Молодой учёный

6 Математика
«Молодой учёный» . № 3 (50) . Март, 2013 г.
Fig. 1. Trajectories
These characteristics defi ne two trajectories for t∈ [ tk-1, tk]
in plane ( t, x ): C (t, x (t)), i =1,2. Each of these trajectories
i i
crosses line t = t in some point ( t k -1
k- 1,
Ai).
We suppose that they are not mutually crossed and therefore 1 2 . A A <
Theorem 1. For smooth solution of equation (1) we have equality
x2 A2
∫ ∫
r( t , x) dx = r( t - , x) dx.
x
k k 1
1 A1
Proof. Defi ne by G the curvilinear quadrangle bounded by lines t = tk, C2, t = tk-1, C1.
And defi ne by Γ1, Γ2, Γ3, Γ 4
the corresponding parts of these lines, which form the boundary Γ = Γ1∪Γ2 ∪Γ3 ∪Γ (see Fig. 2). Introduce also the ex-
4
ternal normal n defi ned at each part of boundary except 4 vertices of quadrangle.
Now use formula by Gauss-Ostrogradskii [16, 17] in the following form:
G
⎛∂r ∂(
ru)
⎞
⎜ + ⎟ dG = ( rr , u) ⋅ndΓ ⎝ ∂t ∂x
⎠
∫ ∫
Γ
where sing ⋅ means scalar product. Since the boundary Γ consists of four parts we calculate the integral over Γ separately
on each line:
( rr , u) ⋅ndΓ= ( rr , u) ⋅ndΓ+ ( rr , u) ⋅ndΓ+ ( rr , u) ⋅ndΓ+ ( rr , u) ⋅ndΓ. ∫ ∫ ∫ ∫ ∫
Γ Γ1 Γ2 Γ3 Γ4
Fig. 2. Integration along boundary
Along the line 1 Γ the external normal equals n = (1,0). Then
x2
∫( rr , u) ⋅ndΓ=-∫ r(
tk , x) dx.
x1
Γ1
Minus appeared in right-hand side because of opposite direction of integration. At arbitrary point (, tx) ∈ C2the
tangent
vector is v(t, x) = (1, u(t, x)). Therefore the external normal equals
n =
1
u 2 +1
Therefore we get
(u, –1). -1).
dΓ
( rr , u) ⋅ndΓ= ( rr , u) ⋅( u,
- 1) = 0.
2
u + 1
∫ ∫
Γ2 Γ2
(4)
(5)
(6)
(7)
(8)
(9)