# 偏微分方程数值习题解答

? (λ ) = J ( x0 + λx) = ( A( x0 + λx), x0 + λx) ? (b, x0 + λx)
= J ( x0 ) + λ ( Ax0 ? b, x) +
1 2

λ2

2 ? ' (λ ) = ( Ax0 ? b, x) + λ ( Ax, x)

( Ax, x)

Ax0 ? b = 0, Ax0 = b

? ' (0) = ( Ax0 ? b, x) + λ ( Ax, x) |λ =0 = 0

b

a b

g1 ( x)? ( x)dx = ? ∫a f ( x)? ' ( x)dx
b b

' ∫a g 2 ( x)? ( x)dx = ? ∫a f ( x)? ( x)dx

( g1 ? g 2 )? ( x) = 0 ? ? ∈ C0∞ ( I ) ∫a
b

| a (u, v) |=| ∫a ( pu 'v ' + quv)dx |≤ M || u ' || . || v ' || + M ' || u || . || v || ≤ 2 M * || u ||1 . || v ||1 ,其中 M * = max{M , M ' } 其中
b

b

a

g ( x)? ( x)dx = (?1) k ∫a f ( x)? ( k ) ( x) dx
b

dk f 广义导数, 广义导数,并记 g ( x) = k dx

|| f n ? f m ||1 = || f n ? f m ||0 + || f n' ? f m' ||0 → 0

|| f n ? f ||1 → 0 (关键证明 g =
df ) dx

| ∫a ( f n ( x ) ? f ( x ))? ( x) |≤|| f n ? f ||0 . || ? ||0
b

| ∫a ( f n' ( x ) ? g ( x ))? ( x) dx |≤|| f n' ? f ' ||0 || ? ||0
b

lim ∫a f n ( x )? ( x )dx = ∫a f ( x )? ( x) dx
b b n →∞

lim ∫a f ( x )? ( x)dx = ∫a g ( x)? ( x )dx
b n →∞ ' n b

b b

b b

|| f n ? f ||1 = || f n ? f ||0 + || f n' ? f ' ||0 → 0

? Lw = f ? Lu0 ? ' ?w(a ) = 0, w (b) = 0

(P)

1 求 w* ∈ C 2 I H E , J ( w* ) = min J * ( w) w∈H
1 E

1 J * ( w) = a (u ? u0 , u ? u0 ) ? ( f ? Lu0 , u ? u0 ) 2 ~ = J (u ) + ( Lu0 , u ) ? a (u0 , u ) + C

1 2

u* ∈ C 2 I H 1 , u (a ) = α

J (u* ) = min J (u )
u∈H 1 u ( a ) =α

Lw = Lu ? Lu0 = f ? Lu0 w( a) = 0, w' (b) = 0
1 等价于: 等价于: ? v ∈ H E

1 2

( Lw, v) ? ( f ? Lu0 , v) = 0

(?
b d b d dw du dw dw dv ( p ), v) = ? ∫a ( p )vdx = ? p v |b + ∫a p dx a dx dx dx dx dx dx dx

a ( w, v) ? ( f ? Lu0 , v) = a (u , v) ? ( f , v) + ( Lu0 , v) ? a (u0 , v) = a (u , v) ? ( f , v) ? p (b) β v(b)

a (u , v) ? ( f , v) ? p (b) βv(b) = 0

d 4u ( Lu ? f , v ) = ( 4 + u ? f , v) = 0 dx 4 4 bd u d u d 3u b b d 3u dv 而 ( 4 , v) = ∫a 4 vdx = 3 v |a ? ∫a 3 . dx dx dx dx dx dx 2 2 2 2 2 bd u d v d u dv b b d u d v =? 2 |a + ∫a 2 2 dx = ∫a 2 2 dx dx dx dx dx dx dx

d 2u d 2 v 上式为 ∫a [ 2 2 + uv]dx = ( f , v) dx dx
b

d 2u d 2 v 为双线性形式. 定义 a(u, v) = ∫a [ 2 2 + uv]dx ,为双线性形式. dx dx 变分问题为: 变分问题为:求 u ∈ H 02 ( I ) , ? v ∈ H 02 ( I )
b

a (u , v ) = ( f , v )

1 -4 1.用 1.用 Ritz ? Galerkin 方法求边值问题
?? u " + u = x 2 0 < x < 1 ? ? u (0) = 0, u (1) = 1

?i ( x) = sin(iπx), i = 1,2,..., n

:(1)边界条件齐次化 边界条件齐次化: 解 :(1) 边界条件齐次化 : 令 u0 = x , w = u ? u0 , 满足齐次边界条件, 则 w 满足齐次边界条件,且
Lw = Lu ? Lu0 = x 2 ? x w(0) = 0, w(1) = 0

i =1

n

∑ a(? ,?
i =1 i

n

j

)ci = ( x 2 ? x,? j )

j = 1,2,..., n

a (?i ,? j ) = ∫0(? ? + ?i? j )dx = ijπ
1 ' i ' j 1

2 1 0

∫ cos(iπx) cos( jπx)dx
ijπ 2

+ ∫0sin(iπx) sin( jπx)dx =

π

cos(ix) cos( jx)d x

+

1 2π

π

sin ix sin jx

? i 2π 2 1 ? a (?i , ? j ) = ? 2 + 2 , i = j ?0, i≠ j ?

2 1

2 ( jπ )
3

[(?1) j ? 1]

?8 ( x 2 ? x, ? j ) ? j为奇数 ? cj = = ? ( jπ )3 (1 + j 2π 2 ) a (? j ,? j ) ?0 j为偶数 ?

[

un ( x) = x +

∑ (2k ? 1) π
k =1

n +1 ] 2

? 8 sin[(2k ? 1)πx] 3 3 [1 + (2k ? 1) 2 ]

2.在题 代替右边值条件, 2.在题 1 中,用 u (1) = 0 代替右边值条件, un (x) 是用 Ritz ? Galerkin 方法求解相应问题的第 n 次 近似, 近似,证明 un (x) 按 L2 (0,1) 收敛到 u ( x) ,并估计误 差. 证明: 对应的级数绝对收敛, 证明: un 对应的级数绝对收敛,由{sin iπx}的完

Rn ≤ 8 π 3n3

3.就边值问题(1.2.28)和基函数 3.就边值问题(1.2.28)和基函数 就边值问题(1.2.28) ?i ( x) = ( x ? a )i (i = 1,2,..., n) ,写出 Ritz ? Galerkin 写 方程 解 : 边 界 条 件 齐 次 化 , 取 u0 = α + β ( x ? a ) , w = u ? u0 , w 对应的微分方程为
Lw = Lu ? Lu0 = f ? Lu0 w( a) = 0, w' (b) = 0

a ( w, v) ? ( f ? Lu0 , v) = 0 Lu0 = ? ? ∫a
b

d du dp ( p 0 ) + qu0 = ? β + q[α + β ( x ? a )] dx dx dx

b dp v = ? p(b)v(b) + ∫a pv ' ( x)dx dx

a ( w, v ) = ( f , v ) + βp (b)v (b) ? ∫a [ β pv ' ( x ) ? qu0 v]dx
b

∑ a(? ,?
j =1 i

n

j

)c j = ( f ,?i ) + β p (b)?i (b) ? ∫a p ( x) β i ( x ? a )i?1 dx + ∫a q ( x)[α + β ( x ? a )]dx
b b

a (?i ,? j ) = ∫a [ p?i'? 'j + q?i? j ]dx
b

n = 1 , a (?1 ,?1 ) = ∫a 1dx = (b ? a )
b

1 1 d1 = (b ? a ) 2 + β (b ? a ) ? β (b ? a ) = (b ? a ) 2 , 2 2 1 1 c1 = (b ? a ) ,即解 u1 = u0 + ( x ? a ) 即解 2 2 n = 2: a (?1 ,?1 ) = (b ? a), a (?1 ,? 2 ) = ∫a 2( x ? a )dx = (b ? a ) 2
b

b 4 a (? 2 ,? 2 ) = ∫a 4( x ? a ) 2 dx = (b ? a )3 3

d 2 = ∫a ( x ? a ) 2 dx + β (b ? a ) 2 ? ∫a 2 β ( x ? a )dx
b b

1 1 = (b ? a )3 + β (b ? a ) 2 ? β (b ? a ) 2 = (b ? a )3 3 3

? b?a ? ? (b ? a ) 2 ? ?1 ? (b ? a) 2 ?? c ? ? (b ? a ) 2 ? ? 1? =?2 ? 4 3 ? (b ? a) ?? c2 ? ? 1 3 ? ?? ? ? (b ? a ) ? 3 ?3 ? ?1? ?1 1 ?? c1 ? ? ? ? 4 ?? ? = ? 2 ? ?1 ?? c2 ? ? 1 ? ? 3 ?? ? ? ? ? 3?

1 3

1 6

1 2

1 2

?y +
"

π2

x, 0 < x < 1 4 2 y (0) = 0, y ' (1) = 0

y = 2 sin

π

"

Galerkin 形式的变分方程为 ( Lu , v ) = ( f , v ) ,

( Lu , v ) = ? ∫0u vdx +
1 " ' 1

1 "

π2
4

∫0uvdx , ( f , v) = ∫02 sin
1 ' ' 1 ' '

1

1

π
2

xv( x) dx

' ' 1

π2
4

uv )dx

x ? xi?1 h

x ? xi ? 1?ξ, ξ = xi ≤ x ≤ xi+1 ? h ? x ? xi?1 ? ?i ( x) = ?ξ , ξ = , xi?1 ≤ x ≤ xi (i = 1,2,3) h ? other ? 0, ? ?

a (?1 ,?1 ) = ∫x + ∫x [? +
x1
0

x2
1

'2 1

π2
4

? 2 ]dx

1? 1 ? 1 1 π2 2 π2 ? ? = h ?∫0[ 2 + ξ ]dξ + ∫0? 2 + (1 ? ξ ) 2 ? dξ ? 4 4 ?h ? ? ? h

1 2

π2
12

1 π π2 a (?1 ,? 2 ) = ∫0[? + hξ (1 ? ξ )dξ = ?2 + h 4 12 1 1 π π 1 1 ( f ,?1 ) = 2h[ ∫0sin (0 + hξ )ξdξ + ∫0sin ( + ξ )(1 ? ξ )dξ 2 2 2 2 1 1 πhξ π (1 + ξ ) = ∫0sin ξdξ + ∫0sin (1 ? ξ )dξ 2 4 1 π 1 1 ( f ,? 2 ) = 2h ∫0sin ( + ξ )ξ dξ 2 2 2

? a (?1 ,?1 ) a (?1 ,? 2 ) ?? y1 ? ? ( f ,?1 ) ? ? ? a (? ,? ) a (? ,? ) ?? y ? = ? ( f ,? ) ? ?? ? ? ? ? ? 1 2 2 2 ?? 2 ? 2 ?

∑ a(? ,?
i =1 i

4

j

)ui = ( f , ? j ) j = 1,2,3,4

1 1 1 π 2h 2 π 2h a (? j ,? j ) = ∫0( + ξ )dξ + ∫0[ + (1 ? ξ ) 2 ]dξ h 4 h 4 1

= ∫0[8 +

1

π2
8

ξ ]dξ = 8 +
2

π2

24

1 π 2h a (? j + ,? j +1 ) = ∫0[ ? + ξ (1 ? ξ )]dξ h 4
1

= ?4 +

π2

16 ∫0

1

ξ (1 ? ξ )dξ = ?4 + π2
96

π2
96

a (? j ?1 ,? j ) = ?4 +

1

π2
96
1

,8 +

π2
24

,?4 +
1 4

π2
96

}

( f ,? j ) = h ∫02 sin[ ( x j + hξ )]ξdξ 2
1

π

+ h ∫02 sin[ ( x j +1 + hξ )](1 ? ξ )dξ 2 2 1 π j ξ 1 1 π j +1 ξ = ∫0sin[ ( + )]ξdξ + ∫0sin[ ( + )](1 ? ξ )dξ 4 2 4 4 2 2 4 4 1 1 π( j +ξ) π ( j +1+ ξ) = ∫0sin[ ] ? sin[ ]ξdξ 2 8 8 1 1 π ( j +1+ ξ) 1 8 π ( j +1+ ξ) 0 + ∫0sin[ ]dξ = × [cos( )] |1 2 8 2 π 8 +
1

π

2.就非齐次第三边值条件 2.就非齐次第三边值条件
u ' (a ) + α1u ( a) = β1 , u ' (b) + α 2u (b) = β 2

(( pu ' ) ' , v) = pu 'v |b ?( pu ' , v ' ) = p (b)v(b)[ β 2 ? α 2u (b)] a ? p (a )v(a )[ β1 ? α1u (a )] ? ( pu ' , v ' )

( pu ' , v ' ) + (qu , v) + α 2 p (b)u (b)v(b) ? α1 p (a )u (a )v(a ) = ( f , v) + p (b) β 2v(b) ? p (a ) β1v(a )

u0 = u (a ), u N = u (b)

A(u , v) = ( pu ' , v ' ) + (qu , v) + α 2 p (b)u (b)v(b) ? α1 p (a )u (a )v(a ) F (v) = ( f , v) + p (b) β 2v(b) ? p (a ) β1v(a )

∑ A(? ,?
i =0 i

N

j

)ui = F (? j ) ( j = 0,1,..., N )

2014-2015学年偏微分方程数值解法试题
2014-2015学年偏微分方程数值解法试题 - 编号 浙江理工大学考试命题稿( A (2014 /2015 学年 第 1 开课学院: 机械与自动控制学院 卷) 学期) 开课年级 课程...