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10 Математика<br />

«<strong>Молодой</strong> <strong>учёный</strong>» . № 3 (50) . Март, 2013 г.<br />

with a discrete function that satisfies the estimate<br />

h<br />

h<br />

2<br />

ξk -1 ≤ck ( - 1) h.<br />

(29)<br />

1<br />

Because of Taylor series in x of r ( tk, x)<br />

in the vicinity of point x i we get equality<br />

xi+<br />

12<br />

k<br />

r( tk, xi) = ∫ r( tk, x) dx h+ e<br />

x<br />

i<br />

i-12<br />

k<br />

where ei<br />

2<br />

≤ ch 1<br />

2<br />

1 ∂ r<br />

with c1 = max ( t , ) .<br />

[0,1]<br />

2 k x<br />

24 x∈<br />

∂x<br />

(30)<br />

Because of Theorem 1<br />

k-1<br />

xi+ 12 Ai+<br />

12<br />

r k = r k-1<br />

x<br />

k 1<br />

i 12 A<br />

-<br />

- i-12<br />

∫ ∫<br />

( t , x) dx h ( t , x) dx h.<br />

Instead of let use its piecewise linear periodical interpolant Then<br />

k-1 Ai+ 12<br />

r( t 1 1, ) k k x dx h -<br />

A<br />

- =<br />

i-12 k-1<br />

Ai+<br />

12<br />

k<br />

r 1 int ( t 1,<br />

)<br />

k<br />

A<br />

k x dx h h<br />

-<br />

- + i<br />

i-12<br />

k<br />

hi<br />

≤ 2<br />

k-1 i+ 12- k-1<br />

i-12 1<br />

2 =<br />

x∈[0,1]<br />

2<br />

∂ r<br />

2 k<br />

∫ ∫<br />

where chA ( A ) with c max ( t , x)<br />

.<br />

8 ∂x<br />

Thus, we get equality<br />

k-1<br />

Ai+<br />

12<br />

k k<br />

k i = k-1<br />

A<br />

int k -1<br />

+ i + i<br />

i-12<br />

r( t , x ) ∫ r ( t , x) dx h h e .<br />

(32)<br />

For we use (21) and (28):<br />

k-1 k-1<br />

Ai 12 A<br />

h + i+<br />

12 h<br />

k i = k-1 int k 1 k 1<br />

A<br />

- + - k 1<br />

i-12 A<br />

-<br />

i-12<br />

∫ ∫ (33)<br />

r ( t , x ) r ( t , x) dx h ξ ( x) dx h<br />

where values of are constructed by piecewise linear periodical interpolation.<br />

Now let subtract (33) from (32), multiply its modulus by h , and sum for all i = 0, 1,...,n–1:<br />

r( t ,) r ( t ,)<br />

⎛<br />

⎜h h<br />

⎝<br />

e h ξ ( x) dx<br />

⎞<br />

⎟.<br />

⎠<br />

Due to Theorem 3 last terms ( ,) ⋅- in brackets ( ,) ⋅ is ≤ evaluated ( + ) by + . Thus<br />

∑ ∫ (34)<br />

k ⋅-<br />

h<br />

k ⋅<br />

h<br />

≤<br />

1<br />

0≤≤ i n-1<br />

k<br />

i +<br />

k<br />

i +<br />

k-1<br />

Ai+<br />

12<br />

k-1<br />

Ai-12<br />

h<br />

k-1<br />

r tk h<br />

r tk h<br />

1<br />

c1 c2 2<br />

h<br />

h<br />

h<br />

ξk -1<br />

1<br />

h<br />

h<br />

2 h<br />

h<br />

k k<br />

1<br />

1 2 k 1<br />

1<br />

r( t ,) ⋅-r ( t ,) ⋅ ≤ ( c + c ) h + ξ - .<br />

(35)<br />

Let put c = c 1 + c 2 then this inequality is transformed with the help (29):<br />

h<br />

r( t ,) ⋅-r ( t ,) ⋅ ≤ ( c + c ) kh<br />

k k<br />

h<br />

1<br />

1 2<br />

that is equivalent to (27).<br />

We can see that at last level we get inequality<br />

h<br />

h<br />

2<br />

( tm,) ( tm,) cT h .<br />

1<br />

2<br />

r ⋅-r ⋅ ≤ t<br />

(36)<br />

In some sense we got a restriction on temporal meshsize t to get convergence. For example, to get first order of convergence,<br />

it is enough to take<br />

t = ch<br />

with any constant c independent of t and h. But this restriction is not such strong up to constant as Courant–Friedrichs–<br />

Lewy (CFL) condition:<br />

(31)

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