Computational Fluid Theory of Dynamics Partial ...gtryggva/CFD-Course2017/Lecture-6-2017.pdf · partial differential equations Computational Fluid Dynamics a ∂f ∂x +b ∂f ∂y

Computational Fluid Dynamics


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

Grétar Tryggvason

Lecture 6February 1, 2017


Theory ofPartial Differential

Equations

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


•  Basic Properties of PDE•  Quasi-linear First Order Equations

- Characteristics- Linear and Nonlinear Advection Equations

•  Quasi-linear Second Order Equations - Classification: hyperbolic, parabolic, elliptic- Domain of Dependence/Influence

OutlineComputational Fluid Dynamics

Examples of equations

�

∂f∂t

+U ∂f∂x

= 0 Advection

�

∂f∂t

−D∂ 2 f∂x 2

= 0

�

∂ 2 f∂t 2

− c 2 ∂2 f

∂x 2= 0

�

∂ 2 f∂x 2

+ ∂ 2 f∂y 2

= 0

Diffusion

Wave propagation

Laplace equation

Evolution in time


�

∂u∂t

+ u∂u∂x

+ v∂u∂y

= −1ρ∂P∂x

+µρ

∂ 2u∂x 2

+∂ 2u∂y2

⎛ ⎝ ⎜ ⎞

⎠ ⎟

Navier-Stokes equations

�

∂u∂x

+∂v∂y

= 0

Diffusion part

Advection part

Laplace part


•  The order of PDE is determined by the highest derivatives

•  Linear if no powers or products of the unknown functions or its partial derivatives are present.

•  Quasi-linear if it is true for the partial derivatives of highest order.

Definitions

02, =+∂∂+

∂∂=

∂∂+

∂∂

xfyf

xf

fyf

yxf

x

222

2

2

2

, fyf

yxf

xfyf

xf

f =∂∂+

∂∂=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂+

∂∂


Quasi-linear first order partial differential equations


a∂f∂x

+ b∂f∂y

= c

a = a(x, y, f )b = b(x, y, f )c = c(x, y, f )

Consider the quasi-linear first order equation

where the coefficients are functions of x,y, and f, but not the derivatives of f:


The solution of this equations defines a single valued surface f(x,y) in three-dimensional space:

f=f(x,y)

f

x

y


dyyf

dxxf

df∂∂+

∂∂=

An arbitrary change in f is given by:

f=f(x,y)

f

x

y

f

dxdy

df


The normal vector to the curve f=f(x,y)

f=f(x,y)f

x

�

∂f∂x

1

-1

�

n = ∂f∂x,∂f∂y,−1

⎛

⎝ ⎜

⎞

⎠ ⎟ Same arguments in the y-direction. Thus


a∂f∂x

+ b∂f∂y

= c

df =∂f∂x

dx +∂f∂ydy

∂f∂x,∂f∂y,−1

⎛ ⎝ ⎜

⎞ ⎠ ⎟ ⋅ (a,b,c) = 0

∂f∂x,∂f∂y,−1

⎛ ⎝ ⎜

⎞ ⎠ ⎟ ⋅ (dx,dy ,df ) = 0

⇒

⇒

The original equation and the conditions for a small change can be rewritten as:

Normal to the surface

Both (a,b,c) and (dx,dy,df) are in the surface!


(dx,dy ,df ) = ds (a,b,c)

dxds

= a;dyds

= b;dfds

= c;

dxdy

=ab

⇒

⇒

Picking the displacement in the direction of (a,b,c)

Separating the components


dxds

= a;dyds

= b;dfds

= c;

dxdy

=ab

The three equations specify lines in the x-y plane

Given the initial conditions:

x = x(s, to ); y = y(s,to ); f = f (s,to);the equations can be integrated in time

slope

Characteristics


x(s); y(s); f (s)

x

y

x(0); y(0); f (0)

dxds

= a;dyds

= b;dfds

= c;


∂f∂t

+ U∂f∂x

= 0

dtds

= 1;dxds

=U;dfds

= 0;

dxdt

=U; df = 0;

Consider the linear advection equation

The characteristics are given by:

or

Which shows that the solution moves along straight characteristics without changing its value


∂f∂t

+ U∂f∂x

= 0

Graphically:

Notice that these results are specific for:

t

f

f

x

�

f (x, t) = ft= 0 x −Ut( )


�

∂f∂t

+U ∂f∂x

= ∂g∂η

−U( ) +U ∂g∂η

= 0

where

Substitute into the original equation

�

f (x, t) = g x −Ut( )

�

g(x) = f (x,t = 0)

�

∂f∂x

= ∂g∂η

∂η∂x

= ∂g∂η

1( )

�

∂f∂t

= ∂g∂η

∂η∂t

= ∂g∂η

−U( )�

η(x, t) = x −Ut

The solution is therefore:

This can be verified by direct substitution:

Set

Then and


Discontinuous initial data

�

∂f∂t

+U ∂f∂x

= 0

Since the solution propagates along characteristics completely independently of the solution at the next spatial point, there is no requirement that it is differentiable or even continuous

t

f

f

x

�

f (x, t) = g x −Ut( )


�

∂f∂t

+U∂f∂x

= − fAdd a source

t

f

f

x

�

dtds

=1; dxds

= U; dfds

= − f ;

�

dxdt

= U; dfdt

= − f

⇒ f = f (0)e−t


or

Moving wave with decaying amplitude


�

∂f∂t

+ f∂f∂x

= 0

�

dtds

=1; dxds

= f ; dfds

= 0;

�

dxdt

= f ; df = 0;

Consider a nonlinear (quasi-linear) advection equation


or

The slope of the characteristics depends on the value of f(x,t).


t

fx

t3

t2

t1

Multi-valued solution


Why unphysical solutions?

02

2

=∂∂−

∂∂+

∂∂

xf

xf

ftf ε

- Because mathematical equation neglects some physical process (dissipation)

Burgers Equation

0→ε

Additional condition is required to pick out the physicallyRelevant solution

Correct solution is expected from Burgers equation with

Entropy Condition


Constructing physical solutions

�

∂f∂t

+ f ∂f∂x

= 0In most cases the solution is not allowed to be multiple valued and the “physical solution” must be reconstructed using conservation of f

The discontinuous solution propagates with a shock speed that is different from the slope of the characteristics on either side


Quasi-linear Second order partial differential equations


a∂ 2 f∂x 2

+ b∂ 2 f∂x∂y

+ c∂ 2 f∂y2

= d

�

a = a(x,y, f , fx, fy )

�

b = b(x,y, f , fx, fy )

�

c = c(x,y, f , fx, fy )

�

d = d(x,y, f , fx, fy )

Consider

where


�

a∂2 f∂x2

+ b∂2 f∂x∂y

+ c∂2 f∂y 2

= d

�

v =∂f∂x

�

w =∂f∂y

�

a∂v∂x

+ b∂v∂y

+ c∂w∂y

= d

�

∂w∂x

−∂v∂y

= 0

�

∂2 f∂x∂y

=∂2 f∂y∂x

First write the second order PDE as a system of first order equations

Define

The second equation is obtained from

and

then


�

a∂2 f∂x2

+ b∂2 f∂x∂y

+ c∂2 f∂y 2

= d

�

v =∂f∂x

�

w =∂f∂y

�

a∂v∂x

+ b∂v∂y

+ c∂w∂y

= d

�

∂w∂x

−∂v∂y

= 0

Thus

Is equivalent to:

where

Any high-order PDE can be rewritten as a system of first order equations!


�

∂v ∂x∂w ∂x⎛ ⎝ ⎜ ⎞

⎠ ⎟ +

b a c a−1 0

⎛ ⎝ ⎜ ⎞

⎠ ⎟ ∂v ∂y∂w ∂y⎛ ⎝ ⎜ ⎞

⎠ ⎟ =

d a0

⎛ ⎝ ⎜ ⎞

⎠ ⎟

�

ux +Auy = s

�

a∂v∂x

+ b∂v∂y

+ c∂w∂y

= d

�

∂w∂x

−∂v∂y

= 0

In matrix form

�

u =vw⎛ ⎝ ⎜ ⎞

⎠ ⎟

Are there lines in the x-y plane, along which the solution is determined by an ordinary differential equation?

or


dvdx

=∂v∂x

+∂v∂y

∂y∂x

=∂v∂x

+α∂v∂y

α =∂y∂x

The total derivative is

where

If there are lines (determined by α) where the solution is governed by ODE’s, then it must be possible to rewrite the equations in such a way that the result contains only α and the total derivatives.

Rate of change of v with x, along the line y=y(x)


l1∂v∂x

+ba∂v∂y

+ca∂w∂y

⎛ ⎝ ⎜ ⎞

⎠ + l2

∂w∂x

−∂v∂y

⎛ ⎝ ⎜ ⎞

⎠ = l1

da

�

l1∂v∂x

+ α ∂v∂y

⎛

⎝ ⎜

⎞

⎠ ⎟ + l2

∂w∂x

+ α ∂w∂y

⎛

⎝ ⎜

⎞

⎠ ⎟ = l1

da

Is this ever equal to

Add the original equations:

For some l’s and α


l1∂v∂x

+ba∂v∂y

+ca∂w∂y

⎛ ⎝ ⎜ ⎞

⎠ + l2

∂w∂x

−∂v∂y

⎛ ⎝ ⎜ ⎞

⎠ = l1

da

l1∂v∂x

+ α∂v∂y

⎛ ⎝ ⎜ ⎞

⎠ + l2

∂w∂x

+α∂w∂y

⎛ ⎝ ⎜ ⎞

⎠ = l1

da

l1ca= l2α

l1ba− l2 = l1α

Compare the terms:

Therefore, we must have:


l1ba− l2 = l1α

l1ca= l2α

b a − α −1c a −α

⎛ ⎝ ⎜ ⎞

⎠ l1l2

⎛ ⎝ ⎜ ⎞

⎠ =00

⎛ ⎝ ⎜ ⎞ ⎠

Or, in matrix form:

Characteristic lines exist if:


b a − α −1c a −α

⎛ ⎝ ⎜ ⎞

⎠ l1l2

⎛ ⎝ ⎜ ⎞

⎠ =00

⎛ ⎝ ⎜ ⎞ ⎠ Rewrite

The original equation is:

�

∂v ∂x∂w ∂x⎛ ⎝ ⎜ ⎞

⎠ ⎟ +

b a c a−1 0

⎛ ⎝ ⎜ ⎞

⎠ ⎟ ∂v ∂y∂w ∂y⎛ ⎝ ⎜ ⎞

⎠ ⎟ =

d a0

⎛ ⎝ ⎜ ⎞

⎠ ⎟

�

b a −1c a 0⎛ ⎝ ⎜ ⎞

⎠ ⎟ l1l2

⎛ ⎝ ⎜ ⎞

⎠ ⎟ −

−α 00 −α

⎛ ⎝ ⎜ ⎞

⎠ ⎟ l1l2

⎛ ⎝ ⎜ ⎞

⎠ ⎟ =

00⎛ ⎝ ⎜ ⎞ ⎠ ⎟

AT

A

as

�

AT −αI( )l = 0


−αba− α⎛

⎝ ⎞ ⎠ +

ca

= 0

α =12a

b ± b2 − 4ac( )

b a −α −1c a −α

⎛

⎝⎜⎜

⎞

⎠⎟⎟

l1l2

⎛

⎝⎜⎜

⎞

⎠⎟⎟= 0

0⎛⎝⎜

⎞⎠⎟

The determinant is:

Or, solving for α

The equation has a solution only if the determinant is zero

�

AT −αI = 0


�

b2 − 4ac = 0�

b2 − 4ac > 0

�

b2 − 4ac < 0

Two real characteristics

One real characteristics

No real characteristics

α =12a

b ± b2 − 4ac( )


Examples


∂ 2 f∂x2

− c2∂ 2 f∂y2

= 0

a∂ 2 f∂x 2

+ b∂ 2 f∂x∂y

+ c∂ 2 f∂y2

= d

Examples

Comparing with the standard form

shows that a = 1; b = 0; c = −c 2; d = 0;

b2 − 4ac = 02 + 4 ⋅1 ⋅ c2 = 4c2 > 0

Hyperbolic


∂ 2 f∂x2

+∂ 2 f∂y2

= 0

a∂ 2 f∂x 2

+ b∂ 2 f∂x∂y

+ c∂ 2 f∂y2

= d


shows that a = 1; b = 0; c = 1; d = 0;

b2 − 4ac = 02 − 4 ⋅1 ⋅1 = −4 < 0

Elliptic

ExamplesComputational Fluid Dynamics

∂f∂x

= D∂ 2 f∂y2

a∂ 2 f∂x 2

+ b∂ 2 f∂x∂y

+ c∂ 2 f∂y2

= d


shows that a = 0; b = 0; c = −D; d = 0;

b2 − 4ac = 02 + 4 ⋅0 ⋅D = 0

Parabolic

Examples


∂ 2 f∂x2

− c2∂ 2 f∂y2

= 0 HyperbolicWave equation

∂f∂x

= D∂ 2 f∂y2

ParabolicDiffusion equation

∂ 2 f∂x2

+∂ 2 f∂y2

= 0 EllipticLaplace equation

ExamplesComputational Fluid Dynamics

α =12a

b ± b2 − 4ac( )α =∂y∂x

Hyperbolic Parabolic Elliptic


Summary


Why is the classification Important?

1.  Initial and boundary conditions2.  Different physics3.  Different numerical method apply


�

∂u∂t

+ u∂u∂x

+ v∂u∂y

= −1ρ∂P∂x

+µρ

∂ 2u∂x 2

+∂ 2u∂y2

⎛ ⎝ ⎜ ⎞

⎠ ⎟

Navier-Stokes equations

�

∂u∂x

+∂v∂y

= 0

Parabolic part

Hyperbolic part

Elliptic equation

SummaryComputational Fluid Dynamics

The Navier-Stokes equations contain three equation types that have their own characteristic behavior

Depending on the governing parameters, one behavior can be dominant

The different equation types require different solution techniques

For inviscid compressible flows, only the hyperbolic part survives

Summary


The Wave Equationand Advection

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

�

∂ 2 f∂t 2

− c 2 ∂2 f∂x 2

= 0

First write the equation as a system of first order equations

�

u = ∂f∂t; v = ∂f

∂x;

�

∂u∂t

− c 2 ∂v∂x

= 0

∂v∂t

− ∂u∂x

= 0

yielding

Introduce

from the pde

since

�

∂ 2 f∂t∂x

= ∂ 2 f∂x∂t

Wave equation


�

∂u∂t

− c 2 ∂v∂x

= 0

∂v∂t

− ∂u∂x

= 0

∂u∂t∂v∂t

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

+ 0 −c2

−1 0⎡

⎣⎢

⎤

⎦⎥

∂u∂x∂v∂x

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

= 0

det AT −αI( ) = det −α −1−c2 −α

⎡

⎣⎢⎢

⎤

⎦⎥⎥= α 2 − c2( ) = 0

�

⇒α = ±c

α =12a

b ± b2 − 4ac( )

�

b = 0; a =1; c = −c 2We can also use

with

To find the characteristics

Wave equationComputational Fluid Dynamics

t

x

P

Two characteristic lines

�

dtdx

= +1c; dtdx

= −1c

�

dtdx

= +1c

�

dtdx

= −1c

Wave equation


l1∂u∂t

− c2∂v∂x

= 0⎛⎝⎜

⎞⎠⎟

+l2∂v∂t

−∂u∂x

= 0⎛⎝⎜

⎞⎠⎟

−α −1−c2 −α

⎡

⎣⎢⎢

⎤

⎦⎥⎥

l1l2

⎡

⎣⎢⎢

⎤

⎦⎥⎥= 0

�

⇒α = ±cTo find the solution we need to find the eigenvectors

Take l1=1

�

−c l1 − l2 = 0 ⇒ l2 = −cFor

�

α = +c

�

+c l1 − l2 = 0 ⇒ l2 = +cFor

�

α = −c


�

1 ∂u∂t

− c 2 ∂v∂x

= 0⎛ ⎝ ⎜

⎞ ⎠ ⎟ − c

∂v∂t

− ∂u∂x

= 0⎛ ⎝ ⎜

⎞ ⎠ ⎟ Add the

equations�

l1 =1 l2 = −cFor

�

α = +c

For

�

α = −c

�

l1 =1 l2 = +c�

∂u∂t

+ c ∂u∂x

⎛ ⎝ ⎜

⎞ ⎠ ⎟ − c

∂v∂t

+ c ∂v∂x

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = 0

�

dudt

− c dvdt

= 0 on dxdt

= +c

�

dudt

+ c dvdt

= 0 on dxdt

= −c

Relation between the total derivative on the characteristicSimilarly:

Wave equation


For constant c�

dudt

− c dvdt

= 0 on dxdt

= +c

�

dudt

+ c dvdt

= 0 on dxdt

= −c

�

dr1dt

= 0 on dxdt

= +c where r1 = u − cv

�

dr2dt

= 0 on dxdt

= −c where r2 = u + cv

r1 and r2 are called the Rieman invariants


The general solution can therefore be written as:

�

f x, t( ) = r1 x − ct( ) + r2 x + ct( )

�

r1 x( ) = ∂f∂t

− c ∂f∂x

⎡ ⎣ ⎢

⎤ ⎦ ⎥ t= 0

r2 x( ) = ∂f∂t

+ c ∂f∂x

⎡ ⎣ ⎢

⎤ ⎦ ⎥ t= 0

where

Can also be verified by direct substitution

Wave equation


Domain of Influence

Domain of Dependence

t

x

P

Two characteristic lines

�

dtdx

= +1c; dtdx

= −1c

�

dtdx

= +1c

�

dtdx

= −1c

Wave equation-generalComputational Fluid Dynamics

Ill-posed problems

Computational Fluid DynamicsIll-posed Problems

∂ 2 f∂t2

= −∂ 2 f∂x2

Consider the initial value problem:

∂ 2 f∂t2

+∂ 2 f∂x2

= 0

This is simply Laplace’s equation

which has a solution if ∂f/∂t or f are given on the boundaries


Here, however, this equation appeared as an initial value problem, where the only boundary conditions available are at t = 0. Since this is a second order equation we will need two conditions, which we may assume are that f and ∂f/∂t are specified at t =0.

Ill-posed Problems


Look for solutions of the type:

�

f = ak (t)eikx

�

d2akdt 2

= k 2ak

Substitute into:∂ 2 f∂t2

+∂ 2 f∂x2

= 0

to get:

Ill-posed Problems

�

f x, t( ) = ak (t)eikx

k∑

The general solution can be written as:where the a’s depend on the initial conditions


)0( ;)0( ikaa�

d2akdt 2

= k 2ak

determined by initial conditions

General solution

�

ak (t) = Aekt + Be−kt

BA,

Therefore: ∞→)(ta as ∞→t

Ill-posed Problem

Ill-posed Problems

Generally, both A and B are non-zero

Computational Fluid DynamicsIll-posed Problems

Long wave with short wave perturbations


∂f∂t

= D∂2 f∂x2 ; D < 0

has solutions with unbounded growth rate for high wave number modes and is therefore an ill-posed problem

Similarly, it can be shown that the diffusion equation with a negative diffusion coefficient

Ill-posed Problems


Ill-posed problems generally appear when the initial or boundary data and the equation type do not match.

Frequently arise because small but important higher order effects have been neglected

Ill-posedness generally manifests itself in the exponential growth of small perturbations so that the solution does not “depend continuously on the initial data”

Inviscid vortex sheet rollup, multiphase flow models and some viscoelastic constitutive models are examples of problems that exhibit ill-posedness.

Ill-posed ProblemsComputational Fluid Dynamics

Stability in terms of Fluxes


We can do a similar analysis for the diffusion equation

ddt(hf j) = Fj−1/ 2 − Fj+1/ 2 Fj+1/2 = −D

fj+1 − f jh

∂f∂t+∂F∂x

= 0

Thus, we update:

The finite volume approximation is

F = −D ∂f∂x

f jn+1 = f j

n +ΔtDh2( f j+1

n − 2 f jn − f j−1

n )

where


f j−1 f j f j+1

Consider the following initial conditions:

1

During one time step, ∆tD/h of f flows into cell j, but nothing flow out of it. Eventually cell j-1 becomes empty and cell j becomes full.

Stability in terms of fluxes

Fj+1/2 = −Dfj+1 − f jh

= 0Fj−1/2 = −Dfj − f j−1h

=Dh


f j−1n+1 f j

n+1 f j+1n+1

1

It seems reasonable to limit ∆t in such a way that we stop when both cells are equally full.


Fj+1/2 = 0Fj−1/2 =Dh

f j−1n +

ΔtDh2( f j

n − 2 f j−1n + f j−2

n ) = f jn +

ΔtDh2( f j+1

n − 2 f jn + f j−1

n )

Since fj-2n=fj-1

n =1 and fjn=fj+1

n =0 we get:

1+ ΔtDh2(0− 2+1) = 0+ ΔtD

h2(0− 0+1) or: ΔtD

h2= 2 as maximum ∆t


Advection Equation:

ddt(hf j) = Fj−1/ 2 − Fj+1/ 2

fj+1 / 2 ≈ 12 ( f j+1 + f j)

Fj+1/2 =Ufj+1/2

∂f∂t+∂F∂x

= 0

Approximate the fluxes by the average values of f to the left and right.

The finite volume approximation is

F =Uf


The fluxes are therefore

Fj+1/2 =12U( f j+1 + f j )

So that

Then approximate the time derivative by

ddt

fj ≈1Δt( fj

n+1 − fjn )

f jn+1 = f j

n −Δth(Fj+1/2

n −Fj−1/2n )


fj+1

Consider the following initial conditions:

1fj−1 fj

Fj−1/ 2 =U2( fj−1

n + fjn ) =1.0 Fj+1/ 2 =

U2( fj

n + f j+1n ) = 0.5

fjn+1 = fj

n −Δth(Fj+1/ 2

n − Fj−1/ 2n ) = 1.0 − 0.5(0.5 −1) =1.25

So cell j will overflow immediately!

By considering the fluxes, it is easy to see why the centered difference approximation is always unstable.


U = 1Δth= 0.5

Computational Fluid Dynamicshttp://www.nd.edu/~gtryggva/CFD-Course/

Summary

Characteristics for 1st order PDEs

Classification of second order PDEs

The wave equation and advection

Ill posed problems

Fluxes and stability


In the next several lectures we will discuss numerical solutions techniques for each class:

Advection and Hyperbolic equations, including solutions of the Euler equations

Parabolic equations

Elliptic equations

Then we will consider advection/diffusion equations and the special considerations needed there.

Finally we will return to the full Navier-Stokes equations

Summary

Documents

Computational Fluid Theory of Dynamics Partial ...gtryggva/CFD-Course2017/Lecture-6-2017.pdf · partial differential equations Computational Fluid Dynamics a ∂f ∂x +b ∂f ∂y