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0.
249
§3 FOURIER HANKEL INTEGRAL
from the equation
=0,
(33)
where c is an arbitrary real number. Let us denote by A1 and A2 successive positive roots of the equation (3.3). From the asymptotic formulas (2.10) and (2.10') it easily follows that for b—p \ "2
(34)
V
Ai
°
—T
/1
By (2 15) and the fact that the boundary value problem is homo
geneous, the functions
=
—
also are eigenfunctions To determme the spectrum of the boundary value problem under study for the interval [0, cc), we need to calculate the limit (x) dx
hin
It follows from the asymptotic formulas (2 10) that for fixed A > 0 P*and large x
(x) 2
cos (sax
—
—
=
s;*
cos
(sax —
—
+ pit) +0
—
— / cos pit) cos
erefore
lim
(x)
{(c —
cos
pit)2 +
sin2 pit)
b-÷cx
= tting, as in §1, Chapter 2,
(c2
—
COS pit +
V. EXAMPLES.
250
(x) dx
we obtain, by virtue of (3.4), A
V
(
8n+1 — b
{* + —
(x) dx
o
—s- + o(1) }
I
sds
-p.'
ç
Consequently,
—*
—
2c1-,, cos
'1
s4P
for X> 0 the spectrum is continuous. If there is no
negative spectrum, then the expansion formula has the form
/ (x)= S
s
ds.\ \/t (ts)} /
—
(ts) cit.
Let us now investigate, for which c there will be a negative spectrum. = icr (cr> 0). From equation (3.3) it follows that Let A (k + 2) and, as is easily seen, for sufficiently large x C=/=O,
and consequently y =zexp(— x2/2) is not square mtegrable over the real line This establishes that the orthogonal Hermite functions 2( — x) of our problem are the umque solutions in The completeness of the orthogonal functions in 2( —
follows
from the general theory discussed in Chapter 2
Expansion in terms of the Legendre polynomials and the associated Legendre functions
1 The Legendre polynomials are defined by the following formula P0 (z)
1,
=
I
d'(x2—I)" dx'
(n
=
1,
2,
.
.
.).
257
§5. LEGENDRE POLYNOMIALS
Obviously the function
is a polynomial of degree n. We will prove that the polynomials form an orthogonal system = (x2 — on the interval [— 1,11. In fact, putting we have
for any nonnegative integer m 0. Consebelongs to =O(x,X) 2(0 quently, the function O1(x,A) +m(X)cci(x,A) also belongs to for ImA > 0, and so it is a multiple of —a2 }, i.e. we have
—
{sin
+m
cos
= C {cos (x
2
sin
— a2}
+ i sin {x
a2)
— a2)].
Hence we obtain
or /
sin a±
rn(A)==—
ens a
If A > a, it follows from (8.8) that %/X2_a2
If — a , cx>), it is necessary and sufficient that condition (3 19) hold ie ( + 0) = f' (+ 0)
Lhf(+0),
=0
§4 Solution of a mixed problem on a fimte interval We now consider a mixed problem for the one-dimensional wave equation on a finite interval of the real line which, without loss of generality, we may assume to be the interval [0, lr]:
ô2u/i9t2 =82u/8x2 —q(x)u,
(4.1)
44.3)
(4.4)
We wifi show that the problem (4.1) —(4.4) can be solved by the method which we already used in the preceding section, without
resorting to an eigenfunction expansion. However, in contrast with the considerations of the preceding section for the case of the haifline, for the case of a finite interval the extension has to be carried out both the left of the point x =0 and to the nght of the point x = Let the function q(x) be contmuous on [0, ir] We extend q(x) to the entire real line, preservmg its contmuity but otherwise arbitranly, and we seek a solution of the problejn (4.1) — (4 5)
u (x, t) =
{/ (x + t) + / (x — t))
+
w (x, t, s) f(s) ds,
where the function f(t) has to be extended outside the interval [0, ir}, Using the boundary conditions (4 3) and (4 4) We will first derive certam necessary conditions which the functions and have to satisfy. It follows from the condition (4 3) that (46)
From condition (4 3) and the second of conditions (4 2) it follows that
=0. Similarly, from condition (4 4) and the second of conditions (4 2) It follows that
290
VI. CAUCHY PROBLEM FOR WAVE EQUATION
(4.8)
—0)
+Hf(,r —0),
(4.9)
=0.
Using
the boundary condition (4.3), we extend the function 1(t) from
the interval (0,7) to the interval (—7,0). Suppose that 1(t), E C2(0, ir). We saw in the preceding section that the conditions (4.6)
and (4.7) are necessary and sufficient for the extended function f(t) to belong to C2( — ir, 7). To extend the function f(t) into the interval (,r, 27) we wifi use the boundary condition (4.4), from which, as we saw above, follow the conditions (4.8) and (4.9). We will show that for the function f(t), after extension to (ir, 27), to belong to C2(0, 27), conditions
(4.8) and (4.9) are not only necessary, but also sufficient. In fact, in the boundary condition (4.4), we obtain
inserting
(4.10)
+w(n,
t, Tv+t)/(n+t)—w(ic, t,
ôw(x,t,
it—t)/(it--—t)
s)
t,
We have obtained an integro-differential equation for the determina4. If this equation is integrated once, then we obtain an integral equation of Volterra type for 1(7 + t). Let us first rewrite (4.10) in the form tion of 1(7 + t) (0 t
t,
(4.11)
(z,t, s)
+
{
(t)
— I'
— t)
ôw(x,t,
+ Hw (it,
— {H + w
t,
t,
s)}
— t))
f(s) ds /
— t)
+Hw(n, t,
or after a change of variables in the integral on the left side
t, it+t)}/(ic+t) (4.11')
ow (x,t, s)
+ 0
Puttmg
{
+ Hw
t,
+
+ a)da=g1
291
§4. MIXED PROBLEM ON FINITE INTERVAL
K1 (t, c) = Ow (x,t, s)
+ Hw (iv, t,
iv
+
integrating equation (4.11') from 0 to t with respect to t, we obtain
/(iv+t)_f(it+O)
t,
after simple transformations 4.12)
f(iv+t)-.—f(ic+0)
Let us consider the integral equation 4.13)
/ (iv + t) + K (t, s) /
+ s) ds = g (t) +
C,
rhere C is an arbitrary constant. It is obvious that for any C a solution (the integral equatiion (4.13) is also a solution of the integro-differential quation (4.11). Therefore for C—f(ir —0) equation (4.13) also has solution, which is in turn a solution of equation (4.11). Putting —0) and t =0 in (4.13), we obtain
f(7r+0) =f(ir—0), the extended function f(t) is continuous at the point t = ir. putting t = 0 in (4.11) and taking into account (2.5), (4.8) fld (4.14), we find that 4.14)
415)
f'(7r+0) rf'(ir—O),
e. the first derivative ot the function f(t) extended according to (4.11)
i continuous at the point t = ir. let us examine the second derivative of the extended function U). To
this end, we differentiate equation (4.11) with respect to t,
staining
f'(iv+t)+f(iv+t){H+w(iv, t,
__
292
VI. CAUCHY PROBLEM FOR WAVE EQUATION
+ / (ic
(x,t, s)
+ t)
+ 11w (it,
t,
+ t)}
{
1c+t
'
t, s)
s)lI(\d
t,
_LH
ôxot
3
J' '
/
t, n—t))
(4.16) t,
t1
dt a)
(:,t,
a)
+H
ow (z,t, a)
f(s) ds.
Putting t =0 and taking account of (2.5), (4.9), (4.14) and (4.15), we obtain it—t)
(4.17) f'(it+O)—/"(it—O)= J(
dw(it, t,
Further, by virtue of (2.9) and (2.10) we have t, dt
dw(it, t, it—t)
— 1.
/
Therefore from (4.17) it finally follows that f"(7 —0)
f"(ir +0), i.e.
the second derivative of the function 1(t), extended according to (4.11),
is continuous at the point t = Thus we have extended the original function f(t) from the interval (0,7) to the intervals (— ir,0) and (ii-, 27). This extension is of class C2( —7,27). The integral equations (3.23) and (4.13) make it possible to extend f(t) to the entire real line. Let us show, for example, how to obtain an extension of f(t) to the interval (— 2ir, —
To do this we have
to assume that t varies in equation (3.23) in the interval (0,27). Since the solution of a Volterra equation is differentiable as many times as is the free term (assuming the kernel is sufficiently differentiable, which is the case here), our extension is of class C2(— 27,0) Since equation (3.23) is a Volterra equation, the original function, which is undefined m (—7,0), does not appear Therefore the extension which we obtam is of class C2(—21r,21r).
If we now assume that t varies from 0 to 37 in equation (4.13), then we obtam an extension of the original function f(t) to the interval
REFERENCES
4ir). This extension is of class C2( — 2ir,
293
Thus step by step we
r±end the original function to the entire real axis. This extension is of oo). If the function thus obtained is inserted in (4.5), lass C2(— hen by virtue of the very process used to extend it we obtain a solution f the problem (4.1) —(4.4). Bibliographical references
§1 Formula (1 14) can be obtained by Riemann's method Indeed is derived in this way in a paper by Povzner [1] The method which we have presented here is also, in essence, close to that of Riemann As concerns the estimate (1 16) and what follows, see the paper by Levitan [14] See also the joint paper by the authors [3] §2 A similar method, but m another connection, was applied in the paper by Gel'fand and Levitan [1] **3, 4 So far as we know, the extension method has been previously apphed only for the case of mixed problems for the simple wave equation
= uxx
CHAPTER 7
EIGENFUNCTION EXPANSION OF A STURM-LIOUVILLE OPERATOR §1. Derivation of auxiliary formulas
We will study the equation y" +[A —q(x)]y =0,
(1.1)
defined
on the entire real line.
Concerning
the function q(x)
we
will
that it is real and summable over every finite interval. Denote A) the solution of (1.1) satisfying the initial conditions =1, (1.2) =0, and by (x, A) the solution of (1.1) satisfying the conditions assume by
(1.2')
4/(0,A) =1.
=0,
Together
with equation (1.1) we will consider the partial differential
equation
= 02u/ôx2 — q(x) u.
(1.3)
Suppose that the function f(x) has a continuous second derivative. We denote by u (x, t; f) the solution of equation (1.3) satisfying the initial conditions (1.3') As was shown in the preceding represented
ou/otIt.o=0. chapter, the function u(x, t; f) can
be
in the form x+t
(1.4)
u(x,
Putting"
t; f) f(x)
t, s)/(s)ds.
{f in
(1.4) equal successively to p(x, A) and
A),
we
obtain the equalities
X) precisely, putting f(x) in the conditions (1.3') equal successively to then solving the problem (1.3)—(1.3') by Fourier's method, from (1.4) we obtain, by the uniqueness of the solution of this problem, (1.5) and (1.5').
294
296
§1. AUXILIARY FORMULAS
(1.5) 5
=
(z, X) cos
iv(x, t, s)p(s, X)ds,
(x + t, X) + (x — t,
X)}
(1.5') X)ds.
t,
Let us denote by g, (t) a function satisfying the following conditions: 1. is even, i.e. g,(t) vanishes outside the interval 3. g, (t)
has a piecewise continuous and piecewise monotonic first
j 4erivative
Further, let us denote by i.e. we put
the Fourier cosine transform of the
g,(t)cos p.tdt
(16)
It follows from condition 3 that for large
one has the estimate
1(17)
and integrating with respect
Multiplying both sides of (1.5) by to t from 0 to we obtain (18)
(x
s)
X)
(x,
s)
dt
the same way, it follows from (1 5') that (1 8')
{g8 (x— s)
s) is defined by (1.9). 2(_ cx) We put Let f(x),g(x)
+
(x, s))
(s, ?) ds,
296
VII. EXPANSION OF A STURM-LIOUVILLE OPERATOR A)dx,
A)dx, (1.10)
g(x)p(x, A)dx, A)dx.
The functions F1(X), F2(X), G1(A) and G2(X) are called the Fourier transforms of the functions 1(x) and g(x). If f(x) and g(x) vanish- outside
some finite interval, then the integrals in formulas (1.10) exist in the ordinary sense. In the contrary case (1.10) is to be understood in a generalized sense (cf. §6, Chapter 2). Further, as is known, there exist monotone functions and which are bounded in every finite interval, and a function 77(X) which is of bounded variation over every finite interval, such that if 1(x), g(x) E 2( — a>, cx>), then one has the Parseval equality
/ (x) g (x) dx = (1.11)
F1 (A)
(A)
(A)
{F1 (A) G2 (A) + F2 (A) G1 (A))
(A)
(A) G2 (A)
(A).
Formulas (1.8) and (1.8') show that for fixed x the functions and are the Fourier transforms of the function which is equal to [g, (x — s) + (x, s)] in the interval (x —
and equal to zero outside this interval. Therefore, replacing x by y in (1.8') and applying the Parseval equality (1.11), we obtain x
A)p(y,
(112)
=1
A)
s))
s))ds,
297
§1. AUXILIARY FORMULAS
is the intersection of the intervals (x — x + and (y —
where
y+
(1.12) assumes a simpler form if we introduce the spectral function of equation (1.1): Formula
o
(x, y, A)
A)p(y, X)dE(A)+[p(x, (y, A)
—
(y,
(p (x, A) p
A) dE (A)
+
A)
A)
(A))
A)
(y,
(A
A)
(A 1 we have the inequality
therefore the inequality (2.3) implies the bound
._!. 2
(2.4)
x,A)
{g8 (x
s) + x. (x, s))2 ds +
(x, x,
The bound (2.4) proves the lemma for x = y. The case x y reduces to the case x = y by means of the mequality (1 16) This completely proves the lemma. I Let us denote by a an arbitrary real number and put X cos at. Since the function g. (t, a) satisfies all the requirements placed on the function g (t), replacing g (t) by g (t, a) in (1 14) we obtain (25)
s, a))2ds,
x,
where
(26)
s, a)_—
w(x, t,
(t, a) dt,
(27)
(28)
H(x, y, a,
y, A)
If the function g (t) is chosçn as before, then from
VII. EXPANSION OF A STURM-LIOUVILLE OPERATOR
300
a) =
-
(s
— t I
cos
at• cos
t dt
there follows for A O),
Similarly, (2.17) can
be
we have 1121
..,i the bounds (3.11) and (3.12) it follows that for ii> r I
w(x, t,
s)
sirtfLt
dt
—
I
proves (3.9) since the number t us prove (3.10). We have
is arbitrary.
VII. EXPANS!ON OF A STURM-LIOUVILLE OPERATOR
308
2r
(x, s, v) dv
= 2ç h (x, s, t) —i--- dt. 0
0
Since s,
s, t) is an entire analytic function of t, and so, in particular, it is differentiable in a neighborhood of t =0. Therefore by a well-known theorem of the theory of Fourier series it follows that
the function h (x,
IL S
Ii.
2
2c h(x, s,
s,
dt
t)
=h(x, s, 0)+o(l)=—O(z, s, —oD)+o(1), as was to be proved. The uniformity of (3.10) follows from the bounded ness of the functions h(x, s, t) and äh(x, S. t)/at. REMARK. From a Tauberian theorem of (cf. Chapter 14 Theorem 4.2, special case) there follows a result which is more than (3.8), namely, (3.13)
I
I urn tOl(x, s;
s; —co).
}L
For x = s the equality (3.13) follows directly from the identity (2.23
and the indicated theorem of the monotonic function P(x,s;M) and
apply
For x
s one has to conside
=O1(x,x;IL) +201(x,s;IL) +O1(s,s;IL)
theorem to it (agam using the identity (2 23)j
§4 Asymptotic behaviour of the spectral function The case of the haifline
Let the equation
y"+[x—q(x)]y=O be defined on the haifline [0, co). We will assume the function summable over every finite interval. We adjoin to equation (4.1) boundary condition at x = 0: (4.1)
(4.2)
y'(O) —hy(0) =0,
309
§4. ASYMPTOTICS OF SPECTRAL FUNCTION. HALFLINE
is an arbitrary real number. The case h = will not be conhere. It is rather completely treated in the paper [15] by Levitan.
h
denote the solution of equation
A)
(4.1)
satisfying the
conditions =h.
=1,
is known, for a given h there
exists at least one function p (A)
is monotonic, bounded in every finite interval, continuous from left, and which generates an isometric mapping of the space
the space
(—
F(X)=lim 2(Q
cxx, cx>)
according
(x)ph(x, X)dx,
2(0,
x>)
to the formula
f(x)=lim
the integrals converge in the metrics of (— ccx) respectively, and one has the Parseval equality
cxx)
and
Fa (A) dp(A). 0
will call the function Oh(x, y,
spectral function of equation (4.1) (for the initial conditions (4.3)). Further, we by (x, y, A) the spectral function of the same oblem for q(x) 0. Let us first assume that h> 0. In this case the y" + Xy = 0
not have a negative spectrum. For x > 0 put x =
Then
Oh(X,y,A) =Oh(X,y,,z2) =Oh,1(x,y,,4,
the arguments 'inptotic
formula
of the preceding section,2> we obtain the
>x>)
Oh,1(X,y,,L) —Uhl(X,Y,/.L) =Oh(X,y, — x>) +o(1).
the Remark followmg the proof of Theorem 3 1
310
VII. EXPANSION OF A
OPERATOR
Let us find the principal term of
0
1(x y,
As is known (cf.
(CoSvx+! Sin vx)(cnc COSVX
COS
p.
2k ç sin (x + y)
_L 2h2 ç sin
v2+h2 d VT
0 p.
(4.6)
d
0
r
2
vx• sin
3
p.
ç cos vx • cos vy v2 + h2
cos vx • cos vy
o
o
sin vz. sin vy
2h2
2
cos vx 5
2k
sin (x +
cos vy dv +
+ h2
2h2ccosV(x_Y)d
2h25cogv(x_y)
v
d
y'
dv+o(1)
vdv
(x+y>O),
0
where
y,
=
5 cos vx
cos vy dv
is the spectral function of the problem (4.1) + (4.2) for q(x)
From the asymptotic formula (4.5) and the relation
follows at once:
0 and h (4.6) th
THEOREM 4.1. Let the function q(x) satisfy the condition (1.19) Chapter 6 For every fixed x 0 and y 0 lim {0A 1(x, y,
(47) )
y'
y
311
§5. REISZ SUMMABILITY OF SPECTRAL FUNCTION
relation (4.7) holds uniformly as x and y vary in any bounded region
>0). x=y=0 we obtain from (4.5) >
.1
—h.
Eiiii IL
us consider the case h
We choose some region u ={ x
STURM-LIOUVILLE OPERATOR
VII.
324
outside Let G0(x,
—h—c
t) denote the Green's function of the Cauchy problem for the parabolic equation öu/öt = — Lu. t) is the kernel of the integral operator i.e. thel Green's function of the Cauchy problem for the operator equation au/at— —Lu, then as can be seen by means of Green's formula, wel
the region S G (x,
(8.10)
t) —
(x,
t) = 0(1)
where c is a constant which does not depend upon On the other hand, as is not difficult to see, G(x,
or t.
h
t)=5
Taking account of this representation, (8.10) can be rewritten in the form (8.11)
(x, is
X)
the spectral function of the "reduced" operator L.
After multiplying the left and right sides of (8.11) by with respect to t from 0 to we obtain
X)ç
çdO (x,
1
3
I
I
and integrating
1 + 0 ()
is a constant which depends only upon c, and h* = Using this last relation, we can establish the following useful where
(x, E; A) =
(8.12)
(x,
A) + G
where G(s) is analytic with respect to s = + jr in the strip
I
and G(0) =0. Putting p(X) =O(x,x; we obtain
o(A)
(A)
(8 13)
(8,9)
on
(A) + G (s)
find that
A2n)=!X+o(1).
the basis of (8.12
§8. EQUATION OF ORDER 2n
325
on the basis of
Tauberian theorem Chapter 14), ,hich makes possible a sharper estimate of the remainder in Ikehara's theorem, we obtain from (8.13)
—
(A+O)—
(X)
P
2
an absolute constant Since the number h can be chosen arbitrarily large, and h, along pith h, is arbitrarily large, taking into account (8.7) and the fact that =0, we obtain the assertion of Theorem 8.1 for the case = x. this, the proof for the case x is carried out by considerakm of the nondecreasing function there A
is
=O(x,x;x) +26(x,E;X) 1'he proof of Theorem 8.2 is technically rather complicated, but can e carried out according to the same scheme [I] on estimates for the We remark that applying results of pectral function of general effiptic operators on the negative spectrum, can also establish Theorems 8 1 and 82 for operators which are Ot semibounded. At least for operators with sufficiently smooth coftIcients, this can be done rather easily. Namely, one has to consider square of the non-semibounded operator L which is now a positive perator. One further has to use Theorem 8.1 or 8.2.
Theorem 8.1 follows: THEOREM 8.3
(ON EQUIcONVERGENcE). Every function 1(x)
with
mpact support which can be expanded in a Fourier integral can also panded as an integral with respect to eigenfunctions, and conversely,
be
We are concerned here with the pointwise convergence of the expansions.
Fhis theorem follows imthediately from Theorem 8.1, since the partial
ins of an eigenfunction expansion and a Fourier integral expansion equal respectively to b
S(x, f;
0(x,
X)
d- consequently
S(x,
(
/; X)__j a
. S'11
(x,
X) /
326
VII. EXPANSION GFASTURM-LIOUVILLE OPERATOR
By this
one. can. also obtain the corresponding asymptotic (x, A), which enabled D/ 0 (x X) = one to write down the asymptotic behaviour of the spectral matrix: behaviour of
derivatives
Bibliographical references
§1. The basic results concerning eigenfunction expansions are pre-j sented in the monographs by Titchmarsh [1], Levitan [1] and The identities (1.12) and (1.14) were obtained in a paper by Levitan [14])
*2. Lemmas 2.1 and 2.2 for the case of the halfline [0, x') are part due to [1,2]. The proof of Lemma 2.1 here is due to Levitan [14]. §3. The asymptotic formula (3.8) was obtained in papers by [14, 15]. Cf. also
[3].
§4. Formula (4.5) was derived in a paper by Levitan [15]; a paper by [3]. §5. The Riesz summabiity of the spectral function was in papers by Levitan [14,15]. 7. The theorem on equiconvergence was first proved by
cf.
[14]; cf. also his paper [15] and [1,2]. The Riesz of an eigenfunction expansion was considered by Levitan [14, 15].
§8 The results of this section are due to
[1,2]
CHAPTER 8
DIFFERENTIATION OF AN EIGENFUNCTION EXPANSION §1. Preliminary estimates of the derivatives of the spectral function
In this chapter we will study the question of the differentiation s expansion with respect to the eigenfunctions of the selfadjoint sd-order differential operator
y" +[x —q(x)]y=0, med on the halfline [0, with the initial
conditions
y'(O) =0.
y(0) =1,
that q (x) is a real function defined summable over every finite interval.
assume
on the halfline
[0,
hroughout the chapter we will assume that the spectrum of the (1.1) —(1.2) is nonnegative or, if a negative spectrum exists, it is bounded from below. The latter case can easily be reduced former. Thus, let A the solution (A> 0). We denote by problem (1.1) —(1.2). s is known, there exists a monotonic function p (,z) which is bounded
finite interval, such that for every function f(x) E has the Parseval equality
every
co)
(FL),
FL)ds.
;t us denote by O(x, s,
the spectral function of equation (1.1) (for
initial data (1.2)), i.e. we put v)p(s, v)dp(v),
O(x, s, p.)= 0,
327
328
VIII. DIFFERENTIATION OF AN EIGENFUNCTION EXPANSION
In this section we will obtain various preliminary estimates for the derivatives of the spectral function O(x, s, As we have seen in the preceding chapter, for arbitrary real t we have I
(1.5)
x+t
t, s)cp(s,
Differentiating (1.5) with respect to x, we obtain
t, (x,
t, x — I
Suppose that the function
x+i
(x
ow(x, t,
t, u.)) s)
p (s,
ds.
satisfies the conditions indicated on and then integrate with
p. 296. If we multiply both sides of (1.6) by respect to t from 0 to we obtain
p.')}g8(t)dt
+
(w (x,
t, x + t) p (x + t,
(1.7)
— w (x,
t, x — t) p (x — t,
(t) dt
Let us transform the right side of (1.7). Consider the first term. Integrating by parts, we find that
=—p(x,
fl. PRELIMINARY E&FIMATES
329
p(x—t, using the evenness of g, (t) with respect to t, we obtain
Ii = — p (x + t,
(t) dt + p (x — t, x)
(s,
—
gç (t) dt
x)
(s,
ds
x
ag (s—x)
ds
We now transform the second term of (1.7). By virtue of (2.9) and 2.10) of Chapter 6 we obtain
t, x+t)p(x+t,
x+B
—4
p (s,
g1 (s— x) {
= —4 —5
w(x,
{g1
q (x+
(s — x)
q (jt)
p (s,
t, x—t)p(x—t,
=45p(x_t,
=4
5
p(s,
5
dt} p(s,
ds
330
VIII. DIFFERENTIATION OF AN EIGENFUNC'rION EXPANSION
Consequently, (1.9)
p(s,
Finally, changing the order of integration in the last integral in (1.7), obtain
we
=
x+s p (s,
X,
(x, s) ds,
where
(x,i, s) dt.
(x, s)
(1.11)
By virtue of (1.8), (1.9) and (1.10), the relation (1.7) assumes the form (x,
(1 12)
= —4 (s — x)
q
p (s,
ds
p.)
p (s,
p.) ds + 4
i.(x, s) p (s
s)—
ts
p)ds
It follows from (1.12) that for fixed x the function be regarded as the generalized Fourier transform (with respect to eigenfunctions (x, of the function which is equal to (113)
s)—
—
s E (x and is equal .to zero outside this interval. Therefore by the Parseval equality we obtain from (1.12)
(1.13')
(p.)
(x,
p.)
(p.) dp (p.)
=
/ (s)
(x, s) ds.
Hence, by the arbitrariness of f(s) and by (1.13) we obtain •(p.)p(s,
p.)dp(p.)=
331
§1. PRELIMINARY ESTIMATES
(x, s) —
(s— x)
x)
(s
J 1
J Obtain
q (t) 6,
X—
Ix—s(>s.
0,
the identity (1.14) with respect to s 1) and using (1.11),
for x> s
ow (z, t, s) x
+
O2g, (s— x)
s)
g, (t)
=1 x —sJ
IX—St>E.
0,
the definition of the spectral function O(x, S.
Le. (1.4), the identity
assumes the form
s)1
+ f
0,
(t)
02:,t, r
—
(s— x)
I
q(s),
Jx—sI>s.
assume here that the function q(x) is absolutely continuous.
332
VIII. DIFFERENTIATION OF AN EIGENFUNCTION EXPANSION
Extending the function O(x, s, in the variable odd function, for a = x we obtarn from (1 15) d
ö20 (x
S
p.)
— g2 t s)
(i., t
( )
so as to obtain an
s)
(x)
J
2 We will now prove some simple lemmas concerning the behaviour of the derivatives of the eigenfunctions LEMMA 1 1
If the function q(x) is absolutely continuous in every finite
interval, then as a —* {a20
(1 17)
have the following estimate
we P.)}
=
(x,
dp (it)
0(a2)
5
This estimate holds uniformly as x varies in any finite interval PROOF From well-known asymptotic formulas for the eigenfunctions it follows that for x in a finite interval and —*
(118)
Hence (1 17) follows from Lemma 2 3 of Chapter 7 One can prove the following lemma in a similar way LEMMA 1 2 If the function q(x) has a derivative of order 2k which is summable over every finite interval then for every fixed x we have as a
a',
a-i-i
s,
(1.19)
a
=5
t.
a
(k=1,2,...). This estimate holds uniformly as x varies over any finite interval.
From Lemma 2.3 of Chapter 7, Lemma 1.1 and the CauchyBunjakovskii inequality follows: LEMMA 1.3. If q(x) has a first derivative which is summable over every finite interval then for every fixed x and s we have as a—* co,
(120)
P1=
a-4--i
§2. ASYMPTOTICS
333
estimate holds uniformly as x and s vary over any finite intervals.
s
!rom Lemma 1.2 and the
Cauchy-Bunjakovskii inequality follows:
LEMMA 1.4. Under the conditions of Lemma 1.2 we have the estimate ct+1
y
A)
{a2k6 (x, S,
(z, " i
(s,
0
I
(1+1=2k). s estimate holds uniformly as x and s mzzy over any finite intervals.
t
denote the spectral function of the problem (1.1) —(1.2)
q(x)
0, i.e. we put s,
the definition of O*(x, 1.5. As
we have the estimate
(k+f=n=O, 1,
àxôsi
a
s
there immediately follows:
.
estimate holds uniformly over the entire halfline (0, ce.), §2. Asymptotic behaviour of the derivatives of the spectral function
ormula (1.14) enables us to study the asymptotic behaviour of the of the spectral functions 8(x,s,,z). In order to study the •;otic behaviour of the derivatives of the spectral function of the oble a (1.1) + (1.2), we wifi compare them with the derivatives of function of a simpler problem, namely the problem
y"+Xy=O y(0) =1,
this end we will transform the
=
X8
0,
—
(A>0), y'(O)
=0.
right side of (1.14). We have
x)
334
VIII. DIFFERENTIATION OF ANEIGENFUNCTION EXPANSION
we can give (2.3) the form
Using the spectral function
(2.3') —x)
Jx — x
0,
i.e. for the spectral function1 Writing down for the function of the problem (2.1) + (2.2), a formula analogous to (2.3'), subtracting1 it from (2.3') and taking into account that for the problem (2.1) + one has w(x, t, s) 0, we obtain {ôO
ôO*(x,s,
(x,s,
} (2.4)
= Ix—sI>e,
0, where
(2.5)
(x,
s) =
(t)
5
ow (x,t, s) dt.
Put Ow(x,t,
(2.6)
5 tx—al
Then by the Parseval equality for the ordinary Fourier integral it
from (1.6), Chapter 7, and (2.6) that (2.7)
5
0
(x, s,
=
Ow
(x,t, s)
gg (t) dt.
5
tx—il
Further, by virtue of the inversion formula we find from (1.6), ChaptE
7,that (2.8)
' 'd
335
§2. ASYMPTOTICS
since the functions and a(x,s,,2) are even with respect it follows from (2.7) and (2.8) that (2.4) can be rewritten in the wing form: s,
}=O. x
2.1. If q(x) is absolutely continuous in every finite interval, for every fixed x and s we have (p cx) s, ax
S.
J
0
asymptotic relation holds uniformly as x and s vary over any finite ervals.
IL
V
s,
v) dv = (i')'
V IL
0
a by (1.20) and (2.9) the Tauberian Theorem 4.1 of Chapter 14 is ?, from which it follows that for every fixed x and s the Riesz hans of order 1 of the function ôO*(x,s,
dO(x,s,
s, v)dv
cosv(s—x)dv i.e. we have
nain bounded as IL
j\
j.L2
d,
{ao
(x,:,
ôÜ*
s
.v) _!
s, a)da
0
cos a (s —x) da}
= 0(1).
the theorem it remains to compute the integrals (1
s, v)dv,
336
EIGENFUNCTION EXPANSION
VIII. 8
'2=
q
(i
(t) d—
x
cos
—
0
(s
— x)
Let us compute the first one. By the definition of the function.i a(X,S,v), i.e. (2.6), we have
= ni\/ I ——- is
ow (x, t, s) cos i ———— Ox r
ç
Changing
J
dt
the order of integration, we obtain
(2.12)
=
s)
Ow
(i {0
IX—81
cos vt dv}
—
dt.
Further, by a well-known theorem (cf. Titchmarsh [3], §7.1) (2.13)
cos vt dv
—
=
172
From the last two formulas we finally obtain
'2 14'
Ow (x, t, s)
I— —
dt
Ox
3
For the second integral we obtain, by (2.13), j
8 18
(s —
(2.15) the theorem follows from (2.11). Similar asymptotic formulas can be obtained for the higher-ordei derivatives of the spectral function O(x,s,ii). However we will not dc this here, restricting ourselves to stating a theorem for the second
derivatives.
If q(x) has a first derivative which is absolutely in every finite interval, then for every fixed x and s we have the asymptotic formula THEOREM 2 2
v)
(216)
O2Q*(x: v)]O(1)
Ths formula holds uniformly as x and s vary over any finite intervals
§3. EQUISUMMABILITY
337
§3. Equisummability of differentiated eigenfunction expansions
this section, using the preliminary estimates obtained in §1 for derivatives of the spectral function O(x, s, hz), we will prove a theorem equisummability of differentiated expansions in the eigenfunctions
Sturm-Liouville operator and in an ordinary Fourier cosine integral Linctions in
will first prove some lemmas. formula (1.13') follows (p.)
(x, p.) dp (p.) =
F (p.)
f(s)
(x, s) ds,
and F (p.)
1
.
i. m f(s)
(s, p.) ds,
A-+co
ow (x,t, s)
(x, s)
g1
(t) di, +
0g. (s— x)
Ix—sf a
—4g1(s—x)
introduce the notation S(x, s, p.)ds.
the definition of the function S(x,,z), (3.1) assumes the form (p.) d
OS
=
s)
(s) {
(t) dt
Ix—'I
+
g. (s — x)
q(t) dt} ds.
function S (x, is a segment of the expansion in the eigenfunctions a Sturm-Liouville operator, and S*(x, is a segment of the expansion ordinary Fourier cosine integral, of the function f(x) E 2(0, cx>). (x, riting for a formula analogous to (3.6), taking into account for this case q(x) 0, w(x,t,s) 0, and subtracting it from (3.6), obtain -
338
VIII. DiFFERENTIATION OF AN EIGENFUNCTION EXPANSION
fôS(x,
r
j
ax
üx
37)
=
f(s)
g5 (t) dt —
g1 (s — x)
q
ds
2(0 If q(x) is absolutely continuous ir LZMMA 3.1. Let f(x) E the every finite interval, then for each fixed x we have, as p.+i
(3.8) This
estimate holds uniformly as x varies in any finite interval.
i.e. by (3.4) we have
PROOF. By the definition of II.
p.)F(p.)dp(p.)
Therefore from the Cauchy Bunjakovsku inequality it follows {OS (x,
}=
I
(x, p.) F (p.) I dp (p.)
( . )
'/,
p.+i
F2 (p.) dp
(x, p.)]2 dp (p))
(p))
By Lemma 1.1 we have (x, p.)]2 dp (p.) = 0
(3 10)
Further, since J'2.
dp (ii)
).
1(x) is absolutely continuous in some neighborhood of
f'(x) is continuous at x0 Ilin
'
dS(x,v)
0'
first derivative of the expansion of f(x) with respect to the eigenions of the Sturm-Liouville operator is summable at the point x0 by rst-order Riesz method to the value f'(x0).
ar theorem holds for the higher-order derivatives. §5. Convergence of a differentiated eigenfunction expansion
5.1. Suppose that q(x) is summable over every finite interval, wppose that f(x) satisfies the following conditions for x 0:
f(x) and
belong to
..J)cosa+f'(O)sina=O. W{ (x, A), f(x) } = 0 for all A from the spectrum.
the eigenf unction expansion of f(x) converges absolutely.
OF. From the hypotheses of the theorem and from Green's follows
=
1
1(x)
p (x,
X) dx =
VIII DIFFERENVATION OF AN EIGENFUNCTION EXPANSION
348
i .m
—
{p (x, A)) dx
1(x)
=—
0
(A) =1.i• m 2 {/ (x)) p (x,
A) dx.
0
From Lemma 2.1, Chapter 7 and the Cauchy-Bunjakovskii inequali
it follows that
Let us consider the integral JF(A)fJp(x, A)Jdp(A).
If we replace A by F(A) this
and put p(A)
F1(u),
integral assumes the form
It further follows from (5.1), Lemma 2.3
of
Chapter 7
and
the
Bunjakovskir inequality that a+l F1
I p1
(x,
dp1
(FL)
J
a+1
where
(z, x, a +1) —
dp1 (p))
( the constant C
does
01
(x,
x, a))
4,
not depend upon a Therefore from 1
p4jdp1(p.)< o
o
a—
349
§6. JUSTIFICATION OF FOURIER'S METHOD
together with (5.2) proves the theorem. purpose of this section is the proof of the following theorem. i 5.2. Suppose that f(x) satisfies the conditions of the previous em and, moreover, that the function g(x) =f"(x) — q(x)f(x) can be nded with respect to the eigenfunctions ce'(x, A) at the point x0
f'(x0) = urn
0.2) Then
(x0, A) dp (A).
F (A)
By hypothesis tim
(A) p (x0, A) dp (A)
F (A)
—urn
AF (A) p (x0, A) dp (X)
(x0, A) dp (A)_q (x0) tim
F (A) p (x0, A) dp (A).
—m e
g(x0) ==f"(x0) —q(x0)f(x0) and urn
F (A) p (x0, A) dp (A)
/(x0),
(5.4) follows from (5.5). §6. Justification of Founer's method for the one-dimensional wave equation
section we will show that the results of §4 of the preceding r, in which we investigated the solution of a mixed problem for Due-dimensional wave equation on a finite interval, together with rem 5.2, enable us to justify Fourier's method for the one-dimensional
equation under weaker conditions on the initial function than ones.
will consider the following mixed problem on an interval: ô2u/ôt2 =ô2u/ôx2 —q(x)u,
ulto =f(x), a follows from Theorem 7.1, Chapter 7, for this it is sufficient that the function satisfy any local conditions which suffice for it to be expandable in a trigonometric er series. Since, by hypothesis, f(x) has a second derivative for x 0, it is for
le sufficient that f" (x) and q(x) be of bounded variation in a neighborhood of hit Xe•
350
VIII. DIFFERENTIATION OF AN EIGENFUNCTION EXPANSION
=0,
(6.3) (6.4)
[ôu/Ox
(6.5)
[Ou/äx
=0, = 0.
(cf. §4, Chapter 6), if f(x) has a continuous derivative and satisfies the conditions f'(+O) —hf(+0) =0, f'(7r —0) +Hf(ir —0) =0, (6.7) then the mixed problem (6.1) —(6.5) has a solution for all t> 0. Let ui denote this solution by u0(x, t). For fixed t this solution satisfies boundary conditions and has a continuous second derivative. by a classical theorem u0(x, t) can for any fixed t be expanded in a1 absolutely and uniformly convergent series with respect to the functions of the Sturin-Liouville operator (6.8) (6.9)
y'(lr) +Hy(7r)
y'(O) —hy(0) =0,
0,
i.e. (6.10)
(x,
c,,
"=1
(t)
p,, (x),
where
(6.11)
(t)
=
u0
(x, t)
(x) dx,
are the eigenfunctions of the problem (6.8) + (6.9). and we will first determme To this end we use the initial conditions (6.2) —(6.3). Putting t =0! (6.11), by virtue of (6.2) we obtain
and the
determine the coefficients
(6.12)
(0) = 1(x)
(x) dx
=
Differentiating (6.11) with respect to t and then putting t = 0, by
(6 11) twice with respect to t, by (6 1) we obtal
351
§6. JUSTIFICATION OF FOURIER'S METHOD
c (t)
ô2u0(x,
t)
(x) dx =
—
q
(x) u0 (x,
(x) dx.
utegrating the last integral by parts twice and using the boundary nditions (6.4) —(6.5) and (6.9), we obtain
(t) =
(x, t)
(x)} dx,
(x) — q (x)
which by (6.8) follows (t)
=
u0 (x, t)
(x) dx
=
(t).
by virtue of (6.12), (6.13) and (6.14) we obtain the following auchy problem for the determination of
=0,
=0.
c,,(0)
solution of this problem has the form Inserting this value of
in (6.10), we obtain
u0(x,
the function f" (x) satisfies a Lipschitz condition with exponent on the the interval then it follows from (4.5), Chapter 6, Theorem that the series (6.10') can be differentiated twice respect to x, i.e. ô2u
0=
cos
tp (x)
the series (6.16) converges uniformly. will prove that the series obtained from (6.10') by differentiating e with respect to t converges absolutely. In fact, put ieorem 5.2 considers the halfline [0, ). However it is easily seen that this theorem .e proved by the same methods for the case of a regular Sturm-Liouville problem. as is evident from the proof, for the regular case Theorem 5.2 follows from a theorem on the equiconvergence of the expansion of a function in a Sturmuville series and the expansion of this function in a Fourier cosine series (cf. §9, —
Ler 1).
352
VIII.
OF AN EIGENFUNCTION EXPANSION
Integrating the last integral by parts twice, by the conditions (6.6)
—
(6.'3
and (69) we obtain
0
=
(x) dx =
/ (x)
from which (6.17)
Differentiating the series (6 10') with respect to t, we obtain ——
(6 18)
(x)
sin
Further, by virtue of (6.17) and the Cauchy-Bunjakovskii we obtain (619)
n=i
n=1
which follows from the fact that f"(x) —q(x)f(x) E
It follows from the bound (6.19) that the series (6.18)
is
convergent.
of the second derivative Let us now clarify the respect to t. By virtue of the equation (6.8) we have —
t
cos
=
(r) cos m=1
(x)
— q(x) n=1
(x) p,, (x))
(x) cos
the convergence of the series (6 10) and (6 16) follows
of the series
(x=x0)
353
REFERENCES
now consider the case
=f(x). u(x,0) =0, u1(x, t) denote the solution of the problem (6.1) + (6.20) + (6.4) Then
u1(x, t)=çu0(x, now put u1(x, t) =
c1(t)==
t)
dx,
n=O
we obtain for
c(t) +
(t) =
sin
t.
rest of this investigation can be carried out analogously to the investigation. Bibliographical references
2,3,4. The results of these sections are due to Sargsjan [1—4]. Theorem 5.2 is due to the authors and was published in their papers [1—2]. Another proof is given in the joint paper [3] by
authors.
A justification of Fourier's method was given in the paper by itan [17]. Cf. also the joint paper [3] of the authors.
CHAPTER 9 SOLUTION OF THE CAUCHY PROBLEM
FOR A ONE-DIMENSIONAL DIRAC SYSTEM §1. Derivation of formulas for the solution of the Cauchy problem
Let p (x) and r(x) be real functions defined on the entire line summable over every finite and let the vector-function f(x) be
=
real and continuously differentiable. Let u1(x,t) u(x,t)=(I\ u2(x, t)
We will consider the Cauchy problem (1.1)
01
p(x)
(0
(1.2)
0
0
ufr, 0)=/(x).
Let us first consider the problem (1.1) +(1.2) for p(x) =r(x) the problem (1.3)
0,
u°(x, 0)=/(x),
where we have put
f10\
B
/01
0
The problem (1.3) is equivalent to two independent problems i t) of the vector-function u°(x, t) and
volving the components
Further on, depending upon the circumstance, other smoothness conditions be impoeed upon the functions p(x) and r(x). 354
355
§1. FORMULAS FOR THE SOLUTION àx2 —
ot2 1.4)
(x, 0) =
àx2 —
àt2
1.5)
(x, 0\
= ., (xj,
(x),
= —i/
= /2(x),
(z).
J
the problems (1.4) and (1.5) by D 'Alembert's method, we obtain —if2(x—t)},
i4(x,t)
t6)
—t) +f2(x —t) —ifi(x+t) +f2(x+t)I the solution of the problem (1.3) is given by
}.
u0(x t)
Liere
Hr" is the transpose of H. Let us now consider an inhomogeneous problem with homogeneous itial data àü
7)
iZ(x, 0)=0. were
g(x,t)
= differentiable vector-function. It is not hard to see that this problem o breaks up into two independent problems involving the components t) of the vector-function ü(x, t), namely: (x, t) and —
ôt2
(x,
.
àx2 —
Ot
02a2
fl)=0,
ox
= —1g1 (x, 0),
0) = 0, Ox2 —
+
.
0g2
iIg1
Ot
Ox
0),
IX. CAUCHY PROBLEM FORA DIRAC SYSTEM
356
As is known,2> the solution of the problem 32v/ôt2 —ô2v/ôx2 =h(x,t),
v(x,0) =0, is given by
t)=4.
v(x,
x-f-(t—t)
z+t
q('r)d'r
h(s,
d'r 5
5
x—(i—-r)
0
applying this formula to the problems (1.8) and (1.9), find that the solutions of these problems are given respectively by Therefore,
_ig1(x-__t—j—t, 'c)—g2(x—t+r, t)}dt, (x,
t) =
{—g1 (x + t — 't, t) — tg2 (x + t —
t)
+g(x—t+;
t)}d.
Thus the solution of (1.7) is given by (1.10)
II (x, t)
+
= —4 5 {Hg (x + t —
(x — t
+
We now turn to the problem (1.1) + (1.2). We will rewrite it in the forni (1.1') (1.2)
where
t),
u(x, O)=/(x), fp(x)
0
o
r(x)
Let us assume that p (x) and r(x) have continuous derivatives. Putting now g(x, t) = — Q(x)u(x, t), assuming g(x, t) to be and then taking into account that the solution of the problem (1.1' s the sum of the solutions of the problems (1 3) and (1 7), vi4ueof (1 1) and 10) we obtam
*çL
kilibert [1] Chapter IV §2 3 be weakened
357
§1. FORMULAS FOR THE SOLUTION
u(x, t)=4{Hf(x+t)+Hnf(x_t)}
+4
{HQ(x+t—c)u(x+t—'c,
r, changing the variable of integration, u(x,
+4
111)
HQ(s)u(s, x+t—s)ds
s—x+t)ds. Thus, we have constructed for the problem (1.1) + (1.2) an equivalent
rstem of integral equations (1.11). These equations are of Volterra rpe and can therefore be solved by the method of successive approxiations. Let u0(x, t) be defined by u0(x,t)
put x+t (x, .12)
t)
=
i 4-
(s, x + t — s)
BQ (s)
+4-i
ds
s—x+t)ds.
from the uniform convergence of the successive approximations, proof we omit, it follows that the solution of equation (1.11) is yen by .13)
+
u(x,t) —u0(x,t) +ui(x,t) +
We wifi show that each of the vector-functions
t), p =
n be represented in the form u,, (x, .14)
t) =
(x, t) / (x + t) + I,,, (x, t) / (x — t) x4-/ 1 5
t, s)/(s)ds,
1,
2, 3,...,
_____
358
IX. CAUCHY PROBLEM FOR A DIRAC SYSTEM
t) and
t, s) are 2 X 2 matrices. Indeed, for p = 1 it follows from (1.12), by virtue of (1.6), that
where
t),
x+t
(x, t)
=
4- i
+4-is
HQ (s) (HI (x + t) + HT/ (2s — x — t)) ds HTQ
(s) (H/(2s— x + t) + HT! (x—t)} ds,
or, changing variables. t)
=(4-t
x+t
HQ(s)Hds)f(x+t)
+ (4-i
(1.15)
z+t
+4
(x --
HTQ (s)
I
from which (1.14) follows if we put K1 (x,
t) = 4-i
(s) H ds,
L1
(x, t) =4- i
11TQ (s) HTds.
(1.16) H1 (x, t,
Let us now assume that (1.14) has been proved for p = 1,2, . ., q — will prove it for p = q. Putting p = q —1 in (1.14) and inserting t expression for Uq_i(x, t) in (1.12), we obtain
We
Uq (x,
t) =4 i
HQ (s) {Kq_i (s, x + t —s) / (x + t)
.x+t—s)/(2s—--x—t)}ds.
s—x+t)/(2s—x+t) s—x+t)/(x—t)}ds
359
§1. FORMULAS FOR THE SOLUTION
x+i
+ 41
HQ (s) x
1
(s,
4 28—x—i
z
x + t — s,
'c)
ds
f (jc)
28—x+t
(s){4
ds
t,
=11+12+13+14. will transform each of the integrals 'k, k = 1,2,3,4. Making a change we obtain f variables in the second part of the integral 1.17')
z+t
+4-i
x+t_s)ds)/(x+t'
HQ(
bnilarly, we obtain 12
= (4 i5 HTQ (s)
s — x + t) ds) f (x — t) +
x+t
+4-i inally, changing the order of integration in the integrals 13 and e obtain
x+t—s, 'c)ds
+ we now put
(s)
(s, s — x
+ t,
't) ds
/
d-:
360
IX. CAUCBY PROBLEM FOR A DIRAC SYSTEM
x+t
x
(1.18)
(1.19)
Lq (x, t)
"
= 4- i
—'
t
(s)
(s, s — x
(x+t+s\L fx+t+s 2
t — s) ds,
2
1
+ t) ds,
'
2
(z4-t+a)
+4-i
x+t—'c, s)ds+
+ 4i
4 HTQ (ic) Hq_j ('c,
—x+
t,
s) d'c,
then (1.17) can be rewritten in the form (.; t) =
Kg (x,
t) / (x + t) + Lq (x, t) / (x — t) x+1
t, s)/(s)ds, which is (1.14) for p =q. Thus (1.14) is proved for any p 1. , m (1 13), Now, msertmg the expression for (x, t), p =0,1,2, find that the solution of the problem (1.1) + (1.2) is given by U
(x,
t) = K (x, t) f (x + t) + L (x,
t) / (x — t)
(1.20)
+4
H(x, t, s)f(s)ds,
where we have put
t)=
K(x, (x,
t),
(x, t, s).
The proof of the uniform convergence of these series is analogous
d
361
§1. FORMULAS FOR THE SOLUTION
he proof of the uniform convergence of the successive approximations, nd will be omitted. t) t) and Let us point out certain properties of the matrices thich will subsequently enable us to explicitly determine them. From he definition of the matrices Q(s) and H it follows that
p(s)+r(s), lip (s)—f-r(s)j
HQ(s)H= (
p(s)+r(s),
121)
=1P(s)+r(s)J( ['herefore by the definition of the matrix K1(x,t), i.e. (1.16), we have K1(x,t) =k1(x,OH,
1.22)
rhere
z+f k1 (x,
t)
[p (s) + r (s) I ds.
rurther,
p(s)+r(s), —i[p(s)+r(s)} p(s)+r(s) \z[p(s)+r(s)J,
/ HQ(s)H ==I. 1.23
onsequently, by virtue of (1.16) we obtain for the matrix L1(x, t) the L24)
L1(x, t)
t)HT,
'here
re
will
show that each of the matrices in the form t)
L25) r__lp(X,t)HT
t)
and
t)
can be
t)H, (p =
1,2,..
for p = 1 these formulas follow from (1.22) and (1.24). Let us that the formulas (1.25) have been proved for p = 1,2, . . ., q — 1. wifi prove them for p = q. Putting p = q —1 in (1.25) and inserting ie expression for Kqi(X, t) in (1.18), we obtain
362
IX. CAUCHY PROBLEM FOR A DIRAC SYSTEM
x+i K9 (x,
t) = 4- 1
HQ (s)
k9_1
(s, x + t — s) H ds
.x+t
Inserting the expression for HQ(s) H from (1.21), we find that x+t
K9(x,
t)=4-
k9_1(s, x+t—s)[p(s)+r(s)]ds. H,
which proves (1.25) for the matrix Kq(X, t), if we put x±t kg(X,t)
=-j-
k9_1(s, z+t—s)[p(s)+r(s)Jds.
The formula (1.25) for Lq(x, t) can be proved analogously by mean of (1.19) and (1.23). Now, inserting the expressions for t) and t) from (1.25) in (1.20) and putting k (x,
t) =
(x, t),
1 (x, t)
(x, t),
the formula (1.20) can be rewritten in the form ii (x,
(1.26)
t) = k (x, t) Hf (x + t) +
+
1
(x, t) HTI (x — t)
H (x, t, s) f(s) ds.
Formula (1.26) yields the solution of the problem (1.1) + (1.2) will play an important role in what follows.
ani
§2. Reduction to the Goursat problem
We again consider the problem (1.1) + (1.2). We will clarify that the functions k (x, t), 1 (x, t) and the matrix H(x, t, i mint satisfy m order that the vector-function u(x t), defined by (1 shall be a solution of the problem (1 1) + (1 2) In this section we matrix H(x, t, s) can be obtained from a Goursat we are able to explicitly determine the functions k (x, t)
of (1
'42,8), (2 10) below) the solution of (1 1) + (1 2) is given by (1 26) By ko(x,t) = 4 = t), and from (1 22), (1 24) and (1 25)
363
§2. REDUCTION TO GOURSAT PROBLEM
blows from the expressions for
t)
and
t)
that
0) =
0)
=0 (p =1,2,...). Therefore k(x,0) =l(x,0)
2.1)
It follows from (1.26) and (2.1) that the initial condition (1.2)
is
mtomatically satisfied.
Let u(x, t; f) denote the solution of (1.1) + (1.2), and suppose we know that the vector-function u(x, t; f) can be represented by (1.26). denote the operator denote the operator T1 = — iIO/ôt, and
In fact, it is not hard (x, t; f) u (x, t; D see that the left and right sides of this relation satisfy the same
We will show that problem
v(x,0)
the required equality follows from the uniqueness of the )lution of a Cauchy problem.
equation (1.1) can be written in the form
t,f) =u(x
12)
By the definition of the operator (x,
1; f)
= —II
[u (x,
and
formula (1 26) we have
t; j)J
t)Hf(x±t)—tk(x, t)H/'(x+t) —
i1(x,
t)
t) t,
t)
H (x,
s)f(s)ds.
the other hand, by the definition of the operator 26) we have
and
t)HB/'(x+t)
u(x, t;
+k
t) / (x
t, x
(x,
t) HQ (x +
t) / (x + t) + 1 (x, t) HTB/' (x — t)
+
formula
IX. CAUCHY PROBLEM FOR A DIRAC SYSTEM
364
± l(x, t)HTQ (x—t)f (x—t)
(x, t,
s) B/'(s)ds
t, s)Q(s)/(s)ds. the first integral in the right side of this relation by — iH, HTB = iHT, this relatioij can be rewritten in the form
and taking into account that HB = a (x, t; Baf)
+
1
(2.4)
-4
(x,
=
(x, t)
Hf' (x
1)
+/c(x, t)HQ(x±t)f(x-{-t) ffT/! (x t) HTQ (x — t) / (x — t) + ii (x, t) — t) + {IJ (x, t, x + t) Bf (x ± t) — H (x, t, X — t) Bf (x -— t, s)B—H(x, t, s)Q(s)}/(s)ds.
.
Since
the vector-function f(x) is arbitrary, it follows by virtue 4
(2.2) that the coefficients of f(x + t) and f(x — t), as well as the in the expressions (2.3) and (2.4), must coincide (the coefficients of the
derivatives I' (x + t) and f' (x — t) cancel each other). Thus, the coefficients of f(x + t) and f(x — t) respectively, we obtain (2.5)
—ik (x, t) H —4 H (x, t, x + t)
=k(x, t)HQ(x+t)+411(x, t, (2.6)
t, x—t) =l(x, t)HTQ(x_t)_4H(.x, t, x—t)B.
We will rewrite (2.5) in the form
kf(x t)H —zk(x,t)HQ(x+t) =
+t) [iB —
I]
The left and right sides of (2 5') are 2 x 2 matrices, and their the equality of their corresponding elements Let
x+t)=
(H11 (z, t, a; ± t)
t,
H12 (a;, t,
a; + t)'\
t, x±t))
365
§2. REDUCTION TO GOURSAT PROBLEM
the values of the matrices H, Q(x +t), B and H(x,t,x +t) (2.5'), we obtain Ikt (x t) + k (x t) r (x + t)
(x t) — ik (x t) p (s ± t)
(x 1)— ik (x t) r (x + t)
(x t) — k (x t) p (x + t) (x,
t, x +
t) — H11 (x, t, x -f- t), t, x -f- t) "12 t) — (a,, t, x --f- t),
t, x ±
iH11 (a,,
—f (x, t, x
±
t, x
4—
t) —
(a,, t, x
t)
t)
—i--
ince, in the matrix on the right, the first element of the first row, then multiplied by i, is equal to the negative of the second element I the first row, the matrix on the left must have the same property. ore
+k(x,t)r(x+tfl.
=
k(x,
r(x+t)}k(x, t).
olving this equation with due regard for the initial condition (2.1), we obtain L8)
k (x, t)
+ r (x +
[p (x +
exp
will now transform equation (2.6). We first write it in the form 6')
t) HTQ(x — t) = — 4-H(x, t x — t)
l( (x, t) Hr —
[iB + I]
r, in greater detail, (l't(x. t) 1 (x,
t)
(x,
t) —
t) +
ill21 (a,,
t)
t)
t,
t)
—iH22 (x, t, x —
(x
t,
t)
H21
t, x — t),
-4-—
t, x — 1)
H12 (x, t, a, —
—4--
(a,'
t, x — t)
since the elements of the first column of the matrix on the right, when by i, are equal to the corresponding elements of the second Olumn, the matrix on the left must have the same property Therefore i{4'(x,t) —ip(x —t)l(x,t) = —il[(x,t) —r(x —t)l(x,t),
366
(29)
IX. CAUCHY PROBLEM FOR A DIRAC SYSTEM
lax,
± r (x— t)) l(x, t).
Integrating this equation wijh due regard for the initial condition (2.1), we obtain the following explicit expression for the function 1 (x, t): (2.10)
1(x,
Further, by virtue of equation (2.7) we obtain from (2.5')
H(x, t, x+t)[iB—IJ =ik(x, t){p(x±t)±r(x±t)JH—2HQ(x±t)}. Since
iB_I=_HT, [p(x±t)+r(x+t)IH_2HQ(x±t)
= I [p (x + t) — r (x + t)]E,
where
it follows from (2.11) that (2.12)
H(x,t,x+t)HT=k(x,t) [p(x+t) —r(x+t)]E.
By (2.9) we obtain from (2.6') (2 13)
H (x, t, x — t) [lB
+ Il
—_—il(x, t){[p(x—t)+r(x—-t)]IJT—2HTQ(x—-t)).
But since by the definition of B, HT and Q(x — t)
IB+I—=H, [p(x_t)+r(x_t)}HT—2HTQ(x—t)
= I [r (x — t) — p (x — t)] E',
f—i 1\ (2.13) assumes the form H(x,t,x—t)H=l(x,t)[r(x---t) —p(x—t)]E'. (2.14)
Finally, equating the mtegrands in (2 3) and (2 4), we obtain for matrii li(x, t,s) the equation (2.15)
t, s) — H8' (x, t, s) B + H(x, t, s) Q(s) = 0.
367
§3. OPERATOR-MATRIX TRANSFORMATIONS
equation, together with (2.12) and (2.14), defines the Goursat roblem for the matrix H(x,t,s). From (2.5") and (2.6") we obtain another relation, which we will further on. Equating, in each of these equalities, the elements of first row and column, taking (2.7) and (2.9) into consideration, then putting t=O, by virtue of (2.1) we obtain
H11(x, 0, x)+iH12(x, 0, x)=±4[p(x)—r(x)], H11(x, 0, x)—tH12(x, 0, x)=4[p(x)—r(x)}, H11(x, 0, x)_—4[p(x)—r(x)],
H12(x, 0, x)=O.
equating the elements of the second row and first column, H21(x, 0,
0,
H21(x, 0,
0, x)==—4[p(x)—r(x)],
H21(x,
0, x)=0, H02(x, 0, x)=—4[p(x)—r(x)]. §3. Operator-matrix transformations
Let A1 and A2 be two linear differential operators, and L1 and L2, linear function spaces. )EFINITION. A continuous Linear operator X, which maps L1 into is called an operator transformation if it satisfies the following two 1. A1X = XA2. 2. The inverse operator X1 exists.
(p1(x)
(
0
d
o)
(p1(x) U
0
368
IX. CAUCHY PROBLEM FOR A DIRAC SYSTEM
/ —
(33)
( P2
421
(x) d
\— r2(x)j
0 1'\ d
1
(p2(x)
0
where p1(x), r1 (x), p2(x) and r2(x) are real functions which are summable
over every finite interval (0 < x b < cx>). We take for L1 the set of all continuously differentiable vector-functions
f(x) = \f2(x)
defined on the interval [0, b) and satisfying the boundary (34)
12(0) —h1f1(0) =0,
where h1 is an arbitrary finite real number. We take for L2 the set of all continuously differentiable vector-function1
g(x) = g2(x)
which are defined on the same interval [0, b) and satisfy the boundar condition g2(0) —h2g1(0) =0,
(35)
where h2 is an arbirary finite real number
We will seek the operator matrix X in the form (f(x) E L1) (3.6)
X [/
(z)J
R (x) / (x)
+ K (x,
s) / (s)
where R(x) and K(x,s) are second-order square matrices. By virtue of (3.2) and (3.6) we have
A1X[/(x)}=BR'(x)/(x)+ BR(x)/'(x)
+ BK (x, x) 1(x) +
(x,
+ Q1 (x) K (x, s)} f(s) d
by virtue of (3 3) and (3 6) II (x) Bf' (x) + R (x) Q2 (x)f (x) K (x, s) {Bf'
+ Q2 (s) /
369
§3. OPERATOR-MATRIX TRANSFORMATIONS
by parts, this relation can be rewritten in the form XA2 [1(x)]
R (x) Bf' (x) + R (x) Q2 (x) 1(x).
+K(x, x)B/(x)—K(x, O)Bf(O) s) B) / (s) ds.
(x,
ince f(x) is an arbitrary vector-function from L1, by virtue of (3.1) coefficients of f(x) and f' (x), and the integrands in the expressions
and (3.8), must be equal. Therefore, equating the coefficients we obtain BR(x)
if we put
\i(x) from (3.9) 5(x) =a(x) and has the form
/
= —fl(x),
i.e.
the matrix
13(x)
ct(x),
us determine the functions a (x) and (x). To do this, by equating coefficients of f(x) in (3.7) and (3.8), we obtain the following equation
the determination of the matrix R(x) 1)
BR'(x) +Q1(x)R(x) —R(x)Q2(x) =K(x,x)B —BK(x,x).
K(x,
(K11(x, s), K12(x, s)' s)
niing the explicit forms of the matrices Q1(x), Q2(x), B and R(x) J.1O)), we can write (3.11) in the following form: (x), (x) + ( —[3' (x) + [p1 (x) — p2 —cu (x) + [p2 (x) — (x)] [3(x), —[3'(x)
— (—[K12
(x, x) + K21(x,
x)
[3(x)
(x) —
T2
(x, x) x)
the matrix on the right side of (3.12) the elements on the principal 1 are equal except for having opposite signs, while the two offonal elements are equal. The same is therefore true of the matrix
DIRAC SYSTEM
370
on the left side of (3. 12), and so it follows that 4-[p1(z)
r2 (x)]
(x) =
(x)
+
(x) —
—P'(x)+
(x)1
— r2(x)] B(x),
i.e.
2a'(x)
(3.13)
2fl'(x) —q(x)a(x)
+q(x)(3(x) =0,
0,
q(x) =p1(x) —p2(x) +ri(x) —r2(x).
the system
From
ta2(x) +fl2(x)
= 0,
(3 13) follows 2a(x)a'(x) +2fl(x)fl'(x) = 0, ie from which we obtain
(315)
+132(x)
Suppose
2(0)
that the vector-function f(x) =
is
continuously differentiable and satisfies the conditions
=1,
(3.16)
12(0) =h1.
satisfies the boundary condition (3.4), and then L1. Further, let
Then f(x) obviously
fore f(x)
X[f(x)] =g(x),
(3.17) where
the vector-function g(x)
=
an element of the space L2 and so satisfies the boundary condit (3.5). Then for x =0 there follows from (3.17) by the definition of I is
operator-matrix X, i.e. by virtue of (3.6) and (3.10),
g1(0) =a(0)fj(0) +fl(0)f2(0),
g2(0) =
+a(0)f2(0).
the first of these equations by h2, subtracting it and takmg into account the boundary condition (3 5) condition (3.16), we find that P (0)
Put
§3. OPERATOR-MATRIX TRANSFORMATIONS
371
a(0) =1. =(h1—h2)/(1+h1h2). (1±hi)2(i_1_h2)2 _.x2 solving the system (3.13) and taking account of the expressions ), (3.18) —(3.20), we obtain the following explicit expressions for and $(x):
cL(x)=x sin
+ arcsin
4 q (t)
P (x) = x cos
+ arcsin 4-
the function q(x) is defined by (3.14), and the number ic is defined (3.20).
equating the integrands in (3.7) and (3.8), we obtain the owing equation for the matrix-kernel K(x, a): K,' (x,s)
B
(x, s) =
K(x,s)Qa(a) — Q1(x)K(x,s).
the term in (3.8) which contains f(O) must be zero in view of absence of a similar term in (3.7). Thus K(x, 0) Bf(0) =0, i.e. (—K12 (x, 0), K11 (x, 0)'\ (x, 0), K21 (x, 0))
/ 0 =0 is equivalent to the system of equations K12 (x, 0)
(0)
K11 (x,
0)/2(0),
(0) = K21 (x, 0)12 K22 (x, by virtue of the boundary condition (3.4) we finally obtain 0)/i
K12(x, 0)
= h1K11(x, 0),
K22(x, 0)
= h1K21(x, 0).
is further put K11(x,0)
K21(x,0)
are for the present arbitrary continuously differand functions. The conditions (3.24) and (3.25) for the matrix-
372
ix. CAUCHY PROBLEM FOR A DIRAC SYSTEM
kernel K(x,s) combine to give the condition p(x), h1p(x)\ (3.26) K(x,
and this condition together with equation (3.23) defines a Cau problem for K(x, s). This problem is Expressing the solution of the problem (3 23) + (3 26) by means ( (1.26) and inserting it in (3.11), we obtain an integral equation for
determination of the vector-function
(2x)
/
The quantities
.
and (0), which also appear in this equation, can be determined I the equation (3.12) for x = 0. Therefore, carrying out the calculations
the opposite order, we can prove that the operator-matrix X, &. by (3.6), in which the matrix-kernel K(x, s) is the solution of the
problem (3.23) + (3.26), satisfies the relation (3.1). The existence the inverse X 1 follows from the form of the operator X. §4. Solution of a mixed problem on the halfline
Let p(x), r(x) and the vector-function f(x) =
satisfy
previous requirements. We extend the functions p (x) and r(x) to negative halfline in such a way that they are still suminable over e finite interval, but otherwise arbitrary, and extend fi(x) and f2 a manner which for the present we leave unspecified (the ms which they are to be extended will be made precise further on). Under these assumptions we will consider the mixed problem (4.1) (4.2)
u(x, O)=/(x),
(4.3)
u2 (0, t) — ha1 (0, t)
where,
=0,
as before, I is the unit matrix,
/ Ot\
/p(x)
o)'
arbitrary finite
0
number.
0
\
§4. MIXED PROBLEM ON HALFLINE
373
Let
u(x,t) = / u1(x,t)
\U2(X,t)
the solution of the problem (4.1) —(4.3). According to the iula (1.26) the solution of the problem (4.1) + (4.2) can be represented the form u (x, t) k (x, t) Hf (x + t) + 1 (x, t) HTf (x — t)
t. s)f(s)ds, by the results of §2 k(x,
I
r(x+
l(x,
Let us now consider the application of operator transformations in to extend the solution of (4.1) to the negative haifline. We will w how to express the solution u(x, t) at the point — x as the result I a linear operator acting on u (a, t) (0 s x). Let the matrices A1 have the form 4 A2==B$—+Q(x)
will assume that the functions p(x) and r(x) have been extended negative halfline in such a way that p (—0) = p ( + 0), r( —0) i.e.
Q(—0) =Q(+0). ier words, we wifi assume that the extension of the matrix Q(x) ;inuous at zero. Further, let us denote by X an operator-matrix isformation which maps the space Lh of continuously differentiable f(x) satisfying the boundary condition (4.3), i.e. f2(0) =0, onto the space Lh. jose that the vector-function U
+1
I A
I
+(
_ \ ui4(x, t)
374
IX CAUCHY PROBLEM FOR A DIRAC SYSTEM
is the solution of the mixed problem (4 1) —(4 3) on the halfime [0, ccx) By virtue of (4 3), for every fixed t the vector-function u+(x, t) belongs
to the space Lh; therefore the operator X can be applied to The extension of the solution t) to the negative halfline be defined by the formula (4.7)
u(x, t) = X
u( — x, t)
t)J,
i.e. by the definition of X (cf. (3.6)) (4.8)
(x, t)] = u
t) = B (x)
(x,
(x, t) + K (x, s)
(s, t) ds.
We will show that the vector-function u — (x, t) satisfies the equatü (4.1) on the negative halfline. By virtue of the definition of the i
A1 we have to show that —
(49)
A1 [u (x, t)]
—ii àu .
Since for every fixed t the solution t) belongs to the space I; by the definition of the operator-matrix transformation X and (3.1 we have (4.10)
t)] =
A1X
t)].
According to (4.7) the left side of (4.10) coincides with the left of (4.9). Let us calculate the right side of (4.10). By virtue of (4 which can be rewritten in the form A2[u+(x,
t)] = X [—ii àu+] = —iIX
It follows in obvious fashion from (4.8) that the operator-matrix and the operator of differentiation ä/8t commute; therefore by the above relation implies that [n+ (x, t)] i.e.
= —il-i- X
(x, t)] =
the right sides of (4.9) and (4.10) coincide, which proves our asseri
We will now show that the extension of the solution u+(x, t) to negative is continuous together with its first derivative respect to x.
375
§4. MIXED PROBLEM ON HALFLINE
fact, putting x =0 in (4.8), we obtain
u(—0,t) definition (cf. (3.10)) (
(0))'
a(0) = 1, and can be computed from (3.19). Since in the .t instance the spaces L1 and L2 which were considered in the iious section coincide, L1 = L2 = Lk, it follows that h1 = h2 = h, and = 0. Hence R(0) = I, and the continuity at zero of the efore follows from (4.11).
of
ther, using the explicit form of the operators A1, A2 and X, by of (4.7) the relation (4.10) can be rewritten in the form B
+ Q (—x) u = R (x) [B
+ Q (x) u÷]
t)]ds.
utting x =0, we obtain B
Ou (—x, t)
+ Q (—0) u (—0, t) t)
àzi
+ Q(+0)u(+0,
R(0) = I, it follows by (4.6) and (4.11) that on (—x, t)
On (x, t)
Ox
Ox
first derivative is continuous at zero. ormula (4.8) can be used to extend the original vector-function e
•
In fact, putting t = 0 in (4.8) and taking account of the initial i (4.2), we obtain
f(—x)=R(x)/(x)+
s)f(s)ds.
continuity at zero of the extension of f(x) and its derivative follows (4.11) and (4.12), if we put t = +0 in them. t us now return to the problem (4.1) —(4.3).
x> t > 0, then the solution of this problem coincides with the
376
IX. CAUCHY PROBLEM FOR A DIRAC SYSTEM
solution of the problem (4.1) + (4.2), and consequently is given (4.4). Now if 0 <x ),
\J2tXi
by the generalized Parseval equality (2.4) we have s;
+ W12(x, s; e)12(s)} ds
(i = 1,
2),
in matrix form 5
-w
0
REMARK The identity (27'), which was derived under the assumption
p(x) and r(x) are continuously differentiable functions, is valid summable p(x) and r(x). This is easily obtained by a passage to limit.
ILEMMA 2.1. If the elements p(x) and r(x) of the matrix Q(x) of the (1 1) are summable over every finite interval and if (x0, x1) is an
finite interval of the real tine, then there exists a constant C 1:C(xo, x1) such that for any x and s from the interval (x0, x1) and for a we have the bounds +1
a+1
(x, s, ')} =
1)11
51
(s, ')l di (1)
). 1. 1(x) E The trector-function f' (x) is continuous at some point x0.
2.
1
431
REFERENCES
Then
/ —
ox
=I
(x0)'
i.e. the first derivative at the point x0 of the expansion of f(x) with respect to the eigen-vector-functions of a one-dimensional Dirac system is summable by the Riesz method of order one to the value f' (x0). BibliographicaJ references
The results of this chapter are due to Sargsjan [6,10,12].
CHAPTER 12
ASYMPTOTIC BEHAVIOUR OF THE NUMBER OF EIGENVALUES OF A STURM-LIOUVILLE OPERATOR An integral equation for the Green's function
1 From the results of Chapter 4 we know that if the function q(4 m the Sturm-Liouville operator
Ly=—y"+q(x)y
(11)
(a<x 1 for all x E (a b)4 We will consider m detail the case of the entire ime, 1 e the case a — b = + The changes required for the case of the halfinie not comphcated, and will be mdicated briefly We put for fixed x and p>O (1.2)
ii
and consider the mtegral equation
(13)G(x,
It will be shown below that for sufficiently large
equation (IJ
can be solved by an iteration method and that the solution is the Green Under the assumption that at least one of the boundary points is singular In connection if one of the boundary points is regular then we have to specify a
condition at this point. 432
§1. INTEGRAL EQUATION FOR THE GREEN'S FUNCTION
433
function of the operator (1.1). (The uniqueness of the Green's function follows from Theorem 2.3, Chapter 2.) 2. Let us denote by X the Banach of numerically valued functions A (x, (— <x, n < + cx) with the norm I!A(x,
and define an operator N in X by (14)
NA (x,
where the function g(x, is given by (1.2). Further on an important role will be played by the following: LEMMA 1.1. Suppose
that
q(x)
satisfies the
conditions:
1. For (1.5)
K> 0 and 0
en .7)
M(x, M2(x,
urther, 2)
For the definition of a Banach space cf. Chapter 13.
p)).
(x,
(i.e.
434
XII. ASYMPTOTIC NUMBER OF EIGENVALUES
a2(x,
(1.8)
g (x,
Ii
—q
(x)
IA
I
{
By virtue of the condition (1.5) we have for x —
1
—xl.
Therefore g(x, (
I
qa(x)g(x,
(K2 (
/
Inserting the value of the function g(x,
from (1.2), we obtain
(x)
Since ic = [q(x) +
]1/2,
we have as M —, qa(x)K_2+e
(1.10) where
0
+
=
since by virtue of 0< a l ow,
just as in the derivation of the estimate (1.12), integrating the we obtain to + inequality with respect to x from —
integrating this inequality with respect to obtain the needed estimate for n; lix
from
by
as
—
—p +
436
XII. ASYMPTOTIC NUMBER OF EIGENVALUES
where C is a constant, and r is an arbitrarily large positive number. A similar estimate holds for Then by virtue of the estimates (1.12) and (1.13), we obtain (1.7) the following estimate for NA (x, ,) as + a': INA(x, which proves the lemma.
REMARK. Let us introduce the Banach spaces 1; r is an arbitrary real number), whose elements are ii valued functions A (x, <x, < cx), with the norms define (— respectively by: (p
(1.14)
A (x,
(1.15)
IA (x,
(1.16)
fJP(p) = sup
=
51 A (x,
5i A(x,
5
= sup
A (x,
dx,
12
{
51 A (x,
12
q2't
dii.
it is easily seen that a lemma analogous to Lemma 1.1 holds f: of these Banach spaces. 3. We will again consider the integral equation (1.3), which ca written, using the definition of the operator N (cf. (1.4)), in the I (1.17)
G(x,i7;1i)
Since for sufficiently large i the operator N is a contraction i of the Banach spaces considered above, it follows that if g(x, belongs to one of these spaces, then equation (1.17) can be solved b
method of iteration and its solution belongs to that space. In the following lemmas we wifi indicate sufficient conditions i function g(x, to belong, respectively, to the spaces X, and (r> 0). LEMMA 1.2. If q(x) >0, then the functions (p > 0), i.e.
and
belong to the space (1.18) (1.18')
sup sup
5
{g.(x,
5{fg(x,
< cx,
437
§1. INTEGRAL EQUATION FOR THE GREEN'S FUNCTION
i.e. from (1.2),
PROOF. From the definition of the function g(x, follows
=
2
whence (1.18) follows at once. The proof of (1.18') is analogous. LEMMA 1.3. If the condition g(x, belongs to the space X, i.e.
{ q(x)
}
312dx
0 we have
CX)
CX)
p.))2 dij =
{g (x,
exp (-.--x j x —
2x3
(1.19) follows from the hypothesis of the lemma. LEMMA 1.4. If for x —
1 the function q(x) satisfies
(1 20)
the
condition
C,
Where C is a constant, then g(x,
p.21)
belongs to the space
ie
sup
PRooF. By the definition of g(x,
=
4 I
+
we
exp (—21 x —
have
J {q
(x) + p.]t/2}
I
4 [q
(—21 x —
From the condition (1.20) follows
I [q (x) + p.]V2)
=a(x)+b(x).
438
XII.ASYM?rOTIC NUMBER OF EIGENVALUES
(1.22)
exp(—2xJuf)du I
1
Therefore, proceeding as we did to obtain the bound (1.22), we which together with (1.22) proves (1.21) and thus
1
(—
2
cx,).
Cexp(—21x Therefore C 5 exp (—2ux)
b (x)
du =
C1e2z
cx'). Consequently a(x) +b(x) so b(x) also belongs to cx'), which proves the lemma. (—
§2. The first derivative of the function G(x,
Differentiating the equation (1.3) formally with respect to ij, we obtain
= ft —
:2.1)
5 g (21,
p.) [g
by K(x,
teplacing OG(x,
— q (x)]
in equation
(2.1), we obtain
he integral equation
K(x,
p.)
2
—
g (x,
p.)
[g
— q (x)} K
p.)
Therefore from Lemma 1 1 (cf Lemma 1 2, Og(x, !2)/8,) E he Remark followmg this lemma, and §1 3) it follows that K(x, x the function K(x,i,, It is also not hard to verify that for E contmuous with respect to
XII. ASYMPTOTIC NUMBER OF EIGENVALUES
440
Integrating (2.2) with respect to
from
—
to
we
obtain
K(x, 2
2'
= x the function ôg(x, /ô,i, and consequently K(x, has a discontinuity of the first kind, which does not prevent the cation of the Newton-Leibnitz formula. The existence of the i For
in the right side of equation (2.2'), and also the validity of interchangir
the order of integration, follow from the fact that K(x, i.e. from the fact that
E
sup
Equation (2.2') coincides with the integral equation (1.3). TI it follows from the uniqueness of the solution of (1.3) that
K(x, —0:)
Differentiating this relation with respect to
K(x,
we
obtain for
= aG(x,
Therefore the integral equation (2.2) can be written in the form —-a- = —
g (x,
[q
— q (x)]
g
(2.3) —
g(x,
loG
ogl
Applying estimates similar to those which we used in the proof the basic Lemma 1.1, one can show that the function l(x, . To do this it is sufficient to the improper integral which defines the function l(x, uniformly with respect to i, which follows easily from (1 6)
is contifluous with respect to
con
§3. SECOND DERIVATIVE OF THE GREEN'S FUNCTION
441
Now, solving the equation (2.3) by iteration, it is not hard to show the continuity with respect to of the function ,2)/a?, — ag(x,,,;
Therefore the function OG(x, has the same discontinuities as the function Og(x, The latter has a unique discontinuity for = x. In fact, if
=
x=
if
Therefore g (x,
(2.4)
Thus the function ÔG(x,
where except for the pomt first kind, and (2.5)
= —L
g (x,
—
is continuous with respect to every= x, at which it has a discontmuity of the
-—G(x,
'ri;
*3 The second derivative of the function G(x, i,,
In the preceding section we proved that the first derivative of G(x, i,; as an element of the space for any p> 1 (for the defimtion cf the Remark followmg the proof of Lemma 1 1 m §1), is of x, and satisfies the integral equation continuous with respect to for exists
g (x,
[q
— q (x)J
Fhis equation can be reduced to the form
there a
2?(x,
a
3.2)
l(x,
IJ.):
\
g(X,
OG
442
XIj. ASYMPTOTIC NUMBER OF EIGENVALUES
Differentiating the equation (3.1) formally with respect to ii, we obtain (3
(3
i)
(3.3)
Further, differentiating (3.2), again formally for the moment, respect to and using (2.4) (cf. also the footnote on p. 000), we
Let
us now consider the integral equation M(x,
(3.4)
\
g(x,
belongs to the space We wifi show that It thei follows from Lemma 1.1 that the solution of (3.4) also belongs to Banach space. We have 11(x,
g(x, —
g (x,
q (x)]
[q
—
From the conditions (1.5) and (1.6) it is not hard to deduce and hence to belongs to the space
Let us now consider '2.
Assuming
that the condition (1.20')
satisfied, we obtam
g(x,
exp { — — g(x,
q(x)t
I
+
+ _IL]"41
§3. SECOND DERIVATIVE OF THE GREEN'S FUNCTION
443
is a constant, depending only upon where Finally, using the condition (1.6), we obtain the following bound for 13: 13
g(x, >1
Therefore
('2 +13)q112(n)
The integral on the right can be bounded as in the proof of Lemma 1.1.
As a consequence it is easily proved that 11(x, space
sup We
belongs
(x,
(ii)
0), X(x) = chKx +hK1shKx. We consider the function (618)
I
g+(x,
+ h)' X
is for
(ii
It is not hard to verify that the function = and the conditions tion
satisfies the equa-
—o —
g
we can construct an With the help of the function equation analogous to equation (1.3), and then prove the existence for the operator (6.16) —(6.17). a Green's function To ultimately see this, we only have to see how to estimate the ii). From (6.18) follows tion
(6.19)
+ (1 +
= 4 4
Since
(6.20)
=o(1) as
e
(1
+
e
(1
+0
{1 + 0 V1))
it follows from (6.19) that +0(1)
+0(1)
______
459
*6. ASYMPTOTIC DISTRIBUTION
The representation (6.20) makes it easy to extend the results obtained earlier to the case of the halfline.
We remark only that for the case of the halfline the coefficient in front of the integral in the asymptotic formulas (6.2) and (6.12) has to be replaced by 1/2ir To conclude this section we will indicate various conditions which imply the inequality (6.1). Suppose that q(x) is monotone increasing, has a first and second derivative and is concave. Then q' (x) > 0, q" (x) > 0. Further, suppose that the condition q(x)q"(x) 0, p" (A) 0 is any constant. Therefore
any s> 0 the' integral (7.4)
exists, and to construct the Fourier series of a function we do not to require that it be square integrable: In this section, using the asymptotic formula (6.14), we will several theorems concerning the convergence and sunimabL.
__________
461
§7. UNBOUNDED INCREASING POTENTIAL
expansions and differentiated expansions, in the eigenfunctions of the problem (7.1) + (7.2), of functions which have polynomial growth at infinity.
We wifi first prove several lemmas. LEMMA 7.1. Suppose that the function q(x) satisfies the conditions of Theorem 6.2 and the condition (7.3). Then for ,j we have the estimate
=
(7.5)
where we have put = A,, (the A,, are the eigenualues of the problem (7.1) — (7.2)), and the numbers are defined by (7.4).
PROOF. By virtue of (7.3) we have (7.6)
u(A)
=
=mes{q(x)
=
Put q,(A) = 10X v'd0r(v), where s is some fixed number. Then, using the
notation of the preceding section, and replacing r +p +1 in (6.14) by p, and 2p by s, we can write (6.14) in the
0, k=1,2,...,2n.
We now introduce the function p
(X)
Ls:(x, A)1'
where the Sk (x, A) are the characteristic roots tor which Im Si (x, A) > 0. As is evident, p (A) is an analytic function in the A = + ir plane slit
along the positive semiaxis. Now let r(t) be defined by
o
The following holds. THEOREM 8 1 Suppose that the operator L satisfies the conditions 1—6, and that (for each t) the coefficient q,(x) of the ith derivative in the operator L1 (x, d/dx) is subordinate to Po (x) in the sense that I
q,(x)
> 0 Further, let the function a (t) defined above satisfy the Tauberian condition ta' (t) au(t) (a > 0)
for some
Then for A—* we have the asymptotic formula L where the are the eigenvalues of the
1
N(A) '—j
This theorem can be proved m accordance with the classical scheme of T Carleman first one finds the asymptotic behaviour of the Green's function K(x, A), the kernel of the resolvent of L, for large A 0 there exists a linear combinati such
that
(L2)
A simple computation shows that
x_kek) (13)
=(x,
e
ek)—
i—i
=11z112+ where ck = (x, ek).
The numbers ck are called the Fourier coeffic of x (relative to the system e1, e2, ...). It follows from the identity (1.3) that the difference x
ABSTRACT HILBERT SPACE
471
has minimum norm when the numbers equal the corresponding Fourier coefficients: = Ck. In this case we obtain from (1.2) and (1.3) (1.4)
x
and since
=11 x112
—
—
J
J2
0 such that if t length of every is less than 6, then any corresponding sum S satisf the inequality S — J is called the integral of the function I with to and is denoted by where
II
(4.1)
From the existence of the integral (4.1) there obviously follows existence of the integral (4.2)
f (X)
x
H,
which is defined as the limit of the sums If in the equality
=Sx.
479
§4. SPECTRAL ANALYSIS
we pass, to the limit, we obtain (4.4)
The
f(A)
x).
right side of this equality is an ordinary Stieltjes integral. If
the function f is continuous for all finite A, then one can define the integral / (A)
(45)
b—÷ + By virtue of the Cauchy criterion and formula (4.4), the integral (4.5) exists if and only if the ordinary Stieltjes integral
as the limit of the integral (4.2) as a—+ —
x)
exists. We then have x),
d
/ (A)
which is obtained from (4.4) by passing to the limit.
3. The basic spectral theorem. THEOREM
4i.
For every selfadjoint operator A there exists a unique
having the following properties: spectral function a) a given vector x belongs to DA if and only if lXI2d(EAx,x) b) when this condition is satisfied, then (4.6)
Ax =
d
A
x).
Conversely, every operator defined by a) and b) by means of some spectral
function EA is a selfadjoint operator. For a proof of this theorem see, for example, Ahiezer and Glazman [1]. Theorem 1.1 of Chapter 2 is easily seen to be a special case of Theorem 4.1, for which A)dp(A),
F(X)
(t, A) dt.
From this same theorem it follows that to every point of Weyl's limit circle there corresponds a selfadjoint extension of the Sturm-
480
XIII. LINEAE OPERATORS IN HILBERT SPACE
Liouvile operator (consequently, in the case of a limit point there exists
only one selfadjoint extension. See also the following section of this chapter). 4. Reducibility. Let M be a closed subspace of H, and P the projector
onM DEFINITION 4.2. The subspace M reduces an operator A, if x E D4 lmplles that Px E DA and APx = PAx THEOREM 4.2. A closed subspace M reduces a selfadjoint operator A i and only if the projector PM commutes with the spectral function of for every value of A.
For the proof see Ahiezer and Glazman [1]. 5. Description of the spectrum of a selfadjoint operator by means of I spectral decomposition. Recall that a number A is called a regular = (A — Al)' exists, is defined of an operator A if the operator is the entire space H and is bounded. In this case the operator the resolvent of the operator A. The set of all nonregular points of A is its spectrum. Obviously every eigenvalue A of A belongs to its spectrum
since in this case the operator (A — A I)' does not exist. The set of a' eigenualues of A is called its discrete spectrum. All other points of spectrum, if such exist, are called points of the continuous spectrum, the set of these points is called the continuous spectrum of the operator. A is selfadjoint, then eigenvectors corresponding to distinct eigenva
of A are mutually orthogonal, and therefore A can have only a or countable set of eigenvalues (in a separable space there cannot an uncountable set of pairwise orthogonal vectors). Thus the dis spectrum of a selfadjoint operator (in a separable Hilbert space) is a finite or countable set of real numbers. We wifi show that every nonreal number A is a regular point of
selfadjoint operator A, and that (4.7)
= (A — Al)—' =
where E,, is the spectral function of A (compare with Theorems 3.1
7.1 of Chapter 2). Indeed, the integral in the right side of (4.7) exists (see of this section). Furthermore, for any x E H we have
§5. EXTENSION OF SYMMETRIC OPERATORS
(4.8)
dE S
2
dE
481
I
Hence this integral is a bounded operator, and it is easy to verify that it is equal to RA. 9 1/ Im A. Moreover, It follows from the inequality (4.8) that it follows from (4.7) that (RA) * The following theorem gives a certain characterization of the spectrum
of a selfadjoint operator A. be its THEOREM 4.3. Let A be a selfadjoint operator in H, and let spectral function. Then a) A real number A0 is a regular point of A if and only if the function EA is constant in some neighborhood of this point. — 0; b) A real number Ao is an eigenvalue of A if and only if in this case
to the
is the projector on the eigenspace of A corresponding
— A0.
For a proof see Naimark [iJ, §12.5 or Ahiezer and Glazman [1], §82. §5. The extension of symmetric operators
1. The deficiency subspaces and deficiency indices. One of the basic problems of the abstract theory of symmetric operators is the description of all symmetric extensions of a given symmetric operator A. A particularly important special case of this problem is that of clarifying conditions under which a given symmetric operator has selfadjoint extensions, and descrIbing in this case all such extensions. If B is a symmetric extension of a symmetric operator A, i.e. A ç B,
then B*
A*. But B is a symmetric operator; therefore B
B*
and hence (5.1)
ACBCB*CA*.
It follows from these inclusion relations that every symmetric extension of A is a restriction of the operator A*. DEFINITION 5.1. A symmetric operator A is said to be maximal if it has no symmetric extensions. It follows from this definition that a selfadjoint operator A is maximal.
In fact, in this case A* =A and (5.1) shows that B =A, i.e. every symmetric extension B of A coincides with A.
482
XHL LINEAR OPERATORS IN HILBERT SPACE
Let us denote by the subspace consistmg of all elements of the form y = (A + ii) x, x E DA, where i = and by the subspace of all elements of the form (A — ii) x, x E DA LEMMA 5.1.
PRooF. Simple and therefore
and
are
calculations
(5.2)
II
closed subspaces, if A is closed. yield
(A ± ii) x
II
if
(A ± ii) x 112 = Ax 112 +11 x 112,
x if.
and that
Let us now assume that y,, = (A ±
Then
and consequently by (5.2) 0. Since H is — Xm —* x0 E H. Thus x0, complete, it follows that E DA, A is a closed operator, we have x0 E DA and Ax0 ix0. It follows that as was to be proved. We denote by the orthogonal complement of II
yn — yrn if
0,
if
if
DEFINITION 5.2. The are called the deficiency subspaces of the operator A Their dimensions are called the deficiency indices of A REMARK 1. Let A denote an arbitrary nonreal number. Similarly to the foregoing one can introduce the two subspaces and prove
that they are closed, define the subspaces N,, and and define the deficiency mdices as the dimensions of the subspaces N,, and The subspaces N,, and depend upon A, but one can prove that theff dimensions m each of the halfspaces Im A >0 and Im A 0 one can find a step function
(x)
such that
(1.3)
It follows from the foregoing that for fixed so large that for H > we have
we can choose Ao =
(1.4)
It therefore follows from (1.3) and (1.4) that for xj > I
1(x) sin Ax dx
[1(x) —
(x)} sin Ax dx +
A0
(x) sin Ax dx
whence the lemma follows by the arbitrariness of REMARK
1. One can similarly prove that Jim a
The same remark holds for the subsequent lemmas of this We will therefore state them only for the first of the integrals (1 1YI
LEMMA 1 2 Let (a, b) be an infinite (on one or both sides) interval an4J let f(x) E (a, b). Then (1.2) holds.
PROOF. For definiteness let us consider an interval of the for! >0 one can find A
(a, + co), where a is finite. For arbitrary
such that (1.5)
By Lemma 1.1, for fixed A we can choose A0
so large that
(16)
for lxi
>x0 It follows from (15) and (16) that for lxi
>x0 one
§1. RIEMANN-LEBESGUE LEMMAS
497
sin
f(x)sin
1(x) dx
from which the lemma follows. The following lemma gives us information on the behaviour of integrals
of the form (1.1) when the function f(x) depends upon a parameter (or parameters). LEMMA 1.3. Suppose that we have an infinite set E = f(x) of functions which satisfy the following conditions: 10. 1(x) where the interval (a,b) can be finite as well as }
infinite.
2°. The set E is relatively compact in the metric space (a, b). 0 one can find Then the relation (1.2) holds uniformly, i.e. for any such that for xl > A0 the inequality = (1.7)
holds for all f(x) E E. PROOF. Let E > 0. From a well-known criterion for the relative compactness of a set in a metric space it follows that one can find a
finite E-set for E, i.e. there exists a finite set of functions E, = }
fk(X)
c E such that for each function f(x) E E there exists a function
E E, such that
(1.8)
For thed we can obviously find A0 =
>0 such that the inequality
sin Axdx
(1.9)
holds for xi > A0 and for all k = 1, . .. n. The lemma follows from (1.8) and (1.9). REMARK 2. If the interval (a, b) is finite and the functions f(x) of the family E are continuous, then for the relative compactness of E in I
498
XIV.THEOREMS OF ANALYSIS
the metric of .1' (a, b) it is sufficient that E be relatively compact in the metric of the space C(a, b), and for this it is sufficient that E be uniformly bounded and equicontmuous 2) If the interval (a, b) is mfinite, then for the relative compactness of E it is sufficient to require that for any > 0 there exist a fimte interval
(a', b') C (a b) such that
for every function f(x) E E, and that the restrictions of the functions in E to the interval (a', b') form a relatively compact family in t (a', b'). In fact, denoting the set of all restrictions mention c
metric of
above by E', it is clear that if the restrictions to (a', b') of f1(x), . . .,f,(x) E E form an -set in E', then these functions will form a 3-set in E; hence E is relatively compact since is arbitrary
A case of interest is that in which the family E is generated by suminable function ={f(x±t)}, xE(a,b), (a,b) finite interval and t belongs to some closed bounded set M on the Let tx+t}E(A,B) and 1(x) E9'(A,B). Then the set E is compac
In fact, let
be an infinite sequence of numbers belonging to J set M If we choose a subsequence { of which converges to number 4, then }
jf(x+
—/(x+ t0)l dx
b+to
a+t,
B
A
§2. Helly's theorems
1. Letuj (X), u2(X),... be an infinite sequence of nondecreasing bouni finite: closed interval [a,b]. FOr defhi functioiis defined on
let us assume that all the functions of this sequence are from the left: —0) = FIRST THEOREM. If the functions
are uniformly
then we eon find a monotone function u(X) and a subsequence which converges to cr(X) at every point of continuity of a(X) 2)
a,,
This is the so called Arzelà Ascoh lemma see for example Ljustermk and Sobol
§2. HELLY'S THEOREMS
499
PROOF. Let Ai, A2,... be a countable everywhere dense subset of the interval [a, b]. By means of the diagonal procedure (see, for example, Smirnov [1], §12) one can find a subsequence which conr,,,(A,) = r,. verges at each of the points Ai, A2, .... Put If 0 the Stieltjes inversion formula assumes the form 1'
(b) — o
(a) = urn — — Im
(z)) da.
a
*4. Tauberian theorems
1. Levitan's Tauberian theorems for Fourier integrals. THEOREM 4.1. Let the function o-(v) iatisfy the following conditions: is of bounded variation in every finite interval. a) (4.1')
c) For every function which has a bounded derivative of order r + 2 and which vanishes outside the interval (— A, A), h (v) da (v) = 0,
Then for
—*
where it (v)
= 4—
(t)
have the estimate
we IL
—
)8 da (v)
(4.1)
= o (pr_a)
(s
0).
PROOF. Let us consider the integral 1
(4.2)
s)
— v2)s
(v).
The proof wifi consist in giving another representation for this integral,
from which the estimate (4.1) follows easily. We put v,
(s>0),
f
s)
By the Fourier mversion formula we have
s)
§4. TAUBERIAN THEOREMS
p
v,
h (t,
s)
505
s)
Inserting this expression in (4.2), we obtain (4.2')
1
s)
=
Ii (t,
da (v)
Let n denote the integer part of s: n = (4.3) by parts n times, we obtain h (t,
(4.4)
s)
[s].
(ii,
s)
Integrating the integral in
v, s)
We will consider the case of integer and noninteger s separately. Suppose
first that s is an integer: s =n, with n ? 1 (the case s =0 will be considered separately). Integrating the right side of (4.4) by parts once more, we obtain h (t,
s)
=
—
v, s)
v,
s)
(4.5) p(n+1)
v, s)
dv.
Let #(t) be a function which has a bounded derivative of order n + 2, and which equals 1 for
t
and zero outside the interval (—1, 1).
Put #A(t) =#(t/A). The function equals zero outside the interval (— A, A). Therefore condition c) of the theorem implies that da (v)
(t) h (t,
s)
o.
Subtracting this relation from (4.2), we obtain
I
s)
=
If we replace here J(M,s) where
(v)
—
(t)] h (t,
s)
by the expression (4.5), we obtain
XIV. THEOREMS OF ANALYSIS
506
I—
a)
'1,
=
2t
$
do
(t)
(v) { .1
Co
=
a)
(EL,
{I
$ do (v)
2it
.-a)
a)
— (_j)fl+1
do (v)
—
f
I
—a)
(t)
S
p(fl+l)
(t) etdt1,
—
J
s)
dt
L
We now put Co
ç I
(4.6)
2ic
J
(t) e
(tt)
—a)
It is easily seéñ that this integral converges absolutely for s> 1, and converges for s =1. Moreover, integrating by parts, it is not hard to show that for v one has the estimate cx
A9(v) =
(4.7)
Interchanging the order of integration in J (,z, s), which is permissible by virtue of the convergence of the integral (4.6), we obtain (4.8)
J(3) (p., s) =
(v)
s)
(p.,
(v +
Further, using the notation (4.6), we obtain a)
(4.9)
s)=
p(8) (EL, v, a)
d,o (v + EL),
$ Aa
2it
—a) a)
(4.10)
(#)
v, a) L=_L
J(2) (p., s) —
A
(v)
a (v — p.).
—a)
It is easily seen that as z — (4.11)
is,
s) I
=
Let us now estimate the integrals (4.9) and (4.10). We will consider (4.9); the estimation of (4.10) is similar. Let N be a positive integer and put
507
§4. TAUBERIAN THEOREMS
Given an arbitrary
> 0, for
>
and
N we will have
'I
j+1
V Therefore (4.12)
Further, (4.13) and a similar estimate holds -for i3:
(4.13')
If we first choose N, and then and (4.13') we obtain for M >
then from the estimates (4.12), (4.13)
(4.14) It
follows from (4.11) that for
L
(4.15)
J J
for
s)
for large
As before, let N be a positive
integer. We put
Just as before, (4.17)
Further, s)
12+ 13)
508
XIV. THEOREMS OF ANALYSIS
It follows from (4.17) that (4.18)
Iifl +
E,
s) (j
jr
+ 1)dE
N±i
since
Let us now consider s)
Let I
v
We have A, (v)
(v
+
be an arbitrary positive number and choose a so large that for I > a one has
