2022 Cpnn ¥êÆ¿m
üÀKµ^5AÛ
•
r¢%á•N ABCD − A1 B1 C1 D1 (b
1)
6 ^¡þ
é
AB = 1 )UXe•ª©¤ 12 ¬£ã1¤:
‚ AC,AB1 , AD1 , C1 B, C1 D, C1 A1 ;
2) •Ä±á•N¥%•º:, þã 6 ^é
3) ù 18 ‡n
N c;
/rá•Nƒ¤
4) z‡o¡N=ÏLÙþá•N
‚9 12 ^cƒ˜•é>
n
/;
12 ¬, z¬´˜‡o¡N, z‡o¡Nkü^c´á•
cÚÙ§o¡Në(.
图1: 磁性几何魔方
ù样˜‡玩具Œ±{ш«/G£ã2¤.
图2: 其它形状的例子
¯: 3ù‡玩ä¤kŒU
/G¥,Ùþü:ƒm(3˜m¥)ål
1
•ŒŠ´õ
º
√
11
√
7+4 2
p
√
13
√
1+2 2
±þÑØé
2
üÀKµm…Œ*
,F§,/ ˜¡ãŒ˜|/þk˜| Þ²âLü§ÚåÜ©+¯Œ*. ·‚òd/wŠ
˜‡î¼²¡§Lü «•¥%PŠ: K.
+¯• A1 , A2 , . . . , An , . . ., ¦‚Uìl A1 m© ^S•gÀ½˜‡Œ*
˜
P1 , P2 , . . . , Pn , . . . ,
IÓž÷v±en‡^‡.
^‡1: An Œ*
˜† K
ålØ u 10 ’§=é?¿ n, KPn ≥ 10 ’¶
^‡2: An Œ*
˜†c¡z‡<
¿ 1 ≤ m ≤ n − 1, Pm Pn ≥ 1 ’.
˜
ålÑI‡Ø
u 1 ’§=é?¿ n ≥ 2 Ú?
^‡3: 3÷v^‡1 Ú^‡2 cJe§An ÀJ† K ¦þ C :?1Œ*§=¦F
" KPn
• ŒU Š. XJÓž÷v^‡1 Ú^‡2 …¦
KPn
• ŒUŠ
Pn ØŽ˜‡§@o An Œ±ÀJÙ¥ ?¿˜‡:.
~X§éu A1 , ¦ÀJ ˜žvk^‡2§ ¦¬ÀJ± K • %§10 ’•Œ»
± C þ ?¿˜:£† K
ålT• 10 ’¤¶éu A2 , ¦•F" P2 † K
ålT
• 10 ’§= P2 •3 C þ. du C þkNõ:† P1 ålØ u 1 ’§¦Œ±ÀJÙ¥
?¿˜:.
(1) ž¯§±e`{=‡
(º
•3 ¢ê c1 , c2 , ¦ é?¿
ê n, ÃØ A1 , A2 , . . . , An , NoÀJ ˜§
þk c1 ≤ KPn ≤ c2 £ü µ’¤¶
•3 ¢ê c1 , c2 , ¦ é?¿
ê n, ÃØ A1 , A2 , . . . , An , NoÀJ ˜§
√
√
þk c1 n ≤ KPn ≤ c2 n £ü µ’¤¶
•3 ¢ê c1 , c2 , ¦ é?¿
ê n, ÃØ A1 , A2 , . . . , An , NoÀJ ˜§
þk c1 n ≤ KPn ≤ c2 n £ü µ’¤¶
•3 ¢ê c1 , c2 , ¦ é?¿
ê n, ÃØ A1 , A2 , . . . , An , NoÀJ ˜§
2
2
þk c1 n ≤ KPn ≤ c2 n £ü µ’¤.
(2) du<´k˜½NÈ §¤±XJ,<Õ3,˜<Œ* À‚´»NCž§1 <Ò
¬ 1˜< 4À‚. ·‚@•§éØÓ i, j, XJ± Pi • %§ 16 ’•Œ»
±†‚
ã KPj ƒ §@o Aj Ò¬ Ai 4À‚ wØ Lü
m.
ž¯§±e=‡`{
(º
k 60 ¶+¯Œ*ž§˜½k<wØ Lü
m¶
k 60 ¶+¯Œ*ž§•3¤k<Ñw Lü m ŒU5§
k 800 ¶+¯Œ*ž§˜½k<wØ Lü
m¶
k 800 ¶+¯Œ*ž§•3¤k<Ñw Lü m ŒU5§
k 10000 ¶+¯Œ*ž§˜½k<wØ Lü
m¶
k 10000 ¶+¯Œ*ž§•3¤k<Ñw Lü m ŒU5.
3
üÀKµ/mm)%0_Ý
S!Ïm§/]cÚÚ0ÚEúiíÑ #S84¹ÄµzÝÚEÑNR˜‡ù•§ù•¥
õke /m0§/)0§/%0¥ ˜±ãY"£Xeã¤
8àü‡/m0§˜‡/)0§˜‡/%0=Œ©à¤•/mm)%0 [4"ù‘¹Ä˜
²íѧҤ•
ù ±§éõ<F"U 8à˜ @"b ù•¥ ãY´Õá‘Å
©Ù £¿…ØUlù• *þ?1«O¤§/m0§/)0§/%0n±ù•Uþ!V
Ç 13 ©Ù"
(1) Â8à˜ @/mm)%0 [4¤I‡ ï ÚEÝê êÆÏ"´õ º
6 13 ;
7 31 ;
8 31 ;
9 31 ;
±þÑØé"
(2) 3½|Ü ±¬?Ø¥§Œ[@• c ãYݘ'~§¬ —3Â8/mm)%0
[4žÂ8
õ /)0Ú/%0§u´&?ŒU U?•Y"PãY/m”!/)0Ú
/%0 ݘ'~•(p, q, r)§@oe¡=«•Ye§Â8à˜ @/mm)%0 [4¤I
‡ ï ÚEÝê êÆÏ"´•
º
(p, q, r) = ( 13 , 13 , 31 );
(p, q, r) = ( 12 , 14 , 41 );
3 3
(p, q, r) = ( 25 , 10
, 10 );
(p, q, r) = ( 34 , 18 , 81 ).
4
y²K
‰ ½ 8 Ü X, e ¼ ê f : X × X → [0, 1] ÷ v µ é ? ¿ > 0, • 3 k • ‡ X ¥
ƒ b1 , b2 , ..., bm ¦ é?¿ t ∈ X, •3,‡ bi(t) ÷vµ
|f (x, t) − f (x, bi(t) )| < , ∀x ∈ X,
K¡ f ´m˜— "aq/§é?¿ > 0, •3k•‡ X ¥
¿ t ∈ X, •3,‡ ai(t) ÷vµ
|f (t, x) − f (ai(t) , x)| < , ∀x ∈ X,
K¡ f ´†˜— "¦yµ†˜— dum˜—"
5
ƒ a1 , a2 , ..., an ¦
é?
y²K
n´
ê, V = Rn ´ n ‘m, k˜|Ä ei = (0, . . . , 0, 1, 0, . . . , 0) (1 ≤ i ≤ n)÷
| {z } | {z }
i−1
n−i
v
(ei , ej ) = δi,j ,
Ù¥
δi,j =
1 if i = j
0 if i 6= j
´Kronecker ÎÒ, (·, ·) ´ V þ SÈ. é V ¥ š"•þ v , ½Â‚5C† sv : V → V •
sv (u) = u −
2(u, v)
v, ∀u ∈ V.
(v, v)
éu0u 0 Ú n ƒm
ê k, P Grk (V ) • V
f˜m W , P [W ] • Grk (V ) ¥ ƒA ƒ.
(vi , vj ) = δi,j ), ½Â s[W ] : V → V •
k ‘f˜m
W
˜|
8Ü. éu V
˜‡ k ‘
5‰Ä {v1 , . . . , vk } (•=,
s[W ] = sv1 · · · svk .
(1) y² s[W ] Ø•6u
5‰Ä {v1 , . . . , vk }
À .
(2) y² s2[W ] = id.
(3) é [W 0 ] ∈ Grk (V ), ½Â
t[W ] ([W 0 ]) = [s[W ] (W 0 )],
Ù¥ s[W ] (W 0 ) ´ W 0 3 s[W ] e
”. ·‚¡ Grk (V )
f8 X •˜‡“k 8”e
t[W ] ([W 0 ]) = [W 0 ], ∀[W ], [W 0 ] ∈ X.
žé
Grk (V ) ¥k 8 •Œ
ƒ‡ê, ¿y²ƒ.
6
)‰Kµa²
¯4
˜ ¯4 3 ‘‚:²¡þ‹I• (n, 0) ?
¯4"¦¤óŠ ¯4ÕK3
: (0, 0) ?"d §¯4 z˜Ú‰Ú•• 1 {ü‘ÅiÄ"3±e ˆ¯¥§Ø”@•
ê n Œu 1"
(1) - P1,n •T¯4
Áy²
3r
TÐ bn1.5 c Úž§ål¯4Õ
†‚ålŒu n2
lim P1,n = 1
n→+∞
(2) - P2,n •T¯4 3c bn1.5 c ÚSQ²£ L¯4Õ VǧÁy²
lim P2,n = 0
n→+∞
(3) - P3,n •T¯4
c 2n ÚSQ²£ L¯4Õ VǧÁy²
lim P3,n = 1
n→+∞
7
Vǧ
)‰K
(1) y²Ø•3÷vXe^‡
±ÏS
sup
a1 , a2 , a3 , . . ., z‘þ• ±1, …é?¿knê θ,
N
X
an e2πinθ < +∞.
N ∈N n=1
(2) y²Ø•3z‘þ• ±1
S
a1 , a2 , a3 , . . . ÷vµé?¿knê θ
sup
N
X
an e2πinθ ≤ 2022 .
N ∈N n=1
(3) Þ~`²µ•3z‘þ• ±1 S
õ‘ −1§…é?¿ θ ∈ Q − Z,
sup
a1 , a2 , a3 , . . . ÷vµÙ¥káõ‘ 1§•ká
N
X
an e2πinθ < +∞.
N ∈N n=1
8
)‰Kµm4ª
!8
O
b \ Ác¬m4ª
üÀ•EâÏn§KI^êÆ•£ïÄ O•Y Ün5"3
m4ª
À!8¥§k˜‡•Y´4˜+dÃ<Å›› 3ŒÔ7˜‡
/G |/
wX"Ï•Ã<Åv
õ£ ´Ø¬P×½öƒE¤§·‚@•Œ±^˜‡VǗݼ
ê ρ(t, v)(≥ 0) 5•xÃ<Å ©Ù"Ï•|/´ ‚/ §¤±·‚Œ±@•„Ý v ∈ R§
§L«Ã<Å ‚„Ý"@o§éu?¿‰½ žm t, Úü‡„Ý v1 < v2 §
Z v2
ρ(t, v) dv
v1
L« NÃ<Å¥„Ý0u v1 Ú v2 ƒm VÇ"
duÃ<Å
$ÄÅn§®•ù‡—ݼê üz÷vXe •§
ρt + (u(t) − v)ρ v = ρvv , v ∈ R, t > 0.
Ù¥ u(t) •Ã<Å •-„Ý"
(1) • ïÄNo‰Ã<ÅÜ·
•-„ݧŒ ‰ üJÆ`§AT4
u(t) = u0 + u1 N (t)
Ù¥ u0 > 0§u1 > 0§
N (t) L«Ã<Å „Ý Ü£v+ = max{0, v}¤ ²þ
Z +∞
Z +∞
N (t) =
v+ ρ(t, v) dv =
vρ(t, v) dv.
−∞
0
´§\õ¿/J2 §XJ u1 > 1 @o N (t) 3üzL§¥Ø¬kþ.§±–ué¯
ÚåÃ<Å
æ"\Œ±y²\ þã(Øíº£• •B?ا·‚ Ñ ρ 9Ù ê
3 |v| → +∞ ž
z"¤
(2) 3æB Œ Ú\ ïÆ § üq3•Äù Ã<ŴĬ3w1¥3
|/þþ
!©Ù§u´·‚I‡•Ä'u„ÝÚ R ˜ éܗݼê p(t, x, v)(≥ 0)"ùp x ∈ [0, 2π]
2π
L«Ã<Å3
þ ƒé ˜§w, 0 p(t, x, v)dx = ρ(t, v)"®•ù‡éܗݼê
üz÷v
pt + vpx + (u(t) − v)p v = pvv , x ∈ [0, 2π], v ∈ R, t > 0.
…duÃ<Å37 w1§3 x ••þXe^‡÷v
p(t, 0, v) = p(t, 2π, v),
v ∈ R,
t > 0.
\ŒÿßÿµÃØЩ©ÙXÛ§Ã<Åé¯Ò3
þ C˜‡þ!©Ù"\Œ±y²½
öy–ù‡·Kíº£• •B?ا·‚ Ñ p 9Ù ê3 |v| → +∞ ž
z"¤
9
2022 Cpnn
11K
K£ ü À K ¤ ^ 5 A Û
¥êÆ¿m£ýÀm¤
•
r¢%á•N ABCD − A1 B1 C1 D1 (b
1)
AB = 1 )UXe•ª©¤ 12 ¬£ã1¤:
6 ^¡þ é ‚ AC,AB1 , AD1 , C1 B, C1 D, C1 A1 ;
2) •Ä±á•N¥%•º:, þã 6 ^é ‚9 12 ^cƒ˜•é> n /;
3) ù 18 ‡n
N c;
/rá•Nƒ¤
12 ¬, z¬´˜‡o¡N, z‡o¡Nkü^c´á•
4) z‡o¡N=ÏLÙþá•N cÚÙ§o¡Në(.
图1: ^5AÛ •
ù ˜‡玩具Œ±{ш«/G£ã2¤.
图2: Ù§/G ~f
¯: 3ù‡玩具¤kŒU /G¥,Ùþü:ƒm(3˜m¥)ål •ŒŠ´õ º
1
√
(A) 11
(B)
p
√
7+4 2
(C)
√
13
√
(D)1 + 2 2
(E) ±þÑØé
1 ‰ Y ÀD"±ABCD − A1 B1 C1 D1 IPá•N
•AÚC1 . @o,ùü‡º:©Oáu6‡ o¡N,
C/Ñ) ¤˜‡>••1 8>/(ŒUòz).
5¿ ù‡
º:.
äd12‡o¡N(l
´àã/)
l‡º:, ¿ m©žÀ½ éº:
B,C,D,D1 ,A1 , B1 ù8‡º:(ÃØNo
¤, Ïd
•ŒŠ
7½´ü‡
o¡N
3˜‡o¡N¥, o‡º:Œ©•3a:
• ˜‡º:3á•N¥éAuA½C1 , ·‚¡ƒ•1
1˜a º:;
• ü‡º:(l §‚ƒm
c)áu(˜m)8>/BCDD1 A1 B1 , ·‚¡ƒ•1
1
• ˜‡º:éAuá•N
¥%, § Ó˜‡
o¡N
,n‡º:
¡ƒ•1
1n aº:.
aº:;
√
3
ålÑ´
, ·‚
2
·‚• ù º:ƒm÷
÷c ål •ŒŠ, ´ eL:
ål
1˜aº:
1 aº:
1naº:
1˜aº:
√
1+2 2
√
2+ 2
√
√
3
1+ 2+
2
1 aº:
√
2+ 2
3
√
2+
3
2
√
l ù‡ äþ?¿ü:ƒm ålج‡L1 + 2 2.
2
1naº:
√
√
3
1+ 2+
√ 2
3
2+
2
√
2+ 3
ù‡Š´Œ±ˆ
, Xe㤫:
12,3K
K£ ü À K ¤ m … Œ *
,F§,/ ˜¡ãŒ˜|/þk˜| Þ²âLü§ÚåÜ©+¯Œ*. ·‚òd/wŠ
˜‡î¼²¡§Lü «•¥%PŠ: K.
+¯• A1 , A2 , . . . , An , . . ., ¦‚Uìl A1 m© ^S•gÀ½˜‡Œ*
˜
P1 , P2 , . . . , Pn , . . . ,
IÓž÷v±en‡^‡.
^‡1: An Œ*
˜† K
ålØ u 10 ’§=é?¿ n, KPn ≥ 10 ’¶
^‡2: An Œ*
˜†c¡z‡<
¿ 1 ≤ m ≤ n − 1, Pm Pn ≥ 1 ’.
˜
ålÑI‡Ø
u 1 ’§=é?¿ n ≥ 2 Ú?
^‡3: 3÷v^‡1 Ú^‡2 cJe§An ÀJ† K ¦þ C :?1Œ*§=¦F
" KPn
• ŒU Š. XJÓž÷v^‡1 Ú^‡2 …¦
KPn
• ŒUŠ
Pn ØŽ˜‡§@o An Œ±ÀJÙ¥ ?¿˜‡:.
~X§éu A1 , ¦ÀJ ˜žvk^‡2§ ¦¬ÀJ± K • %§10 ’•Œ»
± C þ ?¿˜:£† K
ålT• 10 ’¤¶éu A2 , ¦•F" P2 † K
ålT
• 10 ’§= P2 •3 C þ. du C þkNõ:† P1 ålØ u 1 ’§¦Œ±ÀJÙ¥
?¿˜:.
(1) ž¯§±e`{=‡
(º
(A) •3 ¢ê c1 , c2 , ¦ é?¿
ê n, ÃØ A1 , A2 , . . . , An , NoÀJ
þk c1 ≤ KPn ≤ c2 £ü µ’¤¶
˜§
(B) •3 ¢ê c1 , c2 , ¦ é?¿
√
√
þk c1 n ≤ KPn ≤ c2 n £ü
ê n, ÃØ A1 , A2 , . . . , An , NoÀJ
µ’¤¶
˜§
(C) •3 ¢ê c1 , c2 , ¦ é?¿
ê n, ÃØ A1 , A2 , . . . , An , NoÀJ
þk c1 n ≤ KPn ≤ c2 n £ü µ’¤¶
˜§
(D) •3 ¢ê c1 , c2 , ¦ é?¿
ê n, ÃØ A1 , A2 , . . . , An , NoÀJ ˜§
þk c1 n2 ≤ KPn ≤ c2 n2 £ü µ’¤.
(2) du<´k˜½NÈ §¤±XJ,<Õ3,˜<Œ* À‚´»NCž§1 <Ò
±†‚
¬ 1˜< 4À‚. ·‚@•§éØÓ i, j, XJ± Pi • %§ 61 ’•Œ»
ã KPj ƒ §@o Aj Ò¬ Ai 4À‚ wØ Lü
m.
ž¯§±e=‡`{
(º
(A)
k 60 ¶+¯Œ*ž§˜½k<wØ
Lü
(B)
k 60 ¶+¯Œ*ž§•3¤k<Ñw Lü m ŒU5§
k 800 ¶+¯Œ*ž§˜½k<wØ Lü
m¶
(C)
k 800 ¶+¯Œ*ž§•3¤k<Ñw Lü m ŒU5§
k 10000 ¶+¯Œ*ž§˜½k<wØ Lü
m¶
(D)
k 10000 ¶+¯Œ*ž§•3¤k<Ñw Lü m ŒU5.
3
m¶
2 ‰ Y ÀB"
Œ»Š §d Pn
§Ïd
KPn •Ý• dn ’. ˜•¡§·‚± P1 , P2 , . . . , Pn−1 • %§1 ’•
À Œ•ù
2\þ C SÜ7,CX ± K • %§dn •Œ»
π · d2n ≤ (n − 1) · π · 12 + π · 102 ,
dn ≤
√
n + 99 ≤
√
√
100n = 10 n.
,˜•¡§·‚± P1 , P2 , . . . , Pn • %§ 21 ’•Œ»Š §du P1 , P2 , . . . , Pn üüƒm
ålØ u 1 ’§ ù
*d؃ . , §dK¿Œ• KP1 , KP2 , . . . , KPn þA؇
L dn ’£ek 1 ≤ m ≤ n − 1 ¦ KPm •Ý‡L dn ’§K Am Œ±ÀJål•C
Pn :§gñ¤§Ïd§ù
Ñ3± K • %§dn + 12 ’•Œ»
S§Ïd
1
1
π(dn + )2 ≥ n · π · ( )2 ,
2
2
√
dn ≥
n 1
− .
2
2
√
5¿
n = 1 ž§d1 = 10§ é n ≥ 2, 21 < 2 5 n ,
√
√
√
n 2 n
n
dn >
−
=
.
2
5
10
√
√
Ïd§ 10n ≤ dn ≤ 10 n, B À‘ (.
3 ‰Y ÀB"˜•¡§
π
1
1
π
k 60 <Œ*ž§du sin 60
> 10
sin π6 = 20
,
20 sin 60
> 1, ù
`²˜‡S u C
60 >/ >•Ø u 1 ’. Ïd§ P1 , P2 , . . . , P60 TЕù‡
60 >/ ¤kº:ž§ÎÜK8^‡. é?¿ØÓ i, j, : Pi
KPj
ålØ
π
π
1
u 10 sin 30
’§d sin 30
> 15 sin π6 = 10
Œ•ù ålÑØ u 1 ’§ ¤k<ÑUw L
ü
m.
−−→ −−→
−−−−→
,˜•¡§ k 800 <Œ*ž§Š ‚ KP1 , KP2 , . . . , KP800 , XJkü^ ‚-ܣؔ
−−→ −−→
• KPi , KPj -ܤ§ KPj > KPi , @o Aj
Ai 4 À‚.
1
1
b vk-Ü
‚§·‚ky²§XJ•3˜‡ ∠Pi KPj ÷v ∠Pi KPj ≤ 12
KPi +
1
£
ü •lݧ‚ã ü •’¤§@o, Ai , Aj ¥k˜< ,˜< 4 À‚.
KPj
1
Ø”˜„5§·‚b KPi ≤ KPj , d KPi , KPj ≥ 10 • ∠Pi KPj ≤ 60
,
d Pi
KPj Rv u‚ã KPj Sܧ Pi
KPj ål•
KPi sin ∠Pi KPj < KPi · ∠Pi KPj ≤
Aj
Ai
4 À‚.
4
KPi 1
1 1
+
≤ ,
12 KPi KPj
6
§´b
. Ï
−−→ −−→
−−−−→
•Ä ‚ KP1 , KP2 , .. . , KP800 r±
K •º:
1
1
1
´z‡ éA 12
+
Ú
KPi
KPj
±
©¤
800 ‡
§§‚
Ú• 2π§
800
X 1 1
1 1X 1
+
=
12 KPi KPj
6
KPi
≥
=
1
6
1
6
i=1
800
X
√
i=1
899
X
1
(1(1) ¯ (Ø)
i + 99
1
√
m
m=100
899 Z m+1
X
1
1
√ dx (Ï• √ ´~¼ê)
x
x
m
m=100
√
√
Z 900
1
900 − 100
20
1
√ dx =
=
> 2π.
=
6 100
3
3
x
≥
Ïd§•3˜‡
1
6
1
∠Pi KPj ÷v ∠Pi KPj ≤ 12
nþ¤ã§B À‘ (.
5
1
1
KPi + KPj
7,k<wØ Lü
m.
14,5K
K£ ü À K ¤ / m m ) % 0 _ Ý
S!Ïm§/]cÚÚ0ÚEúiíÑ #S84¹ÄµzÝÚEÑNR˜‡ù•§ù•¥
õke /m0§/)0§/%0¥ ˜±ãY"£Xeã¤
8àü‡/m0§˜‡/)0§˜‡/%0=Œ©à¤•/mm)%0 [4"ù‘¹Ä˜
²íѧҤ•
ù ±§éõ<F"U 8à˜ @"b ù•¥ ãY´Õá‘Å
©Ù £¿…ØUlù• *þ?1«O¤§/m0§/)0§/%0n±ù•Uþ!V
Ç 13 ©Ù"
(1) Â8à˜ @/mm)%0 [4¤I‡ ï ÚEÝê êÆÏ"´õ º
(A) 6 13 ;
(B) 7 31 ;
(C) 8 13 ;
(D) 9 31 ;
(E) ±þÑØé"
(2) 3½|Ü ±¬?Ø¥§Œ[@• c ãYݘ'~§¬ —3Â8/mm)%0
[4žÂ8
õ /)0Ú/%0§u´&?ŒU U?•Y"PãY/m”!/)0Ú
/%0 ݘ'~•(p, q, r)§@oe¡=«•Ye§Â8à˜ @/mm)%0 [4¤I
‡ ï ÚEÝê êÆÏ"´•
º
(A) (p, q, r) = ( 13 , 13 , 13 );
(B) (p, q, r) = ( 21 , 14 , 14 );
3 3
(C) (p, q, r) = ( 52 , 10
, 10 );
(D) (p, q, r) = ( 34 , 18 , 18 ).
6
4 ‰ Y ÀB"Œ±ÏLëYžmi\
PoissonL§ ){
˜„ )ÛLˆª"b
_ÝoOkn‡ãY§zgÂ8 ãYi VÇ´pi §±9•ª8I´éuz‡ãYiI‡Â
8ki ‡"PÄgˆ¤ Â8ž ï _Ýêþ´N §8I=OŽE[N ]"·‚òù‡L§i\
ëYžm PoissonL§¥µb k˜‡ëê•1 Poisson:L§§zg&Ò ˆž§ÑÕ
á UìVÇpi Ä ãY"PÊž
Ti = inf{t ∈ R+ : 3žmt Â8
ki ‡ãYi}
T = max Ti .
1≤i≤n
¯¢þkXe(صE[T ] = E[N ]"·‚ky²ù‡(Ø"Œ±òPoisson:L§
&Ò¤I žmPŠτj §@ok
T =
N
X
ˆ
1j‡
τj .
j=1
¯¢þ(τj )j≥1 ´ÕáÓ©Ù
§ëê•1
•ê©Ù"|^^‡Vǧ·‚k
N
X
E[T ] = E[
τj ]
j=1
=
∞
X


k
X
E
τj |N = k  P[N = k]
j=1
k=0
=
∞
X
P[N = k]kE[τ1 ]
k=0
= E[N ].
e5·‚•I‡OŽE[T ]"Äk·‚¦^Fubini½n
Z ∞
E[T ] =
P[T > t] dt
Z0 ∞
=
(1 − P[T ≤ t]) dt
Z0 ∞
=
(1 − P[Ti ≤ t, ∀1 ≤ i ≤ n]) dt.
0
¯¢þ§ëYžm PoissonL§ thinning5Ÿ•·‚ME éõÕá5§¦ OŽC {
ü"'Xd?·‚Œ±rÂ8z˜‡ãYiwŠ´ëê•pi Õá n‡PoissonL§"u´k
!
Z ∞
n
Y
E[T ] =
1−
P[Ti ≤ t] dt.
0
i=1
Óž§P[Ti ≤ t]Œ±† |^Poisson©Ù ÑLˆª
P[Ti ≤ t] = 1 −
kX
i −1
k=0
7
e−pi t
(pi t)k
.
k!
·‚Œ±
•ª
Lˆª
Z ∞
1−
E[N ] =
0
n
Y
1−
kX
i −1
i=1
k=0
(pi t)k
e−pi t
k!
!!
dt.
3·‚ K8¥§n = 3§8IÂ8(k1 , k2 , k3 ) = (2, 1, 1)"·‚Œ±OŽ
E[N ] = 1 + p1 +
2
1
1
+
+
p1 p2 p3
3(p1 , p2 , p3 ) = (1/3, 1/3, 1/3)
−
3
X
i=1
)Û Lˆª
p1
p1
1
−
−
.
2
1 − pi (p1 + p2 )
(p1 + p3 )2
Ï"•7 31 §‰YÀB"
5 ‰ Y ÀC"þ˜¯¥OŽ
•YA Ï"´7 13 §Dw,Ø´˜‡Ð •Y§Ï•Â
8 /%0iãY êÆÏ"´8§®²‡Ñ •YA"¤±Ø%3u• •YB§C"•
YB3ùpš~Î܆*§´˜‡‘kV« €²"|^){2¥
úª
2 1 1
1
1
1
E[τ ] = 1 + p +
+ +
−
+
+
p q r
p+q p+r q+r
p
p
−
.
−
2
(p + q)
(p + r)2
1
B,CéA Ï"©O•7 18
, 6 223
245 "¤±¯¢þ•YC´•Z §ÀC"
8
16K
K£ y ² K ¤
‰ ½ 8 Ü X, e ¼ ê f : X × X → [0, 1] ÷ v µ é ? ¿ > 0, • 3 k • ‡ X ¥
ƒ b1 , b2 , ..., bm ¦ é?¿ t ∈ X, •3,‡ bi(t) ÷vµ
|f (x, t) − f (x, bi(t) )| < , ∀x ∈ X,
K¡ f ´m˜— "aq/§é?¿ > 0, •3k•‡ X ¥
¿ t ∈ X, •3,‡ ai(t) ÷vµ
ƒ a1 , a2 , ..., an ¦
é?
|f (t, x) − f (ai(t) , x)| < , ∀x ∈ X,
K¡ f ´†˜— "¦yµ†˜— dum˜—"
6 ‰Y
•Iy²m˜—íц˜—"?‰ > 0, dm˜—^‡
X ¥k•‡
ƒ b1 , b2 , ..., bm . •ÄN h : X → [0, 1]m , h(x) = (f (x, b1 ), f (x, b2 ), ..., f (x, bm )). d [0, 1]m
;—5§•3k•‡ c1 , c2 , ..., cn ¦ é?¿ x ∈ X, •3,‡ cix ¦ |h(x) − h(cix )| < ,
l
|f (x, bi ) − f (cix , bi )| < , ∀i ∈ {1, 2, ..., m}.
(1)
dm˜—^‡• |f (x, t) − f (x, bi(t) )| < (∀x ∈ X), |f (cix , t) − f (cix , bi(t) )| < , (Ü(1), •é
?¿ t ∈ X,
|f (x, t)−f (cix , t)| ≤ |f (x, t)−f (x, bi(t) )|+|f (x, bi(t) )−f (cix , bii )|+|f (cix , bi(t) )−f (cix , t)| < 3.
Ï• ´?¿ §¤± f ´†˜— "
9
17K
K£ y ² K ¤
n´
ê, V = Rn ´ n ‘m, k˜|Ä ei = (0, . . . , 0, 1, 0, . . . , 0) (1 ≤ i ≤ n)÷
| {z } | {z }
i−1
n−i
v
(ei , ej ) = δi,j ,
Ù¥
δi,j =
1 if i = j
0 if i 6= j
´Kronecker ÎÒ, (·, ·) ´ V þ SÈ. é V ¥ š"•þ v , ½Â‚5C† sv : V → V •
sv (u) = u −
2(u, v)
v, ∀u ∈ V.
(v, v)
éu0u 0 Ú n ƒm
ê k, P Grk (V ) • V
f˜m W , P [W ] • Grk (V ) ¥ ƒA ƒ.
(vi , vj ) = δi,j ), ½Â s[W ] : V → V •
k ‘f˜m
W
˜|
8Ü. éu V
˜‡ k ‘
5‰Ä {v1 , . . . , vk } (•=,
s[W ] = sv1 · · · svk .
(1) y² s[W ] Ø•6u
5‰Ä {v1 , . . . , vk }
À .
(2) y² s2[W ] = id.
(3) é [W 0 ] ∈ Grk (V ), ½Â
t[W ] ([W 0 ]) = [s[W ] (W 0 )],
Ù¥ s[W ] (W 0 ) ´ W 0 3 s[W ] e
”. ·‚¡ Grk (V )
f8 X •˜‡“k 8”e
t[W ] ([W 0 ]) = [W 0 ], ∀[W ], [W 0 ] ∈ X.
žé
Grk (V ) ¥k 8 •Œ
ƒ‡ê, ¿y²ƒ.
10
7 ‰ Y (1) P W ⊥ • W 3 V ¥
Ö. K
s[W ] |W = −1 …s[W ] |W ⊥ = 1.
dd·‚•x
s[W ] ¿y² §† W
5‰Ä {v1 , . . . , vk }
À Ã'.
(2) du
s[W ] |W = −1…s[W ] |W ⊥ = 1,
¤± s2[W ] = id.
(3) (i) ·‚y²
t[W ] ([W 0 ]) = [W 0 ] ⇔ W 0 = (W 0 ∩ W ) ⊕ (W 0 ∩ W ⊥ ).
¿©5´y. 7‡5: b t[W ] ([W 0 ]) = [W 0 ]. Ké?¿ u ∈ W 0 , ·‚k s[W ] (u) ∈ W 0 .
P u = u1 + u2 , Ù¥ u1 ∈ W … u2 ∈ W ⊥ . K −u1 + u2 = s[W ] (u) ∈ W 0 . ¤±, u1 ∈ W 0
…u2 ∈ W 0 . Ïd,
W 0 ⊂ (W 0 ∩ W ) ⊕ (W 0 ∩ W ⊥ ).
w,, (W 0 ∩ W ) ⊕ (W 0 ∩ W ⊥ ) ⊂ W 0 . ù ,
W 0 = (W 0 ∩ W ) ⊕ (W 0 ∩ W ⊥ ).
(ii) P
d(i) ¥
X = {[spanR {ei1 , . . . , eik }] : 1 ≤ i1 < · · · < ik ≤ n}.
OOK, X ´‡“k 8”. ù‡8Ü X
ƒ‡ê• nk .
(iii) é n Š8B, ·‚y²: Grk (V ) ¥k 8 X
ƒ‡ê؇L nk .
n = 1 ž, ù
´w, . b d(Ø3 n < m ž¤á.
n = m.
k = 0 ½ö m ž, (Øw,¤á.
1 ≤ k ≤ m − 1.
[W ] ∈ X. éz‡ ê i (0 ≤ i ≤ k), P
Xi = {[W 0 ] ∈ X : dim W 0 ∩ W = k − i}.
éz‡[W 0 ] ∈ Xi , P
Yi = {[W 0 ∩ W ] : [W 0 ] ∈ Xi } ⊂ Grk−i (W )
Ú
Zi = {[W 0 ∩ W ⊥ ] : [W 0 ] ∈ Xi } ⊂ Gri (W ⊥ ).
d(i) ¥
O O K, Yi ´ Grk−i (W ) ¥ k 8 … Zi ´ Gri (W ⊥ ) ¥
u dim W = k < m … dim W ⊥ = m − k < m, d8Bb ·‚
k
m−k
|Yi | ≤
…|Zi | ≤
.
k−i
i
ù ,
|X| =
X
0≤i≤k
|Xi | ≤
X
0≤i≤k
X k m − k m
|Yi ||Zi | ≤
=
.
k−i
i
k
0≤i≤k
ddy² (Ø.
11
k
8. d
18K
K£ ) ‰ K ¤ a ²
¯4
˜ ¯4 3 ‘‚:²¡þ‹I• (n, 0) ?
¯4"d §¯4 z˜Ú‰Ú•
• 1 {ü‘ÅiÄ"¦¤óŠ ¯4ÕK3 : (0, 0) ?"3±e ˆ¯¥§Ø”@•
ê n Œu 1"
(1) - P1,n •T¯4
Áy²
3r
TÐ bn1.5 c Úž§ål¯4Õ
†‚ålŒu n2
lim P1,n = 1
n→+∞
(2) - P2,n •T¯4 3c bn1.5 c ÚSQ²£ L¯4Õ VǧÁy²
lim P2,n = 0
n→+∞
(3) - P3,n •T¯4
c 2n ÚSQ²£ L¯4Õ VǧÁy²
lim P3,n = 1
n→+∞
12
Vǧ
8 ‰ Y (1) •ÄT¯4
u 12
3r TÐ bn1.5 c Úžål¯4Õ †‚ålŒVÇ
¯‡"dn Ø ª·‚kù˜¯‡7,•¹u±e¯‡ A
A = ¯4 džå ‡:
ålŒu u
u
1
2
?˜ÚŠân Ø ª§XJ¯‡ A u)§K¯4 3– ˜‡‹I••þ£Ä ål–
• n4 "Ø” T••• y-¶••"T¯4 1 k Úž3 y-¶•• £ÄålŒ± w‰
´±eÕáÓ©Ù ‘ÅCþ ¦Úµ
Yk = X1 + X2 + · · · + Xk
Ù¥ P (Xi = 0) = 21 , P (Xi = ±1) = 41 "´„"ù
ª§·‚k
P (A) ≤ 2P
Ybn1.5 c ≥
E(Xi ) = 0, var(Xi ) = 12 §dƒ'ÈÅØ
n
n1.5
≤2× 2
= 16n−0.5 << 1
4
2n /16
(2) †þ˜¯ƒÓ§dn Ø ª·‚kù˜¯‡7,•¹u±e¯‡ A
A = ¯4
3cbn1.5 cÚS ˆLå ‡: ålŒu un
XJ¯‡ A u)§K¯4 3– ˜‡‹I••þ˜½– £ÄL–
n
P (A) ≤ 2P max |Yk | ≥
2
k≤n1.5
5¿
/•
n
2
ål"=
Yk ´˜‡ §ŠâDoob Ø ª
P
max |Yk | ≥
n
k≤n1.5
2
≤
h
i
2
E Ybn
1.5 c
(n/2)2
(3) - Sk ∈ Z2 , k ≥ 0 •¯4 31 k Úž
≤
n1.5
= 2n−0.5 << 1
2(n/2)2
˜"·‚½Â±eÊž
τ = min{k ≥ 0, Sk = 0}
τb = min{k ≥ 0, |Sk | ≥ 2n/4 }
Ïd§éu·‚'% Vǧk
P (τ > 2n ) ≤ P(n,0) (b
τ > 2n ) + P(n,0) (b
τ < τ)
Ù¥ P(n,0) •l (n, 0) :Ñu {ü‘ÅiÄ ; VÇ©Ù"éuþª¥ 1˜‘§dØ
n
C n§·‚k S[2n/2 t] 2− 4 3 t ∈ [0, 1] þÂñ 0 Ñu
‘IOÙK$Ä Bt "éuÙK
$Ä´„
P (|B1 | ≥ 2) = c > 0
13
éu˜ƒ¿©Œ
min
|X0 |≤2n/4
n§(ܲ£ØC5·‚k
PX0 (|S2n/2 | ≥ 2n/4 ) ≥ P0 (|S2n/2 | ≥ 2 × 2n/4 ) ≥ P (|B1 | ≥ 2) −
e¡·‚é¯4
l (0, 0) : 2n/4
l (0, 0) : 2n/4
c
c
≥
2
2
2n ÚU 2n/2
•Ý©ã"XJT¯4 3 2n Ú¥©ª uå
Œ»ƒS§K¦7L3Úê• 2n/2 , 2 × 2n/2 , · · · , [2n/2 ]2n/2 ž©ªå
Œ»ƒS"2d‘ÅiÄ ê¼5§·‚k
h
P(n,0) (b
τ > 2n ) ≤ 1 −
i[2n/2 ] c [2n/2 ]
≤ 1−
PX0 (|S2n/2 | ≥ 2n/4 )
<< 1
2
|X0 |≤2n/4
min
éu1 ‘ p2 = P(n,0) (b
τ < τ ) ùp·‚I‡•Ä
2
kernel)§éu?¿ x ∈ Z •Ä
a(x) =
‘{ü‘ÅiÄ
³Ø(potential
+∞
X
[P0 (Sk = 0) − P0 (Sk = x)]
k=0
dKesten (1987, Lemma 1)1 Υ
1) a(x) ¥ ?ê¦Úé?¿ x ∈ Z2 Âñ…šK
2) éu?¿Ð© ˜Ñu {ü‘ÅiÄ Sk , a(Sk∧τ ) ´˜‡šK
3) •3~ê C0 ¦
(Üþã3 ^±9
|a(x) − log |x|/(2π) − C0 | = O(|x|−2 )
Êž©)½n§·‚k
a((n, 0)) = (1 − p2 ) × 0 + p2 E[a(Sτb)|τ > τb]
Ïd
p2 =
a((n, 0))
E[a(Sτb)|τ > τb]
5¿ dþã13 ^µ
a((n, 0)) log(n)
E[a(Sτb)|τ > τb] n
Ïd§p2 << 1 y."
1
H. Kesten, Hitting probabilities of random walks on Zd , Stochastic Process. Appl. 25 (1987) 165õ184.
14
19K
K£ ) ‰ K ¤
(1) y²Ø•3÷vXe^‡
±ÏS
sup
a1 , a2 , a3 , . . ., z‘þ• ±1, …é?¿knê θ,
N
X
an e2πinθ < +∞.
N ∈N n=1
(2) y²Ø•3z‘þ• ±1
S
a1 , a2 , a3 , . . . ÷vµé?¿knê θ
sup
N
X
an e2πinθ ≤ 2022 .
N ∈N n=1
(3) Þ~`²µ•3z‘þ• ±1 S
õ‘ −1§…é?¿ θ ∈ Q − Z,
sup
a1 , a2 , a3 , . . . ÷vµÙ¥káõ‘ 1§•ká
N
X
an e2πinθ < +∞.
N ∈N n=1
15
9 ‰ Y (1) {Pz := e2πiθ . ^‡y{§b
sup
N
X
•3±Ï•d ∈ N
d±1
¤ S (an )÷v
an z n < +∞,
N ∈N n=1
Ù¥z = e2πiθ , θ ∈ Q.
N •d
ê, N = dM . Ï•(an )´±Ï , ·‚Œ±©|¦Úµ
Md
X
an z n =(a1 z + a2 z 2 + . . . ad z d ) + z d (a1 z + a2 z 2 + . . . ad z d )
n=1
+ · · · + z (M −1)d (a1 z + a2 z 2 + . . . ad z d )
=(a1 z + a2 z 2 + . . . ad z d ) 1 + z d + z 2d + . . . + z (M −1)d .
θ = k/d ∈ Q, Ù¥k ∈ {0, . . . , d − 1}–½. Ï•z d = e2πik/d
Md
X
d
= 1, ¤±
an z n = M (a1 z + a2 z 2 + . . . ad z d ).
n=1
k¦ a1 z + a2 z 2 + . . . ad z d š"Ò
•I
lim
M →+∞
Md
X
an z n =
n=1
gñµ
lim M |a1 z + a2 z 2 + . . . ad z d | = +∞.
M →+∞
ea1 z + a2 z 2 + . . . ad z d ék ∈ {0, . . . , d − 1} (Ù¥z = e2πik/d )þ•0, Kõ‘ªP (X) =
a1 X + a2 X 2 + . . . ad X d kd + 1‡ØÓEŠµ
e2πi0/d , e2πi1/d , . . . , e2πi(d−1)/d
±9 0,
ù´ØŒU "
(2) Ó
^‡y{§b (an )•ù
S "Äk§Ï•
fN :=
N
X
an e2πinθ
n=1
´θ
ëY¼ê§é¤k¢ê•k|fN (θ)| ≤ 2022.
·‚OŽfN
L2
"5¿ én ∈ Z,
(
1
e2πinθ dθ =
0
0
Z 1
16
en = 0,
en 6= 0.
l ·‚k
Z 1
Z 1X
N
2
|fN | dθ =
0
an e
0 n=1
N X
N
X
=
2πinθ
N
X
an e−2πimθ dθ
m=1
Z 1
an am
n=1 m=1
e2πi(n−m)θ dθ
|0
{z
}
š" …= n=m
N
X
=
a2n = N
n=1
ddÒUwÑ
N
X
an e2πinθ ≤ 2022
n=1
ØUé¤k N Úθ ∈ R¤á"ÄK
20222 )§gñœ
R1
R1
{Òk 0 |fN |2 dθ ≤ 20222 , † 0 |fN |2 dθ = N (N >
(3) (•,•ª~Š S ÷v^‡§ ´¿ØÎÜ·‚ ‡¦")
• `²Ž{ d5§·‚kwwθ = 1/3±9N = 3M œ¹"·‚‡éai ¦
!
!
!
M
−1
M
−1
M
−1
X
X
X
sup
a3n+1 · 1 +
a3n+2 · e2πi/3 +
a3n+3 · e4πi/3 < +∞
M ∈N
ùéAu
n=0
n=0
n‡•þ1, e2πi/3 Úe4πi/3 , ,
n=0
OŽÚ
•" Ï•1 + e2πi/3 + e4πi/3 = 0, ·
e2πi/3
1
e4πi/3
Figure 3: θ = 1/3
‚ {4ùn‡•þäkƒÓ
(a3n+i )n≥1 ¥1Ú−1 ©Ù´
ê¤á"
œ¹.
•4§‚¦þ-ž"†óƒ§éi = 1, i = 2±9i = 3§S
Øõ "aquùp 3§ù
5ŸATé¤kŒu1
17
kéõ¥•{ Eù
n
an
Ù¥+1Ñy1!g,
÷v‡¦"
(an ), ˜«Œ1 •{Xe:
1
+1
2
−1
3
−1
4
+1
5
+1
X−1Ñy2!g§,
6
+1
7
+1
8
+1
9
+1
10
−1
11
,
−1
"e¡y²TS
+1Ñy3! = 6g§XdUYe
ék ≥ 0, -f (k) = 0! + 1! + · · · + k!. ^Ik ⊂ NL«÷vf (k) ≤ n < f (k + 1) g,ên. l
n ∈ Ik ž§an = (−1)k . TS
'…A ´ q ≥ 2ž§Ik •Ý´(k + 1)!, l
k¿©
ŒžŒ q Ø"
θ = p/q´knê§Ù¥gcd(p, q) = 1, q > 1. òK¥ n
N
X
an e2πinθ =
n=1
Ú ¤Xe/ªµ
X
X
(−1)k
e2πinp/q .
n∈Ik
n≤N
k≥0
e¡‰ÑTÚØ•6uN ."
P
•ÄÙ¥˜ã§Sk := n∈Ik e2πinp/q (k ≥ 0).
n≤N
(i) ek < q, dýéŠØ ª|Sk | ≤ f (k + 1) − f (k) = (k + 1)!
(ii) ek ≥ q, ·‚|^S
Ey²Ø
• ˜ã± þ•0.
(a) ek ≥ q…f (k + 1) ≤ N , KSk = 0. Ï•Šâ '?ê¦Ú
f (k+1)−1
X
Sk =
e2πip/q
e2πinp/q =
n=f (k)
Ï•k ≥ q, (k + 1)!U q
e2πip/q
(b) ek ≥ q
f (k)+(k+1)!
f (k+1)
− e2πip/q
e2πip/q − 1
f (k)
.
Ø, ©f•0µ
f (k) f (k) e2πip(k+1)!/q − 1 = 0.
− e2πip/q
= e2πip/q
N ∈ Ik (=• ˜ã),·‚k
N
X
Sk =
2πinp/q
e
=
e2πip/q
n=f (k)
N +1
− e2πip/q
e2πip/q − 1
f (k)
.
dýéŠØ ªk
e2πip/q
|Sk | ≤
N
+
e2πip/q
f (k)
e2πip/q − 1
(c) ek ≥ q…N < f (k), KSk = 0§¢SþT㕘"
18
=
2
e2πip/q − 1
.
Ïd
N
X
n=1
an e
2πinθ
≤
q−1
X
|Sk | +
k=0
X
|Sk | ≤ f (q) +
k≥q
=
sup
N
X
an e2πinθ < +∞.
N ∈N n=1
19
2
e2πip/q − 1
.
110K
K£ ) ‰ K ¤ m 4 ª
!8
O
b \ Ác¬m4ª
üÀ•EâÏn§KI^êÆ•£ïÄ O•Y Ün5"3
m4ª
À!8¥§k˜‡•Y´4˜+dÃ<Å›› 3ŒÔ7˜‡
/G |/
wX"Ï•Ã<Åv
õ£ ´Ø¬P×½öƒE¤§·‚@•Œ±^˜‡VǗݼ
ê ρ(t, v)(≥ 0) 5•xÃ<Å ©Ù"Ï•|/´ ‚/ §¤±·‚Œ±@•„Ý v ∈ R§
§L«Ã<Å ‚„Ý"@o§éu?¿‰½ žm t, Úü‡„Ý v1 < v2 §
Z v2
ρ(t, v) dv
v1
L« NÃ<Å¥„Ý0u v1 Ú v2 ƒm VÇ"
duÃ<Å
$ÄÅn§®•ù‡—ݼê üz÷vXe •§
ρt + (u(t) − v)ρ v = ρvv , v ∈ R, t > 0.
Ù¥ u(t) •Ã<Å •-„Ý"
(1) • ïÄNo‰Ã<ÅÜ·
•-„ݧŒ ‰ üJÆ`§AT4
u(t) = u0 + u1 N (t)
Ù¥ u0 > 0§u1 > 0§
N (t) L«Ã<Å „Ý Ü£v+ = max{0, v}¤ ²þ
Z +∞
Z +∞
N (t) =
v+ ρ(t, v) dv =
vρ(t, v) dv.
−∞
0
´§\õ¿/J2 §XJ u1 > 1 @o N (t) 3üzL§¥Ø¬kþ.§±–ué¯
ÚåÃ<Å
æ"\Œ±y²\ þã(Øíº£• •B?ا·‚ Ñ ρ 9Ù ê
3 |v| → +∞ ž
z"¤
(2) 3æB Œ Ú\ ïÆ § üq3•Äù Ã<ŴĬ3w1¥3
|/þþ
!©Ù§u´·‚I‡•Ä'u„ÝÚ R ˜ éܗݼê p(t, x, v)(≥ 0)"ùp x ∈ [0, 2π]
2π
L«Ã<Å3
þ ƒé ˜§w, 0 p(t, x, v)dx = ρ(t, v)"®•ù‡éܗݼê
üz÷v
pt + vpx + (u(t) − v)p v = pvv , x ∈ [0, 2π], v ∈ R, t > 0.
…duÃ<Å37 w1§3 x ••þXe^‡÷v
p(t, 0, v) = p(t, 2π, v),
v ∈ R,
t > 0.
\ŒÿßÿµÃØЩ©ÙXÛ§Ã<Åé¯Ò3
þ C˜‡þ!©Ù"\Œ±y²½
öy–ù‡·Kíº£• •B?ا·‚ Ñ p 9Ù ê3 |v| → +∞ ž
z"¤
20
10 ‰ Y (1) ·‚½Â²þ„Ý
Z +∞
vρ(t, v) dv.
M (t) =
−∞
† OŽ
Z
d
d +∞
vρ(t, v) dv
M (t) =
dt
dt −∞
Z +∞
vρt (t, v) dv
=
−∞
Z +∞
=
−∞
Z +∞
v (−(u(t) − v)ρ + ρv )v dv
((u(t) − v)ρ − ρv ) dv
=
−∞
= u(t) − M (t)
= u0 + u1 N (t) − M (t).
du M (t) ≤ N (t)§qw, N (t) ≥ 0§·‚
d
M (t) ≥ u0 + (u1 − 1)N (t) ≥ u0 > 0.
dt
K§
t → +∞§·‚k M (t) → +∞"
N (t) ≥ M (t)§¤± N (t) •òuÑ
á"
(2) du x •• ±Ï>.^‡§·‚ò•§ ) ¤XeFp“?ê /ª
+∞
1 X
pk (t, v)eikx .
p(t, x, v) =
2π
k=−∞
Ù¥
Z 2π
pk (t, v) =
p(t, x, v)e−ikx dx.
0
@o§·‚I‡y²
k 6= 0 ž§pk (t, v) òP~"
·‚Äk|^Fp“?ê
5§Œ
pk ÷v
∂t pk + ikvpk = −∂v (u(t) − v)pk v + ∂vv pk ,
é ?˜Ú¦)§e¡¦^'u v
t > 0,
Fp“C†"Z +∞
1
p̂k (t, ξ) = √
pk (t, v)e−ivξ dv
2π −∞
21
k ∈ Z,
v ∈ R.
²LOŽ§Œ±
p̂k (t, ξ) ÷v±e •§
∂t p̂k + (ξ − k)∂ξ p̂k = − iξu(t) + ξ 2 p̂k .
e5§·‚¦^A ‚{¦)d•§"•ÄXeA
⥤
d
ξk (t) = ξk (t) − k.
dt
ÙÏ)•
ξk (t) − k = et−s (ξk (s) − k) .
÷XA ‚§·‚
Rt
2
p̂k (t, ξk (t)) = p̂k (0, ξk (0))e− 0 (ξk (s)) +iu(s)ξk (s)ds
2|^A ‚ Ï)§·‚Œ
p̂k (t, ξ) = p̂k (0, (ξ − k)e−t + k)e−Ht,k (ξ−k) ,
Ù¥
Ht,k (z) =
1 − e−2t 2
z + 2k(1 − e−t )z + i
2
, §²LFp“_C†§·‚Œ±
Z t
u(s)e−(t−s) dsz + k 2 t + ik
0
Z t
u(s)ds.
0
Lˆª
pk (t, v)
pk (t, v) = eikv (pt,k ∗ Gt,k )(v).
Ù¥§pt,k (y)
½ÂXe
t
t
pt,k (y) := e p0,k (e y),
−ikv
Z 2π
p0,k (v) := e
pinit (x, v)e−ikx dx.
0
Gt,k (z) ´˜‡EŠ pd¼ê§LˆªXe
1
(z − µ)2
2
Gt,k (z) = √
exp −
exp
(ikΘ)
exp
−k
D
.
2σ 2
2πσ
ùp§
Z t
µ(t) =
e−(t−s) u(s)ds,
σ(t) =
p
1 − e−2t ,
0
1 − e−t
.
1 + e−t
0
·‚5¿ § t > 0 ž§pd¼ê Gt,k duP~Ïf exp −k 2 D
•3òP~§ù‡P
~„Çéu t ¿©Œž´•ê. "• §|^òÈØ ª§·‚Œ±
Θ(t, z) = −
2(z − µ(t))
−
1 − e−t
Z t
u(s)ds,
D(t) = t − 2
kpk (t, v)kL1 (R) = keikv (pt,k ∗ Gt,k )(v)kL1 (R) ≤ kpt,k (g)kL1 (R) kGt,k kL1 (R)
2
= kp0,k (g)kL1 (R) e−k D(t) .
ùy² § k 6= 0 ž§pk (t, v) òׄP~§Ï
˜‡˜m•• þ!©Ù"
22
p(t, x, v) ò¿©
C π1 p0 (t, v)§ùéA
2022 Alibaba Global Mathematics Competition
Single-Choice Problem: Magic Magnetic Cube
Divide a solid cube ABCD − A1 B1 C1 D1 (with AB = 1) into 12 pieces (Figure 1) as follows:
1) Take 6 diagonals of its surfaces AC,AB1 , AD1 , C1 B, C1 D, C1 A1 ;
2) Consider all triangles with the center of the cube as a vertex, and one of the above 6
diagonals and 12 edges as the opposite side;
3) These 18 triangles cut the cube into 12 tetrahedra, and each tetrahedron has two edges
that are cube edges;
4) Each tetrahedron is connected to other tetrahedra only by its two cube edges.
Figure 1: Magic magnetic cube
Such a toy can take on a variety of shapes (Figure 2).
Figure 2: Examples of various shapes
Question: Of all the possible shapes of this toy, what is the maximum distance (in space)
between two points on it?
1
√
11
p
√
7+4 2
√
13
√
1+2 2
None of the above
2
Single-Choice Problem: Onlook With Distance
One day, there is a Street Art Show at somewhere, and there are some spectators around.
We consider this place as an Euclidean plane. Let K be the center of the show. And name
the spectators by A1 , A2 , . . . , An , . . .. They pick their positions
P1 , P2 , . . . , Pn , . . . ,
one by one. The positions need to satisfy the following three conditions simultaneously.
(i) The distance between K and An is no less than 10 meters, that is, KPn ≥ 10m holds
for any positive integer n.
(ii) The distance between An and any previous spectator is no less than 1 meter, that is,
Pm Pn ≥ 1m holds for any n ≥ 2 and any 1 ≤ m ≤ n − 1.
(iii) An always choose the position closest to K that satisfies (i) and (ii), that is, KPn
reaches its minimum possible value. If there are more than one point that satisfy (i)
and (ii) and have the minimum distance to K, An may choose any one of them.
For example, A1 is not restricted by (ii), so he may choose any point on the circle C which
is centered at K with radius 10 meters. For A2 , since there are lots of points on C which are
at least 1 meter apart from P1 , he may choose anyone of them.
(1) Which of the following statement is true?
There exist positive real numbers c1 , c2 such that for any positive integer n,
no matter how A1 , A2 , . . . , An choose their positions, c1 ≤ KPn ≤ c2 always
hold (unit: meter);
There exist positive real numbers c1 , c2 such that for any positive integer n,
√
√
no matter how A1 , A2 , . . . , An choose their positions, c1 n ≤ KPn ≤ c2 n
always hold (unit: meter);
There exist positive real numbers c1 , c2 such that for any positive integer n,
no matter how A1 , A2 , . . . , An choose their positions, c1 n ≤ KPn ≤ c2 n
always hold (unit: meter);
There exist positive real numbers c1 , c2 such that for any positive integer n,
no matter how A1 , A2 , . . . , An choose their positions, c1 n2 ≤ KPn ≤ c2 n2
always hold (unit: meter).
(2) Since human bodies are 3-dimensional, if one spectator’s position is near another spectator’s path of view, then the second one’s sight will be blocked by the first one. Suppose
that for different i, j, if the circle centered at Pi with radius 16 meter intersects with segment
KPj , then Aj ’s sight will be blocked by Ai , and Aj could not see the entire show.
Which of the following statement is true?
If there were 60 spectators, then some of them could not see the entire show;
If there were 60 spectators, then it is possible that all spectators could see the
entire show, but if there were 800 spectators, then some of them could not see
the entire show;
3
If there were 800 spectators, then it is possible that all spectators could see the
entire show, but if there were 10000 spectators, then some of them could not see
the entire show;
If there were 10000 spectators, then it is possible that all spectators could see
the entire show.
4
Single-Choice Problem: Tiger Mystery Box
Brave NiuNiu (a milk drink company) organizes a promotion during the Chinese New Year:
one gets a red packet when buying a carton of milk of their brand, and there is one of the
following characters in the red packet “ m”(Tiger), “)”(Gain), “%”(Strength).
If one collects two “m”, one “)” and one “%”, then they form a Chinese phrases “mm)
%”(Pronunciation: hu hu sheng wei), which means “Have the courage and strength of the
tiger”. This is a nice blessing because the Chinese zodiac sign for the year 2022 is tiger. Soon,
the product of Brave NiuNiu becomes quite popular and people hope to get a collection of
“mm)%”. Suppose that the characters in every packet are independently random, and
each character has probability 31 .
(1) What is the expectation of cartons of milk to collect “mm)%”(i.e. one collects at least
2 copies of “m”, 1 copy of “)”, 1 copy of “%”)?
6 13
7 31
8 31
9 13
None of the above
(2) In a weekly meeting of Brave NiuNiu, its market team notices that one often has to collect
too many “)” and “%”, before getting a collection of “mm)%”. Thus an improved plan
is needed for the proportion of characters. Suppose that the probability distribution of “m”,
“)” and “%” is (p, q, r), then which of the following plans has the smallest expectation
(among the 4) for a collection of “mm)%”?
(p, q, r) = ( 13 , 13 , 31 )
(p, q, r) = ( 12 , 14 , 41 )
3 3
(p, q, r) = ( 25 , 10
, 10 )
(p, q, r) = ( 34 , 18 , 81 )
5
Given a set X and a function f : X × X → [0, 1], we say f is right uniform if for any > 0,
there exist finitely many elements b1 , b2 , ..., bm in X such that for any t ∈ X, the following
holds for some bi(t) :
|f (x, t) − f (x, bi(t) )| < , ∀x ∈ X.
Similarly, we say f is left uniform if for any > 0, there exist finitely many elements
a1 , a2 , ..., an in X such that for any t ∈ X, the following holds for some ai(t) :
|f (t, x) − f (ai(t) , x)| < , ∀x ∈ X.
Prove that f is right uniform if and only if it’s left uniform.
6
Let n be a positive integer and V = Rn be an n-dimensional Euclidean space with a basis
ei = (0, . . . , 0, 1, 0, . . . , 0) (1 ≤ i ≤ n) and with an inner product (·, ·) defined by
| {z } | {z }
i−1
n−i
(ei , ej ) = δi,j
where
δi,j =
1 if i = j
0 if i 6= j
is Kronecker’s symbol. For a nonzero vector v ∈ V , define sv : V → V by
sv (u) = u −
2(u, v)
v, ∀u ∈ V.
(v, v)
For an integer k between 0 and n, write Grk (V ) for the set of k-dimensional subspaces of V .
For a k-dimensional subspace W of V , write [W ] for the corresponding element of Grk (V ).
Choose an orthonomal basis {v1 , . . . , vk } of W , define s[W ] : V → V by
s[W ] = sv1 · · · svk .
(1) Prove that s[W ] is independent of the choice of an orthonomal basis {v1 , . . . , vk }.
(2) Prove that s2[W ] = id.
(3) For another element [W 0 ] ∈ Grk (V ), define
t[W ] ([W 0 ]) = [s[W ] (W 0 )],
where s[W ] (W 0 ) is the image of W 0 under s[W ] . We call a subset X of Grk (V ) a “nice
set” if
t[W ] ([W 0 ]) = [W 0 ], ∀[W ], [W 0 ] ∈ X.
Find the maximal cardinality of a “nice set” in Grk (V ) and prove it.
7
A Busy Courier
A courier picks up a package at coordinate (n, 0) in the two-dimensional lattice, while his
station locates at the origin (0, 0). The courier then does a discrete time simple random walk
on Z2 . In the rest of this question, you may without loss of generality assume n is sufficiently
large.
(1) Let P1,n be the probability that at his bn1.5 cth step, the distance between this courier
and his station is greater than n2 . Prove that
lim P1,n = 1
n→+∞
(2) Let P2,n be the probability that the courier has ever reached the station within his first
bn1.5 c steps. Prove that
lim P2,n = 0
n→+∞
(3) Let P3,n be the probability that the courier has ever reached the station within his first
2n steps. Prove that
lim P3,n = 1
n→+∞
8
(1) Show that there is no periodic sequence a1 , a2 , a3 , . . . of signs an ∈ {±1} such that, for
every rational number θ,
N
X
sup
an e2πinθ < +∞.
N ∈N n=1
(2) Show that there is no sequence a1 , a2 , a3 , . . . of signs an ∈ {±1} such that, for every
rational number θ
N
X
sup
an e2πinθ ≤ 2022 ?
N ∈N n=1
(3) Give an example of a sequence of signs an ∈ {±1} such that, for every θ ∈ Q − Z,
sup
N
X
an e2πinθ < +∞,
N ∈N n=1
and which takes each of the values +1 and −1 an infinite number of times.
9
Program Design For The Opening Ceremony
Suppose you are chosen as a technology assistant by the director of the opening ceremony
for the 2022 Winter Olympics, and your job is to evaluate the program proposals. One of
the backup programs is a skating show of a ensemble of drones dressed as mascots, which
are moving along a circle. Since the number of the drones is sufficiently large, we can use
a probability density function ρ(t, v)(≥ 0) to represent the distribution of the drones. Here,
v ∈ R is the line speed, and for a given time t, and two speeds v1 < v2 ,
Z v2
ρ(t, v) dv
v1
is the probability of find a drone with its speed between v1 and v2 .
Suppose that the dynamics of the density function is governed by
ρt + (u(t) − v)ρ v = ρvv , v ∈ R, t > 0,
where u(t) is the command speed.
(1) To investigate proper choices of the command speed, D.B. propose that we shall choose
u(t) = u0 + u1 N (t)
where u0 > 0, u1 > 0 and N (t) is the average of the positive part of the speed v+ = max{0, v},
i.e.,
Z
Z
+∞
N (t) =
+∞
v+ ρ(t, v) dv =
−∞
vρ(t, v) dv.
0
But you claim that if u1 > 1, N (t) may become unbounded in evolution, such that the
drones may be out of order. Can you prove it? (For simplicity, the contributions of ρ and its
derivatives at |v| → +∞ are neglected.)
(2) After taking those advices, the directly is wondering whether the drones will be evenly
distributed along the circle. Thus, we shall consider the joint density function p(t, x, v)(≥ 0)
of
R 2πposition and speed, where x ∈ [0, 2π] is the position coordinate on the circle. Clearly,
0 p(t, x, v)dx = ρ(t, v). Suppose that the governing equation for p(t, x, v) is
pt + vpx + (u(t) − v)p v = pvv , x ∈ [0, 2π], v ∈ R, t > 0.
Because, the drones are circulating around, the following boundary condition is satisfied
p(t, 0, v) = p(t, 2π, v),
v ∈ R,
t > 0.
You have a feeling that, no matter how the drones are distributed initially, they will become
almost evenly distributed very quickly. Can you prove or disprove this statement? (For
simplicity, the contributions of p and its derivatives at |v| → +∞ are neglected.)
10
2022 Alibaba Global Mathematics Competition
(Qualifying Round)
1 (Single-Choice Problem) Magic Magnetic Cube
Divide a solid cube ABCD − A1 B1 C1 D1 (with AB = 1) into 12 pieces (Figure 1) as follows:
1) Take 6 diagonals of its surfaces AC,AB1 , AD1 , C1 B, C1 D, C1 A1 ;
2) Consider all triangles with the center of the cube as a vertex, and one of the above 6
diagonals and 12 edges as the opposite side;
3) These 18 triangles cut the cube into 12 tetrahedra, and each tetrahedron has two edges
that are cube edges;
4) Each tetrahedron is connected to other tetrahedra only by its two cube edges.
Figure 1: Magic magnetic cube
Such a toy can take on a variety of shapes (Figure 2).
Figure 2: Examples of various shapes
1
Question: Of all the possible shapes of this toy, what is the maximum distance (in space)
between two points on it?
p
√
√
√
√
(A) 11
(B) 7 + 4 2
(C) 13
(D) 1 + 2 2
(E) None of the above
2
1 Answer The answer is D. Label the eight vertices of the cube with ABCD−A1 B1 C1 D1 ,
and let the pair of vertices selected at the beginning be A and C1 . Then, these two vertices
belong to 6 small tetrahedrons, and B, C, D, D1 , A1 , B1 (no matter how they are deformed)
constitute a Hexagon with side length 1 (possibly degenerate).
Note that this toy consists of 12 tetrahedra (and thus convex), so the maximum value must
be the vertices of the two small tetrahedra.
In a tetrahedron, the four vertices can be divided into 3 categories:
• A vertex corresponds to A or C1 in the cube, we call it a vertex of Type I;
• The two vertices (and thus the edges between them) belong to the (spatial) hexagon
BCDD1 A1 B1 , which we call vertices of Type II;
• A vertex corresponds to the center of the
√ cube, and its distance to the other three
3
vertices of the same small tetrahedron is
, we call it a vertex of Type III.
2
We examine the maximum value of the distance along the edges between these vertices,
and it is easy to get the following table:
distance
first kind
second kind
third kind
vertices of Type I
√
1+2 2
√
2+ 2
√
√
3
1+ 2+
2
vertices of Type II
√
2+ 2
3
√
2+
3
2
vertices of Type
√ III
√
3
1+ 2+
√ 2
3
2+
2
√
2+ 3
√
Therefore, the distance between any two points on this toy will not exceed 1 + 2 2. And this
value can be achieved, as shown in the following figure:
3
2,3 (Single-Choice Problem) Onlook With Distance
One day, there is a Street Art Show at somewhere, and there are some spectators around.
We consider this place as an Euclidean plane. Let K be the center of the show. And name
the spectators by A1 , A2 , . . . , An , . . .. They pick their positions
P1 , P2 , . . . , Pn , . . . ,
one by one. The positions need to satisfy the following three conditions simultaneously.
(i) The distance between K and An is no less than 10 meters, that is, KPn ≥ 10m holds
for any positive integer n.
(ii) The distance between An and any previous spectator is no less than 1 meter, that is,
Pm Pn ≥ 1m holds for any n ≥ 2 and any 1 ≤ m ≤ n − 1.
(iii) An always choose the position closest to K that satisfies (i) and (ii), that is, KPn
reaches its minimum possible value. If there are more than one point that satisfy (i)
and (ii) and have the minimum distance to K, An may choose any one of them.
For example, A1 is not restricted by (ii), so he may choose any point on the circle C which
is centered at K with radius 10 meters. For A2 , since there are lots of points on C which are
at least 1 meter apart from P1 , he may choose anyone of them.
(1) Which of the following statement is true?
(A) There exist positive real numbers c1 , c2 such that for any positive integer n,
no matter how A1 , A2 , . . . , An choose their positions, c1 ≤ KPn ≤ c2 always
hold (unit: meter);
(B) There exist positive real numbers c1 , c2 such that for any positive integer n,
√
√
no matter how A1 , A2 , . . . , An choose their positions, c1 n ≤ KPn ≤ c2 n
always hold (unit: meter);
(C) There exist positive real numbers c1 , c2 such that for any positive integer n,
no matter how A1 , A2 , . . . , An choose their positions, c1 n ≤ KPn ≤ c2 n
always hold (unit: meter);
(D) There exist positive real numbers c1 , c2 such that for any positive integer n,
no matter how A1 , A2 , . . . , An choose their positions, c1 n2 ≤ KPn ≤ c2 n2
always hold (unit: meter).
(2) Since human bodies are 3-dimensional, if one spectator’s position is near another spectator’s path of view, then the second one’s sight will be blocked by the first one. Suppose
that for different i, j, if the circle centered at Pi with radius 61 meter intersects with segment
KPj , then Aj ’s sight will be blocked by Ai , and Aj could not see the entire show.
Which of the following statement is true?
(A) If there were 60 spectators, then some of them could not see the entire show;
(B) If there were 60 spectators, then it is possible that all spectators could see the
entire show, but if there were 800 spectators, then some of them could not see
the entire show;
4
(C) If there were 800 spectators, then it is possible that all spectators could see the
entire show, but if there were 10000 spectators, then some of them could not see
the entire show;
(D) If there were 10000 spectators, then it is possible that all spectators could see
the entire show.
2 Answer The answer is B. Suppose the length of KPn is dn meters.
We consider the discs centered at P1 , P2 , . . . , Pn−1 with radius 1 meter. Use the property of
Pn we get that these discs and the interior of C cover the disc centered at K with radius dn ,
so
π · d2n ≤ (n − 1) · π · 12 + π · 102 ,
It follows that
dn ≤
√
n + 99 ≤
√
√
100n = 10 n.
On the other hand, we consider the discs centered at P1 , P2 , . . . , Pn with radius 12 meter.
Since the distance between any two of P1 , P2 , . . . , Pn is not less than 1 meter, all these discs
do not intersect.
Note that every length of KP1 , KP2 , . . . , KPn is not more than dn meter (If the length of
KPm is more than dn meter, then Am can choose Pn , which is closer to K, contradiction.)
So all these discs are inside the circle centered at K with radius dn + 21 , and
1
1
π(dn + )2 ≥ n · π · ( )2 ,
2
2
So
√
dn ≥
n 1
− .
2
2
√
For n = 1, d1 = 10; For n ≥ 2, we have that 12 < 2 5 n , so
√
√
√
n 2 n
n
dn >
−
=
.
2
5
10
√
√
Therefore, 10n ≤ dn ≤ 10 n, (B) is correct.
3 Answer The answer is B. For 60 residents, since sin 60π > 101 sin π6 = 201 , the side length
of a regular 60-gon inscribed in C is not less than 1 meter. Therefore, P1 , P2 , . . . , P60 may
be all vertices of this polygon. For different i, j, the distance from Pi to KPj is not less than
π
π
1
10 sin 30
meter. Since sin 30
> 15 sin π6 = 10
, all residents could see the entire show.
−−→ −−→
−−−−→
On the other hand, if there are 800 residents, we draw rays KP1 , KP2 , . . . , KP800 . If two of
−−→ −−→
them (called KPi , KPj ) coincide, and KPj > KPi , then Aj ’s sight line is blocked by Ai .
5
Suppose
no two
rays coincide, we first prove that if an angle ∠Pi KPj satisfies ∠Pi KPj ≤
1
1
1
12 KPi + KPj (the unit of angle is rad, and the unit of length is meter), then the sight line
of one of Ai , Aj is blocked by the other.
Without loss of generality we suppose KPi ≤ KPj . Since KPi , KPj ≥ 10, we get that
1
∠Pi KPj ≤ 60
, so it is acute. Therefore, the foot point from Pi to KPj is inside segment
KPj , and the distance from Pi to KPj is
KPi sin ∠Pi KPj < KPi · ∠Pi KPj ≤
KPi 1
1 1
+
≤ ,
12 KPi KPj
6
So, Aj ’s sight line is blocked by Ai .
−−→ −−→
−−−−→
Note that KP1 , KP2 , . . . , KP
800 cut theperigon with vertex K to 800 angles, and their sum
1
1
1
is 2π, but the sum of all 12 KP
+ KP
is
i
j
800
X 1 1
1 1X 1
+
=
12 KPi KPj
6
KPi
≥
=
1
6
1
6
1
6
i=1
800
X
√
i=1
899
X
1
(from the conclusion of (1))
i + 99
1
√
m
m=100
Z
899 m+1
X
1
1
√ dx (since √ is a decreasing function)
x
x
m
m=100
√
√
Z 900
1
900 − 100
1
20
√ dx =
=
=
> 2π.
6 100
3
3
x
≥
1
Therefore, there exist an angle ∠Pi KPj satisfies ∠Pi KPj ≤ 12
residents could not see the entire show.
6
1
1
KPi + KPj
, and some of the
4,5 (Single-Choice Problem) Tiger Mystery Box
Brave NiuNiu (a milk drink company) organizes a promotion during the Chinese New Year:
one gets a red packet when buying a carton of milk of their brand, and there is one of the
following characters in the red packet “ m”(Tiger), “)”(Gain), “%”(Strength).
If one collects two “m”, one “)” and one “%”, then they form a Chinese phrases “mm)
%”(Pronunciation: hu hu sheng wei), which means “Have the courage and strength of the
tiger”. This is a nice blessing because the Chinese zodiac sign for the year 2022 is tiger. Soon,
the product of Brave NiuNiu becomes quite popular and people hope to get a collection of
“mm)%”. Suppose that the characters in every packet are independently random, and
each character has probability 31 .
(1) What is the expectation of cartons of milk to collect “mm)%”(i.e. one collects at least
2 copies of “m”, 1 copy of “)”, 1 copy of “%”)?
(A) 6 13
(B) 7 13
(C) 8 13
(D) 9 13
(E) None of the above
(2) In a weekly meeting of Brave NiuNiu, its market team notices that one often has to collect
too many “)” and “%”, before getting a collection of “mm)%”. Thus an improved plan
is needed for the proportion of characters. Suppose that the probability distribution of “m”,
“)” and “%” is (p, q, r), then which of the following plans has the smallest expectation
(among the 4) for a collection of “mm)%”?
(A) (p, q, r) = ( 13 , 13 , 13 )
(B) (p, q, r) = ( 21 , 14 , 14 )
3 3
(C) (p, q, r) = ( 52 , 10
, 10 )
(D) (p, q, r) = ( 34 , 18 , 18 )
7
4 Answer
The answer is B. We can use Poisson process to get the explicit formula
for the general case. Suppose that there are in total n characters. The probability for the
character i is pi §and we aim to collect ki copies of the character i. We denote by N the first
time to realize our collection, and we need to calculate E[N ]. A nice technique is embedding
this model to a Poisson process: we have a Poisson process of density 1. Every time when
the signal arrives, we sample independently pi for the character i. We also denote by
Ti = inf{t ∈ R+ : before t one collects ki copies of the character i},
T = max Ti .
1≤i≤n
We claim that E[T ] = E[N ]. Let us prove this claim. We denote by τj the waiting time for
the j-th signal, then we have
T =
N
X
τj .
j=1
By the property of Poisson process, (τj )j≥1 are i.i.d. exponential random variable. Using
conditional probability we have
N
X
E[T ] = E[
τj ]
j=1
=
∞
X


k
X
E
τj |N = k  P[N = k]
j=1
k=0
=
∞
X
P[N = k]kE[τ1 ]
k=0
= E[N ].
This justifies our claim.
Then it suffices to calculate E[T ]. By Fubini’s lemma
Z ∞
E[T ] =
P[T > t] dt
0
Z ∞
=
(1 − P[T ≤ t]) dt
Z0 ∞
=
(1 − P[Ti ≤ t, ∀1 ≤ i ≤ n]) dt.
0
In fact, the thinning property of the Poisson process creates a lot of independence. We can
treat the collection of the character i as independent Poisson processes of parameter pi . Then
we have
!
Z ∞
n
Y
E[T ] =
1−
P[Ti ≤ t] dt.
0
i=1
8
We write down directly the explicit formula of P[Ti ≤ t] using Poisson distribution
P[Ti ≤ t] = 1 −
kX
i −1
e−pi t
k=0
(pi t)k
.
k!
Finally, it gives us
Z ∞
1−
E[N ] =
0
n
Y
1−
kX
i −1
i=1
e
−pi t (pi t)
k=0
k!
k
!!
dt.
In our setting, n = 3, and the object (k1 , k2 , k3 ) = (2, 1, 1). Thus we have
E[N ] = 1 + p1 +
2
1
1
+
+
p1 p2 p3
−
3
X
i=1
p1
p1
1
−
−
.
2
1 − pi (p1 + p2 )
(p1 + p3 )2
When (p1 , p2 , p3 ) = (1/3, 1/3, 1/3), the expectation is 7 31 .
5 Answer The answer is C. In last question, we know the expectation for Plan A is 7 31 .
Plan D is not a good plan obviously, because the expectation to collect “%0is 8, which is
larger than Plan A. It suffices to calculate Plan B and C. Using the expression
2 1 1
1
1
1
E[τ ] = 1 + p +
+ +
−
+
+
p q r
p+q p+r q+r
p
p
−
−
.
2
(p + q)
(p + r)2
1
The expectation for Plan B and Plan C are respectively 7 18
, 6 223
245 . Plan C is the best one.
9
6 (Proof Question)
Given a set X and a function f : X × X → [0, 1], we say f is right uniform if for any > 0,
there exist finitely many elements b1 , b2 , ..., bm in X such that for any t ∈ X, the following
holds for some bi(t) :
|f (x, t) − f (x, bi(t) )| < , ∀x ∈ X.
Similarly, we say f is left uniform if for any > 0, there exist finitely many elements
a1 , a2 , ..., an in X such that for any t ∈ X, the following holds for some ai(t) :
|f (t, x) − f (ai(t) , x)| < , ∀x ∈ X.
Prove that f is right uniform if and only if it’s left uniform.
6 Answer It suffices to prove that if f is right uniform, then it’s left uniform. Given
> 0, we have finitely many elements b1 , b2 , ..., bm as in the definition of being right uniform.
Consider the map h : X → [0, 1]m , h(x) = (f (x, b1 ), f (x, b2 ), ..., f (x, bm )). Because [0, 1]m is
compact, there are c1 , c2 , ..., cn in X such that for any x ∈ X, |h(x) − h(cix )| < holds for
some cix . Hence
|f (x, bi ) − f (cix , bi )| < , ∀i ∈ {1, 2, ..., m}.
(1)
Now since f is right uniform, we have |f (x, t)−f (x, bit )| < (∀x ∈ X), |f (cix , t)−f (cix , bit )| <
. Combined with (1), it follows that for any t ∈ X,
|f (x, t) − f (cix , t)| ≤ |f (x, t) − f (x, bit )| + |f (x, bit ) − f (cix , bii )| + |f (cix , bit ) − f (cix , t)| < 3.
Since is arbitrary, f is left uniform.
10
7 (Proof Questions)
Let n be a positive integer and V = Rn be an n-dimensional Euclidean space with a basis
ei = (0, . . . , 0, 1, 0, . . . , 0) (1 ≤ i ≤ n) and with an inner product (·, ·) defined by
| {z } | {z }
i−1
n−i
(ei , ej ) = δi,j
where
δi,j =
1 if i = j
0 if i 6= j
is Kronecker’s symbol. For a nonzero vector v ∈ V , define sv : V → V by
sv (u) = u −
2(u, v)
v, ∀u ∈ V.
(v, v)
For an integer k between 0 and n, write Grk (V ) for the set of k-dimensional subspaces of V .
For a k-dimensional subspace W of V , write [W ] for the corresponding element of Grk (V ).
Choose an orthonomal basis {v1 , . . . , vk } of W , define s[W ] : V → V by
s[W ] = sv1 · · · svk .
(1) Prove that s[W ] is independent of the choice of an orthonomal basis {v1 , . . . , vk }.
(2) Prove that s2[W ] = id.
(3) For another element [W 0 ] ∈ Grk (V ), define
t[W ] ([W 0 ]) = [s[W ] (W 0 )],
where s[W ] (W 0 ) is the image of W 0 under s[W ] . We call a subset X of Grk (V ) a “nice
set” if
t[W ] ([W 0 ]) = [W 0 ], ∀[W ], [W 0 ] ∈ X.
Find the maximal cardinality of a “nice set” in Grk (V ) and prove it.
11
7 Answer (1) Write W ⊥ for the orthogonal complement of W in V . Then,
s[W ] |W = 1 and s[W ] |W ⊥ = −1.
This characterizes s[W ] and shows its independence with the choice of an orthonomal basis
of W .
(2) Since
s[W ] |W = 1 and s[W ] |W ⊥ = −1,
then s2[W ] = id.
(3) (i)We show that
t[W ] ([W 0 ]) = [W 0 ] ⇔ W 0 = (W 0 ∩ W ) ⊕ (W 0 ∩ W ⊥ ).
The sufficiency is clear. For the necessarity, suppose that t[W ] ([W 0 ]) = [W 0 ]. Then, for any
u ∈ W 0 , we have s[W ] (u) ∈ W 0 . Write u = u1 + u2 where u1 ∈ W and u2 ∈ W ⊥ . Then,
u1 − u2 = s[W ] (u) ∈ W 0 . Thus, u1 ∈ W 0 and u2 ∈ W 0 . Hence,
W 0 ⊂ (W 0 ∩ W ) ⊕ (W 0 ∩ W ⊥ ).
It is clear that (W 0 ∩ W ) ⊕ (W 0 ∩ W ⊥ ) ⊂ W 0 . Therefore,
W 0 = (W 0 ∩ W ) ⊕ (W 0 ∩ W ⊥ ).
(ii) Put
X = {[spanR {ei1 , . . . , eik }] : 1 ≤ i1 < · · · < ik ≤ n}.
By the above criterion in (i), X is a nice set in Grk (V ). This nice set X has cardinality
k
n .
(iii) We prove by induction on n that any nice set X in Grk (V ) has cardinality at most nk .
When n = 1, this is trivial. Suppose the conclusion holds when n < m. Assume that n = m.
When k = 0 or m, the conclusion is trivial. Assume that 1 ≤ k ≤ m − 1. Choose an element
[W ] ∈ X. For each i (0 ≤ i ≤ k), write
Xi = {[W 0 ] ∈ X : dim W 0 ∩ W = k − i}.
For each [W 0 ] ∈ Xi , put
Yi = {[W 0 ∩ W ] : [W 0 ] ∈ Xi } ⊂ Grk−i (W )
and
Zi = {[W 0 ∩ W ⊥ ] : [W 0 ] ∈ Xi } ⊂ Gri (W ⊥ ).
By the criterion in (i), Yi is a nice set in Grk−i (W ) and Zi is a nice set in Gri (W ⊥ ). Since
dim W = k < m and dim W ⊥ = m − k < m, by induction we get
k−i
i
|Yi | ≤
and |Zi | ≤
.
k
m−k
Therefore,
|X| =
X
|Xi | ≤
0≤i≤k
X
0≤i≤k
X k − i
|Yi ||Zi | ≤
0≤i≤k
This shows the conclusion.
12
k
i
m−k
k
=
.
m
8 A Busy Courier
A courier picks up a package at coordinate (n, 0) in the two-dimensional lattice and then does
a discrete time simple random walk on Z2 . His station locates at the origin (0, 0). In the rest
of this question, you may without loss of generality assume n is sufficiently large.
(1) Let P1,n be the probability that at his bn1.5 cth step, the distance between this courier
and his station is greater than n2 . Prove that
lim P1,n = 1
n→+∞
(2) Let P2,n be the probability that the courier has ever reached the station within his first
bn1.5 c steps. Prove that
lim P2,n = 0
n→+∞
(3) Let P3,n be the probability that the courier has ever reached the station within his first
2n steps. Prove that
lim P3,n = 1
n→+∞
13
8 Answer (1) Consider the event that this courier is of a distance no more than n2 from
the station after bn1.5 c steps. By triangular inequality, such event has to be a subset of event
A defined as follows:
n
from where
2
he picked up the package after bn1.5 c steps
A =the courier is of a distance at least
Again by triangular inequality, if event A happens, then the courier has to move at least
n
4 along at least one coordinate. Without loss of generality, one may assume it is the ycoordinate. Then the displacement of the courier along y-coordinate can be written as the
following i.i.d. sum:
Yk = X1 + X2 + · · · + Xk
where P (Xi = 0) = 21 , P (Xi = ±1) = 41 with E(Xi ) = 0, var(Xi ) = 21 . So by Chebyshev
Inequality,
n1.5
n
≤2× 2
= 16n−0.5 << 1
P (A) ≤ 2P Ybn1.5 c ≥
4
2n /16
(2) Similar to Part (1), by triangular inequality, the event of interest has to be a subset of
event A defined as follows:
A =the courier has ever reached a distance at least n from where
he picked up the package again bn1.5 c steps
If A happens, then the courier has to have wandered at least n2 along at least one coordinate.
I.e.,
n
P (A) ≤ 2P max |Yk | ≥
2
k≤n1.5
Note that Yk forms a martingale, by Doob’s inequality,
h
i
2
E
Y
bn1.5 c
n
n1.5
P max |Yk | ≥
≤
≤
= 2n−0.5 << 1
2
(n/2)2
2(n/2)2
k≤n1.5
(3) Let Sk ∈ Z2 , k ≥ 0 be the location of courier at kth step. Define stopping times as
follows:
τ = min{k ≥ 0, Sk = 0}
τb = min{k ≥ 0, |Sk | ≥ 2n/4 }
Thus for the probability of interest we have,
P (τ > 2n ) ≤ P(n,0) (b
τ > 2n ) + P(n,0) (b
τ < τ)
where P(n,0) stands for the distribution of simple random walk starting from (n, 0). For
n
the first term on RHS of the inequality above, by invariance principle, we have S[2n/2 t] 2− 4 ,
t ∈ [0, 1] converges to a 2D standard Brownian Motion Bt , where
P (|B1 | ≥ 2) = c > 0
14
Thus for all sufficiently large n,
min
|X0 |≤2n/4
PX0 (|S2n/2 | ≥ 2n/4 ) ≥ P0 (|S2n/2 | ≥ 2 × 2n/4 ) ≥ P (|B1 | ≥ 2) −
c
c
≥
2
2
Then we partite the 2n into subintervals of length 2n/2 . If the courier never level the neighborhood of radius 2n/4 within 2n steps, he has to be so at time points 2n/2 , 2×2n/2 , · · · , [2n/2 ]2n/2
as wells. Thus by Markov property,
h
P(n,0) (b
τ > 2n ) ≤ 1 −
i[2n/2 ] c [2n/2 ]
≤ 1−
<< 1
PX0 (|S2n/2 | ≥ 2n/4 )
2
|X0 |≤2n/4
min
Now for the second term p2 = P(n,0) (b
τ < τ ) we need to consider the potential kernel of 2D
2
simple random walk, for any x ∈ Z define
a(x) =
+∞
X
[P0 (Sk = 0) − P0 (Sk = x)]
k=0
By Kesten (1987, Lemma 1)1 , we have
1) a(x) converges for all x ∈ Z2 and is nonnegative
2) For SRW Sk starting from any initial point, a(Sk∧τ ) is a nonnegative martingale
3) There is a constant C0 such that
|a(x) − log |x|/(2π) − C0 | = O(|x|−2 )
Thus by martingale optional stopping theorem, we have
a((n, 0)) = (1 − p2 ) × 0 + p2 E[a(Sτb)|τ > τb]
Thus
p2 =
a((n, 0))
E[a(Sτb)|τ > τb]
Recalling Condition 3 above,
a((n, 0)) log(n)
E[a(Sτb)|τ > τb] n
Thus we have p2 << 1 and complete the proof.
1
H. Kesten, Hitting probabilities of random walks on Zd , Stochastic Process. Appl. 25 (1987) 165-184.
15
9 Questions
(1) Show that there is no periodic sequence a1 , a2 , a3 , . . . of signs an ∈ {±1} such that, for
every rational number θ,
N
X
sup
an e2πinθ < +∞.
N ∈N n=1
(2) Show that there is no sequence a1 , a2 , a3 , . . . of signs an ∈ {±1} such that, for every
rational number θ
N
X
sup
an e2πinθ ≤ 2022 ?
N ∈N n=1
(3) Give an example of a sequence of signs an ∈ {±1} such that, for every θ ∈ Q − Z,
sup
N
X
an e2πinθ < +∞,
N ∈N n=1
and which takes each of the values +1 and −1 an infinite number of times.
16
9 Answer (1) To simplify the notation, we shall write z := e2πiθ . Suppose, by contradiction, that there is a sequence (an ), periodic with period d ∈ N, such that
sup
N
X
an z n < +∞,
N ∈N n=1
whenever z is of the form e2πiθ with θ ∈ Q.
For convenience, take N to be a multiple d, say N = dM . Since (an ) is periodic, we can
group the terms of the above sum as
Md
X
an z n =(a1 z + a2 z 2 + . . . ad z d ) + z d (a1 z + a2 z 2 + . . . ad z d )
n=1
+ · · · + z (M −1)d (a1 z + a2 z 2 + . . . ad z d )
=(a1 z + a2 z 2 + . . . ad z d ) 1 + z d + z 2d + . . . + z (M −1)d .
Take θ = k/d ∈ Q, for some k ∈ {0, . . . , d − 1} to be chosen later. Then z d = e2πik/d
so
Md
X
an z n = M (a1 z + a2 z 2 + . . . ad z d ).
d
= 1,
n=1
To conclude the problem, it suffices to show that a1 z + a2 z 2 + . . . ad z d in nonzero for some
appropriate choice of k, since we would then have
lim
M →+∞
Md
X
n=1
an z n =
lim M |a1 z + a2 z 2 + . . . ad z d | = +∞,
M →+∞
contradicting ??.
This is clear: if a1 z + a2 z 2 + . . . ad z d vanished for all k ∈ {0, . . . , d − 1} (with z = e2πik/d ),
then the polynomial P (X) = a1 X + a2 X 2 + . . . ad X d would have d + 1 distinct complex roots
e2πi0/d , e2πi1/d , . . . , e2πi(d−1)/d
and
0,
which is impossible for a degree d polynomial.
(2) Suppose, by contradiction, that there is such a sequence (an ). At first, since
fN :=
N
X
an e2πinθ
n=1
is a continous function of θ, |fN (θ)| ≤ 2022 for any θ ∈ R.
Now we calculate the L2 norm of fN . Notice that for n ∈ Z,
(
Z 1
1 if n = 0,
e2πinθ dθ =
0 if n 6= 0.
0
17
Then we have
Z 1
|fN |2 dθ =
0
=
Z 1X
N
an e2πinθ
0 n=1
N X
N
X
N
X
an e−2πimθ dθ
m=1
Z 1
an am
0
n=1 m=1
|
e2πi(n−m)θ dθ
{z
}
nonzero only if n=m
=
N
X
a2n = N
n=1
This shows that (an ) cannot have the property that
N
X
an e2πinθ ≤ 2022
n=1
all N and θ ∈ R! If this were Rtrue, we would have |fN (θ)| ≤ 2022 for all N and θ ∈ R.
Rfor
1
1
2
2
2
2
0 |fN | dθ ≤ 2022 , contradicting 0 |fN | dθ = N if N > 2022 .
(3) (We remark that eventually-constant sequences satisfy the inequality by the geometric
series formula, but are explicitely discarded in the statement of the problem.)
To build up some intuition for the problem, let us take θ = 1/3 and N = 3M . We want to
find a sequence ai such that
!
!
!
M
−1
M
−1
M
−1
X
X
X
sup
a3n+1 · 1 +
a3n+2 · e2πi/3 +
a3n+3 · e4πi/3 < +∞
M ∈N
n=0
n=0
n=0
Geometrically, this corresponds to scaling the three vectors 1, e2πi/3 and e4πi/3 , and computing
the norm of the sum of these scaled vectors.
e2πi/3
1
e4πi/3
Figure 3: Case θ = 1/3.
18
Since 1 + e2πi/3 + e4πi/3 = 0, we should try to scale these three vectors by almost the same
amount in order to obtain good cancellation. In other words, the distribution of the number
of 1’s and −1’s in the sequence (a3n+i )n≥1 should be approximately the same for i = 1, i = 2
and i = 3. Of course, a similar property should hold for any natural number (greater that 1
since θ 6∈ Z) in place of 3.
There is a variety of ways to construct such a sequence (an ). One way to do it is as follows:
n
an
1
+1
2
−1
3
−1
4
+1
5
+1
6
+1
7
+1
8
+1
9
+1
10
−1
11
,
−1
where +1 appears 1! time, then −1 appears 2! times, then +1 appears 3! = 6 times, and so
on. Let us now prove that this sequence has the desired property.
For k ≥ 0, let f (k) = 0! + 1! + · · · + k!, and let Ik ⊂ N be the set of integers n with
f (k) ≤ n < f (k + 1). The sequence above is that which takes the value an = (−1)k for
n ∈ Ik . The key to our construction is that, for every integer q ≥ 2, the length of Ik , namely
(k + 1)!, is divisible by q whenever k is sufficiently large.
Let θ = p/q be a rational number, with gcd(p, q) = 1 and q > 1. Then
N
X
an e2πinθ =
n=1
X
X
(−1)k
e2πinp/q .
k≥0
n∈Ik
n≤N
We need to show that this expression is bounded above by a constant that does not depend
on N .
P
Consider, for each fixed k ≥ 0, the inner sum Sk := n∈Ik e2πinp/q .
n≤N
(i) If k < q, we will be able to use the trivial bound |Sk | ≤ f (k + 1) − f (k) = (k + 1)!
(obtained by the triangle inequality). This will be enough since there are only a finite
number of such k.
(ii) If k ≥ q, we will exploit the particular way in which we constructed the sequence (an )
to obtain sufficient cancellation.
(a) If k ≥ q and f (k + 1) ≤ N , we have perfect cancellation, i.e. Sk = 0. Indeed, in
this case,
f (k+1)
f (k)
f (k+1)−1
X
e2πip/q
− e2πip/q
2πinp/q
Sk =
e
=
e2πip/q − 1
n=f (k)
by the geometric series formula. Since k ≥ q, (k + 1)! is divisible by q, so the
numerator vanishes:
f (k)+(k+1)! f (k) f (k) e2πip/q
− e2πip/q
= e2πip/q
e2πip(k+1)!/q − 1 = 0.
(b) If k ≥ q and N ∈ Ik (this happens for exactly one value of k), we have
N
f (k)
N
X
e2πip/q − e2πip/q
2πinp/q
Sk =
.
e
=
e2πip/q − 1
n=f (k)
19
By the triangle inequality, we deduce the bound
e2πip/q
N
|Sk | ≤
+
e2πip/q
f (k)
e2πip/q − 1
=
2
e2πip/q − 1
which is independent of N .
(c) If k ≥ q and N < f (k), we have of course Sk = 0 since the sum is empty.
Therefore,
N
X
n=1
an e
2πinθ
≤
q−1
X
|Sk | +
k=0
This shows that
sup
X
|Sk | ≤ f (q) +
k≥q
N
X
an e2πinθ < +∞.
N ∈N n=1
20
2
e2πip/q − 1
.
10 Program Design For The Opening Ceremony
Suppose you are chosen as a technology assistant by the director of the opening ceremony
for the 2022 Winter Olympics, and your job is to evaluate the program proposals. One of
the backup programs is a skating show of a ensemble of drones dressed as mascots, which
are moving along a circle. Since the number of the drones is sufficiently large, we can use
a probability density function ρ(t, v)(≥ 0) to represent the distribution of the drones. Here,
v ∈ R is the line speed, and for a given time t, and two speeds v1 < v2 ,
Z v2
ρ(t, v) dv
v1
is the probability of find a drone with its speed between v1 and v2 .
Suppose that the dynamics of the density function is governed by
ρt + (u(t) − v)ρ v = ρvv , v ∈ R, t > 0,
where u(t) is the command speed.
(1) To investigate proper choices of the command speed, D.B. propose that we shall choose
u(t) = u0 + u1 N (t)
where u0 > 0, u1 > 0 and N (t) is the average of the positive part of the speed v+ = max{0, v},
i.e.,
Z
Z
+∞
N (t) =
+∞
v+ ρ(t, v) dv =
−∞
vρ(t, v) dv.
0
But you claim that if u1 > 1, N (t) may become unbounded in evolution, such that the
drones may be out of order. Can you prove it? (For simplicity, the contributions of ρ and its
derivatives at |v| → +∞ are neglected.)
(2) After taking those advices, the directly is wondering whether the drones will be evenly
distributed along the circle. Thus, we shall consider the joint density function p(t, x, v)(≥ 0)
of
R 2πposition and speed, where x ∈ [0, 2π] is the position coordinate on the circle. Clearly,
0 p(t, x, v)dx = ρ(t, v). Suppose that the governing equation for p(t, x, v) is
pt + vpx + (u(t) − v)p v = pvv , x ∈ [0, 2π], v ∈ R, t > 0.
Because, the drones are circulating around, the following boundary condition is satisfied
p(t, 0, v) = p(t, 2π, v),
v ∈ R,
t > 0.
You have a feeling that, no matter how the drones are distributed initially, they will become
almost evenly distributed very quickly. Can you prove or disprove this statement? (For
simplicity, the contributions of p and its derivatives at |v| → +∞ are neglected.)
21
10 Answer (1) We define the average speed
Z +∞
vρ(t, v) dv.
M (t) =
−∞
By direct calculations, we find
Z
d
d +∞
M (t) =
vρ(t, v) dv
dt
dt −∞
Z +∞
=
vρt (t, v) dv
−∞
Z +∞
=
−∞
Z +∞
v (−(u(t) − v)ρ + ρv )v dv
((u(t) − v)ρ − ρv ) dv
=
−∞
= u(t) − M (t)
= u0 + u1 N (t) − M (t).
Because M (t) ≤ N (t) and clearly N (t) ≥ 0, we get
d
M (t) ≥ u0 + (u1 − 1)N (t) ≥ u0 > 0.
dt
Then we conclude that when t → +∞, M (t) → +∞.
And since N (t) ≥ M (t), N (t) also diverges to +∞.
(2) Due to the periodic boundary condition in x, we can write the solution in the form of
Fourier series:
+∞
1 X
pk (t, v)eikx .
p(t, x, v) =
2π
k=−∞
where
Z 2π
pk (t, v) =
p(t, x, v)e−ikx dx.
0
And we aim to show for k 6= 0, pk (t, v) shall decay.
Due to the orthogonality, we find pk satisfies
∂t pk + ikvpk = −∂v (u(t) − v)pk v + ∂vv pk ,
t > 0,
k ∈ Z,
Next we apply the Fourier transform in v. Let
Z +∞
1
p̂k (t, ξ) = √
pk (t, v)e−ivξ dv,
2π −∞
and we can derive that p̂k (t, ξ) satisfies
∂t p̂k + (ξ − k)∂ξ p̂k = − iξu(t) + ξ 2 p̂k .
22
v ∈ R.
Next we use the method of characteristics. Consider the characteristic equation
d
ξk (t) = ξk (t) − k,
dt
whose solution is given by
ξk (t) − k = et−s (ξk (s) − k) .
Along the characteristics, we have
Rt
2
p̂k (t, ξk (t)) = p̂k (0, ξk (0))e− 0 (ξk (s)) +iu(s)ξk (s)ds
which leads to
p̂k (t, ξ) = p̂k (0, (ξ − k)e−t + k)e−Ht,k (ξ−k) ,
where
1 − e−2t 2
Ht,k (z) =
z + 2k(1 − e−t )z + i
2
Z t
u(s)e
−(t−s)
Z t
2
dsz + k t + ik
0
u(s)ds.
0
Applying the inverse Fourier transform, we get
pk (t, v) = eikv (pt,k ∗ Gt,k )(v).
where
t
t
pt,k (y) := e p0,k (e y),
−ikv
Z 2π
p0,k (v) := e
pinit (x, v)e−ikx dx.
0
and Gt,k (z) is a complex-valued Gaussian function defined as follows
1
(z − µ)2
exp −
Gt,k (z) = √
exp (ikΘ) exp −k 2 D .
2
2σ
2πσ
Z t
p
µ(t) =
e−(t−s) u(s)ds, σ(t) = 1 − e−2t ,
0
Θ(t, z) = −
2(z − µ(t))
−
1 − e−t
Z t
u(s)ds,
0
D(t) = t − 2
1 − e−t
.
1 + e−t
We observe that when t > 0, Gt,k decays due to the factor exp −k 2 D , and for t sufficiently
large, it decays exponentially fast.
At last, with the convolution inequality, we get
kpk (t, v)kL1 (R) = keikv (pt,k ∗ Gt,k )(v)kL1 (R) ≤ kpt,k (g)kL1 (R) kGt,k kL1 (R)
2
= kp0,k (g)kL1 (R) e−k D(t) .
This shows that for k 6= 0, pk (t, v) decays, and p(t, x, v) is approaching π1 p0 (t, v), which is an
even distribution in space.
23