Analisis kestabilan dan konsistensi (1)

ut + cux

=

0

0 < x < b,

u (x, 0)
u (0, t)

=
=

sin (x)
0

u (b, t)

=

0

t>0

Skema FTBS Eksplisit:

Misalkan

c∆t
∆x

ujn+1 − unj
unj − unj−1
+c
=0
∆t
∆x


c∆t n
un+1

− unj +
uj − unj−1 = 0
j
∆x
= s, maka


un+1
− unj + s unj − unj−1 = 0
j

un+1
= (1 − s) unj + sunj−1
j
√
Analisis Kestabilan, misalkan unj = ρn eiaj , i = −1, a ∈ R.
pemisalan pada persamaan (1)

(1)
Substitusi

ρn+1 eiaj = (1 − s) ρn eiaj + sρn eia(j−1)

(2)

ρ = (1 − s) + se−ia
ρ

=
=
=

(1 − s) + s [cos a − i sin a]

(1 − s) + s cos a − is sin a
[1 + (cos a − 1) s] + i [−s sin a]

Skema numerik (1) akan stabil jika terdapat |ρ| ≤ 1
|ρ|
|[1 + (cos a − 1) s] + i [−s sin a]|
q

2
2
[1 + (cosa − 1) s] + [−s sin a]

≤
≤

1
1
1

[1 + (cosa − 1) s] + [−s sin a]

2

≤
≤

1

2

2

2

2 2

2

2

1 + 2 (cos a − 1) s + (cos a − 1) s + s sin a


1 + 2 (cos a − 1) s + cos2 a − 2 cos a + 1 s2 + s2 sin2 a

≤

≤

1

1 + 2s (cos a − 1) − 2s2 (cos a − 1)


1 + 2s − 2s2 (cos a − 1)

≤
≤

1
1

2

1 + 2 (cos a − 1) s − 2s cos a + s + s

2s (1 − s) (cos a − 1)
(1 − s) (cos a − 1)

1

≤

≤

≤

1
1

0
0

Karena
−2 ≤ (cos a − 1) ≤ 0
−1 ≤ cos a ≤ 1
sehingga
−2 (1 − s) ≤ (1 − s) (cos a − 1) ≤ 0 (1 − s)
−2 + 2s ≤ (1 − s) (cos a − 1) ≤ 0
−2 + 2s

≤

0

2s
s

≤
≤

2
1

Jadi skema (1) stabil dengan syarat s =

c∆t
∆x

≤ 1.

ut + cux

=

0

0 < x < b,

u (x, 0)
u (0, t)

=
=

sin (x)
0

u (b, t)

=

0

t>0

FTCS Implisist,
n+1
un+1
− unj
un+1
j
j+1 − uj−1
+c
= 0
∆t
2∆x


c∆t n+1
un+1

+
= unj
uj+1 − un+1
j
j−1
2∆x


n+1
n
un+1
+ s un+1
j
j+1 − uj−1 = uj

(3)



ρn+1 eiaj + s ρn+1 eia(j+1) − ρn+1 eia(j−1)


ρ + s ρeia − ρe−ia



ρ 1 + s eia − e−ia

=

ρn eiaj

=

1

=

1

=

1

ρ [1 + i2s sin a]

=

1

ρ

=

1
1 + i2s sin a

ρ [1 + s (cos a + i sin a − [cos a − i sin a])]

2

Syarat kestabilan,
|ρ|




1


 1 + i2s sin a 
|1|
|1 + i2s sin a|
1

q
2
1 + (2s sin a)

1
1 + 4s2 sin2 a

≤

1

≤

1

≤

1

≤

1

≤

1

Skema numerik FTCS implisit untuk persamaan transport stabil tanpa syarat.
Konsistensi, truncation term, error estimate, accuracy metode FTBS
eksplisit untuk persamaan transport
un+1
− unj
unj − unj−1
j
+c
=0
∆t
∆x
un+1
j

=

unj−1

=

(4)

∆t2
n
utt |j + · · ·
2
∆x2
n
n
uxx |j + · · ·
unj − ∆x ux |j +
2
n

unj + ∆t ut |j +

(5)
(6)

Substitusikan (5) dan (6) pada persamaan (4) shg diperoleh
n

∆t ut |j +


n
ut | j

∆t2
2

∆t

n

n

utt |j + · · ·

∆t
n
+
utt |j + · · ·
2



+c

∆x ux |j −

+c



n
ux | j

∆x2
2

∆x

n

uxx |j + · · ·

=0



=0

∆x
n
uxx |j + · · ·
−
2



∆x
+ c ux −
uxx + · · ·
2


c∆x
∆t
utt −
uxx + · · ·
[ut + cux ] +
2
2
 ∆t

Error pemotongan pertama = 2 utt − c∆x
2 uxx
Orde error: O (∆t, ∆x)


ut +

∆t
utt + · · ·
2



Skema (4) dikatakan konsisten jika


∆t
c∆x
lim
utt −
uxx = 0
2
2
(∆t,∆x)→0
3

=

0

=

0

epp =

∆t
c∆x
utt −
uxx
2
2

ut + cux

=

0

ut

=

−cux

utt

=

(ut )t

=
=

(−cux )t
−cuxt

=
=
=
=

=

epp

=
=
Kita ingin


−cutx

−c (ut )x
−c (−cux )x

c2 uxx


c∆x
∆t 2
c uxx −
uxx
2
2
c2 ∆t
c∆x
uxx −
uxx
2
 22
c ∆t c∆x
uxx
−
2
2

c2 ∆t c∆x
−
2
2



uxx = 0

maka


c2 ∆t c∆x
−
2
2



=

0

c2 ∆t

=

c∆x

c∆t
c∆t
∆x

=

∆x

=

1

4

(7)