1.differential equations

8/8/2019 1.Differential Equations

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2005 BE Shapiro [Math 351 Spring 2005 Revised 6 Feb 2005] Page 1.1

1. Differential Equations

In this book we will study equations of the form

dy

dt= f(t,y) (1.1)

where f :!2! ! is a function of two variables and the expression dy /dt represents the

derivative ofy with respect to t. We call equation (1.1) an ordinarydifferential equation

(ODE), and we call any function y =!(t) that satisfies !" (t) = f(t,"(t)) a solution of the

differential equation. For the most part, we will restrict equation (1.1) so that f is

bounded, continuous, and differentiable on some rectangle R ! !2. In many cases it will

help us intuitively to think of tas time, and dy /dt as meaning the rate of change ofy

with respect to time. This suggests the following geometric interpretation of (1.1).

Suppose an object is placed at a location y(t0) = y0 on they-axis at a time t = t0 . Then

the object can do one of three things: if f(t0, y

0) = 0 , it will remain where it is, because

dy / dt= 0 ; if f(t0, y

0) > 0 , it will move forward, because dy / dt> 0 ; or if f(t

0, y

0) < 0 ,

it will move backward, because dy / dt< 0 . This observation allows us to predict

(approximately) the location y1

of the object at an incrementally later time t1; and by

applying it recursively, to construct the trajectory of the object for all time. It is precisely

this notion that is used to construct approximate solutions with a computer; this approach

is also used by the existence theory that we will encounter in a later chapter. We will

also study equations involving higher order derivatives,

y(n) = f(t, y, !y , !!y ,...,y(n"1)

) (1.2)


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that we call higher order differential equations. The order of a differential equation is the

order of the highest ordered derivative in the equation; thus (1.1) is a first order

differential equation, and (1.2) is an nth order differential equation.

The vector generalization of (1.1)

dy

dt= f(t,y) (1.3)

where y = (y1 !!y2 !!!!!yn )T !"n and f :! !!n " !n , is called a first order system (ofdifferential equations). It represents the system (or set) of first order differential equations

!y1 =

f1(t, y

1, y

2,...,yn )

!

!yn = fn (t, y1, y2 ,...,yn )

"

#$

%$

(1.4)

Systems of differential equations such as (1.4) have a natural interpretation as a

description of a dynamical system. Suppose that an object moves along a path such that

its location at any time tis (y1(t),!y

2(t),...) . Then its velocity vector at that time is given

by

v =

!y1

!y2!

"

#

$$

%

&

''=

f1(t, y

1, y

2,...)

f2(t, y

1, y

2,...)

!

"

#

$$

%

&

''

(1.5)

Equation (1.5) describes a time-dependent vector field on !n ; an object placed at a point

(y1,!y

2,...) will move in the direction ( !y1, !y2 ,...)

T with a speed ( !y12+ !y2

2+!)

1/2 , and will

follow a path described by a solution of equation (1.3). Furthermore, given any vector

field v on !n , there is a differential equation determined by the vector field, namely


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Any differential equation that cannot be written in the form (1.7) is said to be nonlinear.

Higher order linear equations such as this can always be reduced to a first order linear

system of the form

!y +A(t)y = f(t) (1.8)

where A is a square matrix and f(t) :!! !n is a column vector.. Linear differential

equations have special properties that will, in general, make them easier to study;

nonlinear equations, and especially nonlinear systems, can have very complicated and

often chaotic properties.

Example 1. The equations

dy

dt= y ,

dy

dt= ty + t

2,dy

dt=1+ sin t (1.9)

are all first order linear differential equations;

dy

dt

= ty2,

dy

dt

= t+ siny, ydy

dt

= t (1.10)

are all first order nonlinear differential equations; and the differential equations

(1! t)2 d

2y

dt2! 2t "y +12y = 0 , t2

d2y

dt2+ x

dy

dt+ (x

2! 9)y = 0 (1.11)

are second order linear differential equations. The set of equations

dx / dt= 3x ! 4xy + zdy / dt= xz

2

dz / dt= zcos(x + y)

"#$

%$

(1.12)

is a nonlinear system of three first order equations, equivalent to the first order system


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d

dt

x

y

z

!

"

##

$

%

&&=

3x ' 4xy + zxz

2

zcos(x + y)

!

"

##

$

%

&&

. (1.13)

Generalizations of differential equations that we will not study includepartial differential

equations (PDEs) and functional (or delay) differential equations (FDEs or DDEs). A

PDE is any equation that includes partial derivatives, whether or not ordinary derivatives

are also included in the equation, such as either of the following equations:

!u = uxx + uyy,!!uxx " uyy = f(x,y) (1.14)FDEs are differential equations include references to a function evaluated at more than

one time argument, such us

!y (t) = 3y(t)+ 2y(t" 5) " t4y(t / 2) (1.15)

which includes references to the variabley evaluated at times t, t-5, and t/2.

We will sometimes find it convenient to represent the derivative as an operator D or Dt

where the variable on the bottom of the derivative is clearly understood, as in

D =d

dt(1.16)

The operatorD should be thought of as a shorthand notation for d/dt, as in

Dt(t+ y) = Dt(t)+ Dt(y) =1+dy

dt

(1.17)

TheD operator can also be used to represent higher order derivatives with an exponent,

as in D2 = d2 /dt2, D3 = d3 /dt3, etc. Another notation that we will frequently use is the

prime notation, where !y = dy /dt, !!y = d2y /dt2 , and so forth, as in equation (1.4). We


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will restrict the use of the notation !y (with y replaced by any expression) and the

operatorDwithout a subscriptwill only be used to indicate differentiation with respect to

a variable t. Other popular notations (that we will avoid) include include the "over-dot"

notation, e.g., !y, !!y,!!!y,... to represent differentiating with respect to a time variable

(especially common in physics) and "delta-based" notations such as !xu,!!

xyu,!!

yyu or

!xu,!!

xyu,!!

yyu to represent the partial derivatives !u / !x,!!2u / !x!y,!!2u / !y2 . We will,

however, have need to evaluate partial derivatives from time to time, in which case we

will use the more usual notations such as !u / dx or My .

Example 2. In terms of the operator D, the first order equations in example 1 can be

written as

Dy = y , Dy = ty + t2 , Dy =1+ sin t,

Dy = ty2, Dy = t+ siny, yDy = t (1.18)

and the second order equations are written as

(1!t)2D2y

!2tDy +12y = 0 , t2D2y + xDy + (x2

!9)y = 0 (1.19)

while in terms of the !y notation, these equations become!y =y , !y = ty + t2, !y =1+ sin t,

!y = ty2, !y = t+ siny, y !y = t (1.20)

and

(1! t)2

""y ! 2t "y +12y = 0, t2 !!y +x !y + (x2 " 9)y = 0 (1.21)

respectively.

Example 3. Rewrite the expression D(y + 3tDy) + D2(y + ty) in standard !y notation.

Multiplying through theD's and applying the product rule (D(fg) = fDg + gDfwe get


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D(y + 3tDy)+ D2(y + ty)

= Dy + D(3tDy)+ D2y + D

2(ty)

= Dy + 3Dy + 3tD2y + D

2y + D(y + tDy)

= 4Dy + (3t+1)D2y + Dy + tD

2y + Dy

= 6Dy + (4t+1)D2y

= 6 !y + (4t+1) !!y (1.22)

DEFINITION 1. A function y =!(t) is called a solution of the differential equation

!y = f(t,y) on the open interval I if !" (t) = f(t,"(t)) for all t! I.The concept of a solution is illustrated by the following examples.

Example 4. We can derive a solution of the differential equation dy /dt= y by rewriting

it as

dy

y= dt (1.23)

and integrating

dyy

= dt!! " ln | y |= t+C (1.24)

If y > 0 we have

y = et+C

= eCet=K1e

t (1.25)

where K1 = eC is a positive constant, regardless of the value of C, because the

exponential can never take on a negative or zero value. Furthermore, since the range of

eC is all positive numbers, K1 can take on any positive value. On the other hand, if

y < 0, we have ln | y |= ln(!y) and hence

!y = et+C

= eCet=K1e

t (1.26)


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or equivalently

y =K2et (1.27)

where K2=

!K1 can take on any negative value. Finally, if y

=

0 then we cannot solve

the integral. However, since d(0)/dt= 0 we observe that y = 0 is also a solution of the

differential equation. Combining this observation with our equations (1.25) and (1.27),

we conclude that

y =Ket (1.28)

is a solution for any real value ofK. (In fact it is also a solution for complex values ofK

but we will be mostly concerned with real-valued results in this book).

Example 5. Find a solution of the differential equation !y = "t/y .

Proceeding as we did in example 4, we separate the variables and integrate:

ydy! = " tdt! #1

2y2= "

1

2t2+C# y2 + t2 = 2C (1.29)

which we recognize as the equation of a circle of radius 2C; thus the solution in this

case is only defined on the interval ! 2C, 2C[ ] , and ifC< 0 the solution is not defined

at all on ! .

The solutions we found in examples 4 and 5 depend on a constant. If we plot the

solutions we found in examples 4 and 5 for different values of C, we find that as C is

allowed to range over its entire domain, the solutions seem to fill up the entire real plane.

This illustrates an important result: the solution of a differential equation is a family of


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equations, called a one parameter family of solutions, or the general solution to the

differential equation.

Figure 2. The one parameter family of solutions for the differential equations in

examples 4 (left) and 5 (right) for various values of the parameter. The value of theparameter is indicated on the graph.

DEFINITION 2.A function y =!(t,C) is called a one parameter family ofsolutions, or

the general solution of the differential equation !y = f(t,y) on the open interval I, if

d

dt!(t,C) = f(t,!(t,C)) for all t! I.

If we are given some additional information about the problem such as a particular

point (t0,y0) that the solution passes through we can narrow the solution down to a

single curve, at least if there is a curve that passes through the specified point. For

example if we choose (t0,y0) = (.5, e) in example 4, we pick out the curve

corresponding to K=1; if we choose (t0,y0 ) = (! 5, 5 ) in example 5 we pick out the


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curve corresponding to C= 5. The specific point (t0,y0) is called an initial condition

and is usually written as y(t0) = y0; the initial condition completely determines the value

of the parameter C.

Figure 3. Forcing a solution to go through a particular point reduces the one parameterfamily of solutions to at most a single curve.

DEFINITION 3.An initial value problem is a differential equation combined with an

initial condition,

!y = f(t,y), y(t0) =y0 (1.30)

A function !(t) is called a solution to the initial value problem if !(t0) = y0 and

!" (t) = f(t,"(t)) throughout some open neighborhood D containing (t0,y0) .

Example 6. Solve the initial value problem

!y

= "t/y , y(1)

=

2 .

From example 5, the general solution of the differential equation is y2 + t2 = 2C. If we

substitute the initial condition into the general solution, we obtain 2C= (2)2 + (1)2 = 5 .

Hence the solution of the differential equation is y2 + t2 = 5.


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implicitly suggesting that the solution is unique. In fact, we can even state an algorithm to

solve an initial value problem, based on the examples presented.

ALGORITHM 1.Separation of Variables.To solve the initial value problem

!y = f(t, y),!y(t0 ) = y0 (1.33)(a) Separate the variables, so that all of the t-dependence is on one side of the

equation, and all of the y-dependence is on the other side of the equation.

(b) Integrate both sides of the equation.

(c) Substitute the initial condition to solve for the constant of integration.

The problem with this algorithm is that it usually won't work! Step (a) is only possible if

f(t,y) can be separated into a product of two functions, so that f(t,y) = T(t)Y(y) ; for,

example, if we are faced with an equation like !y = etysin(ty) we are lost. Even if the

variables can be separated we may be faced with an integral that we can't solve, or which

cannot be solved in closed form (consider !y = et2

) or which has singularities in the

integrand that could lead to problems (try !y = 1 / (1" t),!y(1)= 1). Finally, we might notbe able to substitute the initial conditions (the example in the previous sentence suggests

that ln(0)=-1.) And finally, even if the whole process works, how do we know that we

have found the only (unique) solution to the problem? Consider what happens in the

following examples.

Example 8. The initial value problem !y = y ,y(0) = "4 does not have a solution. To see

this we first solve the differential equation by separating the variables:

y!1/ 2

dy = dt# 2y1/ 2 = t+C (1.34)


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from the initial condition, C= 2 !4 and hence there is no real solution.

Example 9. The initial value problem !y = y ,y(0) = 0 has a solution but it is not unique.

To see these, we observe from example 8 that a solution to the differential equation is

2y1/ 2

= t+C; the initial condition gives C=0 and hence

y =1

4t2 (1.35)

Observe however (by direct substitution into the differential equation and initial

condition), that

y =

0, t! 7

1

4(t" 7)2, t# 7

%&

'&(1.36)

is also a solution to the initial value problem. Furthermore we can replace the number 7

in equation (1.36) with any other positive real number, and still satisfy both the initial

condition and the differential equation. Hence there are an infinite number of solutions to

the initial value problem.

Figure 5. Two different solutions of the initial value problem !y = y ,y(0) = 0 .


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Example 10. The initial value problem !y =y2, y(1) =1 can be solved by algorithm 1.1:

y!2dy = dt"" # !

1

y+C= t (1.37)

The initial condition gives C= 2 and therefore

y =1

2 ! t(1.38)

This solution diverges to ! as t! 2, so it is not possible to extend the solution past the

value t=2, and the maximal interval of solution for the given initial condition is

K= (!",2) . Thus we can extend the solution up to t=2, but not beyond t=2. This is not to

say that there are not solutions of this initial value problem for t>2, but we can not

reach them by starting out at the initial point (1,1).

Since we have an algorithm that will work sometimes we know that at least some initial

value problems have solutions; we're not sure yet about uniqueness. Fortunately, we have

the following theorem, which gives us a set of conditions under which a unique solution

exists. There is still some middle ground, because the theorem does not tell us when

solutions do not exist; when the conditions of the theorem fail, a unique solution might

still exist. We are just not guaranteed that one will exist. Fortunately, the conditions cover

a rather broad set of situations that important throughout the sciences.

THEOREM 1. Existence and Uniqueness of Solutions. Let f(t,y) be bounded,

continuous, and differentiable on a closed rectangle R, and that!f /!y is bounded on R.

Then if (t0,y0) is any point on the interior of R, the initial value problem

!y = f(t,y), y(t0) =y0 (1.39)


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has a unique solution y =!(t) in some open neighborhood of (t0,y0) .

Figure 6. Illustration of the conclusion of theorem 1. If the conditions of the theorem aremet within the closed rectangle R, then there is some open neighborhood Nof (t0,y0) in

which a solution y=

!(t) exists to the initial value problem. Note that the curve thatsatisfies the initial value problem is not guaranteed to be a function outside ofN.

The existence of a solution means that corresponding to any point in the plane, it is

possible to draw a solution curve y =!(t) in the (t, y) plane through that point.

Uniqueness means that only one solution curve can go through that point, and that no two

curves in the plot will ever intersect.

Example 11. Show that the initial value problem !y = sin(ty),!y(0) = 0.5 has a uniquesolution in some neighborhood of (0, 0.5).

Since | sin(ty) |! 1 for all real values of t and y, it is bounded. Furthermore,

f(t, y) = sin(ty) is continuous and differentiable for all real values of tandy; hence it is

certainly bounded, continuous and differentiable within any rectangle

R = [!A,A]" [0.5 ! B,0.5 + B] , i.e., any rectangle of width 2A and height 2B centered

on the initial condition. Furthermore, !f / !y = tsinty is also bounded on this rectangle,


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since | !f / !y |=| tsinty |" A . Hence by Theorem 1.4 a unique solution y = !(t) exists to

the initial value problem.

Figure 7. The solution of !y = sin(ty),!y(0) = 0.5 , is plotted as a function of t. Alsoshown are a closed rectangleR on which the conditions of the existence theorem are met,and an open neighborhoodNof the (0,.5) within which the existence theorem ensures a

solution.

Example 12. Show that the initial value problem

dydt

= y(1!

t)

t(1! y), y(1) = 2. (1.40)

has a unique solution, and find that solution.

The function f(t, y) = y(1! t) (t(1! y)) is bound, continuous, and differentiable on any

closed set that does not contain any points on the lines t = 0 or y = 1. Since the initial

condition (1,2) does not lie on either of these lines, we can define a rectangle that

includes the initial condition for which the function is bounded, continuous, and

differentiable, say

R = [1! a,1+ a]" [2 ! b,2 + b],!!0 < a


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If we pick a = .75,!b = 0.5 , for example, we would have R = [0.25,1.75] ! [1.5,2.5].Figure 8. Left: Rectangle chosen for example 12. The choice of rectangle is arbitrary, solong as it does not cross the dashed lines and the initial value is an interior point.

Figure 9. Plots of f(t, y) (left) and !f / !y (right) onR. Both are bounded; the function

takes on its maximum value at the lower right-hand corner ofR, while the derivativetakes on its maximum value at the lower left-hand corner.

Both f(t, y) and its derivative

!f

!y=

!

!y

y(1" t)

t(1" y)=

(1" t)

t(1" y)2

(1.42)


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are bounded onR, because they are quotients of bounded functions and the denominator

does not equal zero onR. Hence a unique solution to (1.40) exists in some open subset of

R that contains the initial value.

Algorithm 1.1 works to solve this problem; rearranging the variables and integrating,

1! y

ydy" =

1! t

tdt" (1.43)

ln | y |!y = ln | t |!t+C (1.44)

Figure 10. The branch ofy / ey = 2tet!1 that goes through (1, 2).

Using the initial condition, C= ln | y0

|!y0! ln | t

0| +t

0= ln 2! 2! ln1+1= ln 2!1, so

that

ln | y / t |= y ! t+ ln 2!1 (1.45)

Ify/t>0, which should be true in the vicinity of the initial condition (1, 2), then

y / t= ey! t+ ln 2!1

= (2 / e)eye!t (1.46)

y / ey= 2te

! t!1 (1.47)


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We proved in the previous example that a unique solution y = !(t) exists that satisfies

the IVP, and that the expression y / ey = 2te! t!1 also satisfies the IVP. Frequently students

will reject the expression y / ey = 2te!t!1 as the solution because they know that it is

impossible to solve this expression for y = !(t) . However, to do so would be wrong,

because the curve in the ty plane that satisfy y / ey = 2tet!1 is a function, even though

there is no closed form expression.

There are, in fact, other branches of this curve, but they are on the other sides of the lines

t=0 andy=1, and hence are not in our rectangle. The actual curve is a function, and that

function is the y = !(t) that solves the initial value problem. The implicit function

theorem formalizes a set of conditions under which an expression gives a functional

relationship.

THEOREM 2.Implicit Function Theorem. Suppose

F(t,y) = 0 (1.48)

in some neighborhood of (t0, y

0) , that the partial derivatives !F / !t and !F / !y exist

and are continuous in this neighborhood, and that

!F(t0 ,y0 )

!y" 0 (1.49)

Then in some neighborhood I of t0

there exists precisely one function ! such that

y = !(t) satisfies equation (1.48)i.e., F(t,!(t)) = 0 for all t in I.


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In other words, the implicit function theorem tells us that even though it is not possible to

solve an equation like (1.47) for y as a function of t, there is a functional relationship

between the variables. The existence of such a functional relationship does not guarantee

the existence of a closed form functional equation in terms of known elementary

functions, but such a relationship does exist (perhaps in terms of some unknown or

undefined functions). For a proof of the implicit function theorem the student can refer to

any text on advanced calculus.

Example 13. Show a functional relationship y = !(t) exists for the locus of points

satisfying y / ey = 2te!t!1 in some neighborhood of the point (1, 2).

Define the function F(t,y) = y / ey ! 2te!t!1 = 0 . The partial derivatives exist and are

continuous, and

!F

!y=

!

!yye

"y " 2te" t"1#$ %& = "ye"y+ e

"y= e

"y(1" y) ' 0,!y ' 1 (1.50)

so as long as we stay above the line y=1, the expression can be written in the form

y = !(t) .

DEFINITION 4. The general nth order initial value problem is

y(n)= f(t, y, !y , !!y ,...,y

(n"1)),!y(t

0) = y0 , !y (t0 ) = y1,...,y

(n"1)(t

0) = yn"1 (1.51)

The existence problem for higher order equations, as it turns out, can be reduced to the

problem for first order equations, for the simple reason that all higher order equations can

be reduced to an equivalent vector first order differential equation

!y = f(t,y),!y(t0 ) = y0 (1.52)


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where

y = y !y ... y(n"1)( )

T

(1.53)

by defining a sequence of new variables

x0= y, x

1= !y ,!x2 = !!y ,..., xn"1 = y(n"1) (1.54)

and then setting y = (x0 !x1 x2 ...!!!xn!1)T as illustrated by the following example.Example 14. Rewrite the third order initial value problem

!!!y

+ !y sint

+

3y

2

t=

t+

cost,!y(t0 )

=y0 ,

!y (t0 )

=y1,

!!y (t0 )

=y2 (1.55)

as a first order vector differential equation.

We make the change of variables

x1 =

y,!x2 = !y ,!x3 = !!y (1.56)

Then equation (1.55) becomes

!x3 + x2 sint+ 3x12t= t+ cost,!x1(t0 ) = y0 ,x2 (t0 ) = y1,x3(t0 ) = y2 (1.57)

and we can form the matrix equation

!x1!x2!x3

"

#

$$

%

&

''=

x2

x3

(x2 sint( 3x12t+ t+ cost

"

#

$$

%

&

'',!

x1(0)

x2(0)

x3(0)

"

#

$$

%

&

''=

y0

y1

y2

"

#

$$

%

&

''

(1.58)

If we defined the vectors x and x0

and the vector function f(t,x) by

x =

x1

x2

x3

!

"

##

$

%

&&,!!!x

0 =

y0

y1

y2

!

"

##

$

%

&&, f(t,x) =

x2

x3

'x2sint' 3x

1

2t+ t+ cost

!

"

##

$

%

&&

(1.59)


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then the original 3rd order initial value problem becomes

dx

dt= f(t,x),!x(t

0) = x

0(1.60)

THEOREM 3. Let f(t,y) and !f / !yi i = 1,2,...,dim(f)be bounded, continuous, and

differentiable on a closed n-dimensional hypercube R. Then if(t0,y

0) is any point on the

interior of R, the initial value problem

!y = f(t,y), y(t0 ) = y0 (1.61)

has a unique solution y = !(t) in some open neighborhood of (t0 ,y0 ) .

Where do differential equations come from? The simple answer is that they arise

naturally in a variety of problems throughout the sciences. One of the most useful

equations for mathematical scientists takes the form of

dy

dt=

K! y

"

,!y(0) = y0

(1.62)

where Kand 0 are known constants, often called the "steady state value" and "time

constant," respectively. Simply stated, this equation says that y decreases (or increases, if


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balloon as air leaks out; the rate at which a potato cooks (Newton's law of cooling); and

the growth of populations (Malthus' theory of population growth), among other problems.

Figure 11. Exponential relaxation. The solutions to equation (1.62) are plotted for

different values of y0 . All solutions relax towards y = K , indicated by the bold line, ast!" .

Example 15.RC Circuits. Electric Circuits are governed by the following rules:

(a)The voltage drop across a resistor (the difference in voltages between the twoends of the resistor) of resistanceR with a current i flowing through it is given by

!Vresistor = iR (Ohms Law)

(b)Current i represents a flow of charge i = dq /dt.(c) If there is a voltage drop across a capacitor of capacitance C, there will be a

charge q = CV on the capacitor, hence (by differentiating), there is a current of

i =CdV /dt.

(d)The voltage drop across an inductor of inductance L is !Vinductor = Ldi /dt.(e)The total voltage drop around a loop must sum to zero.


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(f) The total current through any node (a point where multiple wires come together)must sum to zero (Kirchoffs Law).

Figure 12. An RC-circuit contains a capacitor (C) and a resistor (R) in a loop with a

battery (voltage Vbatt).

Vbatt

C

a

b

If we were to measure the voltage between points a and c in an RC circuit, these rules tell

us first that the capacitor should cause the voltage along the left branch to fluctuate

according to

iC =CdVab

dt(1.64)

where iC is the current through the capacitor. If iR is the current through the resistor,

then the voltage drop through the right side of the circuit is

Vab =Vbatt + iRR (1.65)

Solving for the current through the resistor,

iR =Vab !Vbatt

R

(1.66)

Since the total current around the loop must be zero, iC + iR = 0,

0 = iR + iC =Vab !Vbatt

R+C

dVab

dt(1.67)


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Dropping the ab index,

dV

dt=

Vbatt !V

RC(1.68)

Comparing with equation (1.62), we see that this circuit has a natural time constant of

! = RC (1.69)

and steady state current equal to that of the battery. Therefore the voltage is given by

V = Vbatt

+ (V0!V

batt)e

! t/RC (1.70)

Hence by equation (1.64) the current through the circuit is

i = CdV

dt=

V0!V

batt

Re! t/RC (1.71)

where V0

is the initial voltage of the system.

Example 16. Newton's Laws of Motion. The motion of an object subjected to external

forces F1,F

2,... is given by the solution of the differential equation

d2

dt2(my) = Fi

i

! , y(t0 ) = y0 (1.72)

where m is the mass of the object andy is its vertical position.

Example 17. Falling Objects. Suppose an object of constant mass m is dropped from a

height h, and is subject to two forces: gravity, and air resistance. The force of gravity can

be expressed as

Fgravity = !mg (1.73)

where g = 9.8!meters/ second2 , and the force due to air resistance is


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Fdrag = !CDdy

dt(1.74)

where CD

is a known constant, the coefficient of drag. According to equation (1.72)

m !!y = "mg " CD !y (1.75)

Letting v = !y denote the speed of the object, this simplifies to

!v = "g " (CD / m)v ="gm /CD " v

m /CD(1.76)

This is equivalent to equation (1.62) with K= !gm /CD , ! = m /CD , and v0 = 0 , so the

velocity is given by

v = (gm /CD )(e! tCD /m

!1) (1.77)

In the limit as t!" the speed will approach a magnitude of vterm = gm /CD but the

object will go no faster; this speed ( vterm

) is known as the terminal velocity of an object.

For example the, terminal velocity of a sky diver (which he or she reaches a long time

before landing) is about the equivalent speed one would reach after jumping from a

height of around ten feet, around 10 to 15 miles per hour; without a parachute the

terminal velocity is around 120 miles per hour (terminal velocity is usually terminal). If

the absence of drag (CD= 0 ) the speed can increase without limit.

We can also solve for the position as a function of time by replacing v = dy / dtand

integrating,

y = (gm /CD )(e!tCD /m !1)dt" = (gm /CD )(!(m /CD )e

! tCD /m ! t)+C (1.78)

From the initial condition y(0) = h we have C= h + g(m /CD )2 so


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y = (gm /CD )(!(m /CD )e! tCD /m

! t)+ h + g(m /CD )2

= g(m /CD )2(1! e

!tCD /m )! (gm /CD )t+ h(1.79)

When tbecomes large ( t>> m /CD

) the first term is nearly zero the position decreases at

a constant rate, because the object has reached its terminal velocity.

Figure 13. Height of the falling object as a function of time, where = m /CD

. As the

curve approaches the asymptote (the dashed line) it approaches its terminal velocity.

Example 18. Newton's Law of Heating (or Cooling) says that the rate of change of the

temperature Tof an object (e.g., a potato) is proportional to the difference in temperatures

between the object and its environment (e.g., an oven), i.e.,

dT

dt= k(T

oven! T) (1.80)

Suppose that the oven is set to 350 and a potato at room temperature (70) is put in the

oven at t=0, with a baking thermometer inserted into the potato. After three minutes you

observe that the temperature of the potato is 150. How long will it take the potato to

reach a temperature of 350?

From equation (1.80) the initial value problem we have to solve is


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!T = k(400 " T),!T(0) = 70 (1.81)The solution (by comparison with the previous problems) is

T = 400!330e

!kt

(1.82)

From the observation that T(3) = 150 we conclude that 150 = 400 ! 330e!3k , and hence

k= (1 / 3)ln(330 / 250) ! 0.0925 . Plugging this into (1.82) means that the temperature of

the potato will reach 350 when 350 = 400 ! 330e!0.0925t

or after

t = (1 / .0925)ln(330 / 50) ! 20.4 minutes.

While the differential equation !y = (K" y) / # does arise quite often, usually a more

complicated model is required to describe a phenomenon.

Example 19.Spreading Depression is a phenomenon that occurs during the genesis of

migraine headaches in which waves of potassium ions move across the brain at

approximately one millimeter per minute. In Hodgkin's model of spreading depression,

the concentrationy of potassium ions at a fixed point is given by

!!y "1+ k

0" 2k

1

2!y + (y " k0 )(k1 " y)(y " k2 ) = 0 (1.83)

where k0is the normal concentration of potassium, k

2is the maximum concentration,

and k1

is a known constants that are related to the diffusion constant of potassium.

Unfortunately the model given by equation (1.83) only describes the onset of the wave

(the potassium increase) and not the end of the wave (the subsequent decrease, and more

complicated models are needed to describe this phenomenon. A more complicated model


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that does describe the wave front quite accurately involves a system of 29 differential

equations.

Figure 14. Onset of a spreading depression wave. The potassium concentration is plotted

as a function of time.

Example 20. The dynamics of love (Strogatz). Romeo and Juliet, a young couple we

know, have a strange relationship. Romeo's love discourages Juliet, making her run away

from him whenever he shows any interest. When Romeo gets discouraged and backs off,

however, Juliet begins to find him strangely attractive. Romeo is just the opposite; he

warms up when she is interested, and loses interest when she backs off.

If we define two love functions R(t) and J(t) for Romeo and Juliet then the model just

described becomes

!R (t) = aJ(t),!! !J (t) = "bR(t) (1.84)where a and b are positive constants, say a = 1,!b = .5 . Then if we divide the aboveequations we get


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dR

dJ=

!R (t)

!J (t)= "2

J(t)

R(t)(1.85)

This gives us a differential equation for the one-parameter family of solutions in the RJ-

plane. Cross-multiplying gives RdR = !2JdJ . After integrating both sides of the

differential equation we find that

1

2R

2+ J

2= C (1.86)

The resulting curves in theRJplane are ellipses centered at the origin. The solutions ofR

and Jas a function of time are sine functions, with the star-crossed lovers trapped in a

never-ending cycle of love and hate, only managing to simultaneously love one another

25% of the time.

Figure 15. Left: Romeo (thin line) and Juliet's love (thick line) for one another as afunction of time. Right: Their mutual feelings follow a clockwise path along an ellipse inthe RJ-Plane over the course of time. They are only mutually in love with one anotherwhen both R > 0 and J > 0 , i.e., when the curve on the right is in the first quadrant.

Variations on this theme generalize the system (1.84) to

!R = aR + bJ,! !J = cR + dJ (1.87)where the constants a, b, c and dmay be either positive or negative, or even to nonlinear

models.


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Example 21.Lotka-Volterra Model of Predation. Rabbit populations are known to grow

quickly in the absence of predators; foxes, however, eat rabbits. To survive, the foxes

need to eat rabbits, and rabbits need to avoid the foxes. If the foxes eat too many rabbits,

their food source goes away, and they die, giving the remaining rabbit population a

chance to re-grow. If we let r be the rabbit population andfthe fox population, then

!r = Ar " Brf

!f = "Cf + Drf(1.88)

The first equation says that rabbits are born at a constant rate A proportional to their

population, and die at a rate B when they come into contact with foxes. The second

equation says that foxes die at a constant rate Cproportional to their population in the

absence of food, and grow at a rate D proportional to the rate at which they come into

contact with rabbits. This model leads to oscillatory solutions in the populations of

rabbits and foxes similar to that observed in nature.

Figure 16. Solutions of equations (1.88) for a system where there are initially five timesas many rabbits as foxes, and A = B = C= D =1 . Left: Populations as a function oftime. Dashed line: rabbits; Solid Line: foxes. Right: Population of foxes as a function ofpopulation of rabbits. The plot on the right is called a phase plotof the initial valueproblem, and the curve is called a trajectory in phase space.


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Exercises.

(1) Solve the initial value problem

!y (t) = y(y "1),!!y(0) = 2 (2) Find (and sketch) the one parameter family of solutions of the differential equation

!y =1" y

t"1

(3) Prove that the initial value problem

!y = sin(y3),!!y(0) = 1

has a unique solution. On what domain is this solution valid?


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Historical Notes

Sir Isaac Newton (1643-1721) is arguably the most famous mathematician of all time; if anyone is to "blame" forcreating the field of differential equations, it is he. Among his many accomplishments, he laid the foundations of

calculus, physics, optics, and celestial mechanics. His father was a wealthy yet illiterate farmer who died 3 monthsbefore Newtons birth. Newton was raised by his grandmother and was sent to attend the Free Grammar School at the

age of 10. There he was considered a poor student who was idle and inattentive. Later he moved in with theheadmaster of the school, and he developed a reputation for making models of clocks and windmills. He entered TrinityCollege, Cambridge, in 1661 intending to study the law. After becoming exposed to natural philosophy andmathematics, however, his future was determined. He received his Baccelaurate in 1665. Cambridge was closed for thefollowing two years due to the plague. Unable to continue his studies at school, Newton returned to the family home inLincolnshire. It was during this time that he invented differential and integral calculus (completed in 1671) along with

his theories of gravity and mechanics (the three laws of motion). He returned to Cambridge in 1667, received hisMasters degree the following year, and was appointed Lucasian professor in 1669. He published his first paper onoptics in 1672; the same year he was elected a fellow of the Royal Society, after donating a reflecting telescope to thesociety. Optics occupied much of his time for the next several decades; his text on the subject was not published until1704. He proved that Keplers theory of elliptical motion was a consequence of his own laws of gravity and motion by1679, and published his most famous work, Philosophiae naturalis principia mathematica, in 1687, which presentedhis theories of physics and celestial mechanics in detail. Two years later (1689) he was elected a member of theConvention Parliament that offered the crown of England to William and Mary. He was sent to Parliament again a fewyears later. Newton retired from Cambridge and was appointed Warden (and later Master) of the Royal Mint in 1696, a position in which he became extremely wealthy. He was also intensely fascinated by alchemy, experimentingthroughout his career, seeking to understand the nature of matter. A student of theology and mysticism, and perhapsastrology, he wrote several biblical studies. His later years were plagued by nervous breakdowns, sometimes attributedto a slow poisoning resulting from his experiments in alchemy. Despite a lingering controversy with Liebniz over the

invention of Calculus, Newton was elected president of the Royal Society from 1703 to 1721. He was knighted in 1703in honor of his research, the first scientist ever so-honored.

Vito Volterra (1860-1940) received his Doctorate in Physics from Pisa University in 1882 and became professor ofmechanics the following year. He was appointed Chair of Mathematical Physics in both Turin and Rome in 1900. His

work focused chiefly on the theory and applications of the differential equations of mathematical physics. He inventedthe theory of functional analysis, that is, the study of functions that take other functions as their arguments. During andafter World War One he studied population biology and several models bear his name. He lived in exile after 1931because he refused to swear an oath to the Fascist government.

Alfred James Lotka (1880-1949) was born in the Ukraine, studied in Austria, and emigrated to the US in 1902. He isbest known for his work on population biology. During his career he worked for the General Chemical Company, theUS Patent Office, the National Bureau of Standards, Metropolitan Life Insurance, and Johns Hopkins University. AtJohns Hopkins he publishedElements of Physical Biology (1925). His work on population biology, while performed atnearly the same time as Volterra's, was independent, and the two were probably unaware of each others work at thetime.

Alan L Hodgkin (1914-1998) and Andrew Huxley (b. 1917) shared the 1963 Nobel Prize in Physiology or Medicinewith Sir John Eccles for their mathematical model of electrical conduction in neurons. The model of spreadingdepression in the text is credited to them by Grafstein in a text by Bures et al. The phenomenon of spreadingdepression was first described by the physiologist Aristides Leo in 1944. The connection to migraines was originallybased on comparisons of wave propagation and the migraine auras described in detail by Lashley in 1941, and has sincebeen observed during P.E.T. scans of patients during migraine. The "more accurate" model referenced in the text waspublished by your favorite Math 351 instructor.

Steven Strogratzis currently a professor of Theoretical and Applied Mechanics at Cornell. He is best know for hiswork in nonlinear dynamics and chaos theory, and is quite active in the Mathematical Community, as well as the author

of the beautiful textbook Nonlinear dynamics and chaos: With applications to physics, biology, chemistry, andengineering(1994).

1.differential equations

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