F-VARLAMOV-DIE-15 copy - Project Euclid

Di↵erential and Integral Equations Volume 10, Number 6, November 1997, pp. 1197–1211.

ON SPATIALLY PERIODIC SOLUTIONSOF THE DAMPED BOUSSINESQ EQUATION

Vladimir V. VarlamovDepartamento de Matematicas y Estadistica, Universidad Nacional de Colombia, Bogota, Colombia

(Submitted by: J.A. Goldstein)

Abstract. A classical solution of the damped Boussinesq equation

utt � 2butxx = �↵uxxxx + uxx + �(u2)xx, x 2 R1, t > 0,

with ↵, b = const > 0, � = const 2 R1, ↵ > b2, and small initial data is constructed bymeans of the successive application of the spectral theory and the perturbation one. Itslong-time asymptotic representation is obtained which shows that the major term increaseslinearly with time and the second term contains a combination of the Airy functions of anegative argument. A uniform-in-space estimate of the remainder is given.

1. Introduction. In recent years interest has grown in studying nonlinearevolution equations governing wave propagation in media with dispersion. One ofthe typical representatives of this class of equations is the Boussinesq one ([5])

utt = �↵uxxxx + uxx + �(u2)xx, x 2 R1, t > 0, (1.1)

where ↵,� = const 2 R1,which governs the propagation of one-dimensional longwaves on the surface of shallow water. Here u(x, t) is an elevation of the free sur-face and subscripts denote partial derivatives. Some interesting comments on thederivation of (1.1) can be found in the survey [13].

Among many papers on (1.1) (see, for example, [7–11], [13], [19]) we should liketo mark the note of R. Hirota ([9]), where conservation laws and N -soliton solutionsare examined, and the paper of P. Clarkson ([7]), in which a general approach to acomputation of exact solutions is discussed. Questions of nonexistence of global-in-time solutions to Cauchy problems in Banach spaces for generalizations of (1.1) areconsidered in [10], [11], and [19].

In real processes viscosity, as well as dispersion, plays an important role. There-fore, it is interesting to study the equation

utt � 2butxx = �↵uxxxx + uxx + �(u2)xx, x 2 R1, t > 0, (1.2)

where ↵, b = const > 0, � = const 2 R1, which di↵ers from the classical Boussinesqequation by the second term on the left-hand side accounting for dissipation. In

Accepted for publication May 1996.AMS Subject Classifications: 35Q20, 76B15.

1197

1198 vladimir v. varlamov

[3], [4] P. Biler has deduced time estimates of the operator norms of a solution tothe abstract Cauchy problem for the generalization of (1.2). Strictly speaking, thenonlinearity considered there was di↵erent, namely ||u||pu, where p > 0 and || · || isa norm in L2. The “oscillation condition” examined in [3] in the case of constantcoe�cients takes the form ↵ > b2. It excludes the overdamping phenomenon leadingto the purely parabolic-type behavior of solutions. Su�cient conditions for thepower decay in time of a certain combination of the first derivatives of the solutionswere also obtained in this paper.

In [18], [23] the damped Boussinesq equation was examined from a di↵erent pointof view. The theory of inertial manifolds and global attractors was applied to the(1.2)-type equation in one space dimension and with a more general nonlinearity. Inthe paper [12] the existence and uniqueness of solutions for the damped Boussinesqequation with a linear feedback was studied on a periodic domain. The continuousdependence of the solution on initial data was proved by means of the derivation ofsome conservation laws. However, none of the authors mentioned above has obtaineda long-time asymptotic representation of the classical solution of the initial-value orperiodic problems for either the classical Boussinesq equation or its damped version.

In the present paper the equation (1.2) is considered with small spatially periodicinitial data. Its solution is constructed in the form of a series in small parameterpresent in the initial conditions and the long-time asymptotic representation is ob-tained. It is well known that one of the methods of studying Cauchy problems fornonlinear evolution equations is the inverse scattering transform ([1], [6]). In [15]–[17] P. Naumkin and I. Shishmarev have used another approach based on applyingboth the spectral and perturbation theories. They have examined nonlinear evolu-tion equations of the first order in time and with small initial data. In [20], [21] thismethod has been developed further and adapted for equations of the second orderin t. As a result, the major term of the long-time asymptotics has been obtained inthe explicit form and not in a series in small parameter, as in [15]–[17]. In [15]–[17],[20], [21] a decay of the initial data at infinity in space caused a decrease of solu-tions as t ! +1 and thus facilitated the computation of the asymptotics. In thepresent paper it is shown that, generally speaking, the major term of the long-timeasymptotics of the solution of the classical periodic problem for (1.2) increases lin-early with time as t! +1. The second term of the asymptotics contains the Airyfunctions of a negative argument which account for damped oscillations.

2. Statement of the problem and the main result. We consider the periodicproblem for equation (1.2) adding the initial conditions

u(x, 0) = "2'(x), ut(x, 0) = "2 (x), x 2 R1, (2.1)

where " = const > 0 and the functions '(x) and (x) are 2⇡-periodic on R1.Definition 1. The function u(x, t) defined on x 2 R1, t > 0, is said to be the classi-cal solution of the problem (1.2), (2.1) if it is continuous together with its derivatives

on spatially periodic solutions 1199

included in (1.2), (2.1), 2⇡-periodic in x on R1 together with these derivatives andsatisfies (1.2), (2.1).Definition 2. The function f(x) belongs to the space Cn

2⇡(R1), n � 1, if it isn-times continuously di↵erentiable on R1and 2⇡-periodic on R1 together with itsderivatives through (n� 1)-th order.

Let bf0 =R ⇡�⇡ f(x) dx. Our main result is the following

Theorem. If ↵̇ > b2; b 0 = 0 or � b 0 > 0; '(x) 2 C62⇡(R1), (x) 2 C4

2⇡(R1), thenfor su�ciently small " there exists a classical solution of the problem (1.2), (2.1)represented in the form

u(x, t) =1X

N=0

"N+1u(N)(x, t), (2.2)

where the functions u(N)(x, t) will be defined in the proof (see (3.21), (3.10)). Thissolution has the following asymptotics as t! +1 :

a) if � 2 R1, b 0 = 0, then

u(x, t) = (2⇡)�1{"2 b'0 + e�bt{2Re[eix(A cos�t + B sin�t)] + O(e�bt)}}; (2.3)

b) if � b 0 > 0, then

u(x, t) = (2⇡)�1{"2(b'0 + b 0t) + e�bt{2Re[eix(D Bi(�⇠1(t))+ E Ai(�⇠1(t))] + O(e�bt)}},

(2.4)

where � = (↵ � b2 + 1 + 2"2� b'0)1/2; A, B, D, E are constant complex co-e�cients defined by (3.18), (3.20), (4.11), and (4.12), ⇠1(t) > 0 is a linearfunction of t (see (4.11)), and Ai(�z), Bi(�z), z > 0, are the Airy func-tions of a negative argument of the first and second kind respectively. Theestimates of the remainders in (2.3), (2.4) are uniform in x 2 R1.

3. Proof of the theorem for the case b 0 = 0. We seek the solution of (1.2),(2.1) in the form of a complex Fourier series

u(x, t) = (2⇡)�11X

n=�1einxbun(t), bun(t) =

Z ⇡

�⇡e�inxu(x, t) dx. (3.1)

Expanding the initial data into the Fourier series we write

'(x) = (2⇡)�11X

n=�1einx b'n, b'n =

Z ⇡

�⇡e�inx'(x) dx,

(x) = (2⇡)�11X

n=�1einx b n, b n =

Z ⇡

�⇡e�inx (x) dx.

(3.2)


It follows from the hypothesis that for n 2 Z

|b'n| (1 + |n|)�6, | b n| (1 + |n|)�6, (3.3)

where for convenience unit constants are put in the right-hand sides of (3.3). It ispossible due to the presence of the small parameter "2 in (2.1).

Inserting (3.1), (3.2) into (1.2), (2.1) we obtain the following Cauchy problemsfor determining the Fourier coe�cients:

@2t bun + 2bn2@tbun + (↵n4 + n2)bun = ��n2

1Xq=�1

bun�q(t)buq(t), t > 0,

bun(0) = "2 b'n, @tbun(0) = "2 b n, n 2 Z. (3.4)

We separate three of these problems, namely

@2t bu0 = 0, t > 0,

bu0(0) = "2 b'0, @tbu0(0) = 0(3.5)

and

@2t bum + 2b@tbum + (↵+ 1 + 2�bu0)bum = ��

1Xq=�1,q 6=0,m

bum�q(t)buq(t), t > 0,

bum(0) = "2 b'm, @tbum(0) = "2 b m, m = ±1. (3.6)

The solution of (3.5) is bu0(t) = "2 b'0. In (3.6) we have transferred two termscontaining bumbu0 to the left-hand side of the equation because we intend to obtainsubtle asymptotic estimates of bum(t), m = ±1. The latter ones will contribute tothe major term of the asymptotic expansion of the solution as t! +1 while bu0(t)will form its major part. For the solutions of (3.4) with |n| � 2 we shall find onlyrough bounds in t which will enable us to deduce the estimates of the remaindersin (2.3), (2.4).

Temporarily leaving (3.6) we examine (3.4). Seeking fundamental solutions ofthe homogeneous equation associated with (3.4) in the form e��t we find

�1,2(n) = rn ⌥ i�n, where rn = bn2, �n = |n|(kn2 + 1)1/2, k = ↵� b2 > 0. (3.7)

Setting �n = "b'n, n = " b n we integrate (3.4) in t and obtain

bun(t) = " exp(�rnt){[cos(�nt) + (rn/�n) sin(�nt)]�n + (�n)�1 sin(�nt) n}

� �n2

�n

Z t

0exp(�rn(t� ⌧)) sin(�n(t� ⌧))

1Xq=�1

bun�q(⌧)buq(⌧) d⌧. (3.8)


For solving the integral equation (3.8) we represent bun(t), n 2 Z, as a formal seriesin "

bun(t) =1X

N=0

"N+1v(N)(t). (3.9)

Substituting (3.9) into (3.8) and equating the coe�cients of equal powers of " wededuce that for n 6= 0

v(0)n (t) = exp(�rnt){[cos(�nt) + (rn/�n) sin(�nt)]�n + (�n)�1 sin(�nt) n},

v(N)n (t) = ��n2

�n

Z t

0exp(�rn(t� ⌧)) sin(�n(t� ⌧))·

·1X

q=�1

NXj=1

v(j�1)n�q (⌧)v(N�j)

q (⌧) d⌧, N � 1. (3.10)

Although the formulas (3.10) are valid for all n 6= 0,we shall use them now forestablishing the following estimates for |n| � 2, N � 0, t > 0 :

|v(N)n (t)| cN (N + 1)�5(1 + |n|)�6e�2bt exp(�(n/(N + 1))t), (3.11)

where (p) = bp2/2, p 2 R1. Here and in the sequel we denote by c generic positiveconstants not dependent on n, N, ", x, and t. They may depend on the coe�cientsof the equation and the initial data.

For N = 0 (3.11) follows immediately from (3.3), (3.10) since choosing su�cientlysmall " we have for |n| � 2, t > 0

|v(0)n (t)| "(1 + |n|)�6 exp(�bn2t/2) exp(�bn2t/2)[1 + bk�1/2(�n)�1]

(1 + |n|)�6 exp(�2bt) exp(�(n)t).

Next we use induction on N. Let (3.11) hold for v(s)n (t), 0 s N � 1. Then we

shall show that they are valid for s = N. Using the inequalities

(1 + |n� q|)�6(1 + |q|)�6 26(1 + |n|)�6[(1 + |q|)�6 + (1 + |n� q|)�6],

j�5(N + 1� j)�5 25(N + 1)�5[j�5 + (N + 1� j)�5]

we find that

|v(N)n (t)| 211|�|n2(�n)�1(N + 1)�5(1 + |n|)�6

·1X

q=�1[(1 + |q|)�6 + (1 + |n� q|)�6] (3.12)

·NX

j=1

cj�1cN�j [(N + 1� j)�5 + j�5]S(n, q, t),

S(n, q, t) = exp(�2(n)t)Z t

0exp(L(n, q)⌧) d⌧,

L(n, q) = 2(n)� ((n� q)/j)� (q/(N + 1� j))� 4b.


We estimate a typical term on the right-hand side of (3.12), namely

|I(N)n (t)| |�|(N + 1)�5cN�1(1 + |n|)�6

1Xq=�1

(1 + |q|)�6NX

j=1

(N + 1� j)�5S(n, q, t).

Since (p) is a monotonically increasing function for p 2 R1 the following inequalityholds for n, q 2 Z, N � 1 (see [15], [16]):

(n� q

j) + k(

q

N + 1� j) � (

n

N + 1). (3.13)

Therefore we haveS(n, q, t) exp(�( n

N + 1)t)|L|�1.

On the other hand, if |L| (n), then for N � 1

S(n, q, t) exp[�(n/(N + 1))t](n)

.

Consequently,

S(n, q, t) exp[�( n

N+1)t] exp(�2bt)max[(n), |L(n, q)|] .

Further, we note that

max(, |L|) � 14

max[(n),(q

N + 1� j)] � [(n) + (q)]

8(N + 1� j)2.

Thus,

|I(N)n (t)| |�|cN�1(N + 1)�5(1 + |n|)�6 exp(�2bt) exp[�( n

N + 1)t]

NXj=1

(N + 1� j)�31X

q=�1[(n) + (q)]�6(1 + |q|)�6.

From the last estimate follows (3.11).By analogy with the deduction of (3.11) it may be shown via the integral repre-

sentation (3.8) that for |n| � 1 the following estimates hold:

|v(N)n (t)| cN (N + 1)�5(1 + |n|)�6e�bt/2 exp[�( n

N + 1)t], N � 1, t > 0. (3.14)

These inequalities serve as preliminary rough bounds for deducing subtle asymptoticestimates of the solutions of (3.6).


Now we shall examine (3.6). It su�ces to consider only the case m = 1 becausebu�1(t) = bu1(t), where a bar over the letter denotes complex conjugation. Therefore,

eixbu1(t) + e�ixbu�1(t) = 2Re [eixbu1(t)]. (3.15)

In fact, for all n � 0

bu�n(t) = bun(t). (3.16)

We shall show it by means of (3.9), (3.10). Since from (3.2) b'�n = b'n, b �n =b n, n � 0, it follows from (3.10) that v(0)

�n(t) = v(0)n (t), n � 0. Next we shall use

induction on the number N . Assume that for 0 s N � 1, v(s)�n(t) = v(s)

n (t),n � 0. Then omitting the superscripts for brevity we have

1Xq=�1

v�n�q(t)vq(t) =1X

q=�1vn+q(t) v�q(t) =

1Xq=�1

vn�q(t) vq(t),

and taking the complex conjugation in (3.10) we establish (3.16). Substituting thesolution of (3.4), bu0(t), into (3.5) we find the roots of the characteristic equation

�1,2(1) = b⌥ i�, � = (k + 1 + 2�"2 b'0)1/2.

Note that they are di↵erent from (3.7) with n = 1 and contain the small parameter". Integrating (3.6) in t we reduce it to the integral equation

bu1(t) = "e�bt{[cos(�t) + (b/�) sin(�t)]�1 + ��1 sin(�t) 1}

� �

�

Z t

0e�b(t�⌧) sin(�(t� ⌧))

1Xq=�1q 6=0,1

bu1�q(⌧)buq(⌧) d⌧. (3.17)

Representing bun(t) in the form (3.9), substituting it into (3.17), and equating coef-ficients of equal powers of " we get

v(0)1 (t) = e�bt[A(0) cos(�t) + B(0) sin(�t)],

v(N)1 (t) = e�bt{[A(N) + R(N)

A (t)] cos(�t) + [B(N) + R(N)B (t)] sin(�t)},

A(0) = "b'1, B(0) ="

�(bb'1 + b 1),


A(N) =�

�

Z 1

0eb⌧ sin(�⌧)SN (⌧) d⌧,

B(N) = ��

Z 1

0eb⌧ cos(�⌧)SN (⌧) d⌧,

R(N)A (t) =

�

�

Z 1

teb⌧ sin(�⌧)SN (⌧) d⌧,

R(N)B (t) = ��

�

Z 1

teb⌧ cos(�⌧)SN (⌧) d⌧,

SN (t) =1X

q=�1q 6=0,1

NXj=1

v(j�1)1�q v(N�j)

q (t), N � 1,

(3.18)

where the functions v(s)n (t), s = 0, 1, . . . , N �1, for the calculation of the sum SN (t)

must be taken from (3.10). The expressions for v(N)1 (t) in (3.18) are in accordance

with (3.10), but are used only for obtaining the refined asymptotic estimates ofbu1(t).Next we shall prove that for N � 1, t > 0

|R(N)A,B(t)| cNe�2bt. (3.19)

To this purpose we shall single out two terms, namely with q = �1 and q = 2, fromthe sum SN (t) and estimate them using both (3.11) and (3.14) since (3.11) is validonly for |n| � 2. All the other terms of this sum we shall estimate by means of(3.11). We have

|R(N)A,B(t)| |�|

�

Z 1

teb⌧ [2

NXj=1

v(j�1)�1 (⌧)v(N�j)

2 (⌧) +1X

q=�1q 6=0,±1,2

NXj=1

v(j�1)1�q (⌧)v(N�j)

q (⌧)]d⌧

|�|�

cN�1(N + 1)�5NX

j=1

[j�5 + (N + 1� j)�5]·

· {Z 1

teb⌧e�b⌧/2e�2b⌧ exp[�(1

j)⌧ � ( 2

N + 1� j)⌧ ] d⌧

+1X

q=�1q 6=0,±1,2

Z 1

teb⌧e�4b⌧ exp[�(n� q

j)⌧ � ( 2

N + 1� j)⌧ ] d⌧}.

Making use of (3.13) and estimating the last integral in a manner similar to the oneused for the deduction of (3.11) we obtain (3.19).

Returning to (3.16) and choosing " su�ciently small for convergence of the series,


we get as t! +1

bu1(t) = e�bt[A cos(�t) + B sin(�t)] + O(e�bt)]

A =1X

N=0

"N+1A(N), B =1X

N=0

"N+1B(N),(3.20)

where A(N)and B(N) are computed by means of (3.18), (3.10).For deducing the representation (2.2) of the solution of (1.2), (2.1) we recall the

Fourier series (3.1) with bun(t) described by (3.9), (3.10) and perform the interchangeof summation in the series, that is,

u(x, t) = (2⇡)�11X

n=�1einx

1XN=0

"N+1v(N)n (t) = (2⇡)�1

1XN=0

"N+1u(N)(x, t),

u(N)(x, t) = (2⇡)�11X

n=�1einxv(N)

n (t). (3.21)

This interchange is possible due to the absolute and uniform in x 2 R1, t > 0convergence of the series in N. The last statement in its turn follows from theestimates (3.11), (3.14) with su�ciently small ". With the help of these estimatesit may be verified in a straightforward manner that (3.21) is the solution of (1.2),(2.1).

It remains to prove (2.3). Taking into account (3.11) we can write that

u(x, t) = (2⇡)�1[bu0(t) + eixbu1(t) + e�ixbu�1(t)] + Ru(x, t),

|Ru(x, t)| (2⇡)�1e�2bt1X

N=0

cN"N+1(N + 1)�5·

·1X

n=�1|n|�2

(1 + |n|)�6 exp[�( n

N + 1)t].

(3.22)

Using (3.15) and noting that the exponents in the last series are majorized by theexpression

exp[� 2bt(N + 1)2

] c(N + 1)2

tas t! +1

and "c < 1 we conclude that (2.3) holds uniformly in x 2 R1. The proof of the firstpart of the theorem is complete.

4. Proof of the theorem for the case � b 0 > 0. Su�ce it to consider � >0, b 0 > 0 because the case � < 0, b 0 < 0 may be obtained simply by changing u


to �u in (1.2), (2.1). As a result, �, '(x), and (x) change their signs. The sign of'(x) doesn’t matter in our considerations.

Expanding u(x, t) and the initial data into the complex Fourier series we obtainthe following problem for n = 0:

@2t bu0 = 0, t > 0, bu0(0) = "2 b'0, @tbu0(0) = "2 b 0,

which has the solution bu0(t) = "2(b'0 + b 0t). (4.1)

Now bu0(t) increases linearly with time and the scheme of the proof of the previoussection must be revised.

For all numbers n 6= 0 we get

@2t bun + 2bn2@tbun + [↵n4 + n2 + 2�n2bu0(t)]bun = ��n2

1Xq=�1q 6=0,n

bun�q(t)buq(t), t > 0,

bun(0) = "2 b'n, @tbun(0) = "2 b n, (4.2)

where we have singled out two terms in the right-hand side of the equation and havetransferred them to the left-hand side. We also separate two problems with m = ±1

@2t bum + 2b@tbum + [1 + ↵+ 2�bu0(t)]bum = ��

1Xq=�1q 6=0,m

bum�q(t)buq(t), t > 0,

bum(0) = "2 b'm, @tbum(0) = "2 b m (4.3)

in order to obtain subtle time estimates of their solutions. As in the previous section,bu0(t) will give us the major part of the long-time asymptotics of the solution inquestion while bum(t), m = ±1, will contribute to the next term of these asymptotics.

First we shall examine the problem (4.2). We substitute (4.1) into (4.2) andsolve the corresponding homogeneous equation. Setting bun(t) = wn(t) exp(�rnt),rn = bn2, we obtain that wn(t) satisfies the following equation:

d2wn

dt2+ (an + �nt)wn = 0, an = kn4 + n2(1 + 2"2� b'0) > 0,

k = ↵� b2 > 0, �n = "2n22� b 0 > 0.(4.4)

Introducing the new variable

⇠n(t) = hn(t) + �1/3n t > 0, (4.5)

wherehn = an�

�2/3n =

kn4 + n2(1 + 2"2� b'0)n4/3"4/3(2� b 0)2/3

> 0,


we transform (4.4) into the well-known Airy equationd2wn

d⇠n+ ⇠nwn = 0.

Its fundamental solutions are the Airy functions of the first and second kind, namelyAi(�⇠n) and Bi(�⇠n) (see [2], [22]), which have the following representations:

Ai(�z) =z1/2

3[J�1/3(⇣) + J1/3(⇣)], Bi(�z) = (

z

3)1/2[J�1/3(⇣)� J1/3(⇣)],

where z > 0, ⇣ = (2/3)z2/3 and J⌫(z) are the Bessel functions of the index ⌫. In thesequel we shall also need the estimates of the Airy functions and their derivativesas z ! +1

|Ai(�z)| cz�1/4, |Bi(�z)| cz�1/4,

|Ai0(�z)| cz1/4, |Bi0(�z)| cz1/4,(4.6)

where a prime denotes a derivative of a function of a single variable.Integrating the problem (4.2) in t we reduce it to the integral equation

bun(t) = "2⇡ exp(�rnt){{[rn��1/3n Bi(�hn) + Bi0(�hn)]b'n

+ Bi(�hn)��1/3n

b n]Ai(�⇠n(t))� {[rn��1/3n Ai(�hn)

+ Ai0(�hn)]b'n + Ai(�hn)��1/3n

b n]}Bi(�⇠n(t))]}

+ �n2

Z t

0Gn(t, ⌧)

1Xq=�1q 6=0,n

bun�q(⌧)buq(⌧) d⌧,

(4.7)

where the Green function Gn(t, ⌧) has the following representation:Gn(t, ⌧) = exp[�rn(t� ⌧)]gn(t, ⌧),

gn(t, ⌧) = ��1/3n ⇡[Ai(�⇠n(⌧))Bi(�⇠n(t))�Ai(�⇠n(t))Bi(�⇠n(⌧))],

0 < ⌧ < t, t > 0.(4.8)

For deducing (4.7), (4.8) we have used the expression for the Wronskian of the Airyfunctions (see [2])

W{Ai(z), Bi(z)} =1⇡

.

Setting �n = "b'n, n = " b n, representing bun(t) as a series in small parameter "(see (3.9)) and equating coe�cients of equal powers of " we find that for n 6= 0

v(0)n (t) = ⇡ exp(�rnt){{rn�

�1/3n Bi(�hn) + Bi0(�hn)]�n

+ Bi(�hn)��1/3n n}Ai(�⇠n(t))� {[rn�

�1/3n Ai(�hn)

+ Ai0(�hn)]�n + Ai(�hn)��1/3n n}Bi(�⇠n(t))},

v(N)n (t) = �n2

Z t

0Gn(t, ⌧)

1Xq=�1q 6=0,n

NXj=1

v(j�1)n�q (⌧)v(N�j)

q (⌧) d⌧, N � 1.

(4.9)


Now we shall establish the estimates (3.11) for v(N)n (t) with |n| � 2, N � 0, t > 0.

But this time we must take into account that the arguments of the Airy functionsdepend not only on n, but also on ". Therefore we must make sure that our estimateshold when "! 0.

Thanks to (4.6) we have as "! 0

rn��1/3n |Bi(�hn)| cn�2/3"�1/3, |Bi0(�hn)| ch1/4

n cn�2/3"�1/3,

|Ai(�⇠n(t))| ch�1/4n cn2/3"1/3, |Bi0(�⇠n(t))| cn�2/3"1/3,

|Ai(�⇠n(⌧))Bi(�⇠n(t))| ch�1/2n c"2/3n�4/3.

Consequently, |gn(t, ⌧)| cn2 as "! 0 uniformly in t, ⌧.

Taking into consideration these estimates and (3.3) we obtain from (4.9)

|v(0)n (t)| c"(1 + |n|)�6 exp(�bn2t) (1 + |n|)�6 exp(�2bt) exp(�(n)t)

with |n| � 2 and k(p) = bp2/2, p 2 R1. Applying the procedure described in theprevious section and based on the induction on the number N we can deduce theestimates (3.11) for v(N)

n (t) with |n| � 2 and (3.14) for v(N)n (t) with |n| � 1.

We return to the problems (4.3) and consider the case m = 1. Integrating (4.3)with respect to t we reduce it to the integral equation (4.9) with n = 1, r1 = b and

h1 =k + 1 + 2"2� b'0

"4/3(2� b 0)2/3, �1 = 2"2� b 0 > 0, ⇠1(t) = h1 + �1/3

1 t. (4.10)

The Green function G1(t, ⌧), 0 < ⌧ < t, t > 0, is described by (4.8) with n = 1.Setting �1 = "b'1, 1 = " b 1 and representing bun(t) as a series in small parameter "we obtain

v(0)1 (t) = e�bt[D(0) Bi(�⇠1(t)) + E(0) Ai(�⇠1(t))],

D(0) = "⇡{[b��1/31 Ai(�h1) + Ai0(�h1)]b'1 + Ai(�h1)�

�1/31

b 1]},

E(0) = "⇡{[b��1/31 Bi(�h1) + Bi0(�h1)�

�1/31

b 1},

v(N)1 (t) = e�bt{[D(N) + R(N)

D (t)]Bi(�⇠1(t))� [E(0) + R(N)E (t)]Ai(�⇠1(t))}

D(N) = ��1/31 ⇡

Z 1

0eb⌧Ai(�⇠1(⌧))SmN (⌧) d⌧, (4.11)

E(N) = ��1/31 ⇡

Z 1

0eb⌧Bi(�⇠1(⌧))SmN (⌧) d⌧,

SmN (⌧) =1X

q=�1q 6=0,1

NXj=1

v(j�1)1�q v(N�j)

q (⌧), N � 1,


where the functions v(s)n (t), s = 0, 1, . . . , N � 1, for the calculation of SmN (t) are

defined by (4.9).As in the previous section, using the estimates (3.11), (3.14) and the boundedness

of the Airy functions of a negative argument we can prove that for N � 1, t > 0

|R(N)D,E(t)| cNe�2bt. (4.12)

Thus, we get

bu1(t) = e�bt[D Bi(�⇠1(t)) + E Ai(�⇠1(t)) + O(e�bt)],

D =1X

N=0

"N+1D(N), E =1X

N=0

"N+1E(N).(4.13)

There is no need to study bu�1(t) since we can use (3.15).Finally, the solution is represented in the form (3.21) with v(N)

n (t) described by(4.9). The interchange of summation in the corresponding series is based on thesame estimates as in section 3. The asymptotics of (2.4) follow from (3.22) and(3.15) with bu0(t) and bu1(t) defined by (4.1) and (4.13) respectively. The estimateof the remainder is the same as in (3.22). The proof of the theorem is complete.

5. Remarks and conclusions. We have restricted our attention to the case↵ > b2 which is the most interesting one and corresponds to the existence of aninfinite number of damped oscillations. It also allows us to illustrate clearly themain ideas of the method applied without going into details. The case ↵ = b2 isexamined analogously, but the convergence of the series in n is slightly worse sincek = 0 in (3.7) and (4.5).

There are no major di�culties in using the method proposed above when 0 <↵ < b2, but one must consider several subcases and the calculations are somewhatlengthy. For example, only in the case b 0 = 0, when the roots of the characteristicequation are �1,2(n) = bn2⌥|n|(|k|n2�1)1/2, |k| = b2�↵ > 0, one must examine thefollowing three subcases: 1) n2 < |k|�1, when �1,2 are complex, 2) n2 = |k|�1 = l2,when the roots are multiple, and 3) n2 > |k|�1, when �1,2 > 0. In fact, there is alimited number of damped oscillations here, as well as aperiodic processes. Thereis no need to consider multiple eigenvalues of the linear operator in the subcase 2)if two terms are transferred from the right-hand side of the equation (3.4) to theleft-hand one, so that

@2t bul + 2bl2@tbul + (↵l4 + l2 + 2�l2"2 b'0)bul = ��l2

1Xq=�1q 6=0,l

bul�q(t)buq(t),

and the roots of the characteristic equation become complex: �1,2(l) = bl2 ⌥i|l|"(2� b'0)1/2 (the fundamental solutions are chosen in the form e��t).


Now we shall discuss how the constructed solution is related to the solution of thelinear problem. In fact, the condition � b 0 = 0 covers both the case a) of the theoremand the linear one � = 0, b 0 2 R1. In the latter case the solution is constructedeasily, that is,

ulin(x, t) = (2⇡)�1"2{b'0 + b 0t + 21X

n=1

cos(nx) exp(�rnt)·

· [cos(�nt) + rn(�n)�1 sin(�nt)]b'n + (�nt) b n]}.(5.1)

Taking the limit � ! 0 in (3.21), (3.10) we obtain (5.1) with b 0 = 0. Passing tothis limit in (2.3) gives us the first two terms of this series. In the case b) the verychoice of the fundamental solutions of (4.2) is based on the assumption � b 0 > 0and taking the limit � ! 0 in the final representation is impossible. On the whole,we can say that the case � b 0 � 0 is covered by our investigation.

We must point out that passing to the limit b ! 0 in the asymptotic formulas(2.3), (2.4) is impossible since the constants in the estimates of the remainders tendto infinity as b tends to zero.

Due to the presence of the multiplier exp[�(n/(N +1))t] in the estimates (3.11)they are more subtle than is needed for the convergence of the series in (3.21) andobtaining the estimate of the remainder Ru(x, t).

If � b 0 < 0, then it follows from (4.4) that for su�ciently large t the argument ofthe Airy functions becomes positive. Since Bi(z) ⇠ cz�1/4 exp(2z3/2/3) as z ! +1it causes a strong growth in t of one of the fundamental solutions of (4.2). So inthis case the technique used above for obtaining the long-time asymptotics can’t beapplied. However, when � b 0 � 0 by means of analogous considerations it is possibleto calculate the next terms of the asymptotic expansion of the solution in question.

REFERENCES

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