€¦ · 3 Exercises for Chapter 1 Series tro Inductory De nitions and Examples 101. Determine for...
77
Transcript of €¦ · 3 Exercises for Chapter 1 Series tro Inductory De nitions and Examples 101. Determine for...
Exer ises in Fourier AnalysisWritten by Anders HolstTranslated from Swedish by Kjell ElfströmMar h 2014
ContentsExer ises for Chapter 1 Series 3Exer ises for Chapter 2 Fourier Series 11Exer ises for Chapter 3 Heat Condu tion and Musi 25Exer ises for Chapter 4 The Fourier Transform 35Answers to the Exer ises 43Solutions to Exer ises Marked with an Asterisk 55
3Exer ises for Chapter 1 SeriesIntrodu tory De�nitions and Examples101. Determine for ea h of the series(a) ∞∑
k=0
(1 + x2)−k , (b) ∞∑
k=0
ekx , ( ) ∞∑
k=1
xe−kx , (d) ∞∑
k=0
3k + 4k
xkthe real numbers x for whi h it is onvergent. Compute the sum forthose x.102. Let ∞∑
k=0
ak be a onvergent series with sum s. Show that ∞∑
k=0
(ak + ak+1)is onvergent and determine its sum.Positive Series103. Whi h of the following series are onvergent?(a) ∞∑
k=1
sin (k−2) , (b) ∞∑
k=1
cos (k−2) , ( ) ∞∑
k=1
tan (1/k) ,(d) ∞∑
k=1
k
1 + k3, (e) ∞
∑
k=1
√k + 1−
√k√
k.104. Che k the onvergen e of the series(a) ∞
∑
k=1
k3
2k, (b) ∞
∑
k=1
k!
kk, ( ) ∞
∑
k=1
(2k)!
kk, (d) ∞
∑
k=0
2−√k .105. (a) For whi h positive real numbers a is ∞
∑
k=1
2sin k
ka onvergent?Answer the same question for the series*(b) ∞
∑
k=1
kaak , ( ) ∞∑
k=1
1
ka + a−k, (d) ∞
∑
k=1
ak2
k!,(e) ∞
∑
k=1
(
a− 1
k
)k
, (f) ∞∑
k=1
ka2ak2
.106. Compute ∞∑
j=2
( ∞∑
k=2
k−j
) .
4107. Set ajk = xjk − xj
k+1 − xj+1k + xj+1
k+1 where xk =k
k+1. Show that
∞∑
j=1
( ∞∑
k=1
ajk
)
=1
2and ∞
∑
k=1
( ∞∑
j=1
ajk
)
= −1
2.Hint: First show that
∞∑
k=1
ajk = xj1 − xj+1
1 and ∞∑
j=1
ajk = xk − xk+1 .108. In the harmoni series ∞∑
1
1k, all terms for whi h the integer k ontainsthe digit 9 are deleted. Show that the resulting series is onvergent.Hint: Show that the number of terms 1
kfor whi h k ontains no ninesand 10p−1 ≤ k < 10p is less than 9p.Absolutely Convergent Series109. Whi h of the following series are absolutely onvergent?(a) ∞
∑
k=1
(−1)k2
k2, (b) ∞
∑
k=1
cos k
1 + k2,( ) ∞
∑
k=1
1 + (−1)kk
k2, (d) ∞
∑
n=1
sin n2π/3
n2 − arctann.110. Show the following inequality for absolutely onvergent series.
∣
∣
∣
∣
∣
∞∑
k=1
uk
∣
∣
∣
∣
∣
≤∞∑
k=1
|uk| .111. Set tk = ∞∑
j=2
1jk
and uk =∞∑
j=2
(−1)j
jk. Compute(a) ∞
∑
k=1
t2k , (b) ∞∑
k=1
u2k .
5Conditionally Convergent Series112. For whi h values of α is the series ∞∑
k=1
(−1)k+1
kα onvergent?113. Test the series ∞
∑
k=1
sin π(k + 1k) for absolute onvergen e and onver-gen e.114. Find a number N su h that |s − sn| ≤ 10−3 when n ≥ N for ea h ofthe series (a) ∞
∑
k=1
(−1)k
k2, (b) ∞
∑
k=1
1
k2.Series with Complex Terms115. Whi h of the following series are absolutely onvergent?(a) ∞
∑
k=1
eik
k2, (b) ∞
∑
k=1
1
k + i.116. For whi h omplex numbers z is the series ∞
∑
k=0
2k+ik+2i
zk onvergent?Uniform Convergen e117. Compute ||fn||E if fn(x) = nx1+n2x2 and (a) E = [0, 1], (b) E = [1,∞) .In whi h of these ases is fn uniformly onvergent to 0?118. Determine the limit fun tion f(x), when x ≥ 0, of the fun tion sequen e
fn(x) =x− nx3
1 + nx2and ompute ||fn − f ||E to �nd out if fn is uniformly onvergent to fin the interval E = [0,∞).119. Che k if the fun tion sequen e fn(x) = arctannx , n = 1, 2, 3, . . . , isuniformly onvergent in ea h of the intervals (a) [−1, 1] , (b) [1, 2] and( ) [2,∞).*120. Setfn(x) = (sin x)1/n , 0 ≤ x ≤ π , n = 1, 2, . . . .(a) Compute limn→∞ fn(x) .(b) Che k if the onvergen e is uniform in [0, π] .
6121. Set, for x > 0, fn(x) =√nx arctan 1
nx, n = 1, 2, 3, . . . .(a) Compute limn→∞ fn(x) .(b) Is the onvergen e uniform in the interval x > 0 ?122. Determine whether the fun tion sequen e
fn(x) =
(
x
1 + x
)n1
1 + x, n ≥ 1 ,is uniformly onvergent or not in the interval x ≥ 0.*123. Let s(x) = ∞
∑
k=0
x(1+x)k+1 , x ≥ 0.(a) Show that the series is pointwise onvergent for all x ≥ 0.(b) Show that the series is uniformly onvergent in ea h interval x ≥ d,
d > 0.( ) Show that the onvergen e is not uniform in the interval x ≥ 0.124. Consider the series ∞∑
k=0
x e−kx .(a) Determine the set E of real numbers x for whi h the series is onvergent.(b) Compute the sum s(x) of the series for x ∈ E.( ) Che k if the series is uniformly onvergent in E.125. Show that(a) ∞∑
k=0
(1− x)2xk is uniformly onvergent in the interval 0 ≤ x ≤ 1,(b) ∞∑
k=0
(1−x)xk is not uniformly onvergent in the interval 0 ≤ x ≤ 1.*126. Show that the fun tion series ∞∑
k=1
xk2
k2is uniformly onvergent in theinterval |x| ≤ 1.
7127. Show that the following fun tion series are uniformly onvergent in theintervals stated.(a) ∞∑
0
xk
1 + xk, −1
2≤ x ≤ 1
2, (b) ∞
∑
0
sin 3kx
2k, x ∈ R ,( ) ∞
∑
1
1
kx, x ≥ d > 1 , (d) ∞
∑
1
kx
xk, 2 ≤ x ≤ 3 .128. (a) Let c ≥ 1 be a onstant. Compute
sup0≤x≤1
(1− x)xc .(b) Show that the fun tion series∞∑
k=1
(1− x)xk2is uniformly onvergent in the interval 0 ≤ x ≤ 1.129. (a) Show that the series s(x) =∞∑
k=1
arctan kx1+k2x2 is uniformly onvergentin the interval x ≥ c where c > 0.(b) Show that the sum s(x) is ontinuous for x > 0.*130. Show that the sum of the series
∞∑
k=0
1 +√kx2
1 + k2√xis ontinuous when x > 0.131. Is the sum
s(x) =∞∑
k=1
(
1
k− 1
k + x
) ontinuous when x ≥ 0?132. Set s(x) =∞∑
k=0
cos kx2k
and show that s(x) is di�erentiable for all real x.133. Show that the series ∞∑
k=0
2k sin 3−kx is absolutely onvergent in R andthat its sum s(x) is a di�erentiable fun tion. Also evaluate s′(0).
8134. Show thatd
dx
∞∑
k=1
ln
(
1 +1
k2x
)
= −1
x
∞∑
k=1
1
1 + k2x, x > 0 .135. Compute the sum of the series
∞∑
p=0
(−1)p
2p+ 1.*136. Show that a power series ansatz f(x) =∞∑
0
akxk for the problem
f(x) = (1− x)f(x2) and f(0) = 1leads to a0 = 1 , a2k = ak and a2k+1 = −ak when k ≥ 0 . Thisdetermines the oe� ients ak of the power series uniquely. What is itsradius of onvergen e, i.e. the largest value of R for whi h the series onverges for all x with |x| < R?137. Show that the fun tion s(x) =∞∑
k=0
e−kx
1+k2, x > 0 , is a solution of thedi�erential equation
y′′(x) + y(x) = (1− e−x)−1 , x > 0 .138. Let f(x) =∞∑
k=1
arctan (kx)k2
.(a) Show that f is ontinuous in R.(b) Show that f is di�erentiable when x 6= 0 .( ) Show that f is not di�erentiable at 0.139. (a) Show thats(x) = 1− x3
3!+
x6
6!− · · · =
∞∑
k=0
(−1)kx3k
(3k)!, x ∈ R ,is a solution of the di�erential equations
u′′(x)− u′(x) + u(x) = e−x and u′′′(x) + u(x) = 0 .(b) Compute the sum s(x) of the series for all x.
9140. Show that the series∞∑
k=1
xk
1− xkde�nes an in�nitely di�erentiable fun tion f when |x| < 1 and that1n!f (n)(0) equals the number of positive integers dividing n.141. (a) Show that the fun tion
s(x) =
∞∑
k=1
x
2 + k3xis ontinuous for x ≥ 0.(b) Is the fun tion s(x) ontinuously di�erentiable for x > 0 ?(4 Jan. 85)142. Set s(x) =∞∑
k=1
xk−x3k
k.(a) Evaluate s(x) when |x| < 1, x = 1 and x = −1.(b) Is the onvergen e uniform in the interval 0 ≤ x ≤ 1 ?( ) Can the result of (a) be used to infer that the onvergen e isuniform in the interval −1 ≤ x ≤ 0 ? (4 April 85)143. For whi h real values of a is the following series onvergent?
∞∑
k=0
1
3k + a2ak . (2 Nov. 85)144. Set f(x) =
∞∑
k=0
xk
3k+x2 , |x| < 3. Show that f is a ontinuous fun tion.(4 Jan. 86)145. (a) For whi h positive values of x is the following series onvergent?∞∑
k=1
x2k
xk + 4k.
10 (b) Is the sum s(x) of the series ontinuous for these values of x ?(3 April 86)146. Show, for example by onsidering the ases |x| < 3 and |x| > 2 sep-arately or by omputing the suprema over x ∈ R of the terms, orotherwise, that the sums(x) =
∞∑
k=1
xk
x2k + 4kis ontinuous for all real x. (4 May 86)
11Exer ises for Chapter 2 Fourier Series*201. The fun tion u has period 2π and satis�esu(x) = π − x when 0 ≤ x < 2π .(a) Sket h the graph of u in the interval |x| ≤ 6π.(b) Compute the Fourier oe� ients of u.202. Compute the Fourier oe� ients of u where u(x) is*(a) cos2 x , (b) sin x , ( ) cos4 x , (d) sin2 x cos2 x .203. Let u be a fun tion with the Fourier series ∑∞
−∞ cneinx . Determine c6if u(x) = cos20 x .204. Compute the Fourier oe� ients of the 2π-periodi fun tion u given by
u(x) = |x| when − π ≤ x ≤ π .205. The fun tion u has period 2π and satis�esu(x) =
{
1 when 0 ≤ x < π
0 when π ≤ x < 2π .(a) Determine the Fourier series of u.(b) Compute the sum of the Fourier series at x = 0 and x = π.206. The fun tion u has period 2π and is integrable over an interval oflength 2π. Show the following statements for the Fourier oe� ientsof u.(a) c−n = cn if u is even, i.e. u(−x) = u(x) for all x .(b) c−n = −cn if u is odd, i.e. u(−x) = −u(x) for all x .( ) cn are real if u(−x) = u(x) for all x .(d) cn are real and c−n = cn if u is real and even.
12207. Consider the fun tionsR0(x) = 1, 0 ≤ x < 2π ,
R1(x) =
{
1, 0 ≤ x < π ,
−1, π ≤ x < 2π ,
R2(x) =
{
1, 0 ≤ x < π2, π ≤ x < 3π
2,
−1, π2≤ x < π , 3π
2≤ x < 2π .Show that R0, R1, R2 onstitute an orthonormal system with respe tto the s alar produ t
(u|v) =1
2π
∫ 2π
0
u(x)v(x) dx .208. Show that||u+ v||2 + ||u− v||2 = 2(||u||2 + ||v||2) ,where
||u||2 = (u|u) =1
2π
∫ 2π
0
|u(x)|2 dx .*209. Determine all ontinuous di�erentiable fun tions with period 2π su hthat2u′(x) = u(x+
π
2) − u(x− π
2) for all x .210. Determine all twi e ontinuously di�erentiable 2π-periodi fun tionssu h that
u′′(x) = u(x+ π) .211. Whi h 2π-periodi C1 fun tion has the Fourier oe� ientscn =
{
ne−n , n ≥ 0
0 , n < 0 ?212. Let u be a fun tion that is pie ewise C1, 2π-periodi but not ne essarily ontinuous. Denote its dis ontinuity points in the interval [0, 2π] byx1, x2, . . . , xk and the jumps at these points by s(xj), i.e.
s(xj) = limǫ→+0
u(xj + ǫ) − limǫ→+0
u(xj − ǫ) .
13Show that the Fourier oe� ients of u′ areincn − 1
2π
k∑
j=1
s(xj)e−inxjwhere cn are the Fourier oe� ients of u.Hint: Partition [0, 2π] into subintervals as in the proof of Theorem 2.10and opy this proof. Observe that the boundary terms in the partialintegrations no longer an el ea h other but instead add to the sum inthe expression above.*213. The fun tion u has period 2π and
u(x) =1
2(ex + e−x) when |x| ≤ π .Expand u in Fourier series and ompute
∞∑
k=1
1
k2 + 1.214. Let u be periodi with period 2π and assume that
u(x) = x2 when |x| ≤ π .(a) Show, by hoosing x suitably in the Fourier series of u, that∞∑
1
(−1)n+1
n2= 1− 1
4+
1
9− 1
16+
1
25− · · · = π2
12.(b) What is the sum of the Fourier series at x = 6?215. Expand the fun tion
u(x) = max (cosx, 0)in a Fourier series and ompute∞∑
1
(−1)k
4k2 − 1and ∞
∑
1
1
4k2 − 1.
14216. Show that(a) x sin x = 1− 12cosx − 2
∞∑
k=2
(−1)k(k2 − 1)−1 cos kx , |x| ≤ π ,(b) | sin x| = 2π
(
1 +∞∑
k=1
(
12k+1
− 12k−1
)
cos 2kx
)
.217. The fun tion u is periodi with period 2π andu(x) = cosαx when |x| ≤ πwhere α is a real number that is not an integer. Expand u in Fourierseries and show that
π cot πα =
∞∑
−∞
α
α2 − n2.218. Let u be the 2π-periodi fun tion de�ned by
u(x) =
{
sinxx
when 0 < |x| ≤ π
1 when x = 0 .Show that the Fourier oe� ients of u are given bycn =
1
2π
∫ (n+1)π
(n−1)π
sin x
xdxand use this to ompute the improper integral
∫ ∞
0
sin x
xdx .219. Let u be a C1 fun tion with period 2π and Fourier oe� ients cn. Showthat(a) if u is in Ck, then |nkcn| ≤ Mk whereMk is a onstant independentof n ,(b) if ∞
∑
−∞|nkcn| is onvergent, then u is in Ck ,( ) u is in�nitely di�erentiable if and only if there exist onstants Mksu h that|nkcn| ≤ Mk for all integers n and k = 0, 1, 2, . . . .(We see from this that the rate of onvergen e of the Fourier seriesin reases with in reasing regularity of the fun tion and vi e versa.)
15220. Let g be a C1 fun tion su h that the two series∞∑
−∞g(x+ 2nπ) and ∞
∑
−∞g′(x+ 2nπ)are uniformly onvergent in the interval 0 ≤ x ≤ 2π. Show the Poissonsummation formula:
∞∑
n=−∞g(2nπ) =
∞∑
m=−∞γmwhere
γm =1
2π
∫ ∞
−∞g(x)e−imx dxis assumed to be onvergent.Hint: The numbers γm are the Fourier oe� ients of the 2π-periodi fun tion u(x) =
∞∑
−∞g(x+ 2πn) .221. Let u be a pie ewise ontinuous and 2π-periodi fun tion with Fourier oe� ients cn. SetU(x) =
∫ x
0
(u(t)− c0) dt .Show the following statements.(a) U has the period 2π, is ontinuous and pie ewise C1 .(b) The Fourier series ∞∑
−∞dne
inx of U is absolutely onvergent and hasthe sum U(x) for ea h x .( ) dn = − icnn
when n 6= 0 .(d) d0 = i∑
n 6=0
cnn.Hint: Observe that U(0) = 0 =
∞∑
−∞dn .(e)
∫ x
0
u(t) dt = c0x +∑
n 6=0
cneinx − 1
in.(Hen e, termwise integration of the Fourier series of u is permit-ted even when the series is not known to be onvergent. Theintegrated series is absolutely onvergent.)
16222. When u is a fun tion that has period 2π and is integrable in [0, 2π], wede�ne the symmetri partial sums sN of the Fourier series bysN (x) =
N∑
−N
cneinxwhere cn are the Fourier oe� ients of u. Show that(a)
sN(x) =1
2π
∫ π
−π
u(x− y)DN(y) dywhereDN (y) =
N∑
−N
einy =sin (N + 1
2)y
sin y2
,(b)1
2π
∫ π
−π
DN(y) dy = 1 .Hint to (b): Integrate the sum in the �rst expression for DN term-wise or apply the result of (a) to the fun tion u(x) ≡ 1.223. The arithmeti means σN of the symmetri partial sums in the previousexer ise are de�ned by σN = s0+···+sNN+1
.Show the following statements.(a)σN (x) =
1
2π
∫ π
−π
u(x− y)KN(y) dywhereKN(y) =
(
sin (N + 1)y2
sin y2
)2
· 1
N + 1.(b)
∫ π
−π
KN(y) dy = 1 .Hint: Confer the se ond hint to 222 (b).( )∫
δ<|y|<π
KN(y) dy → 0 as N → ∞for every �xed δ > 0 .
17(d) σN(x) → u(x) as N → ∞ for every �xed x if u is ontinuous.Hint: Observe that (b) implies thatu(x) =
1
2π
∫ π
−π
u(x)KN(y) dyand soσN (x)− u(x) =
1
2π
∫ π
−π
(u(x− y))− u(x))KN(y) dy .Split the integral in the right-hand side into one integral over|y| ≤ δ and one over δ < |y| ≤ π. Choose δ so that the �rstintegral be omes small and then apply ( ) to the se ond integral.224. In a linear spa e endowed with a real s alar produ t we have that
||u− v|| = ||u+ v|| if and only if (u|v) = 0.What is the orresponding statement when the s alar produ t is om-plex?225. Show that(a) when the s alar produ t is real,4(u|v) = ||u+ v||2 − ||u− v||2 ,(b) when the s alar produ t is omplex,
4(u|v) = ||u+ v||2 − ||u− v||2 + i ||u+ iv||2 − i ||u− iv||2 .226. (a) Let en, n = 0, ±1, ±2, . . . be an orthonormal system in a linearspa e with a s alar produ t and setfn =
1√2(en − e−n) , n > 0
e0 , n = 01√2(en + e−n) , n < 0 .Show that the ve tors fn also form an orthonormal system.(b) Set
gn(x) =
√2 sin nx , n > 0
1 , n = 0√2 cos nx , n < 0 .
18 Show that the fun tions gn form an orthonormal system with re-spe t to the s alar produ t(u|v) =
1
2π
∫ π
−π
u(x)v(x) dx .*227. Compute ∞∑
1
1n4 by means of Parseval's formula applied to
u(x) = x2 , |x| ≤ π .228. Compute ∫ 2π
0cos4 x dx by means of Parseval's formula applied to
u(x) = cos2 x .229. Compute ∞∑
1
1k2+1
by means of Parseval's formula applied tou(x) = ex , 0 ≤ x < 2π .230. Compute the Fourier oe� ients of the 2π-periodi fun tion u given by
u(x) = x cosx when − π < x ≤ π .Also determine∞∑
2
n2
(n2 − 1)2.231. Show that
ei|x| =2i
π+ cosx +
4
πi
∞∑
1
cos 2kx
4k2 − 1for all real x with |x| ≤ π. Also ompute∞∑
1
1
(4k2 − 1)2.232. (a) Compute the Fourier oe� ients of the 2π-periodi fun tion ugiven by
u(x) = x(x2 − π2) when |x| ≤ π .
19(b) Determine∞∑
1
1
n6.233. The fun tion u has period 2π and satis�es
u(x) = eiax when 0 ≤ x < 2πwhere a is real but not an integer.(a) Determine the Fourier oe� ients of u.(b) Show that∞∑
−∞
(−1)n
a− n=
π
sin πa.( ) Show that
∞∑
−∞
1
(a− n)2=
π2
sin2 πa.234. The Bessel fun tions Jn(x) are the oe� ients of the Fourier seriesexpansion
eix sin t =∞∑
−∞Jn(x)e
int .Show that the sum ∞∑
−∞|Jn(x)|2 is independent of x and determine itsvalue.235. Expand the fun tion
u(x) =1
r − eix,where r > 1, in a Fourier series and use this to ompute the integral
∫ 2π
0
dx
1− 2r cosx + r2.236. Let u be a C1 fun tion in [0, π] with u(0) = u(π) = 0. Show that
∫ π
0
|u(x)|2 dx ≤∫ π
0
|u′(x)|2 dxwith equality if and only if u is a multiple of sin x .Hint: Extend u to an odd fun tion.
20237. Determine the Fourier oe� ients of the fun tionu(x) = cosn xwhere n is a positive integer. Use the result to show that
n∑
k=0
(
n
k
)2
=
(
2n
n
)
.Hint: Use Euler's formulae.*238. Determine the numbers a, b and c su h that the integral∫ π
−π
|x2 − a− beix − ce−ix|2 dxis as small as possible. Also state the minimum value.239. Determine the trigonometri polynomial p(x) =∑
|n|≤4
aneinx of degreeat most 4 that minimises the integral
∫ π
−π
|u(x)− p(x)|2 dxwhere u(x) = (sin x)8 and state the minimum value of the integral.240. Let u(x) = |x|. Determine the se ond degree trigonometri polynomialw(x) =
∑
|n|≤2
aneinx that minimises the integral
∫ π
−π
|u(x) − w(x)|2 dx .Also state the minimum value.241. Set u(x) = x2 when −π ≤ x ≤ π. What hoi es of a0 minimise thefollowing expressions?(a) |u(0)− a0| ,(b) ∫ π
−π|u(x)− a0| dx ,( ) ∫ π
−π|u(x)− a0|2 dx ,(d) max[−π,π] |u(x)− a0| .Note that the four hoi es of a0 are di�erent although ea h gives thebest approximation of u in some sense.
21*242. Assume that u is a T -periodi fun tion, i.e.u(x+ T ) = u(x) for all x .The Fourier oe� ients cn of u are de�ned as the Fourier oe� ientsof the 2π-periodi fun tion
v(x) = u
(
Tx
2π
)
.(a) Show thatcn =
1
T
∫ T
0
u(x) e−2πinx/T dx .(b) Show thatu(x) =
∞∑
−∞cne
2πinx/Tif u is ontinuous and pie ewise C1.( ) Expand the 2-periodi fun tion u given by u(x) = 1 − |x| when|x| ≤ 1 in Fourier series.(d) What does Parseval's formula look like for a T -periodi fun tion?(e) What is the result when Parseval's formula is applied to the fun -tion in ( )?243. Compute the Fourier oe� ients of the fun tion u if(a) u(x) = x when |x| < 1/2 and u has period 1,(b) u(x) = x when 0 < x < T and u has period T ?*244. Expand u(x) = x(π − x), 0 ≤ x ≤ π, in a(a) osine series,(b) sine series.245. Expand u(x) = sin x, 0 ≤ x ≤ π, in a(a) osine series,(b) sine series.What is the sum of the osine series for −π ≤ x < 0 and π < x ≤ 2π?
22246. Expand u(x) = x, 0 ≤ x ≤ π, in(a) osine series,(b) sine series.247. Expand u(x) = sin3 x, 0 ≤ x ≤ π, in a sine series and use this to ompute∫ π
0
sin6 x dx .248. The fun tion u is given in the interval 0 ≤ x ≤ π byu(x) =
{
x(x− π2) when 0 ≤ x ≤ π
2
0 when π2≤ x ≤ π .Expand u in a sine series and ompute
∞∑
0
(−1)k
(2k + 1)3assuming the identity ∑∞0
1(2k+1)2
= π2
8.249. Show that
(x2 − π2)2 =8
15π4 − 48
∞∑
k=1
(−1)k
k4cos kxfor all real x with |x| ≤ π and ompute(a) ∞
∑
k=1
(−1)kk−4 , (b) ∞∑
k=1
k−4 , ( ) ∞∑
k=1
k−8 .250. (a) Show thatπ − |x| = π
2+
4
π
∞∑
k=0
cos (2k + 1)x
(2k + 1)2for all real x with |x| ≤ π.(b) Compute∞∑
0
sin (2k + 1)x
(2k + 1)3when 0 ≤ x ≤ πby means of (a).
23251. Show that(a) (π2 − 3x2)/12 =∞∑
k=1
(−1)k+1k−2 cos kx when |x| ≤ π ,(b) x(x2 − π2) = 12∞∑
k=1
(−1)kk−3 sin kx when |x| ≤ π ,( ) (3x2 − 6πx+ 2π2)/12 =∞∑
k=1
k−2 cos kx when 0 ≤ x ≤ 2π .252. Show thatπx
4=
∞∑
k=0
(−1)k(2k + 1)−2 sin (2k + 1)x when 0 ≤ x ≤ π
2.253. In the dis ussion of the Gibbs phenomenon the fun tion
w(t) =1
2 tan t2
− 1
tappears. Show that w(t) and w′(t) have limits as t → 0 and evaluatethese limits.254. Determine the partial sums SN(x) =N∑
−N
cneinx of the Fourier series ofthe 2π-periodi fun tion given by
u(x) =
{
−1 when − π ≤ x < 0
1 when 0 ≤ x < π .Evaluate S3(π/3), S4(π/4) and S6(π/6) and ompare with the fa torg = 2
π
∫ π
0sin tt
dt ≈ 1.179. Also evaluate S3(2π/3), S4(2π/4), S6(2π/6)and ompare with the fa tor 2π
∫ 2π
0sin tt
dt ≈ 0.903.
25Exer ises for Chapter 3 Heat Condu tion and Musi *301. Determine a solution of∂tu = ∂2
xu , 0 ≤ x ≤ π , t > 0
u(0, t) = u(π, t) = 0 , t ≥ 0
u(x, 0) = sin x + 2 sin 3x , 0 ≤ x ≤ π .302. Solve the problem∂tu = ∂2
xu , 0 ≤ x ≤ π , t > 0
u(0, t) = u(π, t) = 0 , t ≥ 0
u(x, 0) =
{
x, 0 ≤ x ≤ π2
π − x, π2< x ≤ π .303. Solve the heat transfer problem
∂tu = 3∂2xu , 0 ≤ x ≤ π , t > 0
u(0, t) = u(π, t) = 0 , t ≥ 0
u(x, 0) = sin x cos 4x , 0 ≤ x ≤ π .304. (a) Show that the substitutionv(x, t) = u(x, t) − A
(
1− x
l
)
− Bx
l
g(x) = f(x) − A(
1− x
l
)
− Bx
ltransforms the problem∂tu = a∂2
xu , 0 ≤ x ≤ l , t > 0
u(0, t) = A o h u(l, t) = B , t ≥ 0
u(x, 0) = f(x) , 0 ≤ x ≤ linto the equation∂tv = a∂2
xv , 0 ≤ x ≤ l , t > 0 ,with boundary onditionsv(0, t) = v(l, t) = 0 when t ≥ 0
26 and initial onditionv(x, 0) = g(x) when 0 ≤ x ≤ l .(Observe that the fun tion u(x) = A(1 − x
l) + B x
lsatis�es the ondition u′′(x) = 0. It is therefore a stationary solution of theheat equation, i.e. independent of time, whi h re�e ts the fa t thatthe temperature is stabilised.)(b) Determine the solution of the problem
∂tu = ∂2xu , 0 ≤ x ≤ π , t > 0
u(0, t) = 1 o h u(π, t) = 2 , t ≥ 0
u(x, 0) = x sin x + 1 +x
π, 0 ≤ x ≤ π .305. A large (in�nite) steak with thi kness l and the thermal di�usivity ais put in the oven to grill at the temperature T0. The steak is takenstraight from the refrigerator where the temperature is 0◦ C. The tem-perature distribution in the steak is then des ribed by
∂tu = a∂2xu , 0 ≤ x ≤ l , t > 0
u(0, t) = u(l, t) = T0 , t > 0
u(x, 0) = 0 , 0 ≤ x ≤ l .(a) Determine the temperature distribution u(x, t) in form of a series.Hint: Make use of Exer ise 304 (a). The solution formula of The-orem 3.4 may then be used even though the initial and boundaryvalues are in ompatible at the points (0, 0) and (l, 0).(b) Assume that t is so large that all the terms of the series ex ept the�rst may be negle ted. What is the ooking time for the steak ifit is ready when the temperature everywhere ex eeds T0/4? Howmu h is the ooking time in reased if the thi kness of the steak isdoubled?306. Assume that u is a real solution of the heat equation∂tu = ∂2
xu for 0 ≤ x ≤ π , t > 0that ful�ls the ondition u(0, t) = u(π, t) = 0 for t ≥ 0. Show that∫ π
0
(u(x, t))2 dxis a de reasing fun tion of t. The fun tion u may be assumed to be C2for 0 ≤ x ≤ π, t > 0 and ontinuous for 0 ≤ x ≤ π, t ≥ 0.
27*307. Determine a solution of∂tu = ∂2
xu , 0 ≤ x ≤ π , t > 0
∂xu(0, t) = ∂xu(π, t) = 0 , t ≥ 0
u(x, 0) = x(π − x) , 0 ≤ x ≤ π .308. Determine a solution of∂tu = ∂2
xu , 0 ≤ x ≤ π , t > 0
∂xu(0, t) = ∂xu(π, t) = 0 , t ≥ 0
u(x, 0) = sin2 x cos2 x , 0 ≤ x ≤ π .309. Solve the heat ondu tion problem∂tu = ∂2
xu , 0 ≤ x ≤ 1 , t > 0
∂xu(0, t) = ∂xu(1, t) = 0 , t > 0if u(x, 0) equals (a) 10 , (b) 9 + 2x ,when 0 ≤ x ≤ 1.Also evaluate limt→∞ u(x, t) in the two ases.*310. Solve the problem∂tu = ∂2
xu , 0 ≤ x ≤ π
2, t > 0
u(0, t) = 0 o h ∂xu(π
2, t) = 0 , t ≥ 0
u(x, 0) = sin 3x , 0 ≤ x ≤ π
2.311. Solve the di�erential equation
∂tu = 2∂2xu , 0 ≤ x ≤ π
2, t > 0 ,with boundary values u(0, t) = ∂xu(
π2, t) = 0 for t > 0 and initialvalues u(x, 0) = x for 0 ≤ x ≤ π
2.312. Determine a fun tion u(x, t) su h that
∂tu = ∂2xu , 0 ≤ x ≤ π
2, t > 0
u(0, t) = ∂xu(π
2, t) = 0 , t ≥ 0
u(x, 0) = x(π − x) , 0 ≤ x ≤ π
2.
28313. (a) Solve the equation∂tu = ∂2
xu − au , 0 ≤ x ≤ π , t > 0 ,with boundary values u(0, t) = u(π, t) = 0 when t ≥ 0 and ini-tial values u(x, 0) = sin3 x. (The equation mimi s a rod witha heat-loss through the lateral surfa e proportional to the di�er-en e of the temperature of the rod and the temperature 0◦ of thesurroundings.)(b) For whi h values of the onstant a is limt→∞ u(x, t) = 0?Hint: Set u(x, t) =∞∑
1
bn(t) sinnx .314. If heat sour es o ur, then the (one-dimensional) heat equation be- omescρ∂tu = λ∂2
xu + q(x, t) , 0 ≤ x ≤ l , t > 0 ,where q(x, t) is the in rease in heat per unit time and unit length atthe time t and the position x. (The fun tion q, whi h is allowed to benegative, is assumed to be independent of the temperature u.) Solvethe equation in the ase whereq(x, t)
cρ= e−4t cos 2x,
l = π, λcρ
= 1, and u satis�es the boundary onditions∂xu(0, t) = ∂xu(π, t) = 0 when t > 0and the initial onditionu(x, 0) = cosx when 0 ≤ x ≤ π .315. A ir ular wire (with an insulating lateral surfa e) has the radius 1.If the wire is thin enough, then we may assume that the temperature
u(x, t) is onstant in every ross-se tion of the wire and hen e that usatis�es a one-dimensional heat equation∂tu = h∂2
xu .Here x is the ar length along the wire whi h is assumed to extend overthe interval 0 ≤ x ≤ 2π. Seek a 2π-periodi (in x) solution u of theequation that also satis�es the initial onditionu(x, 0) = x(2π − x) , 0 ≤ x ≤ 2π .Hint: Set u(x, t) =
∞∑
−∞cn(t) e
inx.
29*316. Solve the wave equation∂2t u = 25∂2
xu , 0 ≤ x ≤ 1 , t > 0
u(0, t) = u(1, t) = 0 , t ≥ 0
u(x, 0) = sin 2πx , ∂tu(x, 0) = sin πx − 2 sin 3πx , 0 ≤ x ≤ 1 .317. Determine a solution of the wave equation∂2t u = 4∂2
xu , 0 ≤ x ≤ π , t > 0 ,that satis�es the boundary onditionsu(0, t) = u(π, t) = 0 , t ≥ 0and the initial onditions
u(x, 0) = sin x , ∂tu(x, 0) = x(π − x) , 0 ≤ x ≤ π .318. Find a (weak) solution of the problem∂2t u = ∂2
xu , 0 ≤ x ≤ π , t > 0
u(0, t) = u(π, t) = 0 , t ≥ 0
u(x, 0) = 0 , 0 ≤ x ≤ π ,
∂tu(x, 0) = x , 0 ≤ x ≤ π .319. A string of length l is �xed at its ends. At the time t = 0, it is straightand at rest ex ept in a short portion of length b. This portion is set inmotion at a ertain velo ity v by the strike of a hammer in su h a waythat the initial onditions an be formulated asu(x, 0) = 0 when 0 ≤ x ≤ l ,
∂tu(x, 0) =
{
v when |x− a| < b2
0 else when 0 ≤ x ≤ l .(a) Determine the movement of the string.(b) Where should the hammer hit the string in order that the groundtone be as strong as possible?( ) Where should the hammer hit the string in order that the sixthovertone be as weak as possible?
30320. Use the method of separation of variables in order to �nd a solution ofthe problem of the vibrating string:∂2t u = ∂2
xu , 0 ≤ x ≤ π , t > 0
u(0, t) = u(π, t) = 0 , t ≥ 0
u(x, 0) = f(x) , ∂tu(x, 0) = g(x) , 0 ≤ x ≤ π .321. Determine the fun tion u(x, t) that satis�es∂2t u = 4∂2
xu when t > 0, x ∈ Rand the initial onditionsu(x, 0) = x2 , ∂tu(x, 0) = x .322. Determine the solution of the wave equation ∂2
t u = c2∂2xu in the half-plane t > 0 that has the initial values
u(x, 0) = e−x2
, ∂tu(x, 0) =1
1 + x2.323. The solution in Example 3.11 an be written as
u(x, t) =
2hlx when 0 ≤ x ≤ l
2− ct
h(1− 2ctl) when l
2− ct ≤ x ≤ l
2+ ct
−2hl(x− l) when l
2+ ct ≤ x ≤ lwhen 0 ≤ t ≤ l
2cand satis�es
u(x, t+2l
c) = u(x, t)and
u(x,l
c− t) = −u(x, t) .Expand u in Fourier series and he k that the solution is the same asthat in Example 3.7.324. One end of a string is �rmly an hored at x = 0 while the other endis free to move along a line perpendi ular to the x-axis at x = π
2. If
31the initial velo ity is zero, then we have the following initial-boundaryproblem.∂2t u = c2∂2
xu , 0 ≤ x ≤ π
2, t > 0
u(0, t) = 0 , ∂xu(π
2, t) = 0 , t > 0
u(x, 0) = f(x) , ∂tu(x, 0) = 0 , 0 ≤ x ≤ π
2.Solve this problem, e.g. by letting, for ea h �xed t > 0,
u(x, t) =
∞∑
k=0
bk(t) sin (2k + 1)xwherebk(t) =
4
π
∫ π/2
0
u(x, t) sin (2k + 1)x dx .325. Taking gravity into onsideration, we get the following di�erential equa-tion for the vibrating string:∂2t u = c2∂2
xu − g , 0 ≤ x ≤ l , t > 0 .Here g is the a eleration due to gravity. Solve the equation withboundary onditionsu(0, t) = u(l, t) = 0 , t ≥ 0 ,and initial onditions
u(x, 0) =
x10
when 0 ≤ x ≤ l2
l−x10
when l2≤ x ≤ l .Hint: Set v(x, t) = u(x, t) − gx
2c2(x − l) and try to �gure out whatproblem is solved by v. (Note that u(x, t) = gx
2c2(x− l) (independentlyof t!) is a stationary solution of the equation.)326. The vibration of a string is damped due to air resistan e, whi h is as-sumed to be proportional to the velo ity. This suggests the di�erentialequation
∂2t u = c2∂2
xu − 2h∂tu when 0 ≤ x ≤ l , t > 0 .
32 Assume that 0 < h < πcl, that
u(0, t) = u(l, t) = 0 when t ≥ 0 ,and thatu(x, 0) = f(x) , ∂tu(x, 0) = 0 when 0 ≤ x ≤ l .Show, e.g. by substituting u(x, t) = e−htv(x, t) , that
u(x, t) = e−ht
∞∑
1
An
cos βncos (ωnt− βn) sin
nπx
lwhere An = 2π
∫ π
0f( lx
π) sin nx dx , ωn =
√
(nπcl)2 − h2 and tan βn = h
ωn.(We see that the air resistan e damps the amplitude exponentially,redu es the frequen ies and auses a phase shift.)327. A string, �rmly �xed at x = 0 but elasti ally fastened at x = l, givesrise to an eigenvalue problem of the form
v′′(x) = λv(x) when 0 ≤ x ≤ l , v(0) = 0 , v′(l) + hv(l) = 0where h is a positive onstant.(a) Show that all eigenvalues are negative. Hint: Multiply the equa-tion by v(x) and integrate from 0 to l.(b) Show that λ is an eigenvalue if and only if µ =√−λ satis�es
µ = −h tanµl.328. Let λk be the eigenvalues of the problemv′′(x) − Bv(4)(x) = λv(x) , v(0) = v′′(0) = v(π) = v′′(π) = 0where B > 0 and prove the following statements.(a) λk < 0. Hint: Cf. Exer ise 327 (a).(b) λk = −k2(1+Bk2) , k = 1, 2, 3, . . . , and the orresponding eigen-fun tions are vk(x) = sin kx.329. The Diri hlet problem on the unit dis onsists in �nding a fun tion
u in the dis that satis�es the Lapla e equation ∆u = 0 and attains
33pres ribed values on the boundary of the dis . In polar oordinates theproblem an be written as∂2u
∂r2+
1
r
∂u
∂r+
1
r2∂2u
∂θ2= 0
u(1, θ) = f(θ)where f is a given 2π-periodi ontinuous fun tion of θ. The solution uis required to be C2 in the inner of the dis and ontinuous in the loseddis |x| ≤ 1. (For instan e, u may be interpreted as the stationarytemperature distribution in a ir ular dis that leaks heat only throughthe boundary r = 1, when the distribution on the boundary is givenby f.)(a) Let u(r, θ) =∞∑
−∞cn(r)e
inθ be the Fourier series expansion of u fora �xed r < 1. Show that the di�erential equation above meansthatc′′n(r) +
1
rc′n(r)−
n2
r2cn(r) = 0 , 0 < r < 1 ,for ea h n.(b) Show that cn(r) = Cnr
|n| where Cn is a onstant. (Hint: Substi-tute r = et. Use that u is ontinuous at the origin.)( ) Show that the boundary ondition u(1, θ) = f(θ) implies that Cnis the nth Fourier oe� ient of f .Hen e, a solution u of the problem must satisfyu(x) =
∞∑
−∞Cnr
|n|einθwhere Cn, n = 0 ,±1 , . . . , are the Fourier oe� ients of f , hen euniquely determined.(d) Conversely, show that this fun tion u really is a solution if f is C1.330. Assume that u is a harmoni fun tion (i.e. ∆u = 0) in the unit dis .Show that the mean value1
2π
∫ π
−π
u(r, θ) dθ,where (r, θ) are polar oordinates, is independent of r ∈ [0, 1) and equalto the value of u at the origin.Hint: Use 329 (a) and (b) for n = 0.
34331. Show that separation of variables in the eigenvalue problem∆u = λu in the re tangle 0 < x < a , 0 < y < b ,
u = 0 on the boundary of the re tangle ,leads to solutions of the formu(x, y) = sin
nx
asin
my
bwhereλ = −
(nπ
a
)2
−(mπ
b
)2
, n , m = 1, 2, 3, . . .are the orresponding eigenvalues.
35Exer ises for Chapter 4 The Fourier TransformThe De�nition and the Inversion formula*401. Compute the Fourier transform ofu(x) =
{
(1− |x|)2 , |x| ≤ 1
0 , |x| > 1 .In parti ular, spe ify u(0).*402. (a) Compute the Fourier transform ofu(x) =
{
sin x , 0 ≤ x ≤ 2π
0 , x /∈ [0, 2π] .(b) Show that |u(ξ)| ≤ 4.*403. (a) Determine the Fourier transform ofu(x) = max (1− |x|, 0) .(b) Use (a) to show that
1
π
∫ ∞
−∞
1− cos ξ
ξ2cosxξ dξ =
{
1− |x| , |x| ≤ 1
0 , |x| > 1 .( ) Compute, by means of (b),∫ ∞
−∞
sin2 t
t2dt .*404. The fun tion u is ontinuous and integrable in R. Show that the fol-lowing statements hold for the Fourier transform u.(a) u is even if u is even.(b) u is odd if u is odd.( ) u is real-valued if u(x) = u(−x) for all x ∈ R.(d) u is even and real-valued if u is even and real-valued.Also show that �if� may be repla ed by �if and only if� in these state-ments provided that u is integrable. (Cf. Exer ise 206.)
36Rules of Cal ulationWhen omputing Fourier transforms, it is essential to know the transforms ofsome simple fun tions. At least you should learn that the Fourier transformof e−|x| is 21+ξ2
and that the Fourier transform of e−x2/2 is√2πe−ξ2/2. Knowingthese transforms, it is easy to use the rule of s aling in order to derive thetransforms of e−a|x| and e−ax2 (a > 0). Furthermore, using the inversionformula, one �nds that 11+x2 has the transform πe−|ξ|.*405. Compute the Fourier transform of(a) u(x) = e−5|x| ,(b) u(x) = eix
1+2x2 ,( ) u(x) = xe−x2 .*406. Determine the ontinuous, integrable fun tion that has the Fouriertransform(a) u(ξ) = e2iξ−4|ξ| ,(b) u(ξ) = ξ e−(ξ−3)2 ,( ) u(ξ) = ξ2 e−ξ2/4 .The onvolution Theorem, the Plan herel formula*407. Compute the Fourier transform u of the following fun tion, whi h is a onvolution.u(x) =
∫ ∞
−∞e−|x−y|e−|y| dy .*408. Find a solution u of the onvolution equation
∫ ∞
−∞u(x− y)e−2y2 dy = e−x2
, x ∈ R .*409. State the Plan herel formula for the fun tion (1 + x2)−1 and use thisto evaluate∫ ∞
−∞
1
(1 + x2)2dx .
37Appli ations*410. Use a dire t al ulation to show thatE(x, t) =
1√4πt
e−x2/4tis a solution of the heat equation∂tu = ∂2
xuin the region t > 0, x ∈ R.411. Show that if u is a bounded solution of the heat equation∂tu = k ∂2
xu , t > 0 , x ∈ R ,with initial ondition u(x, 0) = φ(x) where φ is a bounded, ontinuousand odd fun tion, then u is an odd fun tion of x.412. (a) Show that the fun tion E1(x, t), de�ned byE1(x, t) = ∂x
(
1√4πt
e−x2/4t
)
,solves the heat equation∂tu = ∂2
xu , t > 0 , x ∈ R ,and satis�es the initial ondition u(x, 0) = 0, x ∈ R, in the sensethat for every �xed x ∈ R, E1(x, t) → 0 as t → 0, t > 0.(b) Plainly, also u = 0 satis�es the heat equation and equals zeroat t = 0. Both this fun tion and the fun tion E1 are, therefore,solutions of the initial-value problem (at least if the initial-value ondition is interpreted favourably) for the heat equation. Whydoes this not ontradi t the uniqueness in Theorem 4.15 ?Mixed Exer ises413. Compute the Fourier transform ofu(x) =
{
cos πx2, |x| ≤ 1
0 , |x| > 1 .Spe ify, in parti ular, u(ξ) when ξ = ±π2and show that
|u(ξ)| ≤ 4/π for all ξ ∈ R .
38414. Compute the Fourier transform ofu(x) =
{
eiω0x , a ≤ x ≤ a+ T
0 , x /∈ [a, a+ T ] .What happens when a = −T/2 ?415. Compute the Fourier transform of u(x) = max (1− x2, 0) .416. Compute the Fourier transform ofu(x) =
{
e−ax , x ≥ 0
ebx , x < 0where a and b are assumed to be positive, and show thate−|x| =
2
π
∫ ∞
0
cosxξ
1 + ξ2dξfor all real x.417. Determine the ontinuous, integrable fun tion that has the Fouriertransform |ξ|e−|ξ|.418. The fun tion u is ontinuous and integrable over R. Show the followingstatements.(a) If u is even, then
u(ξ) = 2
∫ ∞
0
u(x) cosxξ dx .(b) If u is odd, thenu(ξ) = −2i
∫ ∞
0
u(x) sinxξ dx .419. Let u be a ontinuous and integrable fun tion in R. Show that iff(x) = u(x) cos ax, then
f(ξ) =1
2(u(ξ − a) + u(ξ + a)) .420. Is there any ontinuous, integrable fun tion u in R su h that
u(ξ) = 1− sin ξ ?
39421. Let u and u1, u2, . . . be ontinuous, integrable fun tions in R su h that∫ ∞
−∞|un(x)− u(x)| dx → 0 as n → ∞ .Show that un → u uniformly in R.422. Set, when u is a bounded, integrable and ontinuous fun tion i R,
U(x1, x2) =
{
1π
∫∞−∞ u(t) x2
x22+(x1−t)2
dt , x2 > 0
u(x1) , x2 = 0 .(a) Show that U is bounded and ontinuous in the region x2 ≥ 0 .(b) Show that U is harmoni in x2 > 0 , i.e.∆u = (∂2
x1+ ∂2
x2)u = 0 .(Hint: Di�erentiate under the integral sign.)Remark. The fun tion U is alled the Poisson integral of u for theupper half-plane. For a given bounded, integrable and ontinuous fun -tion u in R, U is the (unique) bounded and harmoni fun tion in theupper half-plane that has ontinuous boundary values u .423. Let u be a ontinuous, integrable and bounded fun tion in R. Showthat
u(x) = limR→∞
1
2π
∫ R
−R
(1− |ξ|R) u(ξ) eixξ dξ .(If u is integrable, this is an almost trivial onsequen e of the inversionformula.)Hint:(a) Set
SR(x) =1
2π
∫ R
−R
(1− |ξ|R) u(ξ) eixξ dξand show that
SR(x) =1
2π
∫ ∞
−∞
(∫ R
−R
eiyξ(1− |ξ|R) dξ
)
u(x− y) dy .(b) Show, using methods in luding partial integration, that the innerintegral equals2
Ry2(1− cosRy) .
40 ( ) Then show thatSR(x) =
1
π
∫ ∞
−∞u(x− s
R)1− cos s
s2dsand that limR→∞ SR(x) may be evaluated by taking the limit un-der the integral sign.(d) Complete the proof using the fa t that
∫ ∞
−∞
1− cos s
s2ds =
∫ ∞
−∞
sin2 t
t2dt = π.(Cf. Exer ise 403.)424. Compute the Fourier transform of
1
x2 + 2ax+ a2 + s2, a, s > 0 ,using the fa t that we know that fs(x) = e−s|x| has the Fourier trans-form 2s
s2+ξ2.425. Compute the Fourier transform of(a) xe−|x| , (b) x
(1 + x2)2, ( ) x2
(1 + x2)2, (d) 1
(1 + x2)2.426. Determine the ontinuous and integrable fun tion u that satis�es
(F(u′))(ξ) = ξe−ξ2/2 .427. The Fourier transform of the fun tion u is u. Determine the Fouriertransform of u(x) sin2 2x.428. Determine the Fourier transform of(a) e−|x| cosx , (b) e−|x| sin |x| , ( ) 1
x4 + 4.429. Let u(x) = 0 when x < 0 and u(x) = xke−αx when x ≥ 0. Here α > 0and k is a positive integer. Determine the Fourier transform of u when(a) k = 1 , (b) k = 2 , ( ) k is arbitrary.430. Determine the Fourier transform of (a) e−x2−2x , (b) e−αx2−βx where
α > 0 and β ∈ R .
41431. Compute u ∗ u, when u(x) = 1π(1+x2)
, making use of the Fourier trans-form of 11+x2 .432. Determine a fun tion u su h that
∫ ∞
−∞u(x− y) u(y) dy = e−x2for all real x.433. Evaluate, for all real values of x, the integral
∫ ∞
−∞
dy
(1 + 4(x− y)2)(1 + y2).434. Find a solution of the integral equation
∫ ∞
−∞u(y) e−|x−y| dy = e−x2
, x ∈ R .435. Show that if u is a ontinuous, bounded and integrable fun tion su hthat∫ ∞
−∞|u(x+ h)− u(x)|2 dx ≤ Chα , 0 ≤ h ≤ 1 ,where α > 1, then u is integrable.Hint:(a) Set vh(x) = u(x+h)−u(x) and show that vh(ξ) = (eihξ−1)u(ξ) .(b) Show, using the Plan herel formula, that
4
∫ ∞
−∞sin2 hξ
2|u(ξ)|2 dξ ≤ 2π Chα , 0 ≤ h ≤ 1 .( ) Then let h = 2−n, where n is a positive integer, and show that
∫
2n−1<|ξ|<2n|u(ξ)|2 dξ ≤ C ′ 2−nα .(d) Make use of the Cau hy�S hwarz inequality to on lude that
∫
2n−1<|ξ|<2n|u(ξ)| dξ ≤ C ′′ 2n(1−α)/2 .(e) Show that it follows from this that u is integrable.
42436. Make use of the Fourier transform to �nd a solution of the ordinarydi�erential equationu′′(x) − u(x) = −e−|x| .Hint: Use 425 (d) to re over u from u .437. Solve the heat ondu tion problem∂tu = ∂2
xu , t > 0 , x ∈ R
u(x, 0) = e−x2
, x ∈ R.438. Solve the heat ondu tion problem∂tu = ∂2
xu , t > 0 , x ∈ R
u(x, 0) = E(x, s) =1√4πs
e−x2
4s , x ∈ Rwhere s > 0 is a �xed onstant. Interpret the result!439. Solve the heat ondu tion problem∂tu = ∂2
xu , t > 0 , x ∈ R
u(x, 0) = cos x , x ∈ R.440. Let u(x, t) be the solution of the heat ondu tion problem∂tu = ∂2
xu , t > 0 , x ∈ R
u(x, 0) = f(x) , x ∈ Rwhere f ∈ C0(R). Show that∫ ∞
−∞u(y, t) dy =
∫ ∞
−∞f(x) dx .What is the interpretation?
43Answers to the Exer isesSolutions to exer ises marked with an asterisk an be found after the answers.Answers for Chapter 1101. (a) x 6= 0 , 1 + 1x2 . (b) x < 0 , 1
1−ex.( ) x ≥ 0 , x
ex−1if x > 0, 0 if x = 0 . (d) |x| > 4, x
x−3+ x
x−4.102. 2s− a0103. (a) and (d) are onvergent, (b), ( ) and (e) are divergent.104. (a), (b) and (d) are onvergent, ( ) is divergent.105. (a) a > 1 . (b) a < 1 . ( ) a 6= 1 . (d) a ≤ 1 . (e) a < 1 .106. 1.109. (a), (b) and (d).111. (a) 3/4 . (b) 1/4.112. α > 0.113. Convergent but not absolutely onvergent.114. (a) E.g. N = 31 . (b) E.g. N = 1000.115. (a) Yes. (b) No.116. |z| < 1.117. (a) ||fn||E = fn(1/n) = 1/2 . Not uniform onvergen e.(b) ||fn||E = fn(1) = n
1+n2 . Uniform onvergen e.118. f(x) = −x. Uniform onvergen e.119. (a) No. (b) Yes. ( ) Yes.120. (a) 1 when 0 < x < π , 0 when x = 0 , x = π.(b) Not uniform onvergen e.121. (a) 0 . (b) Not uniform onvergen e.122. Uniform onvergen e.
44124. (a) x ≥ 0 . (b) x/(1 − e−x) if x 6= 0 , 0 if x = 0 . ( ) Not uniform onvergen e.128. (a) cc
(c+1)c+1 .131. Yes.133. s′(0) = 3.135. π/4.136. R = 1.139. (b) s(x) = 13e−x + 2
3ex/2 cos
√32x .141. (b) Yes.142. (a) s(x) = ln (1 + x+ x2) , |x| < 1 , s(1) = s(−1) = 0 .(b) not uniformly onvergent.( ) no.143. |a| < 3.145. (a) 0 < x < 2 .(b) s(x) is ontinuous.
45Answers for Chapter 2201. (b) c0 = 0 , cn = − in
if n 6= 0.202. (a) c0 =12, c2 = c−2 =
14, the others equal 0.(b) cn = 0 when n 6= ±1 , c1 = − i
2and c−1 = i
2.( ) c0 =
38, c2 = c−2 =
14, c4 = c−4 =
116, the others equal 0.(d) c0 =
18, c4 = c−4 = − 1
16, the others equal 0.203. 2−20
(
207
) .204. cn = (−1)n−1πn2 if n 6= 0, c0 = π
2.205. (a) 1
2+
∞∑
n=−∞n odd einx
iπn.(b) 1
2, 1
2.209. u(x) = c0 + c1e
ix + c−1e−ix where c0, c1 and c−1 are arbitrary omplexnumbers.210. u(x) = c1e
ix + c−1e−ix where c1 and c−1 are arbitrary.211. eix−1
(1−eix−1)2.213. (a) u(x) = eπ−e−π
2π
∑∞−∞
(−1)n
1+n2 einx .(b) −1
2+ π
2eπ+e−π
eπ−e−π .214. (b) (6− 2π)2 .215. (a)u(x) =
1
π+
1
4
(
eix + e−ix)
+∑
|n|≥2
cos πn2
π(1− n2)einx
=1
π+
1
2cos x+
2
π
∞∑
1
(−1)k
1− 4k2cos 2kx .(b) 1
2− π
4.( ) 1
2.217. u(x) = α sinπα
π
∞∑
−∞
(−1)n
α2−n2 einx .
46218. The integral equals π2.224. ||u− v|| = ||u+ v|| if and only if Re (u|v) = 0.227. π4
90.228. 3π
4.229. 1
2
(
π(e2π+1)e2π−1
− 1) .230. cn = n(−1)n
i(n2−1)if n 6= ±1 , c1 = i
4, c−1 = − i
4. The sum equals π2
12+ 1
16.231. π2
16− 1
2.232. (a) cn = −6i(−1)n
n3 if n 6= 0, c0 = 0.(b) π6
945.233. (a) e2πia−1
2πi(a−n).234. The onstant is 1.235. u(x) =
∞∑
0
1rn+1 e
inx . The integral equals 2πr2−1
.237. c2k−n = 2−n
(
nk
) when k = 0, 1, 2, . . . , n, cj = 0 for other valuesof j.238. a = π2
3, b = c = −2. The minimum value is 8π5
45− 16π .239. p(x) = 1
128(14e4ix − 28e2ix + 35− 28e−2ix + 14e−4ix) . The minimumvalue equals 65π214
= 65π16384
.240. w(x) = π2− 2
π(eix + e−ix) . The minimum value is π3
6− 16
π.241. (a) 0. (b) π2
4. ( ) π2
3. (d) π2
2.242. (a) u(x) = 1
2+ 2
π2
∞∑
−∞eπi(2k+1)x
(2k+1)2.(b) ∞
∑
−∞|cn|2 = 1
T
∫ T/2
−T/2|u(x)|2 dx.( ) ∞
∑
0
1(2k+1)4
= π4
96.
47243. (a) c0 = 0 , cn = i(−1)n
2πnwhen n 6= 0.(b) c0 =
T2, cn = iT
2πnwhen n 6= 0.244. (a) u(x) = π2
6−
∞∑
1
cos 2kxk2
.(b) u(x) = 8π
∞∑
0
sin (2k+1)x(2k+1)3
.245. (a) 2π− 4
π
∞∑
1
cos 2kx4k2−1
.(b) sin x . The sum of the osine series is − sin x when −π ≤ x < 0 orπ < x ≤ 2π.246. (a) π2− 4
π
∞∑
0
cos (2k+1)x(2k+1)2
.(b) 2∞∑
1
(−1)n+1
nsinnx .247. 3
4sin x − 1
4sin 3x . The integral equals 5π
16.248. u(x) =
∞∑
1
bn sinnx , bn =
{
(−1)k−12πk3
when n = 2k(−1)k
(2k+1)2− 4
π(2k+1)3when n = 2k + 1 .The sum equals π3
32.249. (a) −7π4
720. (b) π4
90. ( ) π8
9450.Remark: One might wonder if it is possible to he k whether the answerin ( ) is reasonable or not. Sin e 1 ≤
∑∞k=1
1k8
≤∑∞
k=11k4, the answermust lie somewhere between 1 and π4/90 (the answer in (b)). It an ome in handy to know of the approximation π2 ≈ 9.86 ≈ 10, with anerror less than 2%. Using this approximation, we get
π4
90≈ 100
90≈ 1.11 and π8
9450≈ 10000
9450≈ 1.06 ,whi h seems reasonable. It may also be worth to note that π2 ≤ 10and hen e these approximations yield upper bounds for the sums. Amore a urate al ulation shows that
π4
90≈ 1.082 and π8
9450≈ 1.00408 ,when e in the latter ase the �rst two terms 1 + 2−8 ≈ 1.0039 give avery good approximation of the sum of the series.
48250. (b) πx(π − x)
8.253. w(t) → 0 and w′(t) → − 1
12as t → 0.254. S2K+1(x) = 4
π
∑K0
sin (2k+1)x2k+1
, S2K+2 = S2K+1 , K > 0 ,
S3
(π
3
)
=2√3
π≈ 1.103 ,
S4
(π
4
)
=8√2
3π≈ 1.200 ,
S6
(π
6
)
=56
15π≈ 1.188 ,
S3
(
2π
3
)
=2√3
π≈ 1.103 ,
S4
(
2π
4
)
=8
3π≈ 0.849 ,
S6
(
2π
6
)
=8√3
5π≈ 0.882 .
49Answers for Chapter 3301. e−t sin x + 2e−9t sin 3x .302. ∞∑
0
4(−1)k
π(2k+1)2e−(2k+1)2t sin (2k + 1)x .303. 1
2
(
e−75t sin 5x − e−27t sin 3x) .304. (b) 1 + x
π+ π
2e−t sin x− 16
π
∞∑
k=1
k(4k2−1)2
e−4k2t sin 2kx .305. (a) T0
(
1− 4π
∞∑
k=0
12k+1
e−(2k+1)2π2at
l2 sin (2k + 1)πlx
) .(b) The ooking time is approximately l2
aπ2 ln163π. Note that it is pro-portional to the square of the thi kness and inversely proportionalto the thermal di�usivity.307. π2
6−
∞∑
1
e−4k2t cos 2kxk2
.308. 18(1− e−16t cos 4x).309. (a) u(x, t) = 10 .(b) u(x, t) = 10 − 8
π2
∞∑
0
1(2k+1)2
e−(2k+1)2π2t cos (2k + 1)πx .The limit equals 10 in both ases.310. e−9t sin 3x.311. 4
π
∞∑
0
(−1)k
(2k + 1)2e−(2k+1)22t sin (2k + 1)x.312. 8
π
∞∑
0
1
(2k + 1)3e−(2k+1)2t sin (2k + 1)x.313. (a) 3
4e−(1+a)t sin x − 1
4e−(9+a)t sin 3x . (b) a > −1.314. e−t cosx + te−4t cos 2x .315. 2π2
3− 4
∞∑
1
1
n2e−hn2t cosnx .
50316. 1
5sin 5πt sin πx + cos 10πt sin 2πx − 2
15πsin 15πt sin 3πx .317. (cos 2t+ 4
πsin 2t) sin x + 4
π
∞∑
1
1(2k+1)4
sin (2(2k + 1)t) sin (2k + 1)x .318. 2∞∑
1
(−1)n+1
n2sin nt sinnx .319. (a) 4lv
π2c
∞∑
1
1
n2sin
nπb
2lsin
nπa
lsin
nπct
lsin
nπx
l.(b) a = l
2( ) a = k · l7, k = 1, 2, . . . , 6, i.e. at a node point of the eigenmodethat orresponds to the 6th overtone.321. x2 + 4t2 + xt .322. u(x, t) equals
1
2
(
e−(x+ct)2 + e−(x−ct)2 +1
c(arctan (x+ ct) − arctan (x− ct))
)
.324. u(x, t) is equal to∞∑
0
Ak cos ((2k + 1)ct) sin (2k + 1)xwhere Ak = 4π
∫ π/2
0f(x) sin (2k + 1)x dx .325. u(x, t) is the fun tion
gx(x− l)
2c2+
∞∑
0
Ak cos(2k + 1)πct
lsin
(2k + 1)πx
lwhere Ak =2l
5π2
(−1)k
(2k + 1)2+
4gl2
c2π3(2k + 1)3.
51Answers for Chapter 4401.u(ξ) =
{
4ξ3(ξ − sin ξ) when ξ 6= 0
23
when ξ = 0 .402.u(ξ) =
{
1−e−i2πξ
1−ξ2when ξ 6= ±1
∓iπ when ξ = ±1 .403. (a)u(ξ) =
{
2(1−cos ξ)ξ2
when ξ 6= 0
1 when ξ = 0 .(b) See the solution.( ) See the solution.404. See the solution.405. (a) u(ξ) = 1025+ξ2
.(b) u(ξ) = π√2e−|ξ−1|/
√2 .( ) u(ξ) = −1
2iξ√π e−ξ2/4 .406. (a) u(x) = 1
2πˆu(−x) = 4
π(x2 + 4x+ 20)−1 .(b) u(x) = ix+6
4√πe3ix−
x2
4 .( ) u(x) = 2√π(1− 2x2) e−x2 .407. u(ξ) =
(
21+ξ2
)2 .408. 2√πe−2x2 .409. π
2.410. See the solution.411. See the remark after Example 4.20.412. (a) E1(x, t) = ∂xE(x, t) and so, for t > 0,
(∂t − ∂2x)E1 = (∂t − ∂2
x)∂xE = ∂x(∂t − ∂2x)E = 0 ,be ause (∂t − ∂2
x)E = 0 when t > 0 by Exer ise 410.(b) On the urves where x2/t is onstant, −E1 in reases unboundedlyas t → +0. Hen e, the fun tion E1 is not a bounded solution.
52413. 4π cos ξ
π2 − 4ξ2when ξ 6= ±π
2, u(π/2) = u(−π/2) = 1 .414. 1
i(ω0−ξ)ei(ω0−ξ)a (ei(ω0−ξ)T − 1) = 2
ω0−ξei(ω0−ξ)(a+T/2) sin (ω0−ξ)
2T .415. 4
ξ3(sin ξ − ξ cos ξ) when ξ 6= 0 , u(0) = 4/3.416. a+ b
(a+ iξ)(b− iξ).417. 1
π
1− x2
(1 + x2)2.420. No.424. π
se−s|ξ|+iξa .425. (a) −4iξ
(1 + ξ2)2.(b) −iπξ
2e−|ξ| .( ) π
2
d
dξ(ξe−|ξ|) =
π
2(1− |ξ|) e−|ξ| .(d) π
2(1 + |ξ|) e−|ξ| .426. −i√
2πe−ξ2/2 .427. u(ξ)
2− 1
4(u(ξ − 4) + u(ξ + 4)) .428. (a) 2(ξ2 + 2)
ξ4 + 4. (b) 2(2− ξ2)
ξ4 + 4. ( ) π
4(cos ξ + sin |ξ|) .429. (a) 1
(α + iξ)2.(b) 2
(α + iξ)3.( ) (i d
dξ
)k−11
(α + iξ)2=
k!
(α + iξ)k+1.
53430. (a) √π e−(ξ−2i)2/4 . (b) √π
αe−(ξ−iβ)2/4α .431. 2
π(4 + x2).432. π−1/4 21/2 e−2x2
.433. 3π
9 + 4x2.434. 3− 4x2
2e−x2
.436. 1 + |x|2
e−|x| .437. (4t+ 1)−1/2 e−x2/(4t+1) .438. u(x, t) = 1√4π(t+s)
e−x2/(4(t+s)) = E(x, t + s) .439. e−t cosx.
55Solutions to Exer ises Marked with an AsteriskSolutions for Chapter 1105. (b) We begin by applying the ratio test.(k + 1)aak+1
kaak=
(
1 +1
k
)a
a → a as k → ∞ .This shows that the series is onvergent when a < 1 and divergentwhen a > 1. When a = 1, the test gives no information. However,for this value of a the series is ∞∑
1
k whi h is divergent be ause theterms do not tend to zero.Answer: The series is onvergent exa tly when a < 1.120. (a) When 0 < x < π, sin x > 0, and then we have that (sin x)1/n → 1as n → ∞. When x = 0 or x = π, sin x = 0, and then the limitof (sin x)1/n is 0 as n → ∞. In summary,limn→∞
=
{
1 when 0 < x < π
0 when x = 0 or x = π .(b) Sin e all the fun tions fn are ontinuous in the interval [0, π] andthe limit fun tion is not, the onvergen e annot be uniform in[0, π].Answer: (a) The limit fun tion equals 1 when 0 < x < π and equals 0when x = 0 or x = π . (b) The onvergen e is not uniform.123. (a) For x > 0 we have that
s(x) =x
1 + x
∞∑
k=0
(
1
1 + x
)k
=x
1 + x· 1
1− 11+x
= 1(it is a geometri series with the ratio 11+x
and 0 < 11+x
< 1).When x = 0, all the terms are zero and so s(x) = 0. Hen e, theseries is onvergent for all x ≥ 0.(b) When x ≥ d > 0, we have that0 <
x
(1 + x)k+1=
x
1 + x
1
(1 + x)k<
1
(1 + x)k≤(
1
1 + d
)k
.
56 The series∞∑
k=0
(
1
1 + d
)kis a onvergent geometri series. Hen e, by the Weierstrass M-test,the given series is uniformly onvergent in the interval x ≥ d.( ) The terms x(1+x)k+1 are ontinuous fun tions of x in the interval
x ≥ 0. Hen e, if the onvergen e were uniform, then the sums(x) would also be ontinuous in this interval. That is not the ase be ause, a ording to (a), s(x) = 1 when x > 0 but s(0) = 0.Hen e, the series is not uniformly onvergent in the interval x ≥ 0.Alternative solution: The nth partial sum sn(x) of the series maybe al ulated using the formula for a geometri sum. When x > 0, wehave
sn(x) =n∑
k=0
x
(1 + x)k+1= 1 −
(
1
1 + x
)n+1
,whereas sn(0) = 0. With s(x) from (a) above, we get that|sn(x)− s(x)| =
{
(
11+x
)n+1 when x > 0
0 when x = 0 .This yields thatsupx≥d
|sn(x)− s(x)| =(
1
1 + d
)n+1
→ 0 as n → ∞whilesupx≥0
|sn(x)− s(x)| = 1 6= 0 for all n .Hen e, a ording to the de�nition, the series is uniformly onvergentin the interval x ≥ d but not in x ≥ 0.126. We have that∣
∣
∣
∣
∣
xk2
k2
∣
∣
∣
∣
∣
≤ 1
k2when |x| ≤ 1 .Sin e ∞
∑
1
k−2 is onvergent, the fun tion series ∞∑
1
xk2/k2 is uniformly onvergent in |x| ≤ 1 by the Weierstrass M-test.
57130. In order to show that the sum of the series is ontinuous when x > 0,it is enough to show that the series is uniformly onvergent in ea hinterval [d,D] where d > 0. In su h an interval is0 <
1 +√kx2
1 + k2√x
≤ 1 +√kD2
k2√d
≤√k +
√kD2
k2√d
≤ 1 +D2
√d
· 1
k3/2.The uniform onvergen e therefore follows from the Weierstrass M-testand the onvergen e of ∑∞
1 k−3/2 .136. It follows fromf(x) = a0 + a1x+ · · ·+ akx
k + · · ·+ a2kx2k + a2k+1x
2k+1 + · · ·thatf(x2) = a0 + a1x
2 + · · ·+ akx2k + · · ·and hen e,
(1− x)f(x2) = a0 − a0x+ a1x2 − a1x
3 + · · ·+ akx2k − akx
2k+1 + · · · .The ondition f(x) = (1− x)f(x2) therefore means thata2k = ak , a2k+1 = −ak when k ≥ 0 .Furthermore, a0 = f(0) = 1 and hen e
a1 = −a0 = −1, a2 = a1 = −1, a3 = −a1 = 1, a4 = a2 = −1, . . . .Therefore ak = 1 or ak = −1 for all k ≥ 0, when e |ak| = 1 and so∣
∣
∣
∣
ak+1xk+1
akxk
∣
∣
∣
∣
= |x| → |x| as k → ∞ .By the ratio test, the series is therefore onvergent when |x| < 1 anddivergent when |x| > 1, i.e. the radius of onvergen e is 1.Answer: The radius of onvergen e is 1.Remark: It an be shown that the sequen e ak+1
akhas no limit. The�rst twelve terms of the sequen e are
−1 , 1 , −1 , −1 , −1 , 1 , −1 , 1 , −1 , 1 , −1 , −1 .
58Solutions for Chapter 2201. (a) See the �gure.x/π
y/π
1 2 3-1-2-3 1
(b) If n 6= 0, then we havecn =
1
2π
∫ 2π
0
(π − x) e−inx dx
=1
2π
[
(π − x)e−inx
−in
]2π
0
− 1
2π
∫ 2π
0
(−1)e−inx
−indx
=1
2π
( π
in+
π
in
)
+ 0 = − i
n,and c0 = 1
2π
∫ 2π
0(π − x) dx = 0, whi h an be seen dire tly fromthe �gure.202. (a) We wish to determine cn = 1
2π
∫ π
−πcos2 x e−inx dx for ea h n. It isnot parti ularly hard to evaluate these integrals dire tly but it iseven simpler to rewrite cos2 x using Euler's formulae:
cos2 x =
(
eix + e−ix
2
)2
=e2ix + e−2ix + 2
4
=1
4e−2ix +
1
2+
1
4e2ix .Hen e c−2 = 1
4, c0 = 1
2, c2 = 1
4and all other oe� ients are 0.This follows from Theorem 2.2 that says that the oe� ients cn ofa uniformly onvergent series∑∞
−∞ cneinx are equal to the Fourier oe� ients of the sum of that series.
59This is in parti ular true for a �nite sum s(x) =M∑
k=−M
ckeikx sin ethe partial sums sN (x) = N
∑
k=−N
ckeikx (where ck = 0 when |k| > M)are identi ally equal to s(x), and hen e ||sN − s||[0,2π] = 0, when
N > M . Therefore, ||sN − s||[0,2π] → 0 as N → ∞ and this showsthat the onvergen e is uniform.209. By (2.28), the Fourier oe� ients of u(x+ π2) and u(x− π
2) are given by
einπ/2cn and e−inπ/2cn , respe tively. From Theorem 2.10 it follows thatthe Fourier oe� ients of u′ are incn. Hen e the ondition imposed onu implies that
2incn = einπ/2cn − e−inπ/2cn .We get thatncn = cn sin
nπ
2or, equivalently, thatcn
(
n− sinnπ
2
)
= 0 .When |n| > 1, |n − sin nπ2| ≥ |n| − 1 > 0, and hen e cn = 0. When
n = 1, n − sin nπ2
= 1 − sin π2
= 0. Therefore c1 may be hosenarbitrarily. In the same way we see that c0 and c−1 are arbitrary be ausen − sin nπ
2= 0 when n = 0 or n = −1. In on lusion we see that umust be of the form
u(x) = c0 + c1eix + c−1e
−ix ,where c0, c1 and c−1 are omplex numbers.Conversely, it is easy to he k su h fun tions u satisfy the ondition.213. The Fourier oe� ients are equal tocn =
1
2π
∫ π
−π
1
2
(
ex + e−x)
e−inx dx =1
4π
[
ex(1−in)
1− in+
e−x(1+in)
−(1 + in)
]π
−π
=1
4π
(
eπ(−1)n
1− in− e−π(−1)n
1− in− eπ(−1)n
1 + in+
eπ(−1)n
1 + in
)
=(−1)n
4π
(
eπ(
1
1− in+
1
1 + in
)
− e−π
(
1
1− in+
1
1 + in
))
=(−1)n
4π· 2
1− (in)2(
eπ − e−π)
=(−1)n
2π(1 + n2)
(
eπ − e−π)
.
60 Sin e u is C1, the Fourier inversion formula is valid for u, i.e.u(x) =
eπ − e−π
2π
∞∑
−∞
(−1)n
1 + n2einx .In parti ular, when x = π we have that
u(π) =eπ − e−π
2π
∞∑
−∞
(−1)n
1 + n2(−1)n =
eπ − e−π
2π
∞∑
−∞
1
n2 + 1.Re alling that u(π) = eπ+e−π
2, we see that
∞∑
−∞
1
n2 + 1= π
eπ + e−π
eπ − e−π.Sin e the terms orresponding to ±n are equal, we have that
∞∑
−∞
1
n2 + 1= 1 + 2
∞∑
1
1
n2 + 1and hen e ∞∑
1
1
n2 + 1= −1
2+
π
2
eπ + e−π
eπ − e−π.(Using the hyperboli fun tions sinh x = ex−e−x
2, cosh x = ex+e−x
2and
cothx = cosh xsinhx
we an also write the answer as∞∑
1
1
n2 + 1= −1
2+
π
2coth π .)Alternative omputation: For the fun tion u(x) = ex+e−x
2= cosh x wehave that u′(x) = ex−e−x
2= sinh x and u′′(x) = cosh x = u(x). (Inparti ular, v(x) = sinh x is a primitive fun tion of u(x) and vi e versa.In addition, u is an even fun tion (u(−x) = u(x)) and v is an oddfun tion (v(−x) = −v(x)).) These properties are easy to verify alsowithout prior knowledge of the hyperboli fun tions.Sin e u is even, we have by symmetry that
2πcn =
∫ π
−π
u(x)(cosnx− i sinnx) dx =
∫ π
−π
cosh x cosnx dx.
61We an now ompute the left-hand side, whi h we denote by In, bymeans of two partial integrations. We have in fa t thatIn = [sinh x cosnx]π−π −
∫ π
−π
sinh x (−n sin nx) dx
= 2 sinh π cosnπ + [cosh x(n sin nx)]π−π − n2Infrom whi h it follows that(1 + n2)In = 2 sinhπ (−1)n + 0 .Hen e cn = In
2π= sinhπ
π(−1)n
1+n2 whi h is the expression obtained above.227. As in Example 2.4 we �nd that c0 = π2
3and cn = 2(−1)n
n2 when n 6= 0.Parseval's formula yields∞∑
−∞|cn|2 =
(
π2
3
)2
+∑
n 6=0
4
n4=
1
2π
∫ π
−π
x4 dx =1
π
∫ π
0
x4 dx =π4
5and hen e2
∞∑
1
4
n4=
π4
5− π4
9=
4π4
45,i.e. ∞
∑
1
1n4 = π4
90.238. The minimum value is attained for a = c0, b = c1 and c = c−1 where cnare the Fourier oe� ients of u(x) = x2, |x| ≤ π. From the omputationin Example 2.4 we have that c0 = π2
3, c1 = c−1 = −2 .From (2.24) and the pre eeding lines it follows that
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
u−N∑
−N
cnen
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
2
= ||u||2 −∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
N∑
−N
cnen
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
2
= ||u||2 −N∑
−N
|cn|2and hen e the minimum value is∫ π
−π
x4 dx− 2π(|c0|2 + |c1|2 + |c−1|2)
= 2
[
x5
5
]π
0
− 2π(π4
9+ 4 + 4) =
8π5
45− 16π .
62242. (a) Using the hange of variables x = Ty2π
we getcn =
1
2π
∫ 2π
0
v(y)e−iny dy =1
2π
∫ 2π
0
u
(
Ty
2π
)
e−iny dy
=1
T
∫ T
0
u(x) e−2πinx/T dx .(b) The fun tion v(x) = u(
Tx2π
) satis�es the assumptions of the inver-sion formula. Hen eu
(
Tx
2π
)
=
∞∑
−∞cne
inx for all x ,i.e.u(y) =
∞∑
−∞cne
2πiny/T .( ) The fun tion v(x) = u(
xπ
) has period 2π and the Fourier oe�- ients of u are by the de�nition equal to those of v. Hen ecn(u) =
1
2π
∫ π
−π
v(x) e−inx dx =1
2π
∫ π
−π
u(x
π
)
e−inx dx
=1
2π
∫ π
−π
(
1− |x|π
)
e−inx dx
=1
2π
∫ π
−π
(
1− |x|π
)
(cosnx− i sin nx) dx
=1
π
∫ π
0
(
1− x
π
)
cosnx dx
=1
π
([
(
1− x
π
) sinnx
n
]π
0
−∫ π
0
(
−1
π
)
sin nx
n
)
= 0 +1
π2
∫ π
0
sinnx
ndx =
1
n2
[
−cos nx
n2
]π
0
=−(−1)n + 1
n2π2.Here we assumed that n 6= 0. When n = 0 we have
cn = c0 =1
π
∫ π
0
(1− x
π) dx =
1
π
[
1
2(1− x
π)2(−π)
]π
0
=1
2.
63(Those who �nd the onstru tion of the primitive fun tion too ompli ated to be done in one sweep an �rst make the hange ofvariables y = 1− x/π (or x = π(1− y)) and getcn = −
∫ 0
1
y dy =
[
y2
2
]1
0
=1
2.)In on lusion we have that c0 = 1
2, cn = 0 if n 6= 0 is even, and
cn = 2π2(2k+1)2
if n = 2k + 1. Hen eu(x) = v(πx) =
∞∑
−∞cne
πinx =1
2+
2
π2
∞∑
−∞
eπi(2k+1)x
(2k + 1)2.When the method above is used there is no need to know theformula
cn(u) =1
T
∫ T
0
u(x) e−2πinx/T dxfor the Fourier oe� ients of a T -periodi fun tion by heart.(d) With v(x) = u(
Tx2π
) we have∞∑
−∞|cn(u)|2 =
∞∑
−∞|cn(v)|2 =
1
2π
∫ π
−π
|v(x)|2 dx
=1
2π
∫ π
−π
|u(Tx2π
)|2 dx =1
2π
∫ T2
−T2
|u(y)|2 2πT
dy
=1
T
∫ T2
−T2
|u(y)|2 dy .(e) By ( ) and (d) we have that1
4+
4
π4
∞∑
k=−∞
1
(2k + 1)4=
1
2
∫ 1
−1
(1− |y|)2 dy
=
[
−1
3(1− y)3
]1
0
=1
3.Hen e
4
π4
∞∑
k=−∞
1
(2k + 1)4=
8
π4
∞∑
k=0
1
(2k + 1)4=
1
3− 1
4=
1
12and therefore ∞∑
k=0
1
(2k + 1)4=
π4
96.
64244. (a) Sin e u is ontinuous and pie ewise C1 in the interval [0, π] wehave, a ording to Theorem 2.23, thatu(x) =
1
2a0 +
∞∑
1
an cosnxwherean =
2
π
∫ π
0
u(x) cosnx dx .We geta0 =
2
π
∫ π
0
u(x) dx =2
π
∫ π
0
x(π − x) dx =2
π
[
πx2
2− x3
3
]π
0
=2
π
(
π3
2− π3
3
)
=π3
3and when n 6= 0, iterated partial integration yieldsan =
2
π
∫ π
0
(πx− x2) cosnx dx
=2
π
[
(πx− x2)sinnx
n
]π
0
− 2
π
∫ π
0
(π − 2x)sinnx
ndx
= 0− 2
π
[
(π − 2x)cosnx
−n2
]π
0
+2
π
∫ π
0
(−2)cosnx
−n2dx
= 2cosnπ
−n2+ 2
cos 0
−n2+
4
πn3[sinnx]π0 = −2
(−1)n + 1
n2
=
{
0 when n is odd− 1
k2when n = 2k .Thus we have that
u(x) =1
2a0 +
∞∑
1
an cosnx =π2
6−
∞∑
1
cos 2kx
k2and hen e, when 0 ≤ x ≤ π, thatx(π − x) =
π2
6−
∞∑
1
cos 2kx
k2.
65(b) Sin e u(0) = u(π) = 0 we an expand u in a sine series. A ordingto (2.50) and (2.48) we haveu(x) =
∞∑
1
bn sinnxwherebn =
2
π
∫ π
0
u(x) sinnx dx .Integrating by parts, we �nd that (note that n is always positive)bn =
2
π
∫ π
0
x(π − x) sin nx dx
=2
π
[
x(π − x)cosnx
−n
]π
0
− 2
π
∫ π
0
(π − 2x)cos nx
−ndx
= 0 − 2
π
[
(π − 2x)sin nx
−n2
]π
0
+2
π
∫ π
0
(−2)sinnx
−n2dx
= −4
π
[cosnx
n3
]π
0= −4((−1)n − 1)
πn3
=
{
0 when n is even8
π(2k+1)3when n = 2k + 1 .Hen e
u(x) =∞∑
k=0
8
π(2k + 1)3sin (2k + 1)xand therefore, when 0 ≤ x ≤ π ,
x(π − x) =8
π
∞∑
0
sin (2k + 1)x
(2k + 1)3.Remark: In Exer ise 244 we used �ready-made formulae� for the osineseries and sine series expansions and for the oe� ients an and bn.Those who do not want to burden their memory with these formulaemanage �ne doing the simple derivation every time. We illustrate thisfor (b):The graph of the extended 2π-periodi fun tion appears in the �gure (inwhi h the x-axis is graded in multiples of π and the y-axis in multiplesof u(π
2) = π2
4):
66
(Here u(x) = x(π + x) when −π ≤ x ≤ 0.) The Fourier oe� ients ofthe extended fun tion arecn =
1
2π
∫ π
−π
u(x) e−inx dx
=1
2π
(∫ 0
−π
u(x)e−inx dx +
∫ π
0
u(x)e−inx dx
)
=1
2π
∫ π
0
(u(−x)einx + u(x)e−inx) dx
=1
2π
∫ π
0
u(x)(−einx + e−inx) dx
= − i
π
∫ π
0
u(x) sinnx dx = − i
π
∫ π
0
x(π − x) sinnx dxwhi h are equal to 2i((−1)n−1)πn3 when n 6= 0, as was shown above, andfurther c0 = 0. When 0 ≤ x ≤ π, we therefore get that
x(π − x) =∑
n 6=0
cneinx =
∞∑
1
(cneinx + c−ne
−inx)
=
∞∑
1
2icn sinnx =8
π
∞∑
0
sin (2k + 1)x
(2k + 1)3using that c−n = −cn and that c2k = 0.Another alternative is to use our knowledge of (�nite-dimensional) lin-ear algebra in order to re onstru t the formula for the expansion of u.If U is a ve tor in RN and V1, . . . , VN is an orthogonal basis, thenU =
(U, V1)
(V1, V1)V1 + · · ·+ (U, VN)
(VN , VN)VN .
67(If the basis elements are normalised, then all the denominators areequal to 1.) We know that the fun tions vk(x) = sin kx, k = 1, 2, 3, . . . ,form a omplete orthogonal set, whi h is one of the possible generali-sations of the on ept of orthogonal basis in a �nite-dimensional spa e,for the pie ewise ontinuous fun tions with respe t to the s alar prod-u t (u, w) = ∫ π
0u(x)w(x) dx. In analogy with the formula for U above,we haveu(x) =
∞∑
k=1
(u, vk)
(vk, vk)vk(x) =
∞∑
k=1
bk sin kxwherebk =
(u, vk)
(vk, vk)=
∫ π
0u(x) sin kx dx∫ π
0sin2 kx dx
=2
π
∫ π
0
u(x) sin kx dx .In the last step, we used that ∫ π
0sin2 kx dx = π
2.
68Solutions for Chapter 3301. A ording to Theorem 3.4,u(x, t) =
∞∑
1
bne−n2t sinnxis a solution of the problem. Here bn are the oe� ients of the expansionin sine series of the initial values u(x, 0) = f(x) = sin x+ 2 sin 3x. Sin ethe oe� ients are uniquely determined, b1 = 1, b3 = 2 and all the other oe� ients are zero. Hen e,
u(x, t) = e−t sin x + 2e−9t sin 3x .(The theorem also guarantees that there are no other solutions.)Remark: In this ase the initial values f were already given as a(�nite) sine series and so we were able to read o� the oe� ients bn.In general, a al ulation is needed in order to �nd these oe� ients.307. A ording to Theorem 3.5, we an immediately write down the solutionas soon as we have expanded u(x, 0) = x(π − x) in a osine series. Asin Exer ise 244 (a) we getx(π − x) =
π2
6−
∞∑
k=1
cos 2kx
k2when 0 ≤ x ≤ π .(This is where the a tual work is done. Confer the remark to thesolution to Exer ise 301 above.) Hen e,
u(x, t) =π2
6−
∞∑
k=1
e−4k2t cos 2kx
k2is a solution of the problem.310. This time we get the solution using Theorem 3.6:u(x, t) =
∞∑
0
bke−(2k+1)2t sin (2k + 1)xwhere bk are the oe� ients of the expansion
u(x, 0) =
∞∑
0
bk sin (2k + 1)x , 0 ≤ x ≤ π
2.
69(Cf. Theorem 2.27.) In our ase, u(x, 0) = sin 3x is already of this formwith b1 = 1 and bk = 0, k 6= 1, and so we have no use of the formulabk = 4
π
∫ π
0u(x, 0) sin (2k + 1)x dx . Hen e, the solution is
u(x, t) = e−9t sin 3x .316. Using the notation of (3.34)�(3.36), we have c = 5 and l = 1. Set, asin (3.38), U(x, t) = u(
xπ, t5π
) . Then U solves the equation ∂2t U = ∂2
xUand satis�esU(x, 0) = f
(x
π
)
= sin 2x ,
∂tU(x, 0) =1
5πg(x
π
)
=1
5π(sin x− 2 sin 3x) .These initial values are already expressed as sine series, and we get im-mediately (using the notation of Theorem 3.9) that A2 = 1, B1 = 1
5π,
B3 = − 25π
and that all other oe� ients An and Bn are zero. Thetheorem now yields thatU(x, t) =
1
5πsin t sin x + cos 2t sin 2x − 2
15πsin 3t sin 3xand hen e
u(x, t) = U(πx, 5πt)
=1
5πsin 5πt sin x + cos 10πt sin 2πx − 2
15πsin 15πt sin 3πx .
70Solutions for Chapter 4401. We have thatu(ξ) =
∫ 1
−1
(1− |x|)2e−ixξ dx =
∫ 1
0
(1− |x|)2(e−ixξ + eixξ) dx
=
∫ 1
0
(1− 2x+ x2) 2 cosxξ dx .When ξ 6= 0, we integrate by parts and getu(ξ) =
[
(1− 2x+ x2) 2sin xξ
ξ
]1
0
−∫ 1
0
(−2 + 2x) 2sin xξ
ξdx
= 0 − 0 − 4
[
(x− 1)− cosxξ
ξ2
]1
0
− 4
∫ 1
0
cos xξ
ξ2dx
=4
ξ2− 4
[
sin xξ
ξ3
]1
0
=4
ξ2− 4 sinxξ
ξ3=
4
ξ3(ξ − sin ξ) .Furthermore, u(0) = 2
∫ 1
0(1− x)2 dx = 2
[
− (1−x)3
3
]1
0= 2/3.Alternatively, we an evaluate u(0) using that u is ontinuous:
u(0) = limξ→0
u(ξ) = limξ→0
4
ξ3(ξ − sin ξ)
= limξ→0
4
ξ3(ξ − ξ + ξ3/6 + ξ5B5(ξ)) = lim
ξ→0(2
3+ ξ2B5(ξ)).Here B5(ξ) is bounded in a neighbourhood of ξ = 0 and therefore
u(0) = 2/3.402. (a)u(ξ) =
∫ 2π
0
sin x e−ixξ dx =
∫ 2π
0
eix − e−ix
2ie−ixξ dx
=1
2i
∫ 2π
0
(
eix(1−ξ) − e−ix(1+ξ))
dx .When ξ 6= ±1, we have thatu(ξ) =
1
2i
[
eix(1−ξ)
i(1− ξ)− e−ix(1+ξ)
−i(1 + ξ)
]2π
0
=1
2i
(
ei2π(1−ξ) − 1
i(1− ξ)+
e−i2π(1+ξ) − 1
i(1 + ξ)
)
= −1
2
(
e−i2πξ − 1
1− ξ+
e−i2πξ − 1
1 + ξ
)
=1− e−i2πξ
1− ξ2.
71In the next-to-last equality was used that e±i2π = 1. Furthermore,u(1) =
1
2i
∫ 2π
0
(1− e−2ix) dx =1
2i
(
2π −[
e−2ix
−2i
]2π
0
)
=π
i= −iπ .We obtain in the same way that u(−1) = iπ. (Che k this byevaluating limξ→−1 u(ξ) !)(b) Sin e
|u(ξ)| ≤∫ ∞
−∞|u(x)| dx ,we get that
|u(ξ)| ≤∫ 2π
0
| sin x| dx = 2
∫ π
0
sin x dx = 4 .403. Sin e1− |x| ≥ 0 ⇐⇒ |x| ≤ 1 ,we have that
max (1− |x|, 0) =
{
1− |x| when |x| ≤ 1
0 when |x| > 1 .(a) Sin e u is an even fun tion, we get (when ξ 6= 0) thatu(ξ) =
∫ ∞
−∞u(x)(cosxξ − i sin xξ) dx = 2
∫ 1
0
(1− x) cos xξ dx
= 2
[
(1− x)sin xξ
ξ
]1
0
+ 2
∫ 1
0
sin xξ
ξdx
= 2(0− 0) + 2
[
−cos xξ
ξ2
]1
0
= − 2
ξ2(cos ξ − 1)
=2(1− cos ξ)
ξ2.Finally we have that u(0) =
∫ 1
−1u(x) dx = 1, sin e the integralexpresses the area of a triangle with base 2 and height 1.(b) The fun tion u(x) is ontinuous, bounded and integrable over R.Furthermore, u is integrable over R, sin e u(ξ) is ontinuous and
|u(ξ)| ≤ 2(1 + | cos ξ|)/ξ2 ≤ 4/ξ2
72 where the right-hand side is integrable over |ξ| ≥ 1. Consequently,the assumptions of the inversion formula are ful�lled and we getu(x) =
1
2π
∫ ∞
−∞u(ξ) eixξ dξ =
1
2π
∫ ∞
−∞
2(1− cos ξ)
ξ2eixξ dξ
=1
π
∫ ∞
−∞
1− cos ξ
ξ2cos xξ dξ +
i
π
∫ ∞
−∞
1− cos ξ
ξ2sin xξ dξ .The integrand of the se ond integral is odd, hen e that integral iszero. Therefore we have
1
π
∫ ∞
−∞
1− cos ξ
ξ2cosxξ dξ = u(x) =
{
1− |x| , |x| ≤ 1
0 , |x| > 1 .( ) Setting x = 0 in (b), we get1 =
1
π
∫ ∞
−∞
1− cos ξ
ξ2dξ =
1
π
∫ ∞
−∞
2 sin2 ξ2
ξ2dξ
=1
π
∫ ∞
−∞
2 sin2 t
(2t)22dt =
1
π
∫ ∞
−∞
sin2 t
t2dt .Hen e
∫ ∞
−∞
sin2 t
t2dt = π.404. (a) We have that
u(−ξ) =
∫ ∞
−∞u(x) e−ix(−ξ) dx =
∫ ∞
−∞u(−y) e−iyξ dy .In the last equality we made the hange of variables x = −y.Using that u is even, we get that
u(−ξ) =
∫ ∞
−∞u(y)e−iyξ dy = u(ξ) ,i.e. u is even.(b) Analogously, when u is odd we get
u(−ξ) =
∫ ∞
−∞u(−y)e−iyξ dy = −
∫ ∞
−∞u(y)e−iyξ dy = −u(ξ) ,i.e. also u is odd.
73405. (a) With f(x) = e−|x| we have u(x) = f(5x) and by 5. in Theorem 4.7we getu(ξ) =
1
5f(
ξ
5) =
1
5
2
1 + ( ξ5)2
=10
25 + ξ2.(b) Set g(x) = 1
1+x2 . Sin e e−|x| has the Fourier transform 21+ξ2
, theFourier transform of 21+ξ2
is 2πe−|−x| by the inversion formula. By hanging the roles of x and ξ, we get thatg(ξ) =
1
22π e−|ξ| = π e−|ξ|.Now, u(x) = eix · g(x
√2) and hen e, by 4. and 5. in Theorem 4.7,
u(ξ) =1√2g(ξ − 1√
2) =
π√2e−|ξ−1|/
√2 .( ) With h(x) = e−x2 we have that −iu(x) = −ixh(x). Hen e, by 2.in Theorem 4.7,
−iu(ξ) =d
dξh(ξ) =
d
dξ
√π e−ξ2/4 =
√π (−ξ
2)e−ξ2/4 .We an also solve this exer ise by observing that u(x) = −1
2ddxe−x2and using 1. in Theorem 4.7:
u(ξ) = −1
2iξ√π e−ξ2/4 .406. (a) Sin e the Fourier transform of e−|ξ| is 2
1+x2 , 4. and 5. in Theo-rem 4.7 yieldˆu(x) =
1
4· 2
1 + (x−24)2
=8
16 + (x− 2)2=
8
x2 − 4x+ 20.Hen e, it follows from the inversion formula that
u(x) =1
2πˆu(−x) =
4
π
1
x2 + 4x+ 20.(b) The Fourier transform of e−(ξ−3)2 is
√π e−x2/4 · e−3ix
74 by 3. in Theorem 4.7. From 2. in the same theorem it follows thatˆu(x) = i
d
dx(√π e−3ix e−x2/4) = i
√π(−3i− x
2) e−3ix−x2
4 .Hen e, by the inversion formula,u(x) =
1
2πi√π e3ix−
x2
4 (−3i+x
2) =
ix+ 6
4√π
e3ix−x2
4 .( ) The Fourier transform of e−ξ2/4 is 2√π e−x2 by Example 4.8. Us-ing 2. in Theorem 4.7 twi e, we getˆu(x) = i2
d2
dx2(2√π e−x2
) = −2√πd
dx(−2x e−x2
)
= 4√π(e−x2
+ x(−2x)ex2
) = 4√π(1− 2x2)e−x2
.Hen e,u(x) =
1
2πˆu(−x) =
2√π(1− 2x2)e−x2
.407. With v(x) = e−|x| we have u = v ∗ v. Theorem 4.10 yieldsu(ξ) = v(ξ) · v(ξ) = (
2
1 + ξ2)2 .408. Set v(x) = e−2x2 . Then v(ξ) =
√
π2e−ξ2/8. The equation an be writtenas
(u ∗ v)(x) = e−x2
,when e Fourier transformation givesu ∗ v(ξ) = u(ξ) · v(ξ) = u(ξ)
√
π
2e−ξ2/8 =
√π e−ξ2/4 .From this it follows that
u(ξ) =√2 e−ξ2/8 .Now we ould use a Fourier transformation and the inversion formulaon e again in order to �nd u. Here, however, a simpler approa h is to ompare with the �rst line of this solution. We see that
u(ξ) =√2
√
2
πv(ξ) =
2√πv(ξ) .Hen e,
u(x) =2√πe−2x2
.
75409. The fun tion u(x) = 11+x2 is ontinuous, integrable and bounded in R.Hen e, the assumptions of Theorem 4.11 are ful�lled. Furthermore
u(ξ) = π e−|ξ| and so∫ ∞
−∞
dx
(1 + x2)2=
1
2π
∫ ∞
−∞π2e−2|ξ| dξ = π
∫ ∞
0
e−2ξ dξ
= π
[
e−2ξ
−2
]∞
0
=π
2.410. It follows from
E(x, t) =1√4π
(t−1/2 e−x2/4t)that∂tE(x, t) =
1√4π
(−1
2t−3/2 e−x2/4t + t−1/2(−x2
4(−t−2)) e−x2/4t)
=t−3/2e−x2/4t
√4π
(−1
2+
x2
4t) ,
∂xE(x, t) =1√4π
(t−1/2 e−x2/4t) (−2x
4t) =
t−3/2
√4π
(−x
2) e−x4/4t ,
∂2xE(x, t) =
t−3/2
√4π
(−1
2− x
2(− x
2t)) e−x2/4t ,and we see that ∂tE = ∂2
xE .
