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Chapter 20

Complex Series

20.1

Power Series

20.2

Taylor and Laurent Series

20.3

Problems

There are three special cases of Equation (23.1) that lead immediately to a solution. (a) If ( ) is independent of , then = ( ), and the most general solution can ( ) where = ( ). be written as = ( ) = + (b) If ( ) is independent of , then = ( ), and ( )

=

()

+

=0

( )

+

=0

embodies a solution. That is, ( ) = can be solved for in terms of , say ( )= . = ( ), and this will be a solution of the DE. Note that (c) The third special case is really a generalization of the ďŹ rst two. If ( ) = ( )= ( ) and ( ) ( ), then = ( ) ( ) or ()

=

()

(23.2)

is an implicit solution.

Consider a point particle moving under the inďŹ&#x201A;uence of a force depending on position only. Denoting the position by and the velocity by , we have, by Newtonâ&#x20AC;&#x2122;s second law, = ( ). Using the chain rule, = ( )( )= , we obtain =

( )

=

( )

(23.3)

which is easily integrated to =

( )

+

The ( ) = ( ) integral. We can write Equation (23.4) as +

( )+

(23.4)

has been introduced as an indeďŹ nite ( )=

(23.5)

+ ( ) which is the exThus, the integral of Equation (23.3) is ( ) = pression for the energy of the one-dimensional motion of a particle experiencing the potential ( ). If is a solution of Equation (23.3), then ( ) = constant. Since a solution of Equation (23.3) describes a motion of the particle, Equation (23.5) implies that the energy of a particle does not change in the course of its motion. This statement is the conservation of (mechanical) energy.

The diﬀerential is not exact. Let us see if we can ﬁnd a function ( ) such that = , for some ( ). We assume that the domain of the -plane in which is deﬁned is contractable to a point. Then a necessary and suﬃcient condition for the equation above to hold is (

)=

(a) Let us assume that = 2 or = =

(

)

+

+2 =0

(23.7)

is a function of only. Then Equation (23.7) reduces to where = 0. In this case we get 1

where

=

=0

Thus, as long as = 0, any function , with arbitrary , is an integrating factor for = 0. This integrating factor leads to the solution =

= constant

In order to determine the constant, suppose that determines the constant in terms of : 1

= constant

(23.8) =

when

= 1. Then (23.8)

constant =

So, (23.8) becomes =

=

(b) Now let us assume that is a function of only. This leads to the integrating where = 0. In this case = is the integral of the DE, and a factor = general solution is of the form = constant. If we further impose the condition (1) = , we get = constant. Equation (23.8) then yields =

=

as in (a). (c) The reader may verify that =

where

+

(

) = (0 0)

is also an integrating factor leading to the integral = constant

= tan Imposing (1) =

gives

=

, so that

= tan(constant) =

as before.

In an electric circuit with a resistance and a capacitance , Kirchhoﬀ’s law gives rise to the equation + = ( ), where ( ) is the time-dependent voltage and is the (instantaneous) charge on the capacitor. This = , = 1 , and = . The integrating factor is is a simple FOLDE with ( )=

1

1

exp

=

1

which yields ( )=

1 () ()

=

+

1

()

+

()

Recall that an indeﬁnite integral can be written as a deﬁnite integral whose upper limit is the independent variable—in which case we need to use a diﬀerent symbol for the integration variable. For the arbitrary lower limit, choose zero. We then have ( )= Let

=

+

( )

(0) be the initial charge. Then, substituting and the charge at time will be given by ( )=

+

(23.14)

= 0 in (23.14), we get

( )

(23.15)

As a speciﬁc example, assume that the voltage is a constant , as in the case of a battery. Then the charge on the capacitor as a function of time will be ( )=

+

(1

)

, independent of the initial It is interesting to note that the ﬁnal charge ( ) is charge. Intuitively, this is what we expect, of course, as the “capacity” of a capacitor to hold electric charge should not depend on its initial charge.

As a concrete illustration of the general formula derived in the previous example, we ďŹ nd the charge on a capacitor in an circuit when a voltage, cos , is applied to it for a period and then removed. ( ) can thus ()= be written as cos if ( )= 0 if The general solution is given as Equation (23.15). We have to distinguish between two regions in time, and . (a) For , we have (using a table of integrals) ( )=

+

=

+

cos (1

1 ) +

1

+

cos

+

sin

, then only the oscillatory part If , and we wait long enough, i.e., survives due to the large negative exponents of the exponentials. Thus, () The charge (b) For

(1

1 ) +

cos

+

sin

( ) oscillates with the same frequency as the driving voltage. , the integral goes up to beyond which ( ) is zero. Hence, we have

( )=

+

=

+

cos (1

+

) +

cos

+

sin

We note that the oscillation has stopped (sine and cosine terms are merely constants now), and for , the charge on the capacitor becomes negligibly small: If there is no applied voltage, the capacitor will discharge.

In Problem 23.11 you are asked to ďŹ nd the velocity of a falling object when the air drag is proportional to velocity. This is a good approximation at low velocities for small objects; at higher speeds, and for larger objects, the drag force becomes proportional to higher powers of speed. Let us consider the case when the drag force is proportional to . Then the second law of motion becomes =

=

This equation can be written as = Now we rewrite

=

1

=

1 +

1 2

1

multiply both sides of Equation (23.16) by 2 ln

+

ln

and integrate to obtain =2

+ ln

where we have written the constant of integration as ln equation can be rewritten as + Suppose that at

+

+

for convenience. This

=

= 0, the velocity of the falling object is

and

(23.16)

=

, then

= +

=

Now note that 0, and 0 (if we take â&#x20AC;&#x153;downâ&#x20AC;? to be the positive direction). Therefore, the last equation becomes +

+

=

Suppose that ; then we can remove the absolute value sign from the RHS, and since the two sides must agree at = 0, we can remove the absolute value sign , then as well. It follows that on the LHS as well. Similarly, if + Solving for

=

+

( +

)(

)=(

)(

+

)

gives = = =

( + ) ( + )

+ (

)

( + 1) + ( 1) ( 1) + ( + 1) cosh( ) + sinh( ) sinh( ) + cosh( )

(23.17)

It follows from Equation (23.17) that at = 0, the velocity is , as we expect. It , the so-called also shows that, when , the velocity approaches = . This is the velocity at which the gravitational force and the drag force become equal, causing the acceleration of the object to be zero. The terminal velocity can thus be obtained directly from the second law without solving the diďŹ&#x20AC;erential equation.

Figure 23.1 shows the plot of speed as a function of time for the two cases of the drag force being proportional to and with the same proportionality constant. Because of the higher power of speed, the terminal velocity is achieved considerably force than for force. Furthermore, as the ﬁgure shows clearly, more quickly for the terminal speed itself is much smaller in the former case. Since larger surfaces drag force, parachutes that have very large surface are desirable. provide a We consider here some other examples of (nonlinear) FODEs whose solutions are available: + ( ) + ( ) = 0 where and choose appropriately.

(a) : This equation is of the form = 1. This DE can be simpliﬁed if we substitute = In terms of , the DE becomes +

( )

+

( )

=0

The simplest DE—whose solution could be found by a simple integration—would be obtained if the exponent of the last term could be set equal to 1. But this would require to be zero, which is not acceptable. The next simplest DE results if we set the exponent equal to zero, i.e., if = 1 (1 ). Then the DE becomes + (1

) ( ) + (1

) ( )=0

which is a ﬁrst-order

DE whose solution we have already found.

(b)

: This DE is of the form =

To ﬁnd the solution, make the obvious substitution and ( ) + = ( ) =

=

, to obtain ( )

=

=

+

with the solution ln where

=

()

or

= exp

()

is an arbitrary constant to be determined by the initial conditions.

∇ ∇

∇ ∇

∇ ∇

∇ ∇

∇ ∇

path

neighboring paths

function

not all possible

continuous

function

line

tive

ďŹ rst variational derivaextremum local

any

non-interacting

external

ordinary

ďŹ rst-order

Legendre transformation