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Solutions to the Problems in

Numerical Analysis

by David Kincaid and Ward Cheney

John Weatherwax

1

Text copyright©2022 John L. Weatherwax

All Rights Reserved

Please Do Not Redistribute Without Permission from the Author 2

To my family.

3 IntroductionThis is a solution manual forsomeof the problems in the excellent numerical analysis textbook: Numerical Analysis: Mathematics of Scientific Computing by David Kincaid and Ward Cheney This solution manual was prepared form thefirstedition of the textbook. I"m currently working on finishing more of the problems in this book. Inthe meantime I"m publishing my partial results for any student who does not want to wait for the full book to be finished. One of the benefits of this manual is that I use theRstatistical language to perform any of the needed numerical computations (rather than do them "by-hand"). Thus if you work though this manual you will be learning theRlanguage at the same time as you learn statistics. TheRprogramming language is one of the most desired skills for anyone whohopes to use data/statistics in their future career. TheRcode can be found at the following location: As a final comment, I"ve worked hard to make these notes as good as I can, but I have no illusions that they are perfect. If you feel that that there is a better way to accomplish or explain an exercise or derivation presented in these notes; or that one or more of the explanations is unclear, incomplete, or misleading, please tell me. If you find an error of any kind - technical, grammatical, typographical, whatever - pleasetell me that, too. I"ll gladly add to the acknowledgments in later printings the name of the first person to bring each problem to my attention. 4

Chapter 2: Computer ArithmeticStable and Unstable Computations; ConditioningNotes on Numerical InstabilityNow ifxis single precision then whenx1=1

3the absolute error in representingx1in a

computer will be of order 10 -8. Note that in the formula used to computexn+1fromxnand x n-1we are multiplyingxnby13

3at each iteration. Thus the absolute error will grow by a

factor of 13

3at each iteration.

•This means thatx2has an absolute error of?13

3?10-8.

•andx3has absolute error?13 3?

210-8,

•andx4has absolute error?13 3?

310-8.

Continuing this logic we see thatxnhas absolute error of?13 3? n-110-8. To make the absolute error inxn=O(10) which would make the calculation worthless since the actual answer is?1 3? nand we would have a complete loss of all significant digits we would need ?13 3? n-1

×10-8≈101,

or ?13 3? n-1 ≈10+9, or solving this fornwe find (n-1)≈9log10 log?133? = 14.13268. Thusn≈15.13268 so it takes around fifteen iterations before we have an absolute error or O(10) (which is what we see in the example in the book). If we use double precision instead of single precision then absolute error in represengint x 1=1

3in a computer will beO(10-16). This means that the absolute error of iterationnfor

x nwill be so large as to render the computation "worthless" when ?13 3? n-1

×10-16≈101.

Solving this fornwe find

n-1≈17log10 log?133? = 26.69505.

Thus whenn≥27.69505 all precision is lost.

5

Chapter 9: Numerical Solutions of Partial DifferentialEquationsFirst-Order Partial Differential Equations; Characteristic Curves

Problem 1

Part (a):From the given differential equation the method of characteristicswould seek functionsx(s),y(s), andu(s) such that dx ds= 1 dy ds=x du ds= 0, with the initial conditions x(0) = 0 y(0) =r u(0) =f(r). Integrating the differential equation forx(s) givesx(s) =s. Integrating the differential equation foru(s) givesu(s) =u(0) =f(r). Using these in the differential equation fory(s) givesdy ds=s, or y(s) =s2 2+r.

Now fromx=sandy=s2

2+rwe haver=y-x22so thatu=f(r) becomes

u=f? y-x2 2? for the solutionu=u(x,y). Part (b):From the given differential equation the method of characteristicswould seek functions dx ds= 1 dy ds= 2u du ds= 0, 6 with initial conditions given by x(0) = 0 y(0) =r u(0) =f(r). Integrating the differential equation forx(s) we findx(s) =s. Integrating the differential equation foru(s) givesu(s) =u(0) =f(r). Using these in the differential equation fory(s) givesdy ds= 2f(r).

Integrating we get

y= 2f(r)s+C , forCa constant. Using the initial conditiony(0) =rgivesC=rso y= 2f(r)s+r. Thus the solution is represented in parameterized form as x=s y= 2f(r)s+r u=f(r). Part (c):From the given differential equation the method of characteristicswould seek functions dx ds=x dy ds= 2y du ds= 0, with initial conditions given by x(0) = 1 y(0) =r u(0) =f(r).

Integrating these three equations gives

x(s) =C1es y(s) =C2e2s u(s) =C3. Using the initial conditions to evaluate the constantsC1,C2, andC3whens= 0 we get x(s) =es y(s) =re2s u(s) =f(r). 7

The first two of these equations combine to give

y=rx2orr=y x2.

This means that

u=f?y x2? for the solutionu=u(x,y).

Problem 2

We are given the differential equation

u x+yuy= 0, with the initial condition of u(18,3e) =kπ 2. The method of characteristics seeks to find functionsx(s),y(s), andu(s) such that dx ds= 1 dy ds=y du ds= 0, with initial conditions given by x(0) = 18 y(0) = 3e u(0) =kπ 2. If we integrate the above and use these initial conditions we get x(s) =s+ 18 y(s) = 3es+1 u(s) =kπ 2. If we seek to evaluateu(17,3) then we first need to findssuch that s+ 18 = 17

3es+1= 3.

Both of these equations are true whens=-1. Using the solution foru(s) above we see that u(-1) =kπ 2. 8 Problem 3The differential equation in Example 8 is xu x+yuuy=xy , with the boundary-values of u(x,y) = 2xywhenxy= 3.

The solution found there was

u(x,y) =-1 +?

43 + 2xy .

From this note that

u x=2y

2⎷43 + 2xy=y⎷43 + 2xy

u y=2x

2⎷43 + 2xy=x⎷43 + 2xy,

so that xu x+yuuy=xy ⎷43 + 2xy+xyu⎷43 + 2xy=xy⎷43 + 2xy(1 +u) =xy , as it should to be a solution. Next ifxy= 3 note that u(x,y) =-1 +⎷

43 + 6 =-1 + 7 = 6 = 2xy ,

again as it must.

Problem 4

The differential equation in Example 6 is

6ux+xuy=y,

with the boundary-values of u=exsin(y), on the curvey=x3. The method of characteristics seeks to find functionsx(s),y(s), and u(s) such that dx ds= 6 dy ds=x du ds=y , 9 with initial conditions given by x(0) =r y(0) =r3 u(0) =ersin(r3). Integrating the differential equation forx(s) we findx(s) = 6s+CforCa constant. Applying the initial conditionsx(0) =rwe findC=rsox(s) = 6s+r. Using this in the equation fory(s) we find dy ds= 6s+r. Integrating the differential equation fory(s) we findy(s) = 3s2+rs+DforDa constant. Applying the initial conditions fory(s) we findy(s) = 3s2+rs+r3. Using these, we have that the differential equation foru(s) is given by du ds= 3s2+rs+r3.

Integrating this we get

u(s) =s3+rs2

2+r3s+E ,

for a constantE. Applying the initial conditions foru(s) gives u(s) =s3+rs2

2+r3s+ersin(r3).

Thus the solution is represented in parameterized form as x= 6s+r y= 3s2+rs+r3 u=s3+rs2

2+r3s+ersin(r3).

We are told that (1,1) is on the same characteristic as (7,5). This means that

1 = 6s+r

1 = 3s2+rs+r3,

for somerands. Note thats= 0 andr= 1 satisfy the above system. Now if (7,5) is on the same characteristic then there needs to be a value forssuch that

7 = 6s+ 1(1)

5 = 3s2+s+ 1,(2)

Now Equation 1 is satisfied bys= 1 and that this value ofsalso satisfies Equation 2. The value ofuat (x,y) = (7,5) is then given by u=s3+rs2

2+r3s+ersin(r3)????

s=1,r=1= 1 +12+ 1 +esin(1) =52+esin(1). 10

Problem 5Equation 7 in the book is

u x+cuy= 0, which can be written asdquotesdbs_dbs4.pdfusesText_8

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