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“Young Scientist” . #3 (50) . March 2013 Mathematics
1 1 1 1
h h h h
r
0
m = r
0
m-1=
= r
0
1 = r
0
init
∫ ∫ ∫ ∫
( t , x) dx ( t , x) dx ( t , x) dx ( x) dx.
Now we prove a stability of algorithm (17) – (20) in the discrete norm analogous to that of space L 1 ([0,1]) :
h
r = ∑ r(
x ) h.
(23)
1
0≤≤ i n-1
i
Theorem 3. For any intermediate discrete function the solution of (17)–(20)
satisfies the inequality:
h
h
h
h
tk t
1
k 1
1
ξ ( ,) ⋅ ≤ ξ ( - ,) ⋅ .
(24)
Proof. Indeed, because of (18) and (20) we get
x
h
h
h i+
12 h
( k , ⋅ ) = ( , ) ( , )
1
k i =
x
k
i-12
0≤≤ i n-1 0≤≤ i n-1
ξ t ξ t x h ξ t x dx
0≤≤ i n-1
∑ ∑ ∫
∑ ∫ ∫
k-1 k-1
Ai+ 12 A
h n-12
h
ξ k 1 k 1 ξ
- k 1
A
- - k 1
i-12
A
-
-12
= ( t , x) dx ≤ ( t , x) dx.
h
h
Due to periodicity of functions ξ ( t - 1,
x)
and ξ ( tk-1, x)
k-1
An-12
k-1
A-12
1
h h
k-1 =
0
k-1
∫ ∫
ξ ( t , x) dx ξ ( t , x) dx.
Let introduce the basis functions for linear interpolation
k
( -1
]
( )
⎧( xi-x) h if x∈ xi , xi
,
⎪
ji
( x) = ⎨(
x-xi) h if x∈ xi, xi+
1 ,
⎪
⎩
0 else.
Then for piecewise linear interpolant we get
ξ ( t , x) = ∑ j ( x) ξ ( t , x ).
h h
k -1 i k-1 i
0≤≤ i n-1
Then
∫ ∫ ∑ ∑ ∫
1 1
x
h h h
i+
1
ξ (
0
k -1, ) ≤ j ( ) (
0
i ξ k-1, i) = ξ ( k-1, i) ji(
)
xi-1
0≤≤ i n-1 0≤≤ i n-1
∑
= ξ ( t , x ) h= ξ ( t , ⋅)
.
0≤≤ i n-1
t x dx x t x dx t x x dx
h h
k-1 i k-1
h
1
Combine this inequality with (25) and (26) we get (24).
Now evaluate an error of approximate solution in introduced discrete norm.
Theorem 4. For sufficiently smooth solution of problem (1)–(2) we have the following estimate for the constructed
approximate solution:
h
h
2
( k,) ( k,)
0,1, ,
1
r t ⋅-r t ⋅ ≤ckh ∀ k = m
(27)
with a constant c independent of k and h.
Proof. We prove this inequality by induction in k . For k = 0 this inequality is valid because of exact initial condition (2):
Suppose that estimate (26) is valid for some k -1≥ 0 and prove it for k.
So, at time-level tk–1we have decomposition
9
(25)
(26)
(28)