Evolution Equations and Applications
Chapter One
Preamble of the Study
In this section, we recall some de nitions and results from linear functional analysis
De nition 1.1.1 Let X be a linear space over a eld K, where K holds either for R or C. A mapping k.k: X −→ R is called a norm provided that the following conditions hold:
kxk≥ 0 for all x ∈ X, and kxk= 0 ⇔ x = 0
kλxk= |λ|kxk, for all λ ∈ K, x ∈ X
kx + yk≤ kxk+kyk, for arbitrary x, y ∈ X.
If X is a linear space and k.k is a norm on X, then the pair (X, k.k) is called a normed linear space over K.
Should no ambiguity arise about the norm, we simply abbreviate this pair by saying that X is a normed linear space over K.
Example . Let X = C([0, 1]) be the space of all real-valued continuous functions on [0, 1]. Each of the following expressions de nes on the vector space C([0, 1]) a norm which is in common use.
CHAPTER TWO
ABSTRACT LINEAR EVOLUTION EQUATIONS
Most often signicant external forces a sect the evolution of a process. Let X be a real Banach space. In this chapter, we discuss the non-homogeneous Cauchy problem
U0(t) = AU(t) + f(t), t > 0
U(0) = U0
where A : D(A) ⊂ X −→ X is a given linear operator and f : [0, ∞) −→ X is a given function of the time variable only. This equation is called linear evolution equation. Basically we shall study the existence and uniqueness of solutions of the above problem, imposing di erent conditions on f.
Linear Evolution Equations in nite dimensional spaces: Well Posedness
In this section, we examine the linear Cauchy problem in the nite dimensional case. In this case we identify the linear operator A with an N × N matrix. We shall see that if the forcing term is continuous, then there is existence and uniqueness of the solution. So consider the following initial value problem (I.V.P) :
U0(t) = AU(t) + f(t), t > 0
U(0) = U0(2.1)
where A∈ MN (R), f : [0, ∞) −→ RN and U0 ∈ RN . We shall formulate the theory for (2.1). We have
U0(s) = AU(s) + f(s)
⇐⇒ U0(s) − AU(s) = f(s),
⇐⇒ e−sA(U0(s) − AU(s)) = e−sAf(s)
⇐⇒ dds (e−sAU(s)) = e−sAf(s),since,dds (e−sA) = −Ae−sA.
⇐⇒ R t0dds (e−sAU(s))ds =R t0e−sAf(s)ds ⇐⇒ e−tAU(t) − U(0) = R t0e−sAf(s)ds It implies that
U(t) = etAU0 +
Z t 0
e(t−s)Af(s)ds (2.2)
De nition 2.1.1 Let U : [0, T] −→ RN be a function
- a) U is a classical solution of (1) if
- i) U is continuous on [0, T]
- ii) U is di erentiable on (0, T]
iii) U satis es (2.1)
- b) U is a mild solution of (1) if
- i) U is continuous on [0, T]
- ii) U is given by (2), ∀t ∈ [0, T] .
Theorem 2.1.1 (Existence and Uniqueness) Let T > 0 and suppose that f ∈ C([0, T]; RN ). Then (2.1) has a unique classical solution on [0, T] given by (2.2).
Proof:
Existence
Let U be given by (2.2). From the continuity of f we have that the map t 7→R t0e(t−s)Af(s)ds and also the map t 7→ etAU0 is continuous. therefore U given by (2.2) is continuous as the sum of two continuous functions. Moreover
U0(t) = AetAU0 + f(t) + A
Z t 0
e(t−s)Af(s)ds
= A(etAU0 +
Z t 0
e(t−s)Af(s)ds) + f(t)
= AU(t) + f(t)
Also U(0) = e0AU0 = e0U0 = U0
Thus U is a classical solution of (2.2).
Uniqueness:
Suppose that U and V are both classical solutions of (2.1). Then de ne Z : [0, T] −→ RN by Z(t) = U(t) − V (t) .Then Z is continuous and di erentiable on [0,T] and (0,T] respectively as the sum of two continuous and di erentiable functions. Moreover
Z0(t) = U0(t) − V0(t)
= A(U(t) − V (t))
= AZ(t)
So Z(t) = etAZ0 but Z0 = Z(0) = 0, thus Z(t) = 0, ∀t ∈ [0, T] and therefore U = V proving uniqueness and completing the proof of the theorem.
Continuous dependence on the given data:
Consider the following perturbed system from (2.1).
V0(t) = AV (t) + g(t), t > 0
V (0) = V0(2.3)
where A is the matrix given in (2.1).
We are hopeful that the di erence between the solutions U and V of (2.1) and (2.3), respectively, in the sense of the supnorm on C([0, T]; RN ) , for any time interval [0,T] can be controlled by making the error terms su ciently small. In this case we say that the solution depends continuously on the given data. We summarize this in the following proposition.
CHAPTER THREE
SEMI-LINEAR EVOLUTION EQUATIONS
Introduction
In this chapter we study another class of evolution equations in which the forcing term depends on the state of the system at some time t. We consider the following Cauchy problem: u0(t) = Au(t) + f(t, u(t)), t > 0
u(0) = u0(3.1)
where A is the in nitesimal generator of a C0-semigroup denoted by {etA, t ≥ 0} and f : [0, T] × X → X is continuous.
In the linear case we need the forcing term to just be continuous to guarantee the existence of a mild solution. But in this present case, we will require more than continuity on f to have existence of a solution, as we can see in the following example.
Example: In (3.1) above, let A = 0 and X = C0 the Banach space of all real-valued sequences u = {ξn}∞n=1 with limn→∞ ξn = 0 and kuk = supn≥1|ξn|. De ne the function
f : X → X by f(u) = {|ξn|12 + n−1}∞n=1, u = {ξn}∞n=1 ∈ X.
The continuity of the function ξ 7→ ξ12 for ξ ≥ 0 and the de nition of the norm on X imply that f is continous on X. But the initial value problem
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