advanced engineering analysis

8/12/2019 Advanced Engineering Analysis

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Advanced Engineering Analysis - Notes

Ramesh Kadambi

November 14, 2009


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Contents

1 Ordinary Differential Equations

1.1 ODE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Principle of Superposition (Linearity Test) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 General Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4 Generalized Linear ODE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5 Difference Linear ODE and Non Linear DE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.6 Linear Dependence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.7 Establishing Linear Dependence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1.8 Linear Equation Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.8.1 Homogeneous Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.9 First Order ODE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.10 LDE ofnth order with constant coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.11 C ase of Repeated Roots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.12 Complex Roots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.13 Method of Undetermined Coefficients MOUC . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.14 Method of Variation Of Parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.15 Euler Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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4 CONTENT


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Chapter 1

Ordinary Differential Equations

1.1 ODE

definition: An ODE is a statement of functional dependence.

1.2 Principle of Superposition (Lin-earity Test)

f(ax1+bx2) = af(x1) +bf(x2)

1.3 General Solution

The general solution to an ODE is the set of all functional re-lations that satisfy the ODE.

ex: d2y

dx2+y = 0 solution y = c1sin(x) +c2cos(x)

1.4 Generalized Linear ODE

a0(x)dny

dxn+a1(x)

dn1y

dxn1+....+any= f(x)

1.5 Difference Linear ODE and NonLinear DE

A nonlinear DE has nonlinear function of the depended vari-able.ex:

yd2y

dx2+exy2

dy

dx=y

1.6 Linear Dependence

A linear combination of functions are linearly independent overan interval x [a, b] if over the interval no nontrivial linearcombination of the functions is identically 0. This basicallymeans that none of the functions of the set can be expressed interms of the other functions in the set. In mathematical terms,

ci such thatn

i=1ciui(x) = 0, unless ci = 0, x [a, b]

1.7 Establishing Linear Dependence

Consider ui(x) defined in the interval [a, b] such that ui(have n finite derivatives in the interval [a, b]. If ci such th

n

i=1

ciui(x) = 0, then we just differentiate ntimes and since ea

of the equations so obtained must also sum to zero, we have tfollowing,

ni=1

ciui(x) = 0

ni=1

ciu1i (x) = 0

ni=1

ciu2i (x) = 0

n

i=1

ciun1i = 0

u1 u2 u3 unu11 u

12 u

13 u

1n

u21 u22 u

23 u

2n

un11 un12 u

n13 u

n1n

c1c2c3

cn

= 0

For a non trivial solution we have,

W(u1,...,un)(x) =

u1 u2 u3 unu11 u

12 u

13 u

1n

u21 u22 u

23 u

2n

un11 un12 u

n13 u

n1n

= 0

for some x [a, b]

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6 CHAPTER 1. ORDINARY DIFFERENTIAL EQUATION

1.8 Linear Equation Solution

The standard from ODE is given as

dnf(x)

dxn +a1(x)

dn1f(x)

dxn1 + +an1

dy

dx+ an(x)f(x) = h(x)

. The expression

dn

dxn+a1(x)

dn1

dxn1+ +an1(x)

d

dx+ an(x)

is called the linear differential operator. So the equation can berewritten as

Lf(x) = h(x)

. One way to solve the ODE is to integrate things through ntimes. Each integration step will leave you with a constantci,leaving you with n constants. Every possible solution can beobtained by specifying values to these constants. Another pointto note is that the solutions differ from each other only by aconstant.

1.8.1 Homogeneous Solution

The solution of the ODE is broken into two parts. The first sol-tuion is called the homogeneous solution fh(x) solves the DELfh(x) = 0. The second part is called the particular solutionwhich will solve the DE Lfp(x) = h(x). The complete solutionis then given by

f(x) = fh(x) +fp(x)

.

1.9 First Order ODEThe first order ODE is obtained when n = 1. So the generalFODE is given by,

df(x)

dx +a1(x)f(x) = h(x) (1.9.1)

. The solution to this is obtained by solving two parts.

dfh(x)

dx +a1(x)fh(x) = 0

dfh(x)

dx = a1(x)fh(x)

1fh(x)

dfh(x) = a1(x)dx

integrating both sides we get

log(fh(x)) =

a1(x)dx+C

fh(x) = e

R a1(x)dx+C =e

R a1(x)dx +C

In order to obtain the particular solution we do the followinwe are looking for a function P(x) such that,

dP(x)fp(x)

dx =P(x)h(x) (1.9

. In order to do that we first multiply our original ODE P(x) We have,

P(x)df

(x

)dx +P(x)a1(x)f(x) = P(x)h(x) (1.9

This method is called the method of integrating factor. Aintegrating factor is a function that is chosen to facilitate tsolving of a given differential. The integrating factor converan inexact differential to an exact differential. This finds appcations in multivariate calculus. The interesting aspect of thmethod is that it can be used to solve some types of non-linedifferential equations1.We now obtain the differential of d

dxP(x)fh(x)

dP(x)fp(x)

dx =P(x)

dfp(x)

dx +fp(x)

dP(x)

dx (1.9.

comparing the original ODE (1.9.3) with (1.9.4) we see thatdP(x)

dx =P(x)a1(x)

P(x) = eR

a1(x)dx+C =eR

a1(x)dx +C

Now integrating (1.9.2) we obtain,

P(x)fp(x) =

P(x)h(x)dx+C

fp(x) = 1

P(x)

P(x)h(x)dx+C

(1.9.

Example:

1. xdy

dx ky = x2

writing in cannonical form

dy

dx

k

xy= x

we haveP(x) = eR

a(x)dx where a(x) = k

x

P(x) = eR k

xdx+C =C1e

k lnx =C1xk

yp(x) = 1

C1xk

C1x

kxdx+ C2

C1xk

=xkxk+2

k+ 2+ Cxk if k != 2

= x2

2 k+Cxk k != 2

if k = 2 then we go back to our original problem and solve fthat specific case.

fork = 2 we have

yp(x) = x2

1

xdx+Cx2

yp(x) = x2lnx+Cx2

1Wikipedia url:http://en.wikipedia.org/wiki/Integrating factor


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1.10. LDE OFNT H ORDER WITH CONSTANT COEFFICIENTS

1.10 LDE of nth order with constantcoefficients

Consider the differential operator L given by,

L= dn

dxn+a1

dn1

dxn1+ +an1

d

dx+ an

In this section we will look into solving ODE of the form,

Ly= h(x)

, Let us assume a solution of the form y = erx where r is aconstant. We would like to solve the homogeneous equationLy = 0. Differentiating the sulution of the form y = erx andsubstituting we have,

(rn +a1rn1 + +an1r+an)e

rx = 0

since erx

= 0 we have the following polynomial in terms of requal to zero,

rn +a1rn1 + +an1r+an= 0

Let r1, r2, , rn be the roots of the polynomial. The solutionyh is now of the form,

yh= c1er1x +c2e

r2x + +cnernx =

ni=1

cierix

1.11 Case of Repeated RootsSuppose some of the roots are repeated, i.e. r = r1 = r2 areroots then we have,

Lerx =rn +a1rn1 +a2r

n2 +anerx

= (r r1)(r r1)(r r3) (r rn)erx

From the above equations we can see that

r[Lerx]r=r1 = 0

We no switch the differential operator (I dont know the tech-nical restrictions to do so but will do it anyway).

L[erx

r ] = 0

L[xerx]r=r1 = 0

This implies that xer1x is a sulution to the ODE as well. Itfollows from the above argument that if there a root repeats ntimes then xer1x, x2er1x, ...., xner1x are all solutions.

1.12 Complex Roots

Since the coefficients of the Characteristic Equation (CE) aall real, the complex roots will be in complex conjugate paiThe solution is given by Aer1x +Ber2x where r1 = a +ib anr2 = a ib. The solution can then be written as

yH=eax[(A+B)cos bx+ (A+B)sin bx]

.

Example 1: d4y

dx4 k4y= 0

r4 k4 = 0

The roots are r1 = k, r2 = k, r3 = ik,r4 = ik. The solutiy= c1e

kx +c2ekx +c3cos kx+c4sin kx

Example 2: y y y + y= 0 The CE r3 r2 r+ 1 =The CE simplifies to (r 1)(r2 1) = 0 We there fore ha(r 1)2(r+ 1) = 0. The solution is therefore give by:

y= c1ex +c2xe

x c3ex

1.13 Method of Undetermined Coeficients MOUC

The objective is to find the particular solution. This methworks only on systems with constant coefficients. The RHhas have certain kinds of functions in order for this method work. The idea is to guess the solution to be of the same foras the right hand of the inhomogeneous equation.

Homogenious Soln Solution Formkxm, m= 0, 1, 2, knx

n + +kn1xn

keax Ceax

k cos axor k sin ax A cos ax+Bsin axkeax sin bx or k eax cos bx eax[A cos ax+Bcos(n

1 kixi)cos ax or (

n1 kix

i)sin ax eax[cos axn

1 Aixi

+sin axn

1 Bixi]


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8 CHAPTER 1. ORDINARY DIFFERENTIAL EQUATION

Example 1: y +y = sin x

y = yp+yh

we first find the homogeneous solution

y +y = 0

d2 +d= 0

d= i

yh= A cos x+Bsin x

using the method of undetermined coefficients and the fact that

the homogeneous solution is not linearly independent of the RHS.

yp = x[K1cos x+K2sin x]

substituting the particular solution into the DE

x[K1cos x K2sin x] +x[K1cos x+K2sin x]

+ 2K2cos x 2K1sin x= sin x

2K2cos x 2K1sin x= sin x

K1 = 1

2

The complete solutiony = A cos x+Bsin x

x

2cos x

1.14 Method of Variation Of Parameters

1.15 Euler Equation

The Euler Equidimensional Linear D.E is given below., Ly[xn d

n

dxn+b1x

n1 +bn]y= h(x) where, bi are constants. W

will solve this problem by using a change of variables. We wuse the substitution x= ez, i.ez = log x. We now see that tEuler equation gets converted to a constant coefficient ODE

dy

dx=

dy

dz

dz

dx=

1

x

dy

dz

xdy

dx=

dy

dz

d2y

dx2 =

1

x

d

dz

dy

dx=

1

x

d

dz

1

x

dy

dz

=

1

x2dy

dz+

1

x2d2y

dz2

x2d2y

dx2 =

d

dz

d

dz 1

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