mathematics and computation

8/9/2019 Mathematics and Computation

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Basic Mathematics

MSc Petroleum Engineering

University of UMSA

La Paz

April 08-26, 2013

Mesfin BelaynehUoS, Norway


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Contents1 INTRODUCTION ................................................................................................................... 3

2 VECTOR ALGEBRA ............................................................................................................. 4

2.1 Addition of vectors ......................................................................................................... 4

2.2 Multiplication by a scalar ............................................................................................... 4

2.3 Resolving vectors into components ................................................................................ 5

2.4 Dot product ..................................................................................................................... 7

2.5 Cross product .................................................................................................................. 8

3 MATRIX ALGEBRA ........................................................................................................... 15

3.1 Properties of matrices ................................................................................................... 15

3.2 Procedure to solve simultaneous equation with a matrix system ................................. 20

3.3 Determinants ................................................................................................................ 21

3.4 Minors and Cofactors ................................................................................................... 21

3.5 Adjoint matrix .............................................................................................................. 23

3.6 Inverse matrix ............................................................................................................... 23

4 MODELING VIA DIFFERENTIAL EQUATIONS ............................................................ 26

4.1 First order differential equation .................................................................................... 26

4.2 Second order equations and systems ............................................................................ 30 4.3 Numerical solution methods ......................................................................................... 33

4.3.1 Newton-Raphons non-linear root finding ........................................................ 33

4.3.2 Numerical intergration ..................................................................................... 36 4.3.2.1 Simpson ............................................................................................................. 36

4.3.2.2 Trapezium rule ................................................................................................... 37

4.3.3 Numerical differentiation ................................................................................. 39 4.3.3.1 Numerical Eulers solution methods .................................................................. 39

4.3.4 Curve fitting ..................................................................................................... 44 4.3.4.1 Linear , polynomical, exponential and power .................................................. 44


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1 Introduction

All fields of studies (Science and Engineering) use mathematics. The language ofmathematics is a tool to formulate theories based on experimental observations. Therefore, inthis section we will learn the basic of mathematics along with the applications. The topics to

be reviewed are:

Vectors (Dot product Determinants; cross product), Matrices; inverse matrices and determinant Finally modeling based on differential equation and solution methods

This will refresh students knowledge of the basic mathematical tools


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2 Vector algebra

A vector is used to represents a physical quantity that has both a direction and magnitude, forexample, displacement and force. A vector is generally represented by the symbols

U or U

On a diagram, it is generally represented by an arrow. The direction of the arrow represents itsdirection and the length represents its magnitude.

2.1 Addition of vectors

The parallelogram rule shows how the sum of two vectors can be obtained graphically. Itdemonstrates that the sum of two vectors is independent of the order in which the vectors areadded. These rules also lead to the following principles:

Commutative law: U + V = V + U

Associative law: (U + V) + W = U + (V + W)

U + V = V + U = W Fig. 1 Parallelogram rule

2.2 Multiplication by a scalar

The product of a scalar p and a vector U can be written as p U. It follows that:

p(q U) = (pq) U (p + q) U = p U + q U p(U + V ) = p U + p V

A vector - U is called "the negative of the vector U" and it has the same magnitude as vector U

but its direction is in the opposite direction.

UV

V

U

W


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U

A vector U

2U

Multiplication of U with scalar

-U

Negative of the vector U

Fig. 2 Multiplication rule

2.3 Resolving vectors into components

For two dimensions, we can resolve a vector U into vector components Uxand Uy, which are parallel to the x and y axis, respectively. The vector components, Ux and Uy can be expressedas a product of their magnitude and unit vectors, i and j .

U = U xi and Uy j

Similarly, in three dimensional system, we can write

U = U xi + Uy j + Uzk

where i, j and k are the unit vectors defined to point in the positive direction of the x,y and z

axis.

The magnitude for the three dimensional vector is

2z

2y

2x UUUU

Fig. 3: Resolving vectors

Example: Force Vector Characteristics

Since a force can be represented as a vector, we can use the previous vector concepts to helpsolve static problems with multiple forces. Force is a vector quantity, having both magnitudeand direction. When there are several forces acting on a body, these forces can be grouped

and referred upon as a force system. The body is said to be in equilibrium when the appliedforces produces no external effects .

Ux=Uxi

Uy=Uy j

U=Uxi +Uy j


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When there are two concurrent forces applied to a body a single resultant force can be used torepresent the two forces by applying the addition law for vectors (i.e. parallelogram law). Theuse of trigonometry based on the law of sine and cosine can also be helpful in solving

problems involving two force vectors.

Law of sines:

sinc

sin b

sina

Law of cosine:

cosac2ca b 222

Fig. 4 Law of sines and cosine

However, if there are more than two forces present, the standard method to solve a problem isreduce each vector into its rectangular components so that it they are easy to add and subtract.

A force can be resolved into its rectangular components in the x and y axis as shown in thefigure on the left. The resolved force can be written in Cartesian form:

F = F xi + F y j

F = Fcos i + Fsin j

2y

2x FFF

A force that is resolved (decomposed) into its rectangular components in the x, y, and z-axiscan be expressed in the following Cartesian form:

F = F xi + F y j + F zk

F = Fcos xi + Fcos y j + Fcos zk

2z

2y

2x FFFF

a

b

c

A

C

B


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Fig. 5 The resolved force in Cartesian form

2.4 Dot product

The dot product of two vectors U and V is defined to be the product of the magnitude of U

and V and the cosine of the angle between them.

U . V = |U| |V| Cos

The dot product can also be calculated by multiplying the x, y and z terms together as:

U . V = (Uxi + Uy j + Uzk) (Vxi + Vy j + Vzk)

U = UxVx + UyVy + UzVz

Dot product has the following properties:

Commutative law: U . V = V . U Associative law: q( U . V) = (q U) . V = U . (qV) Distributive law: U .(V + W ) = U . V + U . W

U . V = |U| |V| cos

Fig. 6 Dot product of U and V


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Figure 7 Example work done by a force at an angle

W= F 1 .S +F 2

.S = |F 1| |S| cos +|F 2| |S| cos

2.5 Cross product

Let W be the cross product of two vectors U and V such that U x V = W . We define W tohave the following properties:

W points in a direction perpendicular to the plane containing U and V as determined by the right hand rule.

The magnitude of W , |W| = U V sin

Fig. 8 the right hand rule of vector

U x V = W V x U = -W

In general, cross products have the following properties:

Does not follow the commutative law:

S

F1

F2


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U x V = -( V x U)

Satisfies associative law:

q(U x V) = (q U) x V = U x (q V)

Satisfies distributive law:

U x ( V + W ) = U x V + U x W

Example: Vector Representation of a Moment

When an arbitrarily-shaped body is acted upon by a force, there exists a tendency for theforce to rotate the body about a point or axis. This tendency is known as a moment. The pointor axis about which the rotation takes place is referred to as the moment center. Moments areusually described with reference to a point or axis on the body.

The magnitude of a moment, M, can be described as the product of the perpendiculardistance, d, measured from the point or axis where the force is applied. The perpendiculardistance is also known as the moment arm.

In practice, the moment is considered to be positive if the resultant rotation iscounterclockwise. Negative moments generate clockwise rotation. A moment is a vectorquantity, having both magnitude and direction.

Referring to the figure shown on the left, the moment about point O caused by the appliedforce F , can be described by the cross product of the distance vector, r , and the force vector,F .

Mo = r x F

= r F sin 0 180 0

Mo=dFThe direction of the rotation can be determined using the Right Hand Rule

dF

F

d Moment = Fd


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2.5.1 Cross Products of Unit Vectors in Cartesian coordinate system

The cross product of unit directional vectors, i, j , and k , can be written as:

i x i = 0 i x j = k i x k = - j

j x i = -k j x j = 0 j x k = i

k x i = j k x j = - i k x k = 0

For a vector U = U xi + U y j + U zk and a vector V = V xi + V y j + V zk , the determinant of U andV can be written as:

zyx

zyx

VVV

UUU

k ji

UxV

Fig. 9 The cross product of unit directional vectors

2.5.2 Moment about a Point

The force applied at point 2 generates a moment about point 1. This moment can also be

calculated using the formula:

Mo = r x F

In this case r calculated in this manner:

r = r 2-r 1

= (x 2 - x1)i + (y 2 - y1) j + (z 2 - z1)k


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Fig. 10: The force applied at point 2 generates a moment about point 1

In a two dimensional case, there are no components in the k direction. The force and vector r are given in the following:

r = r xi + r y j

F = F xi + F y j

Mo = r x F

0FF0r r

k ji

rxFM

yx

yxo

= (r xFy - r yFx)k

= M zk

Fig. 11: The two dimensional force


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For a three dimensional case, there are three vector components.

zyx

zyxo

FFF

r r r

k ji

rxFM

= (r yFz - r zFy)i + (r zFx - r xFz) j + (r xFy - r yFx)k

= M xi + M y j + M zk

Fig. 12: Three vector components.

The following is an alternative way of presenting moments in the x, y, and z-axis.

Mx = Mcos

My = Mcos

Mz = Mcos

The magnitudes of the moments are:

Fig. 13 The magnitude of the moments


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Example: Moment and Force Equilibrium

A body or an object is said to be in equilibrium when the resultant moment and force vectorsare equal to zero.

Fx = 0 Fy = 0 Fz = 0

Mx = 0 My = 0 Mz = 0Example 1

Determine the moments about point O caused by the two applied forces.

X and Y components of the applied forces:

150 N Force:

Fx = 150 cos 45o = 106.1 N

Fy = 150 sin 45o = 106.1 N

250 N Force:

Fx = -250 sin 70 o = -234.9 N

Fy = 250 cos 70o = 85.51 N

Applied forces in Cartesian vector form:

F150 N = (106.1 i + 106.1 j )

F250 N = (-234.9 i + 85.51 j )

Position vector of point 1 and 2 relative to point O in Cartesian form:

r 1/O= (5.00 i + 10.00 j ) m

r 2/O= (-5.00 i + 10.00 j ) m

Using the following equation:


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0FF

0r r

k ji

rxFM

yx

yxo

= (rxFy - ryFx)k

= Mzk

MO1= [(5.00)(106.1) - (10.00)(106.1)] k= -530.4 k Nm= 530.4 Nm (clockwise)MO2= [(-5.00)(85.51) - (10.00)(234.9)] k

= -2777 k Nm= 2777 Nm (clockwise)

Example 2

Determine the moment about point A induced by the applied force of 1000 lb.

Applied force in Cartesian form:

F = 152 i + 531 j + 834 k

Position vector of point A relative to the origin:r O/A = 500 i + 700 j - 150 k

Using the following equation:

zyx

zyxo

FFF

r r r

k ji

rxFM

= (r yFz - r zFy)i + (r zFx - r xFz) j + (r xFy - r yFx)k

= M xi + M y j + M zk

MAO = [(700)(834) - (-150)(531) i .+ (-150)(152) - (500)(834) j + (500)(531) - (700)(152) k ]lb.in.

MAO = (663.5x103i - 439.8x10 3 j + 159.1x10 3k ) lb-in.


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3 Matrix algebra

In mathematics we have many ways of expressing numbers and each method use its own rulesof manipulation (algebra). For example, vector algebra is quite different but logical and veryuseful in the solution of problems. Matrix theory is another tool that helps us to solve

problems of a particular kind in Engineering and Science and it is also linked to vector theory.The theory is quite difficult to understand and the rules used to manipulate a matrix are quitecomplex. It will help if you have studied vector theory. In order to fully understand matrixtheory, you will need to have studies to quite an advanced level. This section is only a

beginning.

In this section you will be introduced to some of the basics terms and operations that we can perform with them. We shall also look at more advanced ideas and how they are used to solvesimultaneous equations

On the completion of this chapter, you should be able to do the following Explain the general method for solving system of equations Calculate determinants Calculate minor and cofactors Define and form the adjoint matrix Define and form the inverse matrix Solve system of equations

3.1 Properties of matrices

2.1.1 Matrix description

A matrix is an array of numbers or elements arranged in columns and rows. Each element isdesignated by a letter followed a row and column as shown. The matrix shown might belabeled as A and all the elements as a.

The size of the array is always stated as:

The number of rows x number of columnsThe one shown here is a 3x3 matrix

Columns

1 2 3


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3.1.2 Transposed matrix

This is a matrix in which row 1 becomes column 1; row 2 becomes column 2 and so on. Thetranspose of matrix A is designated as A T. An example is shown below.

3

84

7

52

A384

752A T

Note that the transpose of a unit matrix is still a unit matrix

3.1.3 Adding and subtracting matrices

Consider how we add two similar equations (2x+3y+5z) + (3x+4y-2z). We arrange them sothe variables line up in columns and then add the coefficients.

2x+3y+5z3x+4y-2z

Add 5x+7y+3z

To show this a matrix operation we would write the following. In other words we simply addor subtract the elements in the same position in each array. So it follows that the arrays must

be the same size.

zy

x

275zy

x

243zy

x

432

We add and subtract two matrices, we dont need to write in the column vector as it is thesame for each part so we only need to write

275243432

Example

Add and subtract the two arrays as shown


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3.1.4 Multiplication

Scalar

If an equation is multiplied by a constant (scalar) every coefficient will be multiplied by that

constant so a matrix may be easily multiplied or divided by a scalar

333231

232221

131211

333231

232221

131211

aaaaaaaaa

aaaaaaaaa

Vector

Now we will consider how to multiply a matrix by a column vector X to obtain a column

vector b. We are trying to find AX = b.

We want the result to be three equations

a11x+a 12y+a 13z =b 1

a22x+a 22y+a 23z =b 2

a31x+a 32y+a 33z =b 3

It follows that the rule is

(column 1 of A) x (row 1 of X)



It will be seen next that this is a special case of multiplication of arrays in general. For twoarrays the rule is more complex.

Column vectors

X b

A=


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Array multiplication

Consider how we multiply an array A by another B to obtain a new array C

2221

1211

aa

aa

A

2221

1211

b b

b b

B

2221

1211

cc

cc

CAB

Each element in C is obtained as follows

2222122122

2122112121

2212121112

2112111111

ba bac ba bac ba bac ba bac

This is the general rule. Note that a row x column produces a row. For element c 11 , take thefirst row of A and multiply by the first column of B

2112111121

111211 ba ba b

baa

For element c 12, take the first row of A and multiply by the second column of B

2212121122

121211 ba ba b

baa

For element c 21, take the second row of A and multiply by the first column of B

2122112121

112221 ba ba b

baa

For element c 22, take the second row of A and multiply by the second column of B

2222122122

122221 ba ba b

baa

Collecting all

2222122121221121

2212121121121111

ba ba ba ba

ba ba ba baC

The rule may be summarized as follows and extended to larger arrays


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2.xCol2.Row1.xCol2.Row

2.xCol1.Row1.xCol1.RowC

Other rules

It can be shown that the normal rules of multiplication may be observed if the orders ofmatrices A, B, C are mentioned on the left side of the equality sign

A(BC) = (AB)C

(A+B)C = AC+BC

A(B+C)= AB+AC

It can also be shown that (AB) T= B TAT

Example: Consider the matrices

01

10A

1

2B 510C

Evaluate ( AB)C and A( BC ). Check that you get the same matrix.

Answer: We have

21

AB

so

105

21

00

51021

C)AB(

On the other hand, we have

5

10

1

2

0

0)BC(

so

10

5

2

1

0

0

5

10

1

2

0

0

01

10)BC(A


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Example. Consider the matrices

11

21A

11

21B

Then it is easy to check that

AB = I 2 and BA = I 2

The identity matrix behaves like the number 1 not only among the matrices of the form nxn.Indeed, for any nxm matrix A, we have

InA = A and AI m =A

3.2 Procedure to solve simultaneous equation with a matrix system

An important use of matrix theory is the solution problem with more than two unknownvariables and because the method is based on strict rules, it is suitable for use in computer

programs. Consider how three simultaneous equations are presented as a vector. The Matrix Ais called the Coefficient matrix and it is formed from an array of numbers made from thecoefficients a 11 , a12,

The column vector b is made from the coefficients b 1, b 2,.and the column vector X is madeup from the variables, x, y, and z

The column vector b is the product of matrix A and X

a11x+a 12y+a 13z =b 1

a21x+a 22y+a 23z =b 2

a31x+a 32y+a 33z =b 3

We write this as AX = b

Definition:

Suppose that we had a matrix B such that AB = I (The unit matrix), we then could state as:

ABX = IX = Bb , X = Bb

This gives a numerical solution for X. the problem is finding the matrix B such that AB = I.

This matrix is called the inverse matrix and we need to understand the following section inorder to determine the inverse matrix.

Column vectors

X bA


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3.3 Determinants

The determinant of a matrix is a single number that results from performing a specificoperation on the array. It will be used later to solve simultaneous equations. The determinant

of a matrix A is denoted by det A or A. The rule for finding the determinant can onlyapplied to a square matrix and the following is an explanation for it.

For a 2x2 array the determinant is found as follows.

2221

1211

aa

aaAdet

A=a11*a22 - a12*a21

An important point to remember is that the determinant of all unit matrix is 1

For larger square arrays, the rule for finding the determinant is more complicated and it iscrucial to understand the following in order to do it.

3.4 Minors and Cofactors

If we cross out one row and one column of a matrix and find the determinant of the remainingarray, we have the minor. The minor is designated M and the subscript is the number of therow and the column eliminated

333231

232221

131211

aaa

aaa

aaa

3332

23221111 aa

aadetAM

The cofactor is numerically the same as the minor but it changes sign for every position in theraw or column and the change in sign is indicated by the pattern shown. This is designatedwith a letter corresponding to the elements so in this case the minor would be A 32 and fromthe sign patter we find A 32 = M 32


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Consider how we find the determinant of the following 3x3 matrix

333231

232221

131211

aaa

aaa

aaa

A

STEP 1Put a line through row 1 and column 1 leaving the elements shows.Find the determinant of the 2x2 array enclosed in the square. Thisis called the MINOR of a11 and designated by M11The COFACTOR is A 11 = M 11

333231

232221

131211

aaa

aaa

aaa

STEP 2Put a line through row 1 and column 2 leaving the elements shown.From these into a 2x2 array and find the determinant. The result is

the Minor M12. The COFACTOR is A 12 = -M 12 333231

232221

131211

aaa

aaa

aaa

STEP 3Put a line through row 1 and column 3 leaving the elements shown.From these into a 2x2 array and find the determinant. The result isthe Minor M13.The COFACTOR is A 13 = M 13 The determinant of the whole array is now found from:

A=a11A11+a12A12+a13A13 333231

232221

131211

aaa

aaa

aaa

For larger arrays the method is the same but the process is repeated until we left with a 2x2array. The cofactors takes one the sign as indicated by the elements position shown below


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3.5 Adjoint matrix

Another concept used in matrix methods is the adjoint or adjugate matrix. This has veryuseful properties in the solution of problems. This is a matrix formed from all the cofactors ofthe original matrix and then transposed. We designate this adj

If we had 3x3 matrix designated A, the adjoint is given as

T

333231

232221

131211

333231

232221

131211

AAAAAAAAA

adjA,aaaaaaaaa

A

3.6 Inverse matrix

We are in the last stage now and next we will be able to solve system of equations. Supposewe had two matrices A and B such that the product is the unit matrix, i.e

AB = I and it follows that BA = I

Matrix B is the inverse of matrix A so we denote it A -1 and replace B with this, so AA -1 =I

We have already used the relationship A adj A =AI

So equating we have A adj A = AAA-1

This result: A-1= (adj A)/ A

Where, the determinant is not equal to zero.

Example: For a square matrix of order 2, we have

ac bd

a bcd

dc ba

adjT

which gives

ac bd

bcad1

dc ba

1


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Example solving with inverse matrices

Let

22

32A and

112/31

B

One may easily check that 2I10

01BAAB

Hence A is invertible and B is its inverse.

Notation. A common notation for the inverse of a matrix A is A-1. So

AA -1 = A -1 A = I n

Example. Find the inverse of

21

11A

Write

dc

baA 1

Since 21 I

d2 bc2ad bcaAA

we get

1d2 b

0d b

0c2a

1ca

Algebraic manipulations give

32

a 31

b 31

c 31

d

or

3

1

3

131

32

A 1


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The inverse matrix is unique when it exists. So if A is invertible, then A-1 is also invertible and

AA 11

The following basic property is very important: If A and B are invertible matrices, then AB isalso invertible and

111 ABAB


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4 Modeling via Differential Equations

One of the most difficult problems that a scientist deals with in his everyday research is:"How do I translate a physical phenomenon into a set of equations which describes it?''

It is usually impossible to describe a phenomenon totally , so one usually struggle for a set ofequations which describes the physical system approximately and adequately .

In general, once we have built a set of equations, we compare the data generated by theequations with real data collected from the system (by measurement). If the two sets of data"agree'' (or are close), then we gain confidence that the set of equations will lead to a gooddescription of the real-world system. For example, we may use the equations to make

predictions about the long-term behavior of the system. It is also important to keep in mindthat the set of equations stays only "valid" as long as the two sets of data are close. If a

prediction from the equations leads to some conclusions which are by no means close to thereal-world future behavior, then we should modify and "correct" the underlying equations. Asyou can see, the problem of generating "good" equations is not an easy exercise.

Note that the set of equations is called a Model for the system.

How do we build a Model?

The basic steps in building a model are:

Step 1: Clearly state the assumptions on which the model will be based. These

assumptions should describe the relationships among the quantities to be studied.

Step 2: Completely describe the parameters and variables to be used in the model.

Step 3: Use the assumptions (from Step 1) to derive mathematical equations relatingthe parameters and variables (from Step 2).

4.1 First order differential equation

Examples of first order differential linear equations with constant coefficients included:exponential growth, modelling examples including radioactive decay and time delay equation.

A first order linear differential equation has the following form:

)x(qy)x( pdx

dy

The general solution is given by

)x(u

Cdx)x(q)x(uy


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where

dx)x( pexp)x(u

called the integrating factor . If an initial condition is given, it can be used to find theconstant C .

Here are some practical steps to follow:

1. If the differential equation is given as

)x(cy)x( bdx

dy)x(a

Rewrite it in the form

)x(qy)x( pdx

dy

where

)x(a)x( b

)x( p and )x(a

)x(c)x(q

2. Find the integrating factor

dx)x( pe)x(u

3. Evaluate the integral

dx)x(q)x(u

4. Write down the general solution

)x(u

Cdx)x(q)x(uy

5. If you are given an IVP, use the initial condition to find the constant C .

Example: Find the particular solution of:

2)0(y,)x(cosy)xtan(y 2'

Solution: Let us use the steps:


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Step 1: There is no need for rewriting the differential equation. We have

)xtan()x( p

)x(cos)x(q 2

Step 2: Integrating factor

)xsec(eee)x(u ))xln(sec())xln(cos(dx)xtan(

Step 3: We have

)xsin(dx)xcos(dx)x(cos)xsec( 2

Step 4: The general solution is given by

)xcos(C)xsin()xsec( C)xsin(y

Step 5: In order to find the particular solution to the given IVP, we use the initialcondition to find C . Indeed, we have

y(0) = C =2

Therefore the solution is

)xcos(2)xsin(y

Application

1 Radioactive decay modeling

Many radioactive materials disintegrate at a rate proportional to the amount present. Forexample, if X is the radioactive material and Q(t ) is the amount present at time t , then the rateof change of Q(t ) with respect to time t is given by

rQdtdQ

where r is a positive constant ( r >0). Let us call Q(0) = Q 0 the initial quantity of the material X , then we have

rt0eQ)t(Q


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Clearly, in order to determine Q(t ) we need to find the constant r . This can be done usingwhat is called the half-life T of the material X . The half-life is the time span needed todisintegrate half of the material. So, we have Q(T) = Q o. This gives rT = ln(2).

Therefore, if we know T , we can get r and vice-versa. For example, the half-life of Carbon-14is 5568 +/-30years. Therefore, the constant r associated with Carbon-14 is r =1.24410 -4. As aside note, Carbon-14 is an important tool in the archeological research known as radiocarbondating .

Example: A radioactive isotope has a half-life of 16 days. You wish to have 30 g at the endof 30 days. How much radioisotope should you start with?

Solution: Since the half-life is given in days we will measure time in days. Let Q(t ) be theamount present at time t and Q o the amount we are looking for (the initial amount). We know

that

rt0eQ)t(Q

where r is a constant. We use the half-life T to determine r . Indeed, we have

)2ln(16

1)2ln(

T

1r

Hence, since 30r 0eQ30)30(Q

we get g04.110e30Q)2ln(

16

30

0

2 Population Dynamics modeling

Here are some natural questions related to population problems:

What will the population of a certain country be in ten years?

Given that the rate of change of the population is proportional to the existing population. Inother words, if P (t ) measures the population, we have

kPdtdP

where the rate k is constant. It is fairly easy to see that if k > 0, we have growth, and if k


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if k >0, then the population grows and continues to expand to infinity, that is,o Lim P(t) (as r goes to infinity) = +infinity

if k


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Example: Undamped System

If there is no external force applied on the system, f(t) = 0, the system will experience freevibration . Motion of the system will be established by an initial disturbance (i.e. initialconditions). Furthermore, if there is no resistance or damping in the system,c v =0, the

oscillatory motion will continue forever with a constant amplitude. Such a system is termedundamped and is shown in the following figure

Solution for Undamped Systems

The equation of motion derived can be simplified to,

0kxx.m..

With the initial conditions,

o

.o

)0t(x

x)0t(x

This equation of motion is a second order, homogeneous, ordinary differential equation(ODE). If the mass and spring stiffness are constants, the ODE becomes a linearhomogeneous ODE with constant coefficients and can be solved by the CharacteristicEquation method. The characteristic equation for this problem is,

0k s.m 2

which determines the 2 independent roots for the undamped vibration problem. The finalsolution (that contains the 2 independent roots from the characteristic equation and satisfiesthe initial conditions) is,

tsindtcosd)t(x n2n1

tsintcosx)t(x nn

ono

The natural frequency n is defined by,

mk

n

km

xlo


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and depends only on the system mass and the spring stiffness (i.e. any damping will notchange the natural frequency of a system).

Alternatively, the solution may be expressed by the equivalent form,

)tcos(A)t(x ono

where the amplitude A0 and initial phase 0 are given by,

2

n

o2oo xA

no

o1

o xtan

Sample Time Behavior

The displacement plot of an undamped system would appear as,

Please note that an assumption of zero damping is typically not accurate. In reality, there almost always exists some resistance in vibratory systems. This resistance will damp thevibration and dissipate energy; the oscillatory motion caused by the initial disturbance willeventually be reduced to zero. This type of problem will not be studied here.


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4.3 Numerical solution methods

4.3.1 Newton-Raphons non-linear root finding

Root finding in one dimension

Consider the Taylor Series expansion of f ( x) about some point x = x0:

f(x) = f(x 0) + (x-x 0)f'(x 0) + (x-x 0)2f"(x 0) + O(|x-x 0|

3).

Setting the quadratic and higher terms to zero and solving the linear approximation of f ( x) = 0 for x gives

)x(f

)x(f xx

n'

001

Subsequent iterations are defined in a similar manner as

)x(f )x(f

xxn

'n

n1n

Geometrically, xn+1 can be interpreted as the value of x at which a line, passing through the point (xn,f(x n)) and tangent to the curve f(x) at that point, crosses the y axis. Figure provides a

graphical interpretation of this

Figure3: Graphical interpretation of the Newton Raphson algorithm.


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Example 1. Let us approximate the only solution to the equation

x = cos(x)

In fact, looking at the graphs we can see that this equation has one solution.

This solution is also the only zero of the function f(x) = x-cos(x) . So now we see how Newton's method may be used to approximate r . Since r is between 0 and /2, we set x1 = 1.The rest of the sequence is generated through the formula

)xsin(1)xcos(x

x)x(f )x(f

xxn

nnn

n'

nn1n

We havex1 1x2 0,75036386784024400x3 0,73911289091136200x4 0,73908513338528400x5 0,73908513321516100x6 0,73908513321516100x7 0,73908513321516100x8 0,73908513321516100

-3

-2

-1

0

1

2

3

-3 -2 -1 0 1 2 3

x

cos (x)


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Example 2 Let us find an approximation to 5 to ten decimal places.

Note that 5 is an irrational number. Therefore the sequence of decimals which defines 5will not stop. Clearly 5x is the only zero of f ( x) = x2 - 5 on the interval [1,3]. See thePicture.

Let nx be the successive approximations obtained through Newton's method. We have

n

2n

nn

'n

n1nx2

5xx

)x(f

)x(f xx

Let us start this process by taking x1 = 2.

It is quite remarkable that the results stabilize for more than ten decimal places after only 5iterations!

-6

-4

-2

0

2

4

6

0 0.5 1 1.5 2 2.5 3

x1 2

x2 2,25x3 2,23611111111111000x4 2,23606797791580000x5 2,23606797749979000x6 2,23606797749979000


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4.3.2 Numerical intergration

4.3.2.1 Simpson

In numerical analysis, Simpson's rule is a method for numerical integration, the numerical

approximation of definite integrals. Specifically, it is the following approximation:

b

a

b f ba

f a f ab

dx x f )(2

4)(6

)()( .

The method is credited to the mathematician Thomas Simpson

Simpsons 1/3 Rule for Integration- Examples

Example 1 Electrical Engineering

For an oscillator to have its frequency within 5% of the target of 1 kHz, the likelihood of thishappening can then be determined by finding the total area under the normal distribution forthe range in question:

dxe2

1f 2

x9.2

15.2

2

a) Use Simpsons 1/3 Rule to find thefrequency.

Solution

a)

) b(f 2

baf 4)a(f

6

a bf

15.2a

9.2b

37500.02

ba


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2

2

2

1)(

x

e x f

215.2 2

2

115.2 e f

039550.0

29.2 2

2

19.2 e f

0059525.0

2

375.0 2

21375.0 e f

37186.0

) b(f 2

baf 4)a(f

6

a bf

9.237500.0415.26

15.29.2( f f f

0059525.037186.04039550.0605.5

2902.1

4.3.2.2 Trapezium rule

The trapezoidal rule (also known as the trapezoid rule or trapezium rule ) is anapproximate technique for calculating the definite integral

b

a

dx)x(f

Figure: Graphical interpretation of the trapezium rule.


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The trapezoidal rule works by approximating the region under the graph of the function f ( x) asa trapezoid and calculating its area. It follows that

b

a 2

) b(f )a(f )a b(dx)x(f

In general, suppose we need to integrate from x0 to x1. We shall subdivide this interval into n steps of size x=(x 1-x0)/n as shown in figure.

Example 1 Chemical Engineering

A very simplified model of the reaction developed suggests a functional relation in an integralform. To find the time required for 50% of the oxygen to be consumed, the time, sT isgiven by

dx x

xT

6

6

1061.0

1022.1 11

7

10316.2103025.473.6

a) Use single segment Trapezoidal rule to find the time required for 50% of the oxygento be consumed.

b) Find the true error, t E , for part (a).

c) Find the absolute relative true error, t , for part (a).

Solution

a) 2

b f a f ab f , where

61022.1 a

61061.0 b

x x

x f 117

10316.2

103025.473.6)(

11

611

766 100581.3

1022.110316.2

103025.41022.173.61022.1 f

11

611

766 102104.3

1061.010316.2

103025.41061.073.61061.0 f


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2

102104.3100582.31022.11061.0

111166 f

s109119.1 5

b) The exact value of the above integral is,

dx x

xT

6

6

1061.0

1022.1 11

7

10316.2

103025.473.6

s109014.1 5

so the true error is

Valuee Approximat ValueTrue E t

55 109119.1109014.1

2.1056

c) The absolute relative true error, t , would then be

100ValueTrue

Error Truet 100

109014.1

2.10565

%55549.0

4.3.3 Numerical differentiation

4.3.3.1 Numerical Eulers solution methods

Consider a first order differential equation of the form )y,t(f

dt

dy

subject to some boundary/initial condition f (t =t 0) = c. The finite difference solution of thisequation proceeds by discretising the independent variable t to t 0, t 0+ t , t 0+ 2 t , t 0+ 3 t , We shall denote the exact solution at some t = t n = t 0+n t by yn = y (t=t n) and our approximatesolution by Y n. We then look to solve

Y' n = f (t n,Y n)


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The Euler method is the simplest finite difference scheme to understand and implement. Byapproximating the derivative in as

Y' n ~ (Y n+1 - Y n)/ t

in our differential equation for Y n we obtain

Yn+1 ~ Y n + tf(t n,Yn).

Given the initial/boundary condition Y 0 = c, we may obtain Y 1 from Y 0 + tf (t 0,Y 0), Y 2 fromY 1 + tf (t 1,Y 1) and so on, marching forwards through time. This process is shown graphicallyin figure .

Sketch of the function y(t ) (dark line) and the Euler method solution (arrows). Each arrow istangental to to the solution of passing through the point located at the start of the arrow. Notethat this point need not be on the desired y(t ) curve.

Example 1The concentration of salt x in a homemade soap maker is given as a function of time by

x

dt dx

5.35.37

At the initial time, 0t , the salt concentration in the tank is 50 g/L. Using Eulers methodand a step size of min5.1h , what is the salt concentration after 3 minutes?

Solution

xdt dx

5.35.37

x xt f 5.35.37,

The Eulers method reduces to

h xt f x x iiii ,1


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For 0i , 00 t , 500 x

h xt f x x 0001 ,

5.150,050 f

5.1)50(5.35.3750

5.15.13750

g/L25.156

1 x is the approximate concentration of salt at

ht t t 01 5.10 min 5.1

g/L25.1565.1 1 x x

For 1i , 5.11 t , 25.1561 x

h xt f x x 1112 , 5.125.156,5.125.156 f

5.1)25.156(5.35.3725.156 5.138.58425.156

g/L31.720

2 x is the approximate concentration of salt at

ht t t 12 5.15.1 min 3

g/L31.7203 2 x x

Figure 1 compares the exact solution with the numerical solution from Eulers method forstep size of 5.1h .


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Figure 1 Comparing exact and Eulers method

The problem was solved again using smaller step sizes. The results are given below in Table1.

Table 1 Concentration of salt at 3 minutes as a function of step size, h .

stepsize,h

3 x t E %|| t

3483.0

6622.2

2556.5

0.023249

0.010082

Figure 2 shows how the concentration of salt varies as a function of time for different stepsizes.


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Figure 2 Comparison of Eulers method with exact solution for different step sizes

The exact solution of the ordinary differential equation is given by

t et x 5.3286.39714.10)(

The solution to this nonlinear equation at min3t is

g/L 715.10)3( x


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4.3.4 Curve fitting

This section presents both the theory of curve fitting and compute methods using Excel

4.3.4.1 Linear , polynomical, exponential and power

Curve fitting - capturing the trend in the data by assigning a single function across the entirerange.The example below uses a straight line function

A straight line is described by f(x) = ax+b

The goal is to identify the coefficients a and b such that f(x) fitsthe data well Quantifying the error in the curve fit

Assumption

1) Positive or negative error have the same value (data point is above or below the line)

244

233

222

211

sint po.data#

1i

2i ))x(f y())x(f y())x(f y())x(f y()d(error

Our fit is a straight line, so now substitute f(x) = ax + b

s podata

iii

s podata

iii bax y x f yerror

int.#

1

2int.#

1

2 ))(())((

The best fit line has a minimum error between the line and data points. This is called the leastsquare approach, since we minimize the square of the error.

Minimizing the error

n

1iiii 0) baxy(x2a

)error (

(x4 , f(x4))

(x2 , f(x2))(x3 , f(x3))

(x1 , f(x1))

(f(x) = ax+b


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n

1iii 0) baxy(2 b

)error (

Solve for the a and b so that the previous two equations both =0

Rewrite these two equations

iii2i yxx bxa

ii yn* bxa

ii

i2ii

i

yx

y

a

b

xx

xn

We have the data points (xi, yi) for all i = 1..n, so we have all the summation terms in thmatrix form.

The unknowns are a and b

We already knows how to solve matrix

2ii

i

xx

xnA

a

bX

ii

i

yx

yB

AX = B

Using matlab, the coefficients a and b can be solved using matrix inversion

>> X = A -1*B

Note: A, B and X are not the same as a, b and x

Let us test this with an example:

I 1 2 3 4 5 6X 0 0.5 1 1.5 2 2.5Y 0 1.5 3 4.5 6 7.5

First we find values for all the summation terms, n =6

5.7x i , 5.22y i 75.13x 2i , 25.41yx ii

Now plugging into the matrix form gives us:


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25.41

5.22

a

b

75.135.7

5.76

25.415.22

*75.135.75.76

inva

b

The solution is

30

a b

f(x) = 3x+0

This fits the data exactly. That is, the error is zero. Usually this is not the outcome. Usuallywe have data that doesnt exactly fit a straight line.

Example 2

Here is an example with some noisy data

x = [ 0 .5 1 1.5 2 2.5],

y = [-0.4326 -0.1656 3.1253 4.7877 4.8535 8.6909]

6584.418593.20

a b

75.135.75.76

6584.41

8593.20*

75.135.7

5.76inv

a

b

561.3975.0

a b

So our fit is f(x) = 3.561 x 0.975


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The following figures shows data fitting with Excel

Figure: Linear regression (trend) fitting for the data given in Example 2

y = 3.5621x - 0.9761R = 0.9334

-2

0

2

4

6

8

10

0 0.5 1 1.5 2 2.5 3

Y

X

x

Linear (x)

y = 0.3537x 2 + 2.6779x - 0.6814R = 0.9383

-2

0

2

4

6

8

10

0 0.5 1 1.5 2 2.5 3

Y

X

x

Poly. (x)

mathematics and computation

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