Sturm-Liouville Theory and its Applications (Instructor Solution Manual, Solutions)

69 Pages·2007·0.422 MB·English

by  M. A. Al-Gwaiz

SOLUTIONS TO THE EXERCISES
FOR
STURM-LIOUVILLE THEORY
AND ITS APPLICATIONS

M.A. AL-GWAIZ
1
Chapter 1
1.1 (a)0 x=0 x+[0 x+( 0x)]=(0+0) x+( 0 x)=0 x+( 0:x)=0:
(cid:1) (cid:1) (cid:1) (cid:0) (cid:1) (cid:1) (cid:0) (cid:1) (cid:1) (cid:0)
(b)a 0=a 0+[a 0+( a 0)]=a (0+0)+( a 0)=a 0+( a 0)=0:
(cid:1) (cid:1) (cid:1) (cid:0) (cid:1) (cid:1) (cid:0) (cid:1) (cid:1) (cid:0) (cid:1)
(c) ( 1) x+x=( 1+1) x=0 x=0:
(cid:0) (cid:1) (cid:0) (cid:1) (cid:1)
(d) If a=0 then a x=0 implies x=a 1 (a x)=a 1 0=0:
(cid:0) (cid:0)
6 (cid:1) (cid:1) (cid:1) (cid:1)
1.2 (a) Complex vector space, (b) real vector space, (c) not a vector space, (d)
real vector space.
1.3 If x ;:::;x are linearly dependent then there are scalars b ;:::;b ; not all
1 n 1 n
zeros, such that n b x = 0: Assuming b = 0 for some k 1;:::;n ,
i=1 i i k 6 2 f g
wecanmultiplybPyb(cid:0)k1 toobtainxk = i=kaixi,whereai =(cid:0)b(cid:0)k1bi:The
converse is obvious. If the set of vectors 6x ;x ;x ;::: is in…nite , then it
P 1 2 3
is linearly dependent if, and only if, it has a …nite subset which is linearly
dependent, and the desired conclusion follows.
1.4 Assume that x ; ;x and y ; ;y are bases of the same vector
1 n 1 m
f (cid:1)(cid:1)(cid:1) g f (cid:1)(cid:1)(cid:1) g
space with n = m and show that this leads to a contradiction. If m > n;
6
express each y , 0 i n, as a linear combination of x ; ;x : The
i 1 n
(cid:20) (cid:20) (cid:1)(cid:1)(cid:1)
resulting system of n linear equations can be solved uniquely for each x ,
i
0 i n; as a linear combination of y , 0 i n (why?). Since each
i
(cid:20) (cid:20) (cid:20) (cid:20)
vector y ;:::;y is also a linear combination of x ; ;x (and hence
n+1 m 1 n
(cid:1)(cid:1)(cid:1)
of y ; ;y ), this contradicts the linear independence of y ; ;y :
1 n 1 m
(cid:1)(cid:1)(cid:1) f (cid:1)(cid:1)(cid:1) g
Similarly, If m<n then x ;:::;x is linearly dependent. Hence m=n:
1 n
f g
1.5 Assume a xn + ::: + a x + a = 0 for all x in the interval I. We can
n 1 0
di¤erentiate both sides of this identity n times to conclude that a = 0,
n
then n 1 times to obtain a = 0, etc. Therefore all the coe¢cients a
n 1 k
are zero(cid:0)s, and so 1;x;:::;xn :(cid:0)x I is linearly independent for every n:
f 2 g
It follows that the in…nite set 1;x;::::x I is linearly independent.
f 2 g
1.6 It su¢ces to consider the case where both dimX and dimY are …nite.
If is a basis of Y, then lies in X: Since the vectors in are linearly
B B B
independent,dimX cannotbelessthanthenumberofvectorsin ,namely
B
dimY:
1.7 Recallthatadeterminantiszeroif,andonlyif,oneofitsrows(orcolumns)
is a linear combination of the other rows (or columns).
1.8 Since x+y 2 = x 2+2Re x;y + y 2, the equality holds if, and only
k k k k h i k k
if, Re x;y =0: Consider x=(1;1) and y=(i;i) in C2.
h i
2
1.9 (a) Let a(x+y)+b(x y) = (a+b)x+(a b)y=0: Because x and y
(cid:0) (cid:0)
are linearly independent, it follows that a+b=0 and a b=0. But this
(cid:0)
implies a=b=0, hence the linear independence of x+y and x y:
(cid:0)
(b) x+y;x y = x 2+ y;x x;y y 2 = x 2 y 2.Therefore
h (cid:0) i k k h i(cid:0)h i(cid:0)k k k k (cid:0)k k
x+y and x y are orthogonal if x = y .
(cid:0) k k k k
1.10 (a) 0, (b) 2=3, (c) 8=3, (d) p14:
1.11 ' ;' = ' ;' = ' ;' = ' ;' =0: Thus the largest orthogonal
h 1 3i h 1 4i h 2 4i h 3 4i
subset is ' ;' ;' :
f 1 3 4g
1.12 f;f = f = (cid:25)=2; f;f = f =0; f;f = f =p(cid:25)=2:
1 1 2 2 3 3
h i k k h i k k h i k k
1.13 Let a+bx+cx2p=0 for all x [ 1;1]: Setting x=0; x=1; and x= 1;
2 (cid:0) (cid:0)
the resulting three equations yield the only solution a = b = c = 0: The
corresponding orthogonal functions are given by
f (x)=1;
1
x;1
f (x)=x h i =x;
2 (cid:0) 1 2
k k
x2;1 x2;x 1
f (x)=x2 h i h ix=x2 :
3 (cid:0) 1 2 (cid:0) x 2 (cid:0) 3
k k k k
1.14 Let a+bx+c x =0 for all x [ 1;1]: Setting x=0; x=1; and x= 1
j j 2 (cid:0) (cid:0)
yields a=b=c=0: The corresponding orthogonal set is
f (x)=1;
1
x;1
f (x)=x h i =x
2 (cid:0) 1 2
k k
x ;1 x ;x 1
f (x)= x hj j i hj j i = x ;
3 j j(cid:0) 1 2 (cid:0) x 2 j j(cid:0) 2
k k k k
and the normalized set is
f (x) 1
1
= ;
f1 p2
k k
f (x) x
2
= ;
f2 2=3
k k
f (x) 1
3 = p (x 1=2):
f3 p6 j j(cid:0)
k k
The set 1;x; x is not linearly independent on [0;1] because x = x on
f j jg j j
[0;1]:
3
1.15 From the result of Exercise 1.3 we know that f ;:::;f are linearly de-
1 n
pendent if, and only if, there is a number k 1;:::;n such that
2 f g
f = a f on I. By di¤erentiating this identity up to order n 1,
k i=k i i (cid:0)
we arrive6 at the system of equations f(j) = a f(j); 0 j n 1.
P k i=k i i (cid:20) (cid:20) (cid:0)
Writing this system in matrix form, and using t6he properties of determi-
P
nants, we conclude that the system is equivalent to the single equation
det(f(j))=0 on I, where 1 i n and 0 j n 1:
i (cid:20) (cid:20) (cid:20) (cid:20) (cid:0)
1.16 Noting that both ' and ' are even whereas ' is odd, we conclude
1 2 3
that ' ;' = ' ;' = 0: Moreover, ' ;' = 0: The corresponding
h 1 3i h 2 3i h 1 2i
orthonormal set is
' (x) 1
1 = ;
' p2
k 1k
' (x) 45 1
2 = x2 ;
' 8 (cid:0) 3
k 2k r (cid:18) (cid:19)
' (x) 1
3 = ' (x):
' p2 3
k 3k
1.17 Solvingthepairofequations x2+ax+b;x+1 =0and x2+ax+b;x 1 =
h i h (cid:0) i
0 gives a= 1; b=1=6:
(cid:0)
1.18 If f : [a;b] C is a continuous function and f = 0, then it follows
that b f(x)!2dx = 0 and f(x)2 is continuous akndknonnegative on [a;b]:
a j j j j
But this implies f(x) = 0 for all x [a;b]: On the other hand, the non-
R 2
continuous function
0; x [0;1] 1=2
f(x)= 2 nf g
1; x=1=2
(cid:26)
clearly satis…es f =0; but f is not identically 0 on [0;1]:
k k
1.19 From the CBS inequality, f +g 2 = f 2 +2Re f;g + g 2 f 2 +
k k k k h i k k (cid:20) k k
2 f g + g 2 =( f + g )2: The triangle inequality follows by taking
k kk k k k k k k k
the square root of each side.
1.20 1;x =1=2; 1 =1; and x =1=p3: Clearly 1;x < 1 x :
h i k k k k h i k kk k
1.21 Use the de…nition of the Riemann integral, based on Riemann sums, to
show that f; and hence fg; is not integrable on [0;1], whereas f2 =1 and
g2 =1 are both integrable.
1.22 (i) 1=p2, (ii) not in 2(0; ), (iii) 1, (iv) not in 2(0; ):
L 1 L 1
4
1.23 If f = 0 then f, being continuous, is identically 0 and the pair f, g is
k k
linearly dependent. The same is true if g =0. Hence we assume f =0
k k k k6
and g =0: Now
k k6
f g 2 b f2 b g2 b fg
= + 2
(cid:13)kfk (cid:0) kgk(cid:13) Za kfk2 Za kgk2 (cid:0) Za kfkkgk
(cid:13) (cid:13)
(cid:13) (cid:13) =1+1 2=0;
(cid:13) (cid:13) (cid:0)
where we used f;g = f g in the second equality. The implies g =(cid:21)f
h i k kk k
with (cid:21)= g = f :
k k k k
Conversely, if g = (cid:21)f for some positive number (cid:21), then f;g = (cid:21) f 2 =
h i k k
f g :
k kk k
1.24 In general f +g 2 = f 2 + g 2 +2Re f;g : If f +g = f + g
k k k k k k h i k k k k k k
then we must have Re f;g = f g ; and if, furthermore, the functions
h i k kk k
f and g are positive and continuous, then Re f;g = f;g = f g and
h i h i k kk k
(by Exercise 1.23) f and g are linearly dependent.
Conversely, if the functions are linearly dependent then g = (cid:11)f for some
number (cid:11), and f +g = (1+(cid:11))f = 1+(cid:11) f : For the equality
k k k k j jk k
f +g = f + g to hold we must therefore have 1+(cid:11) = 1+ (cid:11) ;
k k k k k k j j j j
which implies (cid:11) 0:
(cid:21)
1.25 The norm x(cid:11) 2 = 1x2(cid:11)dx is …nite if, and only if, 2(cid:11) > 1; that is,
k k 0 (cid:0)
(cid:11)> 1=2:
(cid:0) R
1.26 (cid:11)< 1=2:
(cid:0)
1.27 Suppose lim f(x) = ` = 0, then lim f(x) = ` > 0 and there is
x x
!1 6 !1j j j j
a positive integer n such that, for all x n;
(cid:21)
f ` < ` =2
jj j(cid:0)j jj j j
0< ` =2< f(x) <3 ` =2:
j j j j j j
But this implies 1 f(x)2dx= ; which contradicts the integrability of
n j j 1
f 2 on (0; ):
j j 1 R
1.28 b f(x) dx = f(x) ;1 f 1 = pb a f by the CBS inequality.
a j j hj j i (cid:20) k kk k (cid:0) k k
f(x)=1=px is integrable on (0;1) but f2(x)=1=x is not.
R
1.29 Suppose f(x) M for all x 0: Then
j j(cid:20) (cid:21)
1f2(x)dx M 1 f(x) dx< :
(cid:20) j j 1
Z0 Z0
The function f(x)=(1+x) 1 is bounded on [0; ), lies in 2(0; ), but
(cid:0)
1 L 1
is not integrable on [0; ):
1
5
1.30 Using familiar trigonometric identities,
sin3x=sinx(1 cos2x)
(cid:0)
1
=sinx cosxsin2x
(cid:0) 2
1
=sinx (sin3x+sinx)
(cid:0) 4
3 1
= sinx sin3x:
4 (cid:0) 4
1.31 Therearemanyanswerstothisexercise.Anyoddfunction,suchasf(x)=
ax; where a is a (non-zero) constant satis…es f;x2+1 =0 since x2+1 is
h i
even.Forthischoiceoff,theconstantahastosatisfy ax = a 2=3=2,
k k j j
that is, a =p6: Thus f(x)=p6x is one possible answer.
j j p
1.32 The L2(cid:26)(0;1) norm of a polynomial p is given by 01p2(x)e(cid:0)xdx 1=2: It
is therefore su¢cient to show that the integral I =(cid:2)R 01q(x)e(cid:0)xdx i(cid:3)s …nite
for any polynomial q: This follows from integrating by parts, noting that
R
q(x)e(cid:0)x ! 0 as x ! 1; to obtain I = q(0)+ 01q0(x)e(cid:0)xdx. Similarly,
01q0(x)e(cid:0)xdx=q0(0)+ 01q00(x)e(cid:0)xdx: If q haRs degree n, we can use in-
ductiontoshowthat,inthen-thstep,theintegralisreducedtoaconstant
R R
multiple of 01e(cid:0)xdx=1.
1.33 Using the mRonotonic property of the integral,
b b
f 2 = f(x)2(cid:27)(x)dx f(x)2(cid:26)(x)dx= f 2:
k k(cid:27) j j (cid:20) j j k k(cid:26)
Za Za
Therefore, if f 2(a;b) then f 2(a;b):
2L(cid:26) 2L(cid:27)
1.34 (a) The limit is the discontinuous function
1; x >1
j j
xn 0; x <1
lim =8 j j
n!11+xn >>< 1=2; x=1
unde…ned, x= 1:
(cid:0)
>>:
0; x=0
(b) lim pnx=
n!1 1; x>0:
(cid:26)
(c) lim sinnx does not exist, except when x is an integral multiple of
n
!1
(cid:25):
1.35 (a) Pointwise (not uniform), by Theorem 1.17(i), since the limit (Exercise
1.34(a)) is discontinuous.
6
(b) Uniform, since x1=n 1 for all x [1=2;1] and x1=n 1 1
! 2 (cid:0) (cid:20) (cid:0)
(1=2)1=n 0:
! (cid:12) (cid:12)
(cid:12) (cid:12)
(c) Pointwise (not uniform), since the limit is discontinuous at x = 0 (see
Exercise 1.34(b)).
0; x=0
1.36 f iscontinuousforeveryn;whereaslim f (x)= is
n n!1 n 1; 0<x 1:
(cid:26) (cid:20)
discontinuous, hence the convergence f f is not uniform. Clearly,
n
!
1 1
lim f (x)dx=1= limf (x)dx:
n n
Z0 Z0
1.37 Atx=0,f (0)=0foreveryn:Whenx>0,f (x)=n(1 x)=(n 1) 1:
n n
(cid:0) (cid:0) !
Therefore the sequence f converges pointwise to
n
0; x=0
f(x)=
1 x; 0<x 1:
(cid:26) (cid:0) (cid:20)
Since f is not continuous the convergence is not uniform.
1.38 f (0) = f (1) = 0 and, for all x (0;1); f (x) = nx(1 x2)n n(1
n n n
2 j j (cid:0) (cid:20) (cid:0)
x2)n 0 as n : Hence lim f (x)=0 on [0;1]: Since
n n
! !1 !1
1 n 1
lim f (x)dx= lim = ;
n
n n 2n+2 2
!1Z0 !1
1
whereas lim f (x)dx = 0; the convergence f 0 is not uniform
0 n!1 n n !
(by Theorem 1.17(ii)).
R
x a x
u
1.39 0 0, hence 0 on [0;a]:
(cid:20) n+x (cid:20) n ! n+x !
x
Forx 0,wealsohavelim =0pointwise.Inthiscase,assuming
n
(cid:21) !1 n+x
x
0<"<1;theinequality <"cannotbesatis…edwhenx n"=(1 "),
n+x (cid:21) (cid:0)
hence the convergence is not uniform.
Another approach: Since the statement f (x) f(x) " for all x I is
n
j (cid:0) j (cid:20) 2
equivalent to the requirement that sup f (x) f(x) ", we see that
u x2Ij n (cid:0) j (cid:20)
f f on I if, and only if, sup f (x) f(x) 0 as n : When
n ! x2Ij n (cid:0) j ! ! 1
x [0;a];supf (x)=a=n 0;butwhenx [0; )wehavef (n)=1=2,
n n
2 ! 2 1
hence supf (x) 1=290:
n
(cid:21)
1.40 The sequence f is de…ned by
n
1=n; x n
f (x)= j j(cid:20)
n 0; x >n:
(cid:26) j j
7
Since 0 f (x) 1=n 0 for all x R; the convergence f 0 is
n n
(cid:20) (cid:20) ! 2 !
uniform. 1 fn(x)dx=2; therefore
(cid:0)1
R
1 1
lim f (x)dx= limf (x)dx;
n n
6
Z(cid:0)1 Z(cid:0)1
the reason being that the domain of de…nition of f is not bounded.
n
u
1.41 Suppose f f on [a;b]. Given any ">0; it then follows that there is an
n
!
integer N such that
1
n N f (x) f(x) <" for all x [a;b];
1 n
(cid:21) ) j (cid:0) j 2
u
whichimplies f (x) f(x) 0:Wecanalso…ndanintegerN suchthat
n 2
j (cid:0) j!
n N f (x) f(x) <1 for all x [a;b]:
2 n
(cid:21) ) j (cid:0) j 2
If N =max N ;N , then
1 2
f g
n N f (x) f(x)2 <" for all x [a;b];
n
(cid:21) ) j (cid:0) j 2
which implies f (x) f(x)2 u 0:
n
j (cid:0) j !
1.42 (a) f (x) 1=n2 for all x R. Since 1=n2 converges, f (x) con-
n n
j j (cid:20) 2
verges uniformly on R by the Weierstrasse M-test.
P P
(b) If x ( 1;1) then there is an integer N such that xn < 1=2 for all
2 (cid:0) j j
n N; and hence
(cid:21)
xn xn
j j =2 xn for all n N;
1+xn (cid:20) 1 1=2 j j (cid:21)
(cid:12) (cid:12) (cid:0)
(cid:12) (cid:12)
from which we c(cid:12)onclude(cid:12)that the series converges by comparison with the
(cid:12) (cid:12)
geometric series. It diverges on ( ; 1] [1; ) where f (x) 9 0 (see
n
(cid:0)1 (cid:0) [ 1
Exercise 1.34(a)).
1.43 a sinnx a for all x R: Since a converges, a sinnx con-
n n n n
j j (cid:20) j j 2 j j
verges uniformly.
P P
1.44
(n+1)(cid:25) sinx 1 (n+1)(cid:25)
j jdx sinx dx
x (cid:20) n(cid:25) j j
Zn(cid:25) Zn(cid:25)
1 (cid:25)
= sinxdx
n(cid:25)
Z0
2
= 0 as n 0:
n(cid:25) ! !
8
LetA = (n+1)(cid:25)x 1 sinx dx:Becausex 1 sinx >(x+(cid:25)) 1 sin(x+(cid:25))
n n(cid:25) (cid:0) j j (cid:0) j j (cid:0) j j
for every x>0, we see that A A for all n and A 0. Moreover,
R n (cid:21) n+1 n !
(n+1)(cid:25) sinx k=n (k+1)(cid:25) sinx
dx= ( 1)kj jdx
x (cid:0) x
Z0 k=0Zk(cid:25)
X
k=n
= ( 1)kA :
k
(cid:0)
k=0
X
Hence 01x(cid:0)1sinxdx = 1k=0((cid:0)1)kAk; which converges by the alternat-
ing series test.
R P
On the other hand,
(n+1)(cid:25) sinx 1 (cid:25) 2
j jdx sinxdx=
x (cid:21) (n+1)(cid:25) (n+1)(cid:25)
Zn(cid:25) Z0
1 sinx 1 (n+1)(cid:25) sinx 1 2
j jdx= j jdx = :
) x x (cid:21) (n+1)(cid:25) 1
Z0 n=0Zn(cid:25) k=0
X X
1.45 Letr =R ":Then a xn a rnforallx [ r;r]:Since0<r <R,the
n n
(cid:0) j j(cid:20)j j 2 (cid:0)
numericalseries a rn isconvergent.BytheM-test,withM = a rn,
n n n
j j j j
the series a xn is uniformly convergent in [ r;r]:
n P (cid:0)
1.46 BeingunifPormlyconvergenton[R ";R+"],theseries a xn representsa
n
(cid:0)
continuousfunctionf on[R ";R+"]:Sincethisistrueforevery">0;the
(cid:0) P
series is continuous on ( R;R): Each term f (x)=a xn is di¤erentiable,
n n
(cid:0)
fn0(x) = nanxn(cid:0)1; and the series 1n=1nanxn(cid:0)1 has the same radius of
convergenceRastheoriginalseries(bytheroottest).Nowthepowerseries
P
1n=1nanxn(cid:0)1; by the result of Exercise 1.45, is uniformly convergent on
[R ";R+"]. Therefore Theorem 1.17(iii) applies and we conclude that
P(cid:0)
f0(x)=( 1n=0anxn)0 = 1n=1nanxn(cid:0)1:
1.47 With bn P= (n+1)an+1 wPe can write f0(x) = 1n=0bnxn and repeat the
argument in Exercise 1.46 to conclude that fP00(x) = 1n=1nbnxn(cid:0)1 =
1n=2n(n(cid:0)1)anxn(cid:0)2; and so on to any order of di¤erenPtiation. By induc-
tion we clearly have
P
1 n! (k+1)! (k+2)!
f(k)(x)= a xn k =k!a + a x+ a x2+ :
n (cid:0) k k+1 k+2
(n k)! 1! 2! (cid:1)(cid:1)(cid:1)
n=k (cid:0)
X
At x=0 this yields a =f(k)(0)=k!; k N:
k
2
9
1.48 Let the function f(x) be continuous and di¤erentiable up to any order on
R: According to Taylor’s theorem we can represent such a function at any
x R by
2
f (0) f (0) f(n)(0) f(n+1)(c)
f(x)=f(0)+ 0 x+ 00 x2+ + xn+ xn+1;
1! 2! (cid:1)(cid:1)(cid:1) n! (n+1)!
wherecliesbetween0andx:Iff(x)=ex;thenf(n)(x)=ex andf(n)(0)=
1 for all n. Thus we obtain
x x2 xn xn+1
ex =1+ + + + + ec:
1! 2! (cid:1)(cid:1)(cid:1) n! (n+1)!
Since,foranyx R;xn+1=(n+1)! 0asn ,wearriveatthedesired
2 ! !1
power series representation of ex by taking the limit of the right-hand side
as n :
!1
If f(x) = cosx, then f(n)(0) = 0 if n is odd and f(n)(0) = ( 1)n=2 if n is
(cid:0)
even.Theremaindertermisboundedbyxn+1=(n+1)!whichtendsto0as
n ; and we obtain cosx = 1 ( 1)nx2n=(2n)!. Similarly we arrive
! 1 n=0 (cid:0)
at the given representation for sinx:
P
1.49 Euler’s formula is obtained by replacing x by ix in the power series which
represent ex; cosx; and sinx; and using the equations i2n = ( 1)n and
(cid:0)
i2n+1 =( 1)ni.
(cid:0)
1.50 (a) 1:
(b) 1.
(c) f (0) = f (1) = 0: For every x (0;1); f (x) = nx(1 x)n 0 as
n n n
2 (cid:0) !
n : Therefore f (x) 0 pointwise.
n
!1 !
1
f 0 2 =n2 x2(1 x)2ndx
n
k (cid:0) k (cid:0)
Z0
2n2 1
= x(1 x)2n+1dx
2n+1 (cid:0)
Z0
n2 1
= (1 x)2n+2dx
(2n+1)(n+1) (cid:0)
Z0
n2
= 0 as n :
(2n+1)(n+1)(2n+3) ! !1
2
Hence fn L 0:
!
1.51 (a)convergent,since k 2=3sinkx 2 = sinkx 2 k 4=3 =(cid:25) k 4=3 <
(cid:0) (cid:0) (cid:0)
k k
.
1 (cid:13)P (cid:13) P P
(cid:13) (cid:13)