0387986995Basic TheoryC

138	V. SINGULARITIES OF THE FIRST KIND

Comment. The series F(a, Q, ry, x) is called the hypergeometric series (see, for example, [CL; p. 1351, 101; p. 159], and [IKSYJ).

V-3. For each of the following differential equations, find all formal solutions of

c

the form x'' I 1 + > cmxm] . Examine also if they are convergent.

L	m=1
(i)	xb2y + aby + /3y = 0,
(ii)	b2y + aby + $y = rxmy,

where b = xa, the quantities a, 3, and 7 are nonzero complex constants and m is a positive integer.

V-4. Given the system

(E)	dt = (N + R(t))il,	N = [0 O] , and			R(t) = t-3 I 0 0J ,
show that
(i) (E) has two linearly independent solutions
		,	where	(t)	om!(m+1)!'
	and
	J2(t) = 1612 (t)1 ,	where	02(t) = t +		+00 bmt-'n - ¢1(t)logt,
	(t)				m-1
	with the constants bm determined by b_1 = 1, bo = 0, and
	m(m + 1)bm - b,,,-,	=	2m + 1	(m = 1,2,... ),
			m!(m+ 1)!

(ii) 1 lim t-1(Y(t)-e IN) = O for the fundamental matrix solution Y(t) = [if1(t)12(t)J,

(iii) the limit of e-'NY(t) as t -+ +oo does not exists.

V-5. Let y be a column vector with n entries and let Ax, y-) be a vector with n entries which are formal power series in n + 1 variables (x, yj with coefficients in

C. Also, let u' be a vector with n entries and let P(x, u) be a vector with n entries which are formal power series in n + 1 variables (x, u") with coefficients in C. Find the most general P(x, u") such that the transformation u + xP(x, u7 changes

dy" du"

the differential equation ds = f (X, y-) to = 0.

Hint. Expand xP(x, u) and f (x, u+xP(x, u)) as power series in V. Identify coefficients of d[xP(x, V-)] with those of Ax, u'+zP(x, iX)) to derive differential equations

which are satisfied by coefficients of xP(x, u).

EXERCISES V

139

V-6. Suppose that three n x n matrices A, B, and P(x) satisfy the following conditions:

(a)the entries of A and B are constants,

(b)the entries of P(x) are analytic and single-valued in 0 < [xj < r for some positive number r,

dii

(c) the transformation y' = P(x)u changes the system xfy = Ay" to xjj = Bu.

Show that there exists an integer p such that the entries of xPP(x) are polynomials

in x.

Hint. The three matrices P(x), A, and B satisfy the equation xd) = AP(x) -

+"0

P(x)B. Setting P(x) = >2 xmPm, we must have mPm = APm - PmB for all

m=-oo

integers m. Hence, there exists a large positive integer p such that Pm = 0 for

ImI ? P.
V-7. Let ff be a column vector with n entries {y,... ,	and let A(ye) be an n x n

matrix whose entries are convergent power series in {yj, ... , yn} with coefficients in C. Assume that A(0) has an eigenvalue A such that mA is not an eigenvalue of

A(0 for any positive integer m. Show that there exists a nontrivial vector O(x)

with n entries in C{x} such that y" = b(exp[At]) satisfies the system L = A(y-)y.

Hint. Calculate the derivative of (exp(At]) to derive the system Axe =

for .

N

V-8. Consider a nonzero differential operator P = >2 ak(x)Dk with coefficients

k=O

ak(x) E C([x]], where D = dx. Regarding C([x]] as a vector space over C, de-

fine a homomorphism P : C[[x]] -. C[[x]j. Show that C((xjj/P(C[[x]j] is a finite- dimensional vector space over C.

Hint. Show that the equation P(y) = x"`'4(x) has a solution in C([xjj for any

O(x) E C[[x]] if a positive integer N is sufficiently large. To do this, use the indicial polynomial of P.

N

V-9. Consider a nonzero differential operator P = >2 ak(x)dk with coefficients

		k=0
ak(x) E CI[x]j, where b = x2j, n > 1, and		54 0. Define the indicial poly-
nomial	as in Theorem V-1-3. Show that if the degree of		is n and

if n zeros {A1, ... , A, } of do not differ by integers, we can factor P in the following form:

P = a+,(x)(b - &1(x))(b -

42(x))...(6 - 0n(x)),

140	V. SINGULARITIES OF THE FIRST KIND

where all the functions ¢, (x) (j = 1, 2, ... , n) are convergent power series in x and

4f(O) = Aj.

Hint. Without any loss of generality, we can assume that an(x) = 1. Then, An + n-1

Eak(O)Ak = (A - A1)... (A - An). Define constants {yo... , In-2} by A' ' +

k=0 n-2

,YkAk = (A - A2) ... (A - An). For v] (X) E xC[[xI] and ch(x) E xC([x]l (h =

k=O

0,. . . , n - 2), solve the equation

( n-2

(C)P = (6 - Al -

2(7h+Ch(x))6h

h=0

If we eliminate O(x) by Al + V(x) = 1`n-2 + ct-2(x) - an-1(x), condition (C) becomes the differential equations

6(CO(x)1 = a0(x) + (-Yn-2 + Cn-2(x) -a. -1(x))(1'0 +40(x)),

1 b[ch(x)] = ah(x) + (1'n-2 + Cn-2(x) - an-1(x))(7h +Ch(x)) - (1h-1 + Ch-1(x)),

where h=I,...,n-2.

n

V-10. Consider a linear differential operator P = E atbt, where ao,... , an are t=o

complex numbers and 6 = x. The differential equation

(P)

P[y1= 0

is called the Cauchy-Euler differential equation. Find a fundamental set of solutio of equation (P).

Hint. If we set t = log x, then b = dt

V-11. Find a fundamental set of solutions of the differential equation

(6-06-8)(6-a-8)[yl =xn'Y'

where 6 = xjj and m is a positive integer, whereas a and /3 are complex numbers

such that they are not integers and R[al < 0 < 8t[01.

V-12. Find the fundamental set of solutions of the differential equation

(LGE)

f(1-x2)LJ +a(a+1)y=0

at x = 0, where a is a complex parameter.

EXERCISES V

141

Remark. Differential equation (LGE) is called the Legendre equation (ef. [AS, pp. 331-338] or [01, pp. 161-189]).

V-13. Show that for a non-negative integer n, the polynomial Pn(x) = 1

2nn!

x do [(x2 - 1)'] satisfies the Legendre equation

[(1-x2) ] +n(n+1)Pn=0.

d

Show also that these polynomials satisfy the following conditions:

(1) Pn(-x) = (-1)nPn(x), (2) Pn(1) = 1, (3) JPn(t)j ? 1 for {xj > 1,

+00

E Pn(x)tn, and (5) lPn(x)I < 1 for Jxj < 1.

(4) 1 - 2xt + t = n=o

Hint. Set g(x) = (1 - x2)n. Then, (1 - x2)g'(x) + 2nxg(x) = 0. Differentiate this relation (n + 1) times with respect to x to obtain

(1,-	x2)g(n+2)(x) - 2(n + 1)xg(n+1)(x) - n(n + 1)g(n)(x)
	+ 2nxg(n+1) (x) + 2n(n + 1)g(")(x) = 0

or

(1 -X 2)g(n+2)(x) - 2ng(n+1)(x) + n(n + 1)g(n)(x) = 0.

Statements (1), (2), (3), and (4) can be proved with straight forward calculations. Statement (5) also can be proved similarly by using (4) (cf. [WhW, Chapter XV,

Example 2 on p. 303]). However, the following proof is shorter. To begin with, set

F(x) = Pn(x)2	-x2)((1 -	x2)P''(x))2.
n(n + 1){1

Then, F(±1) = Pn(±1)2 = 1, p(X) _ (P"(x))2 ,and F(x) = Pn(x)2 if (x) n(n + 1)

0. Therefore, for 0 < x < 1, local maximal values of Pn(x)2 are less than F(1) = 1, whereas, for -1 < x < 0, local maximal values of Pn(x)2 are less than F(-1) = 1.

Hence, Pn(x)I < 1 for JxI < 1 (cf. [Sz, §§7.2-7.3, pp. 161-172; in particular,

Theorems 7.2 and 7.3, pp. 161-162]. See also (NU, pp. 45-46].) The polynomials Pn(x) are called the Legendne polynomials.

V-14. Find a fundamental set of solutions of (LGE) at x = oo. In particular, show what happends when a is a non-negative integer.

V-15. Show that the differential equation b"y = xy has a fundamental set of solutions consisting of n solutions of the following form:

(logx)k-1-A

OA(x) (k = 1, ... , n),

A-o (k - 1 - h)!

where the functions Oh(x) (h = 0,... , n -1) are entire in x, 00(0) = 1, and Oh(O) _

0 (h = 1, ... , n -1). Also, find O0 (x).

142	V. SINGULARITIES OF THE FIRST KIND

V-16. Find the order of singularity at x = 0 of each of the following three equa- tions.

(i)x5{66y - 362y + 4y} = y,

(ii)x5{56y - 3b2y + 4y} + x7{b3y - 55y} = y,

(iii)x5y"' + 5x2y" + dy + 20y = 0.

V-17. Find a fundamental matrix solution of the system

x2 dbidx = xyi + y2,

dy2

x2 dx = 2xy2 + 2y3,

x2 dx3 = x3y! + 3y3.

V-18. Let A(x) be an n x n matrix whose entries are holomorphic at x = 0. Also, let A be an n x n diagonal matrix whose entries are non-negative integers. Show that for every sufficiently large positive integer N and every C"-valued function fi(x) whose entries are polynomials in x such that those of xA¢(x) are of degree N, there exists a C"-valued function f (x; A, ¢, N) such that

(a)the entries off are polynomials in x of degree N-1 with coefficients depending on A, N, and ¢,

(b)f is linear and homogeneous in tb,

(c)the linear system

	xAdd9 = A(x)3l +
	z
	has a solution #(x) whose entries are holomorphic at x = 0 and il(x) is linear
	and homogeneous in ',
(d)	fi(x)) = O(xN+i) as x 0.

Hint. See [HSW].

V-19. Let A(x) and A be the same as in Exercise V-18. Assuming that n >

trace(A), show that the system xA dz = A(x)y has at least n - trace(A) linearly independent solutions holomorphic at x = 0.

Hint. This result is due to F. Lettenmeyer [Let]. To solve Exercise V-19, calculate f (x; A, ¢, N) of Exercise V-18 and solve f(x; A, 4, N) = 0 to determine a suitable function 0.

00

V-20. Suppose that for a formal power series ¢(x) _ > cx' E C[]x]], there ex-

m=0
/dl	d
ist two nonzero differential operators P = E ak (x) ` I	k and Q = > bk (x) (d) k
k=0	rk=10

with coefficients ak(x) and bk(x) in C{x} such that P[0] = 0 and Q I = 0. Show

J

that 0 is convergent.

l

EXERCISES V

143

Hint. The main ideas are

(a) derive two algebraic (nonlinear) ordinary differential equations

(E)

F(x,v,v',... ,v(")) = 0

and

G(x,v,v',... ,v(' ) = 0

for v = from the given equations P[¢) = 0 and Q ['01 = 0.

(b)eliminate all derivatives of v from (E) to derive a nontrivial purely algebraic equation H(x, v) = 0 on v. See [HaS1] and [HaS2[ for details.

CHAPTER VI

BOUNDARY-VALUE PROBLEMS OF LINEAR

DIFFERENTIAL EQUATIONS OF THE SECOND-ORDER

In this chapter, we explain (1) oscillation of solutions of a homogeneous secondorder linear differential equation (§VI-1), (2) the Sturm-Liouville problems (§§VI-

2-VI-4, topics including Green's functions, self-adjointness, distribution of eigenvalues, and eigenfunction expansion), (3) scattering problems (§§VI-5-VI-9, mostly focusing on reflectionless potentials), and (4) periodic potentials (§VI-10). The materials concerning these topics are also found in [CL, Chapters 7, 8, and 11],

[Hart Chapter XI], [Copl, Chapter 1], [Be12, Chapter 61, and [TD]. Singular selfadjoint boundary-value problems (in particular continuous spectrum, limit-point and limit-circle cases) are not explained in this book. For these topics, see [CL,

Chapter 9].

VI-1. Zeros of solutions

It is known that real-valued solutions of the differential equation L2 + y = 0 are

linear combinations of sin x and cos x. These solutions have infinitely many zeros on the real line P. It is also known that real-valued solutions of the differential equation

2 - y = 0 are linear combination of e= and a-x. Therefore, nontrivial solutions

has at most one zero on the real line R. Furthermore, solutions of the differential

equation2 + 2y = 0 have more zeros than solutions of L2 + y = 0. In this

section, keeping these examples in mind, we explain the basic results concerning zeros of solutions of the second-order homogeneous linear differential equations. In §§VI-1-VI-4, every quantity is supposed to take real values only.

We start with the most well-known comparison theorem concerning a homogeneous second-order linear differential equation

(VI.1.1)	+ 9(x)y = 0.
	dx2

Theorem VI-1-1. Suppose that

(i)g, (x) and g2(x) are continuous and g2(x) > 91(x) on an interval a < x < b,

(ii)d 2 01 (x) + 91(x)o1(x) = 0 and d2622x) + g2(x)02(x) = 0 on a < x < b,

(iii)S1 and 6 are successive zeros of 01(x) on a < x < b.

Then. q2(x) must vanish at some point £3 between t:l and C2.

144

1. ZEROS OF SOLUTIONS

145

Proof.

Assume without any loss of generality that Sl < 1;2 and 01(x) > 0 on 1;1 < x < £2.

Notice that 41(1::1) = 0, 1 (l:l) > 0, 01(S2) = 0, and (1:2) < 0. A contradiction will be derived from the assumption that 02(x) > 0 on S1 < z < 1;2. In fact, assumption (ii) implies

VI.1. 2	02(x)	d 20, (7)	_ 01 (x) d202 (x)	=	(x) - 91 (x)101 (x)02(x)
				[92
and, hence,
(VI- 1 - 3)	0202) 1 L ( 6) - 02 V 1 )!!LV1 ) =			rE2	(x) - 91 (x)j¢ 1 (x)02 (x)dx.
				E1

The left-hand side of (VI.1.3) is nonpositive, but the right-hand side of (VI.1.3) is positive. This is a contradiction. 0

A similar argument yields the following theorem.

Theorem VI-1-2. Suppose that

(i) g(x) is continuous on an interval a < x < b,

(ii)0i(x) and 02(x) are two linearly independent solutions of (VI.1.1),

(iii)1;1 and 6 are successive zeros of 01 (x) on a < x < b.

Then, 02(x) must vanish at some point 3 between t;1 and 1;2.

Proof.

Assume without any loss of generality that Sl < £2 and .01 (x) > 0 on S1 < x < 1;2.

Notice that 01(11) = 0, 1 (S1) > 0, 01(12) = 0, and X1(1;2) < 0. Note also that if 02(1) = 0 or 02(6) = 0, then 01 and 02 are linearly dependent. Now, a

contradiction will be derived from the assumption that 02(x) > 0 on fi < x < 2
In fact, assumption (u) implies 02(x) a- (x) - p1(z) d2 02(x) = 0 and, hence,
	dS2	dx2
(VI.1.4)	02(2) 1(6) - 4'201)	1(W =0.

Since the left-hand side of (VI.1.4) is positive, this is a contradiction. 0

The following result is a simple consequence of Theorem VI.1.2.

Corollary VI-1-3. Let g(x) be a real-valued and continuous function on the interval Zo = {x : 0 < x < +oo}. Then,

(a)if a nontrivial solution of the differential equation (VI.1.1) has infinitely many zeros on I , then every solution of (VI.1.1) has an infinitely many zeros on

Zo,

(b)if a solution of (VI.1.1) has m zeros on an open subinterval Z = {x : a < x < 3} of Zo, then every nontrivial solution of (VI.1.1) has at most m + 1 zeros on Z.

Denote by W(x) = W(x;01,02) =

I the Wronskian of the set of

02(x)

functions (01(x), 02(x)}. For further discussion, we need the following lemma.

146	VI. BOUNDARY-VALUE PROBLEMS

Lemma VI-1-4. Let g(x) be a real-valued and continuous function on the interval To = {x : 0 < x < +oo}, and let 771(x) and M(x) be two solutions of differential equation (VI 1.1). Then,

(a) W (x; 772 , ri2 ), the Wronskian of {rli (x), 712(x)}, is independent of x,

(b) if we set t;(x) = 771(x), then		4(x) =	c , where c is a constant.
	772(x)	dx	772(x)2
Also, if 77(x) is a nontrivial solution of (VI.1.1) and if we set w(x) =				then
we obtain
(VI.1.5)	dw(x)	+ w(x)2 + g(x) = 0.

Proof.

(a) It can be easily shown that

d	771(x)	772(x)	=	771(x)		772(x)		-g(x) !	712(x)	= 0
dx I77i(x)						772(x)
		772(x)		77i'(x)				771(x) 712(x)
(b) Note that dl;(x} = -				1	I rl' (x)		712(x) !	and that	dw(x)	if'(x)
	dx		712(x)2		771(x)		772(x)1		dx	:l(x)

112 = -g(x) - w(x)2.

((xx))

The following theorem shows the structure of solutions in the case when every nontrivial solution of differential equation (VI.1.1) has only a finite number of zeros

on To= {x:0<x<+oo}.

Theorem VI-1-5. Let g(x) be a real-valued and continuous function on the inter- val I. Assume that every nontrivial solution of differential equation (VI.1.1) has only a finite number of zeros on T. Then, (VI.1.1) has two linearly independent solutions 777(x) and 712(x) such that

(a) lim 771(x)	= 0,
x-+oo 712(x)
(b)
	+00	dx		+00	dx
(VI.1.6)		771( x)2 = +oo	and	1	7r (x) 2 < +00,
	1.0			Jxo

where xo is a sufficiently large positive number,

(c) the solution 712(x) is unbounded on To.

Proof.

(a) Let (1(x) and (2(x) be two linearly independent solutions of (VI.1.1) such that

(, (x) > 0 (j = 1, 2) for x > xo > 0. Using (b) of Lemma VI-1-4, we can derive the

	1. ZEROS OF SOLUTIONS
	(1(x)	= 0, (ii) =U
following three possibilities (1) =limo(2(x)
Urn ( 1( x )>0. Set
x-+oo (2 (x)
(i(x)	in case (i),
711(x) _ (2(x)	in case (ii),	772(x) =

(i(x) - 7'(2(x) in case (iii),

Then, (a) follows immediately.

(b) Conclusion (b) of Lemma VI-1-4 implies that

147

+oo, and (iii)

(2(x)	in case (i),
(i(x)	in cage (ii),
1(2(x)	in case (iii).

1 _Id(Y12(x)	1	_-1d >)1(x)
and		c dx \ 2(x)	,
ql (x)2 ^ c dx . q1(x))	712(x)2

where c is the Wronskian of q1 and rte. Hence, (VI.1.6) follows.

+oc

(c) If q2 is bounded, we must have j()2-= +00. 0

The following theorem gives a simple sufficient condition that solutions of (V1.1.1) have infinitely many zeros on the interval Zo.

Theorem VI-1-6. Let g(x) be a real-valued and continuous function on the in-

terval lo. If f g(x)dx = +oo, then every solution of the differential equation

0

(VI.1.1) has infinitely many zeros on Zo.

Proof.

Suppose that a solution 17(x) satisfies the condition that 77(x) > 0 for x >

xo > 0. Set w(x) = !L(x). Then, w(x) satisfies differential equation (VI.1.5).

77(x)

Hence, lien w(x) = -oo. This implies that :-+co

lim 17(x) = 0 since 17(x) = r7(xo)

:-+oo

x exp Vo

This contradicts (c) of Theorem VI-1-5. O

J

The converse of Theorem VI-1-6 is not true, as shown by the following example.

Example VI-1-7. Solutions of the differential equation

(VI.1.7)2 + 1 A = 0,

z2

where A is a constant, have infinitely many zeros on the interval -oo < x < +oo if

and only if A > 4

Proof

Look at (VI.1.7) near x = oo. To do this, set t = x to change (VI.1.7) to

(VL1.8)	r	462 + 26 + 1 + t,
	L		0,