Chjimin Guo matematika va axborot fanlari maktabi, ÿ Muallif: Zhiming Guo

U(0, y) = U0(y) = u0(y),
ÿz ÿÿ(y)(h(t) ÿ h0)
ÿ
= DA2Vyy + (DB - C)Vy + G(U, V ),
ÿ
2y ÿx2
(y)| ÿ
(h0)). [10] dagi kabi, biz mahalliy mavjudligini isbotlaymiz
U + a
2 , |z
t > 0, 0 < y < h0,
y ÿ [0, h0], t = 0.
=
8
ÿ h˜C[0,T ] ÿ 1},
V (0, y) = V0(y) = y0(y),
ÿt
= B(h(t), y(t)), [1
+ z ÿ(y)(h(t) ÿ h0)]3
Lemma isboti 2.1. Asosiy fikr [4] dan moslashtirilgan. z ÿ C3 ([0, ÿ)) shunday bo'lsinki , |y - h0| bo'lsa,
6 z(y) = 1 bo'lsin. ÿ barcha y
uchun. h0 ni aniqlang ,
|h(t) ÿ h0| ÿ
va
t > 0, 0 < y < h0,
=
x = y + z(y)(h(t) ÿ h0), 0 ÿ y < +ÿ.
Machine Translated by Google

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C
(1+th) , 1+th
C
C
(R)
C
ÿ2
1+th
1
t
C 2
2 (R)
[Uˆy(t, h0) + rVˆy(t, h0)]dt.
0
(R)
C
2 (R)
16
Y. Liu, Z. Guo, M. El Smaily va L. Vang
(1+th),
1+th
th
ÿ2
ÿ2
1+th
2
2
(1+th),
1+th
a
1+th
1+th
0,
th
th
th
th T
2 [0,T ]
1+th
2
1+th
2
1+th
0,
1+th
T
Shunday qilib, biz XT dan o'z ichiga olgan xaritaga egamiz .
(1+th)
,1+th
(R).
ÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿ
ÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿ
= DA2 (h2(t), y(t))V¯yy + (DB(h2(t), y(t)) ÿ C(h2(t), y(t)))V¯y + G,
0 < T ÿ min C
2 [0, T] va khˆÿk
Uˆy(t, 0) = Vˆy(t, 0) = 0,
kUˆ ÿ U0kC(R) ÿ kUˆk
Uˆ(0, y) = U0(y) = u0(y),
= A 2Uˆyy + (B ÿ C)Uˆy + F(U, V ),
ÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿ
ÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿ
T
ÿ C1 va kVˆ k
ÿVˆ
t > 0 va 0 < y < h0 uchun.
Endi biz P ning etarlicha kichik T uchun qisqarish xaritasi ekanligini ko'rsatamiz. (Uˆ i,
Vˆ i, XT i = 1, 2 uchun. Biz U¯ = Uˆ1 ÿ Uˆ2 va V¯ = Vˆ1 ÿ Vˆ2 ni o'rnatamiz. Keyin,
,
t > 0,
ÿt
C 3
ÿ C3, bu erda C3 bog'liq
Uˆ(t, h0) = Vˆ (t, h0) = 0,
C 2
the
Vˆ (0, y) = V0(y) = y0(y),
ÿt
ÿ C2,
= DA2Vˆyy + (DB ÿ C)Vˆy + G(U, V ),
U¯y(t, 0) = V¯y(t, 0) = 0, t > 0,
khˆÿ ÿ h˜kC[0,T ] ÿ khˆÿ k
ˆ
,
t > 0,
= A
2 ÿ 1,
,
bu erda C1 va C2 konstantalari h0 , th, kU0kC2[0,h0] va kV0kC2[0,h0] ga bog'liq .
(48)
hˆ i) ÿ
,
t > 0 va 0 < y < h0 uchun.
2 ÿ C2T
Keyin, hˆÿ (t) = ÿµ(Uˆy(t, h0) + rVˆy(t, h0)) ÿ C µ, r,
h0, a, kU0kC2[0,h0] va kV0kC2[0,h0] da .
t > 0, 0 < y < h0,
U¯(t, h0) = V¯ (t, h0) = 0, t > 0,
2 ÿ C3T
keyin bizda bor
ˆ hˆ). Biz da'vo qilamiz
y ÿ [0, h0],
kUˆk
(h2(t), y(t))U¯yy + [B(h2(t), y(t)) ÿ C(h2(t), y(t))]U¯y + F,
hˆ(t) = h0 - µ Z
ˆ Vˆ ) ÿ C
har qanday th ÿ (0, 1) uchun yagona chegaralangan yechimni qabul qiladi (U,
t > 0, 0 < y < h0,
ÿU¯ 2
2 ÿ C1T
(R) × C
ÿ
Endi aniqlaymiz
ÿV¯
2 ÿ 1,
Endi biz PH xaritalashni joriy qilishga tayyormiz: (U, V, h) ÿ (U, V, bu PH XT
ni o'ziga kichik T uchun moslashtiradi: Haqiqatan ham, agar T ni shunday olsak.
ÿt
2 ÿ 1.
U¯(0, y) = V¯ (0, y) = 0, 0 ÿ y ÿ h0,
,
y ÿ [0, h0],
Standart Lp nazariyasi va Sobolev singdirish teoremasi bo'yicha, har qanday (U, V, h) ÿ XT
uchun boshlang'ich chegaraviy masala ÿUˆ
ÿ
kVˆ ÿ V0kC(R) ÿ kVˆ k
Bundan tashqari,
ÿt
(49)
Machine Translated by Google

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(R)
C 2
h0
C 2
([0,T ])
'
C 2
1
C(R)
'
1
'
2
C 2
'
2
C 2
'
(R)
2 C([0,T ])
(R)
2
C 2
2 C([0,T ]) ÿ C7T
Erkin chegaraga ega Lesli-Gower modeli
17
1
ÿ2
1+th
ÿ2
th
th
1+th
,1+th
1+th
,1+th
1+th
ÿ2
th
1+th
,1+th
1+th
1+th
,1+th
1+th
,1+th
ÿ2
1+th
1+th
,1+th
th
+T
Shunga qaramay, standart Lp taxminlari va Sobolev o'rnatish teoremasidan foydalanib, biz bor
C 1
(D) ÿ C5(U1 - U2C(R) + V1 - V2C(R) + h1 - h2C1[0,T ] ),
2 C[0,T ] ÿ T
u¯
2 UC(R) + V
F : = [A2 (h1(t), y(t)) ÿ A2 (h2(t), y(t))]Uˆ1yy + [(B(h1(t), y(t)) ÿ B(h2 (t),
,
taqqoslash tamoyili orqali barcha t ÿ [0, T] va x ÿ [0, h(t)] uchun y(t, x) ÿ max{M1 + a, y0ÿ} degan
xulosaga kelish.
U¯C(R) + V¯
ÿ2(t, x) = M2[2M(h(t) ÿ x) ÿ M2 (h(t) ÿ x)
(t) = ¯u(1 - u¯) t > 0 uchun,
.
va
(t) > 0 barcha t ÿ (0, T] uchun.
] t ÿ [0, T] & x ÿ [h(t) ÿ Mÿ1 , h(t)] uchun,
2 U¯
y¯
1
y(t))) ÿ (C(h1(t), y(t)) ÿ C(h2(t), y(t))]Uˆ1y + F(U1, V1) ÿ F(U2, V2).
Endi h ni u va y ni quyidagi ikkita
yordamchi funktsiyaga ÿ1(t, x) = M1[2M(h(t) ÿ x) ÿ M2 (h(t) ÿ x)
bilan solishtirishini isbotlashga o‘tamiz.
,
,
C(R) + h¯ÿ
C(R) + h¯ÿ
u¯(0) = u0ÿ.
C 3
ÿ¯h ÿ
h¯'
1
M1 + a
,
C 2
T = min 1, 8(1 +
˜h) 2
” y¯
bu erda C4, C5 va C6 > 0 konstantalari A, B, C va Ci ga bog'liq, i = 1, 2, 3 uchun.
G : = [DA2 (h1(t), y(t)) ÿ DA2 (h2(t), y(t))]Vˆ1yy + [(DB(h1(t), y(t)) ÿ DB(h2 (t),
Endi biz ga murojaat qilamiz
Bu erda C7 := max{C4, C5, C6}. Biz tanlaymiz
U¯C(R) + V¯
] t ÿ [0, T] & x ÿ [h(t) - Mÿ1 , h(t)] uchun.
(50)
ÿ h¯'
2 V¯
ÿ h¯ÿ
2
y¯(0) = y0ÿ,
,
2 h¯
+h 1
U¯
(t) = kyu(1 ÿ ) t > 0 uchun,
va qisqarish xaritalash teoremasini qo'llang, PH XTda yagona sobit nuqtaga ega degan xulosaga
keling . Bu Lemma 2.1 ning isbotini to'ldiradi.
Bizda ham bor
y(t))) ÿ (C(h1(t), y(t)) ÿ C(h2(t), y(t)))]Vˆ1y + G(U1, V1) ÿ G(U2, V2) .
(t) t ÿ (0, T] uchun ÿ L. Bunga erishish uchun biz
Lemma isboti 2.2. Kuchli maksimal printsip barcha t ÿ [0, T] va x ÿ [0, h(t)) uchun u > 0 va y > 0 bo'lishini
beradi. u(t, h(t)) = y(t, h(t)) = 0 boÿlganligi sababli, Hopf Lemma ux(t, h(t)) < 0 va yx(t, h(t)) < ni hosil
qiladi. 0 hamma uchun t ÿ (0, T]. Shunday qilib, h Endi biz quyidagi boshlang'ich qiymat masalasini
ko'rib chiqamiz.
ÿ h¯'
V¯
Taqqoslash tamoyili barcha t ÿ [0, T] va barcha x ÿ [0, h(t)] uchun u(t, x) ÿ u¯(t, x) ÿ max{1, u0ÿ} ekanligini
bildiradi. Xuddi shunday, biz quyidagi muammoni ko'rib chiqamiz
C 7
([0, T]) ÿ C6(U1 - U2C(R) + V1 - V2C(R) + h1 - h2C1[0,T ] ),
qayerda
(51)
,
ÿ h
Yuqoridagilarga asoslanib, agar T ÿ (0, 1] bo'lsa, u holda
ÿ C4(U1 - U2C(R) + V1 - V2C(R) + h1 - h2C1[0,T ] ),
+ T
va
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x
ÿ
x
h0
h0
tÿ+ÿ
M1
'
y
M2
M1
ÿ
u+a
'
M2
'
18
Y. Liu, Z. Guo, M. El Smaily va L. Vang
7. Natijalarni muhokama qilish va xulosa qilish
h0
h0
,
buni olish uchun
Shunday qilib, bizda h
hozir tugallandi.
2
ÿÿÿÿÿÿÿÿÿÿÿÿÿÿ
ÿÿÿÿÿÿÿÿÿÿÿÿÿÿ