Атты студенттердің IV жоо аралық дәстүрлі ғылыми конференциясының ЕҢбектері
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PROBLEM STATEMENT: It is required to find the solutions of Heat Equation with moving (known) boundary conditions:
1 2 2 1 1
u a t u , ) ( ) (
x t , 0
2
2 2 2 2 x u a t u , x t) ( , 0 t
I.C: 0 ) 0 , 0 ( 1 u , ) ( ) 0 , ( 2 x x u (1) ,
0 ) 0 , ( 2
411
B.C: ) ( ) ( 1 1 t bu t u a t x (2),
) ( 2 2 ) ( 1 1 t x t x fu x u e du x u c , 0, 0
f (3)
) ), ( ( ) ), ( ( 2 1
t u t t u
(4); PROBLEM SOLUTION:
Solutions is represented in the polynomial form of Integral Error Functions: 0 1 1 1 2 2 ) ( ) , (
n n n n n t a x erfc i B t a x erfc i A t t x u , 0 2 2 2 2 2 ) ( ) , ( n n n n n n t a x erfc i D t a x erfc i C t t x u ; where , , , n n n n A B C D has to be determined. Using the L’Hospitalle’s rule for (1) represented in the following form 0 2 lim( ) 0 2 n n n t x t C i erfc a t , n n n n n n n n t n n n t D a x n x t a x a t a x erfc i D t a x erfc i D t ) 2 ( ! 2 2 ) 2 ( 2 lim 2 ) ( lim 2 2 2 2 0 2 0
0 2 2 ( ) ! (2 )
n n n n x D x n a ,where 0 ) ( ! ) 0 ( ) (
n n x n x By using Taylor series we determine ) 0 ( ) ( 2 ) ( 2 1 n n n n a D ; From (2) by substitution
we are ready to use the Leibniz’s formula
l l l n n v u l n v u 0 ) ( ) ( ) ( ) * ( , we obtain n l l n l n n n n l l n l n n n a erfc i l n a B a erfc i l n a A 0 ) ( 1 2 1 ) ( 1 1 0 ) ( 1 2 1 ) ( 1 1 2 ) ( ) ( 2 2 ) ( ) ( 2
First of all, we should solve this part, which
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( ) 2 1 (
) 1 0 1 ( ) ( ) 2 l n n n l n l n i erfc l a
0 2 1 1 1 ( ) ...
0 2
n n n i erfc a
2 0 1 1 1 ( ) 2
n n n i erfc n a
. Using Faa Di Bruno’s formula for each part of ( ) 2 1 ( ) 1 0 1 ( ) ( ) 2 l n n n l n l n i erfc l a
, we get
1 1 2 1 2 ) ( )!*
1 ( * a erfc i n n n =
! 1 2 ) ( 2 ) ( ! 1 ) 1 ( )!*
1 ( * 1 1 1 2 1 2 2 1 1 b n b a a erfc i b n n
. . . . . . . . . . . . . . . . . . . n n n a erfc i 1 2 1 1 2 ) ( = 1 1 2 1 2 2 1 2 1 1 ! 2 ) ( 2 ) ( exp 2 ) ( ) 1 ( ...
1 j b n b n n n n j a a a b b n j , where ! !... ! ...
1 1
n b b n b b n . Let’s say that where 2 ( ) can be represented in the following form:
n n n a a a ...
2 1 2 ... 2 ) ( 1 0 1 1 1 2 1 0 1 2 and 1 1 1 ) 1 ( 1 2 ...
2 1 2 ) ( n n a a . Consider n m n n a 0 1 2 1 ) (
By substitution 0 ,we have
' 1 0 2 ) 0 ( a , ' 1 1 2 ) 0 ( a , . . . , ' 1 ) ( 2 ) 0 (
n n 1 1 1 1 1 2 0 1 1 2 1 ( )! ...
! 2 1! b n b a n i erfc b a ( ) 2 1 1 0 0 1 1 1 1 (0)
2 0 * ( 1) exp
... 2 2 ! j b j n n n n j n n a b b a a j By cancelling other parts we get ! 1
2 ! 1 !* * 1 1 1 1 1 0 2 1 1 0 b n b k n n a a erfc i b n A .
For the first part of first boundary condition (2), we obtain
n n n n n a erfc i B a erfc i A b n a a 0 1 0 2 1 0 2 1 2 1 1 2 2 ! ! 4
(5) By using Leibniz’s and Faa Di Bruno’s formula, we get for second part of condition (2),
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n n n n n a erfc i B a erfc i A n b 0 1 0 2 1 0 2 2 2 ! (6)
And by adding (5) and (6) we have k n n n n n a erfc i B a erfc i A n a a 0 1 0 2 1 0 2 2 1 1 2 2 ! 4 + k n n n n n a erfc i B a erfc i A n b 0 1 0 2 1 0 2 2 2 ! = ' 0
For the second boundary condition(3), we obtain
n n n n n n n n n k n k n n n n n n n n n k n a erfc i D a erfc i C n f a erfc i D a erfc i C n a e a erfc i B a erfc i A n d a erfc i B a erfc i A n a c 0 2 0 2 0 2 0 2 2 0 2 0 2 2 1 0 1 0 1 0 1 0 2 1 0 2 0 2 1 1 2 2 ! 2 2 ! 4 2 2 ! 2 2 ! 4 (8)
For the third boundary condition (4)
n n n n n k n n n n n a erfc i D a erfc i C a erfc i B a erfc i A 0 2 0 2 0 0 1 0 1 0 2 2 2 2
(9) From (9) we can find n C 1 2 0 2 0 ) ( 2 1 1 0 1 0 2 2 ) 0 ( 2 2 2 I a erfc i a erfc i a a erfc i B a erfc i A C n n n n n n n n n n , where 0,1, 2,...,
From (8) we can find n A
Let’s take right side as G , then we obtain n A 2 1 0 1 0 2 2 1 1 1 0 1 0 2 2 1 1 2 2 2 2 I a erfc di a erfc i a c a erfc di a erfc i a c B G A n n n n n n
where 0,1, 2,..., n k
From (6) we can find n B
) 0 ( 2 2 ! 2 2 ! 4 ) ( 0 0 1 0 1 0 2 1 0 2 1 0 2 2 2 1 1 n k n k n n n n n n n a erfc i B a erfc i I n b a erfc i B a erfc i I n a a
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1 0 1 0 2 2 1 1 1 0 1 0 2 2 1 1 2 ) ( 2 2 4 ! 2 2 4 ! ) 0 ( a erfc bi a erfc i a a n a erfc bi a erfc i a a n I B n n n n n n , where 0,1, 2,..., n k
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