WPI Computational Fluid Dynamics Iphoenics/SITE_PHOENICS/Apostilas/CFD... · 2003-09-23 · WPIPI Computational Fluid Dynamics I ∂f ∂t + U ∂f ∂x = D ∂2 f ∂x2 We will use

Computational Fluid Dynamics IPPPPIIIIWWWW

ElementaryNumerical Analysis:

Finite Difference Approximations-II

Instructor: Hong G. ImUniversity of Michigan

Fall 2001

Computational Fluid Dynamics IPPPPIIIIWWWW• The linear advection-diffusion equation• Finite difference approximations• Finite Difference Approximation of the linear

advection-diffusion equation• Accuracy• Stability• Consistency• Higher order finite difference approximations in

space• Time integration: implicit versus explicit and higher

accuracy • Finite Volume Approximations

Outline


Solving a partial differential

equation


∂f∂t

+ U∂f∂x

= D∂ 2 f∂x 2

We will use the model equation:

Although this equation is much simpler than the full Navier Stokes equations, it has both an advection term and a diffusion term. Before attempting to solve the equation, it is useful to understand how the analytical solution behaves.

to demonstrate how to solve a partial differential equation numerically.

Model Equations

Computational Fluid Dynamics IPPPPIIIIWWWWFor initial conditions of the form:

)2sin()0,( kxAtxf π==

))(2sin(),(2)2( UtxkAetxf tkD −= − ππ

It can be verified by direct substitution that the solution is given by:

which is a decaying traveling wave

Model Equations

k = 1.0U = 1.0D = 0.05A = 1.0


Finite Difference Approximations of

the Derivatives

Computational Fluid Dynamics IPPPPIIIIWWWWDerive a numerical approximation to the governing equation, replacing a relation between the derivatives by a relation between the discrete nodal values.

h h

f(t,x-h) f(t,x) f(t,x+h)∆ t

∆ t

f (t)

f (t + ∆t)

f (t + 2∆t)

Finite Difference Approximations


The Time Derivative is found using a FORWARD EULER method. The approximation can be found by using a Taylor series

h h

f(x-h) f(x) f(x+h)∆ t



TheTime Derivative



f (t + ∆t) = f (t) +∂f (t)

∂t∆t +

∂ 2 f (t)∂t2

∆t2

2+"

∂f (t)∂t

=f (t + ∆t) − f (t)

∆t+

∂ 2 f (t)∂t 2

∆t2

+"

Solving this equation for the time derivative gives:

Time derivative



TheSpatial Derivative:

First Order



When using FINITE DIFFERENCE approximations, the values of f are stored at discrete points.

h h

f(x-h) f(x) f(x+h)

The derivatives of the function are approximated using a Taylor series:



Subtracting the second equation from the first:

f ( x + h) = f (x) +∂f (x)

∂xh +

∂ 2 f (x)∂x 2

h2

2+

∂ 3 f (x)∂x3

h3

6+

∂ 4 f (x)∂x 4

h4

24+"

f ( x − h) = f (x) −∂f (x)

∂xh +

∂ 2 f (x)∂x 2

h2

2−

∂ 3 f ( x)∂x3

h3

6+

∂ 4 f (x)∂x 4

h4

24−"

!!f (x + h) − f (x − h) = 2 ∂f (x)

∂xh + 2 ∂ 3 f (x)

∂x3h3

6+"

Start by expressing the value of f(x+h) and f(x-h) in terms of f(x)



!!

∂f (x)∂x

=f (x + h) − f (x − h)

2h+

∂ 3 f (x)∂x3

h2

6+"

Rearranging this equation to isolate the first derivative:

f (x + h) − f (x − h) = 2 ∂f (x)∂x

h + 2 ∂ 3 f (x)∂x3

h3

6+"


The result is:


TheSpatial Derivative:

Second Order



!!f ( x + h) = f (x) +

∂f (x)∂x

h +∂ 2 f (x)

∂x 2h2

2+

∂ 3 f (x)∂x3

h3

6+

∂ 4 f (x)∂x 4

h4

24+"

f ( x − h) = f (x) −∂f (x)

∂xh +

∂ 2 f (x)∂x 2

h2

2−

∂ 3 f ( x)∂x3

h3

6+

∂ 4 f (x)∂x 4

h4

24−"

Start by expressing the value of f(x+h) and f(x-h) in terms of f(x)

Adding the second equation to the first:

!!f (x + h) + f (x − h) = 2 f (x) + 2 ∂f 2(x)

∂x2h2

2+ 2 ∂ 4 f (x)

∂x4h4

24+"



!!

∂ 2 f (x)∂x2 =

f (x + h) − 2 f + f (x − h)h2 +

∂ 4 f (x)∂x4

h2

12+"

f (x + h) + f (x − h) = 2 f (x) + 2 ∂f 2(x)∂x2

h2

2+ 2 ∂ 4 f (x)

∂x4

h4

24+"

Rearranging this equation to isolate the second derivative:

The result is:



Solving the partial differential equation



∂f∂t

+ U∂f∂x

= D∂ 2 f∂x 2

A numerical approximation to

Is found by replacing the derivatives by the following approximations


∂f∂t

j

n

+ U ∂f∂x

j

n

= D ∂ 2 f∂x2

j

n

h h

fjn+1

fjn fj +1

nfj −1n ∆t


f jn = f (t, x j )

f jn+1 = f (t + ∆t, x j )

f j +1n = f (t, x j + h)

f j −1n = f (t, x j − h)

Using the shorthand notation


∂f∂t

j

n

=fj

n+1 − fjn

∆t+ O(∆t)

∂f∂x

j

n

=fj +1

n − fj −1n

2h+ O(h2 )

∂ 2 f∂x 2

j

n

=fj +1

n − 2 fjn + fj −1

n

h2 + O(h2 )

gives

Computational Fluid Dynamics IPPPPIIIIWWWWSubstituting these approximations into:

gives

∂f∂t

+ U∂f∂x

= D∂ 2 f∂x 2


f jn+1 − fj

n

∆t+ U

f j +1n − fj −1

n

2h= D

fj +1n − 2 f j

n + f j −1n

h2 + O(∆t, h2)

fjn+1 = fj

n − U∆t2h

( fj +1n − fj −1

n ) + D∆th2 ( fj +1

n − 2 f jn + fj −1

n )

Solving for the new value and dropping the error terms yields


fjn+1 = fj

n − U∆t2h

( fj +1n − fj −1

n ) + D∆th2 ( fj +1

n − 2 f jn + fj −1

n )

The value of every point at level n+1 is given explicitly in terms of the values at the level n

h h

fjn+1

fjn fj +1

nfj −1n ∆t


Thus, given f at one time (or time level), f at the next time level is given by:


Example


A short MATLAB programThe evolution of a sine wave is followed as it is advected and diffused. Two waves of the infinite wave train are simulated in a domain of length 2. To model the infinite train, periodic boundary conditions are used. Compare the numerical results with the exact solution.

Computational Fluid Dynamics IPPPPIIIIWWWW% one-dimensional advection-diffusion by the FTCS schemen=21; nstep=100; length=2.0; h=length/(n-1); dt=0.05; D=0.05;f=zeros(n,1); y=zeros(n,1); ex=zeros(n,1); time=0.0;for i=1:n, f(i)=0.5*sin(2*pi*h*(i-1)); end; % initial conditionsfor m=1:nstep, m, time

for i=1:n, ex(i)=exp(-4*pi*pi*D*time)*...0.5*sin(2*pi*(h*(i-1)-time)); end; % exact solution

hold off; plot(f,'linewidt',2); axis([1 n -2.0, 2.0]); % plot solutionhold on; plot(ex,'r','linewidt',2);pause; % plot exact solutiony=f; % store the solutionfor i=2:n-1,

f(i)=y(i)-0.5*(dt/h)*(y(i+1)-y(i-1))+...D*(dt/h^2)*(y(i+1)-2*y(i)+y(i-1)); % advect by centered differences

end;f(n)=y(n)-0.5*(dt/h)*(y(2)-y(n-1))+...

D*(dt/h^2)*(y(2)-2*y(n)+y(n-1)); % do endpoints forf(1)=f(n); % periodic boundariestime=time+dt;

end;


Evolution forU=1; D=0.05; k=1

N=21∆t=0.05

Exact

Numerical


Accuracy

Computational Fluid Dynamics IPPPPIIIIWWWWIt is clear that although the numerical solution is qualitatively similar to the analytical solution, there are significant quantitative differences.

The derivation of the numerical approximations for the derivatives showed that the error depends on the size of h and ∆t.

First we test for different ∆t.

Number of time steps= T/ ∆t


x

f(x,t)

0 0.5 1 1.5 2

-1

-0.5

0

0.5

1

x

f(x,t)

0 0.5 1 1.5 2

-1

-0.5

0

0.5

1m = 10, time = 0.50Evolution

forU=1; D=0.05; k=1

N=21∆t=0.05

Exact

Numerical

Accuracy


x

f(x,t)

0 0.5 1 1.5 2

-1

-0.5

0

0.5

1

x

f(x,t)

0 0.5 1 1.5 2

-1

-0.5

0

0.5

1m = 20, time = 0.50Repeat with

a smaller time-step∆t=0.025

Exact

Numerical

Accuracy


x

f(x,t)

0 0.5 1 1.5 2

-1

-0.5

0

0.5

1

x

f(x,t)

0 0.5 1 1.5 2

-1

-0.5

0

0.5

1m = 40, time = 0.50Repeat with

a smaller time-step∆t=0.0125

Exact

Numerical

Accuracy


x

f(x,t)

0 0.5 1 1.5 2

-1

-0.5

0

0.5

1

x

f(x,t)

0 0.5 1 1.5 2

-1

-0.5

0

0.5

1m = 1000, time = 0.50

U=1; D=0.05; k=1

N=201∆t=0.0005

Exact

Numerical

Accuracy

Very fine spatial resolution and a small time step


Quantifying the ErrorI. Order of Accuracy


E = h ( fj − fexact )2

j =1

N

∑

n=11; E = 0.1633n=21; E = 0.0403n=41; E = 0.0096n=61; E = 0.0041n=81; E = 0.0022n=101; E = 0.0015n=121; E = 0.0011n=161; E = 9.2600e-04

Exact

Numerical

Order of Accuracy

time=0.50


100 101 10210-4

10-3

10-2

10-1

100

Accuracy. Effect ofspatial resolution

1/ h

dt=0.0005N=11 to N=161 E

Order of Accuracy

time=0.50


E = Ch2 = CLn

2

= CL2n−2

E = ˜ C n−2

ln E = ln ˜ C n−2( )= ln ˜ C − 2ln n

On a log-log plot, the E versus n curve should therefore have a slope -n

If the error is of second order:

Order of Accuracy


100 101 10210-4

10-3

10-2

10-1

100Accuracy. Effect ofspatial resolution

21

1/ h

dt=0.0005

N=11 to N=161E

Order of Accuracy

time=0.50

?


Why is does the error deviate from the line for the highest values of N?

Number of Steps, N = 1/h

Erro

r Total

Roundoff

Truncation


Quantifying the ErrorII. Modified Wave Number


Order of Accuracy

- how mesh refinement improves the accuracy

Modified Wave Number

- how the difference scheme maintains accuracy at high wave numbers



ikxexf =)(

Consider a pure harmonic of period L

where k can take on any of the following values

2/,22/,12/2 NNNnnL

k "+−+−== π


The exact derivative isikxikexf =′ )(

We now ask how the second order central difference computes the derivative of f


.1,2,1,0, −== NjjNLx j "

Discretize the x axis with a uniform mesh

Substituting


NLff

xf jj

j

/,2

11 =∆∆−

= −+

δδ

Finite-difference approximation:

,)( /2 Nnjiikx eexf π==

Computational Fluid Dynamics IPPPPIIIIWWWWModified Wave Number

Modified Wave Number:

∆−=

−+

2

/)1(2/)1(2 NjniNjni

j

eexf ππ

δδ

jj

j

NniNni

fkifNni

fee

′=∆

=

∆−=

−

)/2sin(2

/2/2

π

ππ

∆=′ )/2sin( Nnk π

Computational Fluid Dynamics IPPPPIIIIWWWWModified Wave Number

Summary:

)2sin(;)/2sin(

2;2

NnkNnk

Nnkn

Lk

ππ

ππ

=′∆∆

=′

=∆=


k∆

k'∆

0 1 2 30

1

2

3 Exact2nd Order Central4th Order Central6th Order Central4th Order Pade6th Order Pade


Computational Fluid Dynamics IPPPPIIIIWWWWEnhancing Accuracy - Padé Scheme

4th Order Padé Scheme:v

jjjjjj fhffh

fff30

)(344

1111 +−=′+′+′ −+−+

2nd Order Central Difference:

jjjj fhffh

f ′′′−−=′ −+ 6)(

21 2

11

4th Order Central Difference:

)()88(12

1 42112 hOffff

hf jjjjj +′−+−=′ ++−−

Tridiagonal system of equations for jf ′

Documents

WPI Computational Fluid Dynamics Iphoenics/SITE_PHOENICS/Apostilas/CFD... · 2003-09-23 · WPIPI Computational Fluid Dynamics I ∂f ∂t + U ∂f ∂x = D ∂2 f ∂x2 We will use