Week 15 - Explicit vs Implicit

8/13/2019 Week 15 - Explicit vs Implicit

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Computational Fluid Dynamics

Implicit method19-Dec-13

21

2

2

x

u

t

u

Therefore parabolic equation can be written as:

2

1

1

11

1

1 2

x

uuu

t

uu n

i

n

i

n

i

n

i

n

i

2

1

1

11

1

1 )2(x

truuuuru

n

i

n

i

n

i

n

i

n

i

Difference equation is:

Time (n+1) Time (n)

SCHEME is ORDER (t, x2)

t

uu n

i

n

i

1

2

1

1

11

1 2

x

uuu n

i

n

i

n

i

Requires Matrix Solvers.

Known as the Fully Implicit Method.


Crank Nicholson

Approximating the derivatives at mid-point intime (n+0.5, i) with half the time step, wehave:

LHS:

RHS:

19-Dec-13

22

2

2

x

u

t

u

t

uu n

i

n

i

5.02

5.05.05.05.0

2

1

1

11

1

2

11 22

2

1

x

uuu

x

uuu n

i

n

i

n

i

n

i

n

i

n

i

2

1

11211

211

1121 22

x

uuuuuu n

i

n

i

n

i

n

i

n

i

n

i

2

5.0

1

5.05.0

1 2

x

uuu n

i

n

i

n

i

t

uu n

i

n

i

5.02

1


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Crank Nicholson

Crank Nicholson in 1947 proposed this method to allowgreater time steps in their calculations.

The resulting difference equation is:

In this case we have three unknowns on the RHS of theequation.

Therefore the new value +1 is not given directly in

terms of known values as is the case of the Explicitmethod.

19-Dec-13

23

Time (n+1) Time (n)

n

i

n

i

n

i

n

i

n

i

n

i ruurruruurru 111

1

11

1 )22()22(

2x

tr

SCHEME is ORDER (t2, x2)


Explicit vs Implicit19-Dec-13

24

i-1 i i+1i-1 i i+1

n-1

n

n+

1

i-1 i i+1

Explicit Crank-Nicholson Fully Implicit

O(t, x2) O(t, x2)O(t2, x2)

Computational moleculesshow differences between the schemes

If there is more than one node at timestep n+1 the scheme is implicit

n-1

n

n+1

n-

1

n

n+1


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Comparison

Explicit method

Unknown values atcurrent time step dependonly on known values atprevious time step

Advantage:Easy to implement insoftware and solve

Disadvantage:

Restrictions on time stepfor stability.

Implicit method

Unknown values atcurrent time step dependon known values atprevious time step and oneach other Disadvantage:

Requires solution ofsystem of equations

Advantage:

No restrictions on time step

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25


Example

Consider heat loss from a heated rod subject

to the above equations:

Use the explicit scheme with mesh spacing0.1 and timestep 0.001 to obtain

approximate values for u(x,0.03).

Now change the timestep to 0.01. What

happens?

19-Dec-13

28

ICEICE

0),1(),0( tutu102

2

x

x

u

t

u

xxu sin)0,(


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Solution

The problem is set up as follows

19-Dec-13

29

=sin(*c24)

)2( 111 n

i

n

i

n

i

n

i

n

i uuuruu

Explicit Equation:

= C25+rr*(D25-2*C25+B25)

Name rr

=B22/B21^2


19-Dec-13

30

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10.00 0 0.31 0.59 0.81 0.95 1.00 0.95 0.81 0.59 0.31 0

0.01 0 0.28 0.53 0.73 0.86 0.90 0.86 0.73 0.53 0.28 0

0.02 0 0.25 0.48 0.66 0.77 0.81 0.77 0.66 0.48 0.25 0

0.03 0 0.23 0.43 0.59 0.70 0.73 0.70 0.59 0.43 0.23 0

0.04 0 0.20 0.39 0.54 0.63 0.66 0.63 0.54 0.39 0.20 0

0.05 0 0.18 0.35 0.48 0.57 0.60 0.57 0.48 0.35 0.18 0

0.06 0 0.17 0.32 0.44 0.51 0.54 0.51 0.44 0.32 0.17 0

0.07 0 0.15 0.29 0.39 0.46 0.49 0.46 0.39 0.29 0.15 0

0.08 0 0.14 0.26 0.35 0.42 0.44 0.42 0.35 0.26 0.14 0

0.09 0 0.12 0.23 0.32 0.38 0.40 0.38 0.32 0.23 0.12 0

0.10 0 0.11 0.21 0.29 0.34 0.36 0.34 0.29 0.21 0.11 0

0.11 0 0.10 0.19 0.26 0.31 0.32 0.31 0.26 0.19 0.10 0

0.12 0 0.09 0.17 0.24 0.28 0.29 0.28 0.24 0.17 0.09 0

0.13 0 0.08 0.15 0.21 0.25 0.26 0.25 0.21 0.15 0.08 0

0.14 0 0.07 0.14 0.19 0.22 0.24 0.22 0.19 0.14 0.07 0

0.15 0 0.07 0.13 0.17 0.20 0.21 0.20 0.17 0.13 0.07 0

0.16 0 0.06 0.11 0.16 0.18 0.19 0.18 0.16 0.11 0.06 0

0.17 0 0.05 0.10 0.14 0.17 0.17 0.17 0.14 0.10 0.05 0

0.18 0 0.05 0.09 0.13 0.15 0.16 0.15 0.13 0.09 0.05 0

0.19 0 0.04 0.08 0.11 0.13 0.14 0.13 0.11 0.08 0.04 00.20 0 0.04 0.07 0.10 0.12 0.13 0.12 0.10 0.07 0.04 0

0.21 0 0.04 0.07 0.09 0.11 0.11 0.11 0.09 0.07 0.04 0

0.22 0 0.03 0.06 0.08 0.10 0.10 0.10 0.08 0.06 0.03 0

0.23 0 0.03 0.05 0.08 0.09 0.09 0.09 0.08 0.05 0.03 0

0.24 0 0.03 0.05 0.07 0.08 0.08 0.08 0.07 0.05 0.03 00.25 0 0.02 0.04 0.06 0.07 0.08 0.07 0.06 0.04 0.02 0

0.26 0 0.02 0.04 0.06 0.07 0.07 0.07 0.06 0.04 0.02 0

0.27 0 0.02 0.04 0.05 0.06 0.06 0.06 0.05 0.04 0.02 0

0.28 0 0.02 0.03 0.05 0.05 0.06 0.05 0.05 0.03 0.02 0

0.29 0 0.02 0.03 0.04 0.05 0.05 0.05 0.04 0.03 0.02 0

0.30 0 0.01 0.03 0.04 0.04 0.05 0.04 0.04 0.03 0.01 0


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Results for dt=0.00119-Dec-13

31


Results for dt=0.0119-Dec-13

32


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Example

Solve the same equation as the previousexample, but use the Crank-Nicholson

scheme to obtain approximate values for

with the time step 0.001.

Now change the time step to 0.01. What

happens?

19-Dec-13

33

n

i

n

i

n

i

n

i

n

i

n

i ruurruruurru

11

1

1

11

1

)22()22(

2x

tr


Solution

The problem is set up as follows:

19-Dec-13

34

1111111 )22()22(

1

ni

n

i

n

i

n

i

n

i

n

i rururuurru

ru 2

x

tr

Name FLAG

0: Initialize

1: Iterate (solve)

Implicit Equation

=IF(FLAG=0,0,(rr*B25+(2-2*rr)*C25+rr*D25+rr*B26+rr*D26)/(2+2*rr))

[B26 & D26 are at current time-step]

Name rr

=B22/B21^2

Iteration counter

=IF(FLAG=0,0,E19+1))


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Results

For dt = 0.001, courant number = 0.1 andafter 7 iterations.

19-Dec-13

35


Results

For dt = 0.01, courant number = 1 and after

20 iterations, with convergence.

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36

Documents

Week 15 - Explicit vs Implicit