Élément de corrigé - Informatique

[1,2,3,4,5,6]
[1,2,3,1,2,3]
−

y0(t) = z(t)C2
(S)⇔y0(t) = z(t)
z0(t) = f(y(t))
y0(t) = z(t)
Rti+1
tiy0(t)dt =Rti+1 tiz(t)dt
y(ti+1) = y(ti) + Zti+1
ti
z(t)dt
z0(t) = f(y(t))
Rti+1
tiz0(t)dt =Rti+1 tif(y(t)) dt
z(ti+1) = z(ti) + Zti+1
ti
f(y(t)) dt
((tk+1 −tk).z(tk)≈Rtk+1
tkz(t)dt
(tk+1 −tk).f(y(tk)) ≈Rtk+1
tk.f(y(t))dt
∀k∈[0, n −2] , h =tk+1 −tk
(h.z(tk)≈Rtk+1
tkz(t)dt
h.f(y(tk)) ≈Rtk+1
tkf(y(t))dt
(y(ti+1) = y(ti) + Rti+1
tiz(t)dt ≈y(ti) + h.z(ti)
z(ti+1) = z(ti) + Rti+1
tif(y(t))dt ≈z(ti) + h.f(y(ti))
yiziy(ti)z(ti)

y(ti+1) = y(ti) + h.z(ti)
z(ti+1) = z(ti) + h.f(y(ik))
yizif, y0, z0, i h
tmax, tmin
n h n
−
− −
− −
y00(t) = −ω2.y(t)
y0(t)
y0(t).y00(t) = −ω2.y(t).y0(t)
1
22y0(t).y00(t)= (−ω2)1 2.2y(t).y0(t)
∃E tq 1
2y0(t)2+ω2 2y(t)2=E
g:x7→ ω2
2x2g0=−f
1
2y0(t)2+g(y(t)) = E
E(ti) = 1
2y0(ti)2+g(y(ti)) = 1 2z(ti)2+g(y(ti))
Ei=1
2z2
i+g(yi) = 1
2z2
i+ω2
2(yi)2
Ei+1 =1
2z2
i+1 +ω2
2(yi+1)2

Ei+1 −Ei=1
2z2
i+1 +ω2
2(yi+1)2−1 2z2
i−ω2
2(yi)2
Ei+1 −Ei=1
2(zi−hω2yi)2+ω2 2(yi+hzi)2−1 2z2
i−ω2
2(yi)2
Ei+1 −Ei=1
2h2ω2(z2
i+ω2y2 i)
Ei+1 −Ei=h2ω2Ei
Ei+1 −Ei= 0
EiE Ei=1
2(z2
i+ω2y2 i)z2 i+ω2y2 i=constante
yi=zi= 0
yizi
Ei+1
Ei
Ei+1
Ei
= 1 + h2ω2
yizi
yizi
f, y0, z0, i h
−
−
−

Ei+1 −Ei=1
2z2
i+1 +ω2
2(yi+1)2−1 2z2
i−ω2
2(yi)2
Ei+1 −Ei=1
2(zi+h 2(−ω2yi−ω2(yi+hzi−h2 2ω2yi))2+ω2 2(yi+hzi−h2 2ω2yi)2−1 2z2
i−ω2
2(yi)2
Ei+1 −Ei=h31
4ω4ziyi−1 2ω4ziyi+1 2ω4ziyi
+h41
8ω4z2
i−1
4ω6y2
i+1
8ω6y2
i
+h5−1/8ω6ziyi
+h61
32ω8y2
i
Ei+1 −Ei=O(h3)
zi+1
i+ 1
−