computational fluid dynamics accuracygtryggva/cfd-course2017/lecture-2-2017.pdfcomputational fluid...

Computational Fluid Dynamics


http://www.nd.edu/~gtryggva/CFD-Course/

Grétar Tryggvason

Lecture 2January 23, 2017


Accuracy


It 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 Δt.

First we test for different Δt.

Number of time steps= T/Δt


m=11; time=0.50

2 4 6 8 10 12 14 16 18 20-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

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

N=21Δt=0.05

Exact

Numerical

Accuracy


Repeat with a smaller time-stepΔt=0.025

N=21

2 4 6 8 10 12 14 16 18 20-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

m=21; time=0.50

Exact

Numerical

AccuracyComputational Fluid Dynamics

Repeat with a smaller time-stepΔt=0.0125

N=21

m=41; time=0.50

2 4 6 8 10 12 14 16 18 20-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

Exact

Numerical

Accuracy


How accurate solution can we obtain?

Take Δt = 0.0005 and N=200


m=1001; time=0.50

20 40 60 80 100 120 140 160 180 200-2

-1.5

-1

-0.5

0

0.5

1

1.5

2U=1; D=0.05; k=1

N=200Δt=0.0005

Exact

Numerical

Very fine spatial resolution and a small time step

Accuracy

Computational Fluid DynamicsAccuracy

Examine the spatial accuracy by taking a very small time step, Δt = 0.0005 and vary the number of grid points, N, used to resolve the spatial direction.

The grid size is h = L/N where L = 1 for our case

Quantifying the ErrorOrder 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

Numericaltime=0.50

Accuracy


100 101 10210-4

10-3

10-2

10-1

100Accuracy. Effect ofspatial resolution

1/h

dt=0.0005N=11 to N=161

E

time=0.50


�

E = Ch2 = C 1h⎛ ⎝

⎞ ⎠

−2

�

lnE = ln C 1h⎛ ⎝

⎞ ⎠

−2⎛

⎝ ⎜ ⎞

⎠ ⎟ = lnC − 2ln 1

h⎛ ⎝

⎞ ⎠

On a log-log plot, the E versus (1/h) curve should therefore have a slope -2

If the error is of second order:

Taking the log:

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=161

E

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

Accuracy


Finite difference approximations by Taylor expansion

Approximating a partial differential equation

Showed, by numerical experiments, that accuracy increases as the resolution is increased

Showed that the error behaves in a way that should be predictable

SummaryComputational Fluid Dynamics

Stability


As long as accuracy is reasonable, integration at larger time steps is more efficient and desirable.

Can we increase the time step indefinitely?

Let’s repeat the 1-D advection-diffusion equation with larges time step.

Use Δt = 0.2, instead of Δt = 0.05

StabilityComputational Fluid Dynamics

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

N=21Δt=0.2

Exact

Numerical

Stability


Instead of decaying as it should, the amplitude of the numerical solution keeps increasing.

Indeed, if we continued the calculations, we would eventually produce numbers larger than the computer can handle. This results in an “overflow” or “NaN” (Not a Number).


Ordinary Differential EquationStability

Computational Fluid DynamicsODE—Example

Take: dfdt

= − f

f t( ) = e−tThe exact solution is

with initial condition f (0) = 1

f n+1 = (1 − Δt) f nf n+1 = f n − f nΔtForward Euler

�

f n+1

f n≤1 only if

�

Δt ≤ 2


0 5 10 15 20-5

-4

-3

-2

-1

0

1

2

3

4

5

Δt=0.5Δt=1.5

Δt=2.5

�

f n+1

f n≤1 only if

�

Δt ≤ 2

f n+1 = (1 − Δt) f n

ODE—Example


�

f n+1

f n= 1(1+ Δt)

f n+1 = f n − f n+1ΔtBackward Euler

�

f n+1 = (1−Δt) f n = (1−Δt)2 f n−1

= ………… = (1−Δt)n f 1

However

for all Δt

�

Δt ≤1

�

f n+1

f n≤1

Obviously, f oscillates unless

ODE—ExampleComputational Fluid Dynamics

Stability Analysis of theAdvection-Diffusion Equation:

Von Neumann Method


Generally, stability analysis of the full nonlinear system of equations is too involved to be practical, and we study a model problem that in some way mimics the full equations.The linear advection-diffusion equation is one such model equation, and we will apply von Neumann's method to check the stability of a simple finite difference approximation to that equation.


2

2

xf

Dxf

Utf

∂∂

∂∂

∂∂ =+

21111

1 22 h

fffD

hff

Utff n

jnj

nj

nj

nj

nj

nj −+−++ +−

=−

+Δ−

Consider the 1-D advection-diffusion equation:

In finite-difference form:

Stability

�

f jn

= ε jn

Look at the evolution of a small perturbation


22

21111

1

hD

hU

t

nj

nj

nj

nj

nj

nj

nj −+−++ +−

=−

+Δ− εεεεεεε

The evolution of the perturbation is governed by:

Write the error as a wave (expand as a Fourier series):

∑∞

−∞=

==k

ikxnkj

nnj

jex εεε )(

Dropping the subscriptjikxnn

j eεε =

Stability

�

eikx = coskx + isinkxRecall:


ε jn = εneikx j

ε j+1n = εneikx j+1 = εneik (x j + h) = εneikx j eikh

The error at node j is:

The error at j+1 and j -1 can be written as

ε j−1n = ε neikx j−1 = ε neik (x j −h) = ε neikx j e−ikh

Stability


ε j+1n = ε neikx j eikh

ε n+1eikx j − ε neikx j

Δt+U εn

2h(eikheikx j − e− ikheikx j ) =

Dε n

h2(eikheikx j − 2eikx j + e− ikheikx j )

Substituting

ε j−1n = ε neikx j e−ikh

ε jn+1 − ε j

n

Δt+U

ε j+1n − ε j−1

n

2h= D

ε j+1n − 2ε j

n + ε j−1n

h2

ε jn = εneikx j

into

yields


ε n+1 − εn

Δ t+U

ε n

2h(eikh − e− ikh) = D

ε n

h2(eikh − 2 + e−ikh )

The equation for the error is:

)2()(2

1 2

1ikhikhikhikh

n

n

eehtDee

htU −−

+

+−Δ+−Δ−=εε

Solving for the ratio of the errors:

Stability


)2()(2

1 2

1ikhikhikhikh

n

n

eehtDee

htU −−

+

+−Δ+−Δ−=εε

�

=1−UΔth

isinkh + DΔth2

2(coskh −1)

�

=1− 4 DΔth2

sin2 k h2− iUΔt

hsinkh

Dividing by the error amplitude at n:amplification factor

Stability

�

eikh + e−ikh = 2coskh; eikh − e− ikh = i2sinkh;Using:

�

2sin2θ =1− cos2θ


�

εn+1

εn=1− 4 DΔt

h2sin2 k h

2− iUΔt

hsinkh

ε n+1

ε n≤ 1

Stability requires that

Since the amplification factor is a complex number, and k, the wave number of the error, can be anything, the determination of the stability limit is slightly involved.

We will look at two special cases: (a) U = 0 and (b) D = 0

The ratio of the error amplitude at n+1 and n is:

Stability


ε n+1

ε n= 1− 4

DΔth2

sin2 kh2

Since the amplification factor is always less than 1, and we find that it is bigger than -1 if

sin2 () ≤ 1

(a) Consider first the case when U = 0, so the problem reduces to a pure diffusion

�

DΔth2

≤12

�

−1≤1− 4 DΔth2

≤1


Since the amplification factor has the form 1+i() the absolute value of this complex number is always larger than unity and the method is unconditionally unstable for this case.

�

εn+1

εn=1− iUΔt

hsinkh

(b) Consider now the other limit where D = 0 and we have a pure advection problem.

�

iUΔthsinkh

1

ε n+1

ε n

Stability


For the general case we must investigate the stability condition in more detail. We will not do so here, but simply quote the results:

Notice that high velocity and low viscosity lead to instability according to the second restriction.

Δ tDh2

≤12

U2ΔtD

≤ 2and


ε i, jn = ε nei(kxi +ly j )

41

2 ≤ΔhtD

4|)||(| 2

≤Δ+D

tVUand

�

DΔth2

≤ 16

and

�

(|U |+ |V |+ |W |)2ΔtD

≤ 8

For a two-dimensional problem, assume an error of the form

A stability analysis gives:

For a three-dimensional problem we get:

Stability


Convergence – the solution to the finite-difference equation approaches the true solution to the PDE having the same initial and boundary conditions as the mesh is refined.

Lax’s Equivalence TheoremGiven a properly posed initial value problem and a finite-difference approximation to it that satisfies the consistency condition, stability is the necessary and sufficient condition for convergence.

Stability – Now you know!


The Modified Equation


Using the finite difference approximation, we are effectively solving an equation that is slightly different than the original partial differential equations.

Does the finite difference equation approach the partial differential equation in the limit of zero Δt and h?

ConsistencyComputational Fluid Dynamics

Consider the 1-D advection-diffusion equation

and its finite-difference approximation

�

f jn +1 − f j

n

Δt+U

f j+1n − f j−1

n

2h= D

f j+1n − 2 f j

n + f j−1n

h2

∂ f∂t

+U∂ f∂x

= D∂ 2 f∂x2

The discrepancy between the two equations can be found by deriving the modified equation.

Consistency


Substituting

�

f jn+1 − f j

n

Δt= ∂f (t)

∂t+ ∂ 2 f (t)

∂t 2Δt2

+

�

f j+1n − f j−1

n

2h= ∂f (x)

∂x+ ∂ 3 f (x)

∂x 3h2

6+

�

f j+1n − 2 f j

n + f j−1n

h2= ∂ 2 f (x)

∂x 2+ ∂ 4 f (x)

∂x 4h2

12+

into the finite difference equationfjn+1 − fj

n

Δt+U

fj+1n − fj−1

n

2h= D

f j+1n − 2 fj

n + fj−1n

h2


�

∂f∂t

+U ∂f∂x

−D∂ 2 f∂x 2

= −∂2 f (t)∂t 2

Δt2−U ∂ 3 f (x)

∂x 3h2

6+ D∂ 4 f (x)

∂x 4h2

12+

In this case, the error goes to zero as h→0 and Δt→0, so the approximation is said to be CONSISTENT

Results in the Modified Equation:

Original Equation Error terms

�

∂f∂t

+ U ∂f∂x

−D∂ 2 f∂x 2

= O(Δt,h2 )

Shorthand:

Consistency


Although most finite difference approximations are consistent, innocent-looking modifications can sometimes lead to approximations that are not!

The Frankel-Dufort is an example of an non-consistent scheme.

You will examine it in the homework


HW: Examine the Frankel-Dufort method

�

∂f∂t

= D∂ 2 f∂x 2

Solve the diffusion equation

Using the Leapfrog time integration method and standard finite-difference approximation for the spatial derivative gives:

�

f jn+1 − f j

n−1

2Δt= Dh2

f j+1n − 2 f j

n + f j−1n[ ]

Leapfrog method

Consistency


This gives:�

f jn+1 − f j

n−1

2Δt= Dh2

f j+1n − 2 f j

n + f j−1n[ ]

�

f jn =

12f jn +1 + f j

n−1( )

�

f jn+1 = f j

n−1 + 2Δt D

h2f j+1n − f j

n+1 − f jn−1 + f j−1

n( ).Which is easily solved for f at the new time step

Now modify it slightly:

In the HW, you will examine the error!

Replace by:


The MODIFIED EQUATION is obtained by substituting the expression for the finite difference approximations, including the error terms, into the finite difference equation. For a CONSISTENT finite difference approximation the error terms go to zero as h→0 and Δt→0. The modified equation can often be used to infer the nature of the error of the finite difference scheme. More about that later.

Consistency


Two-Dimensional Advection-

Diffusion Equation


∂f∂t

+U∂f∂x

+ V∂f∂y

= D∂ 2 f∂x 2

+∂ 2 f∂y2

⎛ ⎝ ⎜ ⎞

⎠

We will use the model equation:

to demonstrate how to solve a partial equation (initial value problem) numerically.

The extension to two-dimensions is relatively straight forward, once the one-dimensional problem is fully understood.

Multidimensional Equations


For a two-dimensional flow discretize the variables on a two-dimensional grid

fi , j = f (x,y)

fi+1, j = f (x + h, y)

fi , j+1 = f (x, y + h)(x, y)

i -1 i i+1

j+1

j

j-1

Multidimensional EquationsComputational Fluid Dynamics

fi, jn+1 − fi, j

n

Δt= −U

fi+1, jn − fi−1, j

n

2h

⎛

⎝⎜

⎞

⎠⎟ −V

fi, j+1n − fi, j−1

n

2h

⎛

⎝⎜

⎞

⎠⎟ +

Dfi+1, j

n + fi−1, jn − 2 fi, j

n

h2 +fi, j+1

n + fi, j−1n − 2 fi, j

n

h2

⎛

⎝⎜

⎞

⎠⎟

∂ f∂t

+U∂ f∂x

+V∂ f∂ y

= D∂ 2 f∂x2 +

∂ 2 f∂ y2

⎛

⎝⎜⎞

⎠⎟

The discrete equation is:



fi, jn+1 = fi, j

n + Δt −Ufi+1, j

n − fi−1, jn

2h

⎛

⎝⎜

⎞

⎠⎟ −V

fi, j+1n − fi, j−1

n

2h

⎛

⎝⎜

⎞

⎠⎟ +

⎡

⎣⎢⎢

Dfi+1, j

n + fi−1, jn + fi, j+1

n + fi, j−1n − 4 fi, j

n

h2

⎛

⎝⎜

⎞

⎠⎟⎤

⎦⎥⎥

Solve for fi , jn+1

fi, jn+1 = fi, j

n −ΔtU2h

fi+1, jn − fi−1, j

n( ) − ΔtV2h

fi, j+1n − fi, j−1

n( )+ΔtDh2 fi+1, j

n + fi−1, jn + fi, j+1

n + fi, j−1n − 4 fi, j

n( )

or


Δ tDh2

≤14

(|U | + | V |) 2ΔtD

≤ 4andA stability analysis gives:

Accuracy: O Δt,h2( )


Example


f=1.0

∂f∂x

= 0∂f∂y

= 0

f=0.0

f=0.0

∂ f∂t

+U∂ f∂x

+V∂ f∂ y

= D∂ 2 f∂x2 +

∂ 2 f∂ y2

⎛

⎝⎜⎞

⎠⎟

U=-1V=0

Uniform flow through the domain


i=1 i=2 i=NXj=1j=2

j=NY

fi , jstored at each grid point

�

fi, jn +1 = fi, j

n −ΔtU2h

fi+1, jn − fi−1, j

n( ) +ΔtDh2

fi+1, jn + fi−1, j

n + fi, j+1n + fi, j−1

n − 4 fi, jn( )

∂ f∂n

�

for

given on the boundary



Boundary conditionsWhere f is given, we simply specify its value

Where the normal derivative is specified, we approximate the value at the boundary by one-sided differences

∂ f∂ y

= 0At the i=1 boundary, for example,

∂ f∂ y

≈fi,2

n − fi,1n

h= 0and by using

we find that: fi,2

n = fi,1n


% two-dimensional unsteady diffusion by the FTCS scheme %------------------------------------------------------------ n=32;m=32;nstep=120;D=0.025;length=2.0;h=length/(n-1); dt=1.0*0.125*h*h/D;f=zeros(n,m);fo=zeros(n,m);time=0.0; u=-0.0; v=-1.0; f(12:21,n)=1.0; for l=1:nstep,l,time hold off;mesh(f); axis([0 n 0 m 0 1.5]);pause; fo=f; for i=2:n-1, for j=2:m-1 f(i,j)=fo(i,j)-(0.5*dt*u/h)*(fo(i+1,j)-fo(i-1,j))-...

(0.5*dt*v/h)*(fo(i,j+1)-fo(i,j-1))+... (D*dt/h^2)*(fo(i+1,j)+ fo(i,j+1)+fo(i-1,j)+fo(i,j -1)-4*fo(i,j)); end,end for i=1:n, f(i,1)=f(i,2);end;for j=1:m,f(1,j)=f(2,j);f(m,j)=f(m-1,j);end; time=time+dt; end;



The unsteady evolution of the solution


Multidimensional Boundary Value

Problems(Steady-State)


∂ 2 f∂x2

+∂ 2 f∂y2

= S

This equation has a solution if f or is specified on the boundary

Boundary Value Problems

Consider the Poisson Equation:

Use standard finite differences to discretize:

�

fi+1, j + f i−1, j − 2 fi, jh2

+f i, j−1 + f i, j+1 − 2 f i, j

h2= Si, j

∂f ∂n


�

fi , j

= 14

f i+1, j + fi−1, j + f i, j−1 + f i, j+1 − h2Si, j( )

Solve for fi,j:

fi+1, j + fi−1, j + fi, j−1 + fi, j+1 − 4 fi, j

h2 = S

can be written as

For uniform grids:

fi+1, j + fi−1, j − 2 fi, j

h2 +fi, j−1 + fi, j+1 − 2 fi, j

h2 = Si, j



�

fi , j

α+1 = 14

f i+1, jα + f i−1, j

α + fi, j−1α + f i, j+1

α − h2Si, j( )

Solve for fi,j and use the right hand side to compute a new value. Denote the old values by α and the new ones with α+1

This iteration process—Jacobi iteration—is very robust but many iterations are required to reach an accurate solution.

Boundary Value ProblemsComputational Fluid Dynamics

The iteration must be carried out until the solution is sufficiently accurate. To measure the error, define the residual:

Ri, j =

fi+1, j + fi−1, j + fi, j−1 + fi, j+1 − 4 fi, j

h2 − Si, j

At steady-state the residual should be zero. The point-wise residual or the average absolute residual can be used, depending on the problem. Often, simpler criteria, such as the change from one iteration to the next is used



Although the Jacobi iteration is a very robust iteration technique, it converges VERY slowly.

We therefore seek a way to ACCELERATE the convergence to steady-state, making use of the fact that it is only the steady-state that is of interest.

Here we introduce the Gauss-Seidler method and the Successive Over-Relaxation (SOR) method.


�

fi, jα+1 = 1

4( f i+1, j

α + f i−1, jα+1 + fi, j+1

α + f i, j−1α+1 − h2Si, j )

The Jacobi iteration can be improved somewhat by using new values as soon as they become available.

j

j-1

j+1

i i+1i-1

for j=1:mfor i=1:n iterateendend

From a programming point of view, Gauss-Seidler iteration is even simpler than Jacobi iteration since only one vector with f values is needed.



fi, jα+1 =

β4( fi+1, j

α + fi−1, jα+1 + fi, j+1

α + fi, j−1α+1 − h2Si, j )

+ (1−β ) fi, jα

The Gauss-Seidler iteration can be accelerated even further by various acceleration techniques. The simplest one is the Successive Over-Relaxation (SOR) iteration

The SOR iteration is very simple to program, just as the Gauss-Seidler iteration. The user must select the coefficient. It must be bounded by 1<β<2. β=1.5 is usually a good starting value.


Example


5 10 15 20 25 30 35 40

5

10

15

20

25

30

35

40

f=1.0 f=0.0

�

∂ 2 f∂x 2

+∂ 2 f∂y2

= 0


% two-dimensional steady-state problem by SORn=40;m=40;iterations=5000;length=2.0;h=length/(n-1);T=zeros(n,m);bb=1.7;T(10:n-10,1)=1.0; for l=1:iterations, for i=2:n-1, for j=2:m-1 T(i,j)=bb*0.25*(T(i+1,j)+... T(i,j+1)+T(i-1,j)+T(i,j-1))+(1.0-bb)*T(i,j); end,end % find residual res=0; for i=2:n-1, for j=2:m-1 res=res+abs(T(i+1,j)+... T(i,j+1)+T(i-1,j)+T(i,j-1)-4*T(i,j))/h^2; end,end l,res/((m-2)*(n-2)) % Print iteration and residual if (res/((m-2)*(n-2)) < 0.001), break,endend;contour(T);



Average absolute error: 0.001 Number of iterations

Jacobi: 1989Gauss-Seidler: 986

SOR (β =1.5): 320SOR (β = 1.7): 162SOR (β = 1.9): 91SOR (β = 1.95): 202

The program is easily modified for the Jacobi and the Gauss-Seidler iteration:


5 10 15 20 25 30 35 40

5

10

15

20

25

30

35

40

The converged solution:

�

∂ 2 f∂x 2

+∂ 2 f∂y2

= 0


010

2030

40

0

10

20

30

400

0.2

0.4

0.6

0.8

1


Accuracy—smaller time step and finer resolution should get us the exact solution

Stability—introduced the von Neumann method. Fairly mechanical process, we will provide more insight by the finite volume point of view

The modified equations helps us see how the approximate and the exact equation differ and if the former is consistent with the latter

Multidimensional advection-diffusion equation. Essentially the same as the one-dimensional problem

Iterative methods for boundary value problems. Elementary approaches to steady state problems

Summary

computational fluid dynamics accuracygtryggva/cfd-course2017/lecture-2-2017.pdfcomputational fluid...

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