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Computational Fluid Dynamics
Computational Fluid Dynamics
http://www.nd.edu/~gtryggva/CFD-Course/
Grétar Tryggvason
Lecture 6February 1, 2017
Computational Fluid Dynamics
Theory ofPartial Differential
Equations
http://www.nd.edu/~gtryggva/CFD-Course/
Computational Fluid Dynamics
• 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
Computational Fluid Dynamics
�
∂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
Computational Fluid Dynamics
• 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 =∂∂+
∂∂=⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂+
∂∂
Computational Fluid Dynamics
Quasi-linear first order partial differential equations
Computational Fluid Dynamics
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:
Computational Fluid Dynamics
The solution of this equations defines a single valued surface f(x,y) in three-dimensional space:
f=f(x,y)
f
x
y
Computational Fluid Dynamics
dyyf
dxxf
df∂∂+
∂∂=
An arbitrary change in f is given by:
f=f(x,y)
f
x
y
f
dxdy
df
Computational Fluid Dynamics
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
Computational Fluid Dynamics
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!
Computational Fluid Dynamics
(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
Computational Fluid Dynamics
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
Computational Fluid Dynamics
x(s); y(s); f (s)
x
y
x(0); y(0); f (0)
dxds
= a;dyds
= b;dfds
= c;
Computational Fluid Dynamics
∂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
Computational Fluid Dynamics
∂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( )
Computational Fluid Dynamics
�
∂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
Computational Fluid Dynamics
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( )
Computational Fluid Dynamics
�
∂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
The characteristics are given by:
or
Moving wave with decaying amplitude
Computational Fluid Dynamics
�
∂f∂t
+ f∂f∂x
= 0
�
dtds
=1; dxds
= f ; dfds
= 0;
�
dxdt
= f ; df = 0;
Consider a nonlinear (quasi-linear) advection equation
The characteristics are given by:
or
The slope of the characteristics depends on the value of f(x,t).
Computational Fluid Dynamics
t
fx
t3
t2
t1
Multi-valued solution
Computational Fluid Dynamics
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
Computational Fluid Dynamics
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
Computational Fluid Dynamics
Quasi-linear Second order partial differential equations
Computational Fluid Dynamics
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
Computational Fluid Dynamics
�
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
Computational Fluid Dynamics
�
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!
Computational Fluid Dynamics
�
∂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
Computational Fluid Dynamics
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)
Computational Fluid Dynamics
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
Computational Fluid Dynamics
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:
Computational Fluid Dynamics
l1ba− l2 = l1α
l1ca= l2α
b a − α −1c a −α
⎛ ⎝ ⎜ ⎞
⎠ l1l2
⎛ ⎝ ⎜ ⎞
⎠ =00
⎛ ⎝ ⎜ ⎞ ⎠
Or, in matrix form:
Characteristic lines exist if:
Computational Fluid Dynamics
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
Computational Fluid Dynamics
−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
Computational Fluid Dynamics
�
b2 − 4ac = 0�
b2 − 4ac > 0
�
b2 − 4ac < 0
Two real characteristics
One real characteristics
No real characteristics
α =12a
b ± b2 − 4ac( )
Computational Fluid Dynamics
Examples
Computational Fluid Dynamics
∂ 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
Computational Fluid Dynamics
∂ 2 f∂x2
+∂ 2 f∂y2
= 0
a∂ 2 f∂x 2
+ b∂ 2 f∂x∂y
+ c∂ 2 f∂y2
= d
Comparing with the standard form
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
Comparing with the standard form
shows that a = 0; b = 0; c = −D; d = 0;
b2 − 4ac = 02 + 4 ⋅0 ⋅D = 0
Parabolic
Examples
Computational Fluid Dynamics
∂ 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
Computational Fluid Dynamics
Summary
Computational Fluid Dynamics
Why is the classification Important?
1. Initial and boundary conditions2. Different physics3. Different numerical method apply
Computational Fluid Dynamics
�
∂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
Computational Fluid Dynamics
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
Computational Fluid Dynamics
�
∂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
Computational Fluid Dynamics
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
Wave equationComputational Fluid Dynamics
�
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
Computational Fluid Dynamics
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
Wave equationComputational Fluid Dynamics
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
Computational Fluid Dynamics
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
Computational Fluid Dynamics
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
Computational Fluid Dynamics
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
Computational Fluid Dynamics
)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
Computational Fluid Dynamics
∂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
Computational Fluid Dynamics
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
Computational Fluid Dynamics
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
Computational Fluid Dynamics
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
Computational Fluid Dynamics
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.
Stability in terms of fluxes
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
Computational Fluid Dynamics
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
Computational Fluid Dynamics
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 )
Computational Fluid Dynamics
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.
Stability in terms of fluxes
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
Computational Fluid Dynamics
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