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Published online: May 29, 2017

THE INITIAL VALUE PROBLEM FOR THE SCHR ¨ODINGER

EQUATION INVOLVING THE HENSTOCK–KURZWEIL INTEGRAL

SALVADOR S ´ANCHEZ-PERALES

Abstract. Let Lbe the one-dimensional Schr¨odinger operator defined by

Ly=−y00+qy. We investigate the existence of a solution to the initial value

problem for the differential equation (L−λ)y=g, whenqandgare Henstock– Kurzweil integrable functions on [a, b]. Results presented in this article are generalizations of classical results for the Lebesgue integral.

1. Introduction

Letqbe a real valued function defined on [a, b] and letLbe the one-dimensional Schr¨odinger operator defined by Ly =−y00+qy. It is well known that if q, g are Lebesgue integrable functions on [a, b], then there exists a unique solution f, f0 ∈ AC([a, b]) of the differential equation (Lλ)y =g satisfying the initial condition

f(c) = αand f0(c) =β, wherec ∈ [a, b], λ, α, β ∈ C and AC([a, b]) denotes the

space of all absolutely continuous functions on [a, b]. See, for example, [5]. In this paper, we generalize this result when q, g are Henstock–Kurzweil integrable functions on [a, b].

2. Preliminaries

In this section, the definition of the Henstock–Kurzweil integral and their main properties needed in this paper are presented.

Definition 2.1. Let f : [a, b] → C be a function. We say that f is Henstock–

Kurzweil (shortly, HK-) integrable on [a, b], if there is anA∈Csuch that, for each >0, there exists a functionγ: [a, b]→(0,∞) (named a gauge) for which

n

X

i=1

f(ti)(xixi−1)−A

< ,

2010Mathematics Subject Classification. 26A39, 34B24, 26A45.

Key words and phrases. Henstock–Kurzweil integral, Schr¨odinger operator,ACG∗-functions,

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for any partitionP ={([xi−1, xi], ti)}ni=1 such thatti∈[xi−1, xi] and [xi−1, xi]⊆

[tiγ(ti), ti+γ(ti)] for alli= 1,2, . . . , n. The numberAis called the integral of f over [a, b] and it is denoted by Rb

af.

The set of all Henstock–Kurzweil integrable functions on [a, b] is denoted by HK([a, b]). This set is a vector space and contains the union of L([a, b]), the space of Lebesgue integrable functions on [a, b], and the Cauchy-Lebesgue integrable functions (i.e., improper Lebesgue integrals). It is well known that iffHK([a, b]) then not necessarily|f| ∈HK([a, b]). If bothf and|f|are HK-integrable on [a, b], we say that f is Henstock–Kurzweil absolutely integrable on [a, b]. The space of Henstock–Kurzweil absolutely integrable functions on [a, b] coincides with the space

L([a, b]).

For each fHK([a, b]) and I ⊆ [a, b] the Alexiewicz seminorm of f on I is defined as

kfkI = sup JI

Z

J f

,

where the supremum is taken over all intervalsJ contained inI.

Definition 2.2. Letϕ: [c, d]→Cbe a function. The variation ofϕon the interval

[c, d] is defined as

V[c,d]ϕ= sup

( n

X

i=1

|ϕ(xi)−ϕ(xi−1)|

{xi}

n

i=0 is a partition of [c, d]

)

.

We say that the functionϕis of bounded variation on [c, d] ifV[c,d]ϕ <∞. The

space of all bounded variation functions on [c, d] is denoted byBV([c, d]).

The next theorem shows that absolutely integrable functions are precisely those integrable functions whose indefinite integrals have bounded variation.

Theorem 2.3 ([1, Theorem 7.5]). Let fHK([a, b]). Then |f| is HK-integrable if and only if the indefinite integral F(x) = Rx

a f has bounded variation on [a, b].

In this case

V[a,b]F =

Z b

a |f|.

The space of Henstock–Kurzweil integrable functions is not multiplicative in general. However, the multipliers for HK-integrable functions are the functions of bounded variation.

Theorem 2.4 (Multiplier Theorem, [1, Theorem 10.12]). If g is a real valued function such that gHK([a, b]) andfBV([a, b]) then the productgf belongs to HK([a, b]).

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Theorem 2.5 ([4, Lemma 24]). If g is a real valued function such that g

HK([a, b])andfBV([a, b]), then

Z b

a f g

≤ inf

t∈[a,b]|f(t)|

Z b

a g(t)dt

+kgk[a,b]V[a,b]f.

Theorem 2.6 ([3, Corollary 3.2]). If g is a real valued function such that g

HK([a, b]) and (fn) is a sequence in BV([a, b]) such that V[a,b]fnM for all n∈N, andgng pointwise on [a, b], then

lim

n→∞ Z b

a f gn=

Z b

a f g.

It is well known that the Lebesgue integral may be characterized by the fact that the indefinite integral is absolutely continuous. A similar characterization is possible with the Henstock–Kurzweil integral.

Definition 2.7. Let F : [a, b] → C. We say that F is absolutely continuous in the restricted sense on a set E ⊆ [a, b] (FAC∗(E)), if for every > 0 there exists η >0 such that if {[ui, vi]}s

i=1 is a collection of nonoverlapping intervals

with endpoints inE and such thatPs

i=1(viui)< η, then s

X

i=1

sup{|F(x)−F(y)|:x, y∈[ui, vi]}< .

Moreover,Fis said to be generalized absolutely continuous in the restricted sense on [a, b] (FACG∗([a, b])), ifF is continuous on [a, b] and there is a countable collection (En)∞n=1 of sets in [a, b] with [a, b] = ∪∞i=1En andFAC∗(En) for all n∈N.

Theorem 2.8 (Fundamental Theorem of Calculus, [2]). Let f, F : [a, b] →C be functions andc∈[a, b].

(1) fHK([a, b]) and F(x) = Rx

c f for all x ∈ [a, b] if and only if FACG∗([a, b]),F(c) = 0, andF0=f almost everywhere on [a, b].

IffHK([a, b])andf is continuous at x∈[a, b]then dxd Rcxf =f(x). (2) FACG∗([a, b]) if and only if F0 exists almost everywhere on [a, b], and

Rx

c F

0 =F(x)F(c) for allx[a, b].

3. The existence and uniqueness theorem

In this section qis a real valued function such that qHK([a, b]) and Lis the Schr¨odinger operator defined as

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Lemma 3.1. Letc∈[a, b],λ, α, βCand letA=

0 1

qλ 0

. IfgHK([a, b]), then there exists a unique solutionf,f0∈ACG∗([a, b])of the initial value problem

(Lλ)y=g a.e.

y(c) =α (3.1)

y0(c) =β

if and only if there exists a unique solutionuC([a, b],C2)of the equation

u= Z (·)

c

A(s)u(s)ds+w, (3.2)

wherew: [a, b]→C2 is defined as w(x) =

α β

Z x

c

0

g(s)

ds.

Proof. Letf, withf, f0∈ACG∗([a, b]), be a solution of the initial value problem (3.1). Sincef0∈C([a, b]), it follows thatf(x) =Rx

a f

0(s)ds+f(a) and hence from Theorem 2.3f is of bounded variation on [a, b]. This implies, by Theorem 2.4, that

qfHK([a, b]). We set u=

f f0

, then uC([a, b],C2) andAu=

f0 qfλf

HK([a, b]). Therefore for all x∈[a, b],

Z x

c

A(s)u(s)ds+w(x) = 

Rx

c f

0(s)ds+α

Rx

c [q(s)f(s)−λf(s)]ds+β

Rx

c g(s)ds

= 

Rx

c f

0(s)ds+f(c)

Rx c [f

00(s) +g(s)]ds+f0(c)Rx c g(s)ds

= 

f(c) +Rcxf0(s)ds

Rx

c f

00(s)ds+f0(c)

= 

f(x)

f0(x) 

,

where the last equality is due to Theorem 2.8 (2) because f, f0 ∈ ACG∗([a, b]). Thereforeusatisfies the equation (3.2).

Conversely, let u =

u1 u2

C([a, b],C2) be a solution of equation (3.2); then

for everyx∈[a, b],

u1(x) = Z x

c

u2(s)ds+α,

u2(x) =

Z x

c

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Therefore u01 = u2 on [a, b] and, from Theorem 2.8 (1), u2 is ACG∗ on [a, b] and u02 = (qλ)u1−g almost everywhere on [a, b]. This implies that u1, u01 ∈ ACG∗([a, b]) and (Lλ)u1=g almost everywhere on [a, b].

The uniqueness in any of the situations follows from the above.

Theorem 3.2. Let c ∈[a, b], λ, α, β ∈ C. If gHK([a, b]), then there exists a

unique solutionf,f0∈ACG∗([a, b])of the initial-value problem

(Lλ)y=g a.e.

y(c) =α

y0(c) =β.

Proof. Let (yn), (zn) be two sequences of functions defined on [a, b] asy0(x) =α,

z0(x) =β, and for alln∈N,

yn(x) =α+ Z x

c

zn−1(s)ds and zn(x) =β

Z x

c

g(s)ds+ Z x

c

(q(s)−λ)yn−1(s)ds.

Clearly y0,z0, y1 and z1 are well defined. Suppose thaty2, z2, . . . , yn, zn exist. Sincezn−1∈L([a, b]), from Theorem 2.3,yn is of bounded variation on [a, b] and

hence, by Theorem 2.4, (qλ)ynHK([a, b]). Then zn+1 exists. Also observe that clearlyyn+1 exists. Therefore (yn) and (zn) are well defined.

We claim that for every x∈[a, b] withx > c,

Z x

c

|zn(s)−zn−1(s)|ds

   

  

k(qλ)αgk[a,b]

(xc)k+1

(k+ 1)! kq−λk

k

[a,b], ifn= 2k+ 1; |β|(xc)

k+1

(k+ 1)! kq−λk

k

[a,b], ifn= 2k.

(3.3)

We prove this only fornodd, by induction onk. Fork= 0 we have that

Z x

c

|z1(s)−z0(s)|ds= Z x

c

Z s

c

[(q(t)−λ)αg(t)]dt

ds

Z x

c

k(qλ)αgk[a,b]ds

=k(qλ)αgk[a,b](xc)

for allx > c. Now, suppose that for everyx > c,

Z x

c

|z2k+1(s)−z2k(s)|ds≤ k(qλ)αgk[a,b](xc) k+1

(k+ 1)! kq−λk

k [a,b].

Letx > c, observe that

Z x

c

z2(k+1)+1(s)−z2(k+1)(s)

ds=

Z x

c

Z s

c

(q(t)−λ)[y2(k+1)(t)−y2k+1(t)]dt

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Now, sincey2(k+1)−y2k+1 is of bounded variation on [c, s] and V[c,s](y2(k+1)−y2k+1)≤

Z s

c

|z2k+1(t)−z2k(t)|dt,

it follows by Theorem 2.5 that

Z x

c

Z s

c

(q(t)−λ)[y2(k+1)(t)−y2k+1(t)]dt

ds

Z x

c

inf

t∈[c,s]|y2(k+1)(t)−y2k+1(t)|+V[c,s](y2(k+1)−y2k+1)

kq−λk[a,b]ds

Z x

c

|y2(k+1)(c)−y2k+1(c)|+V[c,s](y2(k+1)y2k+1)kq−λk[a,b]ds

= Z x

c

V[c,s](y2(k+1)−y2k+1)dskq−λk[a,b]

Z x

c

Z s

c

|z2k+1(t)−z2k(t)|dt

dskq−λk[a,b]

Z x

c

k(qλ)αgk[a,b](sc) k+1

(k+ 1)! kq−λk

k [a,b]

dskq−λk[a,b]

≤ k(qλ)αgk[a,b](xc) k+2

(k+ 2)! kq−λk

k+1 [a,b].

Thus (3.3) follows by induction. Similarly we can prove that for every x∈[a, b] withx < c,

Z c

x

|zn(s)−zn−1(s)|ds

   

  

k(qλ)αgk[a,b]

(cx)k+1

(k+ 1)! kq−λk

k

[a,b], ifn= 2k+ 1; |β|(cx)

k+1

(k+ 1)! kq−λk

k

[a,b], ifn= 2k.

(3.4)

For eachkNdefinelk =y2k+1−y2k. Takek∈Nandx∈[a, b]. Ifx < c, then |lk(x)|=|y2k+1(x)−y2k(x)|

=

Z x

c

[z2k(s)−z2k−1(s)]ds

Z c

x

|z2k(s)−z2k−1(s)|ds

≤ |β|(cx) k+1

(k+ 1)! kq−λk

k [a,b].

Therefore

|lk(x)| ≤ |β| kq−λk[a,b]

(ba)k+1kqλkk+1 [a,b]

(k+ 1)! .

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Now, since

∞ X

k=1 |β| kq−λk[a,b]

(ba)k+1kq−λkk+1[a,b]

(k+ 1)! ≤

|β| kq−λk[a,b]e

(b−a)|qλk[a,b]<∞,

it follows that ∞ X

k=1

lk converges uniformly on [a, b]. Again using the equations (3.3)

and (3.4), we obtain that ifhk =y2ky2k−1 then

∞ X

k=1

hk converges uniformly on

[a, b]. Therefore y0+ ∞ X

n=1

[ynyn−1] = y0+ [y1y0] +

∞ X

k=1 lk+

∞ X

k=1

hk converges

uniformly on [a, b]. Thus its sequences of partial sumssn converges uniformly to a limit functiony on [a, b]. But

sn(x) =y0+

n

X

k=1

[yk(x)−yk−1(x)] =yn(x).

In other words, the sequence (yn) converges uniformly toy on [a, b].

On the other hand, from the inequalities

|z2k+1(x)−z2k(x)| ≤ k(qλ)αgk[a,b]

(ba)kkq−λkk [a,b] k!

and

|z2k(x)−z2k−1(x)| ≤ |β|

(ba)kkq−λkk [a,b] k! ,

it follows that z0+

∞ X

k=0

[znzn−1] converges uniformly on [a, b]. If z denotes its

sum thenzn converges uniformly to zon [a, b]. Consequently, for each x∈[a, b],

y(x) = lim

n→∞yn(x) = limn→∞

α+ Z x

c

zn−1(s)ds

=α+ lim

n→∞ Z x

c

zn−1(s)ds

=α+ Z x

c

lim

n→∞zn−1(s)ds =α+

Z x

c

z(s)ds.

It is not possible to apply the above idea to the sequence (yn) because (q−λ)yn−1

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For everyn∈N,

V[c,x]yn=V[c,x]

"

y1+ n−1

X

k=1

[yk+1yk]

#

V[c,x]y1+

n−1

X

k=1

V[c,x][yk+1yk]

V[c,x]y1+ n−1

X

k=1

Z x

c

|zk(s)−zk−1(s)|ds

V[c,x]y1+

∞ X

k=1

Z x

c

|zk(s)−zk−1(s)|ds

=V[c,x]y1+ Z x

c

|z1(s)−z0(s)|ds

+ ∞ X

k=1

Z x

c

|z2k+1(s)−z2k(s)|ds+ ∞ X

k=1

Z x

c

|z2k(s)−z2k−1(s)|ds

k(qλ)αqk[a,b]

kq−λk[a,b]

+ |β|

kq−λk[a,b]

X

k=0

(ba)k+1kqλkk+1 [a,b]

(k+ 1)!

= k(qλ)αqk[a,b]+|β|

kq−λk[a,b] e

(b−a)kqλk.

Therefore, by Theorem 2.6, it follows that

lim

n→∞ Z x

c

(q(s)−λ)yn−1(s) =

Z x

c

(q(s)−λ)y(s)ds.

Thus

z(x) = lim

n→∞zn(x) = limn→∞

β

Z x

c

g(s)ds+ Z x

c

(q(s)−λ)yn−1(s)ds

=β

Z x

c

g(s)ds+ lim

n→∞ Z x

c

(q(s)−λ)yn−1(s)ds

=β

Z x

c

g(s)ds+ Z x

c

(q(s)−λ)y(s)ds.

(3.5)

A similar reasoning shows that the equality (3.5) holds whenx < c.

Consequently,u=

y z

is a solution of equation (3.2). We shall now prove that

this solution is unique. Assume that

y z

is another solution of equation (3.2).

Therefore

y(x) =α+ Z x

c

(9)

and

z(x) =β

Z x

c

g(s)ds+ Z x

c

[q(s)−λ]y(s)ds

for allx∈[a, b]. Observe that for everykN, Z x

c

|z2k(s)−z(s)|ds ≤ k(qλ)ygk[a,b]

(xc)k+1

(k+ 1)! kq−λk

k, ifc < x,

and Z c

x

|z2k(s)−z(s)|ds ≤ k(qλ)ygk[a,b](cx) k+1

(k+ 1)! kq−λk

k

, ifx < c.

From this it follows that

|y2k+1(x)−y(x)| ≤ k(qλ)ygk[a,b] kq−λk

(ba)k+1

(k+ 1)! kq−λk

k+1

and

|z2(k+1)(x)−z(x)| ≤

k(qλ)ygk[a,b] kq−λk

(ba)k+1

(k+ 1)! kq−λk

k+1

for all x∈ [a, b]. Thus y2k+1(x)→ y(x) and z2(k+1)(x)→ z(x), but y2k+1(x)→ y(x) andz2(k+1)(x)→z(x). Thereforey(x) =y(x) andz(x) =z(x).

4. Example

In this section, we give an example for the application of Theorem 3.2.

Example 4.1. Letqbe a function defined on [0,1] as

q(x) =  

 2π

x sin

π

x2

, ifx∈(0,1];

0, ifx= 0,

and letg: [0,1]→Rbe defined by

g(x) =  

(−1)k+12k

k , forx∈[ck−1, ck), k∈N;

0, forx= 1,

where ck = 1− 1

2k, k = 0,1,2, . . . Then q and g are unbounded HK-integrable

functions on [0,1]. Therefore, by Theorem 3.2, the initial value problem

−y00+q(x)y2y=g(x) a.e.

y1

2

= 0

y01

2

= 1

has a solution.

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References

[1] R. G. Bartle, A modern theory of integration, Graduate Studies in Mathematics, vol. 32, American Mathematical Society, Providence, 2001. MR 1817647.

[2] R. A. Gordon,The integrals of Lebesgue, Denjoy, Perron, and Henstock, Graduate Studies in Mathematics, vol. 4, American Mathematical Society, Providence, 1994. MR 1288751. [3] E. Talvila,Limits and Henstock integrals of products, Real Anal. Exchange 25 (1999), 907–918.

MR 1778542.

[4] E. Talvila, Henstock–Kurzweil Fourier transforms, Illinois J. Math. 46 (2002), 1207-1226. MR 1988259.

[5] G. Teschl, Mathematical methods in quantum mechanics; with applications to Schr¨odinger operators, Graduate Studies in Mathematics, vol. 99, American Mathematical Society, Provi-dence, 2009. MR 2499016.

S. S´anchez-Perales

Universidad Tecnol´ogica de la Mixteca, Instituto de F´ısica y Matem´aticas, Km. 2.5 Carretera a Acatlima, 69000 Oaxaca, Mexico.

[email protected]

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