Chapter 3 Fundamental Properties

Class Handout: Chapter 3 Fundamental Properties

2006 Fall

* Lipschitz condition

I. Existence and Uniqueness of Solutions to Initial Value Problems Theorem 3.1 (Local Existence and Uniqueness)

˙

x=f(t, x) f(t, x): piecewise continuous int and satisfies

kf(t, x)−f(t, y)k ≤Lkx−yk, ∀x, y∈B¯r(x0), ∀t∈[t0, t1] with some L >0,r >0 and somex0∈Rⁿ.

Then, ∃δ >0 s.t. the equation

˙

x=f(t, x), x(t0) =x0

has a uniquesol. over [t0, t0+δ].

Note.

A point x^∗ satisfyingx=T(x) is called “fixed point of the mappingT.”

One way to find x^∗: Letx2=T(x1), · · ·,xk+1=T(xk). IfT(·) is contracting mapping, then it is known that xk →x^∗. (Ex. T(x) = 0.5x,x^∗= 0.) In fact, when T is contracting, x^∗ exists and is unique.

This idea can also be applied to a vector xthat is not only in the Euclidean space, but also in the Banach space (e.g., a set of functionsx(t) having a certain properties). Roughly stated, a solution of ˙x=f(t, x),x(t0) =x0 isx(t) =x0+R_t

t0f(s, x(s))ds. Let P(x(·)) orP(x) :=x0+

Z _t

t0

f(s, x(s))ds.

And, show that P(·) is a contraction. Then, we are done.

However, in order to do that, we need to study properties of the set of functions.

(Linear) Vector Space X over the field R

Components: setX, fieldR, + operation, scalar multiplication, 0 element, 1 element Properties: ... (see other book) ... x+y∈ X,cx∈ X, ...

Normed linear space: Linear vector space with a normk · k See the book.

Convergence: A sequence{xk} ∈ X converges to x∈ X ifkxk−xk →0 ask→ ∞.

Closed Set: A setS⊂ X is closed if and only if every convergent sequence with elements in S has its limit inS.

Cauchy Sequence: A sequence such that

kxk−xmk →0 as k, m→ ∞.

Every convergent seq. is Cauchy, but not vice versa.

Banach Space: Complete Normed Linear Space. (‘Complete’ means that every Cauchy seq. inX converges to a vector inX.)

Example B.1

• The set X: C[a, b] = the set of all continuous functionsf : [a, b]→Rⁿ, which is a vector space on R. (x+y is defined by (x+y)(t) =x(t) +y(t),cxis by (cx)(t) =cx(t), zero/one function isf(t) = 0/1.)

• Norm: kxkC := max_t∈[a,b]kx(t)k.

• Completeness: for a given Cauchy seq.{xk}, we need to show that there existsxinC[a, b]

and kxk−xkC→0 ask→ ∞.

If we fix the timet in [a, b], each vectorxk(t),xm(t) satisfies

kxk(t)−xm(t)k ≤ kxk−xmkC→0 as k, m→ ∞

So{xk(t)} is a Cauchy seq. inRⁿ. The setRⁿ is known to be complete. So, at eacht,∃x(t) s.t. xk(t)→x(t).

Is x(t) continuous and kxk−xkC → 0? To see this, we first show that the convergence xk(t)→ x(t) is uniform in t ∈ [a, b]. Indeed, given ² > 0, ∃N s.t. kxk−xmkC < ²/2 for k, m > N. Then, fork > N,

kxk(t)−x(t)k ≤ kxk(t)−xm(t)k+kxm(t)−x(t)k

≤ kxk−xmkC+kxm(t)−x(t)k

With sufficiently large m (which may depend on t), each term can be made less than ²/2.

Thus, kxk(t)−x(t)k < ² for k > N regardless of t (Uniform Convergence). Therefore, kxk−xkC→0 ask→ ∞.

Now, consider

kx(t+δ)−x(t)k ≤ kx(t+δ)−xk(t+δ)k+kxk(t+δ)−xk(t)k+kxk(t)−x(t)k Then, for a given ² > 0, choose N s.t. kxk(t)−x(t)k < ²/3 for k > N (by uniform convergence). Fix any suchk and chooseδs.t. kxk(t+δ)−xk(t)k< ²/3. Then, we have

kx(t+δ)−x(t)k ≤²

with theδ(Continuity).

Theorem B.1 (Contraction Mapping Theorem) S: a closed subset of a Banach spaceX

T: a mapping that mapsS into S, and

kT(x)−T(y)k ≤ρkx−yk, ∀x, y∈S, 0≤ρ <1 Then,∃a unique vectorx^∗∈S satisfyingx^∗=T(x^∗).

Proof. Letx1∈S andxk+1=T(xk). Prove that{xk}is Cauchy sequence.

Because X is a Banach space, there exists x^∗ ∈ X such thatxk →x^∗ as k → ∞. Also, sinceS is closed,x^∗∈S. Prove thatx^∗=T(x^∗).

Finally, prove that x^∗ is the unique fixed point ofT inS.

Proof of Theorem 3.1. Study of the existence and uniqueness of the solution x(t) of

˙

x=f(t, x) withx(t0) =x0 is equivalent to the study of the existence and uniqueness of x(t) =x0+

Z _t

t0

f(s, x(s))ds. (1)

Let the right-hand side be amappingof the continuous functionx: [t0, t1]→Rⁿ, and denote it by (P x)(t), i.e.,

x(t) = (P x)(t).

(Note that (P x)(t) is continuous.) Since a solution of the above is a fixed point of the mapping P, we may use the contraction mapping theory. This requires defining a Banach spaceX and a closed setS⊂ X s.t. P mapsS into S and is a contraction overS.

X :=C[t0, t0+δ], with kxkC= max

t∈[t0,t0+δ]kx(t)k S:={x∈ X :kx−x0kC ≤r}

where rcomes from the assumption, andδ >0 will be chosen so that [t0, t0+δ]⊂[t0, t1].

AlthoughP mapsX intoX by definition, prove thatP mapsSintoSwithδ≤r/(Lr+h) where h= max_t∈[t0,t1]kf(t, x0)k.

Prove thatP is a constraction mapping overS withδ <1/L.

This proves that there exists a unique solutionx(t) inS. Is this also a unique solution in X (which is our original question)? For this, prove that any solution of (1) inX lies in S, which means that uniqueness of the solution inS implies uniqueness in X.

Theorem 3.2 (Global Existence and Uniqueness)

˙

x=f(t, x) f(t, x): piecewise continuous int and satisfy

kf(t, x)−f(t, y)k ≤Lkx−yk, ∀x, y∈Rⁿ, ∀t∈[t0, t1] with some L >0.

Then, the equation

˙

x=f(t, x), x(t0) =x0

has a uniquesol. over [t0, t1].

• This theorem implies that the solution exists on [t0,∞) if the Lipschitz condition holds for t∈[t0,∞). How?

Proof. In the proof of Theorem 3.1, theδis chosen as δ≤min

½

t1−t0, r Lr+h,ρ

L

¾

, ρ <1 where h= maxt∈[t0,t1]kf(t, x0)k.

We note that, for the given initial condition,his determined, and, sincef satisfies global Lipschitz condition, the constantrcan be taken arbitrarily large so that _Lr+h^r ≥_L^ρ.

Then, δis determined by

δ≤min n

t1−t0, ρ L

o .

If δ = t1−t0, the proof is done. If δ = _L^ρ < t1 −t0, then, choose a new δ ≤ _L^ρ so that [t0, t1] is divided byδ. Finally, by applying the local existence and uniqueness theorem repeatedly, we get the result. (Each time of application, the function hand r may need to be re-calculated becausehdepends on the initial condition.)

Theorem 3.3 f(t, x): piecewise continuous int, locally Lipschitz in xfor allt∈[t0,∞) and allxin a domainD⊂Rⁿ.

W: a compact subset ofDandx0∈W.

Every sol. of ˙x=f(t, x) withx(t0) =x0 lies inW. Then, ∃a unique sol. ont∈[t0,∞).

II. Continuous Dependence on Initial Conditions and Parameters

Gronwall-Bellman’s Lemma If a continuous y: [a, b]→Rsatisfies

y(t)≤λ(t) + Z _t

a

µ(s)y(s)ds, a≤t≤b, λ: [a, b]→Ris continuous,

µ: [a, b]→Ris continuous and nonnegative, then,

y(t)≤λ(t) + Z _t

a

λ(s)µ(s) exp

·Z _t

s

µ(τ)dτ

¸ ds.

In particular, ifλ(t) =λ, then,

y(t)≤λexp

·Z _t

a

µ(τ)dτ

¸ . If, in addition,µ(t) =µ≥0, then

y(t)≤λexp[µ(t−a)].

Proof. Letz(t) :=R_t

aµ(s)y(s)dsandv(t) :=z(t) +λ(t)−y(t)≥0. Then,

˙

z(t) =µ(t)y(t) =µ(t)z(t) +µ(t)λ(t)−µ(t)v(t).

Let the state transition matrix

φ(t, s) = exp

·Z _t

s

µ(τ)dτ

¸ . Then, sincez(a) = 0,

z(t) = Z _t

a

φ(t, s)[µ(s)λ(s)−µ(s)v(s)]ds≤ Z _t

a

φ(t, s)µ(s)λ(s)ds.

Since

y(t)≤λ(t) +z(t)≤λ(t) + Z _t

a

φ(t, s)µ(s)λ(s)ds, the proof is done.

Ifλ(t) =λ, we have

y(t)≤λ+z(t)≤λ+λ Z _t

a

µ(s) exp

·Z _t

s

µ(τ)dτ

¸ ds

=λ−λ Z _t

a

d ds

½ exp

·Z _t

s

µ(τ)dτ

¸¾ ds

=λ−λ

½ exp

·Z _t

s

µ(τ)dτ

¸¾ ¯¯

¯¯

¯

s=t

s=a

=λ−λ µ

1−exp

·Z _t

a

µ(τ)dτ

¸¶

and, in addition, if µ(t) =µ, we have

y(t)≤λ+z(t)≤λexp

·Z _t

a

µ(τ)dτ

¸

=λexp [µ(t−a)].

Theorem 3.4

f(t, x): piecewise continuous int, Lipschitz (with Lipschitz constantL) inx, on [t0, t1]×W where W ⊂Rⁿ is an open connected set.

Lety(t) andz(t) be the solutions of

˙

y=f(t, y), y(t0) =y0

and

˙

z=f(t, z) +g(t, z), z(t0) =z0

such that y(t), z(t)∈W for allt∈[t0, t1].

If

kg(t, x)k ≤µ, ∀(t, x)∈[t0, t1]×W for some µ >0, then,

ky(t)−z(t)k ≤ ky0−z0kexp[L(t−t0)] + µ

L{exp[L(t−t0)]−1}, ∀t∈[t0, t1].

Proof.

y(t) =y0+ Z _t

t0

f(s, y(s))ds z(t) =z0+

Z _t

t0

[f(s, z(s)) +g(s, z(s))]ds ky(t)−z(t)k ≤ ky0−z0k+

Z _t

t0

kf(s, y(s))−f(s, z(s))kds+ Z _t

t0

kg(s, z(s))kds

≤γ+µ(t−t0) + Z _t

t0

Lky(s)−z(s)kds where γ=ky0−z0k. By Gronwall-Bellman inequality,

ky(t)−z(t)k ≤γ+µ(t−t0) + Z _t

t0

L[γ+µ(s−t0)] exp[L(t−s)]ds

≤γ+µ(t−t0)−γ−µ(t−t0) +γexp[L(t−t0)] + Z _t

t0

µexp[L(t−s)]ds

=γexp[L(t−t0)] +µ

L{exp[L(t−t0)]−1}

Theorem 3.5

f(t, x, λ): continuous in (t, x, λ), locally Lipschitz in xuniformly int and λ, on [t0, t1]× D×B¯c(λ0) whereD⊂Rⁿ is an open connected set.

Lety(t, λ0) be a solution of ˙x=f(t, x, λ0) withy(t0, λ0) =y0∈D.

Supposey(t, λ0) is defined and ∈Dfort∈[t0, t1].

Then, given ² >0,∃δ >0 s.t. if

kz0−y0k< δ and kλ−λ0k< δ,

there is a unique solution z(t, λ) of ˙x=f(t, x, λ) with z(t, λ) =z0, defined on [t0, t1], and z(t, λ) satisfies

kz(t, λ)−y(t, λ0)k< ², ∀t∈[t0, t1].

Proof. y(t, λ0) is bounded on [t0, t1] because it is continuous intand [t0, t1] is compact.

Let a tube U be {(t, x) : t∈[t0, t1],kx−y(t, λ0)k ≤ ²} so that U ⊂[t0, t1]×D. This is always possible by reducing ²when necessary. (Why?)

InU,f(t, x, λ) is Lipschitz inxwithL.

By continuity off, for any α >0,∃β >0 (with β < c) s.t.

kf(t, x, λ)−f(t, x, λ0)k< α, ∀(t, x)∈U, ∀λ∈Bβ(λ0) Take α < ²andkz0−y0k< α.

By local existence and uniqueness theorem,∃a unique sol. z(t, λ) on [t0, t0+ ∆] with some

∆>0. This sol. starts insideU, and as long as it remains inU, it can be extended. In fact, with a small enough α, it remains in U for allt∈[t₀, t₁]. To see this, let τ be the first time the solution leaves U (be careful..it will be better to sayτ is the time whenz(t, λ) is at the boundary ofU). On the time interval [t0, τ], the conditions of Theorem 3.4 are satisfied with µ=αbecause

˙

z=f(t, z, λ0) + [f(t, z, λ)−f(t, z, λ0)], z(t0, λ) =z0. Thus, we have

kz(t, λ)−y(t, λ0)k ≤αexp[L(t−t0)] +α

L{exp[L(t−t0)]−1}

< α µ

1 + 1 L

¶

exp[L(t−t0)]

Choose

α≤²Lexp[−L(t1−t0)]

1 +L

then, the sol. z(t, λ) cannot leave U during [t0, τ]. This idea confirms that z(t, λ) cannot leaveU during [t0, t1]. Takingδ= min{α, β} completes the proof.

III. Differentiability of Solutions and Sensitivity Equations

˙

x=f(t, x, λ0), x(t0) =x0

where f(t, x, λ) isC⁰ in (t, x, λ), isC¹in (x, λ), for all (t, x, λ)∈[t0, t1]×Rⁿ×R^p, which has a unique sol. x(t, λ0) over [t0, t1].

Question: Inspect the effect of the parameterλon the solutionx(t, λ). We suppose that thesensitivity function

S(t) :=xλ(t, λ) =∂x

∂λ(t, λ)

represents the effect ofλon the solutionx(t, λ). Can we calculateS(t) at a givenλ0? For xλ(t, λ0) to exist,x(t, λ) (as well asx(t, λ0)) need to exist forλ close toλ0. This is guaranteed by Theorem 3.5, because it guarantees that, for allλsufficiently close toλ0, the equation

˙

x=f(t, x, λ), x(t0) =x0

has a unique sol. x(t, λ) over [t0, t1] (that is close tox(t, λ0)).

Then, how can we obtain xλ(t, λ0)? Since x(t, λ) =x0+

Z _t

t0

f(s, x(s, λ), λ)ds, we have

xλ(t, λ) = Z _t

t0

·∂f

∂x(s, x(s, λ), λ)xλ(s, λ) +∂f

∂λ(s, x(s, λ), λ)

¸ ds.

Idea: differentiate the above to obtain a differential equation forS(t) =xλ(t, λ0).

S(t) =˙ A(t, λ0)S(t) +B(t, λ0), S(t0) = 0 which is called sensitivity equation, where

A(t, λ) = ∂f(t, x, λ)

∂x

¯¯

¯¯

x=x(t,λ)

, B(t, λ) =∂f(t, x, λ)

∂λ

¯¯

¯¯

x=x(t,λ)

• Sensitivity functions provide first-order estimates of the effect of parameter variations on solutions

• To getS(t), we need to

1. solve the nominal solutionx(t, λ) first,

2. and then, solve the sensitivity equation forS(t) or, solve them at the same time by the equation (3.7).

• (MATLAB) Differential equation for a matrixS(t):

• (Discussion) About the initial conditionx0. In Example 3.7, “... This pattern is consistent when we solve for other initial states.”

IV. Comparison Principle

• Gronwall-Bellman inequality

• Comparison principle

Two tools for computing bounds on the solution x(t) for ˙x=f(t, x) without computing the solution itself. Frequently, we will arrive at a scalardifferential inequalityof the type

˙

v(t)≤f(t, v(t)).

Then, by the comparison principle, you will conclude that v(t)≤u(t) where u(t) is the solution

˙

u(t) =f(t, u(t)).

Lemma 3.4 (Comparison Lemma) Consider ascalardifferential equation

˙

u=f(t, u), u(t0) =u0

where f(t, u) is continuous int, locally Lipschitz inu, on (t, u)∈[0,∞)×J.

Let [t0, T) (T could be infinity) be the maximal interval of existence of u(t), and suppose u(t)∈J for allt∈[t0, T).

Letv(t) be continuous and satisfy

D⁺v(t)≤f(t, v(t)), v(t0)≤u0

with v(t)∈J for allt∈[t0, T).

Then, v(t)≤u(t) for all t∈[t0, T).

Dini’s Derivative

D⁺v(t) = lim sup

h→0⁺

v(t+h)−v(t)

h , Dini’s upper right derivative, D+v(t) = lim inf

h→0⁺

v(t+h)−v(t)

h ,

D⁻v(t) = lim sup

h→0⁻

v(t+h)−v(t)

h ,

D−v(t) = lim inf

h→0⁻

v(t+h)−v(t) h

• “Non-smooth analysis”

• y= lim sup_n→∞xn means that

1. for every² >0,∃an integerN s.t. n > N impliesxn< y+² (ultimately all terms of the seq. are less than y+²),

2. given² >0 andm >0,∃ an integern > ms.t. xn> y−² (infinitely many terms are greater than y−²).

• If limn→∞xn exists, then limn→∞xn= lim sup_n→∞xn.

• Therefore, if ˙v(t) exists, then ˙v(t) =D⁺v(t).

Proof. Consider

˙

z=f(t, z) +λ, z(t0) =u0, λ >0

On any compact interval [t0, t1] ⊂ [t0, T), from Theorem 3.5 (continuous dependence on parameter), for any ² >0, ∃δ >0 s.t., if λ < δ, there is a unique solutionz(t, λ) on [t0, t1] and

|z(t, λ)−u(t)|< ², ∀t∈[t0, t1]. (2) Claim 1: v(t)≤z(t, λ) on [t0, t1].

If not,∃a, b∈(t0, t1] s.t. v(a) =z(a, λ) andv(t)> z(t, λ) fora < t≤b. So, v(t)−v(a)> z(t, λ)−z(a, λ), ∀t∈(a, b]

and thus,

D⁺v(t)≥z(a, λ) =˙ f(a, z(a, λ)) +λ > f(a, v(a)) which is a contradiction.

Claim 2: v(t)≤u(t) on [t0, t1].

If not,∃a∈(t0, t1] s.t. v(a)> u(a). Let ²= [v(a)−u(a)]/2>0. Then, by (2), v(a)−z(a, λ) =v(a)−u(a) +u(a)−z(a, λ)≥²

which contradicts Claim 1.

Since the above argument holds for any compact interval in [t0, T), the proof is completed.

Example 3.8

˙

x=−(1 +x²)x, x(0) =a We know it has a unique sol. on [0, t1) for somet1>0. (Why?)

Letv(t) =x²(t). Then,

˙

v(t) =−2x²(t)−2x⁴(t)≤ −2v(t), v(0) =a² Hence,

|x(t)|=p

v(t)≤e^−t|a|, ∀t≥0 Example 3.9

˙

x=−(1 +x²)x+e^t, x(0) =a We know it has a unique sol. on [0, t1) for somet1>0. (Why?)

Letv(t) =x²(t). Then,

˙

v(t) =−2x²(t)−2x⁴(t) + 2x(t)e^t≤ −2v(t) + 2p v(t)e^t (still difficult to solve).

Letv(t) =|x(t)|. Then, forx(t)6= 0,

˙

v(t) = d dt

px²(t) = x(t) ˙x(t)

|x(t)| =−|x(t)|(1 +x²(t)) + x(t)

|x(t)|e^t

≤ −v(t) +e^t forx(t) = 0, we have

|v(t+h)−v(t)|

h =

¯¯

¯¯x(t+h) h

¯¯

¯¯= 1 h

¯¯

¯¯

¯ Z _t+h

t

f(τ, x(τ))dτ

¯¯

¯¯

¯

=

¯¯

¯¯

¯f(t,0) + 1 h

Z _t+h

t

[f(τ, x(τ))−f(t, x(t))]dτ

¯¯

¯¯

¯

≤ |f(t,0)|+ 1 h

Z _t+h

t

|f(τ, x(τ))−f(t, x(t))|dτ Thus,

D⁺v(t)≤ |f(t,0)|=e^t

because the second term is zero by L’Hospital’s rule and the continuity off(t, x(t)) int.

Hence, we conclude that

D⁺v(t)≤ −v(t) +e^t, v(0) =|a|

Then, we have

|x(t)| ≤e^−t|a|+1

2|e^t−e^−t|, ∀t≥0 (Whyt1=∞?)