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perry 手册第七版 Chap03 Mathematics,数学
perry 手册第七版
*The contributions of William F. Ames (retired), Georgia Institute of Technology;
Arthur E. Hoerl (deceased), University of Delaware; and M. Zuhair Nashed, Uni-versity
of Delaware, to material that was used from the sixth edition is gratefully
acknowledged.
3-1
3-2MATHEMATICS
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broad perspective on classical and modern mathematical methods thatare useful in
chemical engineering. The references supplement and extend thetreatment given in this
section. Also included are selected references to impor-tant areas of mathematics
that are not covered in the Handbookbut that may beuseful for certain areas of chemical
engineering, e.g., additional topics innumerical analysis and software, optimal
control and system theory, linear oper-ators, and functional-analysis methods.
Readers interested in brief summaries oftheory, together with many detailed examples
and solved prhttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33coblems on
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MATHEMATICS3-7
3-8MATHEMATICS
equations. These efforts have been most fruitful in the area of the lin-ear equations
such as those just given. However, many natural phe-nomena are nonlinear. While there
are a few nonlinear problems thatcan be solved analytically, most cannot. In those
cases, numericalmethods are used. Due to the widespread availability of software
forcomphttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33cuters,
the
engineer has quite good tools available.
Numerical methods almost never fail to provide an answer to anyparticular situation,
but they can never furnish a general solution ofany problem.
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The mathematical details outlined here include both analytic andnumerical techniques
useful in obtaining solutions to problems.
Our discussion to this point has been confined to those areas inwhich the governing
laws are well known. However, in many areas,information on the governing laws is
lacking. Interest in the applica-tion of statistical methods to all types of problems
has grown rapidlysince World War II. Broadly speaking, statistical methods may be
ofuse whenever conclusions are to be drawn or decisions made on thebasis of
experimental evidence. Since statistics could be defined as thetechnology of the
scientific method, it is primarily concerned with the first two aspects of the method,
namely, the http://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33cperformance
of exper-iments and the drawing of conclusions from experiments. Tradition-ally the
field is divided into two areas:
1.Design of experiments.When conclusions are to be drawn ordecisions made on the basis
of experimental evidence, statistical tech-niques are most useful when experimental
data are subject to errors.The design of experiments may then often be carried out
in such afashion as to avoid some of the sources of experimental error andmake the
necessary allowances for that portion which is unavoidable.Second, the results can
be presented in terms of probability state-ments which express the reliability of
the results. Third, a statisticalapproach frequently forces a more thorough
evaluation of the experi-mental aims and leads to a more definitive experiment than
wouldotherwise have been performed.
2.Statistical inference.The broad problem of statistical infer-ence is to provide
measures
of
thehttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33c
uncertainty of conclusions drawnfrom experimental data. This area uses the theory
of probability,enabling scientists to assess the reliability of their conclusions
in termsof probability statements.
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Both of these areas, the mathematical and the statistical, are inti-mately
intertwined when applied to any given situation. The methodsof one are often combined
with the other. And both in order to be suc-cessfully used must result in the numerical
answer to a problem—thatis, they constitute the means to an end. Increasingly the
numericalanswer is being obtained from the mathematics with the aid of com-puters.
MISCELLANEOUS MATHEMATICAL CONSTANTS
Numerical values of the constants that follow are approximate to thenumber of
significant digits given.
π
=3.1415926536e=2.7182818285γ=0.5772156649ln
π=1.1447298858log
π=0.4971498727Radian =57.2957795131°Degree =0.0174532925 radMinute =0.0002908882
radSecohttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33cnd
=0.0000048481 rad
Pi
Napierian (natural) logarithm baseEuler’s constant
Napierian (natural) logarithm of pi, base eBriggsian (common logarithm of pi, base
10
THE REAL-NUMBER SYSTEM
The natural numbers, or counting numbers, are the positive integers:1, 2, 3, 4, 5, ....
The negative integers are ?1, ?2, ?3, ....
A number in the form a/b,where aand bare integers, b≠0, is arational number. A real
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number that cannot be written as the quotientof two integers is called an irrational
number, e.g., 2, ??3, ??5, π, ??3e,??2.
There is a one-to-one correspondence between the set of real num-bers and the set
of points on an infinite line (coordinate line).
MATHEMATICS
a continuous function. Also with a,b>0 and x,yany real numbers, we have
(ab)x=axbxbxby=bx y(bx)y=bxyThe exponential function with base bcan also be defined
as
theinverse
of
logarithhttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33cmic
the
function.
The most common exponentialfunction in applications corresponds to choosing bthe
transcendentalnumber e.
Logarithmslog ab=log a log b, a>0, b>0
log an=nlog alog (an/b) =log a?log blog ??a=(1/n) log aThe common logarithm (base
10) is denoted log aor log10a.The nat-ural logarithm (base e) is denoted ln a(or in
some texts logea).
RootsIf ais a real number, nis a positive integer, then xis calledthe nth root of
aif xn=a.The number of nth roots is n,but not all ofthem are necessarily real. The
principal nth root means the following:(1) if a>0 the principal nth root is the unique
positive root, (2) if a
n?1k=0
3-9
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or, equivalently,
r=1
Α(aa
n
12
???ar)1/r≤neAn
where eis the best possible constant in this inequality.
Cauchy-Schwarz InequalityLet a=(a1, a2,..., an), b=(b1,b2,..., bn), where the ai’s
and bi’shttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33c are real or
complex numbers. Then
???Αab
n
2
kk
≤
k=1
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?Α|a|??Α|b|?
n
n
k
2
k
2
k=1k=1
The equality holds if, and only if, the vectors a, bare linearly depen-dent (i.e.,
one vector is scalar times the other vector).
Minkowski’s InequalityLet a1, a2,..., anand b1, b2,..., bnbeany two sets of complex
numbers. Then for any real number p>1,
?Α
n
|ak bk|p
k=1
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??Α??Α?
1/p
n
≤|ak|p
1/pn
|bk|p
1/p
k=1k=1
H?lder’s InequalityLet a1, a2,..., anand b1, b2,..., bnbe anytwo sets of complex
numbers, and let pand qbe positive numberswith 1/p 1/q=1. Then
?Α??Α??Α?
n
n
akb?k≤|ak|p
1/pn
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|bk|q
1/q
k=1k=1k=1
n(n?http://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33c1)d =?(a ?)Α(a
kd)
=na ?
22
1n
The equality holds if, and only if, the sequences |a1|p,|a2|p,..., |an|p
and |b1|q,|b2|q,..., |bn|qare proportional and the argument (angle) ofthe complex
numbers akb?kis independent of k.This last condition is ofcourse automatically
satisfied if a1,..., anand b1,..., bnare positivenumbers.
Lagrange’s InequalityLet a1, a2,..., anand b1, b2,..., bnbereal numbers. Then
where ?is the last term, ?=a (n?1)d.Geometric Progression
n
a(rn?1)k?1
ar=(r≠1)?Αr?1k=1
Arithmetic-Geometric Progressionn?1
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a?[a (n?1)d]rndr(1?rn?1)k
(a
kd)r= ????Α1?r(1?r)2k=0n
1
k5=?n2(n 1)2(2n2 2n?1)Α12k=1
k=1
?Α??Α??Αb??
n
akbk
2nn
=a2k2
k
k=1k=1k=11≤k≤j≤n
Α
(akbj?ajbk)2
(r≠1)
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ExampleTwo
chemical
enghttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33cineers,
John
and
Mary, purchase stock in thesame company at times t1, t2,..., tn,when the price per
share is respectively p1,p2,..., pn. Their methods of investment are different,
however: John purchasesxshares each time, whereas Mary invests Pdollars each time
(fractional sharescan be purchased). Who is doing better?
While one can argue intuitively that the average cost per share for Mary doesnot exceed
that for John, we illustrate a mathematical proof using inequalities.The average cost
per share for John is equal to
1nTotal money investedi=1
????=?=?ΑpiNumber of shares purchasedni=1nx
The average cost per share for Mary is
nPn
?=?nn
P1??Α??Αi=1pii=1pi
Thus the average cost per share for John is the arithmetic mean of p1, p2,..., pn,
whereas that for Mary is the harmonic mean of these nnumbers. Since the har-monic
mean
is
less
than
or
equal
to
thehttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33c arithmetic mean for
any set of positivenumbers and the two means are equal only if p1=p2=???=pn,we
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conclude thatthe average cost per share for Mary is less than that for John if two
of the pricespiare distinct. One can also give a proof based on the Cauchy-Schwarz
inequal-ity. To this end, define the vectors
?1/2?1/2?1/2
a=(p1, p2,..., pn)
1/21/21/2
b=(p1, p2,..., pn)
Α(2k?1) =nΑ
nn
n
x Αpi
n
2
k=1
1
(2k?1)=?n(4n2?1)
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3
23
k=1
Α(2k?1)
n
=n2(2n2?1)
γ=lim
n→∞
1
?ln n?=0.577215?Α?m
m=1
ALGEBRAIC INEQUALITIES
Arithmetic-Geometric InequalityLet Anand Gndenote
respectively the arithmetic and the geometric means of a set of posi-tive numbers
a1, a2,..., an. The An≥Gn,i.e.,
a1 a2 ??? an://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33cr
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??≥(a1a2???an)1/n
n
The equality holds only if all of the numbers aiare equal.
Carleman’s InequalityThe arithmetic and geometric meansjust defined satisfy the
inequality
r=1
Then a ?b=1
??? 1 =n,and so by the Cauchy-Schwarz inequality
n
1
(a ?b)2=n2≤Α?
i=1pi
i=1
Αp
n
i
with the equality holding only if p1=p2=???=pn. Therefore
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ΑG
n
r
≤neAn
Αpi
ni=1??n
1≤nΑ??i=1pi
n
3-10MATHEMATICS
MENSURATION FORMULAS
RadiusRof Circumscribed Circle
abc
R=???
4??s(?s???a?)(?s???b?)(?s???c?)
Area of Regular Polygon ofnSides Inscribed in a Circle of
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Radius r
A=(nr2/2) sin (360°/n)Perimeter of Inscribed Regular Polygon
P=2nrsin (180°/n)
Area of Regulahttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33cr Polygon
Circumscribed about a Circle ofRadiusr
A =nr2tan (180°/n)Perimeter of Circumscribed Regular Polygon
180°
P=2nrtan ?
nPLANE GEOMETRIC FIGURES WITH CURVED BOUNDARIES
Circle(Fig. 3-5)LetC=circumferencer=radiusD=diameterA=area
S=arc length subtended by θl=chord length subtended by θ
H=maximum rise of arc above chord, r ?H =dθ=central angle (rad) subtended by arc
SC=2πr=πD(π=3.14159...)S=rθ=aDθ
2
r2???d?=
2rsin (θ/2) =
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2dtan (θ/2)l=2??
11θ
?=?lcot ?d
=???4?r2???l2
222Sdlθ=?=2 cos?1?=2 sin?1?
rrDA(circle) =πr2=dπD2
A(sector) =arS=ar2θ
A(segment) =A(sector) ?A(triangle) =ar2(θ?sin θ)
r?H
?rH???=r2cos?1??(r ?H) ?2??H2
r
Ring(area between two circles of radii r1and r2)The circles neednot be concentric,
but
one
of
the
circles
enchttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33close the other.
A=π(r1 r2)(r1?r2)
r1>r2
must
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FIG. 3-3Parallelogram.FIG. 3-4Regular polygon.FIG. 3-5Circle.
MENSURATION FORMULAS3-11
Ellipse.
FIG. 3-6
FIG. 3-7
Parabola.
Ellipse(Fig. 3-6)Let the semiaxes of the ellipse be aand b
A=πab
C=4aE(k)
where e2=1 ?b2/a2and E(e) is the complete elliptic integral of thesecond kind,
π12
E(e) =?
1 ??e2
22
????
???
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? ????].[an approximation for the circumference C=2π??(a2b2)/2
Parabola(Fig. 3-7)
?y22x ??4?x2? ?y2
? ?ln ??4?x2? ?y2Length of arc EFG=??
2xy4
Area of section EFG=?xy
3
Catenary(the curve formed by a cord of uniform weight sus-pended freely between two
points A, B;Fig. 3-8)
y =acosh (x/a)
Length
of
arc
between
points
Aand
Bis
equal
to
2asinh
(L/a).http://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33c Sag ofthe cord is
D =acosh (L/a) ?1.
SOLID GEOMETRIC FIGURES WITH PLANE BOUNDARIES
?,CubeVolume =a3; total surface area =6a2; diagonal =a?3
where a=length of one side of the cube.
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Rectangular ParallelepipedVolume =abc;surface area =
2??, where a, b, care the lengths2(ab
ac
bc); diagonal =??a2? ?b? ?c2
of the sides.
PrismVolume =(area of base) ×(altitude); lateral surface area =(perimeter of right
section) ×(lateral edge).
PyramidVolume =s(area of base) ×(altitude); lateral area ofregular pyramid
=a(perimeter of base) ×(slant height) =a(numberof sides) (length of one side) (slant
height).
Frustum of Pyramid(formed from the pyramid by cutting offthe top with a plane
???A??V=s(A1 A2 ?A1?2)h
where h=altitude and A1, A2are the areas of the base; lateral area of
a
regular
figure
=a(sum
of
the
perimeters
of
base)
heighhttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33ct).
FIG. 3-8Catenary.
FIG. 3-9Cylinder.FIG. 3-10Sphere.
3-12MATHEMATICS
b21 e
×(slant
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Surface area =2πa2 π ?ln ?
e1?e
V=4?3πa2b
1.If a plane area is revolved about a line which lies in its plane but
does not intersect the area, then the volume generated is equal to theproduct of the
area and the distance traveled by the area’s center ofgravity.
2.If an arc of a plane curve is revolved about a line that lies in itsplane but does
not intersect the arc, then the surface area generatedby the arc is equal to the
product of the length of the arc and the dis-tance traveled by its center of gravity.
These theorems are useful for determining volumes Vand surfaceareas Sof solids of
revolution if the centers of gravity are known. If Sand Vare known, the centers of
gravity may be determined.IRREGULAR AREAS AND VOLUMES
Irregular
AreasLet
yhttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33c1,...,
y0,
ynbe
the
lengths of a series ofequally spaced parallel chords and hbe their distance apart.
The areaof the figure is given approximately by any of the following:AT=(h/2)[(y0
yn)
2(y1 y2 ??? yn?1)]As=(h/3)[(y0 yn)
4(y1 y3 y5 ??? yn?1)
2(y2 y4 ??? yn?2)]
(neven, Simpson’s rule)
The greater the value of n,the greater the accuracy of approximation.Irregular
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VolumesTo find the volume, replace the y’s by cross-sectional areas Ajand use the
results in the preceding equations.
(trapezoidal rule)
MISCELLANEOUS FORMULAS
See also “Differential and Integral Calculus.”
Volume of a Solid Revolution(the solid generated by rotatinga plane area about the
xaxis)
V=π
?[f(x)]dx
b
2
a
where y=f(x) is the equation of the plane curve and a≤x≤b.Area of a Surface of
Revolution
S=2π
?y ds
ba
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??http://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33c?y???dxand y =f(x) is
the equation of the planewhere ds=?1 ?(d/d?x)2
curve rotated about the xaxis to generate the surface.
Area Bounded by f(x), the xAxis, and the Lines x=a, x = b
A=
?f(x) dx
ba
[f(x) ≥0]
Length of Arc of a Plane CurveIf y =f(x),
Length of arc s=
If x=g(y),
Length of arc s=
If x=f(t), y=g(t),
Length of arc s=
2
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2
2
dy
??1 ?dx??????dx?
b
2
a
?????????dy
dc
dx
1 ?
dy
2
2
dxdy??? ????????dt??dt??dt
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t1
2
t0
In general, (ds)=(dx) (dy).
Theorems of Pappus(for volumes and areas of surfaces of revo-lution)
FIG. 3-11Irregular area.
ELEMENTARY ALGEBRA
REFERENCES:20, 102, 108, 191, 269, 277.
OPhttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33cERATIONS ON ALGEBRAIC
EXPRESSIONS
An algebraic expression will here be denoted as a combination of let-ters and numbers
such as
3ax?3xy 7x 7x
2
3/2
?2.8xy
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Addition and SubtractionOnly like terms can be added or sub-tracted in two algebraic
expressions.
Example(3x 4xy?x2)
Example(2x
(3x2 2x?8xy) =5x?4xy 2x2.
3xy?4x1/2)
(3x
6x?8xy)
=2x
3x
6x?5xy?4x1/2.MultiplicationMultiplication of algebraic expressions is term byterm,
and corresponding terms are combined.Example(2x 3y?2xy)(3
3y) =6x 9y 9y2?6xy2.
DivisionThis operation is analogous to that in arithmetic.ExampleDivide 3e2x ex 1
by ex 1.
ELEMENTARY ALGEBRA
(2) x2 2xy
y2=(x
(3) x2 ax
b=(x
(4) by2 cy
d=(ey
y
y)2
z)2(6)
c)(x
d) where c
f)(gy
d =a, cd =b
h) where eg =b, fg
x2?y2?z2?2yz=(x
?y
eh =c, fh =d(5) x2 y2 z2 2yz 2xz 2xy=(x
?z)(x
y
z)(7)
x2
y2
z2?http://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33c2xy?2xz
2yz=(x ?y ?z)2(8) x3?y3=(x ?y)(x2 xy
=(x ?y)(x
xn?2y
y)(x2 y2)(11) x5 y5=(x
xn?3y2 ???
y2)(9) (x3 y3) =(x
y)(x4?x3y
y)(x2?xy
y2)(10) (x4?y4)
x2y2?xy3 y4)(12) xn?yn=(x ?y)(xn?1
yn?1)
Laws of Exponents
(an)m=anm; an m=an?am; an/m=(an)1/m; an?m=an/am; a1/m=m?a?; a1/2=??a; ??x2=|x|
(absolute value of x). For x>0, y>0, ?xy?=?x?
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mm/nn
??y; for x>0 ??x=x; ??1?/x=1/??xTHE BINOMIAL THEOREMIf nis a positive integer,
n(n?1)
(a
b)n=an nan?1b ?an?2 b2
2!
n
n(n?1)(n?2)n?33nn?jjn
??ab ??? b=Αab
j3!j=0
3-13
s(n?1)d2s
a =l?(n?1)d=???=??l
n2nl?a2(s?an)2(nl?s)
d=?=?=?
n?1n(n?1)n(n?1)
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2l d ??(2?l? ?d?)2???8?d?s2sl?a
n=? 1 =?=???
2ddl a
The arithmetic mean or averageof two numbers a, bis (a
of
nnumbers
a1,...,
anis
b)/2;
(a1
a2
???
an)/http://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33cn.
A geometric progressionis a succession of terms such that eachterm, except the first,
is derivable from the preceding by the multipli-cation of a quantity rcalled the
common ratio. All such progressionshave the form a, ar, ar2,..., arn?1. With a=first
term, l=last term, r=ratio, n=number of terms, s=sum of the terms, the following
rela-tions hold:
[a (r?1)s](r?1)srn?1
l =arn?1=??=??
rrn?1
a(rn?1)a(1?rn)rl?alrn?l
s=?=?=?=?
r?11?rr?1rn?rn?1log l?log al(r?1)ss?aa=?=?r=?log r=??n?ln
rr?1s?ln?1
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log l?log alog[a (r?1)s]?log an=?? 1 =???
log rlog r
?; of nThe geometric mean of two nonnegative numbers a, bis ?ab
numbers is (a1a2...an)1/n.
ExampleFind the sum of 1
a d ??? 1?64. Here a=1, r=a, n=7. Thus
a(1?64)?1
s=??=127/64
a?1
aarn
s =a
ar
ar2 ??? arn?1=???
1?r1?http://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33cr
If|r|
a
lims=?n→∞1?r
which is called the sum of the infinite geometric progression.
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then
of year kis
??
nn!
where =?=number of combinations of nthings taken jat
jj!(n?j)!a time. n! =1 ? 2 ? 3 ? 4 ???n,0! =1.
ExampleFind the sixth term of (x 2y)12. The sixth term is obtained by
setting j=5. It is
1212?5
x(2y)5=792x7(2y)55
??
??
Example
j=0
Α?j?=(1
1)
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14
n
14
=214.
ExampleThe present worth (PW) of a series of cash flows Ckat the end
If nis not a positive integer, the sum formula no longer applies andan infinite series
results for (a
b)n. The coefficients are obtainedfrom the first formulas in this
case.
x
2
Example(1
x)1/2=1
ax?a?dx2
a?d?3?6
x3???(convergent
for
dishttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33ccussion
Additional
is
under
“Infinite Series.”
PROGRESSIONS
An arithmetic progression is a succession of terms such that eachterm, except the
first, is derivable from the preceding by the additionof a quantity dcalled the common
difference. All arithmetic progres-sions have the form a, a
d, a 2d, a 3d, .... With
a=first term, l=last term, d=common difference, n=number of terms, and s=sum of the
terms, the following relations hold:
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d
l =a (n?1)d=??
2s(n?1)=? ?dn2
nnn
s=?[2a (n?1)d] =?(a
l) =?[2l?(n?1)d]
222
n
Ck
PW =Α?k
k=1(1 i)
where
iis
an
assumed
interest
rate.
(Thus
the
present
worth
always
requiresspecification of an interest rate.) If all the payments are the same,
Ck=R,thepresent worth is
n
1
PW =RΑ?k
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k=1(1 i)
This can be rewritten as
n
1R1Rn?1
http://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33c=PW =?Α???Αk?1
1 ik=1(1 i)1 ij=0(1 i)j
This is a geometric series with r=1/(1
i) and a =R/(1
i). The formulas above
give
R(1 i)n?1
PW (=s) =???
(1 i)ni
The same formula applies to the value of an annuity (PW) now, to provide forequal
payments Rat the end of each of nyears, with interest rate i.
?????????????
d2ds a??
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2
2
A progression of the form a,(a
d)r,(a 2d)r2, (a 3d)r3, etc., isa combined arithmetic
and geometric progression. The sum of nsuchterms is
a?[a (n?1)d]rnrd(1?rn?1)s=?? ??
I?r(1?r)2a
If |r|
n→∞1?r
3-14MATHEMATICS
has three complex roots. If the coefficients are real numbers, then atleast one of
the roots must be real. The cubic equation x3 bx2 cx
substitution
x
=y?(b/3)
to
the
form
y3
py
d=0 may be reduced by the
q=0,
where
p=s(3c?b2),
q=1?27(27d?9bchttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33c
This equa-3/2)(A ?B), tion has the solutions y1=A
B, y2=?a(A
B)
(i??3
2
y3=?a(A
B) ?(i??3/2)(A ?B), where i=?1, A=????q?/2? ?R, ??
3
B=????q?/2???R, and R=(p/3)3 (q/2)2. If b, c, dare all real and if??
2b3).
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R>0, there are one real root and two conjugate complex roots; if R=0, there are three
real roots, of which at least two are equal; if R
q2/4??p3/27
and the upper sign applies if q>0, the lower if q
φ=cos?1
Examplex3 3x2 9x 9 =0 reduces to y3 6y 2 =0 under x =y?1.
?, B=???4?. The desired roots in yareHere p=6, q=2, R=9. Hence A=?2
333333?????4?and ?a(?2???4?) ?(i?3?/2)(?2? ?4?). The roots in xare x =?2y?1.
3
3
The non-zero numbers a, b, c,etc., form a harmonic progression iftheir reciprocals
1/a,1/b,1/c,etc., form an arithmetic progression.ExampleThe progression 1, s, 1?5,
1?7,...,
1?31is
harmonichttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33c since 1, 3,
5,7,..., 31 form an arithmetic progression.
The harmonic meanof two numbers a, bis 2ab/(a
b).PERMUTATIONS, COMBINATIONS, AND
PROBABILITYEach separate arrangement of all or a part of a set of things is called
apermutation.The number of permutations of nthings taken rat atime, written
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n!
P(n, r) =?=n(n?1)(n?2) ???(n ?r 1)
(n?r)!ExampleThe permutations of a, b, ctwo at a time are ab, ac, ba, ca, cb,and bc.The
formula is P(3,2) =3!/1! =6. The permutations of a, b, cthree at atime are abc, bac,
cab, acb, bca,and cba.
Each separate selection of objects that is possible irrespective of theorder in which
they are arranged is called a combination. The numberof combinations of nthings taken
rat a time, written C(n, r) =n!/[r!(n ?r)!].
3 at a time is abc.
??
Exampley3?7y 7 =0. p=?7, q=7, R
xk=?
where
28φ
?cos
??
120k???http://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33c3327φφ=???, ?
=3°37′52″.
283
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ExampleThe combinations of a, b, ctaken 2 at a time are ab, ac, bc;taken
An important relation is r! C(n, r) =P(n, r).
If an event can occur in pways and fail to occur in qways, all waysbeing equally likely,
the probabilityof its occurrence is p/(p
q), andthat of its failure q/(p
q).
ExampleTwo dice may be thrown in 36 separate ways. What is the prob-ability of throwing
such that their sum is 7? Seven may arise in 6 ways: 1 and 6,2 and 5, 3 and 4, 4 and
3, 5 and 2, 6 and 1. The probability of shooting 7 is j.THEORY OF EQUATIONS
Linear EquationsA linear equation is one of the first degree(i.e., only the first
powers of the variables are involved), and theprocess of obtaining definite values
for the unknown is called solvingthe equation. Every linear equation in one variable
is
written
Ax
B=0
or
x=?B/A.Linear
equations
in
nvariables
have
the
form://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33cpar
a11x1 a12x2 ??? a1nxn=b1a21x1 a22x2 ??? a2nxn=b2?
am1x1 am2x2 ??? amnxn=bm
The solution of the system may then be found by elimination or matrixmethods if a
solution exists (see “Matrix Algebra and Matrix Compu-tations”).
Quadratic EquationsEvery quadratic equation in one variableis expressible in the form
ax2 bx
c=0. a≠0. This equation has twosolutions, say, x1, x2, given by
The roots are approximately ?3.048916, 1.692020, and 1.356897.
ExampleMany
equations
of
state
involve
solving
cubic
equations
for
thecompressibility factor Z.For example, the Redlich-Kwong-Soave equation ofstate
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requires solving
Z3?Z2 cZ
d=0,
d
where cand ddepend on critical constants of the chemical species. In this case,only
positive solutions, Z>0, are desired.
Quartic EquationsSee Ref. 118.
General
Polynomials
of
the
nth
DegreeDenote
the
generalpolynomiahttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33cl
equation of degree nby
P(x) =a0xn a1xn?1 ??? an?1x an=0
If n>4, there is no formula which gives the roots of the generalequation. For fourth
and higher order (even third order), the rootscan be found numerically (see
“Numerical Analysis and ApproximateMethods”). However, there are some general
theorems that mayprove useful.
Remainder TheoremsWhen P(x) is a polynomial and P(x) isdivided by x ?auntil a
remainder independent of xis obtained, thisremainder is equal to P(a).
results in P(x) =(x 1)(2x3?2x2?x 8) ?10 where ?10 is the remainder. It iseasy to see
that P(?1) =?10.
ExampleP(x) =2x4?3x2 7x?2 when divided by x 1 (here a=?1)
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??????b???x1b24ac
=??x22a
If a, b, care real, the discriminant b2?4acgives the character of theroots. If
b2?4ac>0, the roots are real and unequal. If b2?4ac
Two
quadratic
equations
in
two
variables
can
in
general
be
solvedonhttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33cly by numerical
methods (see “Numerical Analysis and Approxi-mate Methods”). If one equation is
of the first degree, the other of thesecond degree, a solution may be obtained by
solving the first for oneunknown. This result is substituted in the second equation
and theresulting quadratic equation solved.
Cubic EquationsA cubic equation, in one variable, has the formx3 bx2 cx
d=0. Every
cubic equation having complex coefficients
?
Factor TheoremIf P(a) is zero, the polynomial P(x) has the fac-tor x ?a.In other words,
if ais a root of P(x) =0, then x ?ais a factorof P(x).
If a number ais found to be a root of P(x) =0, the division of P(x) by(x ?a) leaves
a polynomial of degree one less than that of the originalequation, i.e., P(x)
=Q(x)(x ?a). Roots of Q(x) =0 are clearly roots ofP(x) =0.
(x?3)(x2?3x 2). The roots of x2?3x 2 =0 are 1 and 2. The roots of P(x) aretherefore
1, 2, 3.
://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33carExampleP(x) =x3?6x2 11x?6
=0 has the root
3. Then P(x) =
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Fundamental Theorem of AlgebraEvery polynomial of degreenhas exactly nreal or complex
roots, counting multiplicities.
ELEMENTARY ALGEBRA
Every polynomial equation a0xn a1xn?1 ??? an=0 with rationalcoefficientsmay be
rewritten as a polynomial, of the same degree, withintegral coefficientsby
multiplying each coefficient by the least com-mon multiple of the denominators of
the coefficients.
numbers. The least common multiple of the denominators is 2 ×3 =6. There-fore, the
equation is equivalent to 9x4 14x3?5x2 12x?1 =0.
3-15
Example
?
a11a12a13
aa
a21a22a23The minor of a23is M23= 1112
a31a32
a31a32a33
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?
??
ExampleThe coefficients of 3?2x4 7?3x3?5?6x2 2x?j=0 are rational
The
cofactor
Aijof
the
element
aijis
the
signed
minor
aijhttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33cdetermined
by
of
the
rule Aij=(?1)i jMij. The valueof |A| is obtained by forming any of the
nn
equivalent expressions Αj=1aijAij, Αt=1aijAij, where the elements aijmust betaken
from a single row or a single column of A.
Upper Bound for the Real RootsAny number that exceeds allthe roots is called an upper
bound to the real roots. If the coefficientsof a polynomial equation are all of like
sign, there is no positive root.Such equations are excluded here since zero is the
upper bound to thereal roots. If the coefficient of the highest power of P(x) =0 is
nega-tive, replace the equation by ?P(x) =0.
If in a polynomial P(x) =c0xn c1xn?1 ??? cn?1x cn=0, withc0>0, the first negative
coefficient is preceded by kcoefficientswhich are positive or zero, and if Gdenotes
the greatest of the numer-ical values of the negative coefficients, then each real
root is less than
k
1
??G?/c?0.
://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33crA
lower
bound
to
the
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negative roots of P(x) =0 may be found byapplying the rule to P(?x) =0.
?2?=3. P(?x) =?x7?cients are zero), G=32, c0=1. The upper bound is 1
?3
2x5 4x4?8x2?32 =0. ?P(?x) =x73 2x5?4x4 8x2 32 =0. Here k=3, G=4,c0=1. The lower bound
is ?(1
??4) ≈?2.587. Thus all real roots rlie in therange ?2.587
Example
?
a11a12a13
a21a22a23=a31A31 a32A32 a33A33a31a32a33
=a31
?
?aa
1222
aa13aaa
?a321113 a331112a23a21a23a21a22
?????
In general, Aijwill be determinants of order n?1, but they may in turn be
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expanded by the rule. Also,
j=1
Αa
n
ji
Ajk=ΑaijAjk=|A|
0j=1
n
?
i =k
i ≠k
ExampleP(x) =x7 2x5 4x4?8x2?32 =0. Here k=5 (since 2 coeffi-3
Descartes
Rule
of
SignsThe
number
of
positivhttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33ce real roots of
apolynomial equation with real coefficients either is equal to the num-ber vof its
variations in sign or is less than vby a positive even integer.The number of negative
roots of P(x) =0 either is equal to the numberof variations of sign of P(?x) or is
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less than that number by a positiveeven integer.
P(?x) =x4?3x3?x?1. Here v=1; so P(x) has one negative root. The other tworoots are
complex conjugates.
tive roots. P(?x) =x4?x2?10x?4. v=1; so P(x) has exactly one negative root.
ExampleP(x) =x4 3x3 x?1 =0. v=1; so P(x) has one positive root.
Fundamental Properties of Determinants
1.The value of a determinant |A| is not changed if the rows andcolumns are
interchanged.
2.If the elements of one row (or one column) of a determinantare all zero, the value
of |A| is zero.
3.If the elements of one row (or column) of a determinant aremultiplied by the same
constahttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33cnt
factor,
the
value of the determinant ismultiplied by this factor.
4.If one determinant is obtained from another by interchangingany two rows (or
columns), the value of either is the negative of thevalue of the other.
5.If two rows (or columns) of a determinant are identical, thevalue of the determinant
is zero.
6.If two determinants are identical except for one row (or col-umn), the sum of their
values is given by a single determinantobtained by adding corresponding elements of
dissimilar rows (orcolumns) and leaving unchanged the remaining elements.Example
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ExampleP(x) =x4?x2 10x?4 =0. v=3; so P(x) has three or one posi-
Numerical methods are often used to find the roots of polynomials.A detailed
discussion of these techniques is given under “NumericalAnalysis and Approximate
Methods.”
DeterminantsConsider the system of two linear equations
a11x1
a12x2=b1a21x1
a22http://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33cx2=b2
If the first equation is multiplied by a22and the second by ?a12and theresults added,
we obtain
(a11a22?a21a12)x1=b1a22?b2a12
The expression a11a22?a21a12may be represented by the symbola11a12
=a11a22?a21a12
a21a22
This symbol is called a determinant of second order. The value of thesquare array
of n2quantities aij, where i=1,..., nis the row index, j=1,..., nthe column index,
written in the form
?15? ?75?=13
32
6 =19
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42
DirectlyBy rule 6
?85?=35 ?16 =19
72
7.The value of a determinant is not changed if to the elements ofany row (or column)
are added a constant multiple of the correspond-ing elements of any other row (or
column).
8.If all elements but one in a row (or column) are zero, the valueof the determinant
is the product of that element times its cofactor.The evaluation of determinants
usinghttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33c the definition is
quite labori-ous. The labor can be reduced by applying the fundamental propertiesjust
outlined.
The solution of nlinear equations (not all bizero)
a11x1 a12x2 ??? a1nxn=b1a21x1 a22x2 ??? a2nxn=b2????
an1x1 an2x2 ??? annxn=bn
a11???a1na21???a2n
where |A| = ?≠0
an1???ann
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??
a11a12aa22
|A| = 21
?
an1an2
?
a13 ???a1n?????a2n
an3???ann
?
??
is called a determinant. The n2quantities aijare called the elementsof the
determinant. In the determinant |A| let the ith row and jth column be deleted and
a new determinant be formed having n?1rows and columns. This new determinant is called
the minor of aijdenoted Mij.has a unique solution given by x1=|B1|/|A|,
x2=|B2|/|A|,..., xn=|Bn|/|A|, where Bkis the determinant obtained from Aby replacing
itskth
column
by
b1,
b2,...,
bnhttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33c. This technique is
called Cramer’s rule.It requires more labor than the method of elimination and should
notbe used for computations.
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3-16MATHEMATICS
FIG. 3-12Rectangular coordinates.FIG. 3-13Polar coordinates.FIG. 3-14Straight
line.FIG. 3-15Determination of line.
ANALYTIC GEOMETRY
3-17
FIG. 3-16
Graph of y2=(x2 1)/(x2?1)
3-18MATHEMATICS
FIG. 3-17Circle center (0,0) r =a
.FIG. 3-18Circle center (a,0) r=2
acos θ.
FIG. 3-19Circle center (0,a) r=2asin
θ.
FIG. 3-20Ellipse, 0
1.FIG. 3-21Hyperbola,
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e>1, r=ke/(1 ?ecos θ).FIG. 3-22Parabola, e=1.
Circle at (b, β
), radius a: r2?2brcos (θ?β)?b2?a2=0.
Graphs of Polar EquationsThe equation r=0 corresponds to x=0, y=0 regardless of θ.
The same point may be represented in sev-eral different ways; thus the point (2, π/3)
or (2, 60°)http://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33c has the
followingrepresentations: (2, 60°), (2, ?300°). These are summarized in (2, 60°
n360°), n=0, ?1, ?2, or in radian measure [2, (π/3)
2nπ], n=0, ?1, ?2. Plotting
of polar equations can be facilitated by the fol-lowing steps:
1.Find those points where ris a maximum or minimum.2.Find those values of θwhere
r=0, if any.
3.Symmetry: The curve is symmetric about the origin if the equa-tion is unchanged
when θis replaced by θ?π, symmetric about the xaxis if the equation is unchanged
when θis replaced by ?θ, and sym-metric about the yaxis if the equation is unchanged
when θisreplaced by π?θ.
Parametric EquationsIt is frequently useful to write the equa-tions of a curve in
terms of an auxiliary variable called a parameter.For example, a circle of radius
a,center at (0, 0), can be written in theequivalent form x =acos θ, y =asin φwhere
θis
the
parameter.
Sim-ilarly,
x
=acos
φ,
y
=bsin
φare
the
parametrihttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33cc equations of
the ellipsex2/a2 y2/b2=1 with parameter φ.
SOLID ANALYTIC GEOMETRY
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Coordinate SystemsThe commonly used coordinate systemsare three in number. Others
may be used in specific problems (seeRef. 212). The rectangular(cartesian) system
(Fig. 3-25) consists ofmutually orthogonal axes x, y, z.A triple of numbers (x, y,
z) is used torepresent each point. The cylindricalcoordinate system (r,θ, z;Fig.3-26)
is frequently used to locate a point in space. These are essen-tially the polar
coordinates (r,θ) coupled with the zcoordinate. As
FIG. 3-25Cartesian coordinates.
FIG. 3-23Circle.
FIG. 3-24
Cycloid.FIG. 3-26Cylindrical coordinates.
ANALYTIC GEOMETRY
before, x =rcos θ, y =rsin θ, z =zand r2=x2 y2, y/x=tan θ. If risheld constant
and θand zare allowed to vary, the locus of (r,θ, z) is aright circular cylinder
of
radius
ralong
the
zaxis.
The
locus
=Chttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33cisa
of
r
circle,
and
θ=constant is a plane containing the zaxis and makingan angle θwith the xzplane.
Cylindrical coordinates are convenient touse when the problem has an axis of symmetry.
The sphericalcoordinate system is convenient if there is a point of symmetry in the
system. This point is taken as the origin and thecoordinates (ρ, φ, θ) illustrated
in Fig. 3-27. The relations are x=ρsin φcos θ, y=ρsin φsin θ, z=ρcos φ, and
r=ρsin φ. θ=constantis a plane containing the zaxis and making an angle θwith
the xzplane. φ=constant is a cone with vertex at 0. ρ=constant is the sur-face of
a sphere of radius ρ, center at the origin 0. Every point in thespace may be given
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spherical coordinates restricted to the ranges 0 ≤φ≤π, ρ≥0, 0 ≤θ
Lines and PlanesThe distance between two points (x1, y1, z1),
222
(x2, y2, z2) is d=??(x???x? ?(?y???y? ?(?z1???z?1?2)??1?2)??2)?. There is nothing
inthe geometry of threehttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33c
dimensions quite analogous to the slope of aline in the plane case. Instead of
specifying the direction of a line by a trigonometric function evaluated for one angle,
a trigonometricfunction evaluated for three angles is used. The angles α, β, γthat
a line segment makes with the positive x, y,and zaxes, respectively,are called the
direction anglesof the line, and cos α, cos β, cos γare called the direction
cosines.Let (x1, y1, z1), (x2, y2, z2) be on the line. Then cos α=(x2?x1)/d,cos
β=(y2?y1)/d,cos γ=(z2?z1)/d,where d=the distance between the two points. Clearly
cos2α cos2β cos2γ=1. If two lines are specified by the directioncosines (cos α1,
cos β1, cos γ1), (cos α2, cos β2, cos γ2), then the angle θbetween the lines
is cos θ=cos α1cos α2 cos β1cos β2 cos γ1cos γ2.Thus the lines are perpendicular
if and only if θ=90°or cos α1cos α2 cos β1cos β2 cos γ1cos γ2=0. The equation
of
a
line
withdirection
cosines
(cos
α,
cos
β,
cos
γ)http://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33c passing through (x1,
y1, z1) is (x ?x1)/cos α=(y ?y1)/cos β=(z ?z1)/cos γ.
The equation of every plane is of the form Ax
By
Cz
D=0.The numbers
ABC??????, , 2222222
???A?? ?B?? ?C???A?? ?B?? ?C???A?? ?B2 ?C2??are direction cosines of the normal lines
to the plane. The plane
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through the point (x1, y1, z1) whose normals have these as directioncosines is A(x ?x1)
B(y ?y1)
C(z ?z1) =0.
ExampleFind the equation of the plane through (1, 5, ?2) perpendicularto the line
(x 9)/7 =(y?3)/?1 =z/8. The numbers (7, ?1, 8) are called direc-tion numbers.They
are a constant multiple of the direction cosines. cos α=7/114, cos β=?1/114, cos
γ=8/114. The plane has the equation 7(x?1) ?1(y?5)
8(z 2) =0 or 7x ?y 8z 14 =0.
The distance from the point (x1, y1, z1) to the plane Ax
By
Cz
D=0 is
|Ax1 By1 Cz1 D|d=???
?2? ??? ???B2C2?A
FIG. 3-30
3-19
http://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33cFIG. 3-27
Spherical coordinates.
FIG. 3-28
Parabolic cylinder.
FIG. 3-29
y2z2x2
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??Ellipsoid. ?
=1 (sphere if a =b =c)b2c2a
2
y2z2x2
??Hyperboloid of one sheet. ? ?=1b2c2a
2
Space CurvesSpace curves are usually specified as the set of
points whose coordinates are given parametrically by a system ofequations x =f(t),
y =g(t), z =h(t) in the parameter t.
ExampleThe equation of a straight line in space is (x ?x1)/a=(y ?y1)/b=(z ?z1)/c.Since
all these quantities must be equal (say, to t), we may write x =x1 at, y =y1 bt, z
=z1 ct,which represent the parametric equations of theline.
ExampleThe equations z =acos βt, y =asin βt, z =bt, a,β, bpositiveconstants,
represent a circular helix.
SurfacesThe locus of points (x, y, z) satisfying f(x, y, z) =0,broadly speaking, may
be
interpreted
as
a
surface.
Thehttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33c simplest sur-face
is the plane.The next simplest is a cylinder,which is a surfacegenerated by a straight
line moving parallel to a given line and passingthrough a given curve.
straight line parallel to the zaxis passing through y =x2in the plane z=0.
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ExampleThe parabolic cylinder y =x2(Fig. 3-28) is generated by a
FIG. 3-31
y2z2x2
?Hyperboloid of two sheets. ? ??=?122
bc2a
3-20MATHEMATICS
PLANE TRIGONOMETRY3-21
Relations between Functions of a Single Anglesec θ=1/cos θ; csc θ=1/sin θ, tan
θ=sin θ/cos θ=sec θ/csc θ=1/cot θ; sin2θ cos2θ=1; 1
cot2θ=csc2θ. For 0 ≤θ≤90°thefollowing results hold:
??????cos2?θ=cos θtan θsin θ=cos θ/cot θ=?1
tan θ1θθ
=??=??=2 sin ?cos ?22
1? ?t?an??θ??1? ?c?o?t?θ??22
1???2?θ?=??andcos θ=?1??sin
??1? ?t?an2θ??
tan2θ=sec2θ; 1
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cot θsin θ2θ2θ=??==cos?sin???
://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33car221? ?c?o?t2?θtan θ??
FIG. 3-41Graph of y=cos x.
??????
??
The cofunction property is very important. cos θ=sin (90°?θ),
sin θ=cos (90°?θ), tan θ=cot (90°?θ), cot θ=tan (90°?θ), etc.Functions of
Negative Anglessin (?θ) =?sin θ, cos (?θ) =cos θ, tan (?θ) =?tan θ, sec (?θ)
=sec θ, csc (?θ) =?csc θ, cot (?θ) =?cot θ.
FIG. 3-42
Graph of y=tan x.
Identities
Sum and Difference FormulasLet x, ybe two angles. sin (x ?y) =sin xcos y?cos xsin
y;cos (x ?y) =cos xcos y?sin xsin y;tan (x ?y) =(tan x?tan y)/(1 ?tan xtan y); sin
x?sin y=2 sin a(x ?y) cos a(x ?y); cos x cos y=2 cos a(x
sin
a(x
y)
sin
a(x
?y);
sin2x?sin2y=cos2y?cos2x=sin (x
tan
x?tan
y) cos a(x ?y); cos x?cosy=?2
y=[sin
(x
?y)]/(cos
xcosy);
y) sin (x ?y); cos2x?sin2y=cos2y?sin2x=cos (x
y)
cos (x ?y); sin (45° x) =cos (45°?x); sin (45°?x) =cos (45° x); tan (45°?x) =cot
(45°?x). A cos x
://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33cpar2222??B
x=??A? ?B?sin (α x) =??A? ?B?cos (β?x) where tan α=A/B,
sin
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tan β=B/A;both αand βare positive acute angles.
Multiple and Half Angle IdentitiesLet x=angle, sin 2x=2 sin xcos x;sin x=2 sin axcos
ax;cos 2x=cos2x?sin2x=1 ?2 sin2x=2 cos2x?1. tan 2x=(2 tan x)/(1 ?tan2x); sin 3
x=3 sin x?4 sin3x;cos 3x
=4 cos3x?3 cos x.tan 3x=(3 tan x?tan3x)/(1 ?3 tan2x); sin 4x=4 sin xcos x?8 sin3xcos
x;cos 4x=8 cos4x?8 cos2x 1.
x
sin ?=??a?(1???c?o?s?x)
2
????
x
?(1???s??cos ?=?a ?cox)
2xtan ?=
2
1?cos xsin x1?cos x?=?=?????1 cos x1 cos xsin x
Relations between Three Angles Whose Sum Is 180°Let x, y,
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zbe the angles.
xyz
six x sin y sin z=4 cos ?cos ?cos ?
222
??????
xyz
cos x cos y cos z=4 sin ?sin ?sin ? 1
222xyz
sin
x
sin
y?sin
z=4
http://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33csin ?sin ?cos ?
222
??????
??????
sin2x sin2y sin2z=2 cos xcos ycos z 2; tan x tan y tan z=
tan xtan ytan z; sin 2x sin 2y sin 2z=4 sin xsin ysin z.INVERSE TRIGONOMETRIC FUNCTIONS
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y=sin ?1x=arcsin xis the angle ywhose sine is x.
The complete solution of the equation x=sin yis y=(?1)nsin?1x n(180°),?π/2
≤sin?1x≤π/2 where sin?1xis the principal value of the angle whose sine is x.The
range of principal values of the cos?1xis 0 ≤cos?1x≤πand ?π/2 ≤tan?1x≤π/2.
If these restrictions are allowed to hold, the following formulasresult:
Exampley=sin?1a, yis 30°.
3-22MATHEMATICS
x?????x2?1?1?
??=tan?1?sin?1x=
cos?1?1??x2=cot?2
x1???x???
11π
=sec?1?=csc?1?=??cos?1x2
x21???x???
?1???x2??
???=tan?1?cos?1x=sin?1?1??x2
x
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x1
=cot?1?=sec?1?2
x1???x???1πhttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33c
=csc?1?=?sin?1x?2
1???x?2??x1
tan?1x =sin?1?=cos?1?2
???1???x? ?x2???1
1? ?x?1?????=csc?1?=cot?1?=sec?1?1 ?x2
xx
2
FIG. 3-44Right triangle.FIG. 3-45Oblique triangle.
Right Triangle (Fig. 3-44)Given one side and any acute angle α
or any two sides, the remaining parts can be obtained from the fol-lowing formulas:
(c? ?b)(?c????b)?=csin α=btan αa=??
(c? ?a)(?c????a)?=ccos α=acot αb=??
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aba
?, sin α=?, cos α=?, tan α=?, β=90°?αc=??a2???b2
ccb1a2b2tan αc2sin 2α
A=?ab=?=?=?
22 tan α24
Oblique Triangles (Fig. 3-45)There are four possible cases.1.Given b, cand the
included angles α,
b?c1111
tan (β γ)?(β γ) =90°??α; tan (?β?γ) =??b c22221111bsin α
β=?(β γ)
2222sin
?(β?γ); γ=?(β γ) ??(β?γ); a=?
β2.Given
the
three
sides
a,
b,
c);://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33car
r=
(s?a)(s?b)(s?c)?????s
RELATIONS BETWEEN ANGLES
AND SIDES OF TRIANGLES
c,
s=a(a
b
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Solutions of Triangles (Fig. 3-43)Let a, b, cdenote the sidesand α, β, γthe angles
opposite the sides in the triangle. Let 2s =a
b
c, A=area, r=radius of the inscribed
circle, R=radius of the cir-cumscribed circle, and h=altitude. In any triangle α
β γ=180°.
FIG. 3-43Triangle.
Law of Sinessin α/a=sin β/b=sin γ/c.Law of Tangents
a btan a(α β)b ctan a(β γ)a ctan a(α γ)?=??; ?=??; ?=??a?btan a(α?β)b?ctan
a(β?γ)a?ctan a(α?γ)Law of Cosinesa2=b2 c2?2bccos α; b2=a2 c2?2accos β;c2=a2
b2?2abcos γ.
Other RelationsIn this subsection, where appropriate, twomore formulas can be
generated by replacing aby b, bby c, cby a,αby β, βby γ, and γby α. cos α=(b2
c2?a2)/2bc;
a
=bcos
γ
c=cos
β;
sin
α=(2/bc) ??s(?s???a?)(?s???b?)(?s???c?);αsin ?=
2
(s?b)(s?c)αs(s?a)1
?;
A=http://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33c?bh??;
cos ???=??????bc2bc2
1asin βsin γ
=?absin γ=??=??s(s????a)(?s????b)(?s????c)?=rs22 sin α
wherer=
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(s?a)(s?b)(s?c)?????s
2
1r1r1r
tan ?α=?; tan ?β=?; tan ?γ=?
2s?a2s?b2s?c
3.Given any two sides a, cand an angle opposite one of them α,sin γ=(csin
α)/a;β=180°?a?γ; b=(asin β)/(sin α). There may betwo solutions here. γmay have
two values γ1, γ2; γ190°. If α γ2>180°, use only γ1. This case may be
impossibleif sin γ>1.
4.Given any side cand two angles αand β, γ=180°?α?β; a=(csin α)/(sin γ);
b=(csin β)/(sin γ).HYPERBOLIC TRIGONOMETRY
The hyperbolic functions are certain combinations of exponentials exand e?x.
ex e?xex?e?xsinh xex?e?x
cosh x=?; sinh x=?; tanh x=?=?
22cosh xex e?x
ex e?x1cosh x12
coth x=?=?=?; sech x=?=?; x?xx
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e?etanh xsinh xcosh xe e?x21
csch x=?=?x
://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33crsinh xe?e?x
Fundamental Relationshipssinh x cosh x=ex; cosh x?sinh x =e?x; cosh2x?sinh2x=1;
sech2x tanh2x=1; coth2x?csch2x=1;sinh 2x=2 sinh xcosh x;cosh 2x=cosh2x sinh2x=1
sinh2x=2 cosh2x?1. tanh 2x=(2 tanh x)/(1
2
tanh2x); sinh (x ?y) =sinh xcosh y?cosh
xsinh y;cosh (x ?y) =cosh xcosh y?sinh xsinh y;2 sinh2x/2 =cosh x?1; 2 cosh2x/2 =cosh
x 1; sinh (?x) =?sinh x;cosh (?x) =cosh x;tanh (?x) =?tanh x.
When u =acosh x, v =asinh x,then u2?v2=a2; which is the equa-tion for a hyperbola.
In other words, the hyperbolic functions in theparametric equations u =acosh x, v
=asinh xhave the same relationto the hyperbola u2?v2=a2that the equations u =acos
θ, v =asin θhave to the circle u2 v2=a2.
R =a/(2 sin α) =abc/4A; h =csin a =asin γ=2rs/b.
Examplea=5, b=4, α=30°. Use the law of sines. 0.5/5 =sin β/4, sin β=2?5,
β=23°35′, γ=126°25′. So c=sin 126°25′/1?10=10(.8047) =8.05.The relations
given here suffhttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33cice to
solve any triangle. One methodfor each triangle is given.
DIFFERENTIAL AND INTEGRAL CALCULUS
3-23
3-24Callthen
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MATHEMATICS
?ydylim?=??x→0?xdx
f(x ?x)?f(x)dy
lim???=?x→0dx?x
dysin (x ?x)?sin(x)
?=lim??dx?x→0?x
sin xcos ?x sin ?xcos x?sin x
=lim?????x→0?xsin x(cos ?x?1)sin ?xcos x=lim?? lim???x→0?x→0?x?xsin ?x=cos
xsince lim?=1
?x→0?x
dcos x=?sin x dxdtan x=sec2x dxdcot x=?csc2x dxdsec x=tan xsec x dxdcsc x=?cot xcsc
x
dxdsin?1x=(1
?x2)?1/2dxdcos?1x=?(1
?x2)?1/2dxdtan?1x=(1
x2)?1dxdcotx=?(1
x)dxdsec?1x =x?1(x2?1)?1/2dxdcsc?1x =?x?1(x2?1)?1/2dxdsinh x=cosh x dxdcosh x=sinh
x dxdtanh x=sechx dxdcoth x=?csch2x dxdsech x=?sech xtanh x dxdcsch x=?csch xcoth
x dxdsinhx=(x 1)
?1
2
?1/22
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?1
2?1
(3-19)(3-20)(3-21)(3-22)(3-23)(3-24)(3-25)(3-26)(3-27)(3-28)(3-29)(3http://www.m
ianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33c-30)(3-31)(3-32)(3-33)(3-34)(3-35)
(3-36)(3-37)(3-38)(3-39)
dx
(3-40)(3-41)
Using(3-8)(3-19)(3-5), (3-10)
(3-10)
ExampleFind the derivative of y=sin x.
Differential OperationsThe following differential operations
are valid: f, g,...are differentiable functions of x, cis a constant; eisthe base
of the natural logarithms.
dc
(3-5)?=0
dx
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dx?=1dx
ddfdg?(f
g) =? ?dxdxdxddgdf?(f ×g) =f ? g ?dxdxdx
dy1
?=?dxdx/dy
if
dx?≠0dy
(3-6)(3-7)(3-8)(3-9)(3-10)(3-11)(3-12)(3-13)(3-14)
dx
dcosh?1=(x2?1)?1/2dxdtanh?1x=(1 ?x2)?1dxdcoth?1x=?(x2?1)?1dxdsechx=?(1/x)(1 ?x)d
cschx =?x(x 1)
?1
?1
2
?1
2?1/2?1/2
dx
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?cos (1 ?x2).ExampleFind dy/dxfor y=?x
ddyd
??cos (1 ?x2)
cos (1 ?x2) ??x??=?x
dxdxdxdhttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33cd
?cos (1 ?x2) =?sin (1 ?x2) (1 ??x2)dxdx
=?sin (1 ?x2)(0 ?2x)
x1d??
?=?x?1/2dx2
1dy
?=2x3/2sin (1 ?x2)
?x?1/2cos (1 ?x2)dx2
ddf
?fn=nfn?1?dxdx
dfg(df/dx)?f(dg/dx)??=??dxgg2
dfdfdv
?=?×?dxdvdx
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(chain rule)
??
dfgdfdg?=gfg?1? fgln f ?dxdxdxdax
?=(ln a) axdx
ExampleDerive dy/dxfor x2 y3=x
xy
A.
Here
ddddd
?x2 ?y3=?x ?xy ?Adxdxdxdxdx
ExampleFind the derivative of tan xwith respect to sin x.
v=sin xy=tan x
dtan xdydydx
?=?=??dsin xdvdxdv
1dtan x
=??dsinx
dx??
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dx
2
=secx/cos x
Using(3-12)(3-9)
(3-18), (3-20)
dydy
2x 3y2?=1
y
x? 0
dxdx
by rules (3-10), (3-10), (3-6), (3-8), and (3-5) respectively.
dy2x?1?y
://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33carThus?=??
dxx?3y2
Differentials
dex=exdxd(a) =alog a dxdln x=(1/x) dxdlog x=(log e/x)dxdsin x=cos x dx
x
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x
(3-15a)(3-15b)(3-16)(3-17)(3-18)
Very often in experimental sciences and engineering functions and
their derivatives are available only through their numerical values. Inparticular,
through measurements we may know the values of a func-tion and its derivative only
at certain points. In such cases the preced-ing operational rules for derivatives,
including the chain rule, can beapplied numerically.
ExampleGiven the following table of values for differentiable functions fand
g;evaluate the following quantities:
DIFFERENTIAL AND INTEGRAL CALCULUS
3-25
Partial DerivativeThe abbreviation z =f(x, y) means that zis afunction of the two
variables xand y.The derivative of zwith respectto x,treating yas a constant, is
calledhttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33c
the
partial
derivative withrespect to xand is usually denoted as ?z/?xor ?f(x, y)/?xor simply
fx.Partial differentiation, like full differentiation, is quite simple to
apply.Conversely, the solution of partial differential equations is appreciablymore
difficult than that of differential equations.ExampleFind ?z/?xand ?z/?yfor z =yex
xey.
?ex?x?z
?=y? ey?
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?x?x?x=2xyex ey
22
2
?ey?z2?y
?=ex? x??y?y?y
=ex xey
2
Order of DifferentiationIt is generally true that the order of
differentiation is immaterial for any number of differentiations orvariables
provided the function and the appropriate derivatives arecontinuous. For z =f(x, y)
it follows:
?3f?3f?3f
==????y2?x?y?x?y?x?y2
General Form for Partial Differentiation1.Given f(x, y) =0 and x =g(t), y =h(t).
df?fdx?fdyThen?=?? ??
dt?xdt?ydt://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33c
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d2f?2fdx
=???
dt2?x2dt
??
2
?2fdxdy?2fdy 2 ??? ??
?x?ydtdt?y2dt??
2
?fd2x ???xdt2
?fd2y ???ydt2
ExampleFind df/dtfor f =xy, x=ρsin t, y=ρcos t.
df?(xy)dρsin
t?(xy)dρcos
t?=??
??dt?xdt?ydt=y(ρcos
t)=ρ2cos2t?ρ2sin2t
????
Given f(x, y) =0 and x =g(t, s), y =h(t, s).
?f?f?x?f?y
t)
x(?ρsin
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Then?=?? ??
?t?x?t?y?t
?f?f?x?f?y?=?? ???s?x?s?y?x
Differentiation of Composite Function
dy?f/?x?f
Rule 1.Given f(x, y) =0, then ?=? ??≠0.
dx?f/?y?y
Rule 2.Given f(u) =0 where u =g(x), then
dfdu?=f′(u)?dxdx
2.
??
d2fdu
=f″(u) ??2dxdx??
2
d2u
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f′(u)?
dx2
???ExampleFind df/dxfor f=sin2uand u=?1??x2
?dfdsin2ud??1???x2
?=???dxdudx
1://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33c
=2 sin ucos u?(?2x)(1 ?x2)?1/2
2
??
?????u2?1
=?2 sin ucos u?
u
Rule 3.Given f(u) =0 where u =g(x,y), then
?u?f?u?f
?=f′(u)??=f′(u)??x?x?y?y
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3-26MATHEMATICS
?2f?u
=f″??2?x?x
??
2
?2u
f′?
?x2
?2f?u?u?2u?=f″?? f′??x?y?x?y?x?y?2f?u
=f″??2?y?y
Vis specified once Tand pare specified; it is therefore a state
function.
All applications are for closed systems with constant mass. If aprocess is reversible
and only p-Vwork is done, the first law and dif-ferentials can be expressed as follows.
dU =T dS ?p dVdH =T dS
V dpdA =?S dT ?p dVdG =?S dT
V dp
Alternatively, if the internal energy is considered a function of Sand V,then the
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differential is:
?U?U
dU=?dS ?dV
?SV?VS
This
is
the
equivalent
http://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33cof Eq. (3-43) and gives
the following definitions.
?U?UT=?,p=??
?SV?VS
Since the internal energy is a state function, then Eq. (3-44) must besatisfied.
?2U?2U
=???V?S?S?V
??
2
?2u f′?
?y2
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MULTIVARIABLE CALCULUS APPLIED TO THERMODYNAMICS
Many of the functional relationships needed in thermodynamics aredirect applications
of the rules of multivariable calculus. This sectionreviews those rules in the context
of the needs of themodynamics.These ideas were expounded in one of the classic books
on chemicalengineering thermodynamics.151
State FunctionsState functions depend only on the state of thesystem, not on past
history or how one got there. If zis a function oftwo variables, xand y,then z(x,y)
is a state function, since zis knownonce xand yare specified. The differential of
zis
dz =M dx
N dy
The line intehttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33cgral
????
????
?
C
(M dx
N dy)
This is
is independent of the path in x-yspace if and only if
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?M?N?=??y?x
The total differential can be written as
?z?z
dz=?dx ?dy
?xy?yx
and the following condition guarantees path independence.
??z??z??=???y?xy?x?yx
?T?p
=????????V?S
S
V
(3-42)
????
(3-43)
This is one of the Maxwell relations, and the other Maxwell relations
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can be derived in a similar fashion by applying Eq. (3-44).
?T?V?=??pS?Sp?S?p?=??VT?TV
????
?????????S?V
=?????p???T?
T
p
or
?2z?2z
?=??y?x?x?y
ExampleSuppose zis constant and apply Eq. (3-43).
(3-44)
???
Rearrangement gives
?z?z
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0 =?dx ?://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33cpar
?xy?y??dy?
x
z
?z?z?y(?y/?x)z
(3-45)?=???=? ?
?xy?xz?yx(?y/?z)x
Alternatively, divide Eq. (3-43) by dywhen holding some other variable wcon-stant
to obtain
?z?z?z?x
(3-46)?=?? ?
?yw?xv?yw?yx
Also divide both numerator and denominator of a partial derivative by dwwhileholding
a variable yconstant to get
??????
????????
(?z/?w)y
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y
In process simulation it is necessary to calculate enthalpy as a func-tion of state
variables. This is done using the following formulas,derived from the above relations
by considering Sand Has functionsof Tand p.
?V
dH =CpdT V ?T?dp
?Tp
Enthalpy differences are then given by the following formula.
T2p
2?V
H(T2, p2) ?H(T1, p1) =Cp(T, p1) dT V ?T?dp
T1p1?TpT2,p
The
same
manipulations
http://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33ccan be done for internal
energy as a func-tion of Tand V.
(?V/?T)p
dU =CVdT?p
T?dV
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(?V/?p)T
????
??
?
????
??
=?=?????????x?(?x/?w)?w?x?z
?z
?w
y
y
(3-47)
y
Themodynamic State FunctionsIn thermodynamics, the state
functions include the internal energy, U;enthalpy, H;and Helmholtzand Gibbs free
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energies, Aand G,respectively, defined as follows:
H =U
p VA =U ?TS
G =H ?TS =U
pV ?TS =A
pV
Sis the entropy, Tthe absolute temperature, pthe pressure, and Vthevolume. These are
also state functions, in that the entropy is specifiedonce two variables (like Tand
p) are specified, for example. Likewise,
Partial Derivatives of All Thermodynamic FunctionsThevarious partial derivatives of
the
thermodynamic
functions
can
beclassified
into
thhttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33ce
six
groups.
general
In
formulas
below, the variablesU, H, A, Gor Sare denoted by Greek letters, while the variables
V, T,or pare denoted by Latin letters.
Type I(3 possibilities plus reciprocals)
?a?p
General: ?; Specific: ?
?bc?TV
Eq. (3-45) gives
?p?V?p(?V/?T)p?=???=? ??TV?Tp?VT(?V/?p)T
????
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??????
??
Type II(30 possibilities)
?α?G
General: ?; Specific: ?
?bc?T
??
V
DIFFERENTIAL AND INTEGRAL CALCULUS
The differential for Ggives
?G?p?=?S
V??TV?TV
Using the other equations for U, H, A,or Sgives the other possibilities.Type III(15
possibilities plus reciprocals)
?a?V
General: ?; Specific: ?
?bα?TS
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First expand the derivative using Eq. (3-45).
?V?S?V(?S/?T)V?=???=? ??TS?TV?ST(?S/?V)T
Then
evaluate
the
numerator
and
denomihttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33cnator as type II
derivatives.
CV?V????T?VCV?pT????=??V?p=??TST?V????????Tp?VT
?Tp
3-27
????
?f′(x) dx =f(x)
c
where cis an arbitrary constant to be determined by the problem. Byvirtue of the known
formulas for differentiation the following rela-tionships hold (ais a constant):
????
??????
?(du
v
dv
dw) =?du ?dv ?dw?a dv =a?dv
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?vdv =? c(n≠?1)
n 1
n
n 1
(3-48)(3-49)(3-50)(3-51)(3-52)(3-53)(3-54)(3-55)(3-56)(3-57)(3-58)(3-59)(3-60)(3
-61)(3-62)
??
?????
????
dv??=ln |v|
cv
a
?adv =? c
ln a
?edv =e c
?sin v dv=?cos v
v dv=sec v
c?cos v dv=sin v
c?csc vcot v dv=?csc v
c?secv dv=tan v
c
c?cscv dv=?cot v
c?sec vtan
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dv1v??=?tan? cv aaa
v
v
v
v
22
?1
://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33cpar2
2
These derivatives are of importance for reversible, adiabatic processes
(such as in an ideal turbine or compressor), since then the entropy isconstant. An
example is the Joule-Thomson coefficient.
?T1?V?=??V
T??pHCp?Tp
Type IV(30 possibilities plus reciprocals)
?G?α
General: ?; Specific: ?
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?βc?Ap
Use Eq. (3-47) to introduce a new variable.
?????
??
p
??
????????
?G??A?G=?p?T?T
??A(?G/?T)p=?p(?A/?T)p
This operation has created two type II derivatives; by substitution we
obtain
?GS
?=???ApS p(?V/?T)p
Type V(60 possibilities)
?α?G
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General: ?; Specific: ?
?bβ?pA
Start from the differential for dG.Then we get
?G?T
?=?S? V?pA?pA
The
derivative
is
type
III
and
can
be
evaluated
bhttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33cy using Eq. (3-45).
?G(?A/?p)T
?=S? V?pA(?A/?T)p
The two type II derivatives are then evaluated.
?GSp(?V/?p)T
?=?? V?pAS p(?V/?T)p
These derivatives are also of interest for free expansions or isentropicchanges.
Type VI(30 possibilities plus reciprocals)
?α?G
General: ?; Specific: ?
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?βγ?AH
We use Eq. (3-47) to obtain two type V derivatives.
?G(?G/?T)H?=??AH(?A/?T)H
These can then be evaluated using the procedures for Type V derivatives.
????
dv??=sin
a???v???
2
2
2
2
?1
v
? ca
dv1v?a??=?ln ??? cv?a2av adv???=ln |v ??v???a?| c
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v???a???
?sec v dv=ln (sec v tan v)
c?csc v dv=ln (csc v?cot v)
c
2
2
2
2
????
(3-63)(3-64)(3-65)
??
?http://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33c?
formula dav/dv =avln a,or in the more usable form d(av/ln a) =avdv,let f′=avdv;then
f =av/ln aand hence ∫avdv=(av/ln a)
3 ∫x2dx ∫exdx?10 ∫dx=x3 ex?10x
ExampleDerive ∫avdv=(av/ln a)
c.
c(by Eqs. 3-50, 3-53).
c.By reference to the differentiation
ExampleFind ∫(3x2 ex?10) dxusing Eq. (3-48). ∫(3x2 ex?10) dx=ExampleFind ?. Let
v=2 ?3x2; dv=?6x dx2
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x dx
?27?
3x
2
????
??
Thus
?6x dx7x dxx dx7
??=7 ??=? ???2?3x2?3x62?3x
7dv=? ???
6v
2
2
INTEGRAL CALCULUS
Indefinite IntegralIf f′(x) is the derivative of f(x), an antideriv-ative of f′(x)
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is f(x). Symbolically, the indefinite integral of f′(x) is
7
=? ?ln |v|
c
67
=? ?ln |2 ?3x2|
c
6
3-28MATHEMATICS
x dx???=?(3 4x)
1/4
x3is
3x2,
and
http://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33cx3is
therefore the integral of 3x2. However, if f =x3 10, it follows
that f′=3x2,and x3 10 is therefore also the integral of 3x2. For this reason
theconstant cin ∫3x2dx =x3 cmust be determined by the problem conditions,i.e., the
value of ffor a specified x.
Example—Constant of IntegrationBy definition the derivative of
y4?3
?? y3dy4
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??
y
Methods of IntegrationIn practice it is rare when generallyencountered functions can
be directly integrated. For example, the
si?n?xdxwhich appears quite simple has no elementary integrand in ∫??
function whose derivative is ??si?n?x. In general, there is no explicit wayof
determining whether a particular function can be integrated into anelementary form.
As a whole, integration is a trial-and-error proposi-tion which depends on the effort
and ingenuity of the practitioner.The following are general procedures which
http://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33ccan be used to find
theelementary forms of the integral when they exist. When they do notexist or cannot
be found either from tabled integration formulas ordirectly, the only recourse is
series expansion as illustrated later.Indefinite integrals cannot be solved
numerically unless they are rede-fined as definite integrals (see “Definite
Integral”), i.e., F(x) =∫f(x) dx,
x
indefinite, whereas F(x) =∫af(t) dt,definite.
Direct FormulaMany integrals can be solved by transformationin the integrand to one
of the forms given previously.
Thus
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1=?4
?y(y?3) dy
2
4
1y73y3
=????? c4743
11
=?(3
4x)7/4??(3
4x)3/4 c284
GeneralThe number of possible transformations one might use
are unlimited. No specific overall rules can be given. Success in han-dling
integration
problems
depends
primarily
upon
experience
andingenuithttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33cy.
The
following example illustrates the extent to which alter-native approaches are
possible.
ExampleFind ?. Let ex=y;then exdx =dyor dx=1/y dy.x
dx(1/y) dye?2dyy?1
??=??=??=ln ?=ln ?e?1y?1y?yey
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x
x
2
x
?edx
?1
???dx.Let v=3x3 10 for which dv=9x2dx.ExampleFind ∫x2?3x3? ?10
???dx=?(3x 10)?x?3x? ?10
2
3
3
1/2
(x2dx)
1/2
1
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=?91=?9
?(3x 10)
3
(9x2dx)
?v
1/2
dv
[by Eq. (3-50)]
1v3/2
=?? c93?2
Partial FractionsRational functions are of the type f(x)/g(x)where f(x) and g(x) are
polynomial expressions of degrees mand nrespectively. If the degree of fis higher
than g,perform the algebraicdivision—the remainder will then be at least one degree
less
than
thedenominator.
Consider
thehttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33c following types:
Type 1Reducible denominator to linear unequal factors. Forexample,
11
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=?????32
x?x?4x 4(x 2)(x?2)(x?1)
ABC=? ? ?x 2x?2x?1
A(x?2)(x?1) B(x 2)(x?1) C(x 2)(x?2)=??????
(x 2)(x?2)(x?1)x2(A B C) x(?3A B) (2A?2B?4C)=?????
(x 2)(x?2)(x?1)
Equate coefficients and solve for A, B,and C.
A
B
C=0?3A
B=02A?2B?4C=1A=1?12, B=d, C=?s1111
=? ?????32
x?x?4x 412(x 2)4(x?2)3(x?1)
Hence
dxdxdxdx
???=?? ?????x?x?4x 412(x 2)4(x?2)3(x?1)
3
2
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2
=?(3x3 10)3/2 c27
Trigonometric SubstitutionThis technique is particularly welladapted to integrands
in the form of radicals. For these the function istransformed into a trigonometric
form. In the latter form they may bemore easily recognizable relative to the identity
formulas.
These
func-tions
and
their
transformahttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33ctions are
?Let x =asec θx2???a2??
?Let x =atan θx2? ?a2??
?Let x =asin θa2???x2??
22
ExampleFind ??dx.Let x=?sin θ; then dx=?cos θdθ.2
3
3
3
??(2/3?)???x?????θ?22/3?1??sin
? ???dx=3 ????cos θdθ?x(2/3)sinθ?3
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2
2
2
2
2
2
????9?x
? ?4
x
2
cos2θ=3 ?dθ
sin2θ=3 cot2θdθ
=?3 cot θ?3θ cby trigonometric transform
??
PartsAn extremely useful formula for integration is the relationandor
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????9?x23?4
=????3 sin?1?x
cin terms of x
x2
?u dv ?v du
?u dv =uv??v du
uv=
d(uv) =u dv
v du
Algebraic SubstitutionFunctions containing elements of thetype (a
handled by the algebraic transformation yn=a
ExampleFind
??.
Let
bx)1/nare best
bx.
3
=y4;http://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33c
4x
then
4dx=4y3dyand1/4
x dx
?(3
4x)
No general rule for breaking an integrand can be given. Experiencealone limits the
use of this technique. It is particularly useful fortrigonometric and exponential
functions.ExampleFind ?xexdx.Let
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u =xdu =dx
and
dv =exdxv =ex
DIFFERENTIAL AND INTEGRAL CALCULUS
Therefore
3-29
?xedx =xe??edx
x
x
x
=xex?ex c
ExampleFind ?exsin x dx.Let
u =exdu =edx
xx
x
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?esin x dx =?ecos x ?ecos x dx
x
dv=sin x dxv=?cos x
Techniques for determining when integration is valid under theseconditions are
available in the references. However, the following sim-plified rules will, in
general, serve as a guide for most practical appli-cations.
Rule 1For the integral
∞
φ(x)
dx?
://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33cpar0xn
?
Again
x
u =exdu =edx
xx
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?esin x dx =?ecos x
esin x??esin x dx
c
x
x
dv=cos x dx
v=sin x
c
=(ex/2)(sin x?cos x)
?
2
if φ(x) is bounded, the integral will converge for n>1 and not con-verge for n≤1.∞
It is easily seen that ∫0 e?xdxconverges by noting 1/x2>1/ex>0 forlarge x.
Rule 2For the integral
b
φ(x)
dx,?n
a(a?x)
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?
Series ExpansionWhen an explicit function cannot be found,the integration can
sometimes be carried out by a series expansion.
2
ExampleFind ?e?xdx.Since
x4x62
e?x=1 ?x2 ??? ???
2!3!
if φ(x) is bounded, the integral will converge for n
1
1?dxx0??
?
?e
?x2
dx=
xx
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?dx??xdx ??dx???dx ???://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33car
2!3!
2
4
6
will converge (exist) since a=n
PropertiesThe fundamental theorem of calculus states
?f(x) dx =F(b) ?F(a)
ba
x3x5x7
=x ?? ??? ???
35.2!7.3!
for all x
where
b
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dF(x)/dx =f(x)
Definite IntegralThe concept and derivation of the definiteintegral are completely
different from those for the indefinite integral.These are by definition different
types of operations. However, theformal operation ∫as it turns out treats the
integrand in the same wayfor both.
Consider the function f(x) =10 ?10e?2x. Define x1=aand xn=b,and suppose it is
desirable to compute the area between the curve andthe coordinate axis y=0 and bounded
by x1=a, xn=b.Obviously, bya sufficiently large number of rectangles this area could
be approxi-mated as closely as desired by the formula
n?1i=1
://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33cOther
definite integral are
?c[f(x) dx] =c?f(x) dx
b
a
a
?[f(x)
b
b
f(x)] dx=?f(x) dx ?f(x) dx
properties
of
the
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b
a
12
a
1
a
2
?f(x) dx=??f(x) dx
b
a
a
b
Αf(ξ)(x
i
i 1
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?xi) =f(ξ1)(x2?a)
f(ξ2)(x3?x2)
??? f(ξn?1)(b ?xn?1)
xi?1≤ξi?1≤xi
???b???adF(α)?=dα
?f(x) dx=?
babab
c
a
f(x) dx
?f(x) dx
bc
The definite integral of f(x) is defined as
b
n
?f(x) dx=(b ?a)f(ξ) for some ξin (a, b)?f(x) dx =f(b)
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a
?f(x) dx=limΑf(ξ)(x
a
n→∞
ii 1
?xi)
i=1
where the points x1, x2,..., xnare equally spaced. For a rigorous def-inition of the
definite http://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33cintegral the
references should be consulted.Thus, the value of a definite integral depends on the
limits a, b,andany selected variable coefficients in the function but not on thedummy
variable of integration x.Symbolically
?f(x) dx =?f(a)
ba
or
?f(x) dxindefinite integral where dF/dx =f(x)
F(a, b) =?f(x) dxdefinite integralF(α) =?f(x,α) dx
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F(x) =
baba
?f(x,α)??dxif aand bare constant
?α
ba
?dx?f(x,α) dα=?dα?f(x, α) dx
b
d
d
b
a
c
c
a
(3-66)
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when F(x) =
?
b(x)
a(x)
f(x, y) dy
There are certain restrictions of the integration definition, “The func-tion f(x)
must be continuous in the finite interval (a, b) with at most afinite number of finite
discontinuities,”
which
must
be
observedbefore
http://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33cformulas
integration
can
be
generally applied. Two of theserestrictions give rise to so-called improper
integralsand requirespecial handling. These occur when∞
1.The limits of integration are not both finite, i.e., ∫0 e?xdx.
2.The function becomes infinite within the interval of integra-tion, i.e.,1
1?dxx0??
the Leibniz rule gives
dFdbda
?=?f[x, b(x)] ??f[x, a(x)]
ExampleFind
dxdxdx
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?
b(x)
a(x)
?f
?dy?x
?
π/2
sin x dx.
?
?
since
π/2
ππ/2
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sin x dx=[?cos x]0=?cos ??cos 0=1
2
?dcos x/dx=sin x
??
3-30MATHEMATICS
ExampleFind ?. Direct application of the formula would yield2
?(xdx
?1)
2
20
and since φ(α) →0 as α→∞,
c=π/2
http://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33cφ(α) =?tan?1α π/2
the incorrect value
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1dx
??=????(x?1)x?1
2
2
=?2
It should be noted that f(x) =1/(x?1)2becomes unbounded as x→1
and by Rule 2 the integral diverges and hence is said not to exist.Methods of
IntegrationAll the methods of integration availablefor the indefinite integral can
be used for definite integrals. In addi-tion, several others are available for the
latter integrals and are indi-cated below.
Change of VariableThis substitution is basically the same as pre-viously indicated
for indefinite integrals. However, for definite inte-grals, the limits of integration
must also be changed: i.e., for x=φ(t),
IntegrationIt is sometimes useful to generate a double integralto solve a problem.
By this approach, the fundamental theorem indi-cated by Eq. (3-66) can be
used.ExampleFind
Consider
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?
://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33cr
1
01
xb?xα
?dxln x
α
?x
ba
1
dx=?(α>?1)
α 1
Then multiplying both sides by dαand integrating between aand b,
?f(x) dx=?
ba
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t1
t0
f[φ(t)]φ′(t) dt
But also
b
?dα?x
1
α
dx=
?
b
a
dαb 1?=ln ?α 1a 1
b
??
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where
t =t0when x =a
t =t1when x =b
?6???dx.LetExampleFind ?1??x2
?
4
?dα?x
1
a
α
dx=
1
?dx?x
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10
a
α
dα=
?
1
xb?xα
?dxln x
x=4 sin θ(x=0, θ=0)dx=4 cos θdθ(x=4, θ=π/2)
π/20
Therefore
?
b 1xb?xα
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?dhttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33cx=ln ?ln xa 1??
Then
?6????1??x?dx=16 ?
4
2
π/2
cos2θdθ=16[aθ dsin 2θ]0=4π
DifferentiationHere the application of the general rules for dif-ferentiating under
the integral sign may be useful.ExampleFind
φ(α) =
e?αxsin x
?dx(α>0)0x
Since this is a continuous function of α, it may be differentiated under the inte-gral
sign
∞
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dφ
?=?e?αxsin x dx
0dα
?
∞
Complex VariableCertain definite integrals can be evaluated by
the technique of complex variable integration. This is described in thereferences
for “Complex Variables.”
NumericalBecause of the property of definite integrals anothermethod for obtaining
their solution is available which cannot beapplied to indefinite integrals. This
involves
a
numerical
the
previously
outlihttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33cned
summation
definition:
n→∞
limΑf(ξi)(xi 1?xi) =
1
n?1
approxima-tion
based
on
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?f(x) dx
ba
?
where x1=aandxn=b
Examples of this procedure are given in the subsection “NumericalAnalysis and
Approximate Methods.”
=?1/(1
α2)φ(α) =?tan?1α c
INFINITE SERIES
REFERENCES:53, 126, 127, 163. For asymptotic series and asymptotic meth-ods, see Refs.
51, 127.
while valid for any finite value of rand nnow takes on a differentinterpretation.
In this sense it is necessary to consider the limit of Snasnincreases indefinitely:
S=limSn
n→∞
DEFINITIONS
A succession of numbers or terms that are formed according to somedefinite rule is
called a sequence. The indicated sum of the terms of a sequence is called a series.
A series of the form a0 a1(x ?c)
a2(x ?c)2 ??? an(x ?c)n ???is called a power series.
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Considerhttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33c the sum of a
finite number of terms in the geometricseries (a special case of a power series).
Sn=a
ar
ar2 ar3 ??? arn?1
(3-67)
For any number of terms n,the sum equals
1?rn
Sn=a ?
1?r
In this form, the geometric series is assumed finite.
In the form of Eq. (3-67), it can further be defined that the terms inthe series be
nonending and therefore an infinite series.
S =a
ar
ar2 ??? arn ???
However, the defined sum of the terms [Eq. (3-67)]
1?rn
Sn=a r≠1?
1?r
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(3-68)
1?rn
=alim?
n→∞1?r
For this, it is stated the infinite series converges if the limit of Snapproaches
a fixed finite value as napproaches infinity. Otherwise, theseries is divergent.
On this basis an analysis of
1?rn
S =a lim?
n→∞1?r
shows
that
if
ris
less
than
1http://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33c but greater than ?1,
the infinite series isconvergent. For values outside of the range ?1
Consider the divergence of Eq. (3-68) when r=?1 and
S =a
a(?1)
a(?1)2 a(?1)3 ??? a(?1)n ???=a ?a
INFINITE SERIES
a ?a
1. For theformer case r=?1,
a????
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and for which
1?r
S =a lim?
n→∞1?r
1?(?1)n
=a lim?undefined limit (if a≠0)
n→∞1 1
Since the limit sum does not exist, the series is divergent. This isdefined as a
bounded or oscillating divergent series. Similarly for thevalue r= 1,
S =a
a(1)
a(1)2 a(1)3 ??? a(1)n ???S =a
a
a
a
??? a ???
(a≠0)
The series is also divergent but defined as an unbounded divergentseries.
There are also two types of convergent series. Consider the newseries
1111
(3-69)S=1 ?? ??? ??? (?1)n 1? ???
234n
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It
can
be
shown
that
the
series
(http://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33c3-69) does converge to
the value S=log 2. However, if each term is replaced by its absolute value, theseries
becomes unbounded and therefore divergent (unboundeddivergent):
1111
S=1
? ? ? ? ???(3-70)
2345
In this case the series (3-69) is defined as a conditionally convergentseries. If
the replacement series of absolute values also converges, theseries is defined to
converge absolutely.
Series (3-69) is further defined as an alternating series, while series(3-70) is
referred to as a positive series.OPERATIONS WITH INFINITE SERIES
1.The convergence or divergence of an infinite series is unaf-fected by the removal
of a finite number of finite terms. This is a triv-ial theorem but useful to remember,
especially when using thecomparison test to be described in the subsection “Tests
for Conver-gence and Divergence.”
2.If
a
series
is
conditionally
convergent,
its
sums
canhttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33c be made tohave any
arbitrary value by a suitable rearrangement of the series; itcan in fact be made
divergent or oscillatory (Riemann’s theorem). Thisseemingly paradoxical theorem can
be illustrated by the followingexample.
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11111
ExampleS=1 ?? ??? ??? ???
23456
The series is rearranged so that each positive term is followed by two negativeterms:
11111111
t=1 ???? ????? ????? ???
2436851012
Define t3nfor the first 3nterms in the series
11111111
t3n=1 ???? ????? ??? ?????
243682n?14n?24n111111
=??? ??? ??? ???24684n?24n111111
=?1 ?? ??? ??? ???22342n?12n
n
3-31
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3.A series of positive terms, if convergent, has a sum independent
of the order of its terms; but if divergent, it remains divergent how-ever its terms
are rearranged.
4.An
oscillatory
series
can
always
tohttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33c
be
converge
group-ing the terms in brackets.ExampleConsider the series
123451 ?? ??? ??? ???
23456
which oscillates between the values 0.306 and 1.306. However, the series
123451111
1 ?? ??? ??? ???=??????????? Х0.306 ???
234562123056
??????
and
112345611
1 ???????????? ???=1
23456762042
? ? ? ???=1.306 ???
made
by
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??????
5.A power series can be inverted, provided the first-degree termis not zero. Given
y =b1x
b2x2 b3x3 b4x4 b5x5 b6x6 b7x7 ???
then
x =B1y
B2y2 B3y3 B4y4 B5y5 B6y6 B7y7 ???
whereB1=1/b1
3
B2=?b2/b1
52
B3=(1/b1) (2b2?b1b3)
723
b4?5b2)B4=(1/b1)(5b1b2b3?b1
Additional coefficients are available in the references.
6.Two series may be added or subtracted term by term providedeach is a convergent
series.http://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33c The joint sum is
equal to the sum (or dif-ference) of the individuals.
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7.The sum of two divergent series can be convergent. Similarly,the sum of a convergent
series and a divergent series must be diver-gent.
ExampleGiven
n=1∞
=? ? ? ? ???Α??
n?14916
∞
1 n
2
2345
(a divergent series)(a divergent series)1 n 1?n
2
n=1
=? ????? ???Α??n?4916
1?n
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2
123
However,
Α???=Α????Α??
n?nn
1 n
2
1?n
2
1
=2 Α?
n2
(convergent)
8.A power series may be integrated term by term to represent theintegral of the
function within an interval of the region of conver-gence. If f(x) =a0 a1x
then
a2x2 ???,
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?
x2
x1
f(x) dx=
?
x2
x1
a0dx ://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33car
?
x2
x1
a1x dx
?
x2
x1
a2x2dx ???
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??????
9.A power series may be differentiated term by term and repre-sents the function
df(x)/dxwithin the same region of convergence as f(x).
TESTS FOR CONVERGENCE AND DIVERGENCE
??
1=?S2n
2
where S2nis the sum of the first 2nterms of the original series. Thus
1
limt3n=lim?S2nn→∞n→∞2
1t=?S
2
and since lim t3n 2=lim t3n 1=lim t3n, it follows the sum of the series tis (a) S.Hence
a rearrangement of the terms of an alternating series alters the sum ofthe series.
In general, the problem of determining whether a given series willconverge or not
can require a great deal of ingenuity and resourceful-ness. There is no all-inclusive
test which can be applied to all series. Asthe only alternative, it is necessary to
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aphttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33cply one or more of the
devel-oped theorems in an attempt to ascertain the convergence or diver-gence of the
series under study. The following defined tests are givenin relative order of
effectiveness. For examples, see references onadvanced calculus.
1.Comparison Test.A series will converge if the absolute valueof each term (with or
without a finite number of terms) is less than thecorresponding term of a known
convergent series. Similarly, a positiveseries is divergent if it is termwise larger
than a known divergent seriesof positive terms.
3-32MATHEMATICS
Taylor’s Series
x2x3
f(x
h) =f(h)
xf′(h)
?f″(h)
?f′′′(h)
???
2!3!
f″(x0)f′′′(x0)
orf(x) =f(x0)
f′(x0) (x ?x0)
?(x ?x0)2 ?(x ?x0)3 ???
2!3!ExampleFind a series expansion for f(x) =ln (1
f′(x) =(1
thus
x)?1,
x) about x0=0.
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f(0) =0,
f″(x) =?(1
x)?2,
://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33cparf″(0) =?1,
f′′′(x) =2(1
x)?3, etc.f′′′(1) =2, etc.
2.nth-Term Test.A series is divergent if the nth term of the
series does not approach zero as nbecomes increasingly large.
3.Ratio Test.If the absolute ratio of the (n 1) term divided bythe nth term as nbecomes
unbounded approaches
a.A number less than 1, the series is absolutely convergentb.A number greater than
1, the series is divergentc.A number equal to 1, the test is inconclusive
4.Alternating-Series Leibniz Test.If the terms of a series arealternately positive
and negative and never increase in value, theabsolute series will converge, provided
that the terms tend to zero asa limit.
5.Cauchy’s Root Test.If the nth root of the absolute value of thenth term, as nbecomes
unbounded, approaches
a.A number less than 1, the series is absolutely convergentb.A number greater than
1,
the
series
divergenhttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33ctc.A
is
number
equal to 1, the test is inconclusive
6.Maclaurin’s Integral Test.Suppose Α anis a series of positiveterms and fis a
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continuous decreasing function such that f(x) ≥0 for 1 ≤x
∫1f(x) dxeither both converge or both diverge.SERIES SUMMATION AND IDENTITIESSums
for the First nNumbers to Integer Powers
j=1n
f′(0) =1,
xnx2x3x4
ln (x 1) =x ?? ?? ? ??? (?1)n 1? ???
234n
which converges for ?1
Maclaurin’s Series
x2x3
f(x) =f(0)
xf′(0)
?f″(0)
?f′′′(0)
???
2!3!
This is simply a special case of Taylor’s series when his set to zero.Exponential
Series
x2x3xn
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ex=1
x
? ? ??? ? ????∞
2!3!n!
Logarithmic Series
x?11x?1
ln x =? ??
x2x
ΑΑΑΑ
nnn
n(n 1)
j =?=1
2
3
4
??? n
2
n(n 1)(2n 1)
j2=??http://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33c=12 22 32 42 ??? n2
6n2(n 1)2
j3=??=13 23 33 ??? n3
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4
n(n 1)(2n 1)(3n2 3n?1)
j4=????=14 24 34 ??? n4
30
j=1
j=1
??
2
1x?1 ??3x?? ???
3
(x>a)
j=1
x?11x?1
ln x=2 ? ??
x 13x 1
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Trigonometric Series*
x3x5x7
sin x =x ?? ??? ???
3!5!7!
x2x4x6
cos x=1 ?? ??? ???
2!4!6!
????? ???
3
(x>0)
Arithmetic Progression
k=1
Α[a (k?1)d] =a (a
1
=na ?n(n?1)d
d)
(a 2d)
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2
n
?∞
(x2
(a 3d)
??? [a (n?1)]d
Geometric Progression
j=1
Αar
n
x313x5135x7
sin?1x =x
? ????? ??????? ???
62452467111
tan?1x =x??x3 ?x5??x7 ???
357
(x2
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j?http://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33c1
=a
ar
ar2 ar3 ??? arn?11?rn
=a r≠1?
1?r
Taylor SeriesThe Taylor series for a function of two variables,expanded about the
point (x0, y0), is?f
f(x, y) =f(x0, y0)
?
?x
1?2f
??2!?x2
Harmonic Progression
1111111?=? ? ? ? ? ??? ?Αaa da 2da 3da 4da ndk=0a kd
The reciprocals of the terms of the arithmetic-progression series arecalled harmonic
progression. No general summation formulas areavailable for this series.Binomial
Series
n(n?1)
(x
2!
y)=x nxy
?xn?2y2
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n(n?1)(n?2)n!
??xn?3y3 ??? ?xn?ryr ??? yn
3!(n?r)!r!
n
n
n?1
n
??f
(x ?x0)
?x0, y0?y?
x0, y0
(y ?y0) (x ?x0)(y ?y0)
???2f(x ?x0)2 2 ?x0, y0?x?y?
x0, y0
?2f
??y2?
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x0, y0
(y ?y0)2 ???
?
Pahttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33crtial
Sums
of
Infinite Series, and How They GrowCalcu-lus textbooks devote much space to tests for
convergence and diver-gence of series that are of little practical value, since a
convergent
n(n?1)(n?2)n(n?1)
(1 ?x)n=1 ?nx
?x2???x3 ???(x2
2!3!
?
*tan xseries has awkward coefficients and should be computed as
sin x
(sign) ??.
???2?x???sin?1
?
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COMPLEX VARIABLES
3-33
3-34MATHEMATICS
? i ? 2nπ.Examplelog (1
i) =log ?2
?π4?
General powers of zare defined by zα=eαlog z.Since log zis infi-nitely many valued,
so too is zαunless αis a rational number.
DeMoivre’s formula can be derived from properties of ez.
z=r(cos θ isin θ)=r(cos nθ isin nθ)
Thus
(cos θ isin θ)n=cos nθ isin nθ
n
n
n
n
?v/?x=??u/?y.The
last
two
equations
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ahttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33cre
called
the
Cauchy-Riemann equations. The derivative
dw?u?v?v?u?=? i ?=??i ?dz?x?x?y?yIf f(z) possesses a derivative at zoand at every
point in some neighbor-hood of z0, then f(z) is said to be analytic at z0. If the
Cauchy-Riemann
equations are satisfied and
?u?u?v?vu, v,?, , , ???
?x?y?x?yare continuous in a region of the complex plane, then f(z) is analytic inthat
region.
?v/?x=?v/?y=0. These are continuous everywhere, but the Cauchy-Riemannequations hold
only at the origin. Therefore, wis nowhere analytic, but it is dif-ferentiable at
z=0 only.
y,?u/?y =?exsin y,?v/?x =exsin y,?v/?y =excos y.The continuity and Cauchy-Riemann
requirements are satisfied for all finite z.Hence ezis analytic (exceptat ∞) and
dw/?z=?u/?x
i(?v/?x) =ez.
2222
Examplew =zz?=x y. Here u =x y, v=0. ?u/?x=2x,?u/?y=2y,
cipal value (n=0) =e?π/2.
Exampleii=eilogi=ehttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33ci[lo
g|i| i(π/2 2nπ)]=e?(π/2 2nπ). Thus iiis real with prin-?)1 i=e(1 i)log?2?=elog?2?.
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eilog?2?=?2?. (cos log ?2? Example(?2
?) =?2?[cos (0.3466)
isin (0.3466)].isin log ?2
Inverse Trigonmetric Functionscos?1z =?ilog (z ???z2???1);
ii z
?); tan?1z=?log ?. These functions sin?1z =?ilog (iz???1???z2
2i?z
are infinitely many valued.
z2???1); Inverse Hyperbolic Functionscosh?1z=log (z???
1 z1
sinh?1z=log (z???z2? ?1); tanh?1z=?log ?.
21?z
???
Examplew =ez=excos y
iexsin y. u =excos y, v =exsin y.?u/?x =excos
?
Examplew=?=?=??i ?222222
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It is easy to see that dw/dzexists except at z=0. Thus 1/zis analytic except at
z=0.
1zx?iyx yxx yyx y
COMPLEX FUNCTIONS (ANALYTIC)
In the real-number system agreater than b(a >b) and bless than c(b
set of all points withihttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33cn
and on the circle, centered at x=3, y=0 of radius 5.
Singular PointsIf f(z) is analytic in a region except at certainpoints, those points
are called singular points.Example1/zhas a singular point at zero.
Exampletan zhas singular points at z=?(2n 1)(π/2), n=0, 1, 2, ....The derivatives
of the common functions, given earlier, are the sameas their real counterparts.
Example(d/dz)(log z) =1/z,(d/dz)(sin z) =cos z.
Harmonic FunctionsBoth the realand the imaginaryparts ofany analytic function f =u
ivsatisfy Laplace’s equation ?2φ/?x2 ?2φ/?y2=0. A function which possesses
continuous second partialderivatives and satisfies Laplace’s equation is called a
harmonic func-tion.
ecos y,?u/?y =?exsin y,?2u/?y2=?excos y.Clearly ?2u/?x2 ?2u/?y2=0. Sim-ilarly, v
=exsin yis also harmonic.
x
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?≤5, which is theExample|z?3| ≤5. This is equivalent to ??(x???3?)2? ?y2
Example|z?http://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33c1|
≤xrepresents the set of all points inside and on the
parabola 2x =y2 1 or, equivalently, 2x ≥y2 1.
Functions of a Complex VariableIf z =x
iy, w =u
ivand iffor each value of zin some
region of the complex plane one or morevalues of ware defined, then wis said to be
a function of z, w =f(z).Some of these functions have already been discussed, e.g.,
sin z,log z.All functions are reducible to the form w =u(x, y)
iv(x, y), where u,vare
real functions of the real variables xand y.
i(3x2y ?y3).
Exampleez=excos y
Examplez=(x
3
3
3
2
2
iexsin y. u =excos y,?u/?x =excos y,?2u/?x2=
iy)=x 3x(iy)
3x(iy) (iy)=(x?3xy)
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3
3
2
Examplecos z=cos xcosh y ?isin xsinh y.DifferentiationThe derivativeof w =f(z) is
dwf(z ?z)?f(z)
lim???=?z→0dz?z
and
for
the
derivative
to
exist
the
bhttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33ce
limit
the
same
must
no
matterhow ?zapproaches zero. If w1, w2are differentiable functions of z,thefollowing
rules apply:
dw2d(w1w2)dw1dw2d(w1?w2)dw1
?=w2 ? w1 ???=???
dzdzdzdzdzdz
d(w1/w2)w2(dw1/dz)?w1(dw2/dz)
?=???dzw22
ndw1n?1dw1?=nw1?dzdz
For w =f(z) to be differentiable, it is necessary that ?u/?x=?v/?yand
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If w =u
ivis analytic, the curves u(x, y) =cand v(x, y) =kinter-sect at right angles,
if wi(z) ≠0.
y=k.By
implicit
differentiation
there
results,
respectively,
dy/dx=(x2?y2)/2xy,dy/dx=2xy/(y2?x2), which are clearly negative reciprocals, the
condition forperpendicularity.
3
Examplez3=(x3?3xy2)
i(3x2y ?y3). Set u =x3?3xy2=c, v=3x2y ?
and
IntegrationIn much of the work with complex variables a simpleextension of
integration called line or curvilinear integration is of fundamental importance.
Since
any
complex
inthttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33cegral
line
can
be
ex-pressed in terms of real line integrals, we define only real line inte-grals. Let
F(x,y) be a real, continuous function of xand yand cbe anycontinuous curve of finite
length joining the points Aand B(Fig. 3-47). F(x,y) is not related to the curve
c.Divide cup into nsegments,?si,whose projection on the xaxis is ?xiand on the yaxis
is ?yi.Let (εi,ηi) be the coordinates of an arbitrary point on ?si.The limits of
thesums
?si→0
lim
i=1
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ΑF(ε,η) ?s=?F(x, y) ds
n
iii
c
DIFFERENTIAL EQUATIONS3-35
3-36MATHEMATICS
are distinct real roots, r1and r2, say, the solution is y =Aer1x Ber2x,where Aand
Bare arbitrary constants.
Exampley″ 4y′ 3 =0. The characteristic equation is m2 4m 3 =0.The roots are ?3 and ?1,
and the general solution is y =Ae?3x Be?x.
Multiple
Real
RootsIf
r1=r2,
the
solution
of
the
differentialequation
ihttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33cs y =er1x(A
with roots ?2 and ?2. The solution is y =e?2x(A
Bx).
Bx).
ferential equation. If any set of specific values of the constants is cho-sen, the
result is called a particular solution.
Bsin kt,where A, Bare arbitrary constants. A particular solution is x=acos kt 3 sin
kt.
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ExampleThe general solution of (d2x/dt2)
k2x=0 is x =Acos kt
In the case of some equations still other solutions exist called singu-lar solutions.
A singular solutionis any solution of the differentialequation which is not included
in the general solution.
where cis an arbitrary constant; y =x2is a singular solution, as is easily verified.
Exampley″ 4y 4 =0. The characteristic equation is m2 4m 4 =0
Exampley =x(dy/dx) ?d(dy/dx)2has the general solution y =cx?dc2,
ORDINARY DIFFERENTIAL EQUATIONS OF THE FIRST ORDER
Equations
with
Separable
VariablesEvery
differehttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33cntial equa-tion
of the first order and of the first degree can be written in the formM(x, y) dx
N(x,
y) dy=0. If the equation can be transformed so thatMdoes not involve yand Ndoes not
involve x,then the variables aresaid to be separated. The solution can then be
obtained by quadra-ture,which means that y=∫f(x) dx
c,which may or may not
beexpressible in simpler form.
ExampleTwo liquids Aand Bare boiling together in a vessel. Experi-mentally it is found
that the ratio of the rates at which Aand Bare evaporatingat any time is proportional
to the ratio of the amount of A(say, x) to the amountof B(say, y) still in the liquid
state. This physical law is expressible as(dy/dt)/(dx/dt) =ky/xor dy/dx =ky/x,where
kis a proportionality constant. Thisequation may be written dy/y =k(dx/x), in which
the variables are separated.The solution is ln y =kln x ln cor y =cxk.
Exact
EquationsThe
equation
M(x,
y)
dx
N(x,
y)
dy=0
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ihttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33csexact
if
and
only
if ?M/?y=?N/?x.In this case there exists a functionw =f(x, y) such that ?f/?x =M,?f/?y
=N,and f(x, y) =Cis therequired solution. f(x, y) is found as follows: treat yas though
it wereconstant and evaluate ∫M(x, y) dx.Then treat xas though it were con-stant
and evaluate ∫N(x, y) dy.The sum of all unlike terms in thesetwo integrals (including
no repetitions) is f(x, y).
Example(2xy?cos x) dx (x2?1) dy=0 is exact for ?M/?y=2x,?N/?x=2x.∫ M dx=∫ (2xy?cos
x) dx =x2y?sin x,∫ N dy=∫ (x2?1) dy =x2y ?y.Thesolution is x2y?sin x ?y =C,as may
easily be verified.
Linear EquationsA differential equation is said to be linearwhen it is of first degree
in the dependent variable and its derivatives.The general linear first-order
differential equation has the form dy/dx
P(x)y =Q(x). Its general solution is
y =e?∫Pdx
Complex RootsIf the characteristic roots are p ?iq,then thesolution is y
=epx(Ahttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33ccos qx
Bsin qx).
ExampleThe differential equation My″ Ay′ ky=0 represents thevibration of a linear
system of mass M,spring constant k,and damping constant
?, the roots of the characteristic equationA.If A
A
Mm2 Am
2M
k=0 are complex ???i
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?(At/2M)
and the solution is y =e
????????
Ak
???M2M
2
2
?
c1cos
kAkA
??t
csin ?????t?????????????????M2MM2M???
2
2
This solution is oscillatory, representing undercritical damping.
All these results generalize to homogeneous linear differential
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equations with constant coefficients of order higher than 2. Theseequations
(especially of order 2) have been much used because of theease of solution.
Oscillations, electric circuits, diffusion processes, andheat-flow problems are a
few
examples
for
which
such
equahttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33ctions areuseful.
Second-Order Equations: Dependent Variable MissingSuchan equation is of the form
dyd2yFx, , =0??
dxdx2
It can be reduced to a first-order equation by substituting p =dy/dxand dp/dx
=d2y/dx2.
Second-Order Equations: Independent Variable MissingSuch an equation is of the form
dyd2yFy, ?, ?=0
dxdx2
??
??
??Qe
∫Pdx
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dx
C
?
is dissolved. Six gallons of brine containing 4 lb of salt run into the tank perminute.
If mixing is perfect and the output rate is 4 gal/min, what is the amount Aof salt
in the tank at time t?The differential equation of Ais dA/dt [1/(100
general solution is A=2(100
solution is A=2(100
t)
C/(100
t) ? 104/(100
t)]A=4. Its
t). At t=0, ′A=100; so the particular
t).
ExampleA tank initially holds 200 gal of a salt solution in which 100 lb
dyd2ydp=p,=p ??http://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33c?dxdx2d
y
The result is a first-order equation in p,
dp
Fy, p, p ?=0
dySet
??
ExampleThe capillary curve for one vertical plate is given by
d2y4y
=?1
?dx2c2
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Its solution by this technique is
c2
????c2???y2c2???h?x ??0=?
2
where c, h0are physical constants.
????
dy?dx
23/2
ORDINARY DIFFERENTIAL EQUATIONS OF HIGHER ORDER
The higher-order differential equations, especially those of order 2,are of great
importance because of physical situations describable bythem.
Equation y(n)= f(x)Such a differential equation can be solved bynintegrations. The
solution will contain narbitrary constants.
Linear Differential Equations with Constant Coefficientsand Right-Hand Member Zero
(Homogeneous)The solution ofy″ ay′ by=0 depends upon the nature of the roots of
the
chahttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33crac-teristic
equation m2 am
b=0 obtained by substituting the trialsolution y =emxin the equation.
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Distinct Real RootsIf the roots of the characteristic equation
?cosh
?1
cc??cosh?1 ?yh0?
ExampleThe equation governing chemical reaction in a porous catalyst
in plane geometry of thickness Lis
d2cdcD ?=k f(c),?(0) =0,c(L) =c0
2dxdx
where Dis a diffusion coefficient, kis a reaction rate parameter, cis the
con-centration, k f(c) is the rate of reaction, and c0is the concentration at the
bound-ary. Making the substitution gives
dpkp ?=?f(c)dcD
DIFFERENTIAL EQUATIONS
3-37
Example(1 ?x2) ????=x. The homogeneous equation2
dx
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xdx
d2y1dy
???= 0(1?x2) ?
dx2xdx
dpdx?=?px(1?x2)
when we set dy/dx =p.Upon integrating
twice,
y
=c1??x2???1
c2is
the
http://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33csolution.
that the par-ticular solution has the form y =
2
u??x???1 v.The equations for uand vbecomereduces to
x2???1u′=du/dx=??
dv
v′=?=1 ?x2
dx
so that
homogeneous
Now
assume
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1
u=?[x??x2???1?ln (x ??x2???1)]andv =x ?x3/3.
2
The complete solution is
xx31
x2???1 c2 ???????x2???1ln (x ??x2???1).y =c1??
262
d2y1dy
Perturbation MethodsIf the ordinary differential equation hasa parameter that is
small and is not multiplying the highest derivative,perturbation methods can give
solutions for small values of the param-eter.
ExampleConsider the differential equation for reaction and diffusion ina catalyst;
the reaction is second order: c″=ac2, c′(0) =0, c(1) =1. The solutionis expanded
in the following Taylor series in a.
c(x, a) =c0(x)
The
ac1(x)
goal
a2c2(x)
…
is
to
find
equahttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33ctions governing the
functions ci(x) and solve them. Sub-stitution into the equations gives the following
equations:
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2…=a[c0(x)
ac1(x)
a2c2(x)
…]2c0″(x)
a c″1(x)
ac″2(x)
2
c′′(0)
c0(1)
…=00(0)
ac1(1)
ac′1(0)
a2c2(1)
ac2
…=1
Like terms in powers of aare collected to form the individual problems.
c″0=0,c″1=c,
The solution proceeds in turn.
c0(x) =1,
(x2?1)c1(x) =?,
2
5?6x2 x4
c2(x) =??
12
20
c′0(0) =0,c′1(0) =0,
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c0(1) =1c1(1) =0c2(1) =0
c″2=2c0c1,c2′(0) =0,
SPECIAL DIFFERENTIAL EQUATIONS (SEE REF. 1)
Euler’s EquationThe linear equation xny(n) a1xn?1y(n?1) ???
an?1xy′ any =R(x) can be reduced to a linear equation with constantcoefficients by
the change of variable x =et.To solve the homogeneousequation substitute y =xrinto
it, cancel the pohttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33cwers of
x,which are thesame for all terms, and solve the resulting polynomial for r.In case
ofmultiple or complex roots there results the form y =xr(log x)rand y =xα[cos (βlog
x)
isin (βlog x)].
The roots ofr2?r?2 =0 are r=2, ?1. The general solution is y =Ax2 B/x.The equation
(ax
b)ny(n) a1(ax
substitution ax
b)n?1y(n?1) ??? any =R(x) can bereduced to the Euler form by the
b =z.It may be treated with-out change of variable, the homogeneous
equation having solutions of the formy=(ax
b)r.
ExampleSolve x2y″?2y=0. By setting y =xr, xr[r(r ?1) ?2] =0.
Bessel’s EquationThe linear equation x2(d2y/dx2)
(1 ?2α)x(dy/dx)
[β2γ2x2γ
(α2?p2γ2)]y=0 is the general Bessel equation.By series methods, not to be discussed
here, this equation can beshown to have the solution
y =AxαJp(βxγ)
BxαJ?p(βxγ)y =AxαJp(βxγ)
pnot an integer or zeropan integer
BxαYp(βxγ)
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3-38where
MATHEMATIChttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33cSx
Jp(x) =?
2x
J?p(x) =?
2
∞0
??
p
k=0∞
Α??
k!Γ(p k 1)
k=0
∞
(?1)k(x/2)2k(?1)(x/2)
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k
PARTIAL DIFFERENTIAL EQUATIONS
The analysis of situations involving two or more independent variablesfrequently
results in a partial differential equation.
pnot an integer
dimensional conduction of heat.
??Γ(n) =?x
–p
Α??
k!Γ(k 1?p)
edx
n>0
2k
ExampleThe equation ?T/?t =K(?2T/?x2) represents the unsteady one-
n?1?x
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is the gamma function. For pan integer
xp∞(?1)k(x/2)2k
Jp(x) =?Α??
2k=0k!(p k)!
??
ExampleThe equation for the unsteady transverse motion of a uniformbeam clamped at
the ends is
?4yρ?2y
??=0??x4EI?t2ExampleThe
expansion
of
ahttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33c gas behind a piston is
characterized by thesimultaneous equations
?u?uc2?ρ?ρ?ρ?u? u ? ??=0and? u ? ρ ?=0?t?xρ?x?t?x?xExampleThe heating of a
diathermanous solid is characterized by the
equation α(?2θ/?x2)
βe?γz=?θ/?t.
The partial differential equation ?2f/?x?y=0 can be solved by twointegrations
yielding the solution f =g(x)
h(y), where g(x) and h(y)are arbitrary differentiable
functions. This result is an example of thefact that the general solution of partial
differential equations involvesarbitrary functions in contrast to the solution of
ordinary differentialequations, which involve only arbitrary constants. A number of
meth-ods are available for finding the general solution of a partial differen-tial
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equation. In most applications of partial differential equations thegeneral solution
is of limited use. In such applications the solution ofa partial differential equation
must
satisfy
both
the
equahttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33ction and cer-tain
auxiliary conditions called initialand/or boundaryconditions,which are dictated by
the problem. Examples of these include those inwhich the wall temperature is a fixed
constant T(x0) =T0, there is nodiffusion across a nonpermeable wall, and the like.
In ordinary differ-ential equations these auxiliary conditions allow definite
numbers tobe assigned to the constants of integration. In partial differential
equa-tions the boundary conditions demand that the arbitrary functionsresulting from
integration assume specific forms. Except for a fewcases (some first-order equations,
D’Alembert’s solution of the waveequation, and others) a procedure which first
determines the arbitraryfunctions and then specializes them to fit the boundary
conditions isusually not feasible. A more fruitful attack is to determine directly
aset of particular solutions and then combine them so that the bound-ary conditions
are satisfied. The http://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33conly
area in which much analysis hasbeen accomplished is for linear homogeneous partial
differentialequations. Such equations have the property that if f1, f2,...,
∞
fn,...are individually solutions, then the function f=Αi=1fiis also asolution,
provided the series converges and is differentiable up to theorder (termwise) of the
equation.
Partial Differential Equations of Second and Higher OrderMany of the applications
to scientific problems fall naturally into par-tial differential equations of second
order, although there are impor-tant exceptions in elasticity, vibration theory, and
elsewhere.A second-order differential equation can be written as
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?2u?2u?2u
a ? b
c =f???x2?x?y?y2
where a, b, c,and fdepend upon x, y, u,?u/?x,and ?u/?y.This equa-tion is hyperbolic,
parabolic, or elliptic, depending on whether the dis-criminant b2?4acis >0, =0, or
depend on the soluthttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33cion,
the type of equation can be different at dif-ferent xand ylocations. If the equation
is hyperbolic, discontinuitiescan be propagated. See Refs. 11, 79, 105, 159, and 192.
Phenomena of propagationsuch as vibrations are characterized byequations of
“hyperbolic” type which are essentially different in theirproperties from other
classes such as those which describe equilib-rium (elliptic) or unsteady diffusion
and heat transfer (parabolic). Pro-totypes are as follows:
EllipticLaplace’s
equation
?2u/?x2
?2u/?y2=0
and
Poisson’sequation ?2u/?x2 ?2u/?y2=g(x, y) do not contain the variable time
(Bessel function of the first kind of order p)
[Jp(x) cos (pπ)?J?p(x)]
Yp(x) =???
sin (pπ)
(replace right-hand side by limiting value if Pis an integer or zero).The series
converge for all x.Much of the importance of Bessel’sequation and Bessel functions
lies
in
the
fact
that
the
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http://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33csolutions
ofnumerous
linear differential equations can be expressed in terms ofthem.
(9x3?63?4)y=0. Thus α=a, γ=3?2; β=2, p=8?3. The solution is (since p??integer)
y=Ax1/2J8/3(2x3/2)
Bx1/2J–8/3(2x3/2). Tables are available for the evaluationof
many of these functions.
Exampled2y/dx2 [9x?(63/4x2)]y=0. In general form this is x2(d2y/dx2)
ExampleThe heat flow through a wedge-shaped fin is characterized bythe equation
x2(d2y/dx2)
x(dy/dx) ?a2xy=0, where y =T ?Tair, αis a combi-nation of physical
constants, and x=distance from fin end. By comparing thiswith the standard equation,
there results α=0, p=0, γ=a, β2=?4a2or β=2ai.The solution is y =AJ0(2ai??x)
BY0(2ai??x).Legendre’s EquationThe Legendre equation (1 ?x2)y″?2xy′ n(n 1)y=0,
n≥0, has the solution y =Aun(x)
Bvn(x) for nnot an integer where
n(n 1)n(n?2)(n 1)(n 3)
un(x) =1 ??x2 ???x4
2!4!
n(n?2)(n?4)(n
1)(n
http://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33c5)
? ????x6 ???
6!
(n?1)(n 2)(n?1)(n?3)(n 2)(n 4)
3)(n
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vn(x) =x???x3 ???x5?????
3!5!
If nis an even integer or zero, unis a polynomial in x.If nis an oddinteger, then
vnis a polynomial. The interval of convergence for theseries is ?1
un(x)vn(x)
Pn(x) =?(neven or zero),Pn=?(nodd)
un(1)vn(1)
The polynomials Pnare the so-called Legendre polynomials, P0(x) =1,P1(x) =x, P2(x)
=a(3x2?1), P3(x) =a(5x3?3x), ....
Laguerre’s EquationThe Laguerre equation x(d2y/dx2)
(c ?x)(dy/dx) ?ay=0 is
satisfied by the confluent hypergeometric function.See Refs. 1 and 173.
Hermite’s EquationThe Hermite equation y″?2xy′ 2ny=0is satisfied by the Hermite
polynomial of degree n, y =AHn(x) if nis apositive integer or zero. H0(x) =1, H1(x)
=2x, H2(x) =4x2?2, H3(x) =8x3?12x, H4(x) =16x4?48x2 12, Hr 1(x) =2xHr(x) ?2rHr?1(x).
tion.
Exampley″http://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33c?2xy′ 6y=0.
Here n=3; so y =AH3=A(8x3?12x) is a solu-
Chebyshev’s EquationThe equation (1 ?x2)y″?xy′ n2y=0for na positive integer or
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zero is satisfied by the nth Chebyshev poly-nomial y =ATn(x). T0(x) =1, T1(x) =x,
T2(x) =2x2?1, T3(x) =4x3?3x,T4(x) =8x4?8x2 1; Tr 1(x) =2xTr(x) ?Tr?1(x).
2xT5(x) ?T4(x) =2x(2xT4?T3) ?T4=32x6?48x4 18x2?1. Further details onthese special
equations and others can be found in the literature.
Example(1 ?x2)y″?xy′ 36y=0. Here n=6. A solution is y =T6(x) =
DIFFERENTIAL EQUATIONS
explicitly
and
consequently
represent
equilibrium
configurations.Laplace’s
equation is satisfied by static electric or magnetic potentialat points free from
electric charges or magnetic poles. Other impor-tant functions satisfying Laplace’s
equation are the velocity potentialof the irrotational motion of an incompressible
fluid,
used
in
hydrody-namics;
the
steady
temperature
at
points
http://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33cin a homogeneous solid,
andthe steady state of diffusion through a homogeneous body. The gravi-tational
potential
Vat
points
occupied
by
mass
of
density
dsatisfiesPoisson’s
equation ?2V/?x2 ?2V/?y2 ?2V/?z2=?4πd.
ParabolicTheheatequation?T/?t=?2T/?x2 ?2T/?y2repre-sentsnonequilibriumorunsteady
statesofheatconductionanddiffu-sion.
HyperbolicThe wave equation ?2u/?t2=c2(?2u/?x2 ?2u/?y2)represents wave propagation
of many varied types.Quasilinear first-order differential equations are like
?u?ua ? b ?=f?x?y
where a, b,and fdepend on x, y,and u,with a2 b2≠0. This equation
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can be solved using the method of characteristics, which writes thesolution in terms
of a parameter s,which defines a path for the char-acteristic.
dxdydu?=a,?=b,?=fdsdsdsThese equations are integrated from some initial conditions.
For a
specified
value
of
s,the
value
of
xand
yshows
the
location
theshttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33colution
is
where
u.The
equation is semilinear if aand bdepend just on xand y(and not u), and the equation
is linear if a, b,and fall depend onxand y,but not u.Such equations give rise to shock
propagation, andconditions have been derived to deduce the presence of shocks,
Ref.245. For further information, see Refs. 79, 159, 192, and 245.
An example of a linear hyperbolic equation is the advection equa-tion for flow of
contaminants when the xand yvelocity componentsare uand v,respectively.
?c?c?c? u ? v ?=0?t?x?y
The equations for flow and adsorption in a packed bed or chromatog-raphy column give
a quasilinear equation.
?c?cdf?cφ ? φu ? (1 ?φ) ??=0?t?xdc?t
Here n =f(c) is the relation between concentration on the adsorbent
and fluid concentration.
The solution of problems involving partial differential equationsoften revolves
about
an
attempt
to
reduce
the
partial
differentihttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33calequation
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to one or more ordinary differential equations. The solutionsof the ordinary
differential equations are then combined (if possible)so that the boundary conditions
as well as the original partial differen-tial equation are simultaneously satisfied.
Three of these techniquesare illustrated.
Similarity VariablesThe physical meaning of the term “similar-ity” relates to
internal similitude, or self-similitude. Thus, similar solu-tions in boundary-layer
flow over a horizontal flat plate are those forwhich the horizontal component of
velocity uhas the property thattwo velocity profiles located at different coordinates
xdiffer only by ascale factor. The mathematical interpretation of the term similarity
isa transformation of variables carried out so that a reduction in thenumber of
independent variables is achieved. There are essentiallytwo methods for finding
similarity
variables,
“separation
of
variables”(not
the
chttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33classical concept) and
the use of “continuous transformationgroups.” The basic theory is available in Ames
(see the references).
tions θ=0 at x=0, y>0; θ=0 at y=∞, x>0; θ=1 at y=0, x>0 represents
thenondimensional temperature θof a fluid moving past an infinitely wide flat
plateimmersed in the fluid. Turbulent transfer is neglected, as is molecular
transportexcept in the ydirection. It is now assumed that the equation and the
boundaryconditions can be satisfied by a solution of the form θ=f(y/xn) =f(u), where
θ=
3-39
0 at u=∞and θ=1 at u=0. The purpose here is to replace the independentvariables
xand yby the single variable uwhen it is hoped that a value of nexistswhich will allow
xand yto be completely eliminated in the equation. In this casesince u =y/xn,there
results after some calculation ?θ/?x=?(nu/x)(dθ/du),?2θ/?y2=(1/x2n)(d2θ/du2),
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and
when
these
are
substituted
in
the
equation,
?(1/x)nu(dθ/du)
=http://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33c(1/x3n)(A/u)(d2θ/du2
). For this to be a function of uonly,choose n=s. There results (d2θ/du2)
(u2/3A)(dθ/du) =0. Two integrations anduse of the boundary conditions for this
ordinary differential equation give thesolution
θ=
?
∞
u
exp (?u3/9A) du?
?
∞
exp (?u3/9A) du
Group MethodThe type of transformation can be deducedusing group theory. For a
complete exposition, see Refs. 9, 12, and145; a shortened version is in Ref. 106.
Basically, a similarity transfor-mation should be considered when one of the
independent variableshas no physical scale (perhaps it goes to infinity). The
boundary con-ditions must also simplify (and combine) since each transformationleads
to a differential equation with one fewer independent variable.ExampleA similarity
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variable is found for the problem
?c??c
?=?D(c)
?,c(0,t)
=1,c(∞,t)
=0,c(x,0)
=0?http://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33ct?x?x
Note that the length dimension goes to infinity, so that there is no length scalein
the problem statement; this is a clue to try a similarity transformation.
Thetransformation examined here is
β
xc=aγc?=ax,?
With this substitution, the equation becomes
??
t?=aαt,
??c??c?aα?γ?=a2β?γ?D(a?γ c?) ???t??x?x?
Group theory says a system is conformally invariant if it has the same form in thenew
variables; here, that is
γ=0,
The invariants are
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xη=?,
tδ
and the solution is
c(x, t) =f(η)tγ/α
We can take γ=0 and δ=β/α=a. Note that the boundary conditions combinebecause
the point x=∞and t=0 give the same value of ηand the conditions oncat x=∞and t=0
are the same. We thus make the transformation
x
η=?,c(x, t) =f(η)
4?D???0tThe
use
of
the
4
and
D0makes
the
analysis
below
simplerhttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33c. The result is
ddfdf
?D(c) ? 2η?=0,dηdηdη
β
δ=?
α
α?γ=2β?γ,
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or α=2β
??
??
f(0) =1,f(∞) =0
Thus, we solve a two-point boundary value problem instead of a partial differ-ential
equation. When the diffusivity is constant, the solution is the error func-tion, a
tabulated function.
c(x,t) =1 ?erf η=erfc ηerf η=
?
η
e?ξdξ?
2
?
∞
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e?ξdξ
2
ExampleThe equation ?θ/?x=(A/y)(?2θ/?y2) with the boundary condi-
Separation of VariablesThis is a powerful, well-utilizedmethod which is applicable
in certain circumstances. It consists ofassuming that the solution for a partial
differential equation has theform U =f(x)g(y). If it is then possible to obtain an
ordinary differen-tial equation on one side of the equation depending only
http://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33con xand onthe other side
only on y,the partial differential equation is said to beseparable in the variables
x, y.If this is the case, one side of the equa-tion is a function of xalone and the
other of yalone. The two can beequal only if each is a constant, say λ. Thus the
problem
has
againbeen
reduced
equations.ExampleLaplace’s
to
equation
the
solution
?2V/?x2
of
?2V/?y2=0
ordinary
plus
differential
the
boundary
con-ditions V(0, y) =0, V(l, y) =0, V(x,∞) =0, V(x,0) =f(x) represents the
steady-state potential in a thin plate (in zdirection) of infinite extent in the
ydirection
3-40MATHEMATICS
Since the initial condition must be satisfied, we use an infinite series of
thesefunctions.
∞
sin nπx222
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u=ΑAn?e?nπDt/L
Ln=1
At t=0, we satisfy the initial condition.
∞
sin nπx
x?1 =ΑAn?
Ln=1
This
is
done
by
multiplying
the
equation
bhttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33cy
sin mπx?L
and integrating over x: 0→L.(This is the same as minimizing the mean-squareerror
of the initial condition.) This gives
L
AmL
?=(x?1) sin mπx dx
02
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which completes the solution.
and of width lin the xdirection. A potential f(x) is impressed (at y=0) from x=0 to
x=1, and the sides are grounded. To obtain a solution of this boundary-value problem
assume V(x, y) =f(x)g(y). Substitution in the differential equationyields f″(x)g(y)
f(x)g″(y)
g″(y)
=0,
or
?λ2g(y)
g″(y)/g(y)
=0
andf″(x)
=?f″(x)/f(x)
λ2f(x)
=λ2(say).
=0.
The
This
systembecomes
solutions
of
these
ordinarydifferential equations are respectively g(y) =Aeλy Be?λy, f(x) =Csin λx
Dcos λx.Then f(x)g(y) =(Aeλy Be?λy) (Csin λx
Dcos λx). Now V(0, y) =0 sothat
f(0)g(y) =(Aeλy Be?λy) D?0 for all y.Hence D=0. The solution then hasthe form sin
λx(Aeλy
Be?λy)
where
the
multiplicative
constant
Chas
bhttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33ceen elim-inated. Since
V(l, y) =0, sin λl(Aeλy Be?λy) ?0. Clearly the bracketed functionof yis not zero,
for the solution would then be the identically zero solution.Hence sin λl=0 or
λn=nπ/l, n=1, 2,...where λn=nth eigenvalue.
The solution now has the form sin (nπx/l)(Aenπy/l Be?nπy/l). Since V(x,∞)
=0,Amust be taken to be zero because eybecomes arbitrarily large as y→∞. The
solution
then
reads
Bnsin
(nπx/l)e?nπy/l,where
Bnis
the
multiplicative
con-∞?nπy/l
stant. The differential equation is linear and homogeneous so that Αn=1Bnesin
(nπx/l) is also a solution. Satisfaction of the last boundary condition is en-sured
by taking
2Bn=?
l
??
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??
?
?f(x) sin (nπx/l) dx=Fourier sine coefficients of f(x)
l0
Further, convergence and differentiability of this series are established
quiteeasily. Thus the solution is
∞
nπx
V(xhttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33c,
y)
=ΑBne?nπy/lsin ?
ln=1
ExampleThe diffusion problem
?c??c
?=?D(c) ?,c(0, t) =1,c(∞, t) =0,c(x, 0) =0?t?x?x
can be solved by separation of variables. First transform the problem so that
theboundary conditions are homogeneous (having zeroes on the right-hand side).Let
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c(x, t) =1 ?x
u(x, t)
Then u(x, t) satisfies
?2u?u
,?=D ?
?t?x2
u(x, 0) =x?1,
u(0, t) =0,
u(1, t) =0
??
Integral-Transform MethodA number of integral transforms
are used in the solution of differential equations. Only one, the Laplacetransform,
will be discussed here [for others, see “Integral Transforms(Operational
Methods)”]. The one-sided Laplace transform indicated
∞
by L[f(t)] is defined by the equation L[f(t)] =∫0f(t)e?stdt.It hasnumerous important
properties.
The
ones
of
interest
here
are
L[f′(t)http://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33c]
=sL[f(t)]
?f(0);
L[f″(t)]
=s2L[f(t)]
?sf(0)
?f′(0);
L[f(n)(t)]
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=snL[f(t)] ?sn?1f(0) ?sn?2f′(0) ?????f(n?1)(0) for ordinary derivatives.For
partial derivatives an indication of which variable is being trans-formed avoids
confusion. Thus, if
?y
y =y(x, t),Lt?=sL[y(x, t)] ?y(x,0)
?t
??
Assume a solution of the form u(x, t) =X(x) T(t), which gives
d2XdT
X ?=D T ?dtdx2
Since both sides are constant, this gives the following ordinary differential
equa-tions to solve.
1d2X1dT
=?λ??=?λ,??
D TdtXdx2The solution of these is
T =A e?λDt,
?ydLt[y(x, t)]
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Lt?=???xdx
since L[y(x, t)] is “really” only a function of x.Otherwise the resultsare similar.
These facts coupled with the linearity of the transform,i.e., L[af(t)
=aL[f(t)]
bg(t)]
bL[g(t)], make it a useful device insolving some linear differential
equathttp://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33cions.
Its
use
reduces the solutionof ordinary differential equations to the solution of algebraic
equa-tions for L[y]. The solution of partial differential equations is reducedto the
solution of ordinary differential equations. In both situationsthe inverse transform
must be obtained either from tables, of whichthere are several, or by use of complex
inversion methods.whereas
ExampleThe equation ?c/?t =D(?2c/?x2) represents the diffusion in asemi-infinite
medium, x≥0. Under the boundary conditions c(0, t) =c0, c(x,0) =0 find a solution
of the diffusion equation. By taking the Laplace transform ofboth sides with respect
to t,
∞
?2c1∞?st ?ce?st?dt=e?dt?0?x2D0?t
??
?x
Esin ?λ?xX =Bcos ?λ
The combined solution for u(x,t) is
?x
Esin ?λ?x) e?λDtu =A(Bcos ?λ
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Apply the boundary condition that u(0,t) =0 to give B=0. Then the solution is
?
?
://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33cr?x)e?λDtu =A(sin ?λ
where the multiplicative constant Ehas been eliminated. Apply the boundary
condition at x =L.
or
?L)e?λDt0 =A(sin ?λ
This can be satisfied by choosing A=0, which gives no solution. However, it can
also be satisfied by choosing λsuch that
?L=0,sin ?λ?L =nπ?λ
d2FsF
=(1/D)sF ?c(x,0) =??2dxD
where F(x, s) =Lt[c(x, t)]. Hence
d2Fs
??F=0?dx2D
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The other boundary condition transforms into F(0, s) =c0/s.Finally the solutionof
the ordinary differential equation for Fsubject to F(0, s) =c0/sand Fremains
/Dx?.Reference to a table shows that the func-finite as x→∞is F(x, s) =(c0/s)e??s
tion having this as its Laplace transform is
??
n2π2
Thusλ=?
L2
The combined solution can now be written as
sin nπx222
u =A?e?nπDt/L
http://www.mianfeiwendang.com/doc/0f819afb0aaf71ff74e5e33cL
2
c(x, t) =c0 1 ??
??π
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?
?
?t?x/2?D
2
e?udu
?
??
Matched-Asymptotic ExpansionsSometimes the coefficient in
front of the highest derivative is a small number. Special perturbationtechniques
can then be used, provided the proper scaling laws arefound. See Refs. 32, 170, and
180.
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