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Chapter 1. Introduction to Differential Equations

Chapter 1. Introduction to Differential Equations

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2



Numerous references like

Abell and Braselton’s

Mathematica By Example [1]

are also available to beginning

users of Mathematica.



Chapter 1 Introduction to Differential Equations



modeled with differential equations. For example, we will see that the motion of a

pendulum can be modeled by a differential equation. When we solve the problem

of the motion of a pendulum, we use technology to actually watch the pendulum

move. The same is true for the motion of a mass attached to the end of a spring

as well as many other problems. In having this ability, the study of differential

equations becomes much more meaningful as well as interesting.

If you are a beginning Mathematica user and, especially, new to Version 5.0, the

Appendix contains an introduction to Mathematica, including discussions about

entering and evaluating commands, loading packages, and taking advantage of

Mathematica’s extensive help facility.

Although Chapter 1 is short in length, Chapter 1 introduces examples that will

be investigated in subsequent chapters. Also, the vocabulary introduced in Chapter 1 will be used throughout the text. Consequently, even though, to a large extent,

it may be read quickly, subsequent chapters will take advantage of the terminology and techniques discussed here.



1.1 Definitions and Concepts

We begin our study of differential equations by explaining what a differential

equation is.

Definition 1 (Differential Equation). A differential equation is an equation that

contains the derivative or differentials of one or more dependent variables with respect to

one or more independent variables. If the equation contains only ordinary derivatives (of

one or more dependent variables) with respect to a single independent variable, the equation

is called an ordinary differential equation.



EXAMPLE 1.1.1: Thus, dy/ dx x2 / y2 cos y and dy/ dx du/ dx u x2 y

are examples of ordinary differential equations.

The equation y 1 dx x cos y dy 1 is an ordinary differential equation

written in differential form.

Using prime notation, a solution of the ordinary differential equation xy

x2 n2 y 0, which is called Bessel’s equation, is a function y

xy

y x with the property that x d 2 y/ dx2 x dy/ dx x2 n2 y is identically

the 0 function.



1.1 Definitions and Concepts



3



On the other hand,

dx

dt

dy

dt



a



by x

(1.1)

m



nx y



where a, b, m, and n are positive constants, is a system of two ordinary

differential equations, called the predator–prey equations. A solution

consists of two functions x x t and y y t that satisfy both equations.

Predator–prey models can exhibit very interesting behavior as we will

see when we study systems in more detail.

Note that a system of differential equations can consist of more than

two equations. For example, the basic equations that describe the competition between two organisms, with population densities x1 and x2 ,

respectively, in a chemostat are

S



1



x1



x1



x2



x2



m1 S

x1

a1 S

m1 S

1

a1 S

m2 S

1

a2 S



S



See texts like Giordano,

Weir, and Fox’s A First Course

in Mathematical Modeling [12]

and similar texts for detailed

descriptions of

predator–prey models.



See Smith and Waltman’s The

Theory of the Chemostat [24]

for a detailed discussion of

chemostat models.



m2 S

x2

a2 S

(1.2)



where denotes differentiation with respect to t; S

S t , x1

x1 t ,

x2 t . For equations (1.2), we remark that S denotes the conand x2

centration of the nutrient available to the competitors with population

densities x1 and x2 . We investigate chemostat models in more detail in

Chapter 9.



If the equation contains partial derivatives of one or more dependent variables,

then the equation is called a partial differential equation.



EXAMPLE 1.1.2: Because the equations involve partial derivatives of

u

u

and uux u

uyy are

an unknown function, equations like u

t

x

2

2

u

u

partial differential equations. For Laplace’s equation, 2

0a

x

y2

solution would be a function u u x, y such that uxx uyy is identically

the 0 function. A solution u u x, t of the wave equation is a function

2

2

u

u

.

satisfying 2

t

x2



4



Chapter 1 Introduction to Differential Equations



The partial differential equation



u

t



2



u

is known as the heat

x2



equation.

As with systems of ordinary differential equations, systems of partial

differential equations can be considered. With exceptions, their study is

beyond the scope of this text.

Generally, given a differential equation, our goal in this course will most often be

to construct a solution (or a numerical approximation of the solution). The approach

to solving an equation depends on various features of the equation. The first level

of classification, distinguishing between ordinary and partial differential equations,

was discussed above. Generally, equations with higher order are more difficult to

solve than those with lower order.

Definition 2 (Order). The order of a differential equation is the order of the highest-order

derivative appearing in the equation.



EXAMPLE 1.1.3: Determine the order of each of the following differential equations: (a) dy/ dx x2 / y2 cos y ; (b) uxx uyy 0; (c) dy/ dx 4 y x;

and (d) y3 dy/ dx 1.



SOLUTION: (a) The order of this equation is one because the only

derivative it includes is a first-order derivative, dy/ dx. (b) This equation is classified as second-order because the highest-order derivatives,

both uxx , representing 2 u/ x2 , and uyy , representing 2 u/ y2 , are of

order two. Hence, Laplace’s equation, uxx uyy 0, is a second-order

partial differential equation. (c) This is a first-order equation because

the highest-order derivative is the first derivative. Raising that derivative to the fourth power does not affect the order of the equation. The

expressions

dy

dx



4



and



d4y

dx4



do not represent the same quantities: dy/ dx 4 represents the derivative

of y with respect to x raised to the fourth power; d 4 y/ dx4 represents the

fourth derivative of y with respect to x. (d) Again, we have a first-order

equation, because the highest-order derivative is the first derivative.



Linear differential equations are defined in a manner similar to algebraic linear equations that are introduced in algebra and pre-calculus courses.



1.1 Definitions and Concepts



5



Definition 3 (Linear Differential Equation). An ordinary differential equation (of

order n) is linear if it is of the form

an x



dny

dxn



an



1



x



where the functions ai x , i

function.



dn 1y

dxn 1



a2 x



d2y

dx2



a1 x



dy

dx



a0 x y



f x,



(1.3)



0, 1, . . . , n, and f x are given and an x is not the zero



If the equation does not meet the requirements of this definition, then the equation

is said to be nonlinear. If f x is identically equal to the zero function, the linear

equation (1.3) is said to be homogeneous.



EXAMPLE 1.1.4: Determine which of the following differential equations are linear: (a) dy/ dx x3 ; (b) d 2 u/ dx2 u xx ; (c) y 1 dx x cos y dy

x; (e) y x2 y x; (f) x

sin x 0; (g) uxx yuy 0; and

0; (d) y 3 yy

(h) uxx u uy 0.



SOLUTION: (a) This equation is linear, because the nonlinear term

x3 is the function f x of the independent variable in equation (1.3).

(b) This equation is also linear. Using u as the dependent variable name

does not affect the linearity. (c) Solving for dy/ dx we have dy/ dx

1

y / x cos y . Because the right-hand side of this equation includes a nonlinear function of y, the equation is nonlinear in y. However, solving for

dx/ dy, we see that

dx

dy



cos y

x

1 y



or



dx

dy



cos y

x

1 y



0.



This equation is linear in the variable x, if we take the dependent variable to be x and the independent variable to be y in this equation.

(d) The coefficient of the y term is y and, thus, depends on y. Hence,

this equation is nonlinear. (e) This equation is linear. The term x2 is the

x2 of y. (f) This equation, known as the

coefficient function a0 x

pendulum equation because it models the motion of a pendulum, is

nonlinear because it involves a nonlinear function of x, the dependent

variable in this case. (t is assumed to be the independent variable.) For

this equation, the nonlinear function of x is sin x. (g) This partial differential equation is linear because the coefficient of uy is a function of one

of the independent variables. (h) In this case, there is a product of u and

one of its derivatives. Therefore, the equation is nonlinear.



For the linear differential

equation (1.3), f x is called

the forcing function.



A similar classification is

followed for partial

differential equations. In this

case, the coefficients in a

linear partial differential

equation are functions of the

independent variables.



6



Chapter 1 Introduction to Differential Equations



In the same manner that we consider systems of equations in algebra, we can also

consider systems of differential equations. For example, if x and y represent functions of t, we will learn to solve the system of linear equations

dx/ dt



ax



by



dy/ dt



cx



dy



where a, b, c, and d represent constants and differentiation is with respect to t

in Chapter 8. On the other hand, systems (1.1) and (1.2) involve products of the

dependent variables (x and y; S, x1 , and x2 , respectively) so are nonlinear systems

of ordinary differential equations.

We will see that linear and nonlinear systems of differential equations arise naturally in many physical situations that are modeled with more than one equation

and involve more than one dependent variable.



1.2 Solutions of Differential Equations

When faced with a differential equation, our goal is frequently, but not always, to

determine explicit and/or numerical solutions to the equation.

Definition 4 (Solution). A solution to the nth-order ordinary differential equation

F x, y, y , y , . . . , y n



0



(1.4)



on the interval a < x < b is a function Φ x that is continuous on the interval a < x < b

and has all the derivatives present in the differential equation such that

F x, Φ, Φ , Φ , . . . , Φ n



0



on a < x < b.

In subsequent chapters, we will discuss methods for solving differential equations.

Here, in order to understand what is meant to be a solution, we either give both

the equation and a solution and then verify the solution or use Mathematica to

solve equations directly.

EXAMPLE 1.2.1: Verify that the given function is a solution to the corresponding differential equation: (a) dy/ dx 3y, y x

e3x ; (b) u 16u

x

2y y 0, y x

xe .

0, u x

cos 4x; and (c) y



1.2 Solutions of Differential Equations



7



SOLUTION: (a) Differentiating y we have dy/ dx 3e3x so that substitution yields

dy

3y

or

3e3x 3e3x .

dx

4 sin 4x and u

(b) Two derivatives are required in this case: u

16 cos 4x. Therefore,

16u



u



16 cos 4x



16 cos 4x



0.



(c) In this case, we illustrate how to use Mathematica. After defining y,

In[1]:= y x

Out[1]=



x



x Exp



we use ’ to compute y

In[2]:= dy

Out[2]=



x



x

x



e



xe x , naming the resulting output dy.



y x



x



x



x



Similarly, we use ’’ to compute y

output d2y.

In[3]:= d2y

Out[3]=



y



x



xe x , naming the resulting



x



x



2



2e



x



x



Finally, we compute y

2y y

2e x 2 e x xe x

xe x 0. The

result is not automatically simplified so we use Simplify to simplify

the output.

In[4]:= d2y

Out[4]=



2



2dy

x



2



y x

x



x



x



2



In[5]:= Simplify d2y



2dy



x



x



y x



Out[5]= 0



We obtain the same result by entering

In[6]:= Simplify y



x



2y x



y x



Out[6]= 0



which first computes y

2y y and then applies the Simplify command to the result. We graph this solution with Plot. Entering

In[7]:= Plot y x , x, 1, 1



graphs y x



xe



x



on the interval



1, 1 . See Figure 1-1.



If you are a beginning

Mathematica user, see the

Appendix for help getting

started with Mathematica.



8



Chapter 1 Introduction to Differential Equations



-1



-0.5



0.5



1



-0.5

-1

-1.5

-2

-2.5



Figure 1-1 Plot of y x



xe



x



on the interval



1, 1



In the previous example, the solution is given as a function of the independent

variable. In these cases, the solution is said to be explicit. In solving some differential equations, however, we can only find an equation involving x and y that the

solution satisfies. In this case, the solution is said to be implicit.



EXAMPLE 1.2.2: Verify that the given implicit function satisfies the

differential equation.

Function: 2x2

dy

Differential Equation:

dx



Assuming that y

dy

y.

dx



yx,



y2 2xy 5x

2y 4x 5

2y 2x



0



SOLUTION: We use implicit differentiation to compute the derivative

of the equation 2x2 y2 2xy 5y 0:

4x



2y



dy

dx



2x



dy

dx



2y



2y



2x



5

dy

dx

dy

dx



0

2y



4x



5



2y 4x 5

.

2y 2x



Hence, the given implicit solution satisfies the differential equation.

We also illustrate how to use Mathematica to differentiate the equa0 with respect to x. After clearing all prior

tion 2x2 y2 2xy 5x

definitions of x, y, and eq, if any, with Clear we define eq to be the



1.2 Solutions of Differential Equations



9



equation 2x2 y2 2xy 5x 0. Note how we use a double equals sign

(==) to separate the left and right-hand sides of the equation.

In[8]:= Clear x, y

2x2



eq



Out[8]= 5 x



2x



y2



2



2xy



2xy



5x



y2



0

0



Next, we use Dt to differentiate eq with respect to x, naming the resulting output step1. The symbol Dt[y,x] appearing in the result represents dy/ dx; step1 represents the equation 4x 2yy 2xy 2y 5 0.

In[9]:= step1



Dt eq, x



Out[9]= 5



2y



4x



Finally, we obtain y

with Solve.



2 x Dt y, x



2 y Dt y, x



0



dy/ dx by solving the equation step1 for Dt[y,x]



In[10]:= step2

Out[10]=



Solve step1, Dt y, x

5 4x 2y

Dt y, x

2 x y



Generally, to graph an equation of the form f x, y

C, where C is a

constant, we use the ContourPlot command which is used to graph

level curves of surfaces: the graph of f x, y

C is the same as the

graph of the level curve of z

f x, y corresponding to z

C. Thus,

0 is the same as the

the graph of the equation 2x2 y2 2xy 5x

graph of the level curve of z f x, y

2x2 y2 2xy 5x corresponding

to 0. Note how 2x2 y2 2xy 5x, the left-hand side of the equation eq,

is extracted from eq with Part ([[...]]): 2x2 y2 2xy 5x is the first

part of eq.

In[11]:= eq 1

Out[11]= 5 x



2 x2



2xy



y2



Thus, entering

In[12]:= ContourPlot Evaluate eq 1

, x, 7, 2 ,

y, 7, 2 , Frame False, Axes Automatic,

AxesOrigin

0, 0 ,

AxesStyle GrayLevel 0.5 ,

PlotPoints 100, Contours

0 ,

ContourShading False



10



Chapter 1 Introduction to Differential Equations



2



-6



-4



-2



2



-2



-4



-6



Figure 1-2 Graph of 2x2



y2



2xy



5x



0



graphs the equation 2x2 y2 2xy 5x 0 as shown in Figure 1-2 for

7

x

2 and 7

y

2 (the option Contours->{0} instructs

Mathematica to graph only the level curve corresponding to 0). The

option ContourShading->False specifies to not shade the regions

between contours, Frame->False specifies that a frame is not to be

placed around the resulting graphics object, Axes->Automatic specifies that axes are to be placed on the resulting graphics object while the

option AxesOrigin->{0,0} specifies that they intersect at the point

0, 0 and the option AxesStyle->GrayLevel[.5] specifies that they

be drawn in a medium shade of gray. The option PlotPoints->100

instructs Mathematica to increase the number of sample points to 100

(the default is 15), helping assure that the resulting graphics object

appears smooth. Be sure to enclose eq[[1]] in Evaluate as shown:

this ensures that Mathematica evaluates eq[[1]] before sampling

points; if Evaluate is not included error messages result. (Note that an

alternative way to graph equations is to use the ImplicitPlot command which is contained in the ImplicitPlot package located in the

Graphics folder (or directory).) If you are using Version 5.0 (or later)

and wish to avoid using Part, you can select the left-hand side of the

equation, copy it,



1.2 Solutions of Differential Equations



11



and then paste it into the ContourPlot command.



See Getting Started in the

Appendix.



EXAMPLE 1.2.3: On a rectangular membrane, the solution of the wave

equation,

2



w

x2



2



w

y2



1

2

c0



For details, see Graff’s Wave

Motion in Elastic Solids [13].



2



w

t2



(1.5)



12



Chapter 1 Introduction to Differential Equations



takes the form

Amn cos Ωmnt



w



Bmn sin Ωmnt Wmn ,



n 1m 1



where Amn and Bmn are constants, Ωmn 2 c0 2 Ξn 2

mΠ/ b, and the normal modes, Wnm , are given by

Wnm x, y



Ζm 2 , Ξ n



nΠ/ a, Ζm



sin Ξn x sin Ζm y.



cos Ωmnt Wmn x, y and fs x, y, t

sin Ωmnt

(a) Verify that fc x, y, t

Wmn x, y satisfy the wave equation (1.5) on a rectangular membrane.

(b) Plot the first few normal modes of the membrane.

SOLUTION: After defining Ωmn 2 c0 2 Ξn 2 Ζm 2 , Ξn nΠ/ a, Ζm mΠ/ b,

cos ΩmntWmn x, y and fs x, y, t

sin ΩmntWmn x, y .

we define fc x, y, t

In[13]:= Ω m , n



Π c0



In[14]:= Ξ n





a



Ζ m





b



n2

a2



m2

b2



In[15]:= fc x , y , t

Cos Ω m, n t Sin Ξ n x Sin Ζ m y



In[16]:= fs x , y , t

Sin Ω m, n t Sin Ξ n x Sin Ζ m y

2



To verify that fc x, y, t satisfies equation (1.5), we compute

2



and



fc



t2



fc

,

x2



in fcxx, fcyy, and fczz, respectively.



In[17]:= fcxx

Out[17]=



In[18]:= fcyy

Out[18]=



m2

b2



n2

a2



Π t Sin



nΠx

a



Sin



mΠy

b



nΠx

a



Sin



mΠy

b



a2

D fc x, y, t , y, 2



m2 Π2 Cos c0



m2

b2



n2

a2



Π t Sin

b2



In[19]:= fctt

Out[19]=



D fc x, y, t , x, 2



n2 Π2 Cos c0



D fc x, y, t , t, 2



m 2 n2

Π2 Cos c0

b2 a2

nΠx

mΠy

Sin

Sin

a

b



c02



m2

b2



n2

Πt

a2



2



fc

,

y2



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