numerical differentiation & integration

Numerical Differentiation & Integration

Gaussian Quadrature

Numerical Analysis (9th Edition)

R L Burden & J D Faires

Beamer Presentation Slidesprepared byJohn Carroll

Dublin City University

c 2011 Brooks/Cole, Cengage Learning

Introduction Legendre Polynomials Arbitrary Intervals

Outline

1 Gaussian Quadrature & Optimal Nodes

Numerical Analysis (Chapter 4) Gaussian Quadrature R L Burden & J D Faires 2 / 40


Outline


2 Using Legendre Polynomials to Derive Gaussian Quadrature Formulae



Outline



3 Gaussian Quadrature on Arbitrary Intervals



Outline






Gaussian Quadrature: Contrast with Newton-Cotes

Features of a Newton-Cotes Formula




Features of a Newton-Cotes FormulaThe Newton-Cotes formulas were derived by integratinginterpolating polynomials.





The error term in the interpolating polynomial of degree n involvesthe (n + 1)st derivative of the function being approximated, . . .





The error term in the interpolating polynomial of degree n involvesthe (n + 1)st derivative of the function being approximated, . . .

so a Newton-Cotes formula is exact when approximating theintegral of any polynomial of degree less than or equal to n.




Features of a Newton-Cotes Formula (Contd)




Features of a Newton-Cotes Formula (Contd)All the Newton-Cotes formulas use values of the function atequally-spaced points.





This restriction is convenient when the formulas are combined toform the composite rules which we considered earlier, . . .





This restriction is convenient when the formulas are combined toform the composite rules which we considered earlier, . . .

but it can significantly decrease the accuracy of the approximation.



Consider, for example, the Trapezoidal rule applied to determine theintegrals of the functions whose graphs are as shown.

y

x

yy

xa 5 x1 a 5 x1 a 5 x1x2 5 b x2 5 b x2 5 bx

y 5 f (x)y 5 f (x)

y 5 f (x)

It approximates the integral of the function by integrating the linearfunction that joins the endpoints of the graph of the function.



Gaussian Integration: Optimal integration points

But this is not likely the best line for approximating the integral. Linessuch as those shown below would likely give much betterapproximations in most cases.

yyy

x x xa x1 bx2 a x1 bx2 a x1 bx2

y 5 f (x)

y 5 f (x)y 5 f (x)

Gaussian quadrature chooses the points for evaluation in an optimal,rather than equally-spaced, way.



Gaussian Quadrature: Introduction

Choice of Integration Nodes





The nodes x1, x2, . . . , xn in the interval [a,b] and coefficientsc1, c2, . . . , cn, are chosen to minimize the expected error obtainedin the approximation

b

af (x) dx

n

i=1

ci f (xi ).






b

af (x) dx

n

i=1

ci f (xi ).

To measure this accuracy, we assume that the best choice ofthese values produces the exact result for the largest class ofpolynomials, . . .






b

af (x) dx

n

i=1

ci f (xi ).

To measure this accuracy, we assume that the best choice ofthese values produces the exact result for the largest class ofpolynomials, . . .

that is, the choice that gives the greatest degree of precision.




b

af (x) dx

n

i=1

ci f (xi ).

Choice of Integration Nodes (Contd)




b

af (x) dx

n

i=1

ci f (xi ).

Choice of Integration Nodes (Contd)The coefficients c1, c2, . . . , cn in the approximation formula arearbitrary, and the nodes x1, x2, . . . , xn are restricted only by thefact that they must lie in [a,b], the interval of integration.




b

af (x) dx

n

i=1

ci f (xi ).

Choice of Integration Nodes (Contd)The coefficients c1, c2, . . . , cn in the approximation formula arearbitrary, and the nodes x1, x2, . . . , xn are restricted only by thefact that they must lie in [a,b], the interval of integration.

This gives us 2n parameters to choose.




b

af (x) dx

n

i=1

ci f (xi ).

Choice of Integration Nodes (Contd)




b

af (x) dx

n

i=1

ci f (xi ).

Choice of Integration Nodes (Contd)If the coefficients of a polynomial are considered parameters, theclass of polynomials of degree at most 2n 1 also contains 2nparameters.




b

af (x) dx

n

i=1

ci f (xi ).


This, then, is the largest class of polynomials for which it isreasonable to expect a formula to be exact.




b

af (x) dx

n

i=1

ci f (xi ).


This, then, is the largest class of polynomials for which it isreasonable to expect a formula to be exact.

With the proper choice of the values and constants, exactness onthis set can be obtained.



Gaussian Quadrature: Illustration (n = 2)

Example: Formula when n = 2 on [1, 1]Suppose we want to determine c1, c2, x1, and x2 so that the integrationformula

1

1f (x) dx c1f (x1) + c2f (x2)

gives the exact result whenever f (x) is a polynomial of degree2(2) 1 = 3 or less, that is, when

f (x) = a0 + a1x + a2x2 + a3x

3,

for some collection of constants, a0, a1, a2, and a3.




Finding the Formula Coefficients (1/3)Because

(a0 + a1x + a2x2 + a3x

3) dx

= a0

1 dx + a1

x dx + a2

x2 dx + a3

x3 dx

this is equivalent to showing that the formula gives exact results whenf (x) is 1, x , x2, and x3.




Finding the Formula Coefficients (2/3)Hence, we need c1, c2, x1, and x2, so that

c1 1 + c2 1 = 1

11 dx = 2





c1 1 + c2 1 = 1

11 dx = 2

c1 x1 + c2 x2 = 1

1x dx = 0





c1 1 + c2 1 = 1

11 dx = 2

c1 x1 + c2 x2 = 1

1x dx = 0

c1 x21 + c2 x22 = 1

1x2 dx =

23





c1 1 + c2 1 = 1

11 dx = 2

c1 x1 + c2 x2 = 1

1x dx = 0

c1 x21 + c2 x22 = 1

1x2 dx =

23

c1 x31 + c2 x32 = 1

1x3 dx = 0




Finding the Formula Coefficients (3/3)A little algebra shows that this system of equations has the uniquesolution

c1 = 1, c2 = 1, x1 =

33

and x2 =

3

3,





c1 = 1, c2 = 1, x1 =

33

and x2 =

3

3,

which gives the approximation formula

1

1f (x) dx f

(

33

)

+ f

(3

3

)





c1 = 1, c2 = 1, x1 =

33

and x2 =

3

3,

which gives the approximation formula

1

1f (x) dx f

(

33

)

+ f

(3

3

)

This formula has degree of precision 3, that is, it produces the exactresult for every polynomial of degree 3 or less.



Outline






Gaussian Quadrature: Legendre Polynomials

An Alternative Method of Derivation




An Alternative Method of DerivationWe will consider an approach which generates more easily thenodes and coefficients for formulas that give exact results forhigher-degree polynomials.





This will be achieved using a particular set of orthogonalpolynomials (functions with the property that a particular definiteintegral of the product of any two of them is 0).





This will be achieved using a particular set of orthogonalpolynomials (functions with the property that a particular definiteintegral of the product of any two of them is 0).This set is the is the Legendre polynomials, a collection{P0(x),P1(x), . . . ,Pn(x), . . . , } with properties:(1) For each n, Pn(x) is a monic polynomial of degree n.

(2) 1

1P(x)Pn(x) dx = 0 whenever P(x) is a polynomial of degree

less than n.



Gaussian Quadrature: Legendre PolynomialsThe first few Legendre Polynomials

P0(x) = 1, P1(x) = x , P2(x) = x2 1

3

P3(x) = x3 3

5x and P4(x) = x

4 67

x2 +3

35




P0(x) = 1, P1(x) = x , P2(x) = x2 1

3

P3(x) = x3 3

5x and P4(x) = x

4 67

x2 +3

35

The roots of these polynomials are distinct, lie in the interval(1,1), have a symmetry with respect to the origin, and, mostimportantly,




P0(x) = 1, P1(x) = x , P2(x) = x2 1

3

P3(x) = x3 3

5x and P4(x) = x

4 67

x2 +3

35


they are the correct choice for determining the parameters thatgive us the nodes and coefficients for our quadrature method.




P0(x) = 1, P1(x) = x , P2(x) = x2 1

3

P3(x) = x3 3

5x and P4(x) = x

4 67

x2 +3

35


they are the correct choice for determining the parameters thatgive us the nodes and coefficients for our quadrature method.

The nodes x1, x2, . . . , xn needed to produce an integral approximationformula that gives exact results for any polynomial of degree less than2n are the roots of the nth-degree Legendre polynomial.




TheoremSuppose that x1, x2, . . . , xn are the roots of the nth Legendrepolynomial Pn(x) and that for each i = 1,2, . . . ,n, the numbers ci aredefined by

ci = 1

1

n

j=1j 6=i

x xjxi xj

dx




TheoremSuppose that x1, x2, . . . , xn are the roots of the nth Legendrepolynomial Pn(x) and that for each i = 1,2, . . . ,n, the numbers ci aredefined by

ci = 1

1

n

j=1j 6=i

x xjxi xj

dx

If P(x) is any polynomial of degree less than 2n, then

1

1P(x) dx =

n

i=1

ciP(xi )




Proof (1/5)




Proof (1/5)Let us first consider the situation for a polynomial P(x) of degreeless than n.





Re-write P(x) in terms of (n 1)st Lagrange coefficientpolynomials with nodes at the roots of the nth Legendrepolynomial Pn(x).






The error term for this representation involves the nth derivative ofP(x).






The error term for this representation involves the nth derivative ofP(x).

Since P(x) is of degree less than n, the nth derivative of P(x) is 0,and this representation of is exact. So



Gaussian Quadrature: Legendre PolynomialsProof (2/5)



Gaussian Quadrature: Legendre PolynomialsProof (2/5)Therefore

P(x) =n

i=1

P(xi)Li(x) =n

i=1

n

j=1j 6=i

x xjxi xj

P(xi )




P(x) =n

i=1

P(xi)Li(x) =n

i=1

n

j=1j 6=i

x xjxi xj

P(xi )

and 1

1P(x) dx =

1

1

n

i=1

n

j=1j 6=i

x xjxi xj

P(xi )

dx




P(x) =n

i=1

P(xi)Li(x) =n

i=1

n

j=1j 6=i

x xjxi xj

P(xi )

and 1

1P(x) dx =

1

1

n

i=1

n

j=1j 6=i

x xjxi xj

P(xi )

dx

=n

i=1

1

1

n

j=1j 6=i

x xjxi xj

dx

P(xi) =n

i=1

ciP(xi )

Hence the result is true for polynomials of degree less than n.Numerical Analysis (Chapter 4) Gaussian Quadrature R L Burden & J D Faires 20 / 40



Proof (3/5)




Proof (3/5)Now consider a polynomial P(x) of degree at least n but less than2n.





Divide P(x) by the nth Legendre polynomial Pn(x).






This gives two polynomials Q(x) and R(x), each of degree lessthan n, with

P(x) = Q(x)Pn(x) + R(x)







P(x) = Q(x)Pn(x) + R(x)

Note that xi is a root of Pn(x) for each i = 1,2, . . . ,n, so we have

P(xi ) = Q(xi )Pn(xi) + R(xi) = R(xi)







P(x) = Q(x)Pn(x) + R(x)

Note that xi is a root of Pn(x) for each i = 1,2, . . . ,n, so we have

P(xi ) = Q(xi )Pn(xi) + R(xi) = R(xi)

We now invoke the unique power of the Legendre polynomials.




Proof (4/5)




Proof (4/5)First, the degree of the polynomial Q(x) is less than n, so (by theLegendre orthogonality property),

1

1Q(x)Pn(x) dx = 0




Proof (4/5)First, the degree of the polynomial Q(x) is less than n, so (by theLegendre orthogonality property),

1

1Q(x)Pn(x) dx = 0

Then, since R(x) is a polynomial of degree less than n, the openingargument implies that

1

1R(x) dx =

n

i=1

ciR(xi)




Proof (5/5)




Proof (5/5)Putting these facts together verifies that the formula is exact for thepolynomial P(x):





1

1P(x) dx =

1

1[Q(x)Pn(x) + R(x)] dx





1

1P(x) dx =

1


=

1

1R(x) dx





1

1P(x) dx =

1


=

1

1R(x) dx

=n

i=1

ciR(xi)





1

1P(x) dx =

1


=

1

1R(x) dx

=n

i=1

ciR(xi)

=n

i=1

ciP(xi)



Gaussian Quadrature: Roots & Coefficients

The constants ci needed for the quadrature rule can be generatedfrom the equation given in the theorem:

ci = 1

1

n

j=1j 6=i

x xjxi xj

dx

but both these constants and the roots of the Legendre polynomialsare extensively tabulated.

The following table lists these values for n = 2,3,4, and 5.



Gaussian Quadrature Rules: Roots & Coefficients

n Roots rn,i Coefficients cn,i

2 0.5773502692 1.00000000000.5773502692 1.0000000000

3 0.7745966692 0.55555555560.0000000000 0.8888888889

0.7745966692 0.55555555564 0.8611363116 0.3478548451

0.3399810436 0.65214515490.3399810436 0.65214515490.8611363116 0.3478548451

5 0.9061798459 0.23692688500.5384693101 0.47862867050.0000000000 0.5688888889

0.5384693101 0.47862867050.9061798459 0.2369268850




Example (n = 2)

Approximate 1

1ex cos x dx using Gaussian quadrature with n = 3.

SolutionThe entries in the table of roots and coefficients See Table




Example (n = 2)

Approximate 1


SolutionThe entries in the table of roots and coefficients See Table give us

1

1ex cos x dx 0.5e0.774596692 cos 0.774596692 + 0.8 cos 0

+ 0.5e0.774596692 cos(0.774596692)= 1.9333904.




Example (n = 2)

Approximate 1


SolutionThe entries in the table of roots and coefficients See Table give us

1

1ex cos x dx 0.5e0.774596692 cos 0.774596692 + 0.8 cos 0

+ 0.5e0.774596692 cos(0.774596692)= 1.9333904.

Integration by parts can be used to show that the true value of theintegral is 1.9334214, so the absolute error is less than 3.2 105.



Outline






Gaussian Quadrature on Arbitrary Intervals

Transform the Interval of Integration to [1, 1]An integral

ba f (x) dx over an arbitrary [a,b] can be transformed into

an integral over [1,1] by using the change of variables See Diagram :

t =2x a b

b a x =12

[(b a)t + a + b]




Transform the Interval of Integration to [1, 1]An integral

ba f (x) dx over an arbitrary [a,b] can be transformed into

an integral over [1,1] by using the change of variables See Diagram :

t =2x a b

b a x =12

[(b a)t + a + b]

This permits Gaussian quadrature to be applied to any interval [a,b],because

b

af (x) dx =

1

1f(

(b a)t + (b + a)2

)

(b a)2

dt




Example: Comparing FormulaeConsider the integral

3

1x6 x2 sin(2x) dx = 317.3442466.





3

1x6 x2 sin(2x) dx = 317.3442466.

(a) Compare the results for the closed Newton-Cotes formula withn = 1, the open Newton-Cotes formula with n = 1, and GaussianQuadrature when n = 2.





3

1x6 x2 sin(2x) dx = 317.3442466.

(a) Compare the results for the closed Newton-Cotes formula withn = 1, the open Newton-Cotes formula with n = 1, and GaussianQuadrature when n = 2.

(b) Compare the results for the closed Newton-Cotes formula withn = 2, the open Newton-Cotes formula with n = 2, and GaussianQuadrature when n = 3.



Solution: Part (a): Newton-Cotes Formulae (n = 1)Each of the formulas in this part requires 2 evaluations of the functionf (x) = x6 x2 sin(2x). The Newton-Cotes approximations See Formulaeare:

Closed n = 1 :2)2

[f (1) + f (3)] = 731.6054420

Open n = 1 :3(2/3)

2[f (5/3) + f (7/3)] = 188.7856682




Solution: Part (a): Gaussian Quadrature (n = 2)Gaussian quadrature applied to this problem requires that the integralfirst be transformed into a problem whose interval of integration is[1,1]. Using the transformation gives

3

1x6 x2 sin(2x) dx =

1

1(t + 2)6 (t + 2)2 sin(2(t + 2)) dt .




Solution: Part (a): Gaussian Quadrature (n = 2)Gaussian quadrature applied to this problem requires that the integralfirst be transformed into a problem whose interval of integration is[1,1]. Using the transformation gives

3

1x6 x2 sin(2x) dx =

1

1(t + 2)6 (t + 2)2 sin(2(t + 2)) dt .

Gaussian quadrature with n = 2 then gives

3

1x6 x2 sin(2x) dx

f (0.5773502692 + 2) + f (0.5773502692 + 2)= 306.8199344




Solution: Part (b): Newton-Cotes Formulae (n = 2)Each of the formulas in this part requires 3 function evaluations. TheNewton-Cotes approximations See Formulae are:

Closed n = 2 :1)3

[f (1) + 4f (2) + f (3)] = 333.2380940

Open n = 2 :4(1/2)

3[2f (1.5) f (2) + 2f (2.5)] = 303.5912023




Solution: Part (b): Gaussian Quadrature (n = 3)Gaussian quadrature with n = 3, once the transformation has been done,gives

3

1x6 x2 sin(2x) dx

0.5f (0.7745966692 + 2) + 0.8f (2) + 0.5f (0.7745966692 + 2)

= 317.2641516




3

1x6 x2 sin(2x) dx = 317.3442466.




3

1x6 x2 sin(2x) dx = 317.3442466.

Comparison of Results

Newton-CotesClosed Open Gaussian Quadrature

2-Point Rule 731.6054420 188.7856682 306.8199344

3-Point Rule 333.2380940 303.5912023 317.2641516

The Gaussian quadrature results are clearly superior in each instance.


Questions?

Reference Material

Gaussian Quadrature Rules: Roots & Coefficients

n Roots rn,i Coefficients cn,i

2 0.5773502692 1.00000000000.5773502692 1.0000000000

3 0.7745966692 0.55555555560.0000000000 0.8888888889

0.7745966692 0.55555555564 0.8611363116 0.3478548451

0.3399810436 0.65214515490.3399810436 0.65214515490.8611363116 0.3478548451

5 0.9061798459 0.23692688500.5384693101 0.47862867050.0000000000 0.5688888889

0.5384693101 0.47862867050.9061798459 0.2369268850

Return to 11 e

x cos(x) dx Example

Mapping Interval [a, b] onto [1, 1]

t

x

21

1

a b

(a, 21)

(b, 1)

2x 2 a 2 bt 5

b 2 a

Return to Gaussian Quadrature on Arbitrary Intervals (Introduction)

Return to Gaussian Quadrature on Arbitrary Intervals (Example Part (a))

Return to Gaussian Quadrature on Arbitrary Intervals (Example Part (b))

Open & Closed Newton-Cotes Formulae (n = 1)

Closed n = 1: Trapezoidal Rule:

x1

x0f (x) dx =

h2

[f (x0) + f (x1)] h3

12f ()

where x0 < < x1.

Open n = 1:

x2

x1

f (x) dx =3h2

[f (x0) + f (x1)] +3h3

4f ()

where x1 < < x2.

Return to arbitrary intervals Example (where n = 1)

Open & Closed Newton-Cotes Formulae (n = 2)

n = 0: Midpoint Rule:

x1

x1

f (x) dx = 2hf (x0) +h3

3f ()

where x1 < < x1.

Open n = 2 x3

x1

f (x) dx =4h3

[2f (x0) f (x1) + 2f (x2)] +14h5

45f (4)()

where x1 < < x3.

Return to arbitrary intervals Example (where n = 1)

Initial-Value Problems for ODEs

Elementary Theory of Initial-Value Problems

Numerical Analysis (9th Edition)

R L Burden & J D Faires

Beamer Presentation Slidesprepared byJohn Carroll

Dublin City University

c 2011 Brooks/Cole, Cengage Learning

Lipschitz Condition Unique Solution Well-Posed Problems Example

Outline

1 Lipschitz Condition & Convexity

Numerical Analysis (Chapter 5) Elementary Theory of Initial-Value Problems R L Burden & J D Faires 2 / 25


Outline


2 The Existence of a Unique Solution



Outline



3 Well-Posed Problems



Outline




4 Example of a Well-Posed Problem



Outline







Elementary Theory of IVPs: Lipschitz Condition

We begin by presenting some definitions and results from the theory ofordinary differential equations before considering methods forapproximating the solutions to initial-value problems.




We begin by presenting some definitions and results from the theory ofordinary differential equations before considering methods forapproximating the solutions to initial-value problems.

Definition: Lipschitz ConditionA function f (t , y) is said to satisfy a Lipschitz condition in the variable yon a set D IR2 if a constant L > 0 exists with

|f (t , y1) f (t , y2, )| L |y1 y2|

whenever (t , y1) and (t , y2) are in D. The constant L is called aLipschitz constant for f .




ExampleShow that f (t , y) = t |y | satisfies a Lipschitz condition on the intervalD = { (t , y) | 1 t 2 and 3 y 4 }.





SolutionFor each pair of points (t , y1) and (t , y2) in D we have

|f (t , y1) f (t , y2)| = |t |y1| t |y2|| = |t | ||y1| |y2|| 2|y1 y2|







Thus f satisfies a Lipschitz condition on D in the variable y withLipschitz constant 2.







Thus f satisfies a Lipschitz condition on D in the variable y withLipschitz constant 2. The smallest value possible for the Lipschitzconstant for this problem is L = 2, because, for example,

|f (2,1) f (2,0)| = |2 0| = 2|1 0|



Elementary Theory of IVPs: Convex Set

Definition: Convex Set

A set D IR2 is said to be convex if whenever (t1, y1) and (t2, y2)belong to D, then

((1 )t1 + t2, (1 )y1 + y2)

also belongs to D for every in [0,1].

(t1, y1)

(t1, y1)(t2, y2)

(t2, y2)

Convex Not convex




Comment on the Definition




Comment on the DefinitionIn geometric terms, the definition states that a set is convexprovided that whenever two points belong to the set, the entirestraight-line segment between the points also belongs to the set.





The sets we consider in this section are generally of the form

D = { (t , y) | a t b and < y < }

for some constants a and b.





The sets we consider in this section are generally of the form

D = { (t , y) | a t b and < y < }

for some constants a and b.

It is easy to verify that these sets are convex.



Theory of IVPs: Lipschitz Condition & Convexity

Theorem: Sufficient Conditions

Suppose f (t , y) is defined on a convex set D IR2. If a constant L > 0exists with

fy

(t , y)

L, for all (t , y) D

then f satisfies a Lipschitz condition on D in the variable y withLipschitz constant L.



Theory of IVPs: Lipschitz Condition & Convexity

Theorem: Sufficient Conditions

Suppose f (t , y) is defined on a convex set D IR2. If a constant L > 0exists with

fy

(t , y)

L, for all (t , y) D

then f satisfies a Lipschitz condition on D in the variable y withLipschitz constant L.

As the next result will show, this theorem is often of significant interestto determine whether the function involved in an initial-value problemsatisfies a Lipschitz condition in its second variable, and the abovecondition is generally easier to apply than the definition.



Outline







Elementary Theory of IVPs

Theorem: Existence & UniquenessSuppose that D = { (t , y) | a t b and < y < } and thatf (t , y) is continuous on D. If f satisfies a Lipschitz condition on D in thevariable y , then the initial-value problem

y (t) = f (t , y), a t b, y(a) = ,

has a unique solution y(t) for a t b.




Theorem: Existence & UniquenessSuppose that D = { (t , y) | a t b and < y < } and thatf (t , y) is continuous on D. If f satisfies a Lipschitz condition on D in thevariable y , then the initial-value problem

y (t) = f (t , y), a t b, y(a) = ,

has a unique solution y(t) for a t b.

Note: This is a version of the fundamental existence and uniquenesstheorem for first-order ordinary differential equations. The proof of thetheorem, in approximately this form, can be found in Birkhoff, G. andG. Rota, Ordinary differential equations, (4th edition), John Wiley &Sons, New York, 1989.




Example: Applying the Existence & Uniqueness TheoremUse the Existence & Uniqueness Theorem to show that there is aunique solution to the initial-value problem

y = 1 + t sin(ty), 0 t 2, y(0) = 0




Example: Applying the Existence & Uniqueness TheoremUse the Existence & Uniqueness Theorem to show that there is aunique solution to the initial-value problem

y = 1 + t sin(ty), 0 t 2, y(0) = 0

Solution (1/2)Holding t constant and applying the Mean Value Theorem See Theoremto the function

f (t , y) = 1 + t sin(ty)

we find that when y1 < y2, a number in (y1, y2) exists with

f (t , y2) f (t , y1)y2 y1

=

yf (t , ) = t2 cos(t)




f (t , y2) f (t , y1)y2 y1

=

yf (t , ) = t2 cos(t)

Solution (2/2)




f (t , y2) f (t , y1)y2 y1

=

yf (t , ) = t2 cos(t)

Solution (2/2)Thus

|f (t , y2) f (t , y1)| = |y2 y1||t2 cos(t)| 4|y2 y1|

and f satisfies a Lipschitz condition in the variable y with Lipschitzconstant L = 4.




f (t , y2) f (t , y1)y2 y1

=

yf (t , ) = t2 cos(t)

Solution (2/2)Thus

|f (t , y2) f (t , y1)| = |y2 y1||t2 cos(t)| 4|y2 y1|

and f satisfies a Lipschitz condition in the variable y with Lipschitzconstant L = 4.

Additionally, f (t , y) is continuous when 0 t 2 and < y < , so the Existence & Uniqueness Theorem impliesthat a unique solution exists to this initial-value problem.



Outline







Elementary Theory of IVPs: Well-Posed problems

Question




Question

How do we determine whether a particular problem has the propertythat small changes, or perturbations, in the statement of the problemintroduce correspondingly small changes in the solution?




Question

How do we determine whether a particular problem has the propertythat small changes, or perturbations, in the statement of the problemintroduce correspondingly small changes in the solution?

We first need to give a workable definition to express this concept.



Elementary Theory of IVPs: Well-Posed Problems

Definition: Well-Posed ProblemThe initial-value problem

dydt

= f (t , y), a t b, y(a) =

is said to be a well-posed problem if the following 2 conditions aresatisfied:




Definition: Well-Posed Problem (Continued)A unique solution, y(t), to the problem exists, and





There exist constants 0 > 0 and k > 0 such that for any , with0 > > 0, whenever (t) is continuous with |(t)| < for all t in[a,b], and when |0| < , the initial-value problem

dzdt

= f (t , z) + (t), a t b, z(a) = + 0

has a unique solution z(t) that satisfies

|z(t) y(t)| < k for all t in [a,b].

Note: The problem in z, as specified above, is called a perturbedproblem associated with the original problem for y .




Conditions to ensure that an initial-value problem is well-posed.




Conditions to ensure that an initial-value problem is well-posed.

Theorem: Well-Posed ProblemSuppose D = { (t , y) | a t b and < y < }. If f is continuousand satisfies a Lipschitz condition in the variable y on the set D, thenthe initial-value problem

dydt

= f (t , y), a t b, y(a) =

is well-posed.

The proof of this theorem can be found in Birkhoff, G. and G. Rota,Ordinary differential equations, (4th edition), John Wiley & Sons, NewYork, 1989.



Outline








Example: Applying the Theorem on Well-Posed ProblemsShow that the initial-value problem

dydt

= y t2 + 1, 0 t 2, y(0) = 0.5

is well posed on D = { (t , y) | 0 t 2 and < y < }.




Solution (1/3)Because

(y t2 + 1)y

= |1| = 1

the Lipschitz Condition theorem implies that f (t , y) = y t2 + 1satisfies a Lipschitz condition in y on D with Lipschitz constant 1.




Solution (1/3)Because

(y t2 + 1)y

= |1| = 1

the Lipschitz Condition theorem implies that f (t , y) = y t2 + 1satisfies a Lipschitz condition in y on D with Lipschitz constant 1.

Since f is continuous on D, the Theorem on Well-Posed Problemsimplies that the problem is well-posed.




Solution (2/3)




Solution (2/3)As an illustration, consider the solution to the perturbed problem

dzdt

= z t2 + 1 + , 0 t 2, z(0) = 0.5 + 0

where and 0 are constants.




Solution (2/3)As an illustration, consider the solution to the perturbed problem

dzdt

= z t2 + 1 + , 0 t 2, z(0) = 0.5 + 0

where and 0 are constants.

The solutions to the original problem and this perturbed problemare

y(t) = (t + 1)2 0.5et

and z(t) = (t + 1)2 + ( + 0 0.5)et

respectively.




y(t) = (t + 1)2 0.5et

z(t) = (t + 1)2 + ( + 0 0.5)et

Solution (3/3)




y(t) = (t + 1)2 0.5et

z(t) = (t + 1)2 + ( + 0 0.5)et

Solution (3/3)Suppose that is a positive number. If || < and |0| < , then

|y(t) z(t)| = |( + 0)et | | + 0|e2 + || (2e2 + 1)

for all t .




y(t) = (t + 1)2 0.5et

z(t) = (t + 1)2 + ( + 0 0.5)et

Solution (3/3)Suppose that is a positive number. If || < and |0| < , then

|y(t) z(t)| = |( + 0)et | | + 0|e2 + || (2e2 + 1)

for all t .

This implies that the original problem is well-posed withk() = 2e2 + 1 for all > 0.


Questions?

Reference Material

Mean Value Theorem

If f C[a,b] and f is differentiable on (a,b), then a number c existssuch that

f (c) =f (b) f (a)

b a

y

xa bc

Slope f 9(c)

Parallel lines

Slopeb 2 a

f (b) 2 f (a)

y 5 f (x)

Return to Existence & Uniqueness Example

Gaussian Quadrature & Optimal NodesUsing Legendre Polynomials to Derive Gaussian Quadrature FormulaeGaussian Quadrature on Arbitrary IntervalsLecture 19.pdfLipschitz Condition & ConvexityThe Existence of a Unique SolutionWell-Posed ProblemsExample of a Well-Posed Problem

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