Kenneth Kuttler

19.12. ORDINARY DIFFERENTIAL EQUATIONS IN BANACH SPACE 565

19.12 Ordinary Differential Equations in Banach SpaceHere we consider the initial value problem for functions which have values in a Banachspace. Let X be a Banach space.

Definition 19.12.1 Define BC ([a,b] ;X) as the bounded continuous functions f which havevalues in the Banach space X. For f ∈ BC ([a,b] ;X) , γ a real number. Then

∥ f∥γ≡ sup

t∈[a,b]

∥∥∥ f (t)eγ(t−a)∥∥∥ (19.12.73)

Then this is a norm. The usual norm is given by

∥ f∥ ≡ supt∈[a,b]

∥ f (t)∥

Lemma 19.12.2 ∥·∥γ

is a norm for BC ([a,b] ;X) and BC ([a,b] ;X) is a complete normedlinear space. Also, a sequence is Cauchy in ∥·∥

γif and only if it is Cauchy in ∥·∥.

Proof: First consider the claim about ∥·∥γ

being a norm. To simplify notation, letT = [a,b]. It is clear that ∥ f∥

γ= 0 if and only if f = 0 and ∥ f∥

γ≥ 0. Also,

∥α f∥γ≡ sup

t∈T

∥∥∥α f (t)eγ(t−a)∥∥∥= |α|sup

t∈T

∥∥∥ f (t)eγ(t−a)∥∥∥= |α|∥ f∥

so it does what is should for scalar multiplication. Next consider the triangle inequality.

∥ f +g∥γ

= supt∈T

∥∥∥( f (t)+g(t))eγ(t−a)∥∥∥≤ sup

t∈T

(∣∣∣ f (t)eγ(t−a)∣∣∣+ ∣∣∣g(t)eγ(t−a)

∣∣∣)≤ sup

t∈T

∣∣∣ f (t)eγ(t−a)∣∣∣+ sup

t∈T

∣∣∣g(t)eγ(t−a)∣∣∣= ∥ f∥

γ+∥g∥

The rest follows from the next inequalities.

∥ f∥ ≡ supt∈T∥ f (t)∥= sup

t∈T

∥∥∥ f (t)eγ(t−a)e−γ(t−a)∥∥∥≤ e|γ(b−a)| ∥ f∥

≡ e|γ(b−a)| supt∈T

∥∥∥ f (t)eγ(t−a)∥∥∥≤ (e|γ|(b−a)

)2supt∈T∥ f (t)∥=

(e|γ|(b−a)

)2∥ f∥

Now consider the ordinary initial value problem

x′ (t)+F (t,x(t)) = f (t) , x(a) = x0, t ∈ [a,b] (19.12.74)

where here F : [a,b]×X → X is continuous and satisfies the Lipschitz condition

∥F (t,x)−F (t,y)∥ ≤ K ∥x− y∥ , F : [a,b]×X → X is continuous (19.12.75)

Thanks to the fundamental theorem of calculus, there exists a solution to 19.12.74 if andonly if it is a solution to the integral equation

x(t) = x0−∫ t

aF (s,x(s))ds (19.12.76)

Then we have the following theorem.

19.12. ORDINARY DIFFERENTIAL EQUATIONS IN BANACH SPACE 56519.12 Ordinary Differential Equations in Banach SpaceHere we consider the initial value problem for functions which have values in a Banachspace. Let X be a Banach space.Definition 19.12.1 Define BC (a, b] ;X) as the bounded continuous functions f which havevalues in the Banach space X. For f € BC (a,b];X), ya real number. Thenf (te (19.12.73)Illy = sup |te a,b]Then this is anorm. The usual norm is given by[fll = sup [fF ()l|te [a,b]Lemma 19.12.2 ||-||, is a norm for BC ([a,b|;X) and BC ([a,b|;X) is a complete normedlinear space. Also, a sequence is Cauchy in |\-||, if and only if it is Cauchy in ||-|).Proof: First consider the claim about ||-||, being a norm. To simplify notation, letT = [a,b]. It is clear that || f||, = 0 if and only if f = 0 and || f||,, > 0. Also,afl = sup||af (e) er"teT=|a|sup|f (oerteT= la || fllyso it does what is should for scalar multiplication. Next consider the triangle inequality.Isl, = supl(S)+a()er|| < sup (|e )teT teT+ le (er)< sup Pier”teT+ sup |g(r)e"™teT= If lly+llsllyThe rest follows from the next inequalities.Ifsup |if(1)|| =sup fer Pe %)]] < ela I fl,teT teTel1-4)l cup | f (pera)teT< (e9)*suplirii=(e"°) Ir)teTNow consider the ordinary initial value problemx (t)+F (t,x(¢)) = f(t), x(a) = x0, t € [a,b] (19.12.74)where here F': [a,b] x X — X is continuous and satisfies the Lipschitz condition||F (t,x) — F (t,y)|| < K ||x—y||, F : [a,b] x X + X is continuous (19.12.75)Thanks to the fundamental theorem of calculus, there exists a solution to 19.12.74 if andonly if it is a solution to the integral equationx()=x- [ F(s.x(s))ds (19.12.76)Then we have the following theorem.