Computational Fluid Dynamics II

Computational Fluid Dynamics II

Exercise 2

1. Given is the PDE:

u

tt

−

a

2

o

u

xx

= 0

Formulate the CFL-condition for two possible explicit schemes.

2. The Euler equations for 1-dimensional, unsteady flows is discretised in the following form:

U

_i

n

+1

₋

U

_i

n

∆

t

+

F

_i

n

₊₁

₋

F

_i

n

∆

x

= 0

,

U

=







ρ

ρu

ρE







,

F

i

n

=

F

(

U

i

n

)

For which velocities does this scheme fulfill the CFL-condition?

3. Formulate for the linear model equation

u

t

+

au

x

= 0

the following solution schemes:

(a) explicit scheme, central differences

(b) Lax-Wendroff scheme

(c) Mac-Cormack scheme

Determine the stability conditions and truncation errors. Show that the Mac-Cormack scheme and

the Lax-Wendroff scheme are equivalent for this linear model equation.

4. Check the consistency, stability and convergence of an implicit method with backward difference

in time and central differences in space for the following PDE:

u

t

+

au

x

= 0

Specify the solution algorithm for that scheme, with use of the boundary conditions

u

(

x=

0)

=u

∞

Computational Fluid Dynamics II

Exercise 2 (solution)

1. CFL condition: the numerical domain of influence has to enclose the physical one. The

charac-teristic lines define the physical domain of influence, the difference stencil defines the numerical

domain of influence.

explicit scheme 1:

u

n

_i

+1

₋

2

u

n

_i

+

u

n

_i

−

1

∆

t

2

−

a

2

0

u

n

_i

₊₁

₋

2

u

n

_i

+

u

n

_i

₋

₁

∆

x

2

= 0

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₀

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∆

t

∆

x

x,i

t,n

numerical

influence area

physical

influence area

∆

t

∆

x

x,i

t,n

−

physical domain of influence:

dt

dx

|

C

=

±

1

a

0

numerical domain of influence:

dt

dx

_N

=

±

∆

t

∆

x

CFL-condition:

∆

t

∆

x

≤

1

|

a

0

|

explicit scheme 2:

u

n

_i

+1

₋

2

u

n

_i

+

u

n

_i

−

1

∆

t

2

−

a

2

0

u

_i

n

₊₁

−

1

₋

2

u

_i

n

−

1

+

u

n

_i

₋

−

₁

1

∆

x

2

= 0

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1/a

₀

1/a

₀

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numerical

influence area

physical

influence area

−

∆

t

∆

x

x,i

t,n

physical domain of influence:

dt

dx

|

C

=

±

1

a

0

numerical domain of influence:

dt

dx

|N

=

±

2

∆

t

∆

x

CFL-condition:

∆

t

∆

x

≤

1

2

_|

a

0

|

2. 1D-Euler equations:

ρ

t

+ (

ρu

)

x

= 0

(

ρu

)

t

+ (

ρu

2

+

p

)

x

= 0

(

ρE

)

t

+ (

u

(

ρE

+

p

))

x

= 0

with the equation of state for ideal gases:

p

= (

κ

₋

1)(

ρE

₋

ρ

2

u

2

₎

_and

_a

2

₌

_κRT

Differentiate and convert in non-conservative form:

ρ

t

+

uρ

x

+

ρu

x

= 0

uρ

t

+

ρu

t

+

u

(

ρu

)

x

+

ρuu

x

+

p

x

= 0

(

ρE

)

t

+ (

uρE

)

x

+ (

up

)

x

= 0

Insert the continuity equation into the momentum equation and the continuity, momentum, and

equation of state into the energy equation. After simplification the following form can be obtained:

ρ

t

+

uρ

x

+

ρu

x

= 0

u

t

+

uu

x

+

1

ρ

p

x

= 0

p

t

+

up

x

+

ρa

2

u

x

= 0

Characteristic lines:

Ω

t

+

u

Ω

x

ρ

Ω

x

0

0

Ω

t

+

u

Ω

x

1

_ρ

Ω

x

0

ρa

2

Ω

x

Ω

t

+

u

Ω

x

= 0

(Ω

t

+

u

Ω

x

)

3

−

Ω

x

2

a

2

(Ω

t

+

u

Ω

x

) = 0

⇒

−

Ω

t

Ω

x

1

=

dx

dt

1

=

u

_{∧ −}

Ω

t

Ω

x

2

,

3

=

dx

dt

2

,

3

=

u

_±

a

CFL-condition: all characteristic lines have to lie in the numerical domain of influence:

i+1

n+1

n

i

x

t

u−a

u+a

u

∆

t

∆

x

condition (side i):

u

+

a

_≤

0

condition (side i+1):

−

∆

x

∆

t

≤

u

−

a

i.e.:

a

₋

∆

x

3. Linear model equation

u

t

+

au

x

= 0

(a) Explicit, central differences

u

n

_i

+1

₋

u

n

_i

∆

t

+

a

u

n

_i

₊₁

₋

u

n

_i

₋

₁

2∆

x

= 0

with

u

n

_i

+1

=

u

+ ∆

tu

t

+

∆

t

2

2

u

tt

+

∆

t

3

6

u

ttt

+

. . .

u

n

_i

=

u

u

n

_i

₊₁

=

u

+ ∆

xu

x

+

∆

x

2

2

u

xx

+

∆

x

3

6

u

xxx

+

. . .

u

n

_i

₋

₁

=

u

₋

∆

xu

x

+

∆

x

2

2

u

xx

−

∆

x

3

6

u

xxx

+

. . .

follows

∆

tu

t

+

∆

t

2

2

u

tt

+

∆

t

3

6

u

ttt

∆

t

+

a

2∆

xu

x

+ 2

∆

x

3

6

u

xxx

2∆

x

= 0

u

t

+

∆

t

2

u

tt

+

∆

t

2

6

u

ttt

+

a

(

u

x

+

∆

x

2

6

u

xxx

) = 0

Truncation error

u

t

+

au

x

=

−

∆

t

2

u

tt

−

∆

t

2

6

u

ttt

−

a

∆

x

2

6

u

xxx

=

O

(∆

t,

∆

x

2

₎

von Neumann analysis:

Approach for the error function

ǫ:

ǫ

n

_i

=

φ

=

π

X

φ

=0

V

n

(Φ)

e

i

Φ

I

,

Φ =

2

π

∆

x

λ

,

t

=

n

∆

t ,

I

=

√

−

1

apply the approach in the model equation

V

n

+1

e

i

Φ

I

₋

V

n

e

i

Φ

I

∆

t

+

a

V

n

e

(

i

+1)Φ

I

₋

V

n

e

(

i

−

1)Φ

I

2∆

x

= 0

⇒

V

n+1

V

n

−

1

∆

t

+

a

e

Φ

I

₋

e

−

Φ

I

2∆

x

= 0

with

e

Φ

I

= cos(Φ) +

I

sin(Φ)

e

−

Φ

I

= cos(Φ)

₋

I

sin(Φ)

follows

V

n+1

V

n

−

1

∆

t

+

a

2

I

sin(Φ)

2∆

x

= 0

⇔

V

n

+1

V

n

−

1 +

a

∆

t

∆

x

I

sin(Φ) = 0

⇒

G

=

V

n

+1

V

n

= 1

−

a

∆

t

∆

x

I

sin(Φ)

⇒

|

G

|

2

_{= 1 +}

_a

∆

t

∆

x

sin(Φ)

2

for a stable difference scheme it is required that

_|

G

_|

2

_≤

1

_⇒

scheme is unstable !

(b) Lax-Wendroff scheme

u

n

_i

+1

₋

u

n

_i

∆

t

+

a

u

n

_i

₊₁

₋

u

n

_i

−

1

2∆

x

−

a

2

_∆

_t

u

n

i

+1

−

2

u

n

i

+

u

n

i

−

1

2∆

x

2

= 0

with

u

n

_i

+1

,

u

n

_i

,

u

n

_i

₊₁

and

u

n

_i

₋

₁

(see (a)) follows

u

+ ∆

tu

t

+

∆

t

2

2

u

tt

+

∆

t

3

6

u

ttt

−

u

∆

t

+

a

(

u

+ ∆

xu

x

+

∆

x

2

2

u

xx

+

∆

x

3

6

u

xxx

)

−

(

u

−

∆

xu

x

+

∆

x

2

2

u

xx

−

∆

x

3

6

u

xxx

)

2∆

x

−

a

2

∆

t

(

u

+ ∆

xu

x

+

∆

x

2

2

u

xx

+

∆

x

3

6

u

xxx

)

−

(2

u

) + (

u

−

∆

xu

x

+

∆

x

2

2

u

xx

−

∆

x

3

6

u

xxx

)

2∆

x

2

= 0

⇒

u

+ ∆

tu

t

+

∆

t

2

2

u

tt

+

∆

t

3

6

u

ttt

−

u

∆

t

+

a

2∆

xu

x

+ 2

∆

x

3

6

u

xxx

2∆

x

−

a

2

_∆

_t

2

∆

x

2

2

u

xx

2∆

x

2

= 0

u

t

+

∆

t

2

u

tt

+

∆

t

2

6

u

ttt

+

au

x

+

a

∆

x

2

6

u

xxx

−

a

2

∆

t

2

u

xx

= 0

Truncation error

u

t

+

au

x

=

−

∆

t

2

u

tt

−

∆

t

2

6

u

ttt

−

a

∆

x

2

6

u

xxx

+

a

2

∆

t

2

u

xx

with

u

tt

=

a

2

_u

xx

=

₋

a

2

∆

t

2

u

xx

−

∆

t

2

6

u

ttt

−

a

∆

x

2

6

u

xxx

+

a

2

∆

t

2

u

xx

=

₋

∆

t

2

6

u

ttt

−

a

∆

x

2

6

u

xxx

=

O

(∆

t

2

_,

_∆

_x

2

₎

von Neumann analysis:

V

n

+1

e

i

Φ

I

₋

V

n

e

i

Φ

I

∆

t

+

a

V

n

e

(

i

+1)Φ

I

₋

V

n

e

(

i

−

1)Φ

I

2∆

x

−

a

2

_∆

_t

V

n

e

(

i

+1)Φ

I

−

2

V

n

e

i

Φ

I

+

V

n

e

(

i

−

1)Φ

I

2∆

x

2

= 0

V

n+1

V

n

−

1

∆

t

+

a

e

Φ

I

₋

e

−

Φ

I

2∆

x

−

a

2

_∆

_t

e

Φ

I

−

2 +

e

−

Φ

I

2∆

x

2

=

V

n+1

V

n

−

1

∆

t

+

a

2

I

sin (Φ)

2∆

x

−

a

2

_∆

_t

2 cos (Φ)

−

2

2∆

x

2

= 0

⇒

G

=

V

n

+1

V

n

= 1

−

a

∆

t

I

sin(Φ)

∆

x

+

a

2

_∆

_t

2

cos

(Φ)

−

1

∆

x

2

with

k

=

a

∆

t

∆

x

|

G

_|

2

= (1 +

k

2

(cos(Φ)

₋

1))

2

+ (

₋

k

sin(Φ))

2

= 1 + 2

k

2

(cos(Φ)

₋

1) +

k

4

(cos(Φ)

₋

1)

2

+

k

2

sin

2

(Φ)

=

k

4

(1

₋

cos(Φ))

2

+

k

2

(sin

2

(Φ)

₋

2 + 2 cos(Φ)) + 1

=

k

4

(1

₋

cos(Φ))

2

+

k

2

(

₋

1 + 2 cos(Φ)

₋

cos

2

(Φ)) + 1

=

k

4

(1

₋

cos(Φ))

2

₋

k

2

(1

₋

2 cos(Φ) + cos

2

(Φ)) + 1

=

k

4

(1

₋

cos(Φ))

2

₋

k

2

(1

₋

cos(Φ))

2

+ 1

= (

k

4

₋

k

2

)(1

₋

cos(Φ))

2

+ 1

_≤

1

(for stability)

⇒

|

G

_|

2

₋

1 = (

k

4

₋

k

2

) (1

₋

cos(Φ))

2

|

{z

}

≥

0

≤

0

stable if

k

2

(

k

2

₋

1)

_≤

0

_⇒

k

2

_≤

1

_⇒

_|

a

_|

_∆

∆

_x

t

_≤

1

(c) Mac-Cormack scheme

Step 1

˜

u

i

=

u

n

i

−

a

∆

t

∆

x

(

u

n

i

−

u

n

i

−

1

)

Step 2

u

n

_i

+1

=

1

2

(

u

n

i

+ ˜

u

i

)

−

1

2

a

∆

t

∆

x

(˜

u

i

+1

−

u

˜

i

)

with

˜

u

i

=

u

−

a

∆

t

∆

x

(∆

xu

x

−

∆

x

2

2

u

xx

+

∆

x

3

6

u

xxx

)

˜

u

i

+1

=

u

+ ∆

xu

x

+

∆

x

2

2

u

xx

+

∆

x

3

6

u

xxx

−

a

∆

t

∆

x

(∆

xu

x

+

∆

x

2

2

u

xx

+

∆

x

3

6

u

xxx

)

follows

u

n

_i

+1

=

1

2

"

u

+

u

₋

a

∆

t

∆

x

(∆

xu

x

−

∆

x

2

2

u

xx

+

∆

x

3

6

u

xxx

)

#

−

1

₂

a

∆

t

∆

x

"

u

+ ∆

xu

x

+

∆

x

2

2

u

xx

+

∆

x

3

6

u

xxx

−

a

∆

t

∆

x

(∆

xu

x

+

∆

x

2

2

u

xx

+

∆

x

3

6

u

xxx

)

#

−

1

₂

a

∆

t

∆

x

"

−

(

u

₋

a

∆

t

∆

x

(∆

xu

x

−

∆

x

2

2

u

xx

+

∆

x

3

6

u

xxx

))

#

=

1

2

"

2

u

₋

a

∆

t

∆

x

(∆

xu

x

−

∆

x

2

2

u

xx

+

∆

x

3

6

u

xxx

)

#

+

1

2

a

∆

t

∆

x

"

−

u

₋

∆

xu

x

−

∆

x

2

2

u

xx

−

∆

x

3

6

u

xxx

+

a

∆

t

∆

x

(∆

xu

x

+

∆

x

2

2

u

xx

+

∆

x

3

6

u

xxx

)

#

+

1

2

a

∆

t

∆

x

"

u

+

a

∆

t

∆

x

(

−

∆

xu

x

+

∆

x

2

2

u

xx

−

∆

x

3

6

u

xxx

)

#

u

n

_i

+1

=

1

2

"

2

u

₋

a

∆

t

∆

x

(∆

xu

x

−

∆

x

2

2

u

xx

+

∆

x

3

6

u

xxx

)

#

+

1

2

a

∆

t

∆

x

"

−

∆

xu

x

−

∆

x

2

2

u

xx

−

∆

x

3

6

u

xxx

+

a

∆

t

∆

x

∆

x

2

_u

xx

#

=

u

+

1

2

a

∆

t

∆

x

"

−

∆

xu

x

+

∆

x

2

2

u

xx

−

∆

x

3

6

u

xxx

#

+

1

2

a

∆

t

∆

x

"

−

∆

xu

x

−

∆

x

2

2

u

xx

−

∆

x

3

6

u

xxx

+

a

∆

t

∆

x

∆

x

2

_u

xx

#

=

u

+

1

2

a

∆

t

∆

x

"

−

2∆

xu

x

−

2∆

x

3

6

u

xxx

+

a

∆

t

∆

x

∆

x

2

_u

xx

#

with

u

n

_i

+1

=

u

+ ∆

tu

t

+

∆

t

2

2

u

tt

+

∆

t

3

6

u

ttt

u

+ ∆

tu

t

+

∆

t

2

2

u

tt

+

∆

t

3

6

u

ttt

=

u

+

1

2

a

∆

t

∆

x

"

−

2∆

xu

x

−

2∆

x

3

6

u

xxx

+

a

∆

t

∆

x

∆

x

2

_u

xx

#

and

u

tt

=

a

2

u

xx

follows

∆

tu

t

+

∆

t

2

2

a

2

_u

xx

+

∆

t

3

6

u

ttt

=

1

2

a

∆

t

∆

x

"

−

2∆

xu

x

−

2∆

x

3

6

u

xxx

+

a

∆

t

∆

x

∆

x

2

_u

xx

#

u

t

+

a

2

∆

t

2

u

xx

+

∆

t

2

6

u

ttt

=

−

au

x

−

a

∆

x

2

6

u

xxx

+

a

2

∆

t

2

u

xx

Truncation error

u

t

+

au

x

=

−

∆

t

2

6

u

ttt

−

a

∆

x

2

6

u

xxx

=

O

(∆

t

2

_,

_∆

_x

2

₎

Alternative: insert step 1 into step 2:

u

n

_i

+1

=

1

2

(

u

n

i

+

u

n

i

−

a

∆

t

∆

x

(

u

n

i

−

u

n

i

−

1

))

−

1

2

a

∆

t

∆

x

(

u

n

i

+1

−

a

∆

t

∆

x

(

u

n

i

+1

−

u

n

i

)

−

u

n

i

+

a

∆

t

∆

x

(

u

n

i

−

u

n

i

−

1

))

2

u

n

_i

+1

= 2

u

n

_i

₋

a

∆

t

∆

x

(

u

n

i

−

u

n

i

−

1

)

−

a

∆

t

∆

x

(

u

n

i

+1

−

a

∆

t

∆

x

(

u

n

i

+1

−

u

n

i

)

−

u

n

i

+

a

∆

t

∆

x

(

u

n

i

−

u

n

i

−

1

))

2

u

n

_i

+1

= 2

u

n

_i

₋

a

∆

t

∆

x

(

u

n

i

−

u

n

i

−

1

+

u

n

i

+1

−

u

n

i

) +

a

2

∆

t

2

∆

x

2

(

u

n

i

+1

−

u

n

i

−

u

n

i

+

u

n

i

−

1

)

⇒

2

u

n

_i

+1

₋

2

u

n

_i

+

a

∆

t

∆

x

(

u

n

i

+1

−

u

n

i

−

1

)

−

a

2

∆

t

2

∆

x

2

(

u

n

i

−

1

−

2

u

n

i

+

u

n

i

+1

) = 0

u

n

_i

+1

₋

u

n

_i

∆

t

+

a

u

n

_i

₊₁

₋

u

n

_i

₋

₁

2∆

x

−

a

2

_∆

_t

u

n

i

−

1

−

2

u

n

i

+

u

n

i

+1

2∆

x

2

= 0

Truncation error and stability equivalent to that of the Lax-Wendroff scheme.

4.

u

t

+

au

x

= 0

Implicit scheme with backward deduction in time, central differences in space:

u

n

_i

+1

₋

u

n

_i

∆

t

+

a

u

n

_i

₊₁

+1

₋

u

n

_i

₋

+1

₁

2∆

x

= 0

with

u

n

_i

=

u

n

_i

+1

₋

∆

tu

n

_t

+1

+

∆

t

2

2

u

n

+1

tt

+

. . .

u

n

_i

_±

+1

₁

=

u

_i

n

+1

_±

∆

xu

n

_x

+1

+

∆

x

2

2

u

n

+1

xx

±

∆

x

3

6

u

n

+1

xxx

+

. . .

Consistency:

lim

∆

t,

∆

x

→

0

k

L

(

u

)

−

L

∆

(

u

)

k

=

∆

t,

lim

∆

x

→

0

k

τ

(

u

)

k

= 0

L

∆

(

u

) =

u

n

+1

₋

u

n

+1

+ ∆

tu

_t

n

+1

₋

∆

₂

t

2

u

n

_tt

+1

∆

t

+

a

2∆

xu

n

_x

+1

+ 2

∆

₆

x

3

u

n

_xxx

+1

2∆

x

+

O

(∆

x

3

_,

_∆

_t

3

₎

=

u

n

_t

+1

₋

∆

t

2

u

n

+1

tt

+

a

u

n

x

+1

+

∆

x

2

6

u

n

+1

xxx

!

+

_O

(∆

x

3

,

∆

t

2

) = 0

L

(

u

) =

u

n

_t

+1

+

au

n

_x

+1

= 0

lim

∆

t,

∆

x

→

0

k

τ

(

u

)

k

=

∆

t,

lim

∆

x

→

0

∆

t

2

u

n

+1

tt

−

∆

x

2

6

u

n

+1

xxx

+

O

(∆

x

3

,

∆

t

2

)

= 0

⇒

scheme is consistent

Stability:

u

n

_i

+1

₋

u

n

_i

∆

t

+

a

u

n

_i

₊₁

+1

₋

u

n

_i

₋

+1

₁

2∆

x

= 0

von Neumann analysis

V

n

+1

e

I

Φ

i

₋

V

n

e

I

Φ

i

∆

t

+

a

V

n

+1

e

I

Φ(

i

+1)

₋

V

n

+1

e

I

Φ(

i

−

1)

2∆

x

=

V

n

+1

₋

V

n

∆

t

+

a

V

n

+1

(

e

I

Φ

₋

e

−

I

Φ

)

2∆

x

=

1

−

V

n

V

n+1

∆

t

+

a

2

I

sin Φ

2∆

x

= 1

−

V

n

V

n

+1

+

a

∆

t

∆

x

sin Φ

I

= 0

1

G

=

V

n

V

n

+1

= 1 +

a

∆

t

∆

x

sin Φ

I

1

|

G

_|

2

= 1 +

a

∆

t

∆

x

2

sin

2

Φ

⇒

|

G

_|

2

=

1

1 +

a

∆

t

∆

x

2

sin

2

Φ

|

{z

}

≥

0

≤

1

_⇒

stable

Lax’s theorem: consistency + stability

_⇒

convergence

The tridiagonal equation system, with the unknown

u

i

a

i

u

i

−

1

+

b

i

u

i

+

c

i

u

i

+1

=

R

i

i

= 2

, ..., im

−

1

has the following form













a

2

b

2

c

2

a

3

b

3

c

3

. ..

. ..

. ..

a

im

−

1

b

im

−

1

c

im

−

1













·













u

1

u

2

..

.

u

im













=













R

2

R

3

..

.

R

im

−

1













(

im

₋

2)

equations with

im

unknown variables

_⇒

two boundary conditions have to be given.

This tridiagonal system of equations can be solved with the following recursive solution approach

(Thomas algorithm):

u

i

=

E

i

u

i

+1

+

F

i

with

E

i

=

−

c

i

a

i

E

i

−

1

+

b

i

and

F

i

=

R

i

−

a

i

F

i

−

1

a

i

E

i

1

+

b

i

The initial values for

E

1

and

F

1

follow from e.g. a Neumann boundary condition for

u

at

i

= 1

u

1

=

r

1

u

2

+

s

1

=

E

1

u

2

+

F

1

⇒

E

1

=

r

1

F

1

=

s

1

After all coefficients

E

i

,

F

i

are computed, the solution for

u

i

is obtained by back substitution: