**Format****Pages****Chapters**

# Minimum Principle of Pontryagin – Download Project Topics

**Minimum Principle of Pontryagin**

**TABLE OF CONTENTS**

1 Preliminaries 8

1.1 Linear maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

1.1.1 A basic result concerning linear maps . . . . . . . . . . . 8

1.1.2 Bounded Linear Maps . . . . . . . . . . . . . . . . . . . . 9

1.2 Banach spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.3 Hilbert Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

1.4 Differential Calculus in Banach spaces . . . . . . . . . . . . . . . 13

1.5 Convex sets and convex functions . . . . . . . . . . . . . . . . . . 14

1.5.1 Notation and Further definitions . . . . . . . . . . . . . . 17

1.6 Lower Semi-Continuous Functions . . . . . . . . . . . . . . . . . 18

1.7 Existence Result . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

1.8 Optimality condition: . . . . . . . . . . . . . . . . . . . . . . . . 21

1.9 Optimization with equality constrains . . . . . . . . . . . . . . . 23

2 Pontryagin minimum method principle 26

2.0.1 Towards the principle of Pontryagin . . . . . . . . . . . . 28

3 Minimum Principle of Pontryagin: Linear Quadratic Case 30

3.1 Existence and uniqueness of the optimal control . . . . . . . . . . 31

3.2 characterization of the optimal control . . . . . . . . . . . . . . . 35

3.2.1 Riccati equation . . . . . . . . . . . . . . . . . . . . . . . 39

3.3 Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

**CHAPTER ONE**

**Preliminaries**

**1.1 Linear maps**

In this part we dene linear map and present some basic results concerning them.

Definition: Let X and linear spaces over a scalar eld K. A mapping T : X ! Y is said to be a linear map if: T(x + y) = T(x) + T(y) (1.1) for arbitrary x; y 2 X and arbitrary scalars ; 2 K. Some authors use the term linear operator or linear transformation instead of linear map.Condition (1.1) is equivalent to the following two conditions:

(i)T(x + y) = T(x) + T(y)8x; y 2 X (ii)T(x) = T(x)8x 2 X and for each scalar, .

**1.1.1 A basic result concerning linear maps**

We remark rst that since linear functionals are special forms of linear maps, any result proved for linear map holds for linear functionals.

Proposition 1.1.1 Let X and Y be two linear spaces over a scalar eld, K, and let T : X ! Y be a linear map.Then

1. T(0) = 0

2. The rang of T, R(T) = fy 2 Y : T(x) = y for some x 2 Xg is a linear subspace of Y

3. T is one to one if and only if T(x) = 0 implies that x = 0

4. If T is one to one , then T1 exist on R(T) and T1 : R(T) ! X is also a linear map.

Proof. (1) Since T is linear, we have, T(x) = T(x) for each x 2 X and each scalar . Take = 0 and (1) follows immediately.

(2) We need to show that for y1; y2 2 R(T) and ; scalars, y1 + y2 2 R(T).

Now, y1; y2 2 R(T) implies that there exists x1; x2 2 X such that T(x1) = y1; T(x2) = y2. Moreover, x1 + x2 2 X(since X is a linear space). Further- more, by the linearity of T,we have

T(x1 + x2) = T(x1) + T(x2)

Hence y1 + y2 2 R(T), and so R(T) is a linear subspace of Y.

(3) Assume that T is one to one. Clearly T(x) = 0 ) T(x) = T(0) since T is linear (and so T(0) = 0). But T is one-to-one. So, x = 0. (()Assume that whenever T(u) = 0 then u must be 0. We want to prove that T is one-to-one. So, let T(x) = T(y). Then T(x) T(y) = 0 and by the linearity of T; T(x y) = 0.By hypothesis, x y = 0 which implies x = y. Hence T is one to-one.

(4) Assume that T is one-to-one, since the restriction of T on R(T) is always onto, then T is bijective from X into R(T).So T1 exists on R(T). For the linearity, let y1; y2 2 R(T) and a scalar. Then there exists x1; x2 2 X such that y1 = T(x1); y2 = T(x2). so T1(y1 + y2) = T1(T(x1) + T(x2) which is equivalent to: T1(y1 + y2) = T1(T(x1 + x2)) by using the linearity of T; which is also equivalent to T1(y1 + y2) = x1 + x2 = T1(y1) + T1(y2)

Therefore T1 is linear

**1.1.2 Bounded Linear Maps**

Definition: Let X and Y be normed linear spaces over a scalar eld K, and let T : X ! Y be a linear map. Then T is said to be bounded if there exists some constant K 0 such that for each x 2 X, kT(x)k Kkxk; the constant K is called a bound for T and in this case, T is called a bounded linear map. We denote by B(X; Y ), the family of all bounded linear maps from X into Y.

We now turn our attention to linear maps that are continuous. The notion of continuity can be state, for linear maps in several useful equivalent forms. We state these equivalent forms in the following theorem.

Theorem 1.1.1 Let X and Y be normed linear space and let T : X ! Y be a linear map.Then the following are equivalent:

1. T is continuous;

2. T is continuous at the origin(in the sense that if fxng is a sequence of X such that xn ! 0 as n ! +1, then T(xn) ! 0 as n ! +1)

3. T is Lipschitz, ie , there exists a constant K 0 such that,for each x 2 X, kT(x)k Kkxk;

4. If D = fx 2 X : kxk 1g is the closed unit disc in X, then T(D) is bounded(in the sense that there exists a constant M 0 such that kT(x)k M for all x 2 D)

**1.2 Banach spaces**

Definition Let X be a real linear space, and k:kX a norm on X , and dX the corresponding metric dened by dX(x; y) = kxykX8x; y 2 X. The norm linear space (X; k:kX) is a real Banach space if the metric space (X; dX) is complete, i.e if any Cauchy sequence of elements of space (X; k:kX) converges in (X; k:kX:) That is every sequence satisfying the the following Cauchy criterion: 8 > 0; 9n0 2 N : p; q > n0 ) dX(xp; xq) linear space X; the space X of all bounded linear functionals on X is a Banach space and as a linear space, it has its own corresponding . Let X be a Banach space, there exists a natural mapping J : X ! X of X into X dened, for each x 2 X by J(x) = x where x : X ! R is given by x(f) = hf; xi, for each f 2 X. Thus