Freiling G., Yurko V. Lectures on Differential Equations of Mathematical Physics: A First Course

Подождите немного. Документ загружается.

92 G.

Freiling and V. Yurko

Let us

return to the proof of Theorem 2.7.7. It follows from (2.7.45) and Lemma 2.7.5

that

z(x,ρ) = 1 +

ω(x)

iρ

+ o

, |ρ|

→ ∞

, ρ ∈Ω, (2.7.48)

uniformly for x ≥ 0. From (2.7.36), (2.7.38)-(2.7.40), (2.7.43) and (2.7.48) we derive

) −(i

) for ν = 0. Furthermore, (2.7.32) and (2.7.36) imply

(x,ρ) = (iρ)exp(iρx)

1 −

2iρ

∞

(1 + e

xp(2

iρ(t −x)))q(t)z(t,ρ)dt

. (2.7.49)

Using (2.7.49) we get (i

) −(i

) for ν = 1. It is easy to verify by differentiation that

the function e(x,ρ) is a solution of (2.7.30). For real ρ 6= 0, the functions e(x,ρ) and

e(x,−ρ) satisfy (2.7.30), and by virtue of (2.7.33), lim

x→∞

he(x,ρ), e(x,−ρ)i = −2iρ. Since

the Wronskian he(x,ρ),e(x,−ρ)i does not depend on x, we arrive at (2.7.35). Lemma

2.7.5 is proved. 2

Remark 2.7.3. If q(x) ∈ W

, then there exist functions ω

sν

(x) such that for ρ →

∞, ρ ∈Ω, ν = 0,1, 2,

(ν)

(x,ρ) = (iρ)

exp(iρx)

1 +

N+1

∑

s=1

sν

(x)

(iρ)

+ o

N+1

´´

, ω

1ν

(x)

= ω

(x). (2.7.50)

Indeed, let q(x) ∈W

. Substituting (2.7.48) into the right-hand side of (2.7.37) we get

z(x,ρ) = 1 −

2iρ

∞

(1 −e

xp(2

iρ(t −x)))q(t) dt

−

2(iρ)

∞

(1 −e

xp(2

iρ(t −x)))q(t)ω(t)dt + o

, |ρ|

→ ∞

Integrating by parts and using Lemma 2.7.5, we obtain

z(x,ρ) = 1 +

ω(x)

iρ

(x)

(iρ)

+ o

, |ρ|

→ ∞, ρ ∈ Ω, (

2.7.51)

where

(x) = −

q(x)

∞

q(t)

∞

q(s)d

= −

q(x)

∞

q(t) d

virtue of (2.7.36) and (2.7.51), we arrive at (2.7.50) for N = 1, ν = 0. Using induction

one can prove (2.7.50) for all N.

If we additionally assume that xq(x) ∈ L(0, ∞), then the Jost solution e(x,ρ) exists

also for ρ = 0. More precisely, the following theorem is valid.

Theorem 2.7.8. Let (1 +x)q(x) ∈ L(0,∞). Then the functions e

(ν)

(x,ρ), ν = 0,1 are

continuous for Imρ ≥ 0, x ≥ 0, and

|e(x,ρ) exp(−iρx)|≤ exp(Q

(x)), (2.7.52)

Hyperbolic P

artial Differential Equations 93

|e(x,ρ) e

xp(−i

ρx) −1|≤

(x) −Q

x +

|ρ|

´´

xp(Q

(

x)), (2.7.53)

(x,ρ) exp(−iρx)−iρ| ≤ Q

(x)exp(Q

(x)), (2.7.54)

where

(x) :=

∞

(t) dt =

∞

(t −x)|q(t)|dt.

First we prove an auxiliary assertion.

Lemma 2.7.6. Assume that c

≥ 0, u(x) ≥ 0, v(x) ≥ 0, (a ≤ x ≤ T ≤ ∞), u(x) is

bounded, and (x −a)v(x) ∈ L(a,T ). If

u(x) ≤ c

(t −x)v(t)u(t)dt, (2.7.55)

then

u(x) ≤ c

exp

(t −x)v(t)dt

. (2.7.56)

Proof. Denote

ξ(x) = c

(t −x)v(t)u(t)dt.

Then

ξ(T ) = c

, ξ

(x) = −

v(t)u(t) dt, ξ

= v(x)u(x),

and (2.7.55) yields

0 ≤ ξ

(x) ≤ ξ(x)v(x).

Let c

> 0. Then ξ(x) > 0, and

(x)

ξ(x)

≤ v(x).

Hence

(x)

ξ(x)

≤ v(x) −

(x)

ξ(x)

≤ v(x).

Inte

grating

this inequality twice we get

−

(x)

ξ(x)

≤

v(t) d

t, ln

(x)

ξ(T )

≤

(t −x)v(t)d

and

consequently,

ξ(x) ≤ c

exp

(t −x)v(t)dt

According to (2.7.55), u(x) ≤ ξ(x), and we arrive at (2.7.56).

If c

= 0, then ξ(x) = 0. Indeed, suppose on the contrary that ξ(x) 6= 0. Since ξ(x) ≥

0, ξ

(x) ≤ 0, there exists T

≤ T such that ξ(x) > 0 for x < T

, and ξ(x) ≡ 0 for x ∈

,T ]. Repeating these arguments we get for x < T

and sufﬁciently small ε > 0,

ξ(x)

ξ(T

−ε)

≤

−ε

(t −x)v(t) d

t ≤

(t −x)v

(t) dt,

94 G.

Freiling and V. Yurko

which is

impossible. Thus, ξ(x) ≡ 0, and (2.7.56) becomes obvious. 2

Proof of Theorem 2.7.8. For Im ρ ≥ 0 we have

sinρ

xp(iρ)

≤ 1

. (2.7.57)

Indeed, (2.7.57) is obvious for real ρ and for |ρ| ≥ 1, Imρ ≥ 0. Then, by the maximum

principle [14, p.128], (2.7.57) is also valid for |ρ| ≤ 1, Im ρ ≥0.

It follows from (2.7.57) that

1 −exp(2iρx)

2iρ

≤ x for I

mρ ≥0

, x ≥0. (2.7.58)

Using (2.7.38) and (2.7.58) we infer

k+1

(x,ρ)| ≤

∞

t|q(t)z

(t, ρ)|dt, k ≥ 0, Im ρ ≥ 0, x ≥ 0,

and consequently by induction

(x,ρ)| ≤

∞

t|q(t)|d

, k ≥ 0

, Imρ ≥0, x ≥0.

Then, the series (2.7.39) converges absolutely and uniformly for Im ρ ≥ 0, x ≥ 0, and the

function z(x,ρ) is continuous for Imρ ≥0, x ≥0. Moreover,

|z(x,ρ)| ≤ exp

∞

t|q(t)|dt

, Imρ ≥0, x ≥ 0. (2.7.59)

Using (2.7.36) and (2.7.49) we conclude that the functions e

(ν)

(x,ρ), ν = 0,1 are continu-

ous for Imρ ≥0, x ≥0.

Furthermore, it follows from (2.7.37) and (2.7.58) that

|z(x,ρ)| ≤ 1 +

∞

(t −x)|q(t)z(t,ρ)|dt, Imρ ≥0, x ≥0.

By virtue of Lemma 2.7.6, this implies

|z(x,ρ)| ≤ exp(Q

(x)), Imρ ≥0, x ≥ 0, (2.7.60)

i.e. (2.7.52) is valid. We note that (2.7.60) is more precise than (2.7.59).

Using (2.7.37), (2.7.58) and (2.7.60) we calculate

|z(x,ρ) −1| ≤

∞

(t −x)|q(t)|exp(Q

(t)) dt

≤ exp(Q

(x))

∞

(t −x)|q(t)|dt,

and consequently,

|z(x,ρ) −1| ≤ Q

(x)exp(Q

(x)), Imρ ≥0, x ≥ 0. (2.7.61)

Hyperbolic P

artial Differential Equations 95

More precisely

|z(x,ρ) −1| ≤

|ρ|

(t −x)|q(t)|e

xp(Q

(t)) d

t +

|ρ|

∞

|ρ|

|q(t)|e

xp(Q

(t)

)dt

≤ exp(Q

(x))

∞

(t −x)|q(t)|dt −

∞

|ρ|

t −x −

|ρ|

|q(t)|d

(x) −Q

x +

|ρ|

´´

xp(Q

(x)),

i.e.

(2.7.53) is valid. At last, from (2.7.49) and (2.7.60) we obtain

(x,ρ) exp(−iρx)−iρ| ≤

∞

|q(t)|exp(Q

(t)) dt ≤ exp(Q

(x))

∞

|q(t)|dt,

and we arrive at (2.7.54). Theorem 2.7.8 is proved. 2

Remark 2.7.4. Consider the function

q(x) =

(1 + ax)

where a is

complex number such that a /∈ (−∞,0]. Then q(x) ∈ L(0,∞), but xq(x) /∈

L(0,∞). The Jost solution has in this case the form (see Example 2.7.3)

e(x,ρ) = exp(iρx)

1 −

iρ(1 + ax)

i.e. e(x,ρ) has

singularity at ρ = 0, hence we cannot omit the integrability condition in

Theorem 2.7.8.

Theorem 2.7.9. Let (1 + x)q(x) ∈ L(0, ∞). Then the Jost solution e(x,ρ) can be

represented in the form

e(x,ρ) = exp(iρx) +

∞

A(x,t) exp(iρt)dt, Imρ ≥0, x ≥ 0, (2.7.62)

where A(x,t) is a continuous function for 0 ≤x ≤t < ∞, and

A(x,x) =

∞

q(t) d

t. (

2.7.63)

|A(x,t)| ≤

x + t

(x) −Q

x + t

´´

, (2.7.64)

1 +

∞

|A(x,t)|d

t ≤ e

xp(Q

(x)),

∞

|A(x,t)|dt ≤ Q

(x)exp(Q

(x)). (2.7.65)

Moreover, the function A(x

) has ﬁrst derivatives

∂A

∂x

, i = 1,2; the

functions

∂A

)

∂x

+ x

96 G.

Freiling and V. Yurko

are

absolutely continuous with respect to x

and x

, and satisfy the estimates

∂A(x

)

∂x

+ x

≤

+ x

) −Q

+ x

´´

, i = 1,2. (2.7.66)

oof

. According to (2.7.36) and (2.7.39) we have

e(x,ρ) =

∞

∑

k=0

(x,ρ), ε

(x,ρ) = z

(x,ρ) exp(iρx). (2.7.67)

Let us show by induction that the following representation is valid

(x,ρ) =

∞

(x,t)exp(iρt) dt, k ≥ 1, (2.7.68)

where the functions a

(x,t) do not depend on ρ.

First we calculate ε

(x,ρ). By virtue of (2.7.38) and (2.7.67),

(x,ρ) =

∞

sinρ(s −x)

xp(i

ρs)q(s)ds

∞

q(s)

2s−x

xp(i

ρt)dt

ds.

Interchanging the order of integration we obtain that (2.7.68) holds for k = 1, where

(x,t) =

∞

(t+x)/2

q(s)d

Suppose

now that (2.7.68) is valid for a certain k ≥1. Then

k+1

(x,ρ) =

∞

sinρ(s −x)

q(s)ε

(s,ρ)d

∞

sinρ

(s −x)

q(s)

∞

(s,u)e

xp(

iρu)du

∞

q(s)

∞

(s,u)

s+u−x

−s+u+x

xp(iρt) d

ds.

We extend a

(s,u) by zero for u < s. For s ≥x this yields

∞

(s,u)

s+u−x

−s+u+x

exp(iρt)dt

du =

∞

exp(iρt)

t+s−x

t−s+x

(s,u)du

dt.

Therefore

k+1

(x,ρ) =

∞

xp(i

ρt)

∞

q(s)

t+s−x

t−s+x

(s,u)du

Hyperbolic P

artial Differential Equations 97

∞

k+1

(x,t)e

xp(i

ρt)dt,

where

k+1

(x,t) =

∞

q(s)

t+s−x

t−s+x

(s,u)d

s, t ≥ x.

Changing the variables according to u + s = 2α, u −s = 2β, we obtain

k+1

(x,t) =

∞

(t+x)/2

(t−x)/2

q(α −β)a

(α −β,α + β)dβ

dα.

Taking H

(α,β) = a

(α −β,α + β), t + x = 2u, t −x = 2v, we calculate for 0 ≤ v ≤u,

(u,v) =

∞

q(s)d

, H

k+1

(u,v) =

∞

q(α −β)H

(α,β)dβ

dα. (2.7.69)

It can be shown by induction that

k+1

(u,v)| ≤

(u)

(u −v)−Q

(u))

, k ≥ 0, 0 ≤v ≤ u. (2.7.70)

Indeed,

for k = 0

, (2.7.70) is obvious. Suppose that (2.7.70) is valid for H

(u,v). Then

(2.7.69) implies

k+1

(u,v)| ≤

∞

(α)

|q(α −β)|

(α −β) −Q

(α)

)

k−1

(k −1)!

dβ

dα.

Since

the

functions Q

(x) and Q

(x) are monotonic, we get

k+1

(u,v)| ≤

(u)

(k −1)!

∞

(α −v) −Q

(α))

k−1

(α −v)−Q

(α))dα

(u)

(u −v)−Q

(u))

i.e.

(2.7.70)

is proved. Therefore, the series

H(u,v) =

∞

∑

k=1

(u,v)

converges absolutely and uniformly for 0 ≤v ≤u, and

H(u,v) =

∞

q(s)d

s +

∞

(α −β)H(α,β)dβ

dα, (2.7.71)

|H(u,v)| ≤

(u)e

(u −v

) −Q

(u)

. (2.7.72)

Put

A(x,t) = H

t + x

t −x

. (2.7.73)

Then

A(x,t)

∞

∑

k=1

(x,t),

98 G.

Freiling and V. Yurko

the series

converges absolutely and uniformly for 0 ≤x ≤t , and (2.7.62)-(2.7.64) are valid.

Using (2.7.64) we calculate

∞

|A(x,t)|dt ≤ exp(Q

(x))

∞

(ξ)exp(−Q

(ξ))dξ

= exp(Q

(x))

∞

dξ

xp(−Q

(ξ

))

dξ = exp(Q

(x)) −1,

and we arrive at (2.7.65).

Furthermore, it follows from (2.7.71) that

∂H(u,v)

∂u

= −

q(u) −

q(u −β)H(u,β)dβ, (2.7.74)

∂H(u,v)

∂v

∞

q(α −v)H(α,v)dα. (2.7.75)

follo

ws from (2.7.74)-(2.7.75) and (2.7.72) that

∂H(u,v)

∂u

q(u)

≤

|q(u −β)|Q

(u)e

xp(Q

(

u −β) −Q

(u))dβ,

∂H(u,v)

∂v

≤

∞

|q(α −v)|Q

(α)e

xp(

(α −v) −Q

(α))dα.

Since

|q(u −β)|dβ =

u−v

|q(s)|ds ≤ Q

(u −v),

(α −v)−Q

(α) =

α−v

(t) dt

≤

u−v

(t) dt = Q

(u −v) −Q

(u), u ≤α,

we get

∂H(u,v)

∂u

q(u)

≤

(u)e

xp(Q

(u −v

) −Q

(u))

|q(u −β)|dβ

≤

(u −v)Q

(u)e

xp(Q

(u −v

) −Q

(u)), (2.7.76)

∂H(u,v)

∂v

≤

(u)e

xp(Q

(u −v

) −Q

(u))

∞

|q(α −v)|dα

≤

(u −v)Q

(u)e

xp(Q

(u −v

) −Q

(u)). (2.7.77)

By virtue of (2.7.73),

∂A(x,t)

∂x

∂H(u,v)

∂u

−

∂H(u,v)

∂v

∂A(x,t)

∂t

∂H(u,v)

∂u

∂H(u,v)

∂v

Hyperbolic P

artial Differential Equations 99

where

u =

t + x

, v =

t −x

Hence,

∂A(x,t)

∂x

x + t

∂H(u,v)

∂u

q(u)

−

∂H(u,v)

∂v

∂A(x,t)

∂t

x + t

∂H(u,v)

∂u

q(u)

∂H(u,v)

∂v

aking

(2.7.76) and (2.7.77) into account we arrive at (2.7.66). 2

Theorem 2.7.10. For each δ > 0, there exists a = a

≥0 such that equation (2.7.30)

has a unique solution y = E(x,ρ), ρ ∈Ω

, satisfying the integral equation

E(x,ρ) = exp(−iρx) +

2iρ

xp(

iρ(x −t))q(t)E(t,ρ)dt

2iρ

∞

xp(iρ(t −x))q

(t)E(t, ρ)dt. (2.7.78)

The function E(x, ρ), called the Birkhoff solution for (2.7.30) , has the following proper-

ties:

) E

(ν)

(x,ρ) = (−iρ)

exp(−iρx)(1 + o(1)), x → ∞ ν = 0,1, uniformly for |ρ| ≥

δ, Imρ ≥α, for each ﬁxed α > 0;

) E

(ν)

(x,ρ) = (−iρ)

exp(−iρx)(1 + O(ρ

−1

)), |ρ| → ∞ ρ ∈Ω, uniformly for x ≥ a;

) for each ﬁxed x ≥0, the functions E

(ν)

(x,ρ) are analytic for Imρ > 0, |ρ| ≥ δ, and

are continuous for ρ ∈Ω

;

) the functions e(x, ρ) and E(x,ρ) form a fundamental system of solutions for

(2.7.30) , and he(x, ρ),E(x,ρ)i = −2iρ.

) If δ ≥Q

(0), then one can take above a = 0.

Proof. For ﬁxed δ > 0 choose a = a

≥0 such that Q

(a) ≤δ. We transform (2.7.78)

by means of the replacement E(x,ρ) = exp(−iρx)ξ(x,ρ) to the equation

ξ(x,ρ) = 1 +

2iρ

xp(2

iρ(x −t))q(t)ξ(t, ρ)dt

2iρ

∞

q(t)ξ(t,ρ)d

t. (

2.7.79)

The method of successive approximations gives

(x,ρ) = 1,

k+1

(x,ρ) =

2iρ

xp(2

iρ(x −t))q(t)ξ

(t, ρ)dt +

2iρ

∞

q(t)ξ

(t,ρ)d

(x,ρ) =

∞

∑

k=0

(x,ρ).

This yields

|ξ

k+1

(x,ρ)| ≤

2|ρ|

∞

|q(t)ξ

(t, ρ)|d

100 G.

Freiling and V. Yurko

and hence

|ξ

(x,ρ

)| ≤

(a)

2|ρ|

Thus

for x ≥a

, |ρ| ≥ Q

(a), we get

|ξ(x,ρ)| ≤ 2, |ξ(x, ρ) −1| ≤

(a)

|ρ|

follo

ws from (2.7.78) that

(x,ρ) = exp(−iρx)

−iρ +

xp(2

iρ(x −t))q(t)ξ(t, ρ)dt −

∞

q(t)ξ(t,ρ)d

. (

2.7.80)

Since |ξ(x,ρ)| ≤ 2 for x ≥ a, ρ ∈ Ω

, it follows from (2.7.79) and (2.7.80) that

(ν)

(x,ρ)(−iρ)

−ν

exp(iρx)−1|

≤

|ρ|

xp(−2

τ(x −t))|q(t)|dt +

∞

|q(t)|dt

≤

|ρ|

xp(

−τx)

x/2

|q(t)|dt +

∞

x/2

|q(t)|dt

and consequently (i

) −(i

) are proved.

The other assertions of Theorem 2.7.10 are obvious. 2

Properties of the spectrum. Denote

∆(ρ) = e

(0,ρ) −h e(0, ρ). (2.7.81)

By virtue of Theorem 2.7.7, the function ∆(ρ) is analytic for Imρ > 0, and continuous for

ρ ∈ Ω. It follows from (2.7.34) that for |ρ| → ∞, ρ ∈ Ω,

e(0,ρ) = 1 +

iρ

+ o

, ∆(ρ)

(iρ)

1 +

iρ

+ o

´´

, (2.7.82)

where ω

= ω(0), ω

= ω(0) −h. Using

(2.7.36),

(2.7.45) and (2.7.49) one can obtain

more precisely

e(0,ρ) = 1 +

iρ

2iρ

∞

q(t) e

xp(2

iρt)dt + O

∆(ρ)

(iρ)

1 +

iρ

−

2iρ

∞

q(t) e

xp(2

iρt)dt + O

´´











(2.7.83)

Denote

Λ = {λ = ρ

: ρ ∈ Ω, ∆(ρ)

= 0

= {λ = ρ

: Im ρ > 0, ∆(ρ) = 0},

= {λ = ρ

: Im ρ = 0, ρ 6= 0, ∆(ρ) = 0}.

Hyperbolic P

artial Differential Equations 101

Obviously

, Λ = Λ

∪Λ

is a bounded set, and Λ

is a bounded and at most countable set.

Denote

Φ(x,λ) =

e(x,ρ)

∆(ρ)

. (2.7.84)

The

function Φ

(x,λ) satisﬁes (2.7.30) and on account of (2.7.81) and Theorem 2.7.7 also

the conditions

U(Φ) = 1, (2.7.85)

Φ(x,λ) = O(exp(iρx)), x →∞, ρ ∈ Ω, (2.7.86)

where U is deﬁned by (2.7.31). The function Φ(x,λ) is called the Weyl solution for L.

Note that (2.7.30), (2.7.85) and (2.7.86) uniquely determine the Weyl solution.

Denote M(λ) := Φ(0,λ). The function M(λ) is called the Weyl function for L. It

follows from (2.7.84) that

M(λ) =

e(0,ρ)

∆(ρ)

. (2.7.87)

Clearly

(x,λ) = S(x, λ) + M(λ)ϕ(x,λ), (2.7.88)

where the functions ϕ(x,λ) and S(x, λ) are solutions of (2.7.30) under the initial conditions

ϕ(0,λ) = 1, ϕ

(0,λ) = h, S(0,λ) = 0, S

(0,λ) = 1.

We recall that the Weyl function plays an important role in the spectral theory of Sturm-

Liouville operators (see [19] for more details).

By virtue of Liouville‘s formula for the Wronskian [13, p.83], hϕ(x,λ),Φ(x,λ)i does

not depend on x. Since for x = 0,

hϕ(x,λ), Φ(x,λ)i

|x=0

= U(Φ) = 1,

we infer

hϕ(x,λ), Φ(x,λ)i ≡ 1. (2.7.89)

Theorem 2.7.11. The Weyl function M(λ) is analytic in Π \Λ

and continuous in

\Λ. The set of singularities of M(λ) ( as an analytic function ) coincides with the set

:= {λ : λ ≥0}∪Λ.

Theorem 2.7.11 follows from (2.7.81), (2.7.87) and Theorem 2.7.7. By virtue of

(2.7.88), the set of singularities of the Weyl solution Φ(x,λ) coincides with Λ

for all

x ≥ 0, since the functions ϕ(x,λ) and S(x, λ) are entire in λ for each ﬁxed x ≥0.

Deﬁnition 2.7.1. The set of singularities of the Weyl function M(λ) is called the

spectrum of L. The values of the parameter λ, for which equation (2.7.30) has nontrivial

solutions satisfying the conditions U(y) = 0, y(∞) = 0 (i.e. lim

x→∞

y(x) = 0 ), are called

eigenvalues of L, and the corresponding solutions are called eigenfunctions.

Remark 2.7.5. One can introduce the operator

: D(L

) → L

(0,∞), y → −y

+ q(x)y