Mehrpunktflussverfahren

− div(K grad u) = Q Ω
u K Q
K
q = −K grad u
q · n dσ = Q dτ
∂Ωi
Ωi Ω
Ωi
f
f = q · n dσ,
S
Ωi

k2
k1
S
S f
f
n w
n w S
S n
S
S
Q
f = −
S
n T
K grad u dσ = − w · grad u dσ,
S
w w = Kn
K S
w
K n · w = n T Kn > 0 w
n w
K

xi
xi+1 ∆xi ∆xi+1
k
¯x i+1/2
−ux = f/k
ui − ū i+1/2 = f ¯x i+1/2 − xi
ki
ū i+1/2 − ui+1 = f xi+1 − ¯x i+1/2
ki+1
= f ∆xi
,
2ki
= f ∆xi+1
.
2ki+1
ū i+1/2 u ¯x i+1/2
f
ū i+1/2
ui − ui+1 = f i+1/2
1 ∆xi
+
2 ki
∆xi+1
,
ki+1
f i+1/2 f
f i+1/2 = −
1
2
∆xi
ui+1 − ui
ki
+ ∆xi+1
ki+1
= − ui+1 − ui
xi+1
xi
1
k(x) dx
.
K

∆xi
∆xi+1
xi ¯x i+1/2 xi+1
xi
¯x i+1/2 xi+1
wi wi+1
wi
wi+1 x i+1/2 xi
xi+1
kα
K k
u
w w
grad u
w
w
αi ≥ ɛ w
ɛ
xi xi+1 wi wi+1
¯x i+1/2
¯x i+1/2 xi
wi ¯x i+1/2 xi+1 wi+1
wi wi+1
ui − ū i+1/2 = f
Γ i+1/2
ū i+1/2 − ui+1 = f
Γ i+1/2
¯x i+1/2 − xi
2 ,
Kin2
xi+1 − ¯x
i+1/2 2 ,
Ki+1n 2
Γ i+1/2 ū i+1/2
u ¯x i+1/2 f

n2
K
n1
ui − ui+1 = f i+1/2
Γ i+1/2
¯x i+1/2 − xi
2
Kin 2
k2
✻✲
k1
K
xi+1 − ¯x
i+1/2 2
+
,
Ki+1n 2
f i+1/2 f
¯x i+1/2
xi xi+1
wi wi+1
K K
K
K K
K
K
n1 n2 K Kni
j j = i
n T 2 Kn1 = 0.
Kni
ni
K

k2
n2
k1
n1
α2
n2
α1
e2
e1
n1
n1
n2
e1 e2
K
ei ki ni
αi ei
T
− sin α2 k1 0 cos α1
cos α2
0 k2
sin α1
= −k1 cos α1 sin α2 + k2 sin α1 cos α2 = 0,
tan α1
= tan α2
.
k1 k2
α1 = α2 = 0
αi = 0
α
α2 = 0
K
K
n1 ∼
grad u u K grad u
n2 nT 2 Kn1 = 0
K

n2
n1
Kn1
n1
n2
Rx
ν n
xi
xj
Rx x 90 ◦
n
ν
K
x
ν T Kn = 0.
K
0 1
R =
−1 0
x
R 90 ◦
R −1 = R T = −R
2 × 2 A
(RA) T AR = (det A)I.

x1
¯x3
x2
x0
¯x2
¯x1
K
A = K −1
x3
k2
R T K −1 R = (det K) −1 K.
t1 t2 K −1
n1 n2 K ni =
Rti ti = −Rni i = 1, 2
t T 1 K −1 t2 = n T 1 R T K −1 Rn2 = (det K) −1 n T 1 Kn2.
t T 1 K−1 t2 = 0 n T 1 Kn2 = 0
K
K
x0 xi i = 1, 2, 3
¯xi i = 1, 2, 3 ¯xi xi
K
¯xi −x0 ¯xi −xi+1 K
¯xi − x0 ¯xi − xi+1 K −1
(xi+2 − xi+1) T K −1 (¯xi − x0) = 0, i = 1, 2, 3.
¯xi x0
¯xi
¯xi = 1
2 (xi+2 + xi+1)
0 = (xi+2 − x0) − (xi+1 − x0) T K −1 (xi+2 − x0) + (xi+1 − x0)
= (xi+2 − x0) T K −1 (xi+2 − x0) − (xi+1 − x0) T K −1 (xi+1 − x0)
k1

K
K
i = 1, 2, 3
r
(xi − x0) T K −1 (xi − x0) = r 2 , i = 1, 2, 3.
K x0
K
k1/k2 k1
k2 K
K
K
K
K
K
U = (det K) 1/4 K −1/2 .

x ′ 2
x ′ 1
x ′ 0
x ′ 3
x = U −1 x ′
det U = 1
U = (det K) 1/6 K −1/2
x
x ′ x ′ = Ux
U −1
K
K
x2x3 ˆn =
R(x3 − x2) R ˆn
x3 −x2
x2x3
x2
x0
¯x1 − x02 d =
.
KR(x3 − x2)2 x2x3
d = (x2 − x1) TK −1 (x3 − x1)
.
4F
x1
x3

x2
¯x3
x0
¯x1
x1
θ
¯x2
x3
F
x ′ 2
¯x ′ 3
x ′ 1
θ ′
x ′ 0
¯x ′ 1
α ′
¯x ′ 2
x ′ 3
F = 1
2 (x2 − x1) × (x3 − x1) 2 = 1
2 x2 − x1 2 x3 − x1 2 sin θ,
θ x1
x = U −1 x ′ = (det K) −1/4 K 1/2 x ′ .
x ′
det U = 1
F = 1
2 x2 − x12 x3 − x12 sin θ = 1
′
2 x 2 − x ′
′
1 x 2 3 − x ′
1
2 sin θ ′ ,
(x2 − x1) T K −1 (x3 − x1)
= (det K) −1/2 (x ′ 2 − x ′ 1) T K 1/2 K −1 K 1/2 (x ′ 3 − x ′ 1)
= (det K) −1/2 (x ′ 2 − x ′ 1) · (x ′ 3 − x ′ 1)
= (det K) −1/2 x ′ 2 − x ′ 1
2
′
x 3 − x ′
1 cos θ 2 ′ .
cos θ′
d = (det K)−1/2 x ′ 2 − x′ 12 x′ 3 − x′ 12 2 x ′ 2 − x′ 12 x′ 3 − x′ 1 =
2 sin θ′
1
2 (det K)−1/2 cot θ ′ ,
θ ′ cot θ ′
α ′
¯x ′ 1 − x′ 02 = tan α′ x ′ 3 − ¯x′ 12
¯x ′ 1 − x ′ 0 = tan α ′ R(x ′ 3 − ¯x ′ 1).

α ′ + θ ′ = π/2
tan α ′ = tan 1
2 π − θ′ = cot θ ′ .
¯x ′ 1 − x ′ 0 = 1
2 cot θ′ R(x ′ 3 − x ′ 2),
¯x1 − x0 = (det K) −1/4 K 1/2 (¯x ′ 1 − x ′ 0)
A = K 1/2
= 1
2 (det K)−1/4 cot θ ′ K 1/2 R(x ′ 3 − x ′ 2)
= 1
2 cot θ′ K 1/2 RK −1/2 (x3 − x2).
RK −1/2 = (det K) −1/2 RR T K 1/2 R = (det K) −1/2 K 1/2 R,
K 1/2 RK −1/2 = (det K) −1/2 KR.
¯x1 − x0 = 1
2 (det K)−1/2 cot θ ′ KR(x3 − x2).
θ ′ ≤ 1
2π
d = 1
2 (det K)−1/2 cot θ ′ =
¯x1 − x02 .
KR(x3 − x2)2
K
π
θ ′ > 1
2
d = 1
2 (det K)−1/2 cot θ ′ = − ¯x1 − x02 .
KR(x3 − x2)2
L 2
cot θ ′
K
K
K

xi
K
ν
n
xj
K
K
K
K
K
K

K
K
α2 = 0
α1 = 0
10 ◦
1 : 100
K
K
K
K
K

¯x3
x3
x1
¯x2
¯x1
K
xk
¯xi
x1 ¯x1x2 ¯x4x4 ¯x2x3 ¯x3
x4
x2
¯x4

u
¯xi
xi
i = 1, 2, 3
u(x) =
3
uiφi(x)
i=1
ui u(x) i φi(x)
φi(xj) = δi,j
u
grad φi = − 1
2F νi,
F νi
i
νi

ν1
x3
ν2
x2
x1
ν3
νi
n2
xk
¯x2
ν (k)
2
ν (k)
1
¯x1
n1
k
3
νi = 0.
i=1
grad u = − 1
2F
3
i=1
uiνi = − 1
(u2 − u1)ν2 + (u3 − u1)ν3
2F
k xk
¯x1 ¯x2
ν (k)
2 ν(k) 1
ν (k)
i
xk ¯x1 ¯x2
grad u (k) = 1
2Fk
ν (k)
1 (ū1 − uk) + ν (k)
2 (ū2 − uk) ,
ūi = u(¯xi) i = 1, 2 uk = u(xk)
ni
ni
i k f (k)
i

f (k)
1
f (k)
2
= −
Γ1nT 1
Γ2nT
2
Γ1nT 1
= − 1
2Fk
Γ2n T 2
Kk grad u (k)
Kk
ν (k)
1
ν (k)
2
ū1 − uk
ū2 − uk
Γi i
Gk = 1
2Fk
Γ1n T 1
Γ2n T 2
Kk
ν (k)
1
ν (k)
= 2
1
Γ1n
2Fk
T 1
f (k)
1
f (k)
2
= −Gk
Γ2n T 2
ū1 − uk
ū2 − uk
Kkν (k)
1
Kkν (k)
1
,
Γ1n T 1
Γ2n T 2
(k)
Kkν 2
(k)
Kkν 2
Gk
f (1)
1
f (1)
3
f (3)
2
f (3)
3
= −G1
= −G3
ū1 − u1
ū3 − u1
ū2 − u3
u3 − ū3
,
,
f (2)
1
f (2)
4
f (4)
2
f (4)
4
= −G2
= −G4
u2 − ū1
ū4 − u2
u4 − ū2
u4 − ū4
,
.
uk = u(xk) ūi = u(¯xi)
ν (2)
1
ν(3)
2
ν(4)
1
ν(4)
2
ū1 − u2 ū3 − u3 ū2 − u4
ū4 − u4
f1 = f (1)
1
f2 = f (4)
2
f3 = f (3)
3
f4 = f (2)
4
= f (2)
1 ,
= f (3)
2 ,
= f (1)
3 ,
= f (4)
4 .
Gk = g (k)
i,j

x3
¯x3
x1
ν (3)
2
ν (1)
1
ν (2)
2
¯x2
ν (1)
2
ν (3)
n3
1
n2 n4
¯x1
n1
ν (4)
2
x2
x4
¯x4
ν (2)
1
ν (4)
1
f1 = −g (1)
1,1 (ū1 − u1) − g (1)
1,2 (ū3 − u1) = g (2)
1,1 (ū1 − u2) − g (2)
1,2 (ū4 − u2),
f2 = g (4)
1,1 (ū2 − u4) + g (4)
1,2 (ū4 − u4) = −g (3)
1,1 (ū2 − u3) + g (3)
1,2 (ū3 − u3),
f3 = −g (3)
2,1 (ū2 − u3) + g (3)
2,2 (ū3 − u3) = −g (1)
2,1 (ū1 − u1) − g (1)
2,2 (ū3 − u1),
f4 = g (2)
2,1 (ū1 − u2) − g (2)
2,2 (ū4 − u2) = g (4)
2,1 (ū2 − u4) + g (4)
2,2 (ū4 − u4).
ū1 ū2 ū3 ū4
¯x1 ¯x2 ¯x3 ¯x4
Gk k
K
ūi
K
ūi
f
f = [f1, f2, f3, f4] T
u = [u1, u2, u3, u4] T
v = [ū1, ū2, ū3, ū4] T
f = Cv + F u

x3
x4
x2
x1
x6
x5
f3
f2
f4
f1
Av = Bu
v
v v =
A −1 Bu
f = T u,
T = CA −1 B + F .
T
f (l)
1,2 = t(l) 1 u1 + t (l)
2 u2 + t (l)
3 u3 + t (l)
4 u4
f (r)
1,2
= t(r)
1 u1 + t (r)
2 u2 + t (r)
5 u5 + t (r)
6 u6
f1,2 = (t (l)
1
+ t(r) 1 )u1 + (t (l)
2 + t(r) 2 )u2 + t (l)
3 u3 + t (l)
4 u4 + t (r)
5 u5 + t (r)
6 u6.

f1 + f2 − f3 − f4 = V Q,
fi i V
Q
u
A A
A
A
xv = 1
4 (x1 + x2 + x3 + x4),
xi i = 1, . . . , 4
V x dτ
xa = ,
dτ
V
k
Gk
k a (k)
i i = 1, 2
V

x3
x1
xv xa
x4
x2
xv xa
a (k)
2
xk
ν (k)
2
ν (k)
1
a (k)
1
Γini = a (k)
i /2 ν(k)
i = a(k)
i /2
Fk = Vk/8 Vk k
Gk
Gk = 1
T
.
J k = a (k)
1
Vk
, a(k) 2
a (k)
1
a (k)
2
Kk
a (k)
1
a (k)
2
Vk = |det J k|
Gk =
1
|det J k| J T k KkJ k.
Gk
Gk
Kk Gk
k
a (k)
i
T
Kka (k)
j
= 0, i = j,
K
Gk
∆ηk a (k)
1 ∆ξk
a (k)
2
1
Gk = DkHkDk,
∆ξk∆ηk

Hk =
=
1 T
n1 n2 Kk n1 n2
det[n1, n2]
1
det[n1, n2]
n T 1 Kkn1 n T 1 Kkn2
n T 2 Kkn1 n T 2 Kkn2
,
Dk = diag(∆ηk, ∆ξk).
ni a (k)
i
Hk K
Gk
Gk
Gk = 1
V
a T 1 Ka1 a T 1 Ka2
a T 1 Ka2 a T 2 Ka2
=
a c
,
c b
V = det[a1, a2]
f1 = −a(ū1 − u1) − c(ū3 − u1) = a(ū1 − u2) − c(ū4 − u2),
f2 = a(ū2 − u4) + c(ū4 − u4) = −a(ū2 − u3) + c(ū3 − u3),
f3 = −c(ū2 − u3) + b(ū3 − u3) = −c(ū1 − u1) − b(ū3 − u1),
f4 = c(ū1 − u2) − b(ū4 − u2) = c(ū2 − u4) + b(ū4 − u4).
⎡
2a 0 c
⎤
−c
⎡
a + c a − c 0 0
⎤
⎢
A = ⎢ 0
⎣ c
2a
−c
−c
2b
c ⎥
0 ⎦ ,
⎢
B = ⎢ 0
⎣b
+ c
0
0
a − c
b − c
a + c ⎥
0 ⎦ .
−c c 0 2b
0 b − c 0 b + c
A 2 × 2
A
⎡
⎤
A −1 =
1
4(ab − c2 ⎢
) ⎣
2b − c 2 /a −c 2 /a −c c
−c 2 /a 2b − c 2 /a c −c
−c c 2a − c 2 /b −c 2 /b
c −c −c 2 /b 2a − c 2 /b
⎥
⎦

j
4
1
5
3
2
6 i
i
j
3
4
2
1
j
A −1 B = 1
⎡
2 + c/a 2 − c/a −c/a c/a
⎤
⎢ c/a
4 ⎣2
+ c/b
−c/a
−c/b
2 − c/a
2 − c/b
2 + c/a ⎥
c/b ⎦ .
c/b 2 − c/b −c/b 2 + c/b
⎡
−a 0 −c
⎤
0
⎡
a + c 0 0 0
⎤
⎢
C = ⎢ 0
⎣ 0
a
−c
0
b
c ⎥
0 ⎦ ,
⎢
F = ⎢
⎣
0
0
0
0
0
−(b − c)
−(a + c) ⎥
0 ⎦
c 0 0 −b
0 b − c 0 0
.
f = T u
T = CA −1 B + F
⎡
= 1
4
⎢
⎣
2a + c − c2 /b −2a + c + c2 /b −c + c2 /b −c − c2 /b
c + c2 /b c − c2 /b 2a − c − c2 /b −2a − c + c2 /b
2b + c − c2 /a −c + c2 /a −2b + c + c2 /a −c − c2 /a
c + c2 /a 2b − c − c2 /a c − c2 /a −2b − c + c2 /a
6
5
i
⎤
⎥
⎦ .
T = {ti,j}
i

3
3
1
2
1
4
2
4
i
5
6
7
4
1
3
2
8 9
fi = (t1,1 + t2,3)u1 + (t1,2 + t2,4)u2 + t1,4u3 + t1,3u4 + t2,1u5 + t2,2u6
=
a − c2
2b
(u1 − u2) + c
4
1 + c
(u5 − u3) −
b
c
4
1 − c
(u4 − u6).
b
j
j
fj = (t3,1 + t4,2)u1 + (t3,3 + t4,4)u2 + t4,3u3 + t4,1u4 + t3,2u5 + t3,4u6
=
b − c2
2a
(u1 − u2) − c
4
1 − c
(u5 − u3) +
a
c
4
1 + c
(u4 − u6).
a
9
f1 + f2 − f3 − f4 = mi(ui − u1),
m2 = −a + c2
d , m3 = − c
2
i=2
1 + c
, m4 = −b +
d
c2
d , m5 = c
1 −
2
c
d
mi+4 = mi, i = 2, 3, 4, 5.
d = 2ab/(a + b)

x3
β
x1
α
x4
x2
ni
xk k = 1, . . . , 4
x(α, β) = β
αx4 + (1 − α)x3 + (1 − β) αx2 + (1 − α)x1
(α, β) ∈ [0, 1]×[0, 1] xi i = 1, 2, 3, 4
ˆn = n dσ
S
β ≤ 1
2
S α ≤ 1
2

x
∂x
∂α = β(x4 − x3) + (1 − β)(x2 − x1),
∂x
∂β = α(x4 − x2) + (1 − α)(x3 − x1),
∂x/∂α × ∂x/∂β
∂x ∂x
n dσ = × dα dβ.
∂α ∂β
1/2 1/2
∂x ∂x
ˆn = n dσ =
× dα dβ
S
0 0 ∂α ∂β
1/2 1/2
=
β(x4 − x3) + (1 − β)(x2 − x1)
0
0
× α(x4 − x2) + (1 − α)(x3 − x1) dα dβ
= 1
9(x2 − x1) × (x3 − x1) + 3(x2 − x1) × (x4 − x2)
64
+ 3(x4 − x3) × (x3 − x1) + (x4 − x3) × (x4 − x2) .
ˆn
x1 ˆn
Mu = r.

¯x1
x3
x1
¯x4
M
M
M
Gk
M
M
¯xk k = 1, . . . , 4
¯x1 ¯x2 ¯x3 ¯x4
¯x1 ¯x2 ¯x3 ¯x4
¯x3 ¯x4
¯x1 ¯x2
xk k = 1, . . . , 4
Q = 1
4 (x2 − x1) × (x3 − x1) + (x2 − x1) × (x4 − x2)
+(x4 − x3) × (x3 − x1) + (x4 − x3) × (x4 − x2)
.
Q = (¯x2 − ¯x1) × (¯x4 − ¯x3) .
(¯x2−¯x1) = 1
2 (x2−x1)+(x4−x3)
(¯x4 − ¯x3) = 1
2 (x3 −x1)+(x4 −x2)
¯x3
x4
¯x2
x2

¯x3
x3
x1
¯x2
¯x1
x4
x2
¯x4
¯x3
x3
x1
¯x1
x4
x2
¯x4
x → ξ
ξ = [α, β] T
J −T
dx/dξ
¯x1
¯x1 ¯x3 ¯x4
¯x1
¯x3 ¯x4
4 · 3 = 12

¯x1
¯x2 u ¯x2
ū
= −G3
(3)
2 − u3 ,
f (3)
2
f (3)
3
u3 − ū3
f (4)
2
f (4)
4
= −G4
u4 − ū (4)
2
u4 − ū4
.
¯x1
f1 = f (1)
1
f3 = f (3)
3
f4 = f (2)
4
= f (2)
1 ,
= f (1)
3 ,
= f (4)
4 .
¯x1 ¯x3 ¯x4
¯x3 ¯x4
k
u (k) (x) = uk + (x − xk) · grad u (k)
= uk + 1
T
(x − xk) ν
2Fk
(k)
1
ν (k)
2
ū1 − uk
ū2 − uk
.
u (1) (¯x5) = u (3) (¯x5),
u (2) (¯x6) = u (4) (¯x6),

¡ ¡ ¡ ¡
¡ ¡ ¡ ¡
¡ ¡ ¡ ¡
¡ ¡ ¡ ¡
¡ ¡ ¡ ¡
¡ ¡ ¡ ¡
¯x5 = ¯x3
¯x6 = ¯x4
T
v = ū1, ū (3)
2
, ū(4)
2 , ū3, ū4
u = [u1, u2, u3, u4] T
Av = Bu v = A −1 Bu
f = Cv + F u f = [f1, f2, f3] T
T = CA −1 B +F
K
a b c
i j
fi = a(u1 − u2) − 1
4 c[(u3 − u6) + (u4 − u5)],
fj = b(u1 − u2) − 1
4 c[(u6 − u3) + (u5 − u4)],

6 · 4 = 24
M
x1 x2
xi i = 1, 2, 3, 4 x4
x4
x5
x1
x5
x3
x2
x4

x2 x4
x2 x5
¯xi
K
K
ν (k)
i
x3
x1
¯x3
¯x2
¯x4
¯x1
x4
x2

xi i = 1, 2, 3
¯xi i = 1, 2, 3
x1x2x3 ¯xi xi
¯xix0 i = 1, 2, 3
i ni
xj ¯xi νi νi
−νi ni νi
xj ¯xi
¯xix0 i = 1, 2, 3
¯xi
grad u (1) = 2
ν3(ū2 − u1) + ν2(ū3 − u1)
F
,
grad u (2) = 2
F
grad u (3) = 2
F
ν1(ū3 − u2) + ν3(ū1 − u2) ,
ν2(ū1 − u3) + ν1(ū2 − u3) .

−ν2
¯x2
x1
x3
n1
x0
n2
¯x1
n3
¯x3
−ν1
−ν3
x2
νi
uk = u(xk) ūi = u(¯xi) F
x1x2x3 F
xi¯xj ¯xk
i, j, k
x1x2x3 ¯xi
K
i k f (k)
i
x1x2x3
f (1)
2
f (1)
3
f (2)
3
f (2)
1
f (3)
1
f (3)
2
= −G1
= −G2
= −G3
ū2 − u1
ū3 − u1
ū3 − u2
ū1 − u2
ū1 − u3
ū2 − u3
, G1 = 2
F
, G2 = 2
F
, G3 = 2
F
Γ2n T 2
Γ3n T 3
Γ3n T 3
Γ1n T 1
Γ1n T 1
Γ2n T 2
K1 ν3 ν2 ,
K2 ν1 ν3 ,
K3 ν2 ν1 .
Γi i Γi = x0 − ¯xi 2
K
ν T i Kkni = 0 i Kk
Gk
K
Gk

f1 = f (2)
1
f2 = f (3)
2
f3 = f (1)
3
= f (3)
1 ,
= f (1)
2 ,
= f (2)
3 .
ūi
Gk = g (k)
i,j
f1 = −g (2)
2,1 (ū3 − u2) − g (2)
2,2 (ū1 − u2) = −g (3)
1,1 (ū1 − u3) − g (3)
1,2 (ū2 − u3),
f2 = −g (3)
2,1 (ū1 − u3) − g (3)
2,2 (ū2 − u3) = −g (1)
1,1 (ū2 − u1) − g (1)
1,2 (ū3 − u1),
f3 = −g (1)
2,1 (ū2 − u1) − g (1)
2,2 (ū3 − u1) = −g (2)
1,1 (ū3 − u2) − g (2)
1,2 (ū1 − u2).
K
g (k)
1,2
g(k)
2,1
K
f = [f1, f2, f3] T u = [u1, u2, u3] T v =
[ū1, ū2, ū3] T f = Cv +
F u Av = Bu
f = T u T = CA −1 B + F
ū1 = u2 + u3
2
, ū2 = u3 + u1
2
, ū3 = u1 + u2
.
2
fi = − 1
F Γin T i K(ν1u1 + ν2u2 + ν3u3), i = 1, 2, 3.
¯xi¯xjx0
νi = Γjnj − Γknk i, j, k .

i
i, j, k
−fj + fk = 1
F (Γjnj − Γknk) T K(ν1u1 + ν2u2 + ν3u3)
= 1
F νT i K(ν1u1 + ν2u2 + ν3u3)
= 1
F νT i K νj(uj − ui) + νk(uk − ui) .
x0
i
i j
1
F νT i Kνj.
x0

x3
x1
x5
x2
x0
x6
x4
x3
¯x3
x1
x5
¯x1
x2
x0
f1
x6
¯x4
x4
6 · 3 = 18
x5 x6
f1 x0 ¯x1
xi i = 1, 2, 3, 4
f1
x0 ¯x1 u

x0
x0
¯x1 ¯x3 ¯x4
¯x1 ¯x1
¯x3 ¯x4
¯x2
¯x (3)
2 ¯x(4)
2
ū (3)
2 = u(¯x(3) 2 ) ū(4) 2 = u(¯x(4) 2 )
x0
K

f = f 2P + f MP − f 2P
f 2P f MP
f
f MP
M f 2P
N M N
Mu = r.
Nu (k+1) + (M − N)u (k) = r
k
u (k+1) = (I − N −1 M)u (k) + N −1 r.

ρ(I − N −1 M)
Gk
Gk
Gk
K
K
Gk
K
< 20 ◦ κ
10 −2 < κ < 10 2 ρ 0,6
K

K2
K1
u
u H 1+α
α > 0 α
u
q = q · n q = −K grad u n
h
h
u ∈ H 1+α
uh − u L 2 ∼ h min{2,2α} ,
qh − q L 2 ∼ h min{1,α} .
qh − q L 2 ∼ h min{2,α}

− div(K grad u) = Q Ω
u = 0 ∂Ω
u(x) = G(x, ξ)Q(ξ) dτξ,
Ω
G(x, ξ)
G(x, ξ) ≥ 0.
Q ≥ 0 ⇒ u ≥ 0.
T x
x ≥ 0 T x ≥ 0
Mu = r
M −1 ≥ O,
O u = M −1 r
r ≥ 0 ⇒ u ≥ 0.
M −1
L 2

a/b − (c/b) 2
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
M
a b c
a < b
c = 0
K
c/b
a/b
K c = 0
c/b

a/b − (c/b) 2
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
M
a b c
a < b
c = 0
K
c/b
a/b
K c = 0
c/b

a/b − (c/b) 2
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
M
a b c
a < b
c = 0
K
c/b
a/b
K c = 0
c/b

a/b − (c/b) 2
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
M
a b c
a < b
c = 0
K
c/b
a/b
K c = 0
c/b