Create successful ePaper yourself
Turn your PDF publications into a flip-book with our unique Google optimized e-Paper software.
“Young Scientist” . #3 (50) . March 2013 Mathematics<br />
For other two parts of the boundary we use the same way to calculate the integrals and get<br />
<br />
A2<br />
∫( rr , u) ⋅ndΓ= ∫( rr , u) ⋅( -1, 0) dΓ= ∫ r( tk 1,<br />
x) dx,<br />
A<br />
-<br />
1<br />
Γ3 Γ3<br />
<br />
dΓ<br />
( rr , u) ⋅ndΓ= ( rr , u) ⋅( u,<br />
- 1) = 0.<br />
2<br />
u + 1<br />
∫ ∫<br />
Γ4 Γ4<br />
Combine (5) – (7) and (9) – (11):<br />
⎛∂r ∂(<br />
ru)<br />
⎞<br />
<br />
x2 A2<br />
0 = ∫⎜ + ⎟dG<br />
= ( rr , u) ⋅ndΓ=- r( tk, x) dx + r( tk 1,<br />
x) dx.<br />
t x<br />
x1 A<br />
-<br />
∂ ∂ ∫ ∫ ∫ 1<br />
G ⎝ ⎠ Γ<br />
It implies (4).<br />
3. Simple semi-discrete approximation<br />
Now take integer n > 2 and construct uniform mesh in x with nodes x = ih, i = 0, ±1, ± 2,…,<br />
i ..., and meshsize h =1 n. Let we<br />
h<br />
know the (approximate) solution r ( tk- 1,<br />
x)<br />
at time level tk - 1 and construct the approximate solution at time level t k.<br />
Integrals<br />
of solution in a small vicinities of each point xi may be some useful intermediate data. For example, let construct integrals<br />
xi+<br />
12 h<br />
i ≡ ∫ r<br />
x<br />
k<br />
i-12<br />
I ( t , x) dx<br />
[ x , x ].<br />
at each interval i- 12 i+<br />
12<br />
For this purpose in the context of previous section we take two points x1xi- 12, x2xi+ 12<br />
i i<br />
1 and 2.<br />
A and A at time level 1 . k<br />
C C These trajectories produce two points 1 2<br />
i<br />
A2<br />
h<br />
i = ∫ r i<br />
A<br />
k-1<br />
1<br />
I ( t , x) dx.<br />
Fig. 3. Segment for partial integration on grid<br />
7<br />
(10)<br />
(11)<br />
(12)<br />
= = and construct two trajectories<br />
t - Due to Theorem 1 we get<br />
i i<br />
But at previous level we know only integrals on segment [xi–1/2, xi+1/2] which generally do not coincide with segment [ A1, A 2].<br />
For example, let we have situation at level tk - 1 with some integer s as in Fig. 3. So, we need to use some approximation of partial<br />
integrals.<br />
The simple way consists in approximation of solution by piece-wise constant function. Thus we put<br />
1 x<br />
h h i+<br />
12 h<br />
r ( tk-1, x) = r ( tk-1, xi) ≡ r ( tk1, x) dx x [ xi 12, xi 12)<br />
i 0,1, , n 1.<br />
h ∫x<br />
- ∀ ∈ - + ∀ = -<br />
(14)<br />
i-12<br />
But this interpolation is rather rough. It gives accuracy of order O(h) only. Instead of it we take linear interpolation at each<br />
segment [x i , x i+1 ]. For this purpose at fi rst we put<br />
1 x<br />
h i+<br />
12 h<br />
r ( tk-1, xi) ≡ r ( tk1, x) dx i 0,1, , n 1<br />
h ∫x<br />
- ∀ = -<br />
(15)<br />
i-12<br />
and then defi ne<br />
(13)