Modéliser : temps discret ou temps continu

δ∈]0; 1[
µ > 0
unn
un+1 =un+µ.un−δ.un,
r= 1 + µ−δ
n n.h h > 0
µ δ h
h µ(h) = µ(0) + h.m +o(h)δ(h) = δ(0) + h.d +o(h.)
unu(n.h)
u(n.h+h) = u(n.h)+[µ(0)−δ(0)].u(n.h)+(m−d).h.u(n.h)+
o(h). t n h h =t
n
u(t+h) = u(t)+[µ(0) −δ(0)].u(t)
+ (m−d).h.u(t) + o(h)h t n
lim
h→0u(t+h) = u(t) + [µ(0) −δ(0)].u(t),

u t u(t)>0µ(0) = δ(0)
u(t+h) = u(t)+(m−d).h.u(t) + o(h),
u(n.h +h)−u(n.h)
h= (m−d).u(n.h) + o(h) h.
h t
u0(t) = (m−d).u(t).
u(t) = u(0).e(m−d).t
nun+1 =f(un),
u0donn´e.
½y0(t) = f[y(t)],
y(0) donn´e.
(un)n
f
y0(t) = 2.py(t) avec y(0) = 0,
IR+t→t2
y0=
p|1−y2|avec y(0) = 0 [0; a]a < π
2
π
2x7→ cosh ¡x−π 2¢
cosh(x) = ex+e−x
2

y0=f(y) avec y(0) = y0.
f
[a;b]K x y [a;b]
|f(x)−f(y)| ≤ K.|x−y|
f
[a, b]
f
J
J f
z
z0=f(z)
y
z τ y(τ)6=z(τ)t y(t)6=z(t)
a f a
y06=a y
y=a
a

y0=
f(y)f f a b
a < b ]a;b[
y=a y =b
a < y0< b y t a < y(t)< b
y0
yIR β= lim
t→+∞y(t) IR
y b a < y0≤β≤b y0(t) =
f[y(t)] lim
t→+∞y0(t) = f(β)≥0f(β)>0τ
t≥τ y0(t)≥β
2y(t) = y(τ) + Zt
τ
y0(u)du ≥(t−τ).β
2
lim
t→+∞(t)=+∞
β= 0 a < y0≤β≤b β =b y =b
+∞
y=a−∞
y(0) > b y t y(t)> b
IR+
y=b+∞
−∞
IR−
y(0 < a y t y(t)< a
IR+
y=b−∞ +∞
IR+
y0=y(1 −y)
y= 0 y= 1 y(0) 6= 0 y(0) 6= 1
y0
y(1−y)= 1 y0 y−y0 y−1= 1 −ln |y|+ ln |y−1|=t−ln|y(0)|+ln|y(0) −1|
ln ¯
¯
¯1−1
y¯
¯
¯+ ln ¯
¯
¯1−1
y(0) ¯
¯
¯=ty−1
y.y(0) −1 y(0) =et,
y(t) = y(0)
y(0) + [1 −y(0)].e−t.
y(0) = −1
y(t) = 1
1−2e−t
t= ln(2)
]−ln(2) ; +∞[

y(0) = 2
y(t) = 2
2−e−t
]− ∞ ; ln(2)[
0< y(0) <1y(0) −[y(0) −1].et
IR y= 0 +∞
y= 1 −∞
h > 0
yn+1 =yn+h.f(yn)yn+1 =g(yn)g(x) = x+h.f(x)
nun+1 =f(un),
u0donn´e.un+1 −un=f(un)−un
h= 1 u(n.h +h)−u(n.h) = h.f[u(n.h)] −h.u(n.h).
h > 0n n.h =tu(t+h)−u(t)
h=f[u(t)] −u(t),
h u0(t) = f[u(t)] −u(t),
½u0(t) = f[u(t)] −u(t),
u(0) donn´e.
h
ε
y0(t) = −300.y(t) + 60 avec y(0) = 1
5+ε, ε 6= 0 y(t) = 1 5+ε.e−300.t
1
5ε= 0
yn+1 = (1 −300h).yn+ 6h
yn=µy0−1
5¶.(−300h)n+1 5.