1 on the mean-square exponential stability for scalarlinear SDEs.Higham, Mao and Stuart (2003) discussed Q1 and Q2 onthe mean-square exponential stability for multi-dimensionalnonlinear SDEs under the global Lipschitz condition.Some results answer Q1 on the almost sure exponentialstability.No results answer Q2 on the almost sure exponentialstability yet.





Outline








SDE in Rd :

dx(t) = f (x(t))dt + g(x(t))dB(t), t ≥ 0, x(0) = x0.

The Euler-Maruyama (EM) discrete approximation:

X0 = x0,

Xk+1 = Xk + ∆f (Xk ) + g(Xk )∆Bk , ∀k ≥ 0,

where ∆Bk = B((k + 1)∆)− B(k∆).The EM continuous approximation:

X (t) = Xk + f (Xk )(t −∆) + g(Xk )(B(t)− B(k∆)),

for t ∈ [k∆, (k + 1)∆], k = 0,1,2, · · · .





Outline








The SDE is said to be exponentially stable in mean square ifthere is a pair of positive constants λ and M such that withinitial data x0 ∈ Rd

E |x(t)|2 ≤ M|x0|2e−λt , ∀t ≥ 0. (2.1)

We refer to as the rate constant and M as the growth constant.

For a given step size ∆ > 0, the Euler-Maruyama solution issaid to be exponentially stable in mean square on the SDE ifthere is a pair of positive constants γ and N such that with initialdata x0 ∈ Rd

E |Xk |2 ≤ N|x0|2e−γk∆, ∀k ≥ 0. (2.2)





Hypothesis (H1):

|f (x)− f (y)|2 ≤ K1|x − y |2, ∀x , y ∈ Rd ,

|g(x)− g(y)|2 ≤ K2|x − y |2, ∀x , y ∈ Rd ,

f (0) = 0, g(0) = 0.





TheoremUnder (H1) the SDE is exponentially stable in mean square ifand only if the Euler-Maruyama solution is exponentially stablein mean square for some step size ∆ with the rate constant γand the growth constant N satisfying

CeγT (∆ +√

∆) + 1 +√

∆ ≤ e14γT ,

where T = 1 + 4 log(N)/γ and C > 0 is a constant whichdepends only on T ,K1 and K2 (but not on ∆ and ξ) such that

sup0≤t≤2T

E|x(t)− X (t)|2 ≤

(sup

0≤t≤2TE|X (t)|2

)C∆.





In practice, the constant C can be computed by

C = 4T (TK1 + K2)(1 + K1)e4(T +K1+K2),

though this may not be optimal.We emphasize that this Theorem is an “if and only if ”result, and hence has important practical implications. Ifcareful numerical simulations indicate exponential stabilityin mean square, then we may confidently infer that theunderlying SDE has the same property.




Linear scalar SDEsMulti-Dimensional SDEsThe Backward Euler-Maruyama (BEM) Method

Outline








Considerdx(t) = αx(t)dt + σx(t)dB(t) (3.1)

on t ≥ 0 with initial value x(0) = x0 ∈ R, where α and σ are realnumbers. If x0 6= 0, then

limt→∞

1t

log(|x(t)|) = α− 12σ

2 a.s.

That is, the linear SDE is almost surely exponential stable if andonly if α− 1

2σ2 < 0.





The Euler-Maruyama (EM) method

Given a step size ∆ > 0, the EM method is to compute thediscrete approximations Xk ≈ x(k∆) by setting X0 = x0 andforming

Xk+1 = Xk (1 + α∆ + σ∆Bk ), (3.2)

for k = 0,1, · · · , where ∆Bk = B((k + 1)∆)− B(k∆).

Question: If α− 12σ

2 < 0, is the EM method almost surelyexponentially stable for sufficiently small ∆?





Theorem

If α− 12σ

2 < 0, then for any ε ∈ (0,1) there is a ∆1 ∈ (0,1)such that for any ∆ < ∆1, the EM approximate solution has theproperty that

limk→∞

1k∆

log(|Xk |) ≤ (1− ε)(α− 12σ

2) < 0 a.s. (3.3)





ExampleConsider

dx(t) = x(t)dt + 2x(t)dB(t), t ≥ 0.

The following 4 simulations are carried out using ∆ = 0.001with the initial value x(0) = 10. These simulations show clearlythat the EM method reproduces the almost sure exponentialstability of the linear SDE.





0 2 4 6 8 10

010

3050

70

t

X(t)

or x

(t)

true solnEM soln

0 2 4 6 8 10

02

46

810

t

X(t)

or x

(t)

true solnEM soln

0 2 4 6 8 10

05

1015

2025

t

X(t)

or x

(t)

true solnEM soln

0 2 4 6 8 10

02

46

810

t

X(t)

or x

(t)

true solnEM soln





Outline








Consider the nonlinear SDEdx(t) = f (x(t))dt + g(x(t))dB(t), t ≥ 0,x(0) = x0 ∈ Rd ,

(3.4)

where f ,g : Rd → Rd . As before, we assume thatf ,g : Rd → Rd are smooth enough so that the SDE (3.4) has aunique global solution x(t) on [0,∞). The following stabilityresult can be proved in a similar way as we did in Part 3.





Theorem

If

−λ := supx∈Rd ,x 6=0

(〈x , f (x)〉+ 1

2 |g(x)|2

|x |2− 〈x ,g(x)〉2

|x |4

)< 0, (3.5)

then the solution of the SDE (3.4) obeys

lim supt→∞

1t

log(|x(t)|) ≤ −λ a.s. (3.6)





Question: Under condition (3.5), can the EM reproduce thealmost sure exponential stability?

Unlike the linear case, the answer is in general no. However,under the following additional condition, the answer is yes.

AssumptionAssume that there is a K > 0 such that

|f (x)| ∨ |g(x)| ≤ K |x |, ∀x ∈ Rd . (3.7)





The Euler-Maruyama method

Recall that given a step size ∆ > 0, the Euler-Maruyamamethod is to compute the discrete approximations Xk ≈ x(k∆)by setting X0 = x0 and forming

Xk+1 = Xk + f (Xk )∆ + g(Xk )∆Bk , (3.8)

for k = 0,1, · · · , where ∆Bk = B((k + 1)∆)− B(k∆) as before.





Theorem

Let (3.7) and (3.5) hold. Then for any ε ∈ (0, λ), there is a∆∗ ∈ (0,1) such that for any ∆ < ∆∗, the EM approximatesolution has the property that

lim supk→∞

1k∆

log(|Xk |) ≤ −(λ− ε) a.s. (3.9)





ExampleConsider the two-dimensional SDE

dx(t) = f (x(t))dt + Gx(t)dB(t) on t ≥ 0

with initial value x(0) = x0 ∈ R2 and x0 6= 0, where B(t) is aone-dimensional Brownian motion,

f (x) =

(x2 cos x12x1 sin x2

), G =

(3 −0.3−0.3 3

)





It is easy to verify that

−λ = supx∈Rd ,x 6=0

(〈x , f (x)〉+ 1

2 |g(x)|2

|x |2− 〈x ,g(x)〉2

|x |4

)≤ −0.345.

Hencelim sup

t→∞

1t

log |x(t ; x0)| ≤ −0.345 a.s.

which is the same as we obtained in Part 3.

The following simulation uses ∆ = 0.001 and x(0) = (1,1)T . Itshows clearly that the EM method reproduces the a.s.exponential stability.





0 1 2 3 4 5

0.0

0.5

1.0

1.5

2.0

2.5

3.0

t

X1(t)

0 1 2 3 4 5

01

23

45

t

X2(t)





Stable SDE without the linear growth condition:

dx(t) = (x(t)− x3(t))dt + 2x(t)dB(t). (3.10)

Although the coefficients f (x) = x − x3 and g(x) = 2x dosatisfy (3.5) as

supx∈R,x 6=0

(〈x , f (x)〉+ 1

2 |g(x)|2

|x |2− 〈x ,g(x)〉2

|x |4

)

= supx∈R,x 6=0

(x2 − x4 + 2x2

x2 − 4x4

x4

)≤ −1.

An application of the theorem shows that its solution obeys

lim supt→∞

1t

log(|x(t)|) ≤ −1 a.s. (3.11)





We observe that the shift coefficient f (x) does not obey thelinear growth condition (3.7). We may therefore wonder:

Question: If the EM method is applied to the SDE (3.10), will itrecover the property of almost surely exponential stability?





Counter example

Applying the EM to the SDE

dx(t) = (x(t)− x3(t))dt + 2x(t)dB(t).

givesXk+1 = Xk (1 + ∆− X 2

k ∆ + 2∆Bk ).

LemmaGiven any initial value X0 6= 0 and any ∆ > 0,

P(

limk→∞

|Xk | =∞)> 0.





The following 2 simulations show that the EM method does notreproduce the almost sure exponential stability of this nonlinearSDE. Both use ∆ = 0.001 and the first one uses the initial valuex(0) = 30 while the 2nd one uses x(0) = 50. In particular, the2nd one shows that the EM method could blow up very quickly.





0 1 2 3 4 5

05

1015

2025

30

t

X(t)





0 1 2 3 4 5

−1.2

e+26

5−8

.0e+

264

−4.0

e+26

40.

0 e+

00

t

X(t)





Outline








Definition of the BEM

Given a step size ∆ > 0, set Z0 = x0 and compute

Zk+1 = Zk + f (Zk+1)∆ + g(Zk )∆Bk (3.12)

for k = 0,1,2, · · · .

The BE method is implicit as for every step given Zk , equation(3.12) needs to be solved for Zk+1. For this purpose, someconditions need to be imposed on f .





The one-side Lipschitz condition:There is a constant µ ∈ R such that

〈x − y , f (x)− f (y)〉 ≤ µ|x − y |2, ∀x , y ∈ Rd . (3.13)

Under this condition, it is known that equation (3.12) can besolved uniquely for Zk+1 given Zk as long as the step size∆ < 1/(1 + 2|µ|).

We also need a condition on g: there is a K > 0 such that

|g(x)| ≤ K |x |, ∀x ∈ Rd . (3.14)





Theorem

Let (3.14) and (3.13) hold and f (0) = 0. Assume also that

β := supx∈Rd ,x 6=0

( |g(x)|2

|x |2− 2〈x ,g(x)〉2

|x |4)<∞ (3.15)

If µ+ 12β < 0, then for any ε ∈ (0, |µ+ 1

2β|), there is a∆∗ ∈ (0,1/(1 + 2|µ|)) such that for any ∆ < ∆∗,

lim supk→∞

1k∆

log(|Zk |) ≤ µ+ 12β + ε < 0 a.s. (3.16)





Let us return to the SDE (3.10). By setting f (x) = x − x3 andg(x) = 2x for x ∈ R, we have

〈x − y , f (x)− f (y)〉 ≤ |x − y |2

which gives µ = 1 while

β := supx∈R,x 6=0

( |g(x)|2

|x |2− 2〈x ,g(x)〉2

|x |4)

= −4,

whence µ+ 12β = −1. Thus, for any ε ∈ (0,1), there is a

∆∗ > 0 sufficiently small so that if ∆ < ∆∗, then

lim supk→∞

1k∆

log(|Zk |) ≤ −1 + ε a.s.

which recovers property (3.11) very well indeed.





The following 2 simulations show that the BEM method DOreproduce the almost sure exponential stability of the nonlinearSDE (3.10). Both use ∆ = 0.001 and the first one uses theinitial value x(0) = 30 while the 2nd one uses x(0) = 50.





0 1 2 3 4 5

05

1015

2025

30

t

Z(t)





0 1 2 3 4 5

010

2030

4050

t

Z(t)


Documents

Stability of Stochastic Differential Equations › ma › ws2010 › doc › mao_notes.pdf · Stability of Stochastic Differential Equations Part 1: Introduction Xuerong Mao FRSE