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“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)

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