1n3Cpnn
¥êÆ¿mýmܩ땉Y
1,2. 3˜‡J[ -.¥, z‡Ø¬( Ž•vkŒ
AÛ:) •g?Ò• 1, 2, · · · . •
|Â,«¼œ, ù ج‡ «,¼¢§¿35
3y|3*˜ãžm"y3b
±"• S §‡¦1 m ÒجÚ1 n ÒØ
3* |¤´²¡þ ˜‡Œ»• 41
¬ƒm ål dm,n ÷v
(m + n)dm,n ≥ 1,
ùp·‚•Ä
´ ±þ
1 ÀJ K (4©
©) e
ål, •Ò´ü:m
l l•"@o
À‘( ) ÎÜ¢Sœ¹"
A ù‡3*¿•õUNB 8 ‡Ø¬;
B ù‡3*¿UNB ج‡êkŒu 8
C ù‡3*¿Œ±NB?¿õ‡Ø¬"
2 y ² K(6©
©) y²\ Øä"
1
þ•;
R1 ‰ Y . À‘C ÎÜ¢Sœ¹.
R2 ‰ Y . ) { ˜ . ·‚Œ±Ue㕪Sü1 1, 2, . . . Òج
˜. Äk, ?¿Sü
1 1 Òج
˜"é n ≥ 2, e1 1, 2, . . . , n − 1 Òج
˜®² SüÐ, ·‚•
Ä1 n ÒجØU3=
˜"
1
, ·‚• , l1 m Òج
˜m©, ÷^!
éu 1 ≤ m ≤ n − 1, d dm,n ≥ m+n
1
2
_ž ••ˆr m+n ål, ¤/¤ •Ý• m+n
lSÜ´ØŒ±Sü1 n ÒØ
¬ . ù
l o•Ý
2
2
2
n+1
n+2
2n − 1
2n − 1
+
+· · ·+
< 2(ln
+ln
+· · ·+ln
) = 2 ln
< 2 ln 2.
n+1 n+2
2n − 1
n
n+1
2n − 2
n
Ïd, ù
l ¿8 o•Ý؇L 2 ln 2,
‡ ±•• 14 · 2π = π2 . Ù• π2 >
1.5 > 2 ln 2, ù
lØUCX ‡ ±, Ïd1 n ÒجoŒ±ÀJ˜‡Ü·
˜, ¦ ¦†1 1, 2, . . . , n − 1 Òجƒm ålþ÷vK8^‡. dêÆ8B{Œ•§
ù‡ ±Œ±NB?¿õ‡Ø¬.
) { . ·‚± ±
%• :ïᲡ† ‹IX§¿ò1 1, 2, 3, 4 Òج©O
1
1
1
˜3 ( 4 , 0), (− 4 , 0), (0, 4 ), (0, − 41 ) ?, =¦‚ Ë ÌŠ©O• 0, π, π2 , 3π
. dž?¿ü¶
2
1
, dž 4 ¶Ø¬÷vK8^‡.
ج ålØ u 2 · 41 · π4 = π8 > 1+2
·‚¦^êÆ8B{y²e¡·K: é ê k > 2, Œ±ò1 1, 2, . . . , 2k ÒجS˜
u ±þ˜‡S
2k >/ ˆ‡º:?, ¦ §‚pƒƒm(3 ±þ ) ål÷v
K8^‡, …?Ò• 1, 2, . . . , 2k−1 Ò Ø¬3 ±þüü؃ .
þã·Ké k = 2 ¤á. eÙé k ¤á§=c 2k Òج
˜Ñ®(½. •Ä¦‚
ò ±©¤ 2k ãl. ·‚‡ò1 2k + 1, 2k + 2, . . . , 2k+1 Òج˜˜3ù l ¥:.
y35y²Œ±· ˜˜¦
91 2k + 1, 2k + 2, . . . , 2k+1 Òج ålþ÷vK8
^‡.
·‚ò1 2k + 1, 2k + 2 Òج˜˜3†1 2k−1 Òجƒ
˜(=†1 2k−1 ÒØ
2π
¬ Ë
• 2k+1
ål
˜); ò1 2k + 3, 2k + 4 Òج˜˜3†1 2k−1 − 1 Òج
ƒ
˜; · · · ò1 2k + 2a − 1, 2k + 2a Òج˜˜3†1 2k−1 − a + 1 Òجƒ
˜; · · · ò1 2k+1 − 1, 2k+1 Òج˜˜3†1 1 Òجƒ
˜.
k−1
duc 2
Òج3 ±þüü؃ , ù
˜˜´Œ1 . y3•Ä?¿ü¶
ج ål(•I•Ä– ˜ ج´“#” œ/). Ï• ± ©¤ 2k+1 ã, zãl
2π
π
•• 2k+1
· 14 = 2k+2
, éuü ?Ò©O• m > 2k Ú n ج, e§‚ƒm– küã
l, K
π
π
1
2(m + n)
1
dm,n ≥ k+1 > ·
·
>
;
2
2 m+n
2k+1
m+n
e¦‚ƒm ålT•˜ãl•§
n ∈ {2k + 2a − 1, 2k + 2a}, K m > 2k−1 − a + 1, Ï
2
d
(m+n)dm,n ≥
π
3π · 2k−1
3π
k
k−1
k−1
> 1.
·(2
+2a−1+2
−a+1)
=
·(2·2
+a)
≥
=
2k+2
2k+2
2k+1
8
π
¤±§1 1, 2, . . . , 2k+1 Òجüüƒm
Sü?¿õ¶Ø¬.
ålþ÷vK8^‡. dêÆ8B{•, Œ±
3
3,4. 2019c1˜3CpnnêÆ¿m `‘ö‚3ë\8ÔE žÿ, 8Nx‰Ì••
KI< rÔ, ´˜‡k 60 ‡
n /¡ õ¡N"l㥷‚Œ±w , ù‡õ
¡N L¡´ 60 ‡
˜mo>/©
¤ "
˜‡˜m n >/´•d˜‡²¡ n >/÷eZ^é ‚‰· €ò(=3À½ é
‚?/¤·
¡ )
˜mã/"ü‡˜mã/
• ´§‚Œ±Ï
3
L R ¥ ˜‡ åC†
-Ü"˜‡õ¡N• ´˜‡˜mk.«•, Ù>.Œ±
dk•õ‡²¡õ>/÷ú >©
¤"
3
ä K(4©
©) ·‚• 2021 = 43 × 47. @o´Ä•3˜‡õ¡N, §
d 43 ‡
˜m 47 >/©
¤?
4 ¯ ‰ K (6©
©) žé\
ä‰ÑÜ6 )º"
4
L¡Œ±
R3 ‰ Y . Œ±.
R4 ‰ Y . ·‚•I‡Þ˜‡~f=Œ. •Ä˜‡IO ‚¡ T, Ùþ
ê5L«:
T = {θ, ϕ : 0 ≤ θ, ϕ < 2π}.
·‚Œ±@•ù‡‚¡± z-¶•é¡¶: (θ, ϕ) éAu˜m¥
r cos ϕ) sin θ, r sin ϕ).
éu 1 ≤ k ≤ 43, ·‚•Ä‚¡þ «•
Dk = {θ, ϕ :
:Œ±dü‡ë
((R + r cos ϕ) cos θ, (R +
ϕ
2k
ϕ
2(k − 1)
π+3 ≤θ ≤
π + 3 }.
43
86
43
86
†*/`, r‚¡©¤
43 °ƒ , z˜°÷ {ϕ = 0} ƒm, òƒm? ˜ý ±
ØÄ, ,˜ýÛ=˜½ Ý.
y3, r {ϕ = 0} ù‡ C/¤˜‡ 43 >/, ˆ‡º:©OéAu θ = 2k
π. ù
43
Dk ko^“>”(Ù¥ü^ u {ϕ = 0} þ), o‡º:(ü‡ u 43 >/ º:?,
ü‡ u> ¥:?, ·‚„‡IPÑùü‡¥:ƒm
43 >/ º:). P•
2(k − 1)
2k
π, 0), Ck,1 = ( π, 0);
43
43
2k + 3
2k + 1
π, 2π), Dk,1 = (
π, 2π);
Dk,0 = (
43
43
2k + 2
Ek = (
π, 2π).
43
Ck,0 = (
3 ∂Dk
,˜^>þ
21 ‡:, ~X
Ak,i = (
2(k − 1)
ϕ
i
π + 3 π, π), i = 1, . . . , 21.
43
86 11
,
2
7 z-¶^= 43
π
, 21 ‡:, P• Bk,i , i = 1, . . . , 21.
ë ( ‚ ã Ck,0 Ck,1 , Ck,0 Ak,1 , Ck,1 Bk,1 , Ak,i Ak,i+1 , Bk,i Bk,i+1 , Ak,i Bk,i , Ak,i Bk,i+1 (i =
1, . . . , 21), ±9 Ak,21 Dk,0 , Bk,21 Dk,1 , Ak,21 Ek , Bk,21 Ek , Dk,0 Ek Ú Ek Dk,1 . ·‚
˜‡
˜m 47 >/. ù ·‚Ò
43 ‡
(þã E† k Ã') ˜m 47 >/, §‚
U ©Ñ˜‡õ¡N.
` ² : ˜‡;. †Ø´Ø@•ù ˜mõ>/ º:(>) Ñ´õ¡N º:(>), l
Šâ“z^>ÑŽüg” Ú“ 2021 ´Ûê”
“gñ”, dd@• K )‰´Ä½ .
5
5. c§Ü“FÏ•õ^ ¡ ù§8c¦5
ˆ
‰×/݉¡",U§^‡
ó§“ o‹Ü“FéøóŠ" ȯïÄÚ OŽ{^uN!ˆ« ¬ ëê"
ù
ëꘄŒ±ÏL4 z Rn þ ,‡›”¼ê f ¦ "3 o•C ˜‡‘
8¥§ù‡›”¼ê´, ˜‡‘K|Jø ¶ÑuS •ÄÚEâ ϧT‘K|
J±• o‰Ñd¼ê SÜ[!§ •UJø˜‡ •^uOŽ?¿ x ∈ Rn ? ¼
êŠ f (x)"¤±§ o7L=Äu¼êŠ54 z f " …§zgOŽ f ŠÑžÑ
Ø
OŽ] "Ð3T¯K ‘Ý n Ø´ép (10 †m)", §Jø¼ê Ó¯„
w• oØ”kb f ´1w "
ù‡¯K4Ü“FŽå gCÂõ ˜
ÂÂÑÅ"‡3ù ÂÑÅþÂf˜
‡!8§\I‡ %/5£Ø˜‡Nª^ܧӞ5¿ÂÑ J§† ˆ •Z"3
ùL§¥§vk<(ƒ/• ^Ü
ÝÚÂÑ Jƒm ½þ'X´Ÿo"Ü“
FÚ o¿£ §4 z f ØLÒ´N!˜ kõ‡^Ü Åì: Ž– x z˜‡
©þd˜‡^Ü››§ f (x) L«ù Åì ,«5U§•‡·‚5£N z‡^
ܧӞiÀ f Š§ATÒkF"é •Z x"Édéu§ü<˜åJÑ 4
z f ˜‡S“Ž{§¿·¶•“gÄc N Ž{” (Automated Forward/Backward
Tuning§AFBT§Ž{ 1)"31 k gS“¥§AFBT ÏLc N xk ü‡©þ
2n ‡: {xk ± tk ei : i = 1, . . . , n}, Ù¥ tk •Ú•¶, §- yk •ù :¥¼êŠ
•
˜‡§¿u yk ´Ä¦ f ¿©~ ¶e´§ xk+1 = yk §¿òÚ•O ¶Ä
K§- xk+1 = xk ¿òÚ•~Œ"3Ž{ 1 ¥§ ei L« Rn ¥ 1 i ‡‹I•þ§§
1 i ‡©þ• 1, Ù{ • 0¶1(·) ••«¼ê —– e f (xk ) − f (yk ) – • tk ƒ²
•§K 1[f (xk ) − f (yk ) ≥ t2k ] Š• 1§ÄK• 0"
算法 1 gÄc N Ž{ (AFBT)
Ñ\ x0 ∈ Rn , t0 > 0"é k = 0, 1, 2, . . . , ‰1±eÌ‚"
1: yk := argmin {f (y) : y = xk ± tk ei , i = 1, . . . , n}
# OŽ›”¼ê"
2
2: sk := 1[f (xk ) − f (yk ) ≥ tk ]
# ´Ä¿©eü? ´: sk = 1¶ Ä:sk = 0"
3: xk+1 := (1 − sk )xk + sk yk
# •#S“:"
4: tk+1 := 22sk −1 tk
# •#Ú•" sk = 1: Ú•O ¶ sk = 0: Ú•~Œ"
y3§·‚é›”¼ê f : Rn → R ŠÑXeb
b
"
1. f •à¼ê§=é?Û x, y ∈ Rn † α ∈ [0, 1] Ñk
f ((1 − α)x + αy) ≤ (1 − α)f (x) + αf (y).
b
2. f 3 Rn þŒ‡… ∇f 3 Rn þ L-Lipschitz ëY"
b
3. f
Y²8k.§=é?¿ λ ∈ R§8Ü{x ∈ Rn : f (x) ≤ λ}
6
k."
Äub
b
1 †b
b
2§Œ±y²
h∇f (x), y − xi ≤ f (y) − f (x) ≤ h∇f (x), y − xi +
é?Û x, y ∈ Rn ¤á¶b
b 1 †b
b 3 K yf 3 Rn þ
¼ê •õ5ŸŒë•?Û˜ à©Û ‰Ö"
y² K (20©
©) 3b
b
1–3e§éu AFBT§y²
lim f (xk ) = f ∗ .
k→∞
7
L
kx − yk2
2
k•
•
Š f ∗ "à
R5 y ². b f (xk ) 9 f ∗ "Ï {f (xk )} ØO, inf k≥0 [f (xk )−f ∗ ] > 0"Pgk = ∇f (xk ),
K inf k≥0 kgk k > 0 ( x∗ ¦ f (x∗ ) = f ∗ , K f ƒà5 y hgk , xk − x∗ i ≥ f (xk ) − f ∗ ¶Ó
ž§{f (xk )} ƒüN5† f ƒY²8k.5 y {xk − x∗ } k." inf k≥0 kgk k > 0)"
†óƒ§•3 ε > 0 ¦ kgk k ≥ ε é¤k k ≥ 0 ¤á"?‰ k ≥ 0, Œ ik ∈ {1, . . . , n}
÷v
(1)
|hgk , eik i| ≥ κkgk k ≥ κε,
√
Ù¥ κ = 1/ n"
f (yk ) ≤ min{f (xk ± tk eik )} ≤ f (xk ) − tk |hgk , eik i| +
¤±§•‡
tk ≤
·‚Òk f (yk ) ≤ f (xk ) − t2k , l
Lt2k
Lt2
≤ f (xk ) − κεtk + k . (2)
2
2
2κε
,
L+2
(3)
sk = 1, ? k tk+1 = 2tk "dd§´„
tk ≥ t ≡ min t0 ,
κε
L+2
> 0
(4)
é¤k k ≥ 0 ¤á" •3ᇠk ¦
sk = 1 (ÄK tk → 0); éz˜‡ù
2
k, f (xk )−f (xk+1 ) ≥ t "ù† f e•k.ƒgñ" ¤Ø¤á"
8
6. - n •
ê"é?˜
ê k, P 0k = diag{0, . . . , 0} • k × k
| {z }
"Ý "-
k
0n A
At 0n+1
Y =
!
•˜‡ (2n + 1) × (2n + 1) Ý , Ù¥ A = (xi,j )1≤i≤n,1≤j≤n+1 ´˜‡ n × (n + 1) ¢Ý
… At P A =˜Ý §= (n + 1) × n Ý §(j, i) ? ƒ• xi,j .
(i) y ² K (10©
©) ¡Eê λ • k × k Ý
X
˜‡A Š, XJ•3š" •þ
v = (x1 , . . . , xk )t
¦ Xv = λv. y²: 0 ´ Y
¢ê λ ´ AAt A Š"
A
Š… Y
Ù¦A
(ii) y ² K (15©
©) - n = 3 … a1 , a2 , a3 , a4 ´ 4 ‡p؃
a=
sX
√
Š/X ± λ, Ù¥šK
¢ê"P
a2i
1≤i≤4
±9
1
xi,j = ai δi,j + aj δ4,j − 2 (a2i + a24 )aj
a
(
1 if i = j
(1 ≤ i ≤ 3, 1 ≤ j ≤ 4), Ù¥ δi,j =
. y²: Y k 7 ‡p؃
0 if i 6= j
Š"
9
A
R6 y ² : (i) P In = diag{1, . . . , 1} • n × n ðÓÝ . ŠÐ C†Œy
| {z }
n
det(λI2n+1 − Y ) = λ det(λ2 In − AAt ).
√
¤±, 0 ´ Y
A Š… Y
Ù¦A Š/X ± λ, Ù¥ λ ´ AAt
Š.
a2 +a2 a2 +a2 a2 +a2
(ii) P u = (a4 , a4 , a4 ), v = ( 1 a 4 , 2 a 4 , 3 a 4 ). OŽ
šK¢êA
AAt = diag{a21 , . . . , a23 } + ut u − v t v.
f (s) = det(sIn − AAt ) • AAt
A õ‘ª. OŽ
f (a2i ) =
Y
a2i
(a2i − a2j )
2
a 1≤j≤4,j6=i
(∀i ∈ {1, 2, 3, 4}). - a01 , a02 , a03 , a04 • a1 , a2 , a3 , a4 ²-ü
4~S . d f (x2i )
Lˆª : AAt kn‡p؃
A Š b21 , b22 , b33 , Ù¥ b1 , b2 , b3 ´÷v
a01 > b1 > a02 > b2 > a03 > b3 > a04
¢ê. Ïd, d(i)
Y k 7 ‡p؃
A Š.
10
7. éu R þ ëY…ýéŒÈ Eꊼê f (x), ½Â R þ ¼ê (Sf )(x):
Z +∞
(Sf )(x) =
e2πiux f (u)du.
−∞
1
1
(i) ¯ ‰ K (10©
©) ¦ S( 1+x
2 ) Ú S( (1+x2 )2 )
wªLˆª"
(ii) ¯ ‰ K (15©
©) é?¿ ê k, P fk (x) = (1 + x2 )−1−k .
ê c1 , c2 ¦ ¼ê y = (Sfk )(x) ÷v
~‡©•§
xy 00 + c1 y 0 + c2 xy = 0.
11
b
k ≥ 1, é
~
R7 ‰ Y :
(i)
S(
…
S(
1
) = πe−2π|x|
2
1+x
1
π
) = (1 + 2π|x|)e−2π|x| .
2
2
(1 + x )
2
(ii) c1 = −2k … c2 = −4π 2 .
)‰ P V • R þ
Lemma 0.1.
EêŠ!ëY!ýéŒÈ
¼ê|¤ ‚5˜m.
(i) e f (x) ∈ V , f 0 (x) ∈ V … limx→∞ f (x) = 0, K
(Sf 0 )(x) = −2πix(Sf )(x).
(5)
(ii) e f (x) ∈ V … xf (x) ∈ V , K
(Sf )0 = 2πiS(xf (x)).
Ún0.1
y². (i)
(Sf 0 )(x)
Z +∞
e2πiux f 0 (u)du
=
−∞
Z +∞
2πiux
+∞
= e
f (u)|−∞ −
(e2πiux )0 f (u)du
−∞
Z +∞
e2πiux f (u)du
= −2πix
−∞
= −2πix(Sf )(x)
(ii) é?¿
a, b ∈ R (a < b),
Z b
2πiS(xf (x))dx
Z b
Z +∞
=
2πi(
e2πiux uf (u)du)dx
a
a
−∞
Z +∞ Z b
=
(
2πiue2πiux f (u)dx)du
−∞
a
Z +∞
Z +∞
2πibu
=
e
f (u)du −
e2πiau f (u)du
−∞
−∞
12
(6)
= (Sf )(b) − (Sf )(a).
ù
, (Sf )0 = 2πiS(xf (x)).
Ún0.1kXeíØ.
Corollary 0.2.
(i) b
f, f 0 , Sf, x(Sf )(x) ∈ V …
lim f (x) = 0.
x→∞
e (S(Sf ))(x) = f (−x), K (S(Sf 0 ))(x) = f 0 (−x).
(ii) b
f (x), xf (x), Sf, (Sf )0 ∈ V …
lim (Sf )(x) = 0.
x→∞
e (S(Sf ))(x) = f (−x), K S(S(xf (x))) = −xf (−x).
Lemma 0.3.
(i) S((1 + x2 )−1 ) = πe−2π|x| .
(ii) S(πe−2π|x| ) = (1 + x2 )−1 .
Proof. (i)P f (x) = (1 + x2 )−1 . éu x ≥ 0, ·‚k
Z A
(Sf )(x) = lim
A→+∞
e2πiux
du.
2
−A 1 + u
P
CA := {z = u + iv : −A ≤ u ≤ A, v = 0}
[
{z = Aeiθ : 0 ≤ θ ≤ π}.
1
5¿
A > 1 ž, i ´ 1+z
k.«•S •˜4:. d£´È© •{
2 3 CA .½
−2πx
¿- A → ∞, ·‚
(Sf )(x) = πe
. du f (x) ´ó¼ê, ¤± (Sf )(x) •´ó
−2π|x|
¼ê. ù , (Sf )(x) = πe
.
(ii)P g(x) = πe−2π|x| . † OŽ
(Sg)(x)
Z ∞
=
e2πixu πe−2π|u| du
−∞
Z ∞
= π
(e2πixu + e−2πixu )e−2πu du
0
1 e−2π(1+ix)u e−2π(1−ix)u ∞
= − (
+
)|0
2 1 + ix
1 − ix
1
=
.
1 + x2
13
k ≥ 0, Sfk /X (Sfk )x = e−2π|x| gk (|x|), Ù¥ gk ´
Lemma 0.4. (i) éu?¿
‡ k gõ‘ª.
(ii) éu?¿
k ≥ 0, S(S(fk )) = fk … S(S(xfk+1 (x))) = −xfk+1 (x).
Proof. (i)·‚k48úª
fk+1 = fk +
dÚn0.1,
1
xf 0 (x).
2(k + 1) k
48úª:
Sfk+1 = Sfk −
1
(x(Sfk )(x))0 .
2(k + 1)
dd, d8B{·‚ Ñ(Ø.
(ii)5¿ fk0 (x) = −2(k + 1)xfk+1 (x) … (xfk (x))0 = −(1 + 2k)fk (x) + 2(k + 1)fk+1 (x).
d(i)Ü© (Ø(ÜÚn0.1, ·‚• íØ0.2¥ b é¼ê fk (x) Ú xfk+1 (x)(k ≥
0) ¤á. ù , d8B{Œy S(S(fk )) = fk (x) … S(S(xfk+1 (x))) = −xfk+1 (x)(k ≥
0).
£
K8
. (i)3Ún0.3¥®²y² S((1 + x2 )−1 ) = πe−2π|x| . d(5)
S(−2x(1 + x2 )−2 ) = −2π 2 ixe−2π|x| .
2d(6)
S(−2x2 (1 + x2 )−2 ) = −π(1 − 2π|x|)e−2π|x| .
ù
,
S((1 + x2 )−2 ) =
π
(1 + 2π|x|)e−2π|x| .
2
(ii)Ä k,
k ≥ 1 ž, xj fk (x) (0 ≤ j ≤ 2k) Ñ ´ ý é Œ È . ù , d(6),
y = (Sfk )(x) ´ 2k gëYŒ‡¼ê. dÚn0.1ÚÚn0.4 : xy 00 + c1 y 0 + c2 xy = 0
du
c2
(x2 fk0 + 2xfk ) − c1 xfk − 2 fk0 = 0.
4π
Ñ\ fk (x) = (1 + x2 )−1−k , ·‚
c1 = −2k … c2 = −4π 2 .
14
8. ,úiíј‡#
^‡ž§úi ½|Ü€Ø ¬'%T^‡ ¹ •
r o<ê‘žm Cz§•¬é•r+N ˜ A ‰äN NïÚ©Û"·‚
^ n(t, x) L«•r êþ—Ý£±e{¡—ݤ§ùp t L«žm§ x L«•ré
T
^‡ ¦^ž•§@o3 t ž•§éu 0 < x1 < x2 §¦^ž•0u x1 Ú x2 ƒ
Rx
m •rêþ• x12 n(t, x)dx"·‚b §—Ý n(t, x) ‘XžmüzɱeA‡Ïƒ
K•µ
b
1.
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b
2. •r3¦^L§¥§ŒU¬ÊŽ¦^§·‚b
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1 úi \Dµü
žmSÏdO\ <ê´žm ¼ê§^ c(t) L«"
2 P•r \DµP•r¬ÌÄ•gC Ó¯!*l 핦^T
핤õ „Ç‹•r ¦^ž• x k'§PŠ b(x)"
b XJ3,˜ž•, P• t = 0 ž§—ݼ괮•
ѧn(t, x) žmüz÷vXe •§
(
^‡§
§n(0, x) = n0 (x)"Œ±í
∂
∂
n(t, x) + ∂x
n(t, x) + d(x)n(t, x) = 0, t ≥ 0, x ≥ 0,
∂t
R∞
N (t) := n(t, x = 0) = c(t) + 0 b(y)n(t, y)dy.
(7)
ùp N (t) Œ)Ö•#•r O\„Ç"·‚b b, d ∈ L∞
+ (0, ∞)§= b(x) Ú d(x)
…( Ÿ)k."±e§·‚k‰˜‡{zb µ c(t) ≡ 0§=#•r O\•‹P•r
\Dk'"
(i) ¯ ‰ K (10©
©) Šâb
b 1Úb
b 2§/ª/í Ñ(7) ¥ n(t, x) ¤÷v
‡©
•§§I‡3í L§¥•Ñ .b ÚêÆLˆªƒm éA'X"2Šâb
b
3§)º(7) ¥ N (t) ½Â ¹Â"
(ii) ¯ ‰ K (10©
©) ·‚Ž‡ïÄ#•r O\„Ç N (t) Ú핤õ„Ç b(x) ƒm
'X"•d§ží ј‡ N (t) ¤÷v •§§…•§¥••¹ N (t), n0 (x),
b(x), d(x)§ Ø•¹ n(t, x)"¿y²§ N (t) ÷vXe O
kbk∞ t
Z ∞
|N (t)| ≤ kbk∞ e
|n0 (x)| dx,
0
ùp k · k∞ L« L∞ ‰ê"
15
(8)
(iii) y ² K (10©
©) • §·‚Ž‡ïħ3¿©• žmƒ §êþ—ݼê n(t, x)
kŸoìC ª³"du•ro<êŒU˜†3O\§¤±·‚Ø•B† ïÄ
êþ—ݼê n(t, x)§ •AT w˜‡- z
—ݼê"
•d§·‚Äkb Xe
(
ϕ0 (x) + (λ0 + d(x)) ϕ(x) = 0, x ≥ 0,
R∞
ϕ(x) > 0, ϕ(0) = 0 b(x)ϕ(x)dx = 1,
¿…§ éó¯K•k•˜
(
A Š¯Kk•˜) (λ0 , ϕ(x))µ
) ψ(x):
−ψ 0 (x) + (λ0 + d(x)) ψ(x) = ψ(0)b(x),
R∞
ψ(x) ≥ 0,
ψ(x)ϕ(x)dx = 1.
0
x ≥ 0,
, §·‚½Â- z—Ý ñ(t, x) := n(t, x)e−λ0 t "y²§éu?¿à¼ê H :
R+ → R+ ÷v H(0) = 0, ·‚k
d
dt
¿y²
Z ∞
ψ(x)ϕ(x)H
0
ñ(t, x)
ϕ(x)
Z ∞
λ0 t
dx ≤ 0,
Z ∞
ψ(x)n(t, x)dx = e
0
•
∀t ≥ 0,
ψ(x)n0 (x)dx.
0
{zy²§·‚3üŽ¥b½3 ∞ ?
16
>.‘
zÑ´Œ±
Ñ "
R8 ‰ Y : (i) ù‡•§í •ªkéõ"Þü‡~f"
1§A ‚{"du¦^ž•‘žm‚5O•§·‚½ÂA ‚ x(t)§§÷v
dx(t)
= 1.
dt
^XA ‚§ŠâÊŽ„Ç
¹Â§·‚k
d
n(t, x(t)) = −d(x(t))n(t, x(t)).
dt
n= (7)ª•§"
2, ‡ {"•Ä˜‡žm‡
δt 1§Šâb ˜Úb
§·‚k
n(t + δt, x + δt) = n(t, x) − δtd(x)n(x, t) + o(δt).
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z§1 ‘L«ÊŽ •rêþ"ü>ر δt§2
- δt → 0§= d•§"
'u N (t) ½Â§•I‡`²P•rí•
z"éu ½,‡¦^ž• x
P•r§¦‚ü žmS0
#•r êþ• b(x)n(t, x)"• ¦ü žmS¤k
P•r0
#•rêþ§I‡r¤k¦^ž• P•r
z\3˜å§ Lˆ
R∞
• 0 b(y)n(t, y)dy"
(ii) ŠâK¿Ú N (t) ½Â§·‚I‡kr—ݼê n(t, x) ¤ N (t) ÚÙ¦ëê
Lˆª§ùI‡¦)d•§"
5¿ ù´˜‡˜ V-•§§Œ±^A ‚{¦)"ò•§U •
d
n(t + s, x + s) + d(x + s)n(t + s, x + s) = 0,
ds
Rx
KXJ½Â D(x) = 0 d(y)dy§@o
d D(x+s)
e
n(t + s, x + s) = 0.
ds
@o§
(9)
s ≥ max(−t, −x) ž§·‚k
eD(x+s) n(t + s, x + s) = eD(x) n(t, x),
AO §·‚Œ±- x = y§s = −y Œ §
t≥y ž
n(t, y) = N (t − y)e−D(y) .
17
∀x ≥ 0, t ≥ 0.
(10)
2- x = y§s = −t Œ §
t≤y ž
n(t, y) = n0 (y − t)eD(y−t)−D(y) .
•
Ñ N (t) ÷v •§§·‚ò§ Lˆª ©¤üÜ©
Z ∞
t
0
0
b(y)n(t, y)dy.
b(y)n(t, y)dy +
b(y)n(t, y)dy =
N (t) =
Z ∞
Z t
ŠâA ‚Œ•§mà1˜‘ A ‚å u x = 0, t ≥ 0§
u x ≥ 0, t = 0"ò n(t, y) Lˆª©O“\§·‚
Z t
b(y)e
N (t) =
−D(y)
Z ∞
N (t − y)dy +
0
n§=
‘
A
b(y)eD(y−t)−D(y) n0 (y − t)dy.
‚å
(11)
t
N (t) ÷v
•§
Z t
b(t − x)e
N (t) =
1
−D(t−x)
Z ∞
N (x)dx +
b(x + t)eD(x)−D(x+t) n0 (x)dx.
(12)
0
0
•Ä d(x) > 0 ¤± D ´4O¼ê§þª¥ e−D(t−x) , eD(x)−D(x+t) þØŒu 1"
u´,2|^ b(x) k.5§·‚Œ±é N (t) ‰Xe O
Z ∞
Z t
|N (t)| ≤ kbk∞
|N (x)|dx + kbk∞
0
|n0 (x)| dx.
0
• §|^Gronwall Ún§·‚ÒŒ±
–y Ø ª"
(iii) ù´˜‡2 ƒé
O ¯K"
Äk·‚ò•§(7)U ¤- z—ݼê÷v •§
∂
∂
n
e(t, x) +
n
e(t, x) + (λ0 + d(x)) ñ(t, x) = 0.
∂t
∂x
,
§·‚?˜Ú/U §
n
∂ ñ(t, x)
∂ ñ(t, x)
+
= 0,
∂t ϕ(x)
∂x ϕ(x)
?
k
∂
H
∂t
ñ(t, x)
ϕ(x)
∂
+
H
∂x
18
ñ(t, x)
ϕ(x)
= 0.
òA
Š¯KÚÙéó¯K Ü3˜å§·‚k
(
∂
[ϕ(x)ψ(x)] = −ψ(0)b(x)ϕ(x), x ≥ 0,
∂x
R∞
ψ(x) ≥ 0,
ψ(x)ϕ(x)dx = 1.
0
ÏL† OŽ§·‚
ñ(t, x)
∂
ñ(t, x)
ñ(t, x)
∂
ψ(x)ϕ(x)H
+
ψ(x)ϕ(x)H
= −ψ(0)b(x)ϕ(x)H
.
∂t
ϕ(x)
∂x
ϕ(x)
ϕ(x)
P dµ(x) = b(x)ϕ(x)dx§òþªé x 3 R+ þÈ©§Œ ,
d
dt
Z ∞
ψ(x)ϕ(x)H
0
ñ(t, x)
ϕ(x)
Z ∞
dx = −ψ(0)
H
0
·‚5¿ §d½Â• ϕ(0) = 1§… n(t, 0) =
ñ(t, 0)
= ñ(t, 0) =
ϕ(0)
ñ(t, x)
ϕ(x)
R∞
0
dµ(x) + ψ(0)H
ñ(t, 0)
ϕ(0)
.
b(x)n(t, x)dx§@o
Z ∞
Z ∞
b(x)ñ(t, x)dx =
0
0
ñ(t, x)
dµ(x),
ϕ(x)
u´Œ
Z ∞ Z ∞
Z
ñ(t, x)
ñ(t, x)
d ∞
ñ(t, x)
ψ(x)ϕ(x)H
H
dµ(x) .
dx = ψ(0) −
dµ(x) + H
dt 0
ϕ(x)
ϕ(x)
ϕ(x)
0
0
2dJensenØ ª§·‚Œ
d
dt
•
Z ∞
ψ(x)ϕ(x)H
0
§- H(u) = u§= –y
ª"
19
ñ(t, x)
ϕ(x)
dx ≤ 0.
Alibaba Global Mathematics Competition-Quanlifying
1,2. In a fictional world, each resident (viewed as geometric point) is assigned a number:
1, 2, · · · . In order to fight against some epidemic, the residents take some vaccine and
they stay at the vaccination site after taking the shot for observation. Now suppose that
the shape of the Observation Room is a circle of radius 14 , and one requires that the
distance dm,n between the Resident No. m and the Resident No. n must satisfy
(m + n)dm,n ≥ 1.
Where we consider the distance on the circle, i.e., the length of the minor arc between
two points.
1. Multiple-Choice Question (4 points) Which of the following is correct?
A No more than 8 residents can be placed inside the observation room;
B The maximal number of residents that can be placed simultaneously is greater
than 8, but still finite;
C Any number of residents can be placed inside the observation room.
2. Proof Question (6 points) Give a proof of your answer to Question (i).
1
R1 Answer. The Choice C is correct.
R2 Answer. Solution I. We can place the Residents No. 1, 2, . . . according to the following rule. First, put Resident No.1 arbitrarily. For n > 2, if Residents No.1, 2, . . . , n−1
have already been placed, we consider the positions where Resident No. n cannot be
placed.
1
For 1 ≤ m ≤ n − 1, by dm,n ≥ m+n
, we know that the Resident No. n cannot be
2
placed in the arc that is centered at Resident No. m, and of the length m+n
. The total
length of these arcs is
2
2
n+1
n+2
2n − 1
2n − 1
2
+
+· · ·+
< 2(ln
+ln
+· · ·+ln
) = 2 ln
< 2 ln 2.
n+1 n+2
2n − 1
n
n+1
2n − 2
n
Therefore, the total length of the union of these arcs does not exceed 2 ln 2, while the
perimeter of the circle is 41 · 2π = π2 . It is easy to observe that π2 > 1.5 > 2 ln 2, so
these arcs would not cover the whole circle, hence it is always possible to find a place
for Resident No. n such that its distances to Residents No.1, 2, . . . , n − 1 satisfy the
requirement. By induction we conclude that the circle can accomodate any quantity of
residents.
Solution II. We consider the Cartesion coordinate system whose origin is the center of the circle, and place Residents No.1, 2, 3 and 4 at ( 41 , 0), (− 14 , 0), (0, 14 ), (0, − 41 ),
respectively, or in an equivalent way, we say that (the principle values of) their arguments are 0, π, π2 and 3π
2 . Now the distance between any two residents is no less than
1
, so the placement of these 4 residents satisfies the requirement.
2 · 41 · π4 = π8 > 1+2
We prove the following assertion by induction: for any integer k > 2, we can place
Residents No. 1, 2, . . . , 2k at the vertices of a regular 2k -gon inscribed to the circle, such
that their mutual distances fulfill the requirement, and no two residents among the first
2k−1 occupy adjacent vertices.
The assertion holds for k = 2. If it is valid for k, so that the first 2k residents
are placed. They divide the circle into 2k equal arcs. We need to put Residents No.
2k + 1, 2k + 2, . . . , 2k+1 at the midpoints of these arcs. So we just need to prove the
distances related to Residents No. 2k + 1, 2k + 2, . . . , 2k+1 satisfy the requirement.
We put Residents No. 2k + 1, 2k + 2 in positions next to Resident No. 2k−1 (i.e. the
2π
corresponding arguments differ to that of Resident No. 2k−1 by 2k+1
); put Residents No.
k
k
k−1
2 + 3, 2 + 4 next to Resident No. 2
− 1; · · · put Residents No. 2k + 2a − 1, 2k + 2a
next to Resident No. 2k−1 − a + 1; · · · put Residents No. 2k+1 − 1, 2k+1 next to Resident
No.1.
As the first 2k−1 residents do not occupy any consecutive positions, the above placement would not cause any problem. Now we consider distance between two residents
(only the cases where at least one resident in the pair is “new” need to be considered).
2
2π
π
Since the circle is now divided into 2k+1 arcs, each has length 2k+1
· 14 = 2k+2
. For
k
Residents No. m(> 2 ) and n, if they are separated by at least two pieces of arcs, then
dm,n ≥
π
2k+1
>
π
1
2(m + n)
1
·
·
>
;
k+1
2 m+n
m+n
2
If they are just separated by one piece of arc, then for n ∈ {2k + 2a − 1, 2k + 2a}, we
have m > 2k−1 − a + 1, hence
(m+n)dm,n ≥
π
2
·(2k +2a−1+2k−1 −a+1) =
k+2
π
·(2·2k−1 +a) ≥
2k+2
3π · 2k−1
3π
=
> 1.
k+1
8
2
Therefore, the distances between each pair of Residents No. 1, 2, . . . , 2k+1 satisfies the
requirement. Then by induction, we conclude that any number of residents can be
accomodated in that way.
3
3,4. Two years ago, the winners of 2018 Alibaba Global Mathematics Competition
made a paper polyhedron as a present to the organizers. As shown in the photoes, the
polyhedron has 60 equal triangular faces, and its surface can also be divided into 60
congruent non-planar quadrilaterals.
A non-planar n-gon is a non-planar figure obtained from a planar n-gon by folding it
along some diagonals (i.e., to form certain dihedral angle at each of the chosen diagonals).
Two non-planar figures are congruent if and only if each can be obtained from the other
one by a certain isometry of R3 . A polyhedron is a bounded region in R3 whose boundary
is the union of a finite collection of planar polygons along common edges.
3. True-False Question (4 points) We know that 2021 = 43 × 47. Is there a
polyhedron whose surface can be formed by gluing together 43 equal non-planar
47-gons?
4. Question and Answer (6 points) Please justify your answer to Question (i)
with a rigorous argument.
4
R3,Answer. The answer is YES.
R4 Answer. All we need to do is to construct an example. Let’s consider a standard
torus T, whose points can be represented by two parameters:
T = {θ, ϕ : 0 ≤ θ, ϕ < 2π}.
One can view the z-axis as the axis of symmetry of the torus:
((R + r cos ϕ) cos θ, (R + r cos ϕ) sin θ, r sin ϕ).
For 1 ≤ k ≤ 43, we consider the following region on the torus
2(k − 1)
ϕ
2k
ϕ
π+3 ≤θ ≤
π + 3 }.
43
86
43
86
Intuitively, what we do here is to divide the torus into 43 equal parts, then cut every
part along the circle {ϕ = 0}, keep one side of the cut while sliding the other side along
the circle for certain angle.
Now, we deform the circle {ϕ = 0} into a regular 43-gon whose vertices correspond
to θ = 2k
43 π. Then Dk has four “sides” of (two of which lie on {ϕ = 0}), four “corners”
(two of which are adjacent vertices of the 43-gon, while the other two are midpoints of
two sides, we need then mark the vertex of the 43-gon between these two midpoints).
We denote
2(k − 1)
2k
Ck,0 = (
π, 0), Ck,1 = ( π, 0);
43
43
2k + 1
2k + 3
Dk,0 = (
π, 2π), Dk,1 = (
π, 2π);
43
43
2k + 2
Ek = (
π, 2π).
43
Take another “side” of ∂Dk , mark 21 points, e.g.
Dk = {θ, ϕ :
Ak,i = (
2(k − 1)
ϕ
i
π + 3 π, π), i = 1, . . . , 21.
43
86 11
2
Then rotate around z-axis by 43
π to get another 21 points, denote them by Bk,i , i = 1, . . . , 21.
Now we join
Ck,0 Ck,1 , Ck,0 Ak,1 , Ck,1 Bk,1 , Ak,i Ak,i+1 , Bk,i Bk,i+1 , Ak,i Bk,i , Ak,i Bk,i+1 (i = 1, . . . , 21),
and Ak,21 Dk,0 , Bk,21 Dk,1 , Ak,21 Ek , Bk,21 Ek , Dk,0 Ek , Ek Dk,1 . We get a non-planar 47-gon.
Thus we get 43 congruent (the construction above is independent of k)non-planar 47gons, they can be glue together to form a polyhedron.
Remark: A typical mistake would be to think that the vertices(edges) of these nonplanar polygons are all vertices(edges, not part of eadges) of the polyhedron, then deducts from “each edge is counted twice” and “2021 is an odd number” a fake “condtradiction”.
5
5. Last year, Master Zhang got famous with his topology noodle, and he is now a
part-time chef at Alibaba Damo Academy. One day, Xiao Li, a software engineer at
the DAMO Academy, complained about his recent work with Master Zhang. Xiao Li’s
job involves designing algorithms to tune the parameters of various systems, which can
usually be formulated as finding a vector x ∈ Rn that minimizes a certain loss function
on Rn . In his latest project, Xiao Li has to deal with a loss function f that is provided
by another group. For security and technical reasons, the other group cannot reveal
the explicit definition f , but only offers an interface to evaluate f at any given x ∈ Rn .
Hence, Xiao Li needs to minimize f based on its function values. In addition, the evaluation of f is costly. Fortunately, the dimension n of this problem is not high (around 10).
Besides, the group that provides the function informs Xiao Li that he may assume f is
smooth.
Xiao Li’s problem reminds Master Zhang about an antique radio that he owns.
To listen to a program on this radio, you need to carefully tune the frequency knob
forward and backward while monitoring the quality of the sound until finding the best
frequency for receiving the signal. In this process, nobody knows the precise relation
between the angle of the knob and the quality of the sound. After a careful discussion
with Master Zhang, Xiao Li realizes that minimizing f is like tuning a machine with
multiple knobs: just imagine that each component of x is controlled by a knob, and f (x)
represents certain performance of the machine; one should be able to find the best x
by tuning each knob forward and backward while monitoring the value of f . Therefore,
Xiao Li and Master Zhang propose an iterative algorithm for minimizing f , named
Automated Forward/Backward Tuning (AFBT, Algorithm 1). At iteration k, AFBT
considers 2n points {xk ± tk ei : i = 1, . . . , n} by varying each component of xk forward
or backward with a step size tk , sets yk to the one rendering the smallest value of f ,
and checks whether yk achieves a sufficient decrease in f ; if yes, it takes xk+1 = yk
and doubles the step size; otherwise, it sets xk+1 = xk and halves the step size. In the
algorithm, ei denotes the i-th canonical coordinate vector in Rn (the i-th entry is 1 while
all the others are 0); 1(·) is the indicator function, so that 1[f (xk ) − f (yk ) ≥ t2k ] equals 1
if f (xk ) − f (yk ) is at least the square of tk , or else the value is 0.
Algorithm 1 Automated Forward/Backward Tuning (AFBT)
Input x0 ∈ Rn and t0 > 0. For k = 0, 1, 2, . . . , do the following.
1: yk := argmin f (y) : y = xk ± tk ei , i = 1, . . . , n
# Evaluate loss function.
2
2: sk := 1 f (xk ) − f (yk ) ≥ tk
# Sufficient decrease? Yes: sk = 1; No: sk = 0.
3: xk+1 := (1 − sk )xk + sk yk
# Update iterate.
4: tk+1 := 22sk −1 tk
# Update step size. sk = 1: double; sk = 0: halve.
Now we make the following assumptions on the loss function f : Rn → R.
6
Assumption 1. f is convex. This means that
f ((1 − α)x + αy) ≤ (1 − α)f (x) + αf (y)
for all x, y ∈ Rn and α ∈ [0, 1].
Assumption 2. f is differentiable on Rn and ∇f is L-Lipschitz on Rn .
Assumption 3. f is level-bounded, meaning that {x ∈ Rn | f (x) ≤ λ} is a bounded
set for any given λ ∈ R.
Based on Assumptions 1 and 2, we can prove that
h∇f (x), y − xi ≤ f (y) − f (x) ≤ h∇f (x), y − xi +
L
kx − yk2
2
for all x, y ∈ Rn ; Assumptions 1 and 3 ensure that f has a finite minimum f ∗ on Rn .
For more properties of convex functions, see any textbook on convex analysis.
Proof Question (20 points) Under Assumptions 1–3, prove for AFBT that
lim f (xk ) = f ∗ .
k→∞
7
R5 Proof. Assume that f (xk ) 9 f ∗ . Then inf k≥0 [f (xk ) − f ∗ ] > 0 since {f (xk )} is nonincreasing, Denote gk = ∇f (xk ). Then inf k≥0 kgk k > 0 (Pick an x∗ with f (x∗ ) = f ∗ .
Then f (xk ) − f ∗ ≤ hgk , xk − x∗ i by the convexity of f ; meanwhile, {xk } is bounded due
to the monotonicity of {f (xk )} and the level-boundedness of f . Thus inf k≥0 kgk k > 0).
In other words, there exists an ε > 0 such that kgk k ≥ ε for all k ≥ 0. Given any k ≥ 0,
we can pick an ik ∈ {1, . . . , n} satisfying
|hgk , eik i| ≥ κkgk k ≥ κε
(1)
√
with κ = 1/ n. Hence
f (yk ) ≤ min{f (xk ± tk eik )} ≤ f (xk ) − tk |hgk , eik i| +
Lt2k
Lt2
≤ f (xk ) − κεtk + k . (2)
2
2
Therefore, whenever
tk ≤
2κε
,
L+2
(3)
we will have f (yk ) ≤ f (xk ) − t2k , which renders sk = 1 and hence tk+1 = 2tk . It is then
easy to see that {tk } has a positive lower bound, namely
κε
tk ≥ t ≡ min t0 ,
> 0
for each
k ≥ 0.
(4)
L+2
Thus sk = 1 for infinitely many k (otherwise, tk → 0), and f (xk )−f (xk+1 ) ≥ t2 for each
of such k, contradicting the lower-boundedness of f . The proof is complete.
8
6. Let n be a positive integer. For any positive integer k, write 0k = diag{0, . . . , 0} for
| {z }
k
the k × k zero matrix. Let
Y =
0n
A
At 0n+1
!
be a (2n+1)×(2n+1) matrix, where A = (xi,j )1≤i≤n,1≤j≤n+1 is an n×(n+1) real matrix
and At denotes the transpose of A, that is, the (n + 1) × n matrix whose (j, i)-entry is
xi,j .
(i) Proof Question (10 points) A complex number λ is called an eigenvalue of a
k × k matrix X if Xv = λv for some nonzero column vector v = (x1 , . . . , xk )t .
Show that: 0 is an eigenvalue of Y and every other eigenvalue of Y is of the form
√
± λ where λ is a non-negative real eigenvalue of AAt .
(ii) Proof Question (15 points) Let n = 3 and a1 , a2 , a3 , a4 be four distinct positive
real numbers. Put
s X
a=
a2i
1≤i≤4
and
1
xi,j = ai δi,j + aj δ4,j − 2 (a2i + a24 )aj
a
(
1 if i = j
(1 ≤ i ≤ 3, 1 ≤ j ≤ 4), where δi,j =
. Show that: Y has 7 distinct
0 if i 6= j
eigenvalues.
9
R6 Proof. (i) Write In = diag{1, . . . , 1} for the n×n identity matrix. By an elementary
| {z }
n
reduction the characteristic polynomial
det(λI2n+1 − Y ) = λ det(λ2 In − AAt ).
√
Then, 0 is an eigenvalue of Y and every other eigenvalue of Y is of the form ± λ where
λ is a non-negative real eigenvalue of AAt .
a2 +a2 a2 +a2 a2 +a2
(ii) Put u = (a4 , a4 , a4 ) and v = ( 1 a 4 , 2 a 4 , 3 a 4 ). By calculation we have
AAt = diag{a21 , . . . , a23 } + ut u − v t v.
Let f (s) = det(sIn −AAt ) be the characteristic polynomial of AAt . Then, by calculation
one shows that
Y
a2
(a2i − a2j )
f (a2i ) = 2i
a
1≤j≤4,j6=i
for each i ∈ {1, 2, 3, 4}. Let a01 , a02 , a03 , a04 be the descending re-ordering of a1 , a2 , a3 , a4 .
Then, we get: AAt has three distinct eigenvalues b21 , b22 , b33 where b1 , b2 , b3 are positive
real numbers such that
a01 > b1 > a02 > b2 > a03 > b3 > a04 .
Then, by (i) Y has 7 distinct real eigenvalues.
10
7. For a continuous and absolutely integrable complex-valued function f (x) on R, define
a function (Sf )(x) on R by
Z +∞
e2πiux f (u)du.
(Sf )(x) =
−∞
1
1
(i) Question and Answer (10 points) Find explicit forms of S( 1+x
2 ) and S( (1+x2 )2 ).
(ii) Question and Answer (15 points) For any integer k, write fk (x) = (1+x2 )−1−k .
When k ≥ 1, find constants c1 , c2 such that the function y = (Sfk )(x) solves a
second order differential equation
xy 00 + c1 y 0 + c2 xy = 0.
11
R7 Answer. Write V for the space of complex-valued, continuous and absolutely
integrable functions on R.
Lemma 0.1.
(i) If f (x) ∈ V , f 0 (x) ∈ V and limx→∞ f (x) = 0, then
(Sf 0 )(x) = −2πix(Sf )(x).
(5)
(ii) If f (x) ∈ V and xf (x) ∈ V , then
(Sf )0 = 2πiS(xf (x)).
Proof. (i)
(Sf 0 )(x)
Z +∞
e2πiux f 0 (u)du
=
−∞
= e
2πiux
f (u)|+∞
−∞ −
Z +∞
= −2πix
Z +∞
(e2πiux )0 f (u)du
−∞
e2πiux f (u)du
−∞
= −2πix(Sf )(x)
(ii) For any a, b ∈ R with a < b,
Z b
2πiS(xf (x))dx
Z b
Z +∞
=
2πi(
e2πiux uf (u)du)dx
a
−∞
a
Z +∞ Z b
=
(
2πiue2πiux f (u)dx)du
−∞
a
Z +∞
=
e
2πibu
Z +∞
f (u)du −
−∞
e2πiau f (u)du
−∞
= (Sf )(b) − (Sf )(a).
Thus, (Sf )0 = 2πiS(xf (x)).
Lemma 0.1 implies the following corollary immediately.
Corollary 0.2.
(i) Assume that f, f 0 , Sf, x(Sf )(x) ∈ V and
lim f (x) = 0.
x→∞
If (S(Sf ))(x) = f (−x), then (S(Sf 0 ))(x) = f 0 (−x).
12
(6)
(ii) Assume that f (x), xf (x), Sf, (Sf )0 ∈ V and
lim (Sf )(x) = 0.
x→∞
If (S(Sf ))(x) = f (−x), then S(S(xf (x))) = −xf (−x).
Lemma 0.3.
(i) S((1 + x2 )−1 ) = πe−2π|x| .
(ii) S(πe−2π|x| ) = (1 + x2 )−1 .
Proof. (i) Write f (x) = (1 + x2 )−1 . For x ≥ 0, we have
Z A 2πiux
e
(Sf )(x) = lim
du.
A→+∞ −A 1 + u2
Put
CA := {z = u + iv : −A ≤ u ≤ A, v = 0}
[
{z = Aeiθ : 0 ≤ θ ≤ π}.
1
Note that, i is the only pole of 1+z
2 inside the domain bounded by CA whenever A > 1.
Using the trick of contour integral and letting A → ∞, we get (Sf )(x) = πe−2πx . Since
f (x) is an even function, so is (Sf )(x). Then, (Sf )(x) = πe−2π|x| .
(ii) Write g(x) = πe−2π|x| . By direct calculation
(Sg)(x)
Z ∞
=
e2πixu πe−2π|u| du
−∞
Z ∞
= π
(e2πixu + e−2πixu )e−2πu du
0
1 e−2π(1+ix)u e−2π(1−ix)u ∞
+
)|0
= − (
2
1 + ix
1 − ix
1
.
=
1 + x2
Lemma 0.4. (i) For any k ≥ 0, Sfk is of the form (Sfk )x = e−2π|x| gk (|x|) where gk
is a polynomial of degree k.
(ii) For any k ≥ 0, S(S(fk )) = fk and S(S(xfk+1 (x))) = −xfk+1 (x).
Proof. (i) We have a recursive relation
fk+1 = fk +
1
xf 0 (x).
2(k + 1) k
By Lemma 0.1, we get a recursive relation for Sfk :
Sfk+1 = Sfk −
1
(x(Sfk )(x))0 .
2(k + 1)
13
Then, one proves the conclusion by induction.
(ii) Note that fk0 (x) = −2(k+1)xfk+1 (x) and (xfk (x))0 = −(1+2k)fk (x)+2(k+1)fk+1 (x).
By the conclusion of part (i) and Lemma 0.1, we see that the assumptions in Corollary
0.2 are all satisfied for fk (x) and xfk+1 (x) (k ≥ 0). Then, one shows by induction that
S(S(fk )) = fk (x) and S(S(xfk+1 (x))) = −xfk+1 (x) for any k ≥ 0.
Back to the problem. (i) It is shown in Lemma 0.3 that S((1 + x2 )−1 ) = πe−2π|x| .
By (5), we get
S(−2x(1 + x2 )−2 ) = −2π 2 ixe−2π|x| .
By (6), we get
S(−2x2 (1 + x2 )−2 ) = −π(1 − 2π|x|)e−2π|x| .
Then,
S((1 + x2 )−2 ) =
π
(1 + 2π|x|)e−2π|x| .
2
(ii) Firstly, xj fk (x) (0 ≤ j ≤ 2k) are all absolutely integrable when k ≥ 1. Then, by
(6), y = (Sfk )(x) is a 2k-th order continuous differentiable function. By Lemma 0.1 and
Lemma 0.4, xy 00 + c1 y 0 + c2 xy = 0 is equivalent to
(x2 fk0 + 2xfk ) − c1 xfk −
c2 0
f = 0.
4π 2 k
Inputting fk (x) = (1 + x2 )−1−k , we get c1 = −2k and c2 = −4π 2 .
14
8. When a new social networking software appears on the market, the software company
cares about how the number of the active customers increases in time, and also would
like to investigate how certain traits of the customers evolve over time. We denote
the population density of the customers by n(t, x), where t is the time variable and x
represents how long an active customer has been using this software . We assume that
the time evolution of the density n(t, x) is governed by the following factors:
Assumption 1. When a customer keeps using the software, his or her usage time
length x increases linearly in time.
Assumption 2. When a customer uses the software, he or she may stop using it with
a stopping rate d(x) > 0. Here, we assume the stopping rate only depends on x.
Assumption 3. There are two sources of new customers:
1. The software company promotes this software by commercial advertisements.
The rate of change in the number of customers due to the advertisements is
denoted by c(t).
2. The existing customers may recommend this software to their family or friends. The effective recommendation rate is denoted by b(x), which is related to
the customer’s usage time length.
We assume at t = 0, the population density is given, n(0, x) = n0 (x). We can derive
that the time evolution of n(t, x) is given by
(
∂
∂
x) = 0, t ≥ 0, x ≥ 0,
∂t n(t, x) + ∂x n(t, x) + d(x)n(t,
R∞
(7)
N (t) := n(t, x = 0) = c(t) + 0 b(y)n(t, y)dy.
Here, N (t) is interpreted as the increasing rate of the new customers. We assume
b, d ∈ L∞
+ (0, ∞), that is, b(x) and d(x) are positive and (essentially) bounded. From this
point on, we also assume c(t) ≡ 0 for simplicity of analysis.
(i) Question and Answer (10 points) With Assumption 1 and Assumption 2,
formally derive the partial differential equation that n(t, x) satisfies as in (7), and
clearly indicate the correspondence between the assumptions and the mathematical
expressions during the derivation. Also, explain the meaning of N (t) as given in
(7).
(ii) Question and Answer (10 points) We aim to establish the connection between
N (t) and b(x). To fulfil this task, derive an equation that N (t) satisfies, such that
the equation only contains N (t), n0 (x), b(x) and d(x), but n(t, x) does not appear
in this equation. Prove that N (t) satisfies the following estimate
Z ∞
kbk∞ t
|N (t)| ≤ kbk∞ e
|n0 (x)| dx,
(8)
0
15
where k · k∞ denotes the L∞ norm.
(iii) Proof Question (10 points) Finally, we aim to explore the long time asymptotic behavior of the population density n(t, x). Since the total number might be
increasing, it is more convenient to work with a normalized density function.
We further assume the following eigenvalue problem has a unique solution (λ0 , ϕ(x))
(
ϕ0 (x) + (λ0 + d(x)) ϕ(x) = 0, x ≥ 0,
R∞
ϕ(x) > 0, ϕ(0) = 0 b(x)ϕ(x)dx = 1,
and its dual problem has a unique solution ψ(x):
(
−ψ 0 (x) + (λ0 + d(x)) ψ(x) = ψ(0)b(x),
R∞
ψ(x) ≥ 0,
0 ψ(x)ϕ(x)dx = 1.
x ≥ 0,
We define the normalized density ñ(t, x) := n(t, x)e−λ0 t . Prove that for any convex
H : R+ → R+ with H(0) = 0, we have
Z
ñ(t, x)
d ∞
ψ(x)ϕ(x)H
dx ≤ 0, ∀t ≥ 0,
dt 0
ϕ(x)
and show that
Z ∞
ψ(x)n(t, x)dx = e
λ0 t
Z ∞
ψ(x)n0 (x)dx.
0
0
For simplicity, we assume that the related boundary terms at ∞ can be neglected.
16
R8 Answer. (i). There are many ways to derive the equation.
Approach 1: characteristic method. Since the usage time length grows linearly in
time, we can define the characteristic x(t) satisfying
dx(t)
= 1.
dt
Along the characteristic and according to the meaning of the stopping rate, we have
d
n(t, x(t)) = −d(x(t))n(t, x(t)),
dt
which leads to the PDE.
Approach 2: infinitesimal method. For an infinitesimal time δt 1, by Assumption
I and Assumption II, we have
n(t + δt, x + δt) = n(t, x) − δtd(x)n(x, t) + o(δt).
where the first term on the right hand side is from the time shifting, and the second
term accounts for the stopping rate. If we divide each side by δt and let δt → 0, then
the PDE is derived.
For N (t), it suffices to explain the contribution from the existing customers. For
customers with a certain usage time length, their contribution to the increase rate of the
new customers is b(x)n(t, x). To obtain the increase rate from all existing customers, we
R∞
should add up all their contributions, so the expression is given by 0 b(y)n(t, y)dy.
(ii). We need to first solve the PDE to get an equation for N (t).
This is a first order hyperbolic equation, which can be solved by the method of
characteristics. We rewrite the equation as
d
n(t + s, x + s) + d(x + s)n(t + s, x + s) = 0,
ds
Rx
then if we define D(x) = 0 d(y)dy, we get
i
d h D(x+s)
e
n(t + s, x + s) = 0.
ds
(9)
When s ≥ max(−t, −x), we have
eD(x+s) n(t + s, x + s) = eD(x) n(t, x),
∀x ≥ 0, t ≥ 0.
In particular, we can let x = y and s = −y to get when t ≥ y,
n(t, y) = N (t − y)e−D(y) .
Let x = y, s = −t, we get when t ≤ y,
n(t, y) = n0 (y − t)eD(y−t)−D(y) .
17
(10)
In order to derive an equation for N (t), we split its expression into two parts
Z ∞
Z t
Z ∞
b(y)n(t, y)dy.
b(y)n(t, y)dy +
b(y)n(t, y)dy =
N (t) =
t
0
0
Clearly, the characteristics of the first term originate from x = 0, t ≥ 0, and those of the
second term originate from x ≥ 0, t = 0. With the expression for n(t, x), we get
Z ∞
Z t
−D(y)
b(y)eD(y−t)−D(y) n0 (y − t)dy,
(11)
b(y)e
N (t − y)dy +
N (t) =
t
0
which leads to
Z t
−D(t−x)
b(t − x)e
N (t) =
Z ∞
N (x)dx +
b(x + t)eD(x)−D(x+t) n0 (x)dx.
(12)
0
0
Since d(x) > 0, so D is an increasing function, and e−D(t−x) , eD(x)−D(x+t) in the
expression above are less than 1. By the boundedness of b(x), we get
Z t
Z ∞
|N (t)| ≤ kbk∞
|N (x)|dx + kbk∞
|n0 (x)| dx.
0
0
And with Gronwall’s lemma, we can obtain the desired inequality.
(iii). This is a generalized relative entropy estimate.
First, we rewrite the original equation as an equation that the normalized density
satisfies
∂
∂
n
e(t, x) +
n
e(t, x) + (λ0 + d(x)) ñ(t, x) = 0.
∂t
∂x
We further rewrite
∂ ñ(t, x)
∂ ñ(t, x)
+
= 0,
∂t ϕ(x)
∂x ϕ(x)
and then we get
∂
H
∂t
ñ(t, x)
ϕ(x)
∂
+
H
∂x
ñ(t, x)
ϕ(x)
= 0.
With the eigenvalue problem and the dual problem, we get
(
∂
= −ψ(0)b(x)ϕ(x), x ≥ 0,
∂x [ϕ(x)ψ(x)]
R∞
ψ(x) ≥ 0,
0 ψ(x)ϕ(x)dx = 1.
By direct calculations, we get
∂
ñ(t, x)
∂
ñ(t, x)
ñ(t, x)
ψ(x)ϕ(x)H
+
ψ(x)ϕ(x)H
= −ψ(0)b(x)ϕ(x)H
.
∂t
ϕ(x)
∂x
ϕ(x)
ϕ(x)
Let dµ(x) = b(x)ϕ(x)dx, and integrate the equation above over R+ w.r.t. x, we get
Z
Z ∞ d ∞
ñ(t, x)
ñ(t, x)
ñ(t, 0)
dx = −ψ(0)
dµ(x) + ψ(0)H
.
ψ(x)ϕ(x)H
H
dt 0
ϕ(x)
ϕ(x)
ϕ(0)
0
18
We notice that ϕ(0) = 1, n(t, 0) =
ñ(t, 0)
= ñ(t, 0) =
ϕ(0)
R∞
0
b(x)n(t, x)dx, and thus
Z ∞
Z ∞
b(x)ñ(t, x)dx =
0
0
ñ(t, x)
dµ(x),
ϕ(x)
then we get
Z ∞ Z ∞
Z
d ∞
ñ(t, x)
ñ(t, x)
ñ(t, x)
ψ(x)ϕ(x)H
dx = ψ(0) −
H
dµ(x) + H
dµ(x) .
dt 0
ϕ(x)
ϕ(x)
ϕ(x)
0
0
By Jensen’s inequality, we get
Z
d ∞
ñ(t, x)
ψ(x)ϕ(x)H
dx ≤ 0.
dt 0
ϕ(x)
Finally by setting H(u) = u, we get the desired identity.
19