Mechanical Engineers Handbook 2011 Part 13 pdf

Appendix DifferentialEquations and Systems of Differential Equations 1.1 Ordinary Differential Equations: Introduction 716 1.2 Integrable Types of Equations 726 1.3 On the Existence, Uni

contour Therefore, we will assume the clockwise traversal of a contour to bepositive and the area enclosed within the contour to be on the right.

We consider the transfer function

where A is a constant and z1 de®nes a zero of Y …s†,

s ˆ z1‡ Rejj;where R, j are variables,

s ÿ z1ˆ Rejj:

If the contour C in the s-plane encircles the zero z1, this is equivalent to arotation of the (s±z1) vector by 2p in the case when the correspondingcontour D in the Y …s†-plane encircles the origin in a clockwise direction Ifthe contour C does not encircle the zero z1, the angle of s±z1 is zero whenthe traversal is in a clockwise direction along the contour and the contour Ddoes not encircle the origin (Fig A.3.2)

Now, we consider the transfer function

where p1 de®nes a pole of Y …s†

s ˆ p1‡ Rejj;Then

We assume that the contour C encircles the pole p1 Considering thevectors s as shown for a speci®c contour C (Fig A.3.3) we can determine theangles as s traverses the contour Clearly, as the traversal is in a clockwisedirection along the contour, the traversal of D in the Y …s†-plane is in theopposite direction When s traverses along C a full rotation of 2p rad, for the

Y …s†-plane we will have an angle ÿ2p rad (Fig A.3.4)

We can generalize these results If a contour C in the s-plane encircles Zzeros and P poles of Y …s† as the traversal is in a clockwise direction along thecontour, the corresponding contour D in the Y …s†-plane encircles the origin

of the Y …s†-plane

N ˆ Z ÿ Ptimes in a clockwise direction The resultant angle of Y …s† will be

One of the most interesting mapping contours in the s-plane is from theNyquist contour The contour passes along the jo-axis from ÿj1 to ‡j1and is completed by a semicircular path of radius r (Fig A.3.7) We choosethe transfer function Y1…s† as

From (A3.4) we have

Y1…s† ˆ k

When s traverses the semicircle C with r ! 1 from o ˆ ‡1 to

o ˆ ÿ1, the vector Y1…s† with the magnitude k=tR has an angle changefrom ÿp=2 to ‡p=2

When s traverses the positive imaginary axis, s ˆ jo …0 < o < 1†, themapping is represented by

Y …s† ˆ1 ‡ jotk ˆ k1 ‡ o1 ÿ jot2t2; …A:3:8†which represents a semicircle with diameter k (Fig A.3.7b, the solid line).The portion from o ˆ ÿ1 to o ˆ 0ÿ is mapped by the function

Y1…s†jsˆÿjoˆ Y1…ÿjo† ˆ k 1 ‡ jot

1 ‡ o2t2: …A:3:9†Thus, we obtain the complex conjugate of Y1…jo†, and the plot for theportion of the polar plot from o ˆ ÿ1 to o ˆ 0ÿis symmetrical to the polarplot from o ˆ ‡1 to o ˆ 0‡ (Fig A.3.7b)

A.4 The Signal Flow Diagram

A signal ¯ow diagram is a representation of the relationship between thesystem variables The signal ¯ow diagram consists of unidirectional opera-tional elements that are connected by the unidirectional path segments Theoperational elements are integration, multiplication by a constant, multi-plication of two variables, summation of several variables, etc (Fig A.4.1)

Figure A.4.1 Signal ¯ow diagram elements

These functions are often suf®cient to develop a simulation model of asystem.

For example, we consider a dynamic model described by the differentialequation

x ‡ a1_x ‡ a2x ˆ u: …A:4:1†

Using the notations

x ˆ x1_x ˆ x2;

Eq (A.4.1) can be rewritten as

_x1ˆ x2_x2ˆ ÿa1x2ÿ a2x1‡ u: …A:4:2†

The signal ¯ow diagram of (A.4.1) is presented in Fig A.4.2 The diagramhas two representations, one for time-domain variables and one for theLaplace transform representation

1 R J Schilling, Fundamentals of RoboticsÐAnalysis and Control PrenticeHall, Englewood Cliffs, NJ, 1990

2 P G Darzin, Nonlinear Systems Cambridge University Press, 1992

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4 S CaÏlin, Automatic Regulators Ed Did-Ped, Bucharest, 1967

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7 T Yoshikawa, Foundation of Robotics M.I.T Press, Cambridge, MA, 1990

8 G I Thaler, Automatic Control Systems West Publishing, St Paul, MN, 1990

9 R G Dorf, Modern Control Systems 6th ed Addison-Wesley, Reading, MA,1992

10 B W Niebel, Modern Manufacturing Process Engineering McGraw-Hill,New York, 1989

11 S C Jacobsen, Control strategies for tendon driven manipulators IEEEControl Systems, Vol 10, Feb., 23±28 (1990)

12 I I E Slotine and Li Weiping, Applied Nonlinear Control Prentice-HallInternational, New York, 1991

13 M Asada and I I E Slotine, Robot Analysis and Control Wiley-Interscience,New York, 1986

14 H BuÈhler, ReÂglage par mode de glissement Presses PolytechniquesRomandes, Lausanne, 1986

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16 R C Rosenberg and D C Karnopp, Introduction to Physical System Design.McGraw-Hill, New York, 1986

17 R I Smith and R C Dorf, Circuits, Devices and Systems, 5th ed John Wiley &Sons, New York, 1991

18 W L Brogan, Modern Control Theory Prentice Hall, Englewood Cliffs, NJ,1991

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20 C L Phillps and R D Harbor, Feedback Control Systems Prentice Hall,Springer Verlag, New York, 1988

21 R L Wells, Control of a ¯exible robot arm IEEE Control Systems, Vol 10, Jan.,9±15 (1990)

Appendix Differential

Equations and Systems of

Differential Equations

1.1 Ordinary Differential Equations: Introduction 716

1.2 Integrable Types of Equations 726

1.3 On the Existence, Uniqueness, Continuous Dependence on a Parameter, and Differentiability

of Solutions of Differential Equations 766

1.4 Linear Differential Equations 774

2 Systems of Differential Equations 816

2.1 Fundamentals 816

2.2 Integrating a System of Differential Equations by the Method of Elimination 819

2.3 Finding Integrable Combinations 823

2.4 Systems of Linear Differential Equations 825

2.5 Systems of Linear Differential Equations with Constant Coef®cients 835

References 845

715

1 Differential Equations

1.1 Ordinary Differential Equations: Introduction

1.1.1 BASIC CONCEPTS AND DEFINITIONS

Operatorial Equation

Let X , Y be arbitrary sets and f : X ! Y a function de®ned on X with values

in Y If y02 Y is given, and x 2 X must be found so that

then it is said that an operatorial equation must be solved A solution of

Eq (1.1) is any element x 2 X that satis®es Eq (1.1) The sets X , Y can havedifferent algebraical and topological structures: linear spaces, metricalspaces, etc If f is a linear function, that is, f …ax1‡ bx2† ˆ af …x1† ‡ bf …x2†,and if X and Y are linear spaces, then Eq (1.1) is called a linear equation If

Eq (1.1) is a linear equation and y0ˆ yY (the null element of space Y ), then

Eq (1.1) is called a linear homogeneous equation

Differential Equation

An equation of the form (1.1) for which X and Y are sets of functions iscalled a functional equation A functional equation in which is implied anunknown function and its derivatives of some order is called a differentialequation The maximum derivation order of the unknown function is calledthe order of the equation When the unknown function depends on a singleindependent variable, the equation is termed an ordinary differentialequation (or, more brie¯y; a differential equation) If the unknown functiondepends on more independent variables, the corresponding equation iscalled a partial differential equation The general form of a differentialequation of order n is

F …t; x; x0; x00; ; x…n†† ˆ 0; …1:2†where t is the independent variable, x ˆ x…t† is the unknown function, and

F is a function de®ned on a domain D  Rn‡2(R is the set of real numbers)

It is called a solution of Eq (1.2) on the interval I ˆ …a; b†  R, a function

j ˆ j…t† of Cn…I † class [i.e., j…t† has continuous derivatives until n-order],which has the following properties:

1.…t; j…t†; j0…t†; ; j…n†…t†† 2 D; 8t 2 I

2.F…t; j…t†; j0…t†; ; j…n†…t†† ˆ 0; 8t 2 I

If the function F can be explicated with the last argument, it then yields

x…n†ˆ f …t; x; x0; ; x…nÿ1††; …1:3†which is called the normal form of the n-order equation

on the independent variable and on n arbitrary independent constants

j ˆ j…t; c1; c2; ; cn†, and which satis®es the conditions

1.j…t; c1; c2; ; cn† is a solution for Eq (1.2) on an interval Ic

2.For any initial conditions (1.4), there could be determined the values

A solution obtained from the general solution for particular constants

c1; c2; ; cn is called a particular solution A singular solution is a solutionthat cannot be obtained from the general solution

EXAMPLE 1.1 Consider the following equation:

hence two equations of normal form Let us consider the ®rst equation This

is of the form of Eq (1.3) with n ˆ 1, the right-hand side function being

Ik ˆ …ÿ…p=2† ‡ 2kp; …p=2† ‡ 2kp†, k 2 Z (Z is the set of integer numbers)

Then, on each of these intervals the function j…t† ˆ sin t is the solution forequation x0ˆp1 ÿ x2

Now, consider the functions family j…t; c† ˆsin…t ‡ c†, c 2 R Let us set the interval Icˆ …ÿ…p=2 ÿ c; …p=2† ‡ c† For

t 2 Ic, t ‡ c 2 …ÿ…p=2†; …p=2†† and j…t; c† ˆ sin…t ‡ c† is the solution on Icfor the equation x0ˆp1 ÿ x2

If …t0; x0† 2 G ˆ R  …ÿ1; 1† settled from thecondition j…t0; c† ˆ x0, then sin…t0‡ c† ˆ x0 and the value of c obtainedfrom this condition and denoted by c0 is c0ˆ arcsin x0ÿ t0 The function

j…t; c0† ˆ sin…t ‡ c0† is a solution for the equation and satis®es the initialcondition, so it is the general solution m

Remark 1.1The constant functions j1…t†  1 and j2…t†  ÿ1 are solutions on R for theequation x0ˆp1 ÿ x2

, but they cannot be obtained from the generalsolution for any particular constant c, and hence there are singularsolutions m

1.1.2 SYSTEMS OF DIFFERENTIAL EQUATIONS

A system of differential equations is constituted by two or more differentialequations A system with n differential equations of the ®rst order in normalform is a system of the form

fi: I  D  R  Rn A solution of the system of equations (1.7) on the interval

J  I is an assembly of n functions …j1…t†; j2…t†; ; jn…t†† derivable on Jand that, when substituted with the unknowns x1; x2; ; xn, satis®es Eqs.(1.7) in any t 2 J

are called initial conditions for the system of equations (1.7)

Vectorial Writing of the System of Equations (1.7)

If the column vector X …t† is written as

is a solution for the system of equations (1.15), then the function u…t† ˆ x1…t†

is a solution for Eq (1.12) Conversely, if u…t† is a solution of Eq (1.12), then

®eld direction The Cauchy's problem for Eq (1.16) with the condition

y…x0† ˆ y0 means to ®nd the integral curve that passes through the point

M0…x0; y0† The isocline is the locus of points M …x; y† in the de®nition domain

of function f for which the tangents to the integral curves have the samedirection The equation

The Differential Equation of a Curve Family

Let us consider the curve family

where a is a real parameter It can be considered that Eq (1.18) representsthe complete integral of the differential equation that must be found.Derivation of Eq (1.18) with respect to x yields

In the case when the curves family depends on n parameters

F…x; y; a1; a2; ; an† ˆ 0;

the differential equation of this family is obtained by eliminating the meters a1; a2; ; anfrom the above equation and the other n ÿ 1 equationsobtained by derivation with respect to x, successively until the nth-order

para-The differential equation of a curve family has as a general solution the givenfamily itself

j Find the differential equation of the family Eq (1.23)

j In that differential equation, substitute y0 by …y0ÿ k†=…1 ‡ ky0†, if

a 6ˆ p=2 or, ÿ1=y0when a ˆ p=2

j Find the general solution of the differential equation obtained; this isthe isogonal family of Eq (1.23)

EXAMPLE 1.2 Consider the differential equation y0ˆ ÿx=2y

(a) Make an approximation construction of the integral curves, using theisoclines

(b) Find the general solution (complete integral)

(c) Find the orthogonal family of the complete integral mSolution

(a) The differential equation is of the form y0ˆ f …x; y†, with f …x; y† ˆÿx=2y The de®nition domain of function f is D ˆ R  …ÿ1; 0† [ R

…0; 1† The equation of isoclines (1.17) gives ÿx=2y ˆ k; hence

y ˆ ÿx=2k; k 2 Rnf0g The corresponding isoclines to the values of k

(b) The equation can be written as

x ‡ 2yy0ˆ 0or

xy0ˆ 2y; x dy ˆ 2y dx; dyy ˆ 2dxx ; d…ln jyj† ˆ 2 d…ln jxj†:This yields ln jyj ˆ 2 ln jxj ‡ c The arbitrary constant c could be chosen inthe form c ˆ ln b, b > 0 The isogonal curve family is jyj ˆ bx2, whichrepresents a family of parabolas In Fig 1.2 are represented two isogonalfamilies

1.1.4 PHENOMENA INTERPRETED BY MEANS OF DIFFERENTIAL EQUATIONS

Differential equations make it possible to study some ®nite-determinist anddifferentiable phenomena Determinist phenomena are those processeswhose future evolution state is uniquely determined by the state of present

conditions Finite phenomena are those processes that need a ®nite number

of parameters for their correct description Differentiable phenomena arethose processes in which the functions used for their description arederivable (up to some order) Next, some examples of modeling by means

of differential equations are presented

Phenomenon of Growth (or Decay)

In the study of some phenomena of growth from economy, biology, etc [Forinstance, the growth (decay) of production, population of a race, materialquantity], there appear differential equations of the form

df …t†

where k…t† > 0 in the case of a growth phenomenon and k…t† < 0 when it isthe curve of a decaying phenomenon For example, if in a study of theevolution of a certain species, we denote by f …t† the number of individuals atthe moment t, by n and m the coef®cient of birth rate and death rate,respectively, then given the assumption that the population is isolated (i.e.,there is no immigration or emigration), the variation rate of the population

A Mathematical Model of Epidemics

Let us consider a population of n individuals and a disease spreading bydirect touch At a moment t, the population is composed of three categories:

j x…t† ˆ the number of individuals not infected

j y…t† ˆ the number of infected individuals that are not isolated (arefree)

j z…t† ˆ the number of infected individuals that are isolated (underobservation)

It is natural to presume that the infection rate ÿx0…t† is proportional to x
yand the infected individuals become isolated at a rate that is proportional totheir number, y This yields the system

The Dog Trajectory

A man walks on a line Oy with the uniform velocity v At the moment t ˆ 0

he is at point O, and he calls his dog, which at that moment is at point A atthe distance OA ˆ a The dog runs to the master with the uniform velocity

v1ˆ kv (k > 0), the velocity being always oriented to the master (Fig 1.3).Find the equation of the dog trajectory and the time at which it will reach itsmaster Discuss

The Basic Dynamical Equations

Consider that M is a material point of mass m that moves in R3 under theaction of a force F [which usually depends on time, on the position r …t† of

the point M , and on the motion velocity dr =dt Applying Newton's secondlaw of dynamics, we obtain

where f1, f2, f3 are the components of F

The Problem of the Second Cosmic Velocity

The goal of this problem is to ®nd the velocity v0of the vertical launching of abody as it escapes the in¯uence of the earth's gravitational attraction If weuse the law of universal attraction and Newton's second law, the equation ofmotion is

m r…t† ˆ ÿkrmM2…t†; …1:30†

where r …t† is the distance from the center of the earth to the center of thebody, m is the mass of the body, M is the mass of the earth, and k is theconstant of universal attraction

Equation of an Electrical Oscillatory Circuit

Let us consider an electrical circuit composed of an inductance L, a resistor R,and a capacitor C, having a voltage U The laws of electricity yield thedifferential equation

LI00…t† ‡ RI0…t† ‡C1I …t† ˆ f …t†; …1:31†

where I …t† is the intensity and f …t† ˆ U0…t†

Equation of a Mechanical Oscillator

The equation of motion of a material point with mass m, which moves on the

Ox axis under the action of an elastic force F ˆ ÿo2x, is

Directions of the Normal Stresses in a Plane Problem of Elasticity Theory

These directions are de®ned by the differential equation

dydt

 2

‡sxtÿ sy

xy

dydt

1.2 Integrable Types of Equations

A differential equation is integrable by quadratures if the general solution ofthe equation can be expressed in an explicit or implicit form that may containquadratures (i.e., inde®nite integrals)

1.2.1 FIRST-ORDER DIFFERENTIAL EQUATIONS OF THE NORMAL FORM

Equations with Separable Variables

An equation with separable variables is a ®rst-order differentiable equation

of the form

where q is a continuous function de®ned on the interval …a1; a2†  R, and p

is a continuous nonzero function, on the interval …b1; b2†  R If we separatethe variables …dividing by p…x†† and integrate, the solution is of the form

is given by the equality

EXAMPLE 1.3 Consider the equation dx=dt ˆ ata…xb1‡ bxb2†; a, b, a 2 R, b1, b22 Q The

equation is written in the form dx=…xb1‡ bxb2† ˆ ata; …xb1‡ bxb2 6ˆ 0† If b1,

b22 Z, in the ®rst term is necessary to integrate by decomposing a rationalfunction into simple fractions If b1, b2 2 Q, b1ˆ n0

1=n1, b2ˆ n0

2=n2, we canmake the replacement x ˆ yr, where r is a common multiple of numbers n1and n2 Finally, we will obtain a rational function m

5.b1ˆ 2, b2ˆ 0, b < 0,

…dx

6.b1ˆ 1, b2ˆ12

Making the replacement x ˆ y2 yields

APPLICATION 1.1The relaxation phenomenon (the decrement in time of the stresses of a piece

under a constant deformation and a constant temperature) is described bythe differential equation _s=E ‡ s=Z ˆ 0, in which s represents the stress inthe transverse sections of the piece; _s ˆ ds=dt is the derivative of stress withrespect to time; E is the modulus of elasticity of the material (constant); and Z

is the viscidity coef®cient (constant) Determine the solution s ˆ s…t† mSolution

The equation is separable and can be written as

s…t† ˆ c1eÿ…E=Z†t:Considering the condition s…0† ˆ s0 yields c1 ˆ s0, and the solution is

After the replacement x=l ˆ y, dx ˆ l dy,

 3

‡ab xl 2ÿ2ab22 xl ‡2a9b33ln 3b xl ‡ a

‡ C :The condition j…l† ˆ 0 yields

C ˆ ÿ18EIpl3

0 1 ‡abÿ2ab22‡2a9b33ln…3b ‡ a†

;and then the solution is

j…x† ˆ18EIpl3

0

xl

 3ÿ1 ‡ab xl22ÿ 1

ÿ2ab22xl ÿ 1‡2a9b33ln

3bxl ‡ a3b ‡ a

26

375: m

APPLICATION 1.3Equation of Radioactive Disintegration

The disintegration rate of a radioactive substance is proportional to themass of that substance at the time t, namely, x…t† The differential equation ofdisintegration is x0…t† ˆ ÿax…t†, where a is a positive constant that depends

on the radioactive substance Determine the disintegration law and thehalving time

SolutionThe differential equation dx=dt ˆ ÿax…t† is a separable equation Separatingthe variables, dividing by x…t†, and integrating gives dx=x ˆ ÿa dt;

„dx=x ˆ ÿa„ dt ‡ C Then, ln x…t† ˆ ÿat ‡ ln c1 (C was chosen as ln c1)and the general solution x…t† ˆ c1eÿat From the initial condition, x…t0† ˆ x0,

x0ˆ c1eÿat 0, which yields c1ˆ x0eat 0 The solution is

x…t† ˆ x0eÿa…tÿt 0 †;where x0is the substance quantity at the time t0 The halving time is the timeperiod T after which the substance quantity is reduced by half,

ÿaT ˆ ÿ ln 2;

T ˆ1aln 2:

Remark 1.4Radioactive disintegration is a decaying phenomenon m

APPLICATION 1.4Newton's Law of Cooling

The rate at which a body is cooling is proportional to the difference ofthe temperatures of the body and the surrounding medium It is known thatthe air temperature is U1ˆ 10C and that during T ˆ 15 minutes the body iscooled from U2ˆ 90C to U3ˆ 50C Find the law for the changing bodytemperature with respect to time

Solution

If we denote the time by t and the body temperature by U …t†, then

dU =dt ˆ k…U ÿ U1†, where k is the proportionality factor Separating thevariables gives dU =…U ÿ U1† ˆ k dt Taking integrals of the left- and right-hand sides gives

… dU

U ÿ U1ˆ k

dt ‡ C ;

or ln…U ÿ U1† ˆ kt ‡ ln c1 Hence, U ÿ U1ˆ c1ekt Then U ˆ U1‡ c1ekt To

®nd the constants c1 and k, we use the conditions of the problem,

U …0† ˆ U2 and U …T † ˆ U3:Hence, U2ˆ U1‡ c1 Then c1ˆ U2ÿ U1 and U3ˆ U1‡ …U2ÿ U1†ekT.Thus, ekT ˆ …U3ÿ U1†=…U2ÿ U1†, or ek ˆ ‰…U3ÿ U1†=…U2ÿ U1†Š1=T and

U …t† ˆ U1‡ …U2ÿ U1†‰…U3ÿ U1†=…U2ÿ U1†Št=T Substituting the values U1ˆ

10C, U2ˆ 90C, U3ˆ 50C, T ˆ 15 min gives

U ˆ 10 ‡ 80…1

2†t=15: mRemark 1.5

In Application 1.4, it was assumed that the proportionality factor is constant.Sometimes it is supposed that it depends linearly on time, k ˆ k0…1 ‡ at† Inthis case,

dU

dt ˆ k0…1 ‡ at†…U ÿ U1†or

U …t† ˆ U1‡ c1ek 0 ‰t‡…at 2 =2†Š:Using

U …0† ˆ U2yields

U2ˆ U1‡ c1;or

c1ˆ U2ÿ U1:Next is used

U …T † ˆ U3;or

U3ˆ U1‡ …U2ÿ U1†ek 0 ‰T ‡…aT 2 =2†Š:Thus,

ek0‰T ‡…aT 2 =2†ŠˆU3ÿ U1

U2ÿ U1;or

ek 0ˆ U3ÿ U1

U2ÿ U1

 1=‰T ‡…aT2 =2†Š

:Finally,

U …T † ˆ U1‡ …U2ÿ U1† U3ÿ U1

U2ÿ U1

 …2t‡at2 †=…2T ‡aT 2 †

: m

APPLICATION 1.5The Emptying of a Vessel

Study the law of leakage of water from a vessel that has the shape of arotation surface about a vertical axis, with a hole A in the bottom part Studythe following particular cases:

(a) The vessel has a hemisphere shape of radius R

(b) The vessel has a truncated cone shape with the small base as bottom,the radii R1, R2, and height H

(c) The vessel has a truncated cone shape with the large base as bottom,the radii R1, R2, and height H

(d) The vessel has a cone shape with the vertex at bottom

(e) The vessel has a cylinder shape m

In hydrodynamics is deduced an expression of the form v ˆ kph thatdetermines the leakage velocity through a hole at depth h from the freesurface of the liquid The equation of median curve of the form r ˆ r …h† isassumed to be known The volume of water that leaks in elementary time dt

is evaluated in two different ways The liquid leaks through the hole and ®lls

a cylinder with base A and height v dt; hence, dV ˆ Av dt ˆ Akphdt Onthe other side, the height of liquid in the vessel will descend by dh; thedifferential volume that leaks is dV ˆ ÿpr2dh Introducing into equationsthe two expressions of dV gives the differential equation with separablevariables ÿpr2dh ˆ Akphdt Separating the variables yields

…r2…h†



h

p dh ‡ C :

From the condition h…0† ˆ H , the constant C is determined

(a) In the case of a spherical shape (Fig 1.4), it can be written that

C ˆ pAk

vessel with the

small base at the

bottom

and substituting this in the expression of t, after integration, yields

 

R1…R2ÿ R1†

H h3=2‡

25

The condition h…T † ˆ 0 implies

T ˆp



Hp

vessel with the

large base at the

bottom

Remark 1.6

If in the expression of r from case (b) we replace R1 by R2, we ®nd theexpression of r from case (c) Consequently, the expressions of t and T forcase (c) will be obtained from the corresponding expressions obtained in (b),

in which R1 is replaced by R2and R2 by R1:

Ak …R22ÿ R2

1†:

(d) It is obtained from case (b), taking R1ˆ 0, R2 ˆ R Hence,

t ˆ5AkH2pR22…H5=2ÿ h5=2† and T ˆ2pR5Ak2pH:(e) It is obtained from case (b), taking R1ˆ R2 ˆ R Then,

Equations That Can Be Reduced to Separable Equations (Equations withSeparable Variables)

Equations of the form

EXAMPLE 1.4 Solve the equation x0…2t ‡ 2x ‡ 1† ˆ …t ‡ x ÿ 1†2 m

SolutionLet us make the notation t ‡ x ˆ u and then 1 ‡dxdt ˆdudt, or

APPLICATION 1.6A body with mass m is acted on by a force proportional to time (the

proportionality factor is equal to k1) In addition the body experiences acounteraction by the medium that is proportional to the velocity of the body(the proportionality factor being equal to k2) Find the law of the body'smotion m

SolutionThe differential equation of motion is mdvdt ˆ k1t ÿ k2v Denoting

k1t ÿ k2v ˆ u, we ®nd (after derivation with respect to t) k1ÿ k2dvdt ˆdudt.Multiplying by m and taking into account the replacement k1m ÿ k2u ˆ

mdudt, which is an equation with separable variables, du=…k1m ÿ k2u† ˆ1

mdt After integration,

k1m ÿ k2uˆ

1m

dt ‡ C ) ÿk1

2ln jk1m ÿ k2uj ˆmt ‡ C :The initial condition v…0† ˆ 0 results in u…0† ˆ 0; hence, ÿk1

2ln jk1mj ˆ C Replacing the value of C yields ÿk1

2ln jk1m ÿ k2uj ˆmt ÿk1

2ln jk1mj.Multiplying by (ÿk2), ln jk1m ÿ k2uj ˆ ln jk1mj ÿkm2t; hence, k1m ÿ k2u ˆ

k1meÿk2t=m) k2u ˆ k1m ÿ k1meÿk2t=m Replacing u by its expressiondepending on v gives

To ®nd the dependence of displacement on time, we use the equality

C ˆ s0‡k1m2

k3 2

and s…t† ˆ s0‡k1m2

k3 2

and making the notation f …1; u† ˆ j…u†, x ˆ tu, dxdt ˆ u ‡ tdudt and

Eq (1.45) becomes u ‡ tdudt ˆ j…u†, which is an equation with separablevariables, du

dt ˆ …j…u† ÿ u†=t This equation is de®ned on domains of theform …ÿ1; 0†  …u1; u2† or …0; 1†  …u1; u2†, where u1, u2 are two consecu-tive zeros of function j…u† ÿ u Separating the variables and integrating,

du

j…u† ÿ uˆ

dt

t :The general solution is

The equation is written in the form ydxdy ˆ x ‡px2‡ y2

, ordx

APPLICATION 1.7Parabolic Mirror

Find a mirror such that light from a point source at the origin O isre¯ected in a beam parallel to a given direction

SolutionConsider the plane section of the mirror (Fig 1.7) Consider that the ray oflight OP strikes the mirror at M and is re¯ected along MR, parallel to the x-axis If MT is the tangent in M and a, i and r are the angles indicated, i ˆ r

by the optical law of re¯ection, and r ˆ a by geometry Hence, a ˆ i andjOT j ˆ jOM j; jOT j ˆ jPT j ÿ x, MP=PT ˆ tan a ˆ y0) jMPj ˆ y0jPT j )

y ˆ y0jPT j ) jOT j ˆ jy=y0j ÿ x; jOM j ˆpx2‡ y2

The differential tion is

y…ÿ1† ˆ 0, we obtain c ˆ1

2, and the solution is x ˆ1

4y2ÿ 1 m

APPLICATION 1.8The Problem of the Swimmer

To cross a river, a swimmer starts from a point P on the bank He wants

to arrive at a point Q on the other side The velocity v1of the running water is