exampl2, Kolokwium 2
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[1], [2]
T = 20(x
2
−y
2
)
o
â„„
q = 100
2
A(3, 0.5)
[2]
[1]
[2]
0
@
1
A
0.5−0.5 0
−0.5 1−0.5
0−0.5
K
1
= K
2
=
0.5
0
1
0
@
141.4
1
0
100
100
@
A
A
F
1
=
F
2
=
0
141.1
0
@
1
A
0
@
1
A
0.5 0 0−0.5
0 1.5−0.5 −1
0−0.5
100
0
141.4
241.4
K =
F =
0.5
0
−0.5 −1
0
1.5
KU = F
0
1
0
@
u
1
=−80
1
0
@
1000
1
0.5 0 0−0.5
0 1.5−0.5 −1
0−0.5
@
A
A
A
u
2
= 80
u
3
= 320
u
4
0
141.4
241.4
=
0.5
0
−0.5 −1
0
1.5
T
[
o
]
U =
−80
80
320
187.6
(1)
(x, y) = y/2,
(2)
(x, y) = (x+y−4)/(2+0−4),
(3)
(x, y) = (x−2)/(4−2)
(1)
(x, y)+
(2)
(x, y)+
(3)
(x, y) = 1 8x, y
T
h
(x, y) = u
(1)
(1)
(x, y)+u
(2)
(2)
(x, y)+u
(3)
(3)
(x, y) = 120x+53.81y−160
300
2.5
250
2
200
1.5
150
1
100
0.5
50
0
0
−0.5
−50
0
0.5
1
1.5
2
2.5
3
3.5
4
T
h
(A) = 226.9
o
,
T
h
(C) = 195.8
o
(1)
(C) =
(2)
(C) =
(3)
(C) = 1/3)T
h
(C) =
u
(1)
+ u
(2)
+ u
(3)
3
y
2
[1]
x
[2]
[3]
0
5x−10
P
(e)
N
i
·T Ds,
T =
=
,
x2[0, 2], y = 1
i
@e∩@
N
y
[1]
x
2
=
2
,
3
= y−
2
1
= 1−y,
6
x
2
0
1−y
0
0
1−y
N
1
=
N
2
=
N
3
=
. . . ,
U
h
=
N
i
(x, y)u
i
i=1
P
(1)
0
N
1
·T Dx = 0
2
=
1
P
(1)
0
N
2
·T Dx =
2
2
=
0
(1−1)(5x−10) Dx = 0
2
P
(1)
2
0
N
3
·T Dx = 0
=
3
P
(1)
0
N
4
·T Dx =
2
2
0
x
2
(5x−10) Dx =−3
3
=
4
P
(1)
0
N
5
·T Dx = 0
2
=
5
P
(1)
0
N
6
·T Dx =
2
0
(1−
2
)(5x−10) Dx =−6
3
2
=
6
P
(1)
2
0
P
(1)
2
0
P
(1)
+ P
(1)
2
0
=
3
t
y
Dx,
=
3
t
y
Dx,
t
y
Dx
=
4
6
4
6
8
<
−AEu
′′
= q 8x2(0, 1)
u(0) = 0
:
u(1) = 0.01
AE = 1000
q = AE(6x−4)/100
x
i
u
i
R = q + AEu
′′
h
= 60x−40
â„„
1/2
R
2
Dx = 9.344
2
1
= h
1
0
1
1/2
R
2
Dx = 3.536
2
2
= h
2
l||q||
0
= 20
2
u(x) = x
2
(2−x)/100
e(x) = u(x)−u
h
(x)
â„„
8
<
:
[x
2
(2−x)−
4
x]/100
dla x2[0, 0.5]
e(x) =
[x
2
(2−x)−
4
x−
4
]/100
dla x2[0.5, 1]
AE(e
′
)
2
Dx
||e||
E,(0,0.5)
= 0.08165
2
,||e||
E,(0.5,1)
= 0.02041
2
Â
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