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Relativistic Hydrodynamics BAS/FS17/LectNote-5&6
________________________________ LN_5&6 / S. 1
Taylor expansion:
Let ƒ(x) be an infinitely differentiable real function.
At any point x in the neighbourhood of x=x0, the function ƒ(x)
can be represented as a power series of the following form:
(n)2 30 0 0 0
0 0 0 0 0
0
f (x ) f ( ) f ( ) f ( )f(x)= ( ) f( ) ( ) ( ) ( )
! 1! 2! 3!
n
n
x x xx x x x x x x x x
n
where n! stands for the factorial of n and f(n)(x0) for the n-th derivative of “ f ” at x =x0.
Examples:
The function f(x)= ex in the neighborhood of x=0 has the following Taylor series:
2 3 4x x x1+x +
2 6 24
xe
Q:
3 5
If the TE of f(x)= sin(x) around 0 is: sin(x) = x -3! 5!
What is the TE of f(x)= around x=0?sin( )
x
x xx
e
x
f(x)
f(a)
x
X0
Hint: Let sin( )
xe
x has an expansion of the form:
2 3
1 2 3 3
3 52 3
1 2 3 3
c +c x +c x +c xsin( )
c +c x +c x +c x x-3! 5!
x
x
e
x
x xe
Relativistic Hydrodynamics BAS/FS17/LectNote-5&6
________________________________ LN_5&6 / S. 2
Interpolation techniques
By interpolation we mean a procedure for approximating the value of the function at the
point x which is located inside the interval.
By extrapolation
we mean a procedure for approximating the value of the function f(x) at an exterior point
:
Linear interpolation can be used for both procedures:
2 1 1
2 1 1
2 12 1 1
1
11 2 1
2 1
y(x ) - y(x ) y(x) - y(x ) =
y(x ) - y(x ) = y(x) - y(x )
y(x) = y(x ) - y(x ) - y(x )
x x x x
x x
x x
x x
x x
1x
2xx
(
1x
2x
f(X1) f(X2) F = ?
f(X1) f(X2) f = ?
1 2 1
2
If for x < x <x the functions f (x )
and f (x ) are used to evaluate f (x),
then the procedure is called interpolation.
1 2 1 2
1 2
If for x < x < x or x < x < x the
functions f (x ) and f (x ) are used to
evaluate f (x), then the procedure is
called EXTRA-polation.
Relativistic Hydrodynamics BAS/FS17/LectNote-5&6
________________________________ LN_5&6 / S. 3
0 0 1 1
Polynomial interpolation:
Assume we are given the following n+1 points:
(x ,y ), (x ,y ), , (x ,y ).
With the help of these points, we want to
have an approximation for the additional point (x, y).
The s
n n
n n-1
n n-1 0
trategy here is to construct a polynomial of degree n, such that
y(x) = a x + a x + ... a .
As this polynomial goes through the above n+1 points, it therefore must
satisfy the following equa
1
n 0 n-1 0 0 01
1 0 0 0
n 1 n-1 1 0 1 1
1 1 1
1
1n n-1 0 n
n
`
tions:
a x + a x + ... a = yx x x 1
a x + a x + ... a = yx x x 1
1
a x + a x + ... a = yx x x 1
n n
n n
n n
n n
n n
n nn n
n n
0
1 1
0
X
n
n
n
a y
a y
a y
0 0
1 1
2 2
2 : Assume we are given the following points:
(x ,y ) = (1,2) Use these points to find a
(x ,y ) = (2,3) polynomial approximation for
y at x = 5/2? (x ,y ) = (3,1)
Q
Q1: what does it mean, when the matrix X is singular? Give examples!
Relativistic Hydrodynamics BAS/FS17/LectNote-5&6
________________________________ LN_5&6 / S. 4
0 0 1 1
Lagrange interpolation:
Assume we are given the following points:
(x ,y ), (x ,y ), , (x ,y ).
Based on these points, Lagrange suggested the following approximation
for y at x:
y(x) =
n n
j jN
j j kj=0 j
j k
31 20
0 1 0 2 0 3
y =y(x )
y = y (x), where x - x(x)
x -x
Example
are the following set of points (x,y):
1 2 3 4
8 7 6 5
x - xx - x x - x 1(x) ( 2
x -x x -x x -x 6
n
k j
Given
x
y
x
0 321
1 0 1 2 1 3
0 312
2 0 2 1 2 3
0 13
3 0 3 1
)( 3)( 4)
x - x x - xx - x 1(x) ( 1)( 3)( 4)
x -x x -x x -x 2
x - x x - xx - x 1(x) ( 1)( 2)( 4)
x -x x -x x -x 2
x - x x - x(x)
x -x x -x
x x
x x x
x x x
2
3 2
x - x 1 ( 1)( 2)( 3)
x -x 6
1 1 y(x) = 8 ( 2)( 3)( 4) 7 ( 1)( 3)( 4)
6 2
1 1 + 6 ( 1)( 2)( 4) 5 ( 1)( 2)( 3)
2 6
x x x
x x x x x x
x x x x x x
Relativistic Hydrodynamics BAS/FS17/LectNote-5&6
________________________________ LN_5&6 / S. 5
j j-1 j j-1finite space
x
j j-1
23
j-1 j-1 j j x xx
Backward difference (u > 0):
T(x ) - T(x ) T - TT ΔTT =
x Δx x - x h
And how good is this approxation? Taylor expansion
hT = T(x )= T(x -h) = T h T + T + O(h )
2
Subtit
j
2j j-1 x xx
x
x=x
Truncation error
ute this expression:
T - T T (T- hT +(h /2) T )ΔT= T + O(h)
Δx h h
The scheme is first order accurate also.
Finite difference discretization:
How to represent
Tu
x in finite space?
j+1 j j+1 jfinite space
x
j+1 j
23
j+1 j+1 j j x xx
Forward difference (u < 0):
T(x ) - T(x ) T - TT ΔTT = =
x Δx x - x h
But how good is this approxation? Taylor expansion
hT = T(x )= T(x h) = T h T + T + O(h )
2
Subtitut
j
Truncation e
2j+1 j x xx
x
x=x
rror
ing this expression:
T - T T hT +(h /2) T TΔT= T + O(h)
Δx h h
The scheme is first order accurate.
Relativistic Hydrodynamics BAS/FS17/LectNote-5&6
________________________________ LN_5&6 / S. 6
j j+1
2
xx 2
j j-1 j+1 jfinite space
j j-1
j j-1 j+1 j
j j+1
x=x x=x
Higher order derivatives:
T T TT ( ) , where
x x x
F(x ) - F(x ) F - FΔF=
x Δx x - x h
T - T T - TΔT ΔTF , F
Δx h Δx h
Subtitute these exp
F
F Fx x
F
2j j-1
j j-1 j+1 j j-12 2
23
j 1 j 1 j j x xx
j
ression:
F - FT ΔF 1 1= F - F T - 2T T
x Δx h h h
But how good is this approxation? Taylor expansion
h T = T(x )= T(x h) = T h T + T + O(h )
2
Δ ΔT 1 ( ) TΔx Δx h
2
+1 j j-1 xx - 2T T = T + O(h )
The scheme is second order accurate.
Relativistic Hydrodynamics BAS/FS17/LectNote-5&6
________________________________ LN_5&6 / S. 7
The one-dimensional heat equation:
Consider the following heat diffusion equation:
Assume that both edges of the metal rod are kept at certain constant temperatures Tu and Td.
Let the rod be heated at the center for a certain period of time. How do the initial,
intermediate and final profiles of the temperature look like?
Intuitively (without performing analytical or numerical calculation) we may expect that: the BCs and the ICs may play an essential role in determining the form and evolution of the
solution at any time t.
There are two types of problems: Initial value and boundary value problems. The strength of
dependence on the ICs and BCs determine the type of the problem.
2
2.
T T
t x
In the absence of transport, the heat diffuses in a symmetric
manner, provided the diffusion coefficient, χ, and the BC are
symmetric too. If advection (transport) is included, then this symmetry will be
broken as there is a preferable direction for heat transport, which is exemplified in
the following two figures.
u > 0
Relativistic Hydrodynamics BAS/FS17/LectNote-5&6
________________________________ LN_5&6 / S. 8
The heat equation: discretization:
The temperature depends on t as well as on x, i.e, T = T(t,x). Thus
for each value of “x” and value of “t” there is a suitable value for T
(hopefully a unique value).
1
1 11
1
1 1
2
( )x x x
(pointwise)(1 2 ) ,
Time Explicit
where s = discretizationx
n n n n n n
j j j j j jn n
j j
n n n n
j j j j
T T T T T TT f T
t
T sT s T sT
t
t
x
T
tn
xj
T(tn,xj) = Tnj
An explicit discretization of this equation yields the following form:
2
2
T T
t x
Relativistic Hydrodynamics BAS/FS17/LectNote-5&6
________________________________ LN_5&6 / S. 9
2
2
T T
t x
22
a 2L (T) = + O( t) + O( x )
T T
t x
Truncation error
Consistency: The finite space representation ( )aL T of the equation t xxT = χ T is
said to be consistent, if the truncation error goes to zero as , x 0.t (local analysis)
Stability: The finite space representation ( )aL T of the equation t xxT = χ T is
said to be numerically stable, if accumulated errors do not grow with time (non-local).
Prediction power of explicit procedures:
Accuracy ↔ more point to include
↔ larger domain
(like weather forecast)
The weather in Heidelberg: time versus domain
Relativistic Hydrodynamics BAS/FS17/LectNote-5&6
________________________________ LN_5&6 / S. 10
Weak and strong solutions of Navier-Stokes equations The equations describing the evolution of incompressible viscous fluid flows are called
the Navier-Stokes equations (that were formulated around 1830s) and read:
3
x y z
0
= (u ,u ,u ), , = viscosity, p = pressure, = domain in R , = boundary,
I =(0, T) and u = initial value.
twhere u u u
For a given data, it was proven by Leray (1934) that the Navier-Stokes equations have at least one weak
solution. However, it is still not clear, whether the weak solution is unique or not.
Now by a weak solution we mean a solution that satisfies the above partial differential equations on
average, but not necessary in a pointwise manner. For example, the derivatives may not exist at certain
points.
However, a strong solution is said to satisfy the equations everywhere and at each point of the domain
For numerical mathematicians, the above-mentioned theorems implies the following:
Different conservative numerical methods may yield different solutions, in most cases the number
of grid points here is relatively small.
Different conservative numerical methods that are numerically stable and consistent must converge
to the same weak solution if the number of grid points is relatively large.
Lax-Wendroff theorem: The numerical solution q is said to be a weak solution for the analytical
equation ( )L q , if the numerical scheme employed is conservative, consistent and the numerical
procedure converges.
Conservation convergence weak solution
Lax theorem:
If q has been obtained using a conservative and consistent scheme, then q converges to a weak
solution of the analytical problem when j , i.e., when the number of grid points goes to infinity.
Relativistic Hydrodynamics BAS/FS17/LectNote-5&6
________________________________ LN_5&6 / S. 11
You cannot claim to have found a weak solution for the physical problem, unless you carried out the
caculations with sufficiently large number of grid points, beyond which doubling the number of grid
points yield no noticeable improvement.
From Fletcher : “Computational techniques ... (1990)
From Fletcher : “Computational techniques ... (1990)
Relativistic Hydrodynamics BAS/FS17/LectNote-5&6
________________________________ LN_5&6 / S. 12
2
2: Given is the heat equation: in the domain D=[t] [x] [0,1] [0,1] together with
the IC and BC: T(t=0) = 1, T(t,0) = 2, T(t,1)=2.
Solve the equation using the FTCS formulation,
T TQ
t x
1
1 1
2
x
i.e., (1 2 ) , where
s = t / , =1 and N ( umber of grid points in x-direction) 100 for the
following s-values: s = 0.1, 0.2, 0.4, 0.8, 1.2.,
Plot th
n n n n
j j j jT sT s T sT
x n
e solutions at times: t = 0.1, 0.2, 0.4 and 1.0.