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

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22 G.

Freiling and V. Yurko

2.2. The

Mixed Problem for the Equation of the Vibrating

String

This section deals with mixed problems for the equation of the vibrating string

on a ﬁnite interval. For studying such problems we use the method of separa-

tion of variables, which is also called the method of standing waves. Subsections

1-3 are related to the standard level for a course of partial differential equations of math-

ematical physics. Subsection 4 describes a general scheme of separation of variables, and

it contains more complicated material. This subsection can be recommended for advanced

studies and can be omitted ”in the ﬁrst reading”.

1. Homogeneous equation

We consider the following mixed problem

= a

, 0 < x < l, t > 0, (2.2.1)

|x=0

= u

|x=l

= 0, (2.2.2)

|t=0

= ϕ(x), u

t|t=0

= ψ(x). (2.2.3)

This problem describes the oscillation process for a ﬁnite string of the length l, ﬁxed at

the ends x = 0 and x = l, provided that the initial displacement ϕ(x) and the initial speed

ψ(x) are known. Other mixed problems (including physical motivations) can be found in

Section 6.2.

0 l

u(x,t)

Figure

2.2.1.

Denote D = {

(x,t) : 0 ≤x ≤l, t ≥0}.

Deﬁnition 2.2.1. A function u(x,t) is called a solution of problem (2.2.1)-(2.2.3) if

u(x,t) ∈C

(D), and u(x,t) satisﬁes (2.2.1)-(2.2.3).

First we consider the following auxiliary problem. We will seek nontrivial (i.e. not

identically zero) particular solutions of equation (2.2.1) such that they satisfy the boundary

conditions (2.2.2) and admit separation of variables, i.e. they have the form

u(x,t) = Y (x)T (t). (2.2.4)

Hyperbolic P

artial Differential Equations 23

For

this purpose we substitute (2.2.4) into (2.2.1) and (2.2.2). By virtue of (2.2.1) we have

Y (x)

T (t) = a

(x)T (t).

Here and below, ”

” denotes differentiation with respect to x, and ” ·” denotes differenti-

ation with respect to t. Separating the variables, we obtain

(x)

Y (x)

T (t)

Since x and t are

independent

variables, the equality is here valid only if both fractions

are equal to a constant; we denote this constant by −λ . Therefore,

(x)

Y (x)

T (t)

= −λ. (2.2.5)

Moreo

er, the boundary conditions (2.2.2) yield

0 = u(0,t) = Y (0)T (t),

0 = u(l,t) = Y (l)T (t).

)

(2.2.6)

It follows from (2.2.5) and (2.2.6) that

T (t) + a

λT (t) = 0, (2.2.7)

(x) + λY (x) = 0, (2.2.8)

Y (0) = Y (l) = 0. (2.2.9)

Thus, the function T (t) is a solution of the ordinary differential equation (2.2.7), and the

function Y (x) is a solution of the boundary value problem (2.2.8)-(2.2.9) which is called

the associated Sturm-Liouville problem. We are interested in nontrivial solutions of prob-

lem (2.2.8)-(2.2.9), but it turns out that they exist only for some particular values of the

parameter λ.

Deﬁnition 2.2.2. The values of the parameter λ for which the problem (2.2.8)-(2.2.9)

has nonzero solutions are called eigenvalues, and the corresponding nontrivial solutions are

called eigenfunctions. The set of eigenvalues is called the spectrum of the Sturm-Liouville

problem.

Clearly, the eigenfunctions are deﬁned up to a multiplicative constant, since if Y (x) is a

nontrivial solution of the problem (2.2.8)-(2.2.9) for λ = λ

(i.e. Y (x) is an eigenfunction

corresponding to the eigenvalue λ

), then the function CY (x) , C−const, is also a solution

of problem (2.2.8)-(2.2.9) for the same value λ = λ

Let us ﬁnd the eigenvalues and eigenfunctions of problem (2.2.8)-(2.2.9). Let λ = ρ

The general solution of equation (2.2.8) for each ﬁxed λ has the form

Y (x) = A

sinρx

+ B cosρx (2.2.10)

24 G.

Freiling and V. Yurko

(in particular

, for ρ = 0 (2.2.10) yields Y (x) = Ax + B ). Since Y (0) = 0, in (2.2.10)

B = 0, i.e.

Y (x) = A

sinρx

. (2.2.11)

Substituting

(2.2.11)

into the boundary condition Y (l) = 0, we obtain

sinρl

= 0.

Since

seek nontrivial solutions, A 6= 0, and we arrive at the following equation for the

eigenvalues:

sinρl

= 0. (2.2.12)

The

roots

of equation (2.2.12) are ρ

nπ

and

consequently, the eigenvalues of problem

(2.2.8)-(2.2.9) have the form

nπ

, n = 1,2,3,

.. , (2.2.13)

and the corresponding eigenfunctions, by virtue of (2.2.11), are

(x) = sin

nπ

x, n = 1,2,3,

.. (2.2.14)

(up to a multiplicative constant). Since nontrivial solutions of problem (2.2.8)-(2.2.9) exist

only for λ = λ

of the form (2.2.13), it makes sense to consider equation (2.2.7) only for

λ = λ

(t) + a

(t) = 0.

The general solution of this equation has the form

(t) = A

sin

anπ

t + B

cos

anπ

t, n ≥1, (2.2.15)

where A

and B

are

arbitrary

constants. Substituting (2.2.14) and (2.2.15) into (2.2.4) we

obtain all solutions of the auxiliary problem:

(x,t) =

sin

anπ

t + B

cos

anπ

sin

nπ

x, n = 1,2,3,

.. (2.2.16)

The solutions of the form (2.2.16) are called standing waves (normal modes of vibrations).

We will seek the solution of the mixed problem (2.2.1)-(2.2.3) by superposition of

standing waves (2.2.16), i.e. as a formal series solution:

u(x,t) =

∞

∑

n=1

sin

anπ

t + B

cos

anπ

sin

nπ

x. (2.2.17)

construction

the function u(x,t) satisﬁes formally equation (2.2.1) and the boundary

conditions (2.2.2) for all A

and B

. Choose A

and B

such that u(x,t) satisﬁes also

the initial conditions (2.2.3). Substituting (2.2.17) into (2.2.3) we calculate

ϕ(x) =

∞

∑

n=1

sin

nπ

Hyperbolic P

artial Differential Equations 25

ψ(x) =

∞

∑

anπ

sin

nπ

Using

the

formulae for the Fourier coefﬁcients [4, Chapter 1], we obtain

ϕ(x)sin

nπ

anπ

ψ(x)sin

nπ











(2.2.18)

Let us now show that the function u(x,t), deﬁned by (2.2.17)-(2.2.18), is the solution of

the mixed problem (2.2.1)-(2.2.3). For this purpose we need some conditions on ϕ(x) and

ψ(x).

Theorem 2.2.1. Let ϕ(x) ∈C

[0,l] , ψ(x) ∈C

[0,l] , ϕ(0) = ϕ(l) = ϕ

(0) = ϕ

(l) =

ψ(0) = ψ(l) = 0. Then the function u(x,t), deﬁned by (2.2.17)−(2.2.18) , is the solution

of the mixed problem (2.2.1) −(2.2.3) .

Proof. From the theory of the Fourier series [11, Chapter 1] we know the following

facts:

1) The system of functions

sin

nπ

n≥1

complete

and orthogonal in L

(0,l).

2) Let f (x) ∈ L

(0,l), and let

f (x) sin

nπ

f (x) cos

nπ

be the Fourier coefﬁcients for the function f (x). Then

∑

|α

+ |β

< ∞. (2.2.19)

Moreover, we note:

If numbers γ

are such that

∑

|γ

< ∞, then

∑

|γ

< ∞. (2.2.20)

This

follo

ws from the obvious inequality

|γ

≤

+ |γ

Let

show that

∞

∑

n=1

(|A

|+ |B

|) < ∞. (2.2.21)

26 G.

Freiling and V. Yurko

Indeed, inte

grating by parts the integrals in (2.2.18), we get

ϕ(x)sin

nπ

x =

πn

(x)cos

nπ

= −

(x)sin

nπ

x = −

000

(x)cos

nπ

Similarly, we calculate

= −

aπ

(x)sin

nπ

Hence, using proposition (2.2.20) and the properties of the Fourier coefﬁcients (2.2.19), we

arrive at (2.2.21). It follows from (2.2.21) that the function u(x,t), deﬁned by (2.2.17)-

(2.2.18), has in D continuous partial derivatives up to the second order (i.e. u(x,t) ∈

(D) ), and these derivatives can be calculated by termwise differentiation of the series

(2.2.17). Obviously, u(x,t) is a solution of equation (2.2.1) and satisﬁes the boundary con-

ditions (2.2.2). Let us show that u(x,t) satisﬁes also the initial conditions (2.2.3). Indeed,

it follows from (2.2.17) for t = 0 that

u(x,0) =

∞

∑

n=1

sin

(x,0)

∞

∑

n=1

πn

sin

nπ

and

consequently

u(x,0) sin

nπ

anπ

(x,0) sin

nπ

Comparing these relations with (2.2.18) we obtain for all n ≥ 1 :

(u(x,0) −ϕ(x))sin

nπ

x = 0,

(x,0) −ψ(x))sin

nπ

x = 0.

By virtue of the completeness of the system of functions

sin

nπ

n≥1

in L

(0,l), we

conclude

that

(x,0) = ϕ(x), u

(x,0) = ψ(x),

i.e. the function u(x,t) satisﬁes the initial conditions (2.2.3). 2

Hyperbolic P

artial Differential Equations 27

2. Uniqueness

of the solution of the mixed problem

and the energy integral

Theorem 2.2.2. If the solution of the mixed problem (2.2.1) −(2.2.3) exists, then it is

unique.

Proof. Let u

(x,t) and u

(x,t) be solutions of (2.2.1)-(2.2.3). Denote

u(x,t) := u

(x,t) −u

(x,t).

Then u(x,t) ∈C

(D) and

= a

|x=0

= u

|x=l

= 0, u

|t=0

= u

t|t=0

= 0.

)

(2.2.22)

We consider the integral

E(t) =

(x,t)

)

(x,t))

dx,

which is called the energy integral. We calculate

E(t) =

(x,t)u

(x,t)

+ u

(x,t)u

(x,t)

dx,

where

E(t) :=

E(t). Inte

grating by parts the second term in the last integral we obtain

E(t) =

(x,t)

(x,t) −u

(x,t)

dx +

(x,t)u

(x,t)) .

By virtue of (2.2.22),

(

x,t) −u

(x,t) ≡ 0,

and consequently,

E(t) = (u

)

|x=l

−(u

)

|x=0

According to (2.2.22) u(0,t) ≡ 0 and u(l,t) ≡ 0. Hence u

t|x=0

= 0 and u

t|x=l

= 0. This

yields

E(t) ≡ 0, and consequently, E(t) ≡ const. Since

E(0) =

(x,0)

(x,0

)

dx = 0,

we get E(t) ≡ 0. This implies

(x,t) ≡ 0, u

(x,t) ≡ 0,

i.e. u(x,t) ≡ const. Using (2.2.22) again we obtain u(x,t) ≡ 0, and Theorem 2.2.2 is

proved. 2

28 G.

Freiling and V. Yurko

3. Non-homogeneous

equation

We consider the mixed problem for the non-homogeneous equation of the vibrating string:

= a

+ f (x,t), 0 < x < l, t > 0,

|x=0

= u

|x=l

= 0,

|t=0

= ϕ(x), u

t|t=0

= ψ(x).











(2.2.23)

Let the function f (x,t) be continuous in D, twice continuously differentiable with respect

to x, and f (0,t) = f (l,t) = 0. Let the functions ϕ(x) and ψ(x) satisfy the conditions of

Theorem 2.2.1. We expand the function f (x,t) into a Fourier series with respect to x :

f (x,t) =

∞

∑

n=1

(t) sin

nπ

where

(t)

f (x,t) sin

nπ

We will seek the solution of the problem (2.2.23) as a series in eigenfunctions of the Sturm-

Liouville problem (2.2.8)-(2.2.9):

u(x,t) =

∞

∑

n=1

(t) sin

nπ

x. (2.2.24)

Substituting

(2.2.24)

into (2.2.23) we get

∞

∑

n=1

¨u

(t)sin

nπ

x = −a

∞

∑

n=1

πn

(t) sin

nπ

x +

∞

∑

n=1

(t) sin

nπ

∞

∑

n=1

(0)sin

nπ

x = ϕ(x),

∞

∑

n=1

˙u

(0)sin

nπ

x = ψ(x),

and

consequently

, we obtain the following relations for ﬁnding the functions u

(t) , n ≥1 :

¨u

(t) +

aπn

(t)

= f

(t),

) = B

, ˙u

(0) =

aπn











(2.2.25)

where

the

numbers A

and B

are deﬁned by (2.2.18). The Cauchy problem (2.2.25) has a

unique solution which can be found, for example, by the method of variations of constants:

(t) = A

sin

anπ

t + B

cos

anπ

(τ)sin

anπ

(t −τ) dτ. (2.2.26)

easy to check that the function u(x,t), deﬁned by (2.2.24) and (2.2.26), is a solution

of problem (2.2.23).

Hyperbolic P

artial Differential Equations 29

4. A

general scheme of separation of variables

This subsection is recommended for advanced studies, and it can be omitted ”in the ﬁrst

reading”.

4.1. The method of separation of variables can be used for a wide class of differential

equations of mathematical physics. In this section we apply the method of separation of

variables for studying the equation of a non-homogeneous vibrating string. We consider

the following mixed problem

ρ(x)u

k(x)u

−q(x)u, 0 < x < l, t > 0, (2.2.27)

−hu)

|x=0

= 0, (H

+ Hu)

|x=l

= 0, (2.2.28)

|t=0

= Φ(x), u

t|t=0

= Ψ(x). (2.2.29)

Here x and t are independent variables, u(x,t) is an unknown function, the functions

ρ,k,q, Φ,Ψ are real-valued, and the numbers h, h

,H, H

are real. We assume that

q(x) ∈ L

(0,l), ρ(x),k(x),Φ(x),Ψ(x) ∈ W

[0,l] , ρ(x) > 0, k(x) > 0 , and Φ,Ψ satisfy

the boundary conditions (2.2.28). We remind that the condition f (x) ∈W

[0,l] means that

f (x) and f

(x) are absolutely continuous on [0,l], and f

(x) ∈ L

(0,l) .

First we consider the following auxiliary problem. We will seek nontrivial (i.e. not

identically zero) particular solutions of equation (2.2.27) such that they satisfy the boundary

conditions (2.2.28) and admit separation of variables, i.e. they have the form

u(x,t) = Y (x)T (t). (2.2.30)

For this purpose we substitute (2.2.30) into (2.2.27) and (2.2.28). By virtue of (2.2.27) we

have

ρ(x)Y (x)

T (t) =

³³

k(x)Y

(x)

−q(x)Y (x)

T (t).

Separating the variables, we obtain

k(x)Y

(x)

−q(x)Y (x)

ρ(x)Y (x)

T (t)

Since x and t are

independent

variables, the equality is here valid only if both fractions

are equal to a constant; we denote this constant by −λ . Therefore,

k(x)Y

(x)

−q(x)Y (x)

ρ(x)Y (x)

T (t)

= −λ. (2.2.31)

Moreo

er, the boundary conditions (2.2.28) yield

(0) −hY (0)

T (t) = 0,

(l) + HY (l)

T (t) = 0.







(2.2.32)

30 G.

Freiling and V. Yurko

It follo

ws from (2.2.31) and (2.2.32) that

T (t) + λT (t) = 0, (2.2.33)

−

k(x)Y

(x)

+ q(x)Y (x) = λρ(x)Y (x), 0 < x < l, (2.2.34)

(0) −hY (0) = 0, H

(l) + HY (l) = 0. (2.2.35)

Thus, the function T (t) is a solution of the ordinary differential equation (2.2.33), and

the function Y (x) is a solution of the boundary value problem (2.2.34)-(2.2.35) which is

called the Sturm-Liouville problem associated to the mixed problem (2.2.27)-(2.2.29). We

are interested in nontrivial solutions of problem (2.2.34)-(2.2.35), but it turns out that they

exist only for some particular values of the parameter λ.

Deﬁnition 2.2.3. The values of the parameter λ for which the problem (2.2.34)-

(2.2.35) has nonzero solutions are called eigenvalues, and the corresponding nontrivial

solutions are called eigenfunctions. The set of eigenvalues is called the spectrum of the

Sturm-Liouville problem.

medskip We note that Deﬁnition 2.2.3 is a generalization of Deﬁnition 2.2.2, since it is

related to the more general problem (2.2.34)-(2.2.35) than problem (2.2.8)-(2.2.9).

Clearly, the eigenfunctions are deﬁned up to a multiplicative constant, since if Y (x)

is a nontrivial solution of the problem (2.2.34)-(2.2.35) for λ = λ

(i.e. Y (x) is an

eigenfunction corresponding to the eigenvalue λ

), then the function CY(x) , C −const,

is also a solution of problem (2.2.34)-(2.2.35) for the same value λ = λ

4.2. In this subsection we establish properties of the eigenvalues and the eigenfunctions

of the boundary value problem (2.2.34)-(2.2.35). For deﬁniteness, let below h

6= 0. The

other cases are considered analogously and can be recommended as exercises.

First we study the particular case when

ρ(x) = k(x) ≡ 1, l = π, (2.2.36)

and then we will show that the general case can be reduced to (2.2.36). In other words,

one can consider (2.2.36) without loss of generality. Thus, we consider the Sturm-Liouville

problem L in the following form:

−y

(x) + q(x)y(x) = λy(x), 0 < x < π, (2.2.37)

(0) −hy(0) = 0, y

(π) + Hy(π) = 0, (2.2.38)

where q(x) ∈ L

(0,π), h and H are real. Denote

`y(x) := −y

(x) + q(x)y(x),

U(y) := y

(0) −hy(0),

V (y) := y

(π) + Hy(π).

Let ϕ(x,λ) and ψ(x,λ) be solutions of (2.2.37) under the initial conditions

ϕ(0,λ) = 1, ϕ

(0,λ) = h, ψ(π, λ) = 1, ψ

(π,λ) = −H.

Hyperbolic P

artial Differential Equations 31

For

each ﬁxed x, the functions ϕ(x,λ) and ψ(x,λ) are entire in λ. Clearly,

U(ϕ) = ϕ

(0,λ) −hϕ(0,λ) = 0,

V (ψ) = ψ

(π,λ) + Hψ(π,λ) = 0.

)

(2.2.39)

Denote

∆(λ) = hψ(x, λ),ϕ(x, λ)i, (2.2.40)

where

hy(x),z(x)i := y(x)z

(x) −y

(x)z(x)

is the Wronskian of y and z. By virtue of Liouville‘s formula for the Wronskian,

hψ(x,λ),ϕ(x,λ)i does not depend on x. The function ∆(λ) is called the characteristic

function of L. Substituting x = 0 and x = π into (2.2.40), we get

∆(λ) = V (ϕ) = −U(ψ). (2.2.41)

The function ∆(λ) is entire in λ, and consequently, it has an at most countable set of zeros

{λ

} (see [12]).

Theorem 2.2.3. The zeros {λ

} of the characteristic function coincide with the eigen-

values of the boundary value problem L. The functions ϕ(x, λ

) and ψ(x,λ

) are eigen-

functions, and there exists a sequence {β

} such that

ψ(x,λ

) = β

ϕ(x,λ

), β

6= 0. (2.2.42)

Proof. 1) Let λ

be a zero of ∆(λ). Then, by virtue of (2.2.39)-(2.2.41), ψ(x, λ

) =

ϕ(x,λ

), and the functions ψ(x,λ

),ϕ(x, λ

) satisfy the boundary conditions (2.2.38).

Hence, λ

is an eigenvalue, and ψ(x, λ

),ϕ(x, λ

) are eigenfunctions related to λ

2) Let λ

be an eigenvalue of L, and let y

be a corresponding eigenfunction. Then

U(y

) = V (y

) = 0. Clearly y

(0) 6= 0 (if y

(0) = 0 then y

(0) = 0 and, by virtue of the

uniqueness theorem for the equation (2.2.37), y

(x) ≡ 0 ). Without loss of generality we

put y

(0) = 1. Then y

(0) = h, and consequently y

(x) ≡ ϕ(x,λ

). Therefore, (2.2.41)

yields

∆(λ

) = V (ϕ(x,λ

)) = V (y

(x)) = 0,

and Theorem 2.2.3 is proved. 2

Note that we have also proved that for each eigenvalue there exists only one (up to a

multiplicative constant) eigenfunction.

Denote

(x,λ

)dx. (2.2.43)

The numbers {α

} are called the weight numbers, and the numbers {λ

,α

} are called

the spectral data of L.

Lemma 2.2.1. The following relation holds

= −

∆(λ

), (2.2.44)