ECE257 Numerical Methods and Scientific Computing...Dept. of Electrical and Computer Engineering...

ECE257 Numerical Methods andECE257 Numerical Methods andScientific ComputingScientific Computing

Ordinary Differential EquationsOrdinary Differential Equations

ECE 257 Numerical Methods and Scientific Computing

Fall 2004

Lecture 19

John A. Chandy

Dept. of Electrical and Computer Engineering

University of Connecticut

TodayToday’’s class:s class:

•• Boundary Value ProblemsBoundary Value Problems

•• Eigenvalue ProblemsEigenvalue Problems


Fall 2004

Lecture 19

John A. Chandy



Boundary Value ProblemsBoundary Value Problems

•• Auxiliary conditionsAuxiliary conditions

–– nn-th order equation requires -th order equation requires nn conditions conditions

–– Initial value conditionsInitial value conditions•• All All nn conditions are specified at the same conditions are specified at the same

value of the independent variablevalue of the independent variable

–– Boundary value conditionsBoundary value conditions•• The The nn conditions are specified at different conditions are specified at different

values of the independent variablevalues of the independent variable


Fall 2004

Lecture 19

John A. Chandy



Initial value conditionsInitial value conditions

From Numerical Methods for Engineers, Chapra and Canale, Copyright © The McGraw-Hill Companies, Inc.


Fall 2004

Lecture 19

John A. Chandy



Boundary value conditionsBoundary value conditions

From Numerical Methods for Engineers, Chapra and Canale, Copyright © The McGraw-Hill Companies, Inc.


Fall 2004

Lecture 19

John A. Chandy



Types of Boundary ConditionsTypes of Boundary Conditions

)( ayay =

)()(

α=⋅+⋅ aybdx

ady a

)(

α=== dx

ady

dx

dy

ax

• Simple B.C (Dirichlet B.C): The value of theunknown function is given at the boundary.

• Neumann B.C.: The derivative of the unknownfunction is given at the boundary.

• Mixing B.C.: The combination of the unknownfunction’s value and derivative is given at theboundary.


Fall 2004

Lecture 19

John A. Chandy



Boundary Value ProblemsBoundary Value Problems

•• Shooting MethodShooting Method

–– Convert the boundary value problem into anConvert the boundary value problem into anequivalent initial value problem by choosingequivalent initial value problem by choosingvalues for all dependent variables at onevalues for all dependent variables at oneboundaryboundary

•• Relaxation MethodRelaxation Method

–– Choose values along the range of theChoose values along the range of theintegration and gradually adjust values sointegration and gradually adjust values sothat they satisfy the integral and boundarythat they satisfy the integral and boundaryvaluesvalues


Fall 2004

Lecture 19

John A. Chandy



Shooting MethodShooting Method

•• Solve an Solve an NNth order ODE from th order ODE from xx11 toto x x22

•• Assume Assume nn11 boundary conditions at boundary conditions at xx11 and and nn22boundary conditions at boundary conditions at xx22

•• There are There are nn22=N-n=N-n11 freely specifiable starting values freely specifiable starting valuesat at xx11

•• Guess at those Guess at those nn22 values and then solve the values and then solve theresulting initial value ODE at resulting initial value ODE at xx22

•• Check if the boundary conditions are satisfied at Check if the boundary conditions are satisfied at xx22..If not, adjust the If not, adjust the nn22 values at values at xx11 and try again. and try again.


Fall 2004

Lecture 19

John A. Chandy



Shooting Method ExampleShooting Method Example

•• Convert to first-orderConvert to first-order

€

d2ydx 2

+ 0.01 20 − y( ) = 0 y 0( ) = 40,y 10( ) = 200

€

dydx

= z y 0( ) = 40,y 10( ) = 200

dzdx

+ 0.01 20 − y( ) = 0

•• One free unspecified value at x=0 - guess aOne free unspecified value at x=0 - guess avalue for z(0)value for z(0) €

z 0( ) =10

•• Use RK method or any other method toUse RK method or any other method tosolve the ODE at x=10solve the ODE at x=10


Fall 2004

Lecture 19

John A. Chandy




•• Using RK method with an initial guess of Using RK method with an initial guess of z(z(00)=)=1010 we get we gety(y(1010)=)=168.3797168.3797

•• The initial condition was The initial condition was y(y(1010)=)=200,200, so try again so try again

•• Using RK method with an initial guess of Using RK method with an initial guess of z(z(00)=2)=200 we get we gety(y(1010)=)=285.8980285.8980

•• With two guesses and a linear ODE, we can interpolateWith two guesses and a linear ODE, we can interpolateto find the correct valueto find the correct value

€

z 0( ) =10 +20 −10

285.8980 −168.3797200 −168.3797( )

=12.6907


Fall 2004

Lecture 19

John A. Chandy





Fall 2004

Lecture 19

John A. Chandy




•• If the ODE is non-linear, then you have to recast theIf the ODE is non-linear, then you have to recast theproblem as solving a series of root problemsproblem as solving a series of root problems

•• Think of the guessed values as a Think of the guessed values as a nn22 size vector size vector VV

•• That vector That vector VV will generate a will generate a yy(x(x22)) vector after vector aftersolving the ODEsolving the ODE

•• The The yy(x(x22)) vector should equal the boundary value vector should equal the boundary valueconditions at conditions at xx22


Fall 2004

Lecture 19

John A. Chandy




€

discrepancy = BV2 x2,y( ) − y

•• Since Since y y is a function of is a function of VV, you can create a new, you can create a newdiscrepancy function discrepancy function FF that is dependent on that is dependent on VV

•• The problem then becomes to zero the discrepancyThe problem then becomes to zero the discrepancy

•• We donWe don’’t know t know F F so we cant use normal root-so we cant use normal root-solving methodssolving methods

•• But we can approach it in a Newton-RaphsonBut we can approach it in a Newton-Raphsonmanner by iteratively looking at the finite differencemanner by iteratively looking at the finite differencederivativesderivatives

€

discrepancy = F V( )


Fall 2004

Lecture 19

John A. Chandy




•• The problem then becomes to zero the discrepancy by solvingThe problem then becomes to zero the discrepancy by solvingthe following linear systemthe following linear system

€

α[ ] ⋅ δV = −F

α ij =∂Fi∂Vj

∂Fi∂Vj

≈Fi V1,K,V j + ΔV j ,K,Vn2( ) − Fi V1,K,V j ,K,Vn2( )

ΔV j

•• The vector The vector VV can be adjusted with the results of the solution can be adjusted with the results of the solution

€

Vnew = Vold + δV

•• However, you may need multiple cycles to solve for However, you may need multiple cycles to solve for VV


Fall 2004

Lecture 19

John A. Chandy



Shooting MethodShooting MethodGuess initial values of vGuess initial values of vii (i=1 to n (i=1 to n22))for iter=1 to nitersfor iter=1 to nitersyy22 = Solve ODE at x = Solve ODE at x22F=BV-yF=BV-y22for j=1 to nfor j=1 to n22

vvjj= v= vjj++∆∆vvjjyy’’ = Solve ODE at x = Solve ODE at x22

FF’’=BV-y=BV-y’’for i=1 to nfor i=1 to n22

a[i][j] = (Fa[i][j] = (F’’[i]-F[i])/[i]-F[i])/∆∆vvjjendforendforrestore vrestore vjj

endforendforlinear solve (alinear solve (a••dv=-F)dv=-F)v = v + dvv = v + dvcheck errorcheck error

endforendfor


Fall 2004

Lecture 19

John A. Chandy



Finite-Difference Method (FDM)Finite-Difference Method (FDM)

€

y' '= p(x)y '+q(x)y + r(x) y(a) =α and y(b) = β

Approximate the derivatives using numerical difference schemes.

)(2

')('

),(2

'')(''

. segments. equaln into range theDividing

211

22

11

hOh

yyyxy

hOh

yyyyxy

n

abxh

iiii

iiiii

+−

==

++−

==

−=Δ=

−+

−+

x0 =a x1 xn=b

h

xixi-1xi+1


Fall 2004

Lecture 19

John A. Chandy



FiniteFinite-Difference Method (FDM) (continued)-Difference Method (FDM) (continued)

• Substituting the derivatives by approximations and then we have a set of discretized equations

€

At i =1,2,.......n −1

yi+1 − 2yi + yi−1h2

= piyi+1 − yi−12h

+ qiyi + ri

Multiply both sides by h2.

(1+ pih2)yi-1 - (2 + qih

2)yi + (1− pih2)yi+1 = h2ri


Fall 2004

Lecture 19

John A. Chandy



Finite-Difference Method (FDM) (continued)Finite-Difference Method (FDM) (continued)

€

For i =1.

-(2 + q1h2)y1 + (1− p1

h2)y2 = h2r1 − (1+ p1

h2)α

For i = 2,3,......,n − 2,

(1+ pih2)yi-1 - (2 + qih

2)yi + (1− pih2)yi+1 = h2ri

For i = n −1,

(1+ pih2)yn−2 - (2 + qih

2)yn−1 = h2rn−1 − (1− pn−1h2)β


Fall 2004

Lecture 19

John A. Chandy



Matrix Form of Matrix Form of EquationEquation SystemSystem

€

If h = a-b7

and n = 7, noticing that y(a) =α and y(b) = β.

-(2 + q1h2) 1− p1

h2

0 0 0 0

1+ p2h2

-(2 + q2h2) 1− p2

h2

0 0 0

0 1+ p3h2

-(2 + q3h2) 1− p3

h2

0 0

0 0 1+ p4h2

-(2 + q4h2) 1− p4

h2

0

0 0 0 1+ p5h2

-(2 + q5h2) 1− p5

h2

0 0 0 0 1+ p6h2

-(2 + q6h2)

y1y2y3y4y5y6

=

h2r1 − (1+ p1h2)α

h2r2h2r3h2r4h2r5

h2r6 − (1− p6h2)β


Fall 2004

Lecture 19

John A. Chandy



FDM ExampleFDM Example

€

d2ydx 2

+ 0.01 20 − y( ) = 0 y 0( ) = 40,y 10( ) = 200

€

p(x) = 0,q(x) = 0.01,r(x) = −0.2α = 40,β = 200

•• Set n=5, h=2Set n=5, h=2

€

−2.04 1 0 01 −2.04 1 00 1 −2.04 10 0 1 −2.04

y1y2y3y4

=

−40.8−0.8−0.8−200.8


Fall 2004

Lecture 19

John A. Chandy



Eigenvalue problemsEigenvalue problems

•• Eigenvalue problems are a special class ofEigenvalue problems are a special class ofboundary-value problemboundary-value problem

•• Common in engineering systems withCommon in engineering systems withoscillating behavioroscillating behavior

–– SpringsSprings

–– ElasticityElasticity

–– RLC circuitsRLC circuits


Fall 2004

Lecture 19

John A. Chandy



Physics of EigenvaluesPhysics of Eigenvalues & Eigenvectors & Eigenvectors

A system of two degrees offreedom.

m1

m2

k2(x2-x1) k3x2

k2(x2-x1)k1x1

To simplify the problem,we assume thatk1= k2= k3= k.

)(

)(

12222

2

2

12121

2

1

xxkkxdt

xdm

xxkkxdt

xdm

−−−=

−+−=


Fall 2004

Lecture 19

John A. Chandy



Physics of Physics of Eigenvalues Eigenvalues (continued)(continued)

Both masses (m1& m2) are oscillating w.r.t. their mean positions.

€

xi = Bi sin(ωt) i = 1 or 2.

d2xi

dt 2= −ω 2Bi sin(ωt) i = 1 or 2.

Substituting the above equations into the motion equations and dividing both sides by

€

−ω 2m1B1 = −kB1 + k(B2 − B1)

−ω 2m2B2 = −kB2 − k(B2 − B1)

€

2km1

−ω 2 −km1

−km2

2km2

−ω 2

B1B2

= 0

. )sin t(ω


Fall 2004

Lecture 19

John A. Chandy



Physics of Physics of Eigenvalues Eigenvalues (continued)(continued)

€

2km1

−km1

−km2

2km2

B1B2

=ω 2 B1B2

€

A[ ] x{ } = λ x{ }

Hence, the eigenvalues are related to the natural frequencies or periods which are crucial to the design of structures.The eigenvectors are related to the motion of a structure related to the corresponding natural frequency.For most engineering problems, the largest or the smallestfrequencies are the most important ones to know.


Fall 2004

Lecture 19

John A. Chandy



Steps of Solving Steps of Solving EigenEigen-value Problems-value Problems

• leads to an algebraic equation of .

•Solving the equation for . - Polynomial method (find roots of an equation) - Other methods (more efficient): Power Method.

•Given the eigenvalues , obtain the correspondingeigenvectors .

€

det A − λI[ ] = 0 . λ

λ

λ{ }x


Fall 2004

Lecture 19

John A. Chandy



Eigenvalue ExampleEigenvalue Example

€

m1 = m2 = 40,k = 200

€

10 −ω 2 −5−5 10 −ω 2

B1B2

= 0

€

det A − λI[ ] = 10 −ω 2( ) 10 −ω 2( ) − −5( ) −5( )

= ω 2( )2− 20ω 2 + 75 = 0

€

ω 2 =15 or 5

€

At ω 2 =15, B1 = −B2 and at ω 2 = 5, B1 = B2


Fall 2004

Lecture 19

John A. Chandy



Eigenvalue ExampleEigenvalue Example


Fall 2004

Lecture 19

John A. Chandy



Boundary value eigenvalue problemsBoundary value eigenvalue problems

•• Polynomial MethodPolynomial Method

–– Similar to Finite-Difference MethodSimilar to Finite-Difference Method

•• Power MethodPower Method

–– HotellingHotelling’’s Methods Method


Fall 2004

Lecture 19

John A. Chandy



Power MethodPower Method

The Power Method is an iterative procedure fordetermining the dominant (largest) eigenvalue of thematrix A and the corresponding eigenvector.

Example

€

3.556 −1.778 0−1.778 3.556 −1.7780 −1.778 3.556

x1x2x3

= λ

x1x2x3

1. Guess a solution for the eigenvector , x1 =1, x2 =1, x3 =1.


Fall 2004

Lecture 19

John A. Chandy



Example of Power Method (Continued)Example of Power Method (Continued)

{ } { } 2. and 1 stepsrepeat and

guess theas 101r eigenvecto normalizedcurrent theUsing.3

1

0

1

778.1

778.1

0

778.1

.1-or 1either becomes valueabsolutelargest theofelement that the

so side),-hand-right (ther eigenvecto eapproximat theNormalize .2

778.1

0

778.1

1

1

1

556.3778.10

778.1556.3778.1

0778.1556.3

Eq. of side -hand-left theintor eigenvecto thengsubstituti

=

=

=

−

−−

−

Tx


Fall 2004

Lecture 19

John A. Chandy




€

substituting the eigenvector into the left -hand - side of Eq.

3.556 −1.778 0−1.778 3.556 −1.7780 −1.778 3.556

101

=

3.556−3.5563.556

= 3.5561−11

4. Examine the relative error of λ,

εa =λi+1 − λiλi+1

=3.556 −1.778

3.556= 50%.

if εa is greater than the tolerance error, the iteration continues.

Otherwise, the eigenvalue λ, and the eigenvector x{ } are obtained.


Fall 2004

Lecture 19

John A. Chandy




€

Third Iteration

3.556 −1.778 0−1.778 3.556 −1.7780 −1.778 3.556

1−11

=

5.333−7.1115.333

= −7.111−0.751

−0.75

εa =−7.111− 3.556

−7.111=150%

€

Fourth Iteration

3.556 −1.778 0−1.778 3.556 −1.7780 −1.778 3.556

−0.751

−0.75

=

−4.4446.222−4.444

= 6.222−0.7141

−0.714

εa =6.222 − −7.111( )

6.222= 214%


Fall 2004

Lecture 19

John A. Chandy




€

Fifth Iteration

3.556 −1.778 0−1.778 3.556 −1.7780 −1.778 3.556

−0.7141

−0.714

=

−4.3176.095−4.317

= 6.095−0.7081

−0.708

εa =6.095 − 6.222

6.095= −2.08%

€

The sixth iteration

3.556 −1.778 0−1.778 3.556 −1.7780 −1.778 3.556

−0.7081

−0.708

=

−4.2966.074−4.296

= 6.074−0.7071

−0.707

εa =6.074 − 6.095

6.074= 0.35%


Fall 2004

Lecture 19

John A. Chandy



Inverse Power MethodInverse Power Method

• The Power Method can be modified to provide thesmallest (lowest) (magnitude) eigenvalue and itseigenvector. The modified method is known as theinverse power method

• The smallest eigenvalue λ in the matrix A corresponds to1/ λ the largest eigenvalue in the inverse matrix A-1.


Fall 2004

Lecture 19

John A. Chandy



General I-P MethodGeneral I-P Method

The combination of Shifted Power Method and I-P methodcan be use to determine all the eigenvalues and theirrelated eigenvectors.

•Deflation Hotelling’s Method for symmetric matrices•First, normalize the largest eigenvector found by thesum of the squares of the elements in the eigenvector

€

X{ } =X{ }

xk2

k=1

n

∑


Fall 2004

Lecture 19

John A. Chandy



General I-P MethodGeneral I-P Method

•• Then create new AThen create new A22 matrix matrix

€

A[ ]2 = A[ ]1 − λ1 X{ }1 X{ }1T

€

A[ ]2 X{ } j = A[ ]1 X{ } j − λ1 X{ }1 X{ }1T X{ } j

€

A[ ]2 X{ } j = λ j X{ } j j ≠1A[ ]2 X{ }1 = A[ ]2 X{ }1 − λ1 X{ }1 j =1

•• The new AThe new A22 matrix has the same matrix has the sameeigenvalues as before except that theeigenvalues as before except that thelargest eigenvalue has been replaced with 0largest eigenvalue has been replaced with 0


Fall 2004

Lecture 19

John A. Chandy



HotellingHotelling’’s Methods Method

€

A =

3.556 −1.778 0−1.778 3.556 −1.7780 −1.778 3.556

,λ1 = 6.074,x1 =

−0.7071

−0.707

€

Modified A2

A2 =

3.556 −1.778 0−1.778 3.556 −1.7780 −1.778 3.556

− 6.074−0.7071

−0.707

−0.707 1 −0.707{ }


Fall 2004

Lecture 19

John A. Chandy



HotellingHotelling’’s Methods Method

€

A2 =

3.556 −1.778 0−1.778 3.556 −1.7780 −1.778 3.556

− 6.0740.5 −0.707 0.5

−0.707 1 −0.7070.5 −0.707 0.5

€

A2 =

0.519 2.516 −3.0372.516 −2.518 2.516−3.037 2.516 0.519


Fall 2004

Lecture 19

John A. Chandy



Exam 2Exam 2

•• Parts 4, 6, and 7 of bookParts 4, 6, and 7 of book

•• OptimizationOptimization

•• Numerical Integration and DifferentiationNumerical Integration and Differentiation

•• Ordinary Differential EquationsOrdinary Differential Equations


Fall 2004

Lecture 19

John A. Chandy



Next LectureNext Lecture

•• Partial Differential EquationsPartial Differential Equations

•• Read Chapter PT8, 29Read Chapter PT8, 29

•• HW6 due 11/16HW6 due 11/16

•• Exam 2 11/9Exam 2 11/9

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