07军事模型(数学建模)
合集下载
相关主题
- 1、下载文档前请自行甄别文档内容的完整性,平台不提供额外的编辑、内容补充、找答案等附加服务。
- 2、"仅部分预览"的文档,不可在线预览部分如存在完整性等问题,可反馈申请退款(可完整预览的文档不适用该条件!)。
- 3、如文档侵犯您的权益,请联系客服反馈,我们会尽快为您处理(人工客服工作时间:9:00-18:30)。
y = rx p(r) xp(r) xr r xr ⇒ r x = f (y ) r
Lanchester
p(r) f (y ) x0 M g (x)
y0
Lanchester
N x0 /N N
x = f (y ) g (x)
Lanchester
B y ym A
y0
O
x0 /N
x0
xm
x
Lanchester
Lanchester
2 g=h=0 x(t) = y (t) = 0 3 g, h = 0 y ˙ = h, x(t), y (t) (19) x(t), y (t)
x(t), y (t)
x ˙ = g,
Lanchester
2 g=h=0 x(t) = y (t) = 0 3 g, h = 0 y ˙ = h, x(t), y (t) 4 x(t) = 0 x ˙ = ky x(t) g=0 (19) x(t), y (t)
Lanchester
αβ − kl = 0 λ=
(19)
−(α + β ) ± [(α + β )2 − 4(αβ − kl)]1/2 2 −(α + β ) ± [(α − β )2 + 4kl]1/2 = 2 αβ − kl > 0 (x0 , y0 )
αβ − kl < 0 (x0 , y0 )
x(t), y (t)
x ˙ = g,
Lanchester
(19)
Lanchester
(19) ˙ = AX + R X A ⇔
Lanchester
αβ − kl = 0 λ=
(19)
−(α + β ) ± [(α + β )2 − 4(αβ − kl)]1/2 2 −(α + β ) ± [(α − β )2 + 4kl]1/2 = 2
Lanchester
x(t), y (t) x ˙ (t) = −f (x, y ) − αx + u(t), α > 0 y ˙ (t) = −g (x, y ) − βy + v (t), β > 0 (1)
fwk.baidu.com
g
Lanchester
f (x, y ).
Lanchester
f (x, y ).
2
px = 0.1 sx = 0.1 (17) >
rx
ry
2 · 0.1 · 0.1 × 106 = 100 2 · 1 · 100 10
(18)
Lanchester
8 6
Lanchester
J
H
Engel
Lanchester
J
H
Engel
:
Lanchester
Richardson, 1939
cy 2 − 2bx = n
2 − 2bx0 n = cy0
(14) (15)
:
Lanchester
n>0
2
n<0
n=0
y0 x0 b = r x px c = ry
sry sx
>
2b cx0
(16)
y0 x0
2
>
2rx px sx ry sry x0
(17)
Lanchester
(17) x0 = 100 sry = 1 y0 x0 y0 /x0 > 10
t k>0
k a
> 0,
:
Lanchester
(6) a b y0 x0
2
>
b r x px = a r y py
(7)
Lanchester
(7) 4 4
y0 /x0 2 rx px ry p y y0
2
Lanchester
Lanchester
sx
f = cxy ry sry
c py sx
c = r y py = r y
1 k
x=0 1 y1 = k x ˙ (22)
Lanchester
1933
1 k
1936 α =3 . k = 0.3.
Lanchester
α.
g = 0, y = 0
(19)
x ˙ = −αx
x(t) = x(t0 ) · e−α(t−t0 ) 1 ) = x(t0 ) · e−1 α
1 α
f a
y
f = ay
a ry
a = r y py ) py
Lanchester
g = bx px x ˙ (t) = −ay − αx + u(t) y ˙ (t) = −bx − βy + v (t)
b = rx p x
rx (1) (2)
Lanchester
x0
y0
(2) x ˙ = −ay y ˙ = −bx x(0) = x0 ,
x(t0 + (23)
1 e
(23)
1 α
α
Richardson α = 0.2
Lanchester
−αx + ky + g = 0 lx − βy + h = 0
(20)
Lanchester
Lanchester
1 (19) (19) k l g, h g h k, l
Lanchester
2 g=h=0 x(t) = y (t) = 0 (19) x(t), y (t)
2 − bx2 k = ay0 0
(4)
(5)
(6)
Lanchester
(5) x(t) y (t) y k<0 t1 , x(t1 ) = 0,y (t1 ) = k=0
t k>0
k a
> 0,
Lanchester
(5) x(t) y (t) y k<0 t1 , x(t1 ) = 0,y (t1 ) = k=0
Lanchester
Lanchester
Lanchester
x(t) 1. f (x, y ) 2. 3.
y (t)
t
g (x, y )
u(t)
v (t)
Lanchester
x(t), y (t) x ˙ (t) = −f (x, y ) − αx + u(t), α > 0 y ˙ (t) = −g (x, y ) − βy + v (t), β > 0 (1)
Lanchester
Richardson, 1939
1. 2. 3.
Lanchester
x(t), y (t) 1. x(t) 2. x(t) 3. y (t) x(t) x(t)
Lanchester
x(t), y (t) 1. x(t) 2. x(t) 3. y (t) x(t) x(t) x ˙ = −αx + ky + g y ˙ = lx − βy + h (k, l, α, β, g, h 0) (19)
Mathematic Modeling
Lanchester
Lanchester
Lanchester
1. 2. 3.
Lanchester
Lanchester
x
y
Lanchester
x x f x = f (y ) x x0 x0 y = g (x) M
y y x > f (y )
xm
ym
sry sx
Lanchester
g = dxy (1)
d = rx px = rx ssrx . y (8)
x ˙ (t) = −cxy − αx + u(t) y ˙ (t) = −dxy − βy + v (t)
Lanchester
αx
βy ˙ x y ˙ x(0)
u=v=0 = −cxy = −dxy = x0 ,
Lanchester
αβ − kl > 0 αβ > kl (21)
(x0 , y0 ) (21)
>
Lanchester
αβ − kl > 0 αβ > kl (21) (21) Richardson
(x0 , y0 ) (21)
> α, β, k, l.
Lanchester
g = 0, y = y1 x ˙ = ky1 ⇒ (22) y1
(8)
(9) y (0) = y0
Lanchester
(3) cy − dx = m m = cy0 − dx0 (10) m>0 m<0
(9) (10) (11) m=0
Lanchester
y (t) m > 0,
m = 0,
m < 0, m/c
O
−m/d
x(t)
:
Lanchester
y0 d rx srx sx > = x0 c ry sry sy y0 /x0 s
(12)
y0
Lanchester
Lanchester
f = cxy
g = bx ˙ = −cxy x y ˙ = −bx x(0) = x0 ,
(13) y (0) = y0
Lanchester
cy 2 − 2bx = n
2 − 2bx0 n = cy0
(14) (15)
Lanchester
Lanchester
y ym
M M M
y0
O
x0
xm
x
Lanchester
x = f (y ) y = g (x)
Lanchester
r y r xp(r) x = f (y ) x = f (y ) y = g (x) y = g (x) xp(r) x0 p(r) > 0 x0 x xr y = rx y = rx x = ry
(3) y (0) = y0
Lanchester
(3) (3) dy bx = dx ay (4)
Lanchester
(3) (3) dy bx = dx ay ay 2 − bx2 = k (4)
(5)
Lanchester
(3) (3) dy bx = dx ay ay 2 − bx2 = k (3)