MMJ 1113 Computational Methods for Engineers - @let@token Numerical...

Faculty of Mechanical EngineeringEngineering Computing Panel

MMJ 1113 Computational Methods for Engineers

Numerical Differentiation

Abu Hasan Abdullah

Feb 2013

abu.hasan.abdullah�dev.null MMJ 1113 Computational Methods for Engineers Numerical Differentiation 1 / 41

Outline

1 Introduction


Outline

1 Introduction

2 Sample Engineering Problems


Outline

1 Introduction


3 Definition


Outline

1 Introduction


3 Definition

4 Finite Difference Approximation

Taylor Series Expansions

Approximating Derivatives

Centred Difference Formulae

Improving Derivative Estimates

Finite Difference in PDE


Outline

1 Introduction


3 Definition







5 Polynomial Interpolation

Direct Fit Polynomials

Lagrange Polynomials

Divided Difference Polynomialsabu.hasan.abdullah�dev.null MMJ 1113 Computational Methods for Engineers Numerical Differentiation 2 / 41

Outline

1 Introduction


3 Definition







5 Polynomial Interpolation

Direct Fit Polynomials

Lagrange Polynomials

Divided Difference Polynomials

6 Bibliographyabu.hasan.abdullah�dev.null MMJ 1113 Computational Methods for Engineers Numerical Differentiation 2 / 41

IntroductionWhy/When?

1 Finite Difference Approximation—complex differential equations in many

engineering applications, e.g. wave equation

1

c2

∂2u

∂t2=

∂2u

∂x2+

∂2u

∂y2+

∂2u

∂z2= ∇

2u

may require numerical solutions due to presence of complex geometry, complicated

boundary conditions, nonlinearity, large number of simultaneous differential

equations, etc.

2 Polynomial Interpolation—differentiation of discrete data, e.g. from experimentalresults tabulated below, requires an approximate numerical procedure.

x 0.00 0.12 0.32 0.44 0.54 0.64 0.70 0.81y 0.2000 1.3097 1.7434 2.8430 3.5073 3.1819 2.3630 0.0036

3 Exact Differentiation—of known functions through differential calculus, e.g.

distance traversed at a constant acceleration

s = ut + 12at2 where

ds

dt= v and

d2s

dt2=

dv

dt= aabu.hasan.abdullah�dev.null MMJ 1113 Computational Methods for Engineers Numerical Differentiation 3 / 41

IntroductionFinite Difference Approximation

Differential equations in many engineering applications require numericalsolutions due to

presence of complex geometry and/or complicated boundary conditionsnonlinearity and/or large number of simultaneous differential equations, etc.

Derivatives in these differential equations are replaced by their discrete forms,

known as finite-difference approximations. Use of finite-differences transforms an

ordinary differential equation into an algebraic equation

Taylor series used in computing derivatives using discrete methods is cumbersome

for approximations of increasingly high accuracy or higher-order derivatives


IntroductionPolynomial Interpolation

If a function is very complicated or

known only from values in a table (e.g.

from experiments), it may be necessary

to resort to numerical differentiation.

In cases especially where data are

obtained experimentally it is best to

perform a least squares curve fit to

data and then find the resulting

polynomial.

Formulae for numerical differentiation

may easily be obtained by

differentiating interpolation

polynomials. The essential idea is that

the derivatives f ′, f ′′, . . . of a function

are represented by the derivatives P′

n,

P′′

n , . . . of the interpolating polynomial

Pn—more on this towards the end!Figure 1 : Interpolating polynomial.abu.hasan.abdullah�dev.null MMJ 1113 Computational Methods for Engineers Numerical Differentiation 5 / 41

IntroductionErrors in Numerical Differentiation

Numerical differentiation is avoided wherever it is possible because of several inherent

difficulties:

1 Integration describes an overall property of a function, whereas differentiation

describes the slope of a function at a point. Integration is not sensitive to minor

changes in the shape of a function, whereas differentiation is. Any small changes in

a function can easily create large changes in its slope in the neighborhood of the

change. Compared to integration, numerical differentiation is less reliable.

2 Large errors may also occur in numerical differentiation based on Taylor series,

which is cumbersome for approximations of high accuracy or higher-order

derivatives.


Sample Engineering ProblemsProblem 1

Problem Statement:

The barrel and the recoil

mechanism of a cannon, Figure 2,

have a mass of 500 kg with a recoil

spring stiffness of 10000 N/m. The

response of the cannon, with a

critically damped recoil

mechanism, is given by

x(t) = (c1 + c2t)−ωnt(1)

where c1 = x0, c2 = x0 + ωnx0, x0 is

the initial displacement, x0 is the

initial velocity, and ωn =p

k/m is

the undamped natural frequency

of the system.

Find the recoil velocity of the cannon using

numerical differentiation with x0 = 0 and

x0 = 5 m/s.

Figure 2 : Recoil mechanism of a cannon.



Problem Statement:

The load carrying capacity of a tilt thrust bearing (P), shown in Figure 3, is given by

P =6µ(U1 − U2)lb

2

h22

1

(m − 1)2

»

ln m −2(m − 1)

m + 1

–ff

(2)

where

U1 = surface velocity of the supporting plate, h1 = film thickness at entrance,

U2 = surface velocity of the pad, h2 = film thickness at exit,

m = h1/h2 = tilt ratio, µ = viscosity of the lubricant,

b = length of the lubricant flow path, l = length of the bearing.

The change in the load due to a small change in the tilt ratio, given by dP/dm, will be ofinterest from a design point of view. Find the value of dP/dm for the following data:

b = 3.0 in l = 1.5 in

h1 = 0.10 in h2 = 0.05 in

P = 5000 lb µ = 1.95 × 10−6 reyn

U1 − U2 = 150 in/secabu.hasan.abdullah�dev.null MMJ 1113 Computational Methods for Engineers Numerical Differentiation 8 / 41


Figure 3 : Tilt thrust bearing.



Problem Statement:

The pressure (p)–specific volume (v) relationship of superheated water vapour at 350◦C

is given by van der Waals equation

p =RT

v − b−

a

v2(3)

where

T = absolute temperature = 623.15 K a = 1.7048

R = specific gas constant = 0.461889 kJ/kg-K b = 0.0016895

Expand the pressure in Taylor’s series and estimate the value of p at

v = 0.051, 0.054, and 0.060 assuming that the value of p and its derivatives are known

at v = 0.05.


Derivative Defined

Derivative of a function f(x) is defined as

df(x)

dx

˛˛˛˛x

= f′(x) = lim

h→0

f(x + h) − f(x)

h

(4)

h does not approach zero but remains a

finite quantity. As h = ∆x, Eq. (4) can

also be written as

f′(x) ≈

f(x + ∆x) − f(x)

∆x(5)

Figure 4 : Derivative defined.


Finite Difference ApproximationTaylor Series Expansions

Given a function f(x), we wish to approximate its first and second derivatives at a point

x.

For a given step size +∆x, the forward Taylor series expansion gives

f(x + ∆x) = f(x) + f′(x)(+∆x) +

f ′′(x)

2!(+∆x)2 +

f ′′′(x)

3!(+∆x)3 + . . . (6)

For a negative step size −∆x, backward Taylor series expansion gives

f(x − ∆x) = f(x) + f′(x)(−∆x) +

f ′′(x)

2!(−∆x)2 +

f ′′′(x)

3!(−∆x)3 + . . . (7)


Finite Difference ApproximationApproximating First Derivative to O(h) Accuracy

Solving Eq. (6) for (df(x)/dx)i,j

df(x)

dx= f

′(x) =f(x + ∆x) − f(x)

∆x| {z }

finite difference

−f ′′

2!(∆x) −

f ′′′

3!(∆x)2

− . . .| {z }

truncation error, O(∆x)

=f(x + ∆x) − f(x)

∆x+ O(∆x)

For ∆x = h and ignoring

O(∆x), we get a forward

difference formula

f′(x) ≈

f(x + h) − f(x)

h(8)

which is first-order accurate.

Figure 5 : Forward difference formulation.abu.hasan.abdullah�dev.null MMJ 1113 Computational Methods for Engineers Numerical Differentiation 13 / 41

Finite Difference ApproximationApproximating First Derivative to O(h) Accuracy

Solving Eq. (7) for (df(x)/dx)i,j

df(x)

dx= f

′(x) =f(x − ∆x) − f(x)

∆x| {z }

finite difference

−f ′′

2!(∆x) −

f ′′′

3!(∆x)2

− . . .| {z }

truncation error, O(∆x)

=f(x − ∆x) − f(x)

∆x+ O(∆x)

For −∆x = −h and ignoring

O(∆x), we get a backward

difference formula

f′(x) ≈

f(x) − f(x − h)

h(9)

which is first-order accurate.

Figure 6 : Backward difference formulation.abu.hasan.abdullah�dev.null MMJ 1113 Computational Methods for Engineers Numerical Differentiation 14 / 41

Finite Difference ApproximationApproximating First Derivative to O(h2) Accuracy

Substract Eq. (7) from Eq. (6) to

give centred difference formula for

f ′(x)

f′(x) =

f(x + h) − f(x − h)

2h

−f ′′′(x)

3!h

2 + . . .

≈f(x + h) − f(x − h)

2h(10)

which is second-order accurate.

Figure 7 : Central difference formulation.


Finite Difference ApproximationApproximating Second Derivative to O(h2) Accuracy

Add Eq. (7) to Eq. (6) to give the centred difference formula for f ′′(x)

f′′(x) =

f(x + h) − 2f(x) + f(x − h)

h2−

f iv(x)

4!h

2 + . . .

≈f(x + h) − 2f(x) + f(x − h)

h2(11)

The above approximation is second-order accurate.

We could now improve the accuracy of Eqs. (8) (9) and (10) by retaining the

second-derivative term, f ′′(x), and substituting it with Eq. (11) above.


Finite Difference ApproximationCentred Difference Formulae of Orders O(h2) and O(h4)

Centred Difference Formulae of Order O(h2)

f′(x) ≈

f(x + h) − f(x − h)

2h(12)

f′′(x) ≈

f(x + h) − 2f(x) + f(x − h)

h2(13)

f′′′

(x) ≈

f(x + 2h) − f(x + h) + 2f(x − h) − f(x − 2h)

2h3(14)

f(4)

(x) ≈

f(x + 2h) − 4f(x + h) + 6f(x) − 4f(x − h) + f(x − 2h)

h4(15)

Centred Difference Formulae of Order O(h4)

f′(x) ≈

−f(x + 2h) + 8f(x + h) − 8f(x − h) + f(x − 2h)

12h(16)

f′′(x) ≈

−f(x + 2h) + 16f(x + h) − 30f(x) + 16f(x − h) − f(x − 2h)

12h2(17)

f′′′

(x) ≈

−f(x + 3h) + 8f(x + 2h) − 13f(x + h) + 13f(x − h) − 8f(x − 2h) + f(x − 3h)

8h3(18)

f(4)

(x) ≈

−f(x + 3h) + 12f(x + 2h) − 39f(x + h) + 56f(x) − 39f(x − h) + 12f(x − 2h) − f(x − 3h)

6h4(19)abu.hasan.abdullah�dev.null MMJ 1113 Computational Methods for Engineers Numerical Differentiation 17 / 41

Finite Difference ApproximationImproving Derivative Estimates–Richardson Extrapolation

Two approaches normally used to improve derivative estimates

decrease step size huse higher-order formula that employs more points

There is also a third approach–Richardson extrapolation–a method often used to

improve the results of a numerical method, from a method of order O(hn) it gives

us a method of order O(hn+1). Richardson extrapolation uses two derivatives to

compute a third, more accurate approximation.

In integration, Richardson extrapolation provides a means to obtain an improved

integral estimate I by formula

I ≈ I(h2) +1

„h1

h2

«2

− 1

[I(h2) − I(h1)] (20)

where I(h1) and I(h2) are integral estimates using two step sizes h1 and h2.

Eq. (20) is usually written for the case where h2 = 12h1 to give

I ≈4

3I(h2) −

1

3I(h1) (21)abu.hasan.abdullah�dev.null MMJ 1113 Computational Methods for Engineers Numerical Differentiation 18 / 41

Finite Difference ApproximationImproving Derivative Estimates–Richardson Extrapolation

In similar fashion, Eq. (21) can be written for derivatives, D, as

D ≈4

3D(h2) −

1

3D(h1) (22)

for centred-difference approximations with O(h2). The application of this formula

will yield a new derivative estimate of O(h4)


Finite Difference ApproximationCentred Difference Formulae–Example 1

Problem Statement:

Let f(x) = sin(x), where x is measured in radians.

1 Calculate approximations to f ′(0.8) using the centred difference formula of order

O(h2) with ∆x = h = 0.1, ∆x = h = 0.01, ∆x = h = 0.001. Carry eight or nine

decimal places.

2 Compare with the value f ′(0.8) = cos(0.8).

Solution:

Work through the example.


Finite Difference ApproximationCentred Difference Formulae–Example 2

Problem Statement:

Compare the numerical differentiation by centred difference formulae of order O(h2)and order O(h4). Let f(x) = x3 and find approximations for f ′(2).

1 Use order O(h2) with ∆x = h = 0.05

2 Use order O(h4) with ∆x = h = 0.05

Solution:



Finite Difference ApproximationCentred Difference Formulae–Example 3 (Programming)

Problem Statement:

1 Take a look at ex1s71.f to study sample computation of the derivative of a

function using forward difference formula

2 Take a look at ex2s71.f to study sample computation of the derivative of a

function using central difference formula


Finite Difference ApproximationCentred Difference Formulae–Example 4 (Richardson extrapolation)

Problem Statement:

Using the function

f(x) = −0.1x4− 0.15x

3− 0.5x

2− 0.25x + 1.2

estimate the first derivative at x = 0.5 employing step sizes h1 = 0.5 and h2 = 0.25.

Compute an improved estimate with Richardson extrapolation. As a comparison, the

true value is −0.9125.

Solution:



Finite Difference ApproximationFinite Difference in PDE

So far we dealt with total or ordinary derivatives where function f depends on a

single independent variable x.

In many engineering problems, function f depends on two or more independent

variables.

We use notation (i, j) to designate the pivot point if there are two independent

variables, (i, j, k) for three–see Figure 8–and so on as the respective counters in x, y

and z directions.

(a) (b)

Figure 8 : Two- and three-dimensional grid systems.abu.hasan.abdullah�dev.null MMJ 1113 Computational Methods for Engineers Numerical Differentiation 24 / 41


Consider f(x, y). Finite-difference approximation for the partial derivative

∂f(x, y)

∂xat (x = xi, y = yj)

can be found by fixing value of y at yj and treat f(x, yj) as a one-variable function.

Forward-difference approximation of ∂f/∂x is

∂f

∂x

˛˛˛˛i,j

≈f(xi + ∆x, yj) − f(xi, yj)

∆x(23)

Backward-difference approximation of ∂f/∂x is

∂f

∂y

˛˛˛˛i,j

≈f(xi, yj) − f(xi − ∆x, yj)

∆x(24)

Central-difference approximation of ∂f/∂x is

∂f

∂y

˛˛˛˛i,j

≈f(xi + ∆x, yj) − f(xi − ∆x, yj)

2∆x(25)abu.hasan.abdullah�dev.null MMJ 1113 Computational Methods for Engineers Numerical Differentiation 25 / 41


Central-difference approximations of ∂2f/∂x2, ∂2f/∂y2 and ∂2f/∂x∂y are

∂2f

∂x2

˛˛˛˛i,j

≈f(xi + ∆x, yj) − 2f(xi, yj) + f(xi − ∆x, yj)

(∆x)2(26)

∂2f

∂y2

˛˛˛˛i,j

≈f(xi, yj + ∆y) − 2f(xi, yj) + f(xi, yj − ∆y)

(∆y)2(27)

∂2f

∂x∂y

˛˛˛˛i,j

≈

“f(xi+∆x,yj+∆y)−f(xi+∆x,yj−∆y)

−f(xi−∆x,yj+∆y)+f(xi−∆x,yj−∆y)

”

(4∆x∆y)(28)


Polynomial Interpolation

The function f(x), which is to be differentiated, may be

a known function, ora set of discrete data.

In general, known functions can be differentiated exactly.

Differentiation of discrete data, however, requires an approximate numerical

procedure. To perform numerical differentiation, an approximating polynomial is

fit to the discrete data, or a subset of the discrete data, and the approximating

polynomial is differentiated.

We’ll look at differentiation of1 direct fit polynomials,2 Lagrange polynomials, and3 divided difference polynomials

which can be applied to both unequally spaced data and equally spaced data.


Polynomial InterpolationDirect Fit Polynomials

Based on fitting the data directly by a polynomial and differentiating the

polynomial. The direct fit polynomial is given by

Pn(x) = a0 + a1x + a2x2 + . . . + anx

n(29)

where Pn(x) is determined by one of the following methods:1 Given N = n + 1 points, [xi, f(xi)], determine the exact nth-degree polynomial that

passes through the data points.2 Given N > n + 1 points, [xi, f(xi)], determine the least squares nth-degree polynomial

that best fits the data points.

Derivatives obtained by differentiating the approximating polynomial

f′

≈ P′

n = a1 + 2a2x + 3a3x2 + . . . (30a)

f′′

≈ P′′

n = 2a2 + 5a3x + . . . (30b)


Polynomial InterpolationLagrange Polynomials

Can be used for both unequally spaced data and equally spaced data and is based

on differentiating a Lagrange polynomial.

This method uses the second degree Lagrange polynomial:

P2(x) =(x − b)(x − c)

(a − b)(a − c)f(a) +

(x − a)(x − c)

(b − a)(b − c)f(b) +

(x − a)(x − b)

(c − a)(c − b)f(c) (31)

Differentiating Eq. (31) yields

f′

≈ P′

2(x) =2x − (b + c)

(a − b)(a − c)f(a) +

2x − (a + c)

(b − a)(b − c)f(b) +

2x − (a + b)

(c − a)(c − b)f(c)

(32a)

f′′

≈ P′′

2 (x) =2f(a)

(a − b)(a − c)+

2f(b)

(b − a)(b − c)+

2f(c)

(c − a)(c − b)(32b)


Polynomial InterpolationDivided Difference Polynomials

Can be used for both unequally spaced data and equally spaced data and is based on

differentiating a divided difference polynomial

Pn(x) = f(0)i + (x − x0)f

(1)i + (x − x0)(x − x1)f

(2)i

+ (x − x0)(x − x1)(x − x2)f(3)i + . . . (33)

Differentiating Eq. (33) yields

f′

≈ P′

n(x) = f(1)i + [2x − (x0 + x1)]f

(2)i

+ [3x2− 2(x0 + x1 + x2)x + (x0x1 + x0x2 + x1x2)]f

(3)i (34a)

f′′

≈ P′′

n (x) = 2f(2)i + [6x − 2(x0 + x1 + x2)]f

(3)1 + . . . (34b)


Polynomial InterpolationExample 1

Problem Statement:

Evaluate the derivatives by numerical differentiation formulae developed by fitting

1 direct fit polynomial,

2 Lagrange polynomial, and

3 divided difference polynomial

to the following set of discrete data:

x 3.4 3.5 3.6

y 0.294118 0.285714 0.277778

The exact derivatives at x = 3.5 are f ′(3.5) = −0.081633 . . . and

f ′′(3.5) = 0.046647 . . .



Solution:

Fit the quadratic polynomial, P2(x) = a0 + a1x + a2x22, to the three data points:

0.294118 = a0 + a1(3.4) + a2(3.4)2(35a)

0.285714 = a0 + a1(3.5) + a2(3.5)2(35b)

0.277778 = a0 + a1(3.6) + a2(3.6)2(35c)

Solving for a0, a1, and a2 by Gauss elimination gives a0 = 0.858314, a1 = −0.245500,

and a2 = 0.023400. Substituting these values into Eqs. (5.7a) and (5.7b) and

evaluating at x = 3.5 yields the solution for direct fit polynomial:

P′

2(3.5) = −0.245500 + (0.04680)(3.5) = −0.081700 (35d)

P′′

2 (3.5) = 0.046800 (35e)



Solution: continued. . .

Substituting the tabular values into Eqs. (32a) and (32b) and evaluating at x = 3.5yields the solution for the Lagrange polynomial

P′

2(3.5) =2(3.5) − (3.5 + 3.6)

(3.4 − 3.5)(3.4 − 3.6)(0.294118) +

2(3.5) − (3.4 + 3.6)

(3.5 − 3.4)(3.5 − 3.6)(0.285714)

+2(3.5) − (3.4 + 3.5)

(3.6 − 3.4)(3.6 − 3.5)(0.277778) = −0.087100 (36a)

P′′

2 (3.5) =2(0.294118)

(3.4 − 3.5)(3.4 − 3.6)+

2(0.283714)

(3.5 − 3.4)(3.5 − 3.6)

+2(0.277778)

(3.6 − 3.4)(3.6 − 3.5)= 0.046800 (36b)




Construct a divided difference table for the tabular data to use the divided difference

polynomial:

xi f(0)i f

(1)i f

(2)i

3.4 0.294118

-0.084040

3.5 0.285714 0.023400

-0.079360

3.6 0.277778

Substituting these values into Eqs. (34a) and (34b) yields the solution for the divided

difference polynomial:

P′

2(3.5) = −0.084040 + [2(3.5) − (3.4 + 3.5)](0.023400) = −0.081700 (37a)

P′′

2 (3.5) = 2(0.023400) = 0.046800 (37b)abu.hasan.abdullah�dev.null MMJ 1113 Computational Methods for Engineers Numerical Differentiation 34 / 41



The results obtained by the three procedures are identical since the same three points

are used in all three procedures.

The error in f ′(3.5) is

Error = f′(3.5) − P

′

2(3.5)

= −0.081700− (−0.081633)

= −0.000067

The error in f ′′(3.5) is

Error = f′′(3.5) − P

′′

2 (3.5)

= 0.046800− 0.046647

= 0.000153


Polynomial InterpolationExample 2: Using Matlab

Problem Statement:

An experiment yields the following tabulated x-y pairs:

x 0.000 0.100 0.200 0.300 0.400 0.500

y -0.447 1.978 3.280 6.160 7.080 7.340

x (cont. . . ) 0.600 0.700 0.800 0.900 1.000

y (cont. . . ) 7.660 9.560 9.480 9.300 11.20

Perform a curve fit to the data and find the approximating polynomial. Determine the

derivative at x = 0.5.



Solution:

To find a polynomial that fits a set of data we use the command polyfit(x,y,n),where x is a vector containing x-axis values, y is a vector containing y-axis values and nis the order of the polynomial that we want to fit:

Matlab session>> x = [ 0.000 0.100 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00℄;>> y = [-0.447 1.978 3.28 6.16 7.08 7.34 7.66 9.56 9.48 9.30 11.2℄;>> n = 2;>> P = polyfit(x,y,n)Next we differentiate the polynomial

Matlab session>> dP = polyder(P)and compute the slope at x = 0.5

Matlab session>> slope_of_poly = polyval(dP,0.5)abu.hasan.abdullah�dev.null MMJ 1113 Computational Methods for Engineers Numerical Differentiation 37 / 41


Solution: continued . . .

We could easily plot the approximating polynomial over the original x-y points to seeeffect of the fitting on the approximation of derivatives

Matlab session>> xi = [0.0:0.01:1.0℄;>> yi = polyval(P,xi);>> plot(x,y,'o',xi,yi,'r-')



Problem Statement:

An experiment yields unequally spaced data in x and y

x y x y x y

0.00 0.2000 0.36 2.0749 0.64 3.18190.12 1.3097 0.40 2.4560 0.70 2.36300.22 1.3052 0.44 2.8430 0.80 0.23200.32 1.7434 0.54 3.5073 0.81 0.0036

Determine the differences between adjacent elements of both x and y vectors and

compute divided-difference approximations of the derivative.



Solution:

To differentiate unequally spaced data in x and y we use the command diff(x) anddiff(y), where x is a vector containing x values, y is a vector containing y values

Matlab session>> diff(x);>> diff(y);To compute divided-difference approximations of the derivative we perform vectordivision of y differences by x differences

Matlab session>> dydx = diff(y)./diff(x)abu.hasan.abdullah�dev.null MMJ 1113 Computational Methods for Engineers Numerical Differentiation 40 / 41

Bibliography

1 SINGIRESU S. RAO (2002): Applied Numerical Methods for Engineers and Scientists, ISBN0-13-089480-X, Prentice Hall

2 STEVEN C. CHAPRA, RAYMOND P. CANALE (2006): Numerical Methods for Engineers, 5ed,ISBN 007-124429-8, McGraw-Hill

3 DAVID KINCAID AND WARD CHENEY (1991): Numerical Analysis: Mathematics of ScientificComputing, ISBN 0-534-13014-3, Brooks/Cole Publishing Co.

4 STEVEN C. CHAPRA (2005): Applied Numerical Methods with MATLAB for Engineers andScientists, ISBN 007-124484-0, McGraw-Hill

5 WILLIAM J. PALM III (2005): Introduction to Matlab 7 for Engineers, ISBN 007-123262-1,McGraw-Hill

6 JOHN D. ANDERSON, JR. (1995): Computational Fluid Dynamics–The Basics withApplications, ISBN 007-113210-4, McGraw-Hill


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