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

Endi belgilaymiz
ÿV
Qisqartirish teoremasidan foydalanib, h˜
= ÿµ(Uÿ (h0)+rV bilan belgilaymiz. T ni shunday qilaylikki, 0 < T ÿ funksiya bo‘shliqlari
kiritilsin.
(t)z(y)
|h
X1T := {U ÿ C(R) : U(0, y) = U0(y), U ÿ U0C(R) ÿ 1},
t > 0,
ÿy
ÿt
(45)
Keyin, XT = X1T × X2T × X3T fazo metrikasi bilan to'liq metrik fazodir.
Keyin muammo (4) bo'ladi
1
Yuqorida aytib o'tilganidek, biz mahalliy yechim mavjudligini isbotlash uchun XT dan XT ga
qisqarish xaritasini quramiz. Biz bu qurilishni hozir boshlaymiz. 0 ÿ t ÿ T sifatida A, B va C
koeffitsientlari chegaralangan va A2 ikkita musbat doimiylar orasida.
qayerda
h
V
(46)
d((U1, V1, h1), (U2, V2, h2)) = U1 ÿ U2C(R) + V1 ÿ V2C(R) + h
ÿt
= C(h(t), y(t)). 1
+ z ÿ(y)(h(t) ÿ h0)
,
X2T := {V ÿ C(R) : V (0, y) = V0(y), V ÿ V0C(R) ÿ 1},
t > 0,
ÿh
h0 , (x, t) ÿÿ (y, t) [0, +ÿ) dan diffeomorfizmdir.
(t) = ÿµ(Uy(t, h0) + rVy(t, h0)),
ÿU
= A(h(t), y(t)), 1
+ z ÿ(y)(h(t) ÿ h0)
ÿ h
2C[0,T ] .
h0 (s)|ds ÿ T(1 + h˜) ÿ
R = {(t, y) : 0 ÿ t ÿ T, 0 < y < h0}.
.
Keyin ÿy ÿx ni
hisoblaymiz
(47)
Bizda ... bor
= A 2Uyy + (B - C)Uy + F(U, V ),
U(t, y(t)) = u(t, x), V (t, y(t)) = y(t, x), F(U, V ) = U(1ÿUÿdV ) va G(U, V ) = kV 1 -
4 , z(y) = 0, agar |y - h0| >
8
X3T := {h ÿ C1 [0, T], h
y ÿ [0, h0], t = 0,
E'tibor bering, |h(t) - h0| ekan ÿ dan [0,
+ÿ) gacha). Bundan tashqari,