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

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72 G.

Freiling and V. Yurko

∞

−∞

−u

−q(x)u](x,t −τ)g(τ) dτ,

and consequently

(x,t) = w

(x,t) −q(x)w(x,t), (x,t) ∈ ∆

. (2.6.31)

Furthermore, (2.6.25) and (2.6.29) yield

w(0,t) =

−x

r(t −τ)g(τ)dτ,

(0,t) = h

−x

r(t −τ)g(τ)dτ = 0, t ∈ [−x

Therefore, according to (2.6.24), we have

w(0,t) = w

(0,t) = 0, t ∈ [−x

]. (2.6.32)

Since the Cauchy problem (2.6.31)-(2.6.32) has only the trivial solution, we arrive at

(2.6.26).

Denote u

(x,t) := u

(x,t). It follows from (2.6.30) that

(x,0) = 2g(x) +

−x

−∞

(x,τ)g(τ)dτ +

∞

(x,τ)g(τ)dτ

= 2(g(x) +

∞

(x,τ)g(τ)dτ) = 2(g(x) +

(x,τ)g(τ)dτ).

Taking (2.6.26) into account we get

g(x) +

(x,τ)g(τ)dτ = 0, 0 ≤x ≤x

This integral equation has only the trivial solution g(x) = 0, and consequently Theorem

2.6.2 is proved. 2

Let r and ˜r be the traces for the boundary value problems B(q(x),h) and B( ˜q(x),

respectively. We agree that if a certain symbol α denotes an object related to B(q(x),h) ,

then

α will denote the analogous object related to B( ˜q(x),

h) .

Theorem 2.6.3. If r(t) = ˜r(t), t ≥ 0, then q(x) = ˜q(x), x ≥ 0 and h =

h. Thus, the

speciﬁcation of the trace r uniquely determines the potential q and the coefﬁcient h.

Proof. Since r(t) = ˜r(t), t ≥0, we have, by virtue of (2.6.22),

F(x,t) =

F(x,t).

Therefore, Theorem 2.6.2 gives us

G(x,t) =

G(x,t), 0 ≤t ≤ x. (2.6.33)

Hyperbolic P

artial Differential Equations 73

By virtue

of (2.6.18),

q(x) = 2

x,x), h = G(0, 0) = −r

(0). (2.6.34)

Together with (2.6.33) this yields q(x) = ˜q(x), x ≥0 and h =

h. 2

The Gelfand-Levitan equation (2.6.21) and Theorem 2.6.2 yield ﬁnally the following

algorithm for the solution of Inverse Problem 2.6.1.

Algorithm 2.6.1. Let r(t), t ≥ 0 be given. Then

(1) Construct the function F(x,t) using (2.6.22).

(2) Find the function G(x,t) by solving equation (2.6.21).

(3) Calculate q(x) and h by the formulae (2.6.34).

Let us now formulate necessary and sufﬁcient conditions for the solvability of Inverse

Problem 2.6.1.

Theorem 2.6.4. For a function r(t), t ≥ 0 to be the trace for a certain boundary value

problem B(q(x),h) of the form (2.6.1)−(2.6.2) with q ∈D

, it is necessary and sufﬁcient

that r(t) ∈ D

N+2

, r(0) = 1, and that for each ﬁxed x > 0 the integral equation (2.6.21)

is uniquely solvable.

Proof. The necessity part of Theorem 2.6.4 was proved above, here we prove the

sufﬁciency. For simplicity let N ≥ 1 (the case N = 0 requires small modiﬁcations).

Let a function r(t), t ≥ 0, satisfying the hypothesis of Theorem 2.6.4, be given, and let

G(x,t), 0 ≤t ≤ x, be the solution of (2.6.21). We deﬁne G(x,t) = G(x,−t), r(t) = −r(−t)

for t < 0, and consider the function

u(x,t) :=

(r(t + x)

+ r(t −x))

−x

r(t −τ)G(x, τ)dτ, (2.6.35)

−∞ < t < ∞, x ≥0.

Furthermore,

construct q and h via (2.6.34) and consider the boundary value problem

(2.6.1)-(2.6.2) with these q and h . Let ˜u(x,t) be the solution of (2.6.1)-(2.6.2), and let

˜r(t) := ˜u(0,t). Our goal is to prove that ˜u = u, ˜r = r.

Differentiating (2.6.35) and taking (2.6.20) into account, we get

(x,t) =

(a(t + x)

+ a

(t −x)) + G(x,t)

−x

a(t −τ)G(x,τ)dτ, (2.6.36)

(x,t)

(a(t + x) −a(t −x))

(r(t −x)G(x, x)

+ r(t + x)G(x,−x))

−x

r(t −τ)G

(x,τ) dτ. (2.6.37)

74 G.

Freiling and V. Yurko

Since a(0+) = a

(0−), it follows from (2.6.36) that

(x,t) =

(t + x)

+ a

(t −x)

+ G

(x,t)

−x

(t −τ)G(x,τ)dτ. (2.6.38)

Inte

gration

by parts yields

(x,t) =

(t +x)

+ a

(t −x)

) + G

(x,t)

−

(a(t −τ)G(x,τ)|

−x

+ a(t −τ)G(x,τ)|

)

−x

a(t −τ)G

(x,τ) dτ =

(t +x)

+ a

(t −x))

+ G

(x,t)

(a(t + x)G(x, −x) −a(t −x)G(x,x)

)

−x

(t −τ)G

(x,τ) dτ.

Inte

grating

by parts again and using (2.6.20) we calculate

(x,t) =

(t +x)

+ a

(t −x)

+ G

(x,t)

a(t + x)G(x, −x) −a(t −x)G(x,x)

−

r(t −τ)G

(x,τ)

−x

+ r(t −τ)G

(x,τ)

−x

r(t −τ)G

(x,τ

)dτ =

(t +x)

+ a

(t −x)

a(t + x)G(x, −x) −a(t −x)G(x,x)

r(t + x)G

(x,−x) −r(t −x)G

(x,x)

−x

r(t −τ)G

(x,τ) dτ. (

2.6.39)

Differentiating (2.6.37) we obtain

(x,t) =

(t + x)

+ a

(t −x)

a(t + x)G(x, −x) −a(t −x)G(x,x)

r(t + x)

G(x,−x)

+ r(t −x)

(x,x)

Hyperbolic P

artial Differential Equations 75

r(t + x)G

(x,−x)

+ r(t −x)G

(x,x)

−x

r(t −τ)G

(x,τ

)dτ.

Together with (2.6.35), (2.6.39) and (2.6.34) this yields

(x,t) −q(x)u(x,t) −u

(x,t) =

−x

r(t −τ)g(x,τ)dτ, (2.6.40)

−∞ < t < ∞, x ≥0,

where

g(x,t)

= G

(x,t) −G

(x,t) −q(x)G(x,t).

Let us show that

u(x,t) = 0, x > |t|. (2.6.41)

Indeed, it follows from (2.6.36) and (2.6.21) that u

(x,t) = 0 for x > |t|, and consequently

u(x,t) ≡C

(x) for x > |t|. Taking t = 0 in (2.6.35) we infer like above

(x) =

(r(x)

+ r(−x))

−x

r(−τ)G(x, τ)dτ = 0,

i.e.

(2.6.41)

holds.

It follows from (2.6.40) and (2.6.41) that

−x

r(t −τ)g(x, τ)dτ = 0, x > |t|. (2.6.42)

Dif

erentiating (2.6.42) with respect to t and taking (2.6.20) into account we deduce

(r(0+)g(x,t) −r(0−)g(x,t)

)

−x

a(t −τ)g(x,τ)dτ = 0,

g(x,t)

F(t,τ

)g(x,τ) dτ = 0.

According to Theorem 2.6.2 this homogeneous equation has only the trivial solution

g(x,t) = 0, i.e.

= G

−q(x)G, 0 < |t|< x. (2.6.43)

Furthermore, it follows from (2.6.38) for t = 0 and (2.6.41) that

0 =

(x)

+ a

(−x)

+ G

(x,0

) +

−x

(−τ)G(x,τ)dτ.

Since a

(x)

= −a

(−x), G(x,t)

= G(x, −t), we infer

∂G(x,t)

∂t

t=0

= 0. (2.6.44)

According

(2.6.34) the function G(x,t) satisﬁes also (2.6.18).

76 G.

Freiling and V. Yurko

It follo

ws from (2.6.40) and (2.6.43) that

(x,t) = u

(x,t) −q(x)u(x,t), −∞ < t < ∞,x ≥ 0.

Moreover, (2.6.35) and (2.6.37) imply (with h=G(0,0))

|x=0

= r(t), u

x|x=0

= hr(t).

Let us show that

u(x,x) = 1, x ≥ 0. (2.6.45)

Since the function G(x,t) satisﬁes (2.6.43), (2.6.44) and (2.6.18), we get according to

(2.6.17),

˜u(x,t) =

(˜r(t + x)

+ ˜r(t −x))

−x

˜r(t −τ)G(x,τ) dτ. (2.6.46)

Comparing

(2.6.35)

with (2.6.46) we get

ˆu(x,t) =

(ˆr(t + x)

+ ˆr(t −x))

−x

ˆr(t −τ)G(x,τ) dτ,

where ˆu = u − ˜u, ˆr = r − ˜r. Since

the

function ˆr(t) is continuous for −∞ < t < ∞, it

follows that the function ˆu(x,t) is also continuous for −∞ < t < ∞, x > 0. On the other

hand, according to (2.6.41), ˆu(x,t) = 0 for x > |t|, and consequently ˆu(x,x) = 0. By

(2.6.2), ˜u(x,x) = 1, and we arrive at (2.6.45).

Thus, the function u(x,t) is a solution of the boundary value problem (2.6.1)-(2.6.2).

By virtue of Theorem 2.6.1 we obtain u(x,t) = ˜u(x,t), and consequently r(t) = ˜r(t). The-

orem 2.6.4 is proved. 2

2.7. Inverse Spectral Problems

1. Inverse problems on a ﬁnite interval. Uniqueness theorems

Let us consider the boundary value problem L = L(q(x), h,H) :

`y := −y

+ q(x)y = λy, 0 < x < π, (2.7.1)

U(y) := y

(0) −hy(0) = 0, V (y) := y

(π) + Hy(π) = 0. (2.7.2)

Here λ is the spectral parameter; q(x),h and H are real; q(x) ∈L

(0,π). We shall subse-

quently refer to q as the potential. The operator ` is called the Sturm-Liouville operator.

In Section 2.2 we established properties of the eigenvalues and the eigenfunctions of L.

In this section we study inverse problems of spectral analysis for the Sturm-Liouville op-

erators. Inverse spectral problems of this type consist in recovering the potential and the

coefﬁcients of the boundary conditions from the given spectral characteristics. Such prob-

lems often appear in mathematics, mechanics, physics, geophysics and other branches of

natural sciences. Inverse problems also play an important role in solving nonlinear evolu-

tion equations of mathematical physics (see Section 2.9). There are close connections of

inverse spectral problems and inverse problems for equations of mathematical physics; this

Hyperbolic P

artial Differential Equations 77

is the

reason why we study this topic below in further details. We will use the notations and

facts from Section 2.2.

In this subsection we give various formulations of the inverse problems and prove the

corresponding uniqueness theorems. We present several methods for proving these theo-

rems. These methods have a wide area for applications and allow one to study various

classes of inverse spectral problems.

The ﬁrst result in the inverse problem theory is due to Ambarzumian [23]. Consider the

boundary value problem L(q(x),0, 0), i.e.

−y

+ q(x)y = λy, y

(0) = y

(π) = 0. (2.7.3)

Clearly, if q(x) = 0 a.e. on (0, π), then the eigenvalues of (2.7.3) have the form λ

, n ≥0. Ambarzumian proved the inverse assertion:

Theorem 2.7.1. If the eigenvalues of (2.7.3) are λ

= n

, n ≥ 0, then q(x) = 0 a.e.

on (0,π).

Proof. It follows from (2.2.49) that ω = 0, i.e.

q(x)dx = 0. Let y

(x) be an

eigenfunction for the ﬁrst eigenvalue λ

= 0. Then

(x) −q(x)y

(x) = 0, y

(0) = y

(π) = 0.

According to Sturm’s oscillation theorem, the function y

(x) has no zeros in the interval

x ∈ [0, π]. Taking into account the relation

(x)

get

0 =

(x)dx =

(x)

x =

(

(x)

Thus, y

(x) ≡ 0

, i.e. y

(x) ≡ const, q(x) = 0 a.e. on (0,π). 2

Remark 2.7.1. Actually we have proved a more general assertion than Theorem 2.7.1,

namely:

If λ

q(x)d

x, then q

(x) = λ

a.e. on (0,π).

Ambarzumian‘s result is an exception from the rule. In general, the speciﬁcation of the

spectrum does not uniquely determine the operator. Below we provide uniqueness theorems

where we point out spectral characteristics which uniquely determine the operator.

Consider the following inverse problem for L = L(q(x),h,H) :

Inverse Problem 2.7.1. Given the spectral data {λ

,α

}

n≥0

, construct the potential

q(x) and the coefﬁcients h and H of the boundary conditions.

Let us prove the uniqueness theorem for the solution of Inverse Problem 2.7.1. For

this purpose we agree that together with L we consider here and in the sequel a boundary

78 G.

Freiling and V. Yurko

value

problem

L = L( ˜q(x),

H) of the same form but with different coefﬁcients. If a

certain symbol γ denotes an object related to L, then the corresponding symbol

γ with

tilde denotes the analogous object related to

L, and

γ := γ −

γ.

Theorem 2.7.2. If λ

, α

, n ≥ 0, then L =

L, i.e. q(x) = ˜q(x) a.e. on

(0,π), h =

h and H =

H. Thus, the speciﬁcation of the spectral data {λ

,α

}

n≥0

uniquely

determines the potential and the coefﬁcients of the boundary conditions.

We give here two proofs of Theorem 2.7.2. The ﬁrst proof is due to Marchenko [15]

and uses the transformation operator and Parseval‘s equality (2.2.60). The second proof is

due to Levinson [24] and uses the contour integral method.

Marchenko‘s proof. Let ϕ(x,λ) be the solution of (2.7.1) under the initial conditions

ϕ(0,λ) = 1, ϕ

(0,λ) = h. In Section 2.6 we obtained the following representation for the

solution ϕ(x,λ) :

ϕ(x,λ) = cos ρx +

G(x,t) cos ρt dt, (2.7.4)

where λ = ρ

, G(x,t) is a real continuous function, and (2.6.18) holds. Similarly,

ϕ(x,λ) = cos ρx +

G(x,t) cos ρt dt.

In other words,

ϕ(x,λ) = (E + G)cos ρx,

ϕ(x,λ) = (E +

G)cos ρx,

where

(E + G) f (x) = f (x) +

G(x,t) f (t)dt,

(E +

G) f (x) = f (x) +

G(x,t) f (t)dt.

One can consider the relation

ϕ(x,λ) = (E +

G)cos ρx as a Volterra integral equation (see

[14] for the theory of integral equations) with respect to cos ρx. Solving this equation we

get

cosρx =

ϕ(x,λ) +

H(x,t)

ϕ(t, λ)dt,

where

H(x,t) is a continuous function which is the kernel of the inverse operator:

(E +

H) = (E +

−1

H f (x) =

H(x,t) f (t) dt.

Consequently

ϕ(x,λ) =

ϕ(x,λ) +

Q(x,t)

ϕ(t, λ)dt, (2.7.5)

where Q(x,t) is a real continuous function.

Let f (x) ∈ L

(0,π). It follows from (2.7.5) that

f (x)ϕ(x, λ)dx =

g(x)

ϕ(x,λ) dx,

Hyperbolic P

artial Differential Equations 79

where

g(x) = f (

x) +

Q(t, x) f (t)dt.

Hence, for all n ≥ 0,

f (x)ϕ(x,λ

)dx,

g(x)

ϕ(x,λ

)dx.

Using Parseval‘s equality (2.2.60), we calculate

|f (x)|

dx =

∞

∑

n=0

∞

∑

n=0

∞

∑

n=0

|g(x)|

i.e.

kf k

= k

. (2.7.6)

Consider the operator

A f (x) = f (x) +

Q(t, x) f (t)dt.

Then A f = g. By virtue of (2.7.6), kA f k

= kf k

for any f (x) ∈ L

(0,π). Conse-

quently, A

∗

= A

−1

, but this is possible only if Q(x,t) ≡ 0. Thus, ϕ(x,λ) ≡

ϕ(x,λ), i.e.

q(x) = ˜q(x) a.e. on (0,π), h =

h, H =

Levinson‘s proof. Let f (x), x ∈[0,π] be an absolutely continuous function. Consider

the function

(x,λ) = −

∆(λ)

ψ(x,λ)

f (t)

ϕ(t, λ)d

t + ϕ(x, λ

)

f (t)

ψ(t, λ)dt

and the contour integral

(x) =

2πi

(x,λ) dλ.

The

idea

used here comes from the proof of Theorem 2.2.7 but here the function Y

(x,λ)

is constructed from solutions of two boundary value problems.

Repeating the arguments of the proof of Theorem 2.2.7 we calculate

(x,λ) =

f (x)

−

(x,λ)

where

(x,λ)

∆(λ)

f (x)

[ϕ

(x,λ)(

(x,λ) −ψ

(x,λ))

−ψ(x,λ)(

(x,λ) −ϕ

(x,λ))] +

h f (0)ψ(x,λ) +

H f (π)ϕ(x,λ)

+ψ(x,λ)

(

(t, λ) f

(t) + ˜q(t)

ϕ(t, λ) f (t))dt

+ϕ(x,λ)

(

(t, λ) f

(t) + ˜q(t)

ψ(t, λ) f (t))dt

80 G.

Freiling and V. Yurko

The asymptotic

properties for

ϕ(x,λ) and

ψ(x,λ) are the same as for ϕ(x,λ) and ψ(x, λ).

Therefore, by similar arguments as in the proof of Theorem 2.2.7 one can obtain

(x) = f (x) + ε

(x), lim

N→∞

max

0≤x≤π

|ε

(x)| = 0.

On the other hand, we can calculate I

(x) with the help of the residue theorem:

(x) =

∑

n=0

−

∆(λ

)

´³

ψ(x,λ

)

f (t)

ϕ(t, λ

+ϕ(x,λ

)

f (t)

(t, λ

)dt

It follows from Lemma 2.2.1 and Theorem 2.2.6 that under the hypothesis of Theorem 2.7.2

we have β

. Consequently,

(x) =

∑

n=0

ϕ(x,λ

)

f (t)

ϕ(t, λ

If N → ∞ we

get

f (x) =

∞

∑

n=0

ϕ(x,λ

)

f (t)

ϕ(t, λ

ogether with (2.2.59) this gives

f (t)(ϕ(t, λ

) −

ϕ(t, λ

))dt = 0.

Since f (x) is an arbitrary absolutely continuous function we conclude that

ϕ(x,λ

) =

ϕ(x,λ

) for all n ≥ 0 and x ∈ [0,π]. Consequently, q(x) = ˜q(x) a.e. on

(0,π), h =

h, H =

H. 2

The Weyl function. Let Φ(x, λ) be the solution of (2.7.1) under the conditions

U(Φ) = 1, V (Φ) = 0. We set M(λ) := Φ(0, λ). The functions Φ(x,λ) and M(λ) are

called the Weyl solution and the Weyl function (or Weyl-Titchmarsh function) for the bound-

ary value problem L, respectively. The Weyl function was introduced ﬁrst (for the case of

the half-line) by H. Weyl. For further discussions on the Weyl function see, for example,

[25]. Clearly,

Φ(x,λ) = −

ψ(x,λ)

∆(λ)

= S(x,λ)

+ M(λ)

ϕ(x,λ), (2.7.7)

hϕ(x,λ), Φ(x,λ)i ≡ 1, (2.7.8)

where S(x,λ) is the solution of (2.7.1) under the initial conditions S(0, λ) = 0, S

(0,λ) = 1.

In particular, for x = 0 this yields

M(λ) = −

∆

(λ)

∆(λ)

, (2.7.9)

where ∆

(λ) := ψ(0,λ) is

the

characteristic function of the boundary value problem

= L

(q(x),H) for equation (2.7.1) with the boundary conditions y(0) = V (y) = 0. The

Hyperbolic P

artial Differential Equations 81

eigenv

alues {λ

}

n≥0

of L

are simple and coincide with zeros of ∆

(λ). Moreover, simi-

larly to (2.2.57) one can get

∆

(λ) =

∞

∏

n=0

−λ

(n + 1/2)

Thus,

the

Weyl function is meromorphic with simple poles in {λ

}

n≥0

and with simple

zeros in {λ

}

n≥0

Theorem 2.7.3. The following representation holds

M(λ) =

∞

∑

n=0

(λ −λ

)

. (2.7.10)

oof

. Since ∆

(λ) = ψ(0, λ), it follows from (2.2.46) that |∆

(λ)| ≤

C exp(|τ|π). Then, using (2.7.9) and (2.2.53), we get for sufﬁciently large ρ

∗

> 0,

|M(λ)| ≤

|ρ|

, ρ ∈G

, |ρ|

≥ ρ

∗

. (

2.7.11)

Further, using (2.7.9) and Lemma 2.2.1, we calculate

Res

λ=λ

M(λ) = −

∆

(λ

)

∆(λ

)

= −

∆(λ

)

. (2.7.12)

Consider

the

contour integral

(λ) =

2πi

M(µ)

λ −µ

, λ ∈int Γ

By virtue of (2.7.11), lim

N→∞

(λ) = 0. On the other hand, the residue theorem and (2.7.12)

yield

(λ) = −M(λ)+

∑

n=0

(λ −λ

)

and

Theorem

2.7.3 is proved. 2

We consider the following inverse problem:

Inverse Problem 2.7.2. Given the Weyl function M(λ), construct

L(q(x),h, H).

Let us prove the uniqueness theorem for Inverse Problem 2.7.2.

Theorem 2.7.4. If M(λ) =

M(λ), then L =

L. Thus, the speciﬁcation of the Weyl

function uniquely determines the operator.

Proof. Let us deﬁne the matrix P(x, λ) = [P

(x,λ)]

j,k=1,2

by the formula

P(x,λ)

ϕ(x,λ)

Φ(x,λ)

(x,λ)

ϕ(x,λ) Φ(x,λ)

(x,λ) Φ

(x,λ)

. (2.7.13)