Introduction to Differential Equations

Introduction to Differential Equations

CHAPTER 1

Lecturer: Prof. Hsin-Lung Wu Ch1_2

Chapter Contents

1.1 Definitions and Terminology1.2 Initial-Value Problems1.3 Differential Equations as Mathematical Methods


1.1 Definitions and Terminology

Introduction: differential equations means that equations contain derivatives, eg:

dy/dx = 0.2xy (1)

Ordinary DE: An equation contains only ordinary derivates of one or more dependent variables of a single independent variable. eg: dy/dx + 5y = ex, (dx/dt) + (dy/dt) = 2x + y (2)

An equation contains the derivates of one or more dependent variables with respect to one or more independent variables (DE).

Definition 1.1.1 Differential equation


Partial DE: An equation contains partial derivates of one or more dependent variables of tow or more independent variables.

(3)

Notations: Leibniz notation dy/dx, d2y/ dx2

prime notation y’, y”, ….. subscript notation ux, uy, uxx, uyy, uxy , ….

Order: highest order of derivatives

tu

tu

xu

yu

xu

2 ,0 2

2

2

2

2

2

2

2

xeydydx

dxyd

45

3

2

2

second order first order


General form of n-th order ODE: (4)

Normal form of (4) (5)

eg: normal form of 4xy’ + y = x, is y’ = (x – y)/4x

Linearity: An n-th order ODE is linear if F is linear in y, y’, y”, …, y(n). It means when (4) is linear, we have

(6)

0) , ,' , ,( )( nyyyxF

) , ,' , ,( )1( nn

n

yyyxfdx

yd

)()()()( 011

1

1 xgyxadxdy

xadx

yda

dxyd

xa n

n

nn

n

n


The following cases are for n = 1 and n = 2 and

(7)

Two properties of a linear ODE:(1) y, y’, y”, … are of the first degree.(2) Coefficients a0, a1, …, are at most on x

Nonlinear examples:

)()()( 01 xgyxadxdy

xa

)()()()( 012

2

2 xgyxadxdy

xadx

ydxa

')1( yy ysin 2y


That is, a solution of (4) is a function possesses at least n derivatives and

F(x, (x), ’(x), …, (n)(x)) = 0 for all x in I, where I is the interval is defined on.

Any function , defined on an interval I, possessing at least n derivatives that are continuous on I, when replaced into an n-th order ODE, reduces the equation into an identity, it said to be a solution of the equation on the interval.

Definition 1.1.2 Solution of an ODE


Example 1 Verification of a Solution

Verify the indicated function is a solution of the given ODE on (－ , ) (a) dy/dx = xy1/2; y = x4/16 (b)

Solution:(a) Left-hand side:

Right-hand side: then left = right

(b) Left-hand side:

Right-hand side: 0 then left = right

xxeyyyy ;02

4164

33 xxdxdy

4416

322/142/1 xx

xx

xxy

0)(2)2(2 xxxxx xeexeexeyyy


Note: y = 0 is also the solution of example 1, called trivial solution


Example 2 Function vs. Solution

y = 1/x, is the solution of xy’ + y = 0, however, this function is not differentiable at x = 0. So, the interval of definition I is (－ , 0), (0, ).

Fig 1.1.1 Ex 2 illustrates the difference between the function y = 1/x and the solution y = 1/x


Explicit solution: dependent variable is expressed solely in terms of independent variable and constants.Eg: solution is y = (x).

G(x, y) = 0 is said to be an implicit solution of (4) on I,

provided there exists at least one function y = (x)

satisfying the relationship as well as the DE on I.

Definition 1.3 Implicit solution of an ODE


Example 3 Verification of an Implicit Solution

x2 + y2 = 25 is an implicit solution of dy/dx = −x/y (8)

on the interval -5 < x < 5.

Since dx2/dx + dy2/dx = (d/dx)(25)

then 2x + 2y(dy/dx) = 0 and dy/dx = -x/ysolution curve is shown in Fig 1.1.2


Fig 1.1.2 An Implicit solution and two explicit solutions in Ex 3


Families of solutions: A solution containing an arbitrary constant represents a set G(x, y) = 0 of solutions is called a one-parameter family of solutions. A set G(x, y, c1, c2, …, cn) = 0 of solutions is called a n-parameter family of solutions.

Particular solution: A solution free of arbitrary parameters. eg: y = cx – x cos x is a solution of xy’ – y = x2 sin x on (-, ), y = x cos x is a particular solution according to c = 0. See Fig 1.1.3.


Fig 1.1.3 Some solution of xy’-y=x2 sinx


Example 4 Using Different Symbols

x = c1cos 4t and x = c2 sin 4t are solutions of x + 16x = 0we can easily verify that x = c1cos 4t + c2 sin 4t is also a solution.


Example 5 A piecewise-Defined Solution

We can verify y = cx4 is a solution of xy – 4y = 0 on (-, ). See Fig 1.1.4(a).

The piecewise-defined function

is a particular solution where we choose c = −1 for x < 0 and c = 1 for x 0. See Fig 1.1.4(b).

0 ,

0 ,4

4

xx

xxy


Fig 1.1.4 Some solution of xy’-4y=0 in Ex 5


Singular solution: A solution can not be obtained by particularly setting any parameters. y = (x2/4 + c)2 is the family solution of dy/dx = xy1/2 , however, y = 0 is a solution of the above DE.

We cannot set any value of c to obtain the solution y = 0, so we call y = 0 is a singular solution.


System of DEs: two or more equations involving of two or more unknown functions of a single independent variable.

dx/dt = f(t, x, y) dy/dt = g(t, x, y)(9)


1.2 Initial-value Problems

IntroductionA solution y(x) of a DE satisfies an initial condition.

ExampleOn some interval I containing xo, solve:

subject to:

(1)This is called an Initial-Value Problem (IVP).y(xo) = yo , y(xo) = y1 ,are called initial conditions.

) , ,' , ,( )1( nn

n

yyyxfdx

yd

10)1(

1000 )( , ,)(' ,)( n

n yxyyxyyxy

10)1( )(

nn yxy


First and Second IVPs

(2)

and

(3)

are first and second order initial-value problems, respectively. See Fig 1.2.1 and 1.2.2.

00 )( :

) ,( :

yxytosubject

yxfdxdy

solve

1000

2

2

)(' ,)( :

)' , ,( :

yxyyxytosubject

yyxfdx

ydsolve


Example 1 First-Order IVPs

We know y = cex is the solutions of y’ = y on (-, ). If y(0) = 3, then 3 = ce0 = c. Thus y = 3ex is a solution of this initial-value problem

y’ = y, y(0) = 3.If we want a solution pass through (1, -2), that is y(1) = -2, -2 = ce, or c = -2e-1. The function y = -2ex-1 is a solution of the initial-value problem

y’ = y, y(1) = -2.


In Problem 6 of Exercise 2.2, we have the solution of y’ + 2xy2 = 0 is y = 1/(x2 + c). If we impose y(0) = -1, it gives c = -1. Consider the following distinctions.

1) As a function, the domain of y = 1/(x2 - 1) is the set of all real numbers except x = -1 and 1. See Fig 1.2.4(a).

2) As a solution, the intervals of definition are (-, 1), (-1, 1), (1, )

3) As a initial-value problem, y(0) = -1, the interval of definition is (-1, 1). See Fig 1.2.4(b).

Example 2 Interval / of Definition of a Solution


Fig 1.2.4 Graph of the function and solution of IVP in Ex 2


Example 3 Second-Order IVP

In Example 4 of Sec 1.1, we saw x = c1cos 4t + c2sin 4t is a solution of x + 16x = 0Find a solution of the following IVP:

x + 16x = 0, x(/2) = −2, x(/2) = 1 (4)

Solution:

Substitute x(/2) = − 2 into x = c1cos 4t + c2sin 4t, we find c1 = −2. In the same manner, from x(/2) = 1 we have c2 = ¼.


Existence and Uniqueness:

Does a solution of the IVP exist?If a solution exists, is it unique?


Example 4 An IVP Can Have Several Solution

Since y = x4/16 and y = 0 satisfy the DEdy/dx = xy1/2 , and also initial-value y(0) = 0, this DE has at least two solutions, See Fig 1.2.5.


Let R be the region defined by a x b, c y d that contains the point (x0, y0) in its interior. If f(x, y) and f/y are continuous in R, then there exists some interval I0: (x0- h, x0 + h), h > 0, contained in [a, b] and a unique function y(x) defined on I0 that is a solution of the IVP (2).

Theorem 1.2.1 Existence of a Unique Solution


Fig 1.2.6 Rectangular region R

The geometry of Theorem 1.2.1 shows in Fig 1.2.6.


Example 5 Example 3 Revisited

For the DE: dy/dx = xy1/2 , inspection of the functions

we find they are continuous in y > 0. From Theorem 1.2.1, we conclude that through any point (x0, y0), y0 > 0, there is some interval centered at x0 on which this DE has a unique solution.

2/12/1

2 and ) ,(

yx

yf

xyyxf


Interval of Existence / UniquenessSuppose y(x) is a solution of IVP (2), the following sets may not be the same:

the domain of y(x), the interval of definition of y(x) as a solution, the interval I0 of existence and uniqueness.


1.3 DEs as Mathematical Models

Introduction Mathematical models are mathematical descriptions of something.

Level of resolutionMake some reasonable assumptions about the system.

The steps of modeling process are as following.


AssumptionsMathematicsformulation

Check model Predictionswith know

facts

Obtainsolution

Express assumptions in terms of differential equations

If necessary,alter assumptions

or increase resolutionof the model

Solve the DEs

Display model predictions,e.g., graphically


Population DynamicsIf P(t) denotes the total population at time t, then

dP/dt P or dP/dt = kP(1)where k is a constant of proportionality, and k > 0.

Radioactive DecayIf A(t) denotes the substance remaining at time t, then

dA/dt A or dA/dt = kA(2)where k is a constant of proportionality, and k < 0.

A single DE can serve as a mathematical model for many different phenomena.


Newton’s Law of Cooling/WarmingIf T(t) denotes the temperature of a body at time t, Tm the temperature of surrounding medium, then

dT/dt T - Tm or dT/dt = k(T - Tm)(3)

where k is a constant of proportionality.


Spread of a DiseaseIf x(t) denotes the number of people who have got the disease and y(t) the number of people who have not yet, then

dx/dt = kxy (4)where k is a constant of proportionality.From the above description, suppose a small community has a fixed population on n, If one inflected person is introduced into this community, we have x + y = n +1, and

dx/dt = kx(n+1-x) (5)


Chemical ReactionsInspect the following equation

CH3Cl + NaOH CH3OH + NaClAssume X is the amount of CH3OH, and are the amount of the first two chemicals, then the rate of formation is

dx/dt = k( - x)( - x)(6)


MixturesSee Fig 1.3.1. If A(t) denotes the amount of salt in the tank at time t, then

dA/dt = (input rate) – (output rate) = Rin - Rout

(7)

We have Rin = 6 lb/min, Rout = A(t)/100 (lb/min), then dA/dt = 6 – A/100 or dA/dt + A/100 = 6 (8)


Fig 1.3.1 Mixing tank


Draining a TankReferring to Fig 1.3.2 and from Torricelli’s Law, if V(t) denotes the volume of water in the tank at time t,

(9)

From (9), since we have V(t) = Awh, then

(10)

ghAdtdV

h 2

ghA

A

dtdh

w

h 2


Fig 1.3.2 Water draining from a tank


Series CircuitsLook at Fig 1.3.3.From Kirchhoff’s second law, we have

(11)

where q(t) is the charge and dq(t)/dt = i(t), which is the current.

)(1

2

2

tEqCdt

dqR

dtqd

L


Fig 1.3.3 Current i(t) and charge q(t) are measured in amperes (A) and coulumbs (C)


Falling BodiesLook at Fig1.22.From Newton’s law, we have

(12)

Initial value problem

(13)

or 2

2

mgdt

sdm g

dt

sd 2

2

002

2

)0(' ,)0( , vsssgdt

sd


Fig 1.3.4 Position of rock measured from ground level


Falling Bodies and Air ResistanceFrom Fig 1.3.5.We have the DE

(14)

and can be written as

or (15)

kvmgdtdv

m

dtds

kmgdt

sdm 2

2

mgdtds

kdt

sdm 2

2


Fig 1.3.5 Falling body of mass m


A Slipping ChainFrom Fig 1.3.6.We have

(16)or 2

32 2

2

xdt

xdL 064

2

2

xLdt

xd


Fig 1.3.6 Chain slipping from frictionless peg


Suspended Cables From Fig1.25.We have

dy/dx = W/T1 (17)


Fig 1.3.7 Cables suspended between vertical supports


Fig 1.3.8 Element of cable

Fig 1.3.8 explains the Element of cable.


Evaluation

One midterm 40%Final 40%Quiz & Homework 20%


Some information

Textbook: Advanced Engineering Mathematics 4th-ed by Zill and Wright

Teaching Website: Please link 數位學苑

Introduction to Differential Equations

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