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

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32 G.

Freiling and V. Yurko

where

the numbers β

are deﬁned by (2.2.42) , and

∆(λ) =

dλ

∆(λ).

oof

. Since

−ψ

(x,λ) + q(x)ψ(x,λ) = λψ(x,λ),

−ϕ

(x,λ

) + q(x)ϕ(x,λ

) = λ

ϕ(x,λ

we get

ψ(x,λ)ϕ

(x,λ

) −ψ

(x,λ)ϕ(x, λ

) = (λ −λ

)ψ(x,λ)ϕ(x, λ

and consequently,

hψ(x,λ

),ϕ(x, λ

)i = (λ −λ

)ψ(x,λ)ϕ(x, λ

After integration we obtain

(λ −λ

)

ψ(x,λ)ϕ(x, λ

)dx = hψ(x, λ),ϕ(x,λ

and hence with the help of (2.2.41),

(λ −λ

)

ψ(x,λ)ϕ(x, λ

)dx

= ϕ

(π,λ

) + Hϕ(π,λ

) + ψ

(0,λ) −hψ(0,λ) = −∆(λ).

For λ → λ

, this yields

ψ(x,λ

)ϕ(x,λ

)dx = −

∆(λ

Using (2.2.42) and (2.2.43) we arrive at (2.2.44). 2

Theorem 2.2.4.The eigenvalues λ

and the eigenfunctions ϕ(x,λ

) and ψ(x,

) of L are real. All zeros of ∆(λ) are simple, i.e.

∆(λ

) 6= 0. Eigenfunctions related to

different eigenvalues are orthogonal in L

(0,π).

Proof. Let λ

and λ

( λ

6= λ

) be eigenvalues with eigenfunctions y

(x) and y

(x)

respectively. Then integration by parts yields

(x)y

(x)dx = −

(x)y

(x)dx +

q(x)y

(x)y

(x)dx

(x)y

(x) −y

(x)y

(x)

−

(x)y

(x)dx +

q(x)y

(x)y

(x)dx.

The substitution vanishes since the eigenfunctions y

(x) and y

(x) satisfy the boundary

conditions (2.2.38). Therefore,

(x)y

(x)dx =

(x)`y

(x)dx,

Hyperbolic P

artial Differential Equations 33

and hence

(

x)y

(x)dx = λ

(x)y

(x)dx,

(x)y

(x)dx = 0.

This proves that eigenfunctions related to distinct eigenvalues are orthogonal in L

(0,π).

Further, let λ

= u+iv, v 6= 0 be a non-real eigenvalue with an eigenfunction y

(x) 6= 0.

Since q(x),h and H are real, we get that

= u −iv is

also

the eigenvalue with the

eigenfunction

(x). Since λ

6= λ

, we

deri

ve as before

(x)

(x)d

x = 0

which is impossible. Thus, all eigenvalues {λ

} of L are real, and consequently the eigen-

functions ϕ(x,λ

) and ψ(x,λ

) are real too. Since α

6= 0, β

6= 0, we get by virtue of

(2.2.44) that

∆(λ

) 6= 0. 2

Example. Let q(x) = 0, h = 0 and H = 0, and let λ = ρ

. Then one can check easily

that

ϕ(x,λ) = cos ρx, ψ(x,λ) = cos ρ(π −x),

∆(λ) = −ρ sinρπ, λ

= n

(n ≥ 0), ϕ(x,λ

) = cos nx, β

= (−1)

Lemma 2.2.2. For |ρ| → ∞, the following asymptotic formulae hold

ϕ(x,λ) = cos ρx + O

|ρ|

xp(|τ

|x)

(x,λ) = −ρsin ρx + O(exp(|τ|x)),











(2.2.45)

ψ(x,λ) = cos ρ(π −x) + O

|ρ|

xp(|τ

|(π −x))

(x,λ) = ρsinρ(π −x) + O(exp(|τ|(π −x))),











(2.2.46)

uniformly with respect to x ∈[0,π]. Here and in the sequel, λ = ρ

, τ = Imρ, and o and

O denote the Landau symbols (see [13], §4).

Proof. Let us show that

ϕ(x,λ) = cos ρx + h

sinρx

sinρ(x −t)

q(t)ϕ(t,λ)d

t. (

2.2.47)

Indeed, the Volterra integral equation

y(x,λ) = cos ρx + h

sinρx

sinρ(x −t)

q(t)y(t,λ)d

has

a unique solution (see [14]). On the other hand, if a certain function y(x, λ) satisﬁes

this equation, then we get by differentiation

(x,λ) = −ρsin ρx +hcos ρx +

cosρ(x −t)q(t)y(t,λ)dt,

34 G.

Freiling and V. Yurko

(x,λ) = −ρ

cosρ

x −hρsin ρx + q(x)y(x, λ)

−

ρsin ρ(x −t)q(t)y(t,λ)dt,

and consequently,

(x,λ) + ρ

y(x,λ) = q(x)y(x,λ),

y(0,λ) = 1, y

(0,λ) = h,

i.e. y(x,λ) ≡ ϕ(x,λ), and (2.2.47) is valid.

Differentiating (2.2.47) we calculate

(x,λ) = −ρsin ρx + hcos ρx +

cosρ(x −t)q(t)ϕ(t,λ)dt. (2.2.48)

Denote

µ(λ) = max

0≤x≤π

(|ϕ(x,λ)|exp(−|τ|x)).

Since |sin ρx| ≤ exp(|τ|x) and |cosρx| ≤ exp(|τ|x), we have from (2.2.47) that for |ρ| ≥

1, x ∈[0,π],

|ϕ(x,λ)|exp(−|τ|x) ≤ 1 +

|ρ|

h + µ(λ)

|q(t)|d

≤C

|ρ|

µ(λ),

and

consequently

(λ) ≤C

|ρ|

µ(λ)

µ(λ) ≤C

1 +

|ρ|

−1

sufﬁciently large |ρ|, this yields µ(λ) = O(1), i.e. ϕ(x, λ) = O(exp(|τ|x)). Substi-

tuting this estimate into the right-hand sides of (2.2.47) and (2.2.48), we arrive at (2.2.45).

Similarly one can derive (2.2.46).

We note that (2.2.46) can be also obtained directly from (2.2.45). Indeed, since

−ψ

(x,λ) + q(x)ψ(x,λ) = λψ(x,λ),

ψ(π,λ) = 1, ψ

(π,λ) = −H,

the function

ϕ(x,λ) := ψ(π −x,λ) satisﬁes the following differential equation and the ini-

tial conditions

−

(x,λ) + q(π −x)

ϕ(x,λ) = λ

ϕ(x,λ),

ϕ(0,λ) = 1,

(0,λ) = H.

Therefore, the asymptotic formulae (2.2.45) are also valid for the function

ϕ(x,λ). From

this we arrive at (2.2.46). 2

Theorem 2.2.5. The boundary value problem L has a countable set of eigenvalues

{λ

}

n≥0

. For n → ∞,

= n +

πn

, {κ

}

∈ l

, (

2.2.49)

Hyperbolic P

artial Differential Equations 35

ϕ(x,λ

)

= cos nx +

(x)

, |ξ

(x)|

≤C, (

2.2.50)

where

ω = h + H +

q(t) d

Here

and everywhere below one and the same symbol {κ

} denotes various sequences

from l

, and the symbol C denotes various positive constants which do not depend on x,λ

and n. We remind that {κ

} ∈ l

means

∑

|κ

< ∞ (see e.g. [13]).

Proof. 1) Substituting the asymptotics for ϕ(x,λ) from (2.2.45) into the right-hand

sides of (2.2.47) and (2.2.48), we calculate

ϕ(x,λ) = cos ρx + q

(x)

sinρx

2ρ

q(t) sin ρ(x −2t)d

t + O

xp(|τ|x)

(x,λ)

= −ρ sin ρx + q

(x)cos ρx +

q(t) cos ρ(x −2t)d

t + O

xp(|τ|x)

where

(x)

= h +

q(t) d

According

to (2.2.41), ∆(λ) = ϕ

(π,λ) + Hϕ(π,λ). Hence,

∆(λ) = −ρ sinρπ + ωcos ρπ + κ(ρ), (2.2.51)

where

κ(ρ) =

q(t) cos ρ(π −2t)d

t + O

xp(|τ

|π)

Denote

= {ρ : |ρ −k| ≥ δ, k = 0,±1, ±2,. ..}, δ > 0.

Let us show that

|sin ρπ| ≥C

exp(|τ|π), ρ ∈ G

, (2.2.52)

|∆(λ)| ≥C

|ρ|exp(|τ|π), ρ ∈G

,|ρ| ≥ ρ

∗

, (2.2.53)

for sufﬁciently large ρ

∗

= ρ

∗

(δ).

Let ρ = σ + iτ. It is sufﬁcient to prove (2.2.52) for the domain

= {ρ : σ ∈

−

, τ ≥0, |ρ|

≥ δ}

Denote

θ(ρ) = |sinρπ|exp(−|τ|π).

Let ρ ∈D

. For τ ≤ 1, θ(ρ) ≥C

. Since

sinρπ =

exp(iρπ)−exp(−iρπ)

36 G.

Freiling and V. Yurko

we ha

ve for τ ≥1 ,

θ(ρ) =

1 −e

xp(2

iσπ)exp(−2τπ)

≥

Thus,

(2.2.52)

is proved. Further, using (2.2.51) we get for ρ ∈G

∆(λ) = −ρ sinρπ

1 + O

¶¶

and

consequently

(2.2.53) is valid.

3) Denote

= {λ : |λ| = (n + 1/2)

By virtue of (2.2.51),

∆(λ) = f (λ) + g(λ), f (λ) = −ρsinρπ, |g(λ)|≤C exp(|τ|π).

According to (2.2.52), |f (λ)| > |g(λ)|, λ ∈ Γ

, for sufﬁciently large n (n ≥ n

∗

). Then

by Rouch

e‘s theorem, the number of zeros of ∆(λ) inside Γ

coincides with the number of

zeros of f (λ) = −ρ sinρπ, i.e. it equals n + 1. Thus, in the circle |λ| < (n + 1/2)

there

exist exactly n + 1 eigenvalues of L : λ

,. .. ,λ

. Applying now Rouch

e‘s theorem to the

circle γ

(δ) = {ρ : |ρ −n| ≤ δ}, we conclude that for sufﬁciently large n, in γ

(δ) there

is exactly one zero of ∆(ρ

), namely ρ

√

. Since δ > 0 is

arbitrary

, it follows that

= n + ε

, ε

= o(1), n → ∞. (2.2.54)

Substituting (2.2.54) into (2.2.51) we get

0 = ∆(ρ

) = −(n + ε

)sin(n + ε

)π + ωcos(n + ε

)π + κ

and consequently

−nsin ε

π + ωcos ε

π + κ

= 0. (2.2.55)

Then

sinε

π = O

i.e.

= O

Using

(2.2.55)

once more we obtain more precisely

πn

i.e.

(2.2.49)

is valid. Using (2.2.49) we arrive at (2.2.50), where

(x) =

h +

q(t) d

t −x

−xκ

sinnx

q(t) sin n(x −2t)d

t + O

Hyperbolic P

artial Differential Equations 37

Consequently |ξ

(x)| ≤C, and

Theorem 2.2.5 is proved. 2

By virtue of (2.2.42) with x = π,

= (ϕ(π, λ

))

−1

Then, using (2.2.43), (2.2.44) and (2.2.50) one can calculate

, β

(−1

)

. (2.2.56)

Since ∆(λ) has

only

simple zeros, we have

sign

∆(λ

) = (−1)

n+1

for n ≥ 0.

Theorem 2.2.6. The speciﬁcation of the spectrum {λ

}

n≥0

uniquely determines the

characteristic function ∆(λ) by the formula

∆(λ) = π(λ

−λ)

∞

∏

n=1

−λ

. (2.2.57)

oof

. It follows from (2.2.51) that ∆(λ) is entire in λ of order 1/2, and consequently

by Hadamard‘s factorization theorem, ∆(λ) is uniquely determined up to a multiplicative

constant by its zeros:

∆(λ) = C

∞

∏

n=0

1 −

(2.2.58)

(the

case

when ∆(0) = 0 requires minor modiﬁcations). Consider the function

∆(λ) := −ρ sinρπ = −λπ

∞

∏

n=1

1 −

Then

∆(λ)

= C

λ −λ

πλ

∞

∏

n=1

∞

∏

n=1

1 +

−n

−λ

aking

(2.2.49) and (2.2.51) into account we calculate

lim

λ→−∞

∆(λ)

= 1,

lim

λ→−∞

∞

∏

n=1

1 +

−n

−λ

= 1,

and

hence

C = πλ

∞

∏

n=1

Substituting

this

into (2.2.58) we arrive at (2.2.57). 2

38 G.

Freiling and V. Yurko

Now

we are going to prove that the system of the eigenfunctions of the Sturm-Liouville

boundary value problem L is complete and forms an orthogonal basis in L

(0,π). This the-

orem was ﬁrst proved by Steklov at the end of XIX-th century. We also provide sufﬁcient

conditions under which the Fourier series for the eigenfunctions converges uniformly on

[0,π]. The completeness and expansion theorems are important for solving various prob-

lems in mathematical physics by the Fourier method, and also for the spectral theory itself.

In order to prove these theorems we apply the contour integral method of integrating the

resolvent along expanding contours in the complex plane of the spectral parameter (since

this method is based on Cauchy‘s theorem, it sometimes called Cauchy‘s method).

Theorem 2.2.7. (i) The system of eigenfunctions {ϕ(x, λ

)}

n≥0

of the boundary value

problem L is complete in L

(0,π) .

(ii) Let f (x), x ∈[0,π] be an absolutely continuous function. Then

f (x) =

∞

∑

n=0

ϕ(x,λ

), a

f (t)ϕ(t, λ

t, (

2.2.59)

and the series converges uniformly on [0,π].

(iii) For f (x) ∈ L

(0,π), the series (2.2.59) converges in L

(0,π), and

|f (x)|

dx =

∞

∑

n=0

(Parseval‘s equality) . (2.2.60)

Proof. 1) Denote

G(x,t,λ) = −

∆(λ)







ϕ(x,λ)ψ(t,λ), x ≤t,

ϕ(t, λ)ψ(x,λ), x ≥t,

and

consider

the function

Y (x,λ) =

G(x,t, λ) f (t)dt

= −

∆(λ)

ψ(x,λ)

ϕ(t, λ) f (t)d

t + ϕ

(x,λ)

ψ(t, λ) f (t)dt

The function G(x,t,λ) is called Green‘s function for L. G(x,t,λ) is the kernel of the in-

verse operator for the Sturm-Liouville operator, i.e. Y (x,λ) is the solution of the boundary

value problem

`Y −λY + f (x) = 0,

U(Y ) = V (Y ) = 0;

)

(2.2.61)

this is easily veriﬁed by differentiation. Taking (2.2.42) into account and using Theorem

2.2.4 we calculate

Res

λ=λ

Y (x,λ) = −

∆(λ

)

ψ(x,λ

)

ϕ(t, λ

) f (t) d

+ϕ

(x,λ

)

ψ(t, λ

) f (t) dt

Hyperbolic P

artial Differential Equations 39

and consequently

Res

λ=λ

Y (x,λ) = −

∆(λ

)

ϕ(x,λ

)

f (t)ϕ(t, λ

virtue of (2.2.44),

Res

λ=λ

Y (x,λ) =

ϕ(x,λ

)

f (t)ϕ(t, λ

t. (

2.2.62)

2) Let f (x) ∈ L

(0,π) be such that

f (t)ϕ(t, λ

)dt = 0, n ≥ 0.

Then in view of (2.2.62),

Res

λ=λ

Y (x,λ) = 0,

and consequently (after extending Y (x,λ) continuously to the whole λ - plane) for each

ﬁxed x ∈ [0,π], the function Y (x,λ) is entire in λ. Furthermore, it follows from (2.2.45),

(2.2.46) and (2.2.53) that for a ﬁxed δ > 0 and sufﬁciently large ρ

∗

> 0 :

|Y (x, λ)| ≤

|ρ|

, ρ ∈G

, |ρ|

≥ ρ

∗

Using

the maximum principle and Liouville‘s theorem we conclude that Y (x,λ) ≡0. From

this and (2.2.61) it follows that f (x) = 0 a.e. on (0,π). Thus (i) is proved.

3) Let now f ∈ AC[0,π] be an arbitrary absolutely continuous function. Since ϕ(x,λ)

and ψ(x,λ) are solutions of (2.2.37), we transform Y (x, λ) as follows

Y (x,λ) = −

λ∆(λ)

ψ(x,λ)

(−ϕ

(t,λ)

+ q

(t)ϕ(t,λ)) f (t)dt

+ϕ(x,λ)

(−ψ

(t,λ) + q(t)ψ(t,λ)) f (t)dt

Integration of the terms containing second derivatives by parts yields in view of (2.2.40),

Y (x,λ) =

f (x)

−

(x,λ)

+ Z

(x,λ)

, (

2.2.63)

where

(x,λ) =

∆(λ)

ψ(x,λ)

g(t)ϕ

(t, λ)d

+ϕ(x,λ

)

g(t)ψ

(t, λ)dt

, g(t) := f

(t),

(x,λ) =

∆(λ)

f (0

)ψ(x,λ) + H f (π)ϕ(x, λ)

+ψ(x,λ)

q(t)ϕ(t,λ) f (t))dt + ϕ(x,λ)

q(t)ψ(t,λ) f (t))dt

40 G.

Freiling and V. Yurko

Using (2.2.45),

(2.2.46) and (2.2.53), we get for a ﬁxed δ > 0, and sufﬁciently large ρ

∗

0 :

max

0≤x≤π

(x,λ)| ≤

|ρ|

, ρ ∈G

, |ρ|

≥ ρ

∗

. (

2.2.64)

Let us show that

lim

|ρ|→∞

ρ∈G

max

0≤x≤π

(x,λ)| = 0. (2.2.65)

First we assume that g(x) is absolutely continuous on [0,π]. In this case another integra-

tion by parts yields

(

) =

∆(λ)

(

)

(

)

(

)

(

)

(

)

(

)

−ψ(x,λ)

(t)ϕ(t,λ)d

t −ϕ

(x,λ)

(t)ψ(t,λ)dt

By virtue of (2.2.45), (2.2.46) and (2.2.53), we infer

max

0≤x≤π

(x,λ)| ≤

|ρ|

, ρ ∈G

, |ρ|

≥ ρ

∗

Let

now g(t) ∈ L(0, π). Fix ε > 0 and choose an absolutely continuous function g

(t)

such that

|g(t) −g

(t)|dt <

where

= max

0≤x≤π

sup

ρ∈G

|∆(λ)|

|ψ(x,λ)|

|ϕ

(t, λ)|d

+|ϕ

(x,λ)|

|ψ

(t, λ)|dt

Then, for ρ ∈ G

, |ρ| ≥ ρ

∗

, we have

max

0≤x≤π

(x,λ)| ≤ max

0≤x≤π

(x,λ; g

+ max

0≤x≤π

(x,λ; g −g

)| ≤

C(ε)

|ρ|

Hence,

there

exists ρ

> 0 such that

max

0≤x≤π

(x,λ)| ≤ ε

for |ρ| > ρ

. By virtue of the arbitrariness of ε > 0, we arrive at (2.2.65).

Consider the contour integral

(x) =

2πi

Y (x,λ) dλ,

Hyperbolic P

artial Differential Equations 41

where Γ

= {λ : |λ| = (N + 1

/2)

} (with counterclockwise circuit). It follows from

(2.2.63)-(2.2.65) that

(x) = f (x) + ε

(x), lim

N→∞

max

0≤x≤π

|ε

(x)| = 0. (2.2.66)

On the other hand, we can calculate I

(x) with the help of the residue theorem. By virtue

of (2.2.62),

(x) =

∑

n=0

ϕ(x,λ

where

f (t)ϕ(t, λ

Comparing

this with (2.2.66) we arrive at (2.2.59), where the series converges uniformly on

[0,π], i.e. (ii) is proved.

4) Since the eigenfunctions {ϕ(x,λ

)}

n≥0

are complete and orthogonal in

(0,π), they form an orthogonal basis in L

(0,π), and Parseval‘s equality (2.2.60) is

valid. Theorem 2.2.7 is proved. 2

We note that the assertions of Theorem 2.2.7 remain valid also if h

= 0 and/or H

= 0

provided f (0) = 0 and/or f (π) = 0 .

Let us show that the Sturm-Liouville problem (2.2.34)-(2.2.35) can be reduced to the

case (2.2.36).

Step 1. It follows from (2.2.34) that

−k(x)Y

(x) −k

(x)Y

(x) + q(x)Y (x) = λρ(x)Y (x),

and consequently,

−Y

(x) + p(x)Y

(x) + q

(x)Y (x) = λr(x)Y (x), 0 < x < l, (2.2.67)

where

p(x) = −

(x)

k(x)

, q

(x)

k(x)

, r(x)

(x)

k(x)

> 0.

make the substitution

Y (x) = exp

p(s)d

x). (2.2.68)

Differentiating (2.2.68) we calculate

(x) = exp

p(s)d

(x) +

p(x)

Z(x)

(x)

= e

p(s)d

(x)

+ p(x)Z

(x) +

(x)

Z(x)