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The Cauchy problem for the linearized Boltzmann equation
THE
CAUCHY
PROBLEV
FOR
SOLTZIMAXN A. A.
ARSEN’ Most
(Received
26
THE
LINEARIZED
EQUATION* EV
ow
January
1965)
A number of papers have recently appreared on the Cauchy problem for the linearized Boltzmann equation in the kinetic theory of gases. The question of the behaviour of the solution as t + a and of the construction of the asymptotics of the solution with respect to E as E + 0 is not dealt with in them, however (E is the ratio of the length of the free path of the molecule to tne characteristic measurement of the problem). In this paper a determination of the solution of the Cauchy problem is given for the Boltzmann equation in the whole space, the limit of the solution as t + m is studied and its asymptotics as E + 0 are investigated. It appears that the last two questions are closely connected one with one another and can be solved by investigating the resolvent of the collision operator, which is perturbed by the multiplication operator by i(ck).A large part of this paper is devoted to this investigation. The basic results of the paper are formulated in Theorems 1 - 3. The group of physical problems to which our results are applicable is obviously limited to the molecular acoustics of ideal monatomic gases, but there is some interest in the strict solution of at least the very simple problem of investigating how in this case hypotheses and assumptions, which are usually used in the kinetic theory of gases and derived from the Boltzmann equation of aerodynamics, are justified.
1. ille general formulation
of the problem
Suppose that at the instant t = 0 we can represent the distribution function of the number of particles in phase space in the form
l
Zh. uychisl.
Mat.
mat.
Fiz.
5,
5,
110
664
- 662,
1965.
The
linearized
F(c, x, 0) =
.3oltrmann
111
equation
(i )“i++ e+‘p qf(c,
x, O),
where \?I << 1, c is the dimensionless speed and x a dimensionless coordinate. Tnen to within terms of the first order in 11the function f(c* x, t) will satisfy the linearized 8oltzmann equation [II $c$=Lf,
where L is a linear
operator
(1.1)
x =(q%,~3),
which does not depend on x.
1. We shall say that the solution of the equation (1.1) is in the square f(c, x. t) which for any t > 0 is integrable to c and x
Definition the function with respect
If(c, x, t> pkdx
s and for
c = (Ci, cz, c3),
f]t=o=f(c,W,
which the Fourier
<
transformation
f (c, k, t) = c
00
with respect
k = (‘k,, ii,, k8),
eikxf (c, x, t)dx,
for almost all k is a solution of the following (understood in the sense of [2, P. ~311):
ai at= - i (c, k) 3 +
to x
abstract
Cauchy problem
Lf = A (k) f: (1.2)
i It-II = f^o(~7 4. A f(c,
Here the function x, t) is considered as an element of the space L?(c) depending on k and t. The scalar product < f, g > and the norm 11 11 will be understood in the sense of this space Lz(c). We now make the following operator L.
assumption
about the properties
of the
1. tf= - v(c)f + G’f, where v(c) is integrable in the square on any bounded set and depends only on (cl, and G is a completely continuous integral operator with a symmetrical kernel &, cl ‘), where %c, cl ‘) depends only 2.
on (c1*1,
Lf>=
ICIand
(c,
f>
cl’). Lf>=Owhenandonlywhen
112
A.A.
Arsen’ev
Lf = 0, and Lf = 0 when and only when following five functions:
3.
for
Positive all c.
4.
constants
f
is a linear
vg > fl and a > 0 such that
sIG(
combination
v(c)
of the
> vo + ale/
GO does not depend on c.
c,c')Idc'
These assumptions are satisfied, for example, for the model of hard spheres. Note that we do not require completeness of the eigenfunctions of the operator L, and this is the essential difference between our method and that of [3 - 41.
2. The existence, uniqueness and properties of the solution of problem (1.2) We shall tion
prove that,
with our assumptions,
?
T(t, V&c,
semigroup
(1.2)
has a unique solu-
k),
of class
Co with an infinite-
In fact the operator A(k), A(k)f = - i(&f - v(clf + Gf, considered as an operator from Ll(c) into iq(c), is a linear unbounded closed operator with an everywhere complete region of definition D(A(k)). It can be put in the form of the sum of two operators A(k) = .4,(k)+ G, where Al(k) is the multiplication operator AI( = - i(c, k)f - v(c)f, and G It is easy to see that Al(k) generates a completely continuous operator. a strongly continuous semigroup T1( t, k) of class Co
Ti(t, k)f = exp {- Ii (c,9 + y (c)l Qf, and hence, by virtue of Theorem 13.2. I of [21, it follows that the operator A(k) generates a strongly continuous semigroup of class CO, where
The
linearized
Boltznann
113
equation
II T(t, k) II < exp {(II G II - vo)t).
(2.1)
Lemma 1 A unique solution 6 f,,(c, k) =-W(k)).
of problem (1.2)
exists
for any initial
function
This is a direct consequence of the fact that T( t, k) is a strongly continuous semigroup of class CO (see [21, Theorem 23.8.1). We shall prove that the semigroup T(t,
k) is contracting.
Lemma 2 For any k and t > 0 the inequality
11Rt,
k, 11<< 1 is satisfied.
Proof. Let u E D(A(k)). We shall construct the element T( t. k)u. As is well-known [2, Theorem 10.3.31 GTU=ATU=
- Y(C) Tu - i(c, k) Tu -j- GTu.
(2.2)
Passing in (2.2) to the complex-conjugate values we obtain --$Tu)’ Multiplying
(2.2)
;(Tu)
= -Y(C)
(Tu)’ + i(c, k) (Tu)” + G(Tu)‘.
(2.3)
by (Tu)+, and (2.3) by Tu and adding we obtain
(Tu)’
= - 2v(c) (Tu) (Tu)‘+
TuG(Tu)’
- (TuJ’G(Tu).
(2.4)
that in (2.2) and (2.3) the derivative is understood in the sense of the metric of Ls(cl, and in (2.4) in the sense of the metric of Ll(c). From (2.4) it follows that
Note
(T(t2)4 (TM 7~)l
=
s ti
(T(h)u)
(T(t2)u)’
[- 2v(c) (Tu) (Tu)‘+
=
(2.5)
TuG(Tu)‘+(Tu)‘GTu]&.
We integrate (2.5) with respect to c and on the right-hand side we change the order of integration with respect to c and T. (The change of order of integration is not difficult to justify). Considering property 2 of
114
A.A.
A rsen’ev
the operator L we obtain % 11~(~~)~~12 - liZ’(ti)ulIa = l Z(Tu, L2?u) dz < 80. ti Since (2.6) is valid for WY complete, we have
Let t >.to
> 0. Tien
II <
IIT
II.
by virtue of (2.1)
IIT(t)ll ~4 llT(t~)lI Since to is arbitrary
D(A(kI),and ~(~(k)~ is everywhere
u E
IIT
(2.6)
<
exp
(2.7)
from (2.7) we obtain
{(--WI+
Wll)~~}.
it follows from (2.8) that f\T(t)
(2.W f! <
1.
3. The existence and uniqueness of the solution of problem (1.1). The bebaviour of the solution as t+aJ We shall briefly
discuss the basic idea of the following
calculations.
Let &A, k) = W - A(k))-'and u E &A(k)). Since ‘Qt, It)is a convoluted semigroup of class CO. it can be expressed as follows in terms of the resolvent of its infinitesimal operator: y+io T(t,
l exp(ht)R(h,
k)u==&lim
k)udh
(y > 0).
(3.4)
-y-i0 First of all we shall explain the analytical properties of R(h, k, in the left-hand half-plane of h, transfer the contour of integration there, find those residues at the singular points which are essential for the asymptotics (thes will depend on k) and evaluate the others. Then we shall integrate the solution obtained with respect to k and find .ffc,x, tl. The whole question is consequently reduced to the investigation of the resolvent R(h, k) as a function of the variables h and k. We now introduce the operator K(h, k) =
1
G. h-t-i(c, k)+y(c)
The
It
linearized
Boltznann
115
equation
is not difficultto show that R(~,k)e=(E-K(h,k))-'~+i(e
;j+y(eJ n.
9
(3.2)
From (3.2) it followsthat in the half-planeRe k 2 - vu + 6 (hereand further on we shall denote by 6 8 fixed positivenumber,So small that - vo f 6 < 0) only singularitiesof the function(F:- I((& k))-’ can be singularitiesof ,?(A,k). We Shall study the behaviourof this function. The followinglemma is obvious. Lemma 3 If we are given 8n operatoru"l(a),where 81(a) + 0 as a -)11in strong operationaltopology,and the operatorJ!&is completelycontinuous,then C(a) =Sl(a)E, + 0, a -*0 in uniformoperationaltopology. Lemma 4 If Re A'> - Ve + 6, then 11It'(h, kt I\ + 0 8s d = J+/Kf-;2+
m
and
COnSeqUentlYa number oe(6) exists such that tt,q(h, k) 11< 0.S if d > 00(S).
Proof. Since G is completelycontinuousit is sufficientto show that for any functionu E Lz
sTh+i(c,be> I”
k)+v(c)f*'
as d + coand this is obvious. Lemma
5
For any fixed k the functionRO, k) is analyticwith respectto h everywherein the half-planeRe h > - vg + 6 except for a finite number of poles <&j(k) 1. 2.
The poles Aj(k) satisfythe followingrelations:
(a) Re Aj(kI\7 h = 0;
0
and the equalityapplieswhen and only when IkI =
(b)IIm hi(k) I < 00(6 ), where oo(6) is the constantdefined in Lemma 4, dependingonly on 6;
116
A.A.
Arsen’ev
(c) for Ikl ’ ~~(61 the function .&A, k) has no singular points in the half-plane Re A > - vo + 5.
3. The point ho can be a singular point for the function .?A, k) in the half-plane Re A > - vo + 5 when and only when it is singular for the function (E - K(A, k))-’ and belongs to the discrete spectrum of the operator A(k). Proof. From (3.2) it follows that the point ho can be singular for the function Rl’h, k) in the half-plane Re h >, - vg + 6 only in the case where it is singular for the function (E - K(h, k)J-‘. We shall denote by Z(S, k) the set of those points of the half-plane Re A 2 - v0 + 6, for which the equation u = K(A, k)u
(3.3)
has a non-trivial solution. By virtue of Lemma4 a constant wo(6) exists such that if 1~) > oe(6) theri for all k IIR(o + i-r, k) II < 0.5, if a,, - vg + 6. Consequently, for no k does the set Z(5, k) coincide with the whole half-plane Re A > - vc + 6. But the lfunction K(A, k) is analytic in the half-plane, as is the function h for any fixed k, end hence it follows that (E - K(h, kj1-l is analytic for Se A > - vu + 6 everywhere, except for points of the set Z(5, k, and each point of the set ‘Z(6, k) is an isolated pole of (E - Kth, k))-’ [51. Let ho E (6, k). Then for some u E IQ(C) 1
Hence hou + i(ck)u + Y(C)U = Gu, hou + i(c, k)u = Lu, Mu, u> + i( (c, k)u, u> = (Lu, u>.
(3.4)
By virtue of the symmetryof the operator L we obtain from this Rsho = (22,Lu) < 0,
Im&=
-i( (c, k)u, u>.
let Re A0 = 0. Then < u, Lu > = 0 and f+u = 0, where u is a linear combination of the functions u,(c)
Now
The
linearized
Boltzmann
u= i
117
equation
a,(k)u,(c).
-0
Substituting this equality in (3.4) and considering the explicit form of the functions u,(c), we find that the equality (3.4) is possible only in the case where either all the a,,, are zero or A = (k( = 0. Consequently for points of the set X(6, k) property 2, (a) holds. Properties 2, (b) and 2, (c) for points of the set X(6. k) are obvious consequences of Lemma4. But from this it follows that all points of the set X(6, k) are situated inside the closed rectangle (1x1h( < oo(6). - vg + F - vu + 6 the singular points of the function (8 - X(h, k))-’ are the only points of the set I(&, k) and only singular points of the function (E - K(h, k))-’ can be singular points of R(h, k), it remains to be proved that each point of the set Z(6, k) belongs to the discrete spectrum of the operator A(k). But this is a corollary of the equality of (3.4) which can be rewritten as (A& - A(k))u
= 0,
IlulC= I
Lemma 6
The function (E - K(A, k))-’ is continuous in uniform operational topology with respect to the set of variables (A, k) at those points (A, k) with Re A 2 - vu + 6, at which it exists. Proof. It is sufficient to show that the function &(A, k) is continuous in uniform operational topology. By virtue of Lemma3 this is a consequence of the fact that for any function u E L*(c) 1
h+Ah+i(c,k+Ak)+v(e)
2
1
-
h+i(c,k)+v(c)
I
&CO
if d = T/lAhlz + IAkl”-+O. Lemma
inf p(k), where p(k) is the distance from the spectrum Ikt>r F ’ 0. of the operator A(k) to the imaginary axis. Then pa(r) > 0 if Let
PO(~)
=
118
A.A.
Proof.
From Lemma5 it
follows p(k)
Arsen’ev
that >
if
IkI > oe(S)
we always have
I-vo+SI.
Therefore po(r)>min
inf
{l-v0+6l;
P(k))
W8)2IkI>r
and it
is sufficient
to prove that
p’(r) =
inf P(k) > o,(6)> Ik 1>r
0.
Let p’(r) = 0. Then a sequence {k,,,) exists. which converges to some point ko, where r & 'ko / < oo !a); and a sequence of numbers .Ah,= u,,, + iT,,, such that uR1+ 0, an;l each point A, is a pole of the operator MA, k,,,,. But if A, is a pole of the operator R(h, k,) and o,,, > - vo + 6, then without fail 1~1 < 00(b). Since the function T(T) = 1) (E - X(0 + it, ko))-’ is continuous with respect to T by virtue of Lemmas 5 and 6, it is bounded on the segment by some constant Cl. BY virtue of Lemma6 (CJ! q and EZ > 0 exist such that if
ITI < oo(6)-
for
any
T
E
[-@o(T),
0"(T)]
It(E - K(u + iz, k))-i is satisfied
II < 2Ci,
and so also
tlR(a -I- iz, k)
It <
F <
It must also be satisfied for all k, and h, dicts the choice of the sequence k, and A,.
00.
for
n>mo,
which contra-
Lemma 9 For any r > 0 a constant C(r) exists such that on the line Re h = - 0.5 pa(r) < 0 for all k such that lkl > r, the norm of the resolvent II R(A, k) !’ r. of F;o;{. I .
If
d =~~lh12+
IIR(h,
lkl"2 m(6)
k) II < ;. II (E -
Therefore it is sufficient lie inside the compactum
to consider
and Re A 2
- vo + 6 by virtue
K)-i 11< 216-Cl O”. only those
points
(A, k) which
The linearized
Boltznann
119
equation
II (E - K(h, k))-i 11is continuous with On this compactumthe function respect to the set of varianles (A, k) ani therefore is bounded. In addition, the function 9th. k) is also bounded. The analyticity of R(A, k) in the half -plane Re h >, - 0.5 pa(r) follows from the definition of PO(~) and Lemma5. Lemma 9 Let h = c + iT; c > - VQ + 6. Then for any u E f-2(c) IIR(o + ir, k)u II+O,
1~1--t 00.
for any fixed k uniformly with respect to u 2 - vg + 6. Proof. Let t-rI > w(6).
Then
I” ’ & .+. 0 __---- be) s 6*+(Z+(W)* ’
G4 Lemma 10 For
a,ny u E
Lz(cJ
and any k such that IkI ’ 0, lim IIT(t, k)ull = 0.
t-NW
Proof.
Let
u ED(A)
nD(Az).
initially.
Then by virtue of (3.I)
we
have
If
IL/ >F
integrate
’ 0, we can use contour integration along the rectangle [Y --9
Y-t b,
-0.5po(r)
$- io,
in this integral.
-0.5po(r)
We
- iw],
where pot r) is defined in Lemma8. The integrals along the upper and lower sides of the rectangle approach zero as o -, w, and since R(h, k) is analytic inside it
120
A.A.
Arsen’ev
-0.5pdt)+io
T(t, k)
exp(U)R(h, &lim 5 w-PO0 -0.5p0(r)--io
u’=
k)udh.
Let t > to i 0. Then lim l T(t,k)u=& o*oD
R(h, k)ud( 7)
= &lim 5 c(-&‘(A, CD-PC%7
k))u&.
=
-o.5po+20
1 exp(ht) = -1im s 2ni “-+O” -0.5p,--i&l t But R(h, k) = (E f AR) /A. If - 0.5 pa(r) we have ,,R2u,, < 11412 +
.
@(A, k) u &.
(3.5)
then on the line Re h =
u E D(A2),
2CIL44l + C21M241 . A2
Therefore from (3.51 it follows that -0.5p+ioo
IITuII<
Gexp
{-0.5po(r)t) <
1 _,.,JG
CSew {-O,5
G P(F) t) *
t-too.
0,
According to Lemma2 we have IIT(t, k) II < 1. Since D(A) fl D(A2) is everywhere dense in L*(c), because A(k) is an infinitesimal operator of the semigroup of class Co, by the Banach - Steinhaus theorem for any u lim IIT(t, k)ull=
t-NW Theorem
0.
1
Let fo(c, xl be of integrable square with respect to c and x, and the Fourier transform of this function with respect to x be the function &(c, k) =D(A(k)) for almost all k. Then problem (1.1) has a unique solution flc. x, tl in the sense of definition 1 and
1 If(c,
x, t) pdx-+o,
t-t
00.
By virtue of Lemma 1 a unique function f 0 Proof.
iS
The linearized
co
R (a,
“k)fo0%
Boltzmann
1
1
L h + i (0, k) + y (c) ‘8,(0, k). G
h+i(c,k)+~(C)
i=.
121
equation
1
is strongly measurable with respect Consequently R(A, k)f;,(e, k) a strong limit of strongly measurable functions. Hence
to k as
-f+io f (c, k, t) = &
lim \ exp (b 0 R (A, k) h (0, k) dh o+(x)U-Al
k3, Chap. 3, Theorem 171 for any is also stronglv measurable. Therefore t > 0 the function. F(icc, k, t) is i~easurable on the product of the spaces c @ k. Rut for any k ancl t > 0 by Lemma 2 \ I i (0, k, t) I2 dc -G \ I io (c, k) la do. Therefore
\dk\If(c,k,Wdc<\~\/h(~~Wl’k
of Fubini’s
theorem it
s Therefore
for
almost all
de
s
for
c and all
almost all
We can prove that all
k bs Lemma 10
s
t > 0
c there
of problem theorem
1 G\jf(o,
that
k, t)ladk
f (0, x, t) = & which is a solution virtue of Parseval’s
from this
If (c, k, t) J”dk< cm.
sli(c, consequently function
follows
exists
s
(as an element of Lz(x))
e-ikxj(c,
(1.1)
k, t) la dkdc
-Ip(c, k, t)j2dkdo+0
a
k, t) dk,
in the sense of definition
1. By
= !I#@, x, t)I’dxdc.
as
t*
00.
In fact
for
almost
122
Arsen’ev
A.A.
s A
If(c,k,t)12dc-+0 as
and for
any t ’ 0 s
Therefore
(j (c, k, t) la dc <
by virtue
of Lebesgue’s s
This proves
lf(c,
s
[ io (0, k) la dc=Ll
(k).
theorem
k, t)j2dcdk-+0
as
t--too.
Theorem 1.
‘vote. It is easy to see that Theorem 1 to be satisfied is
4.
t--too
a condition
which is sufficient
for
The refinement of the asymptotic solutions as t + 00
Theorem 1 gives no information the given point x. To obtain this about the initial functions fo(c, a more accurate evaluation of the operator Uh, kJ
Q((h,k)u=
about the behaviour of f(c, x, t) at we must make additional assumptions x). First of all, however, we obtain operator T( t, k). We shall define an i
h+i(c,k)+v(e)U'
Lemma 11
If
h = u + i-r and a >
- vg + 6, then
o+im j
IlQGQIIdh
<
C(d)
(i+
lkl).
o-im Proof.
Let
11u 11 = 1 and y = RG Pu. Tllen
1a+i(c, I.k)+y(c)
Ml= <
( I sup ICI
s
G(c, ci) a+i(ci,k)+v(q
1 a+i(c,k)+y(C)
U(Ci)&
II<
(4.11)
2
I)
7:~
s
I G(c, ci) I dci <
Gy2(b
k) ,
‘he
linearized
Boltrnann
equation
12 3
where
Using the evaluation
Y(C) >
YOj- CZI c 1, we can show that
Ll+iCO ~~‘yWWWWW+
PI).
(4.2)
u-ico From
(4.1)
and (4.2)
the statement
of the lemma follows.
Lemma 12 For any r > 0 and t > 0 if
Ik\ > r
II W, k) II < C(r) (1 + lkl WY where
a(r)
PFOOf.
> 0 if
r > 0.
We have the equation
(see
equation
3.2)
Ru=!2ufS-JGRu. Hence
Ru=Qu$QGQu+QGQGRu. Substituting
this
in (3.1)
we obtain
(u E D(A))
Y+ie
T(t,k)n=&lim
l crop(At) [Ou + QGQu + QGQGRu] dh = -c0 y--i0 = W) + W) + b(t).
(4.3)
For Ii(t)=&-
lim~exp(ht)S2udh O-+O” y-i0
we have the evaluation III,(t) II < of which we can be convinced evaluation
eWl4l,
by calculating
(4.4) 11 in explicit
exp(ht)QGQudh
form. For the
124
A.A.
A rsen’ev
we transfer the contour of integration to the line Re h = a = - vu + S and apply Lemma 11 to the expression under the integral. We obtain
II h(t)
II < C(6) (1 + Ikl)e-utllz41.
(4.5)
With the integral VW lim 1 .exp (it) OGQGRu dh o+oo y-_io we proceed similarly, only we transfer the contour of integration line Re A = - 0.5 PO(F) < 0 and use Lemma 8. We obtain
to the
III,(t) II G G(r) (1 + Jkl)e-a(r)fllull.
(4.6)
Gathering together all the evaluations (4.3) - (4.6) and taking into account the fact that D(A) is everywhere dense, we prove the lemma. Lemma 12 gives an evaluation of the operator T(t, k) for IkI 2 F > 0. Now we need to obtain an evaluation of this operator in the neighbourhood of the point k = 0. We shall previously refine the behaviour of the resolvent in the neighbourhood of the point A = o,k = 0. Lemma 13
An (I)
F,J
>
0
and au > exists
11R(--o,,
+
iT, k) II <
such that
C(r0, 000) < 00
for all
k such that lkl <
ro;
(2) in the half-plane Re h > -CTOy R(h, k) for any fixed k there is a finite number of poles of common multiplicity 5, hj(k), I< j
(3)
the projectors J’j(k) onto the invariant spaces of the operator answering to the eigenvalues of hi(k), are analytic functions of of the vector k = noIkI the parameter IJ = IkI for any fixed direction
A(k),
Pj(k)
=
iP#(n,)
IkIm.
m==O
PFOOf. 1. As we have explained
the noles
of the function
(4.8)
R(h, k) in
The
the half-plane tion
- VII + 6 are those
Re h >
u= h +i has a non-trivial bringing the axis
equalities
points
A, for which the equa-
sG(c, ci)u(ci)dci
1 (c, k) + Y(C)
1
(4.9)
s
-
G(gc, c(f)) u(c(‘))o?c(f).
v(c)+ic+(-G
out the cnange of variables y(gc)
125
equation
We fix the vector k. Let ,g be a rotation solution. of z into the direction of the vector k. Then
zz(gc) = Carrying
Boltzmann
linearized
=:Y(c),
u(gc) = -
c(l) = =
G(gc, gcW)
1
Y(C)+ hlkl-+
h
gc(l)’
and considering
G(c, &I’)., we obtain
s
G (c, 13’)‘)u (g&j’) ddk
It is clear that if the equation (4.9) equation (4.10) for the same h has the tions. Rut those h for which equation depend only on lkl, alid so the same is
= -Y(C)
+ G -
(4.10)
has a non-trivial solution the same number of non-trivial solu(4.10) has a non-trivial solution true also for equation (4.9).
Let no be a unit vector, collinear wii;h k. We shall Ikl and shall consider the operator A(k) as a function and the complex parameter ~1
A(k)
the
&t(w)
= Ao -
assume that P = of the vector no
@Ai.
Rut
and if P = 0 the point A = 0, by virtue of the condition, is an isolated pole for the function 8th. k) of multiplicity 5. Since A(k) is closed and symmetrical about the line Re CI= 0, in view of one of Kate’s theorems [71 constants E > 0 and ~2 > 0 exist such that inside the circle R(A, u) have only a finite number t ~11~ ~2, the functions of poles {hi(k)). and the common multipli.city is 5 and every one is of the first order and is an analytic function of n. The projectors Pj(k) are also analytic functions of CI in the same neighbourhood of ~1 (with a fixed direction for no).
IAI
Let p = 0 and the distance
from the point
A = 0
UP to the rest of the
126
A.A.
Arsen’ev
points of the spectrum of the operator .A, = - v(c) + s be ci. This distance is positive by virtue of Lemma ri. We assume that cl < 1 -vg + 6 1 In the opposite
case we shall
denote
by d the quantity
I- vo + 61). We choose a number ~(6) as was consider the closed region 2 consisting of a lines Im h = oo(6). Im h = - 00(6), Re h = 0, a circle is cut out of radius !h d with centre
d’
= min{d;
done in Lemma4. We shall rectangle bounded by the Se A = - d//2, from which at the point h = 0. If
Ikl = 0 the function (1 (E - K(h, k))-’ It is continuous with respect h in this closed region and bounded there by a constant c. Since )( (E - &A,
k))-’
a neighbourhood 11(E -
K(h,
of
k))-‘11
to
11 is continuous with respect to the set of variables, the region 3 and r ’ > 0 can be found such that <
2C
if
h E O(D);
Ikl <
r.
This reasoning
shows
that if IkI< r’ the function 2(X’, k) is bounded on the line Re h = -d/2, and all its poles, lying to the right of the line Re.A = -d/2 lie inside IhI < d/4. We shall assume that FO = min {r; EP), a2 = -d/2. the circle These numbers will satisfy all the conditions of the lemma. p,,,‘j’(no) can be The coefficients a, (j) in (4.7) and the projectors calculated by using the ordinary methods of perturbation theory. The calculations are quite cumbersome since we are here concerned with a We here quote only those results whicn five-fold degenerate eigenvalue. the properties we shall be using later on. Note that in their derivation of symmetry of the operator i, with respect to the rotation were essentially used. Lemma 14 The following 1) a,(j) =
(i)-nh,tj),
2) aE!+i = -
are valid: are real
a,(j)
(2) azn
0,
,r\
=
statements
=
(2) azn+i
(3) a2n,
=
-
(4) azn
(3) a2n+i,
azn+i,
3) A,(i) =
0, a&2) =
4) 1&j) >
0,
0,
&i(3)
1 < j < 5;
=
0, n2(4!
A&4) =
zzx
1%;
g2’1
j&5) +
=
1/5/s;
2h2(2)];
5
5) p,Cj)f =
(q@,
f) rp&j), &j) =
2 c,(j&. a=1
The numbers
cc,(j)
are given by the following
table:
=
(5) azn ,
f4) knti
=
The
1
-r
1/z
2 3
-ccoscp
sin cp Cos8 r
v Iv
-sin0
4
Boltrnann
v2
sincp
0
sin cp T cos cp CQS0
0 sin 8
FT sin 8 cos cp -rFsinf3
Note that AZ(‘) is proportional thermal conductivity of the gas.
127
equation
0
0
0 0
4
linearized
Cosq
to the viscosity
-v0 0
+s*
3 3
+
-+ose
&
and AZ’ ’ ) to the
Lemma 15
Let k be such that u E D(A(kIJ. Then T(t, k)u =
Jkl <.ro,
where ro is the same a~ in Lemma 13, and
~exp(hj(k)t)Pj(k)u+exp{-a(r,)t}B(k,
t)u,
(4.11)
j=i
where a(~,) kor t.
>
0
and
ljB (k, t) II < L’ (T-,-J < CW, C(Q)
does not depend M
Proof. In (3.1) we transfer the contour of integration with respect to A in the left half-plane onto the line Re A = - 00; on this line, by virtue of Lemma 13, the resolve& is bounded for kI
2
If the initial function io(c, x) is such that its Fourier transformation with respect to x (the function fo(c, k) satisfies the conditions
128
A.A.
then for
any fixed
point
Arsen’eu
x1
CJI.lim Pzf(c, x, t) = t-tw
w(c),
where W(C) = -
1
1
5
4n3 1
-&
““exp (-
C
c2/2) ( c2 -
5/2) 1 exp (-
c2/2) X
ciexp(-e2/2)X
x(c3-5/2)~0(c,x)dcdx+~(&)~‘*~
i=i X 1 ciexp(-
Proof.
Therefore
Let II E Lz( c).
Then
(u, j(c,
(u, T(k, t&c,
for
k, t)> =
follows
-
1
8n3 s
llNi&,
G
i
that
for x (and not for
) (c,
eikx (u,
k, ~D/dII4I~llh(~~
k)ll.
k)IIdk<~. almost
k, t)) dk =
all
,-i’=j
x) the integral (c, k, t) “k>
exists. Consequently for any x and t > 0 a function - an element of the space &(c) 1 f(c, x, t) = s +‘+(c, 8n3 We shall
x)dcdx].
any t ’ 0 \clkl(u,
Hence it
k))
c2/2)f0(c,
divide
the integral
s
cikxj (c,
of
(4.12)
f(c,
.
x, t! exists
k, t)dk.
into
(4.12)
two
k, t) dk +
IkKro
e-ikxj (c, k, t) dk = 1, (t) + 1, (t). s IW>ro
+& Using the value
II12(OK
of the operator
\ IIW Ikl >r.
T(t,
k)llllfo(c7 k)lIdk\
s
k given
in Lemma 12 we obtain
C-MJU
x
x (I + Ikl)llf^,(c,k))Idk = 0 (exp (-- ad>>.
The
linearized
Boltzmann
129
equation
Let u E Lz ( C) . Using Lemma 15 we find
& \
(u,
=
(4.13)
k, t)dk> =
e-ib](c,
IWOo
1
=
<
“=
X [j$le~~
_-
1
s Ikl
%” j=l
e-ikxT (t, k) j. (c, k) dk> = Q, ,k,
{hj (k) t> Pj (k) fo (~7 k) + exp (-
5 21
\
e-iL* x
lU
aat) B (k, t) 70 (c, k)] dk)
ikx + hj (k) t} (u, Pj (k) io
exp {-
\
&
(c,
k)) dk + 0 (exp (-
7~
~zt)).
To evaluate the integral in (4.13) we use the formulae (4.7), (4.3) and Lemma 15, which gives us some information about the behaviour of Aj(kJ in the neighbourhood of the point k = 0. Note that only this neighbourhood makes an essential contribution to the integral of (4.13) since Re Aj (k) < 0 if IkI > 0. After straightforward calculations which are usual in the saddle-point method we obtain
(u, Ii(t)> =
t;2 (u, w(c) > + 0 ( f-).
T
This proves the theorem. We can verify that of the theorem will be satisfied if the initial fies the following requirements:
I)
Ifo(c, 5) Idxdc+
2)
1
[s1x1
Ifok
1 IA2fo(c,
x) kq2do
x) 12dxdc<
<
5.. The construction respect
conditions function
(1) and (2) fo(c, x) satis-
00;
CQ.
of the to E as
asyrptotics E + 0
with
In this paragraph we shall study the asymptotics with respect E + 0 of the solution of the Cauchy problem for the equation
f&c;== fLf,
f 1t=o =
We shall assume that the operator L satisfies Section 1 and that the initial function fo(c,
fo(c, x).
to E as
(5.1)
the conditions 1 - 3 of x) is such that
130
A.A.
fo(e,
k) E NAM
);
11io
Arscn’ev
k) 11 is bounded on any compactum and
s+IklN)IIh (I
We shall accuracy,
k)IIdk
assume the number J to be sufficiently large, but not fixed so as not to complicate the argument unnecessarily.
Since for of Section I ator L, then formation of
tne operator are satisfied, on the basis the solution
in
(I/E)L for any E > 0 all the conditions 1 - 5 as soon as they are satisfied for the operof the previous results the Fourier transof equation (5.1) can be found by the formula Y+iw
f (c, k, t, E) = &
lim 1 exp (W R (A, k, e) jO (o, k) dh, o-%X3 y-iw
where RCA, k, E) is the resolvent
of the operator
R(h, k, E)=(A.E--A(k,
A(k, e)u= It
is not difficult
-i(c,
to verify
E))-i,
k)u+(c)u++Gu. that
R(h, k, e) =
d?(eh, ek, 1).
exp,(ht) R (eh, ek) .=
10 (c,
k)
dh
Hence
=
T (+ , ek) f,, (c, k).
Therefore
f(C,
x7 t,
E) =
&
Se-“‘T
(;,
ek)jO(c,
z +$&,-WET
(5.2)
k)dk =
(f , k)j,
(c,+)dk.
We shall divide the region of integration with respect to k in the integral (5.2) into two parts; into some neighbourhood of 0 and its compleof the operator T( t, k), which ment, For small Ik\ we use the evaluation is given by Lemma 15, and for large Ikl we shall evaluate the operator T( t. kl by means of Lemma 12. We choose ro as in Lemma I5 and obtain
1 1 f(c, x, t, E) = mE3
s
e-ikXIEY’ ($,
k)j,(c,
;)dk=
The
+
We evaluate
According
Hence it
tinearited
1 1 ~8n3 E”
the integml
to Lemma 15, if
follows
s
Bo2tzmann
@‘=/ET
($
equation
7 k) j. (c, f)
131
dk = 11 (t> + 41 (t>-
lkj> PO ‘l,(t),
using Lemma 12
Ik I =zro
that
Therefore
f5.3) X Pj (ek) fa (c, k) dk + 0 (~XP {-$})m Re hj(k) > 0 if IkI > 0, the quantity FO in (5.3) does not play an important role (we can take any smaller positive number). On the basis of (5.3) we construct asymptutics of the function f
Since
132
A.A.
Arsen’ev
respect to E, such that the remaining terms will not only be small relative to E as E + 0 uniformly for t E [to, cc), but will also approach zero as t + co so that their order in t will be larger, the higher the power of E we consider. Lemma 16 Let h+“](p) tion Aj(U):
be the n-th segment of the Taylor
hjLnl
(/.I,)
=
$
series
for
if
lkl
the fllnc-
am(j)pn.
m=i
Let the function
y(k) be such that
1 (1 +Ikln+i) Then for
b(k) Pk <
sufficiently
small
O°F
b@(k)
I<&,
r > 0, e > 0 and any t ‘0
if
n>2
Since
Proof.
analytic function of CI in the neighbourhood 1~1< rg and al ‘j) for all j is either purely imaginary or zero, and a,(j) < 0, we can find an r E (0, rol such that for all IJ E (0, rl is
an
(5.4) and where b > 0, we can assume that
Let us consider I(e, t)=
)
the integral )
{exp
[x’F]-q~[
hjrn’(~k)t
]}rp(k)&
1=
IklGUe (5.5)
=
I
,kja,.exP[
The
linearized
We now apply to the function of the mean
Op----
[
In view of the analyticity
from (5.6),
If all
I
-
0 < -
and from (5.7)
it
I <
considering exp{-
s Id4r/e
133
braces
of
tne theorem
(5.5)
t] x
(5.W
q+(k)dk].
hj (p)
hj(P.>
CiPn+i~
(5.4)
I PI < r-
we obtain
b&t lk12+Ci&ntlkIn+i]Ikln+ilg(k)
Idk(
small
Ikl <
(0 <
r <
follows
(b/2C)1N*-2),
E<
1)
for
r / E,
betIkl2+Ct@Ikln+i
<
--;et(kj2
that
I(&, t) < ctsn
exp{--~d[k12}
1
Ikln+j I+(k)
Idk.
(5.8)
O
shall
now evaluate
Z(c,t)=
the integral
s exp{--;etlk~2} Ikldrle =
in (5.8).
S w(-
have
We
Iki”+‘I$(k)Idk=
l exp(-~lk12}Ikln+iltp:k) Ikla
+
Fk2)
Idk+
Ikln+‘lq(k)
Idk <
i
S exp{-~c?}
. (5.7)
r and E are sufficient1.v k such that
& of
I&j[“‘(P)
I (E, t) > cit.9
in the
&‘n’ (Ek) - hj (Ek)
X exp
Therefore
enclosed
y)t][*k)-6kj(6k)
) ,kL_;xp [
I(&$)’
equation
Boltzmann
an+Bda+e-be’/2~
Ikl”++j(k)Idk<
0 G
+
s 1
+
(Et)
(n-w’2
C3e-bet12
<
” 1
+
(Et)
(n+4)‘2
134
A.A.
In view of this
evaluation
it
Arsen’ev
follows
that
C&s”
]I(& t)I< This proves
from (5.8)
1 + (et) (n+W’
the Lemma.
We now‘proceed (5.3) we have
to the construction
of the asyaptotics.
According
to
5 1
-1
Pi (sk) jo
(c,
k) dk + (54
+ 0 (e-a’/“) = -8;3
i \ exp [-ikx+m] j=l 1,+&f
8
x
x [ 2 P,(j) (no) (E 1k j)m] f. (c, k) dk + 0 (e-u’/E) + m=o
x f,, We also
(c,
k) dk = I,(t)
+ I2 (t) + 0 (e-atie).
have
j=i
I kl
, pj(Ek) - i P,(j)(no) (elkl)m 11 m=o
/Ix (5.10)
x Il.&c, k) I!dk < G { exp [ IklG-/e
We now transform
Idi) =
11(t):
k3 i s
ew
[
?tij
IkJ=Zr/e
The
hj[“I (Ek) t -
csp
E
linearized
ikx ]}[
Boltzaann
135
equation
~p,,ij)in,)(Elki)“]j,(c,k)=rl(t)+lz(t). nG==O
By virtue of Lemma16; for the integral
i,(t)
we have the evaluation (5.12)
From (5.9)
- (5.12) Theorem 3 follows.
Theorem 3
If fo(c, x) satisfies the conditions enumerated at the beginning of this paragraph then for the solution of problem (5.1) the following evaluation holds :
I!&, x, t, E)--2ifn, l(C,x7 t, E) IIGC
C
$-+i
&%
-+ I + ( &t)f*+h)‘2
1 + (Et)@+Qn
I+o
(e---aq,
where
qh,
1
(5.13)
(c, x7 4 8) =
x [ i Gus m=Q
felkl)m]j,je,k)dk. *-
Unfortunately space does not permit us to dwell in detail on the physical meaning of the results obtained or on the method of calculating the quantities occurring in formulae (5.13). We only note that by means of (5.13) we can establish Stokes’s formula for the absorption of ultrasonic waves in gases. I wish to express my deep indebtedness to A.N. Acknowledgements. Tikhonov for nis interest in my work and to A.4 Samarskii for discussing my work and giving me valuable advice. Translated
by H.F.
Cleaves
135
Arscn’ev
A.A.
REFERENCES
1.
KARLEMAN, T. Mathematical Foreign
2.
HILL,
Literature
E. and PHILLIPS,
Foreign
Literature
GRAD, H. Asymptotic 6, 2, 147 - 161,
4.
SIROVICH, L. Dispersion
tions 5.
theory 1963.
KATO, T. math.
analysis
Rouse,
of
Soltzmann
the
theory
of
gases.
1960. and
Moscow,
Soltzmann
Integral
lndag.
equation.
equations
math.
24,
DUNFORD, N. and SC~ARTZ, York,
7.
P. Functional
kinetic
moscow,
semigroups
.
1962.
equation.
Phys.
Fluids,
relations in rarefied gas dynamics. Phys. solutions 1, 10 - 20, 1963; and Formal and asymptotic theory. Ibid., 2, 216 - 230; Generalized normal solu-
the
SEELEY, R.T. meter.
6.
of
in the
Rouse,
Publishing
3.
Fluids, 6, in kinetic
problems
Publishing
4,
Ibid.,
depending
434
D.T.
- 442,
Linear
10,
1423
-
analytically
1439.
on a para-
1962.
Operators.
Interscience.
New
1956. On the SOC.
perturbation
Japan,
4,
theory
323 - 337,
of closed 1952.
linear
operators.
J.
CAUCHY
PROBLEV
FOR
SOLTZIMAXN A. A.
ARSEN’ Most
(Received
26
THE
LINEARIZED
EQUATION* EV
ow
January
1965)
A number of papers have recently appreared on the Cauchy problem for the linearized Boltzmann equation in the kinetic theory of gases. The question of the behaviour of the solution as t + a and of the construction of the asymptotics of the solution with respect to E as E + 0 is not dealt with in them, however (E is the ratio of the length of the free path of the molecule to tne characteristic measurement of the problem). In this paper a determination of the solution of the Cauchy problem is given for the Boltzmann equation in the whole space, the limit of the solution as t + m is studied and its asymptotics as E + 0 are investigated. It appears that the last two questions are closely connected one with one another and can be solved by investigating the resolvent of the collision operator, which is perturbed by the multiplication operator by i(ck).A large part of this paper is devoted to this investigation. The basic results of the paper are formulated in Theorems 1 - 3. The group of physical problems to which our results are applicable is obviously limited to the molecular acoustics of ideal monatomic gases, but there is some interest in the strict solution of at least the very simple problem of investigating how in this case hypotheses and assumptions, which are usually used in the kinetic theory of gases and derived from the Boltzmann equation of aerodynamics, are justified.
1. ille general formulation
of the problem
Suppose that at the instant t = 0 we can represent the distribution function of the number of particles in phase space in the form
l
Zh. uychisl.
Mat.
mat.
Fiz.
5,
5,
110
664
- 662,
1965.
The
linearized
F(c, x, 0) =
.3oltrmann
111
equation
(i )“i++ e+‘p qf(c,
x, O),
where \?I << 1, c is the dimensionless speed and x a dimensionless coordinate. Tnen to within terms of the first order in 11the function f(c* x, t) will satisfy the linearized 8oltzmann equation [II $c$=Lf,
where L is a linear
operator
(1.1)
x =(q%,~3),
which does not depend on x.
1. We shall say that the solution of the equation (1.1) is in the square f(c, x. t) which for any t > 0 is integrable to c and x
Definition the function with respect
If(c, x, t> pkdx
s and for
c = (Ci, cz, c3),
f]t=o=f(c,W,
which the Fourier
<
transformation
f (c, k, t) = c
00
with respect
k = (‘k,, ii,, k8),
eikxf (c, x, t)dx,
for almost all k is a solution of the following (understood in the sense of [2, P. ~311):
ai at= - i (c, k) 3 +
to x
abstract
Cauchy problem
Lf = A (k) f: (1.2)
i It-II = f^o(~7 4. A f(c,
Here the function x, t) is considered as an element of the space L?(c) depending on k and t. The scalar product < f, g > and the norm 11 11 will be understood in the sense of this space Lz(c). We now make the following operator L.
assumption
about the properties
of the
1. tf= - v(c)f + G’f, where v(c) is integrable in the square on any bounded set and depends only on (cl, and G is a completely continuous integral operator with a symmetrical kernel &, cl ‘), where %c, cl ‘) depends only 2.
on (c1*1,
Lf>=
ICIand
(c,
f>
cl’). Lf>=Owhenandonlywhen
112
A.A.
Arsen’ev
Lf = 0, and Lf = 0 when and only when following five functions:
3.
for
Positive all c.
4.
constants
f
is a linear
vg > fl and a > 0 such that
sIG(
combination
v(c)
of the
> vo + ale/
GO does not depend on c.
c,c')Idc'
These assumptions are satisfied, for example, for the model of hard spheres. Note that we do not require completeness of the eigenfunctions of the operator L, and this is the essential difference between our method and that of [3 - 41.
2. The existence, uniqueness and properties of the solution of problem (1.2) We shall tion
prove that,
with our assumptions,
?
T(t, V&c,
semigroup
(1.2)
has a unique solu-
k),
of class
Co with an infinite-
In fact the operator A(k), A(k)f = - i(&f - v(clf + Gf, considered as an operator from Ll(c) into iq(c), is a linear unbounded closed operator with an everywhere complete region of definition D(A(k)). It can be put in the form of the sum of two operators A(k) = .4,(k)+ G, where Al(k) is the multiplication operator AI( = - i(c, k)f - v(c)f, and G It is easy to see that Al(k) generates a completely continuous operator. a strongly continuous semigroup T1( t, k) of class Co
Ti(t, k)f = exp {- Ii (c,9 + y (c)l Qf, and hence, by virtue of Theorem 13.2. I of [21, it follows that the operator A(k) generates a strongly continuous semigroup of class CO, where
The
linearized
Boltznann
113
equation
II T(t, k) II < exp {(II G II - vo)t).
(2.1)
Lemma 1 A unique solution 6 f,,(c, k) =-W(k)).
of problem (1.2)
exists
for any initial
function
This is a direct consequence of the fact that T( t, k) is a strongly continuous semigroup of class CO (see [21, Theorem 23.8.1). We shall prove that the semigroup T(t,
k) is contracting.
Lemma 2 For any k and t > 0 the inequality
11Rt,
k, 11<< 1 is satisfied.
Proof. Let u E D(A(k)). We shall construct the element T( t. k)u. As is well-known [2, Theorem 10.3.31 GTU=ATU=
- Y(C) Tu - i(c, k) Tu -j- GTu.
(2.2)
Passing in (2.2) to the complex-conjugate values we obtain --$Tu)’ Multiplying
(2.2)
;(Tu)
= -Y(C)
(Tu)’ + i(c, k) (Tu)” + G(Tu)‘.
(2.3)
by (Tu)+, and (2.3) by Tu and adding we obtain
(Tu)’
= - 2v(c) (Tu) (Tu)‘+
TuG(Tu)’
- (TuJ’G(Tu).
(2.4)
that in (2.2) and (2.3) the derivative is understood in the sense of the metric of Ls(cl, and in (2.4) in the sense of the metric of Ll(c). From (2.4) it follows that
Note
(T(t2)4 (TM 7~)l
=
s ti
(T(h)u)
(T(t2)u)’
[- 2v(c) (Tu) (Tu)‘+
=
(2.5)
TuG(Tu)‘+(Tu)‘GTu]&.
We integrate (2.5) with respect to c and on the right-hand side we change the order of integration with respect to c and T. (The change of order of integration is not difficult to justify). Considering property 2 of
114
A.A.
A rsen’ev
the operator L we obtain % 11~(~~)~~12 - liZ’(ti)ulIa = l Z(Tu, L2?u) dz < 80. ti Since (2.6) is valid for WY complete, we have
Let t >.to
> 0. Tien
II <
IIT
II.
by virtue of (2.1)
IIT(t)ll ~4 llT(t~)lI Since to is arbitrary
D(A(kI),and ~(~(k)~ is everywhere
u E
IIT
(2.6)
<
exp
(2.7)
from (2.7) we obtain
{(--WI+
Wll)~~}.
it follows from (2.8) that f\T(t)
(2.W f! <
1.
3. The existence and uniqueness of the solution of problem (1.1). The bebaviour of the solution as t+aJ We shall briefly
discuss the basic idea of the following
calculations.
Let &A, k) = W - A(k))-'and u E &A(k)). Since ‘Qt, It)is a convoluted semigroup of class CO. it can be expressed as follows in terms of the resolvent of its infinitesimal operator: y+io T(t,
l exp(ht)R(h,
k)u==&lim
k)udh
(y > 0).
(3.4)
-y-i0 First of all we shall explain the analytical properties of R(h, k, in the left-hand half-plane of h, transfer the contour of integration there, find those residues at the singular points which are essential for the asymptotics (thes will depend on k) and evaluate the others. Then we shall integrate the solution obtained with respect to k and find .ffc,x, tl. The whole question is consequently reduced to the investigation of the resolvent R(h, k) as a function of the variables h and k. We now introduce the operator K(h, k) =
1
G. h-t-i(c, k)+y(c)
The
It
linearized
Boltznann
115
equation
is not difficultto show that R(~,k)e=(E-K(h,k))-'~+i(e
;j+y(eJ n.
9
(3.2)
From (3.2) it followsthat in the half-planeRe k 2 - vu + 6 (hereand further on we shall denote by 6 8 fixed positivenumber,So small that - vo f 6 < 0) only singularitiesof the function(F:- I((& k))-’ can be singularitiesof ,?(A,k). We Shall study the behaviourof this function. The followinglemma is obvious. Lemma 3 If we are given 8n operatoru"l(a),where 81(a) + 0 as a -)11in strong operationaltopology,and the operatorJ!&is completelycontinuous,then C(a) =Sl(a)E, + 0, a -*0 in uniformoperationaltopology. Lemma 4 If Re A'> - Ve + 6, then 11It'(h, kt I\ + 0 8s d = J+/Kf-;2+
m
and
COnSeqUentlYa number oe(6) exists such that tt,q(h, k) 11< 0.S if d > 00(S).
Proof. Since G is completelycontinuousit is sufficientto show that for any functionu E Lz
sTh+i(c,be> I”
k)+v(c)f*'
as d + coand this is obvious. Lemma
5
For any fixed k the functionRO, k) is analyticwith respectto h everywherein the half-planeRe h > - vg + 6 except for a finite number of poles <&j(k) 1. 2.
The poles Aj(k) satisfythe followingrelations:
(a) Re Aj(kI\7 h = 0;
0
and the equalityapplieswhen and only when IkI =
(b)IIm hi(k) I < 00(6 ), where oo(6) is the constantdefined in Lemma 4, dependingonly on 6;
116
A.A.
Arsen’ev
(c) for Ikl ’ ~~(61 the function .&A, k) has no singular points in the half-plane Re A > - vo + 5.
3. The point ho can be a singular point for the function .?A, k) in the half-plane Re A > - vo + 5 when and only when it is singular for the function (E - K(A, k))-’ and belongs to the discrete spectrum of the operator A(k). Proof. From (3.2) it follows that the point ho can be singular for the function Rl’h, k) in the half-plane Re h >, - vg + 6 only in the case where it is singular for the function (E - K(h, k)J-‘. We shall denote by Z(S, k) the set of those points of the half-plane Re A 2 - v0 + 6, for which the equation u = K(A, k)u
(3.3)
has a non-trivial solution. By virtue of Lemma4 a constant wo(6) exists such that if 1~) > oe(6) theri for all k IIR(o + i-r, k) II < 0.5, if a,, - vg + 6. Consequently, for no k does the set Z(5, k) coincide with the whole half-plane Re A > - vc + 6. But the lfunction K(A, k) is analytic in the half-plane, as is the function h for any fixed k, end hence it follows that (E - K(h, kj1-l is analytic for Se A > - vu + 6 everywhere, except for points of the set Z(5, k, and each point of the set ‘Z(6, k) is an isolated pole of (E - Kth, k))-’ [51. Let ho E (6, k). Then for some u E IQ(C) 1
Hence hou + i(ck)u + Y(C)U = Gu, hou + i(c, k)u = Lu, Mu, u> + i( (c, k)u, u> = (Lu, u>.
(3.4)
By virtue of the symmetryof the operator L we obtain from this Rsho = (22,Lu) < 0,
Im&=
-i( (c, k)u, u>.
let Re A0 = 0. Then < u, Lu > = 0 and f+u = 0, where u is a linear combination of the functions u,(c)
Now
The
linearized
Boltzmann
u= i
117
equation
a,(k)u,(c).
-0
Substituting this equality in (3.4) and considering the explicit form of the functions u,(c), we find that the equality (3.4) is possible only in the case where either all the a,,, are zero or A = (k( = 0. Consequently for points of the set X(6, k) property 2, (a) holds. Properties 2, (b) and 2, (c) for points of the set X(6. k) are obvious consequences of Lemma4. But from this it follows that all points of the set X(6, k) are situated inside the closed rectangle (1x1h( < oo(6). - vg + F
= 0,
IlulC= I
Lemma 6
The function (E - K(A, k))-’ is continuous in uniform operational topology with respect to the set of variables (A, k) at those points (A, k) with Re A 2 - vu + 6, at which it exists. Proof. It is sufficient to show that the function &(A, k) is continuous in uniform operational topology. By virtue of Lemma3 this is a consequence of the fact that for any function u E L*(c) 1
h+Ah+i(c,k+Ak)+v(e)
2
1
-
h+i(c,k)+v(c)
I
&CO
if d = T/lAhlz + IAkl”-+O. Lemma
inf p(k), where p(k) is the distance from the spectrum Ikt>r F ’ 0. of the operator A(k) to the imaginary axis. Then pa(r) > 0 if Let
PO(~)
=
118
A.A.
Proof.
From Lemma5 it
follows p(k)
Arsen’ev
that >
if
IkI > oe(S)
we always have
I-vo+SI.
Therefore po(r)>min
inf
{l-v0+6l;
P(k))
W8)2IkI>r
and it
is sufficient
to prove that
p’(r) =
inf P(k) > o,(6)> Ik 1>r
0.
Let p’(r) = 0. Then a sequence {k,,,) exists. which converges to some point ko, where r & 'ko / < oo !a); and a sequence of numbers .Ah,= u,,, + iT,,, such that uR1+ 0, an;l each point A, is a pole of the operator MA, k,,,,. But if A, is a pole of the operator R(h, k,) and o,,, > - vo + 6, then without fail 1~1 < 00(b). Since the function T(T) = 1) (E - X(0 + it, ko))-’ is continuous with respect to T by virtue of Lemmas 5 and 6, it is bounded on the segment by some constant Cl. BY virtue of Lemma6 (CJ!
ITI < oo(6)-
for
any
T
E
[-@o(T),
0"(T)]
It(E - K(u + iz, k))-i is satisfied
II < 2Ci,
and so also
tlR(a -I- iz, k)
It <
F <
It must also be satisfied for all k, and h, dicts the choice of the sequence k, and A,.
00.
for
n>mo,
which contra-
Lemma 9 For any r > 0 a constant C(r) exists such that on the line Re h = - 0.5 pa(r) < 0 for all k such that lkl > r, the norm of the resolvent II R(A, k) !’
If
d =~~lh12+
IIR(h,
lkl"2 m(6)
k) II < ;. II (E -
Therefore it is sufficient lie inside the compactum
to consider
and Re A 2
- vo + 6 by virtue
K)-i 11< 216-Cl O”. only those
points
(A, k) which
The linearized
Boltznann
119
equation
II (E - K(h, k))-i 11is continuous with On this compactumthe function respect to the set of varianles (A, k) ani therefore is bounded. In addition, the function 9th. k) is also bounded. The analyticity of R(A, k) in the half -plane Re h >, - 0.5 pa(r) follows from the definition of PO(~) and Lemma5. Lemma 9 Let h = c + iT; c > - VQ + 6. Then for any u E f-2(c) IIR(o + ir, k)u II+O,
1~1--t 00.
for any fixed k uniformly with respect to u 2 - vg + 6. Proof. Let t-rI > w(6).
Then
I” ’ & .+. 0 __---- be) s 6*+(Z+(W)* ’
G4 Lemma 10 For
a,ny u E
Lz(cJ
and any k such that IkI ’ 0, lim IIT(t, k)ull = 0.
t-NW
Proof.
Let
u ED(A)
nD(Az).
initially.
Then by virtue of (3.I)
we
have
If
IL/ >F
integrate
’ 0, we can use contour integration along the rectangle [Y --9
Y-t b,
-0.5po(r)
$- io,
in this integral.
-0.5po(r)
We
- iw],
where pot r) is defined in Lemma8. The integrals along the upper and lower sides of the rectangle approach zero as o -, w, and since R(h, k) is analytic inside it
120
A.A.
Arsen’ev
-0.5pdt)+io
T(t, k)
exp(U)R(h, &lim 5 w-PO0 -0.5p0(r)--io
u’=
k)udh.
Let t > to i 0. Then lim l T(t,k)u=& o*oD
R(h, k)ud( 7)
= &lim 5 c(-&‘(A, CD-PC%7
k))u&.
=
-o.5po+20
1 exp(ht) = -1im s 2ni “-+O” -0.5p,--i&l t But R(h, k) = (E f AR) /A. If - 0.5 pa(r) we have ,,R2u,, < 11412 +
.
@(A, k) u &.
(3.5)
then on the line Re h =
u E D(A2),
2CIL44l + C21M241 . A2
Therefore from (3.51 it follows that -0.5p+ioo
IITuII<
Gexp
{-0.5po(r)t) <
1 _,.,JG
CSew {-O,5
G P(F) t) *
t-too.
0,
According to Lemma2 we have IIT(t, k) II < 1. Since D(A) fl D(A2) is everywhere dense in L*(c), because A(k) is an infinitesimal operator of the semigroup of class Co, by the Banach - Steinhaus theorem for any u lim IIT(t, k)ull=
t-NW Theorem
0.
1
Let fo(c, xl be of integrable square with respect to c and x, and the Fourier transform of this function with respect to x be the function &(c, k) =D(A(k)) for almost all k. Then problem (1.1) has a unique solution flc. x, tl in the sense of definition 1 and
1 If(c,
x, t) pdx-+o,
t-t
00.
By virtue of Lemma 1 a unique function f
iS
The linearized
co
R (a,
“k)fo0%
Boltzmann
1
1
L h + i (0, k) + y (c) ‘8,(0, k). G
h+i(c,k)+~(C)
i=.
121
equation
1
is strongly measurable with respect Consequently R(A, k)f;,(e, k) a strong limit of strongly measurable functions. Hence
to k as
-f+io f (c, k, t) = &
lim \ exp (b 0 R (A, k) h (0, k) dh o+(x)U-Al
k3, Chap. 3, Theorem 171 for any is also stronglv measurable. Therefore t > 0 the function. F(icc, k, t) is i~easurable on the product of the spaces c @ k. Rut for any k ancl t > 0 by Lemma 2 \ I i (0, k, t) I2 dc -G \ I io (c, k) la do. Therefore
\dk\If(c,k,Wdc<\~\/h(~~Wl’k
of Fubini’s
theorem it
s Therefore
for
almost all
de
s
for
c and all
almost all
We can prove that all
k bs Lemma 10
s
t > 0
c there
of problem theorem
1 G\jf(o,
that
k, t)ladk
f (0, x, t) = & which is a solution virtue of Parseval’s
from this
If (c, k, t) J”dk< cm.
sli(c, consequently function
follows
exists
s
(as an element of Lz(x))
e-ikxj(c,
(1.1)
k, t) la dkdc
-Ip(c, k, t)j2dkdo+0
a
k, t) dk,
in the sense of definition
1. By
= !I#@, x, t)I’dxdc.
as
t*
00.
In fact
for
almost
122
Arsen’ev
A.A.
s A
If(c,k,t)12dc-+0 as
and for
any t ’ 0 s
Therefore
(j (c, k, t) la dc <
by virtue
of Lebesgue’s s
This proves
lf(c,
s
[ io (0, k) la dc=Ll
(k).
theorem
k, t)j2dcdk-+0
as
t--too.
Theorem 1.
‘vote. It is easy to see that Theorem 1 to be satisfied is
4.
t--too
a condition
which is sufficient
for
The refinement of the asymptotic solutions as t + 00
Theorem 1 gives no information the given point x. To obtain this about the initial functions fo(c, a more accurate evaluation of the operator Uh, kJ
Q((h,k)u=
about the behaviour of f(c, x, t) at we must make additional assumptions x). First of all, however, we obtain operator T( t, k). We shall define an i
h+i(c,k)+v(e)U'
Lemma 11
If
h = u + i-r and a >
- vg + 6, then
o+im j
IlQGQIIdh
<
C(d)
(i+
lkl).
o-im Proof.
Let
11u 11 = 1 and y = RG Pu. Tllen
1a+i(c, I.k)+y(c)
Ml= <
( I sup ICI
s
G(c, ci) a+i(ci,k)+v(q
1 a+i(c,k)+y(C)
U(Ci)&
II<
(4.11)
2
I)
7:~
s
I G(c, ci) I dci <
Gy2(b
k) ,
‘he
linearized
Boltrnann
equation
12 3
where
Using the evaluation
Y(C) >
YOj- CZI c 1, we can show that
Ll+iCO ~~‘yWWWWW+
PI).
(4.2)
u-ico From
(4.1)
and (4.2)
the statement
of the lemma follows.
Lemma 12 For any r > 0 and t > 0 if
Ik\ > r
II W, k) II < C(r) (1 + lkl WY where
a(r)
PFOOf.
> 0 if
r > 0.
We have the equation
(see
equation
3.2)
Ru=!2ufS-JGRu. Hence
Ru=Qu$QGQu+QGQGRu. Substituting
this
in (3.1)
we obtain
(u E D(A))
Y+ie
T(t,k)n=&lim
l crop(At) [Ou + QGQu + QGQGRu] dh = -c0 y--i0 = W) + W) + b(t).
(4.3)
For Ii(t)=&-
lim~exp(ht)S2udh O-+O” y-i0
we have the evaluation III,(t) II < of which we can be convinced evaluation
eWl4l,
by calculating
(4.4) 11 in explicit
exp(ht)QGQudh
form. For the
124
A.A.
A rsen’ev
we transfer the contour of integration to the line Re h = a = - vu + S and apply Lemma 11 to the expression under the integral. We obtain
II h(t)
II < C(6) (1 + Ikl)e-utllz41.
(4.5)
With the integral VW lim 1 .exp (it) OGQGRu dh o+oo y-_io we proceed similarly, only we transfer the contour of integration line Re A = - 0.5 PO(F) < 0 and use Lemma 8. We obtain
to the
III,(t) II G G(r) (1 + Jkl)e-a(r)fllull.
(4.6)
Gathering together all the evaluations (4.3) - (4.6) and taking into account the fact that D(A) is everywhere dense, we prove the lemma. Lemma 12 gives an evaluation of the operator T(t, k) for IkI 2 F > 0. Now we need to obtain an evaluation of this operator in the neighbourhood of the point k = 0. We shall previously refine the behaviour of the resolvent in the neighbourhood of the point A = o,k = 0. Lemma 13
An (I)
F,J
>
0
and au > exists
11R(--o,,
+
iT, k) II <
such that
C(r0, 000) < 00
for all
k such that lkl <
ro;
(2) in the half-plane Re h > -CTOy R(h, k) for any fixed k there is a finite number of poles of common multiplicity 5, hj(k), I< j
(3)
the projectors J’j(k) onto the invariant spaces of the operator answering to the eigenvalues of hi(k), are analytic functions of of the vector k = noIkI the parameter IJ = IkI for any fixed direction
A(k),
Pj(k)
=
iP#(n,)
IkIm.
m==O
PFOOf. 1. As we have explained
the noles
of the function
(4.8)
R(h, k) in
The
the half-plane tion
- VII + 6 are those
Re h >
u= h +i has a non-trivial bringing the axis
equalities
points
A, for which the equa-
sG(c, ci)u(ci)dci
1 (c, k) + Y(C)
1
(4.9)
s
-
G(gc, c(f)) u(c(‘))o?c(f).
v(c)+ic+(-G
out the cnange of variables y(gc)
125
equation
We fix the vector k. Let ,g be a rotation solution. of z into the direction of the vector k. Then
zz(gc) = Carrying
Boltzmann
linearized
=:Y(c),
u(gc) = -
c(l) = =
G(gc, gcW)
1
Y(C)+ hlkl-+
h
gc(l)’
and considering
G(c, &I’)., we obtain
s
G (c, 13’)‘)u (g&j’) ddk
It is clear that if the equation (4.9) equation (4.10) for the same h has the tions. Rut those h for which equation depend only on lkl, alid so the same is
= -Y(C)
+ G -
(4.10)
has a non-trivial solution the same number of non-trivial solu(4.10) has a non-trivial solution true also for equation (4.9).
Let no be a unit vector, collinear wii;h k. We shall Ikl and shall consider the operator A(k) as a function and the complex parameter ~1
A(k)
the
&t(w)
= Ao -
assume that P = of the vector no
@Ai.
Rut
and if P = 0 the point A = 0, by virtue of the condition, is an isolated pole for the function 8th. k) of multiplicity 5. Since A(k) is closed and symmetrical about the line Re CI= 0, in view of one of Kate’s theorems [71 constants E > 0 and ~2 > 0 exist such that inside the circle R(A, u) have only a finite number t ~11~ ~2, the functions of poles {hi(k)). and the common multipli.city is 5 and every one is of the first order and is an analytic function of n. The projectors Pj(k) are also analytic functions of CI in the same neighbourhood of ~1 (with a fixed direction for no).
IAI
Let p = 0 and the distance
from the point
A = 0
UP to the rest of the
126
A.A.
Arsen’ev
points of the spectrum of the operator .A, = - v(c) + s be ci. This distance is positive by virtue of Lemma ri. We assume that cl < 1 -vg + 6 1 In the opposite
case we shall
denote
by d the quantity
I- vo + 61). We choose a number ~(6) as was consider the closed region 2 consisting of a lines Im h = oo(6). Im h = - 00(6), Re h = 0, a circle is cut out of radius !h d with centre
d’
= min{d;
done in Lemma4. We shall rectangle bounded by the Se A = - d//2, from which at the point h = 0. If
Ikl = 0 the function (1 (E - K(h, k))-’ It is continuous with respect h in this closed region and bounded there by a constant c. Since )( (E - &A,
k))-’
a neighbourhood 11(E -
K(h,
of
k))-‘11
to
11 is continuous with respect to the set of variables, the region 3 and r ’ > 0 can be found such that <
2C
if
h E O(D);
Ikl <
r.
This reasoning
shows
that if IkI< r’ the function 2(X’, k) is bounded on the line Re h = -d/2, and all its poles, lying to the right of the line Re.A = -d/2 lie inside IhI < d/4. We shall assume that FO = min {r; EP), a2 = -d/2. the circle These numbers will satisfy all the conditions of the lemma. p,,,‘j’(no) can be The coefficients a, (j) in (4.7) and the projectors calculated by using the ordinary methods of perturbation theory. The calculations are quite cumbersome since we are here concerned with a We here quote only those results whicn five-fold degenerate eigenvalue. the properties we shall be using later on. Note that in their derivation of symmetry of the operator i, with respect to the rotation were essentially used. Lemma 14 The following 1) a,(j) =
(i)-nh,tj),
2) aE!+i = -
are valid: are real
a,(j)
(2) azn
0,
,r\
=
statements
=
(2) azn+i
(3) a2n,
=
-
(4) azn
(3) a2n+i,
azn+i,
3) A,(i) =
0, a&2) =
4) 1&j) >
0,
0,
&i(3)
1 < j < 5;
=
0, n2(4!
A&4) =
zzx
1%;
g2’1
j&5) +
=
1/5/s;
2h2(2)];
5
5) p,Cj)f =
(q@,
f) rp&j), &j) =
2 c,(j&. a=1
The numbers
cc,(j)
are given by the following
table:
=
(5) azn ,
f4) knti
=
The
1
-r
1/z
2 3
-ccoscp
sin cp Cos8 r
v Iv
-sin0
4
Boltrnann
v2
sincp
0
sin cp T cos cp CQS0
0 sin 8
FT sin 8 cos cp -rFsinf3
Note that AZ(‘) is proportional thermal conductivity of the gas.
127
equation
0
0
0 0
4
linearized
Cosq
to the viscosity
-v0 0
+s*
3 3
+
-+ose
&
and AZ’ ’ ) to the
Lemma 15
Let k be such that u E D(A(kIJ. Then T(t, k)u =
Jkl <.ro,
where ro is the same a~ in Lemma 13, and
~exp(hj(k)t)Pj(k)u+exp{-a(r,)t}B(k,
t)u,
(4.11)
j=i
where a(~,) kor t.
>
0
and
ljB (k, t) II < L’ (T-,-J < CW, C(Q)
does not depend M
Proof. In (3.1) we transfer the contour of integration with respect to A in the left half-plane onto the line Re A = - 00; on this line, by virtue of Lemma 13, the resolve& is bounded for kI
2
If the initial function io(c, x) is such that its Fourier transformation with respect to x (the function fo(c, k) satisfies the conditions
128
A.A.
then for
any fixed
point
Arsen’eu
x1
CJI.lim Pzf(c, x, t) = t-tw
w(c),
where W(C) = -
1
1
5
4n3 1
-&
““exp (-
C
c2/2) ( c2 -
5/2) 1 exp (-
c2/2) X
ciexp(-e2/2)X
x(c3-5/2)~0(c,x)dcdx+~(&)~‘*~
i=i X 1 ciexp(-
Proof.
Therefore
Let II E Lz( c).
Then
(u, j(c,
(u, T(k, t&c,
for
k, t)> =
follows
-
1
8n3 s
llNi&,
G
i
that
for x (and not for
) (c,
eikx (u,
k, ~D/dII4I~llh(~~
k)ll.
k)IIdk<~. almost
k, t)) dk =
all
,-i’=j
x) the integral (c, k, t) “k>
exists. Consequently for any x and t > 0 a function - an element of the space &(c) 1 f(c, x, t) = s +‘+(c, 8n3 We shall
x)dcdx].
any t ’ 0 \clkl(u,
Hence it
k))
c2/2)f0(c,
divide
the integral
s
cikxj (c,
of
(4.12)
f(c,
.
x, t! exists
k, t)dk.
into
(4.12)
two
k, t) dk +
IkKro
e-ikxj (c, k, t) dk = 1, (t) + 1, (t). s IW>ro
+& Using the value
II12(OK
of the operator
\ IIW Ikl >r.
T(t,
k)llllfo(c7 k)lIdk\
s
k given
in Lemma 12 we obtain
C-MJU
x
x (I + Ikl)llf^,(c,k))Idk = 0 (exp (-- ad>>.
The
linearized
Boltzmann
129
equation
Let u E Lz ( C) . Using Lemma 15 we find
& \
(u,
=
(4.13)
k, t)dk> =
e-ib](c,
IWOo
1
=
<
“=
X [j$le~~
_-
1
s Ikl
%” j=l
e-ikxT (t, k) j. (c, k) dk> = Q, ,k,
{hj (k) t> Pj (k) fo (~7 k) + exp (-
5 21
\
e-iL* x
lU
aat) B (k, t) 70 (c, k)] dk)
ikx + hj (k) t} (u, Pj (k) io
exp {-
\
&
(c,
k)) dk + 0 (exp (-
7~
~zt)).
To evaluate the integral in (4.13) we use the formulae (4.7), (4.3) and Lemma 15, which gives us some information about the behaviour of Aj(kJ in the neighbourhood of the point k = 0. Note that only this neighbourhood makes an essential contribution to the integral of (4.13) since Re Aj (k) < 0 if IkI > 0. After straightforward calculations which are usual in the saddle-point method we obtain
(u, Ii(t)> =
t;2 (u, w(c) > + 0 ( f-).
T
This proves the theorem. We can verify that of the theorem will be satisfied if the initial fies the following requirements:
I)
Ifo(c, 5) Idxdc+
2)
1
[s1x1
Ifok
1 IA2fo(c,
x) kq2do
x) 12dxdc<
<
5.. The construction respect
conditions function
(1) and (2) fo(c, x) satis-
00;
CQ.
of the to E as
asyrptotics E + 0
with
In this paragraph we shall study the asymptotics with respect E + 0 of the solution of the Cauchy problem for the equation
f&c;== fLf,
f 1t=o =
We shall assume that the operator L satisfies Section 1 and that the initial function fo(c,
fo(c, x).
to E as
(5.1)
the conditions 1 - 3 of x) is such that
130
A.A.
fo(e,
k) E NAM
);
11io
Arscn’ev
k) 11 is bounded on any compactum and
s+IklN)IIh (I
We shall accuracy,
k)IIdk
assume the number J to be sufficiently large, but not fixed so as not to complicate the argument unnecessarily.
Since for of Section I ator L, then formation of
tne operator are satisfied, on the basis the solution
in
(I/E)L for any E > 0 all the conditions 1 - 5 as soon as they are satisfied for the operof the previous results the Fourier transof equation (5.1) can be found by the formula Y+iw
f (c, k, t, E) = &
lim 1 exp (W R (A, k, e) jO (o, k) dh, o-%X3 y-iw
where RCA, k, E) is the resolvent
of the operator
R(h, k, E)=(A.E--A(k,
A(k, e)u= It
is not difficult
-i(c,
to verify
E))-i,
k)u+(c)u++Gu. that
R(h, k, e) =
d?(eh, ek, 1).
exp,(ht) R (eh, ek) .=
10 (c,
k)
dh
Hence
=
T (+ , ek) f,, (c, k).
Therefore
f(C,
x7 t,
E) =
&
Se-“‘T
(;,
ek)jO(c,
z +$&,-WET
(5.2)
k)dk =
(f , k)j,
(c,+)dk.
We shall divide the region of integration with respect to k in the integral (5.2) into two parts; into some neighbourhood of 0 and its compleof the operator T( t, k), which ment, For small Ik\ we use the evaluation is given by Lemma 15, and for large Ikl we shall evaluate the operator T( t. kl by means of Lemma 12. We choose ro as in Lemma I5 and obtain
1 1 f(c, x, t, E) = mE3
s
e-ikXIEY’ ($,
k)j,(c,
;)dk=
The
+
We evaluate
According
Hence it
tinearited
1 1 ~8n3 E”
the integml
to Lemma 15, if
follows
s
Bo2tzmann
@‘=/ET
($
equation
7 k) j. (c, f)
131
dk = 11 (t> + 41 (t>-
lkj> PO ‘l,(t),
using Lemma 12
Ik I =zro
that
Therefore
f5.3) X Pj (ek) fa (c, k) dk + 0 (~XP {-$})m Re hj(k) > 0 if IkI > 0, the quantity FO in (5.3) does not play an important role (we can take any smaller positive number). On the basis of (5.3) we construct asymptutics of the function f
Since
132
A.A.
Arsen’ev
respect to E, such that the remaining terms will not only be small relative to E as E + 0 uniformly for t E [to, cc), but will also approach zero as t + co so that their order in t will be larger, the higher the power of E we consider. Lemma 16 Let h+“](p) tion Aj(U):
be the n-th segment of the Taylor
hjLnl
(/.I,)
=
$
series
for
if
lkl
the fllnc-
am(j)pn.
m=i
Let the function
y(k) be such that
1 (1 +Ikln+i) Then for
b(k) Pk <
sufficiently
small
O°F
b@(k)
I<&,
r > 0, e > 0 and any t ‘0
if
n>2
Since
Proof.
analytic function of CI in the neighbourhood 1~1< rg and al ‘j) for all j is either purely imaginary or zero, and a,(j) < 0, we can find an r E (0, rol such that for all IJ E (0, rl is
an
(5.4) and where b > 0, we can assume that
Let us consider I(e, t)=
)
the integral )
{exp
[x’F]-q~[
hjrn’(~k)t
]}rp(k)&
1=
IklGUe (5.5)
=
I
,kja,.exP[
The
linearized
We now apply to the function of the mean
Op----
[
In view of the analyticity
from (5.6),
If all
I
-
0 < -
and from (5.7)
it
I <
considering exp{-
s Id4r/e
133
braces
of
tne theorem
(5.5)
t] x
(5.W
q+(k)dk].
hj (p)
hj(P.>
CiPn+i~
(5.4)
I PI < r-
we obtain
b&t lk12+Ci&ntlkIn+i]Ikln+ilg(k)
Idk(
small
Ikl <
(0 <
r <
follows
(b/2C)1N*-2),
E<
1)
for
r / E,
betIkl2+Ct@Ikln+i
<
--;et(kj2
that
I(&, t) < ctsn
exp{--~d[k12}
1
Ikln+j I+(k)
Idk.
(5.8)
O
shall
now evaluate
Z(c,t)=
the integral
s exp{--;etlk~2} Ikldrle =
in (5.8).
S w(-
have
We
Iki”+‘I$(k)Idk=
l exp(-~lk12}Ikln+iltp:k) Ikla
+
Fk2)
Idk+
Ikln+‘lq(k)
Idk <
i
S exp{-~c?}
. (5.7)
r and E are sufficient1.v k such that
& of
I&j[“‘(P)
I (E, t) > cit.9
in the
&‘n’ (Ek) - hj (Ek)
X exp
Therefore
enclosed
y)t][*k)-6kj(6k)
) ,kL_;xp [
I(&$)’
equation
Boltzmann
an+Bda+e-be’/2~
Ikl”++j(k)Idk<
0 G
+
s 1
+
(Et)
(n-w’2
C3e-bet12
<
” 1
+
(Et)
(n+4)‘2
134
A.A.
In view of this
evaluation
it
Arsen’ev
follows
that
C&s”
]I(& t)I< This proves
from (5.8)
1 + (et) (n+W’
the Lemma.
We now‘proceed (5.3) we have
to the construction
of the asyaptotics.
According
to
5 1
-1
Pi (sk) jo
(c,
k) dk + (54
+ 0 (e-a’/“) = -8;3
i \ exp [-ikx+m] j=l 1,+&f
8
x
x [ 2 P,(j) (no) (E 1k j)m] f. (c, k) dk + 0 (e-u’/E) + m=o
x f,, We also
(c,
k) dk = I,(t)
+ I2 (t) + 0 (e-atie).
have
j=i
I kl
, pj(Ek) - i P,(j)(no) (elkl)m 11 m=o
/Ix (5.10)
x Il.&c, k) I!dk < G { exp [ IklG-/e
We now transform
Idi) =
11(t):
k3 i s
ew
[
?tij
IkJ=Zr/e
The
hj[“I (Ek) t -
csp
E
linearized
ikx ]}[
Boltzaann
135
equation
~p,,ij)in,)(Elki)“]j,(c,k)=rl(t)+lz(t). nG==O
By virtue of Lemma16; for the integral
i,(t)
we have the evaluation (5.12)
From (5.9)
- (5.12) Theorem 3 follows.
Theorem 3
If fo(c, x) satisfies the conditions enumerated at the beginning of this paragraph then for the solution of problem (5.1) the following evaluation holds :
I!&, x, t, E)--2ifn, l(C,x7 t, E) IIGC
C
$-+i
&%
-+ I + ( &t)f*+h)‘2
1 + (Et)@+Qn
I+o
(e---aq,
where
qh,
1
(5.13)
(c, x7 4 8) =
x [ i Gus m=Q
felkl)m]j,je,k)dk. *-
Unfortunately space does not permit us to dwell in detail on the physical meaning of the results obtained or on the method of calculating the quantities occurring in formulae (5.13). We only note that by means of (5.13) we can establish Stokes’s formula for the absorption of ultrasonic waves in gases. I wish to express my deep indebtedness to A.N. Acknowledgements. Tikhonov for nis interest in my work and to A.4 Samarskii for discussing my work and giving me valuable advice. Translated
by H.F.
Cleaves
135
Arscn’ev
A.A.
REFERENCES
1.
KARLEMAN, T. Mathematical Foreign
2.
HILL,
Literature
E. and PHILLIPS,
Foreign
Literature
GRAD, H. Asymptotic 6, 2, 147 - 161,
4.
SIROVICH, L. Dispersion
tions 5.
theory 1963.
KATO, T. math.
analysis
Rouse,
of
Soltzmann
the
theory
of
gases.
1960. and
Moscow,
Soltzmann
Integral
lndag.
equation.
equations
math.
24,
DUNFORD, N. and SC~ARTZ, York,
7.
P. Functional
kinetic
moscow,
semigroups
.
1962.
equation.
Phys.
Fluids,
relations in rarefied gas dynamics. Phys. solutions 1, 10 - 20, 1963; and Formal and asymptotic theory. Ibid., 2, 216 - 230; Generalized normal solu-
the
SEELEY, R.T. meter.
6.
of
in the
Rouse,
Publishing
3.
Fluids, 6, in kinetic
problems
Publishing
4,
Ibid.,
depending
434
D.T.
- 442,
Linear
10,
1423
-
analytically
1439.
on a para-
1962.
Operators.
Interscience.
New
1956. On the SOC.
perturbation
Japan,
4,
theory
323 - 337,
of closed 1952.
linear
operators.
J.