Computational Fluid Dynamics
http://www.nd.edu/~gtryggva/CFD-Course/!
Theory of!
Partial Differential
Equations!
Computational Fluid Dynamics
Outline!
• 
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!
Grétar Tryggvason!
Spring 2011!
Computational Fluid Dynamics
Examples of equations!
∂f
∂f
+U =0
∂t
∂x
∂f
∂2 f
−D 2 =0
∂t
∂x
2
∂ f
∂2 f
− c2 2 = 0
2
∂t
∂x
∂ f ∂ f
+
=0
∂x 2 ∂y 2
2
Navier-Stokes equations!
Diffusion part!
Advection!
Evolution
in time!
Diffusion!
Wave
propagation!
2
Laplace equation!
Computational Fluid Dynamics
Definitions!
• 
• 
∂u
∂u
∂u
1 ∂P µ ⎛⎜ ∂ 2 u ∂ 2 u ⎞⎟
+ u +v =−
+
+
∂t
∂x
∂y
ρ ∂x ρ ⎝ ∂x 2 ∂y 2 ⎠
Advection part!
∂u ∂v
+
=0
∂x ∂y
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.!
x
• 
Computational Fluid Dynamics
∂f
∂f
+y
= f,
∂x
∂y
∂f ∂f
+ + 2 xf = 0
∂x ∂y
Quasi-linear if it is true for the partial derivatives of
highest order. !!
2
f
∂ 2 f ⎛ ∂f ⎞
+⎜ ⎟ = f ,
∂x 2 ⎜⎝ ∂y ⎟⎠
x2
∂f
∂f
+ y2
= f2
∂x
∂y
Quasi-linear first order!
partial differential equations!
Computational Fluid Dynamics
Consider the quasi-linear first order equation!
∂f
∂f
a +b =c
∂x
∂y
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The solution of this equations defines a single valued
surface f(x,y) in three-dimensional space:!
where the coefficients are functions of x,y, and f, but not
the derivatives of f:
f
a = a(x, y, f )
b = b(x, y, f )
c = c(x, y, f )
y
f=f(x,y)
x
Computational Fluid Dynamics
The original equation and the conditions for a small
change can be rewritten as:!
An arbitrary change in f is given by:!
∂f
∂f
df =
dx + dy
∂x
∂y
f
f
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a
df
∂f
∂f
+b =c
∂x
∂y
df =
dy
⎛ ∂f ∂f
⎞
⎜ , ,− 1⎟ ⋅ (dx, dy, df ) = 0
⎝ ∂x ∂y
⎠
Normal to the surface!
f=f(x,y)
x
⎛ ∂f ∂f
⎞
⎜ , ,− 1⎟ ⋅ (a,b, c) = 0
⎝ ∂x ∂y
⎠
∂f
∂f
dx +
dy ⇒
∂x
∂y
dx
y
⇒
Both (a,b,c) and (dx,dy,df) are in the surface!!
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The normal vector to the curve f=f(x,y)
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Picking the displacement in the direction of (a,b,c)!
⇒
1!
(dx, dy,df ) = ds (a,b, c)
f=f(x,y)
f
-1!
∂f
∂x
x
Separating the components!
dx
dy
⇒
= a;
= b;
ds
ds
dx a
=
dy b
Same arguments in the y-direction. Thus
⎛ ∂f ∂ f
⎞
n = ⎜ , ,−1⎟
⎝ ∂x ∂ y ⎠
df
= c;
ds
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dx
= a;
ds
The three equations specify lines in the x-y plane!
dx
= a;
ds
dy
= b;
ds
df
= c;
ds
y
dy
= b;
ds
df
= c;
ds
x(s); y(s); f (s)
Characteristics!
Given the initial conditions:!
x = x(s, to ); y = y(s,t o );
f = f (s,t o );
x(0); y(0); f (0)
the equations can be integrated in time!
slope!
dx a
=
dy b
x
Computational Fluid Dynamics
Consider the linear advection equation!
∂f
∂f
+U
=0
∂t
∂x
dx
= U;
ds
f (x,t) = f t= 0 ( x − Ut )
Graphically: !
t
f
The characteristics are given by:!
dt
= 1;
ds
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df
= 0;
ds
f
x
or!
dx
= U; df = 0;
dt
Notice that these results are specific for: !
∂f
∂f
+U
=0
∂t
∂x
Which shows that the solution moves along straight
characteristics without changing its value!
Computational Fluid Dynamics
The solution is therefore:!
f (x,t) = g( x − Ut ) where !
Discontinuous initial data!
g(x) = f (x,t = 0)
This can be verified by direct substitution: !
Set ! η(x,t) = x − Ut
Then!
∂f ∂g ∂η ∂g
=
=
(−U ) and!
∂t ∂η ∂t ∂η
Computational Fluid Dynamics
∂f ∂g ∂η ∂g
=
=
(1)
∂x ∂η ∂x ∂η
∂f
∂f
+U =0
∂t
∂x
f (x,t) = g( x − Ut )
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
Substitute into the original equation!
∂f
∂f ∂g
∂g
+U =
(−U ) + U = 0
∂t
∂x ∂η
∂η
f
x
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Add a source !
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∂f
∂f
+U
=−f
∂t
∂x
Consider a nonlinear (quasi-linear) advection equation!
∂f
∂f
+f
=0
∂t
∂x
The characteristics are given by:!
dt
= 1;
ds
dx
= U;
ds
or! dx = U;
dt
df
=−f
dt
⇒ f = f (0)e −t
Moving wave with !
decaying amplitude!
The characteristics are given by:!
df
= − f;
ds
dt
dx
df
= 1;
= f;
= 0;
ds
ds
ds
or!
t
f
f
x
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dx
= f;
dt
df = 0;
The slope of the characteristics depends on the
value of f(x,t).
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Why unphysical solutions?!
t
Multi-valued solution
t3
- Because mathematical equation neglects !
some physical process (dissipation)!
∂f
∂f
∂2 f
+f
−ε 2 = 0
∂t
∂x
∂x
t2
Burgers Equation!
Additional condition is required to pick out the physically!
Relevant solution!
t1
f
x
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Correct solution is expected from Burgers equation
with ε → 0
Entropy Condition
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Constructing physical solutions!
∂f
∂f
+ f
=0
∂t
∂x
In 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!
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Computational Fluid Dynamics
First write the second order PDE
as a system of first order equations!
Consider !
∂ f
∂ f
∂ f
+b
+c 2 = d
∂x 2
∂x∂y
∂y
2
a
2
2
Define!
where !
v=
and!
w=
∂f
∂y
then!
a = a(x, y, f , f x , f y )
a
b = b(x, y, f , f x , f y )
c = c(x, y, f , f x , f y )
∂2 f
∂2 f
∂2 f
+b
+c 2 = d
∂x 2
∂x∂y
∂y
∂2 f
∂2 f
=
∂x∂y ∂y∂x
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Thus!
a
∂v
∂v
∂w
+b +c
=d
∂x
∂y
∂y
∂w ∂v
−
=0
∂x ∂y
Computational Fluid Dynamics
∂2 f
∂2 f
∂2 f
+b
+c 2 = d
∂x 2
∂x∂y
∂y
a
∂v
∂v
∂w
+b +c
=d
∂x
∂y
∂y
∂w ∂v
−
=0
∂x ∂y
In matrix form!
Is equivalent to:!
∂v
∂v
∂w
a +b +c
=d
∂x
∂y
∂y
∂w ∂v
−
=0
∂x ∂y
v=
a
The second equation is obtained from!
d = d(x, y, f , f x , f y )
where!
∂f
∂x
∂f
∂x
w=
Any high-order
PDE can be
rewritten as a
system of first
order equations!!
or!
ux + Auy = s
⎛v ⎞
u=⎜ ⎟
⎝ w⎠
Are there lines in the x-y plane, along which the
solution is determined by an ordinary differential
equation?!
∂f
∂y
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The total derivative is!
dv ∂v ∂v ∂y ∂v
∂v
=
+
=
+α
dx ∂x ∂y ∂x ∂x
∂y
⎛ ∂v ∂x ⎞ ⎛ b a c a⎞⎛ ∂v ∂y ⎞ ⎛ d a⎞
⎜
⎟+ ⎜
⎟⎜
⎟ =⎜
⎟
⎝∂w ∂x ⎠ ⎝ −1
0 ⎠⎝ ∂w ∂y ⎠ ⎝ 0 ⎠
Add the original equations:!
where!
∂y
α=
∂x
Rate of change of v with x, along the line y=y(x)!
⎛ ∂v b ∂v c ∂w ⎞
⎛ ∂w ∂v ⎞
d
l1 ⎜ +
+
+l ⎜
−
=l
⎝ ∂x a ∂y a ∂y ⎠ 2 ⎝ ∂x ∂y ⎠ 1 a
Is this ever equal to!
If there are lines (determined by α) where the
solution is governed by ODEʼs, then it must be
possible to rewrite the equations in such a way
that the result contains only α and the total
derivatives. !
⎛ ∂v
⎛ ∂w
∂v ⎞
∂w ⎞
d
l1⎜ + α ⎟ + l2 ⎜
+ α ⎟ = l1
∂y ⎠
∂y ⎠
a
⎝ ∂x
⎝ ∂x
For some lʼs and α
Computational Fluid Dynamics
Compare the terms:!
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Characteristic lines exist if:!
b
− l = lα
a 2 1
c
l1 = l2α
a
⎛ ∂v b ∂v c ∂w ⎞
⎛ ∂w ∂v ⎞
d
l1 ⎜ +
+
+l ⎜
−
=l
⎝ ∂x a ∂y a ∂y ⎠ 2 ⎝ ∂x ∂y ⎠ 1 a
⎛ ∂v
⎛ ∂w
∂v ⎞
∂w ⎞
d
l1 ⎜ + α
+l ⎜
+α
=l
⎝ ∂x
∂y ⎠ 2 ⎝ ∂x
∂y ⎠ 1 a
Or, in matrix form:!
⎛⎜ b a − α
⎝ ca
b
l1 − l2 = l1α
a
c
l1 = l2α
a
Therefore, we
must have:!
Computational Fluid Dynamics
The original equation is:!
A!
Rewrite!
as!
⎛⎜b a −1⎞⎟⎛⎜ l1 ⎞⎟ ⎛⎜ −α 0 ⎞⎟⎛⎜ l1 ⎞⎟ ⎛⎜ 0⎞⎟
−
=
⎝c a 0 ⎠⎝ l2 ⎠ ⎝ 0 −α ⎠⎝ l2 ⎠ ⎝ 0⎠
−1 ⎞ ⎛ l1 ⎞ ⎛ 0⎞
⎜
=⎜
−α ⎠ ⎝ l2 ⎠ ⎝ 0⎠
T
(A
− α I )l = 0
Computational Fluid Dynamics
(
−1 ⎞ ⎛ l1 ⎞ ⎛ 0 ⎞
⎟⎜
⎟=
−α ⎟⎠ ⎜⎝ l2 ⎟⎠ ⎜⎝ 0 ⎟⎠
The determinant is:! A − αI = 0
T
1
b ± b 2 − 4ac
2a
Computational Fluid Dynamics
⎛ b a −α
⎜
⎜⎝
ca
⎛⎜ b a − α
⎝ ca
α=
−1 ⎞ ⎛⎜ l1 ⎞ ⎛⎜ 0⎞
=
−α ⎠ ⎝ l2 ⎠ ⎝ 0⎠
The equation has a solution only if the determinant is zero!
⎛ ∂v ∂x ⎞ ⎛ b a c a⎞⎛ ∂v ∂y ⎞ ⎛ d a⎞
⎜
⎟+ ⎜
⎟⎜
⎟ =⎜
⎟
⎝∂w ∂x ⎠ ⎝ −1
0 ⎠⎝ ∂w ∂y ⎠ ⎝ 0 ⎠
AT!
l1
)
b 2 − 4ac > 0
Two real characteristics!
b 2 − 4ac = 0
One real characteristics!
b 2 − 4ac < 0
No real characteristics!
c
⎛b
− α ⎝ − α ⎞⎠ + = 0
a
a
Or, solving for α !
α=
(
1
b ± b 2 − 4ac
2a
)
Computational Fluid Dynamics
Examples!
Computational Fluid Dynamics
Examples!
Computational Fluid Dynamics
Examples!
2
∂2 f
2 ∂ f
=0
2 −c
∂x
∂y 2
∂2 f ∂ 2 f
+
=0
∂x 2 ∂y 2
Comparing with the standard form!
Comparing with the standard form!
∂2f
∂2 f
∂2f
a 2 +b
+c 2 = d
∂x
∂x∂y
∂y
shows that! a = 1;
b = 0; c = −c ; d = 0;
2
b − 4ac = 0 + 4 ⋅1⋅ c = 4c > 0
2
2
2
2
a
shows that!
∂2f
∂2 f
∂2f
+c 2 = d
2 +b
∂x
∂x∂y
∂y
a = 1; b = 0; c = 1; d = 0;
b − 4ac = 0 − 4 ⋅1⋅1 = −4 < 0
2
2
Hyperbolic!
Elliptic!
Computational Fluid Dynamics
Examples!
∂f
∂2f
=D 2
∂x
∂y
Computational Fluid Dynamics
Examples!
Wave equation!
2
∂2 f
2 ∂ f
=0
2 −c
∂x
∂y 2
Hyperbolic!
Diffusion equation!
∂f
∂2f
=D 2
∂x
∂y
Parabolic!
Laplace equation!
∂2 f ∂ 2 f
+
=0
∂x 2 ∂y 2
Elliptic!
Comparing with the standard form!
a
∂2f
∂2 f
∂2f
+c 2 = d
2 +b
∂x
∂x∂y
∂y
shows that! a = 0;
b = 0; c = − D; d = 0;
b − 4ac = 0 + 4 ⋅0 ⋅ D = 0
2
2
Parabolic!
Computational Fluid Dynamics
α=
∂y
∂x
α=
(
1
b ± b 2 − 4ac
2a
Computational Fluid Dynamics
)
Wave Equation!
Hyperbolic!
Parabolic!
Elliptic!
Computational Fluid Dynamics
Wave equation!
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Wave equation!
To find the characteristics!
∂u 2 ∂v
−c
=0
∂t
∂x
∂v ∂u
− =0
∂t ∂x
∂ f
∂ f
− c2 2 = 0
∂t 2
∂x
2
2
First write the equation as a system of first order equations!
Introduce!
yielding!
u=
∂f
∂f
; v= ;
∂t
∂x
∂u 2 ∂v
−c
=0
∂t
∂x
∂v ∂u
− =0
∂t ∂x
since!
∂2 f ∂2 f
=
∂t∂x ∂x∂t
α=
−1 ⎤
2
2
⎥= α −c =0
−α ⎥⎦
)
We can also use!
(
)
1
b ± b 2 − 4ac with!
2a
Computational Fluid Dynamics
Wave equation!
Two characteristic lines!
⎤
⎡ ∂u ⎤
⎥ ⎡
⎢
⎥
2 ⎤
⎥ + ⎢ 0 −c ⎥ ⎢ ∂ x ⎥ = 0
⎥ ⎣ −1 0 ⎦ ⎢ ∂ v ⎥
⎥
⎢
⎥
⎦
⎣ ∂x ⎦
∂u
∂t
∂v
∂t
⎡ −α
det A T − α I = det ⎢
2
⎢⎣ −c
⇒ α = ±c
(
from the pde!
⎡
⎢
⎢
⎢
⎢
⎣
(
b = 0; a = 1; c = −c 2
Computational Fluid Dynamics
Wave equation!
dt
1 dt
1
=+ ;
=−
dx
c dx
c
t
dt
1
=+
dx
c
P
dt
1
=−
dx
c
To find the solution we need
to find the eigenvectors!
∂v
⎛ ∂u
⎞
l1 ⎜
− c2
= 0⎟
⎝ ∂t
⎠
∂x
⎛ ∂ v ∂u
⎞
+l2 ⎜ −
= 0⎟
⎝ ∂t ∂ x
⎠
⇒ α = ±c
⎡ −α
⎢
2
⎢⎣ −c
−1
−α
⎤ ⎡ l1 ⎤
⎥=0
⎥⎢
⎥⎦ ⎢⎣ l2 ⎥⎦
Take l1=1!
For! α = +c
−c l1 − l2 = 0 ⇒ l2 = −c
For! α = −c
+c l1 − l2 = 0 ⇒ l2 = +c
x
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Wave equation!
For!
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Wave equation!
α = +c
l1 = 1 l2 = −c
⎛ ∂u 2 ∂v
⎞ ⎛ ∂v ∂u
⎞
1⎜ − c
= 0⎟ − c⎜ − = 0⎟
⎝ ∂t
⎠ ⎝ ∂t ∂x
⎠
∂x
⎛ ∂u
∂ u ⎞ ⎛ ∂v
∂v ⎞
⎜ + c ⎟ − c⎜ + c ⎟ = 0
⎝ ∂t
∂ x ⎠ ⎝ ∂t
∂x ⎠
du
dv
−c
= 0 on
dt
dt
Similarly:!
For! α = −c
dx
= +c
dt
l1 = 1 l2 = +c
du
dv
dx
+c
= 0 on
= −c
dt
dt
dt
)
Add the
equations!
Relation
between the
total derivative
on the
characteristic!
du
dv
−c
= 0 on
dt
dt
dx
= +c
dt
du
dv
+c
= 0 on
dt
dt
dx
= −c
dt
For constant c!
dr1
dx
= 0 on
= +c where r1 = u − cv
dt
dt
dr2
dx
= 0 on
= −c where r2 = u + cv
dt
dt
r1 and r2 are called the Rieman invariants!
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Wave equation!
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Wave equation-general!
The general solution can
therefore be written as:!
f ( x,t ) = r1 ( x − ct ) + r2 ( x + ct )
where!
⎡∂f
∂f ⎤
r1 ( x ) = ⎢ − c ⎥
⎣ ∂t
∂x ⎦t= 0
⎡∂f
∂f ⎤
r2 ( x ) = ⎢ + c ⎥
⎣ ∂t
∂x ⎦t= 0
Can also be verified by direct substitution!
Computational Fluid Dynamics
Two characteristic lines!
dt
1 dt
1
=+ ;
=−
dx
c dx
c
t
dt
1
=+
dx
c
P
dt
1
=−
dx
c
x
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Why is the classification Important?!
Summary!
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Summary!
1.  Initial and boundary conditions!
2.  Different physics!
3.  Different numerical method apply!
Computational Fluid Dynamics
Summary!
Navier-Stokes equations!
Parabolic part!
∂u
∂u
∂u
1 ∂P µ ⎛⎜ ∂ 2 u ∂ 2 u ⎞⎟
+ u +v =−
+
+
∂t
∂x
∂y
ρ ∂x ρ ⎝ ∂x 2 ∂y 2 ⎠
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!
Hyperbolic part!
∂u ∂v
+
=0
∂x ∂y
Elliptic equation!
For inviscid compressible flows, only the hyperbolic part
survives!
Computational Fluid Dynamics
Summary!
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!