Lecture 14: Introduction to (Review of) Differential Equations [See Chapter 8,courses.washington.edu/ph227814/228/ellis/Phys_228_09... · 2014-07-29 · Physics 228 Lecture 14 1 Winter

Physics 228 Lecture 14 1 Winter 2009

Lecture 14: Introduction to (Review of) Differential Equations [See Chapter 8,

Sections 1 to 7, in Boas. We will return for the rest of Chapter 8 shortly.]

As we have noted several times, most of the physical systems that we are interested in

can be described by (systems of) differential equations. Here we want to begin to

develop the tool set we need to systematically consider the issue of solving

differential equations. To begin let’s review some terminology. Differential

equations are labeled ordinary or partial depending on whether the derivatives that

appear in the equation are ordinary, e.g., , implying a single independent

variable, or partial, e.g., , implying more than 1 independent variable (the issue

is – who is being varied and who is held fixed – we’ll return to this issue). The order

of a differential equation defines the highest derivative present in the equation. For

example, a first order equation involves only first derivatives, (note the

fairly standard notation that a derivative with respect to x is represented by a prime,

while derivatives with respect to time are represented by dots), and a 2nd

order

equation involves (and possibly ). A linear differential equation is

one where each term in the equation involves the dependent variable, or its

derivatives, to no more than the first power, e.g., is a linear second

order ordinary differential equation. (Note that nonlinear differential equations have

the special feature of exhibiting isolated singular solutions that are not contained in

the usual general solutions with arbitrary constants. Nonlinear equations can also

exhibit chaotic behavior.) If every term in the linear equation has a single power of

the dependent variable, or its derivatives, the equation is said to be homogeneous (if

you multiply y by a constant, you are multiplying every term in the equation by the

same constant – but note that the label homogeneous is used in several related ways,

see Eq. 8.4.11 and exercise 4.13.1 in Boas). Thus is a linear second

order homogeneous ordinary differential equation.

A first order linear ordinary (but inhomogeneous) differential equation can often be

solved simply by integration

0

0 .

x

x

dyy x f x dy y x dx

dx

y x dx y x

y x dx f x C dx f x y x

(14.1)

d dx

x

y x dy dx

2 2y x d y dx y

y by cy f x

0y by cy


This equation illustrates the expected result that, at least in principle, we solve

differential equations by doing integrals. Note also that in the last expression we

have been careful to use a “dummy” variable of integration that is explicitly different

from the independent variable x, which now appears as the upper limit of the integral.

We have also made explicit the point that the constant of integration C is just the

desired function evaluated at the lower limit of the integral. Since the derivative of

an integral with respect to the upper limit is just the integrand evaluated at the upper

limit (see Leibniz’ rule in Eq. 4.12.13 of Boas),

0

0 ,

x

x xx

ddx f x y x f x f x

dx

(14.2)

we are guaranteed to have a solution of the original differential equation satisfying

the boundary condition at . Thus the middle expression in the last

line of Eq. (14.1) can be thought of as the general solution, while the last expression

is the particular solution satisfying the specific boundary conditions (and, perhaps,

other conditions).

Note that, to solve a differential equation by simple integration, it is necessary that we

can separate all the explicit dependence on the independent variable from the

dependence on the dependent variable. Thus for a first order linear homogeneous

equation of the form (where is a constant), we must first divide by y

before integrating, putting all of the y dependence on the LHS,

(14.3)

The final expression has used the boundary value at x = 0. This (necessary)

procedure of separating the variables on the two sides of the equation is called, no

surprise, separation of variables and equations that can be solved this way are called

separable. The most general separable linear first order equation looks like

0y y x0x x

y y

ln

0 .

y x x

x

dyy x y x dx

y

dydx y x x C

y

y x y e


, which we can solve using the above ideas. For the homogenous

case we can proceed as above (but now with not a constant),

(14.4)

As expected there is a single arbitrary constant (of integration), which we use here to

fit the boundary condition at x = 0. Next we return to the original inhomogeneous

problem. This is most easily addressed if we consider the quantity and

perform some manipulations. In particular, we find

(14.5)

We recognize the second term on the right-hand-side as the (previously found)

solution to the homogeneous equation (including the boundary condition), while the

first term is a particular solution to the inhomogeneous problem.

A general non(easily)-separable version of the first order equation can be written in

the form (note the dependence on both x and y, i.e., the P and Q functions are

different from above)

(14.6)

This expression is written in a form so as to remind us of the 2-D Green’s theorem

and the curl-free theorems of Chapter 6 in Boas. In the case that (the

curl-free case) we must be able to find another function such that

, ,, , ,

0.

F x y F x yP QP x y Q x y

y x x y

Pdx Qdy dF

(14.7)

y P x y Q x

0Q

00 0 0 .

x

dxP xdxP x I xy P x y y x Ce y e y e

I xy x e

0 0

0 : .

I x I x I x I x

I x I x

x xI x I x I x

dy x e e y yI e y yP x e Q x

dx

y x e dx e Q x C

y x e dx e Q x y e I x dx P x

,, , 0.

,

P x yy x P x y dx Q x y dy

Q x y

P y Q x

,F x y


In this case the original differential equation is (at least implicitly) solved by

(14.8)

This is a more powerful result than it may first seem because, even if it is not initially

true that , we may be able to find a factor to multiply by in Eq. (14.6) so

that the new functions, P and Q , do satisfy this relation. For example, consider

, , 0,

, , , , , , 0,

.

P QP x y dx Q x y dy

y x

f x y P x y dx f x y Q x y dy P x y dx Q x y dy

P Q

y x

(14.9)

The factor is called an integrating factor, which is just the role played by

above.

ASIDE: To see this connection explicitly we put the equation from Eq. (14.5) in

the language of Eq. (14.6). We have

,0 ,

, 1

, ,0.

P x y yP x Q xdy yP x Q x dx

Q x y

P x y Q x yP x

y x

Hence we need an integrating factor, which is just the familiar exponential factor.

We then have

0

,

,

.

I xx

I x

I x

P x y e yP x Q xI x dxP x

Q x y e

P QP x e

y x

, constant.F x y

P y Q x

,f x y I x

e


Thus the corresponding (potential) function F(x,y) is given by

0

0

, ,

, constant 0 0 .

I x

xI x I x

I x

xI x I x I x

FP e yP x Q x

xF x y e y dxQ x e

FQ e

y

F x y y y x y e e dxQ x e

As expected, we obtain again our previous result.

As in Boas consider the example equation

0 , 1 1.P Q

xdy ydx P y Q xy x

(14.10)

If we choose an integrating factor 21f x , we have

2 2

1 1,

, ,

0.

y P QP Q

x x y x x

yF x y c y x cx

x

xdy ydx xcdx cxdx

(14.11)

As discussed in Boas, equations which do not fall into one of the above simple forms

can often be put in separable, integrable form by an appropriate change of variable.

To illustrate these techniques let us consider familiar problems from mechanics.

Note that, while we typically think of mechanics as described by 2nd

order equations

(i.e., Newton), we can often use conserved quantities, i.e., symmetries, to obtain a 1st

order differential equation. Consider 1-D motion in a uniform (vertical) gravitational

field (motion near the surface of the earth), which is a conservative system with

constant (conserved) total energy (z is the dependent variable and t is the independent

one),


2 2

.2

TOT TOT

m dzE T V z mgz z E mgz

dt m (14.12)

For a given constant total energy we can solve this problem (as above) by separation

of variables. We have

0

0 ,2 2

z

zTOT TOT

dz dzdt t t

E mgz E mgzm m

(14.13)

where we have defined 0 0z t z . Integrating the LHS we find

0 0

0 0

1 2

2

1 2 2.

zz

TOT

z zTOT

TOT TOT

dzE mgz

g mE mgz

m

E mgz E mgz t tg m m

(14.14)

We can simplify this expression if we evaluate the total energy at 0t and define the

velocity at that time to be 0 0z t z , 2

0 02TOTE mz mgz . Thus we have

2

0 0 0 0

0

2 22 2 TOT TOTz mgz mgz E mgz z E mgz

m m

g t t

22

0 0 0 0

22 2

0 0 0 0

2

0 0 0 0

22

2

.2

TOTE mgz z g z z z g t tm

z gz t t g t t

gz z z t t t t

(14.15)


This final result should be familiar from introductory physics. We can, of course,

obtain it directly by integrating the second order Newton equation using separation of

variables twice,

0 0

0 0

0 0

2

0 0 0 0 0 0 .2

z t

z t

z t

z t

dzmz m mg dz g dt z z g t t

dt

gdz z z dt z g t t z t t t t

(14.16)

ASIDE: We can also use similar methods for motion in the 3-D, conservative

gravitation potential of the form V r GMm r in spherical coordinates (where G

is Newton’s gravitational constant, M is the mass of the earth and m is the mass of

the test particle). For purely radial motion ( 0 ) conservation of energy now

yields

2

2

2 2.

TOT

TOT

m GMmE r

r

GM Er

r m

Again this equation can be separated and integrated (although with a bit more effort

than in 1-D). But note that we can already see the general structure of the results.

If 0TOTE , then any trajectory with 0 0r t (the plus sign above) will continue

to have 0r t even as r , i.e., the mass escapes from the earth. For 0TOTE

(which is now possible) and 0 0r t there will be a turning point where 0r ,

TOTr GMm E , and the mass subsequently falls back towards the earth. For the

intermediate situation 0TOTE , the mass “just” escapes from the earth with 0r

as and t r . For more general motion (non-zero angular motion) we can still

simplify the problem by using the fact that angular momentum is conserved in a

central potential.

As a final comment on first order equations, we note that nonlinear equations like 21y y can exhibit both general solutions with an arbitrary constant, e.g.,


0siny x x , and also singular (isolated) solutions like 1y that do not

correspond to a value of the constant of integration 0x . The singular solutions form

the tangent (boundary) to the general solutions.

Returning to the general issue of differential equations in physics, as noted above we

are typically interested in 2nd

order linear differential equations, including those that

cannot be reduced to a 1st

order form. We will focus first on systems with only a

single independent variable, usually time, where only ordinary derivatives occur

(with respect to the independent variable). Further we will deal first with the general

behavior of solutions to 2nd

order linear ordinary differential equations with

coefficients that are constants. A simple example is 1-D motion in gravity as in Eq.

(14.16). This situation becomes somewhat more interesting if we add viscous

(velocity-dependent) damping (due to air resistance),

.mz bz mg (14.17)

We can often proceed by treating the 2nd

order equation as a 1st

order equation. With

the change of variables z v we obtain

b

v v gm

(14.18)

This is an equation amenable to an integrating factor as in Eq. (14.5) above. We

identify P b m and Q g . Thus we find

0 ,

I I

bI t t t

m

dve ge

dt

0

0

0

0

0 0 0

0

01 .

tI t t t b m I t

t

tI t t t b m I t

t

I t I t

v t ge dt e v e v t v

mge e v e

b

mge v e

b

(14.19)


Hence, if we release an object at rest, 0 0v , the large time (

0, 0I t

t t e

) or

terminal velocity is just v mg b and the 2 exponentials have gone to zero.

Due to the damping the falling object reaches a maximum velocity. The result is, in

fact, true for any initial velocity, 0

I tv t v v v e

. To find z t we can

integrate once more.

As mentioned in previous lectures (recall especially Lecture 5), more familiar

examples of 2nd

order linear differential equations (with constant coefficients) arise in

the study of electric circuits and the motion of a mass on a spring in 1-D (the

harmonic oscillator), including damping. By the rule of Feynman, that the same

equations have the same solutions, we can study the general case and then customize

the coefficients to match the specific system. Due to the importance of these systems

we will (once more) discuss the solutions systematically and with full generality. The

simplest and most familiar case is the homogeneous case with no “right-hand-side”

for the differential equation, i.e., no external driving force or voltage. Taking x to

label the dependent variable or coordinate (the location of the mass, or the charge on

a capacitor), the general form of the equation of interest is

(14.20)

where are know constants (> 0), which we take to be real for now, and

, etc. As we have previously discussed, we make use of the simple behavior

of exponentials under the derivative operation and try the Ansatz . Then

Eq. (14.20) becomes

(14.21)

With D representing the derivative with respect to the independent variable we can

also write this equation in the notation of Boas,

(14.22)

0,ax bx cx

, ,a b c

x dx dt

0

tx t x e

2

0

2

2

1,2 2

0

0

.2 4

ta b c x e

a b c

b b c

a a a

1 2 0.ax bx cx a D D x


Thus the general solution to the homogenous equation, also called the complementary

solution, can be written

(14.23)

where are constants to be fit to the initial conditions, . Note that the

exponents can be complex, and that the will be pure imaginary if there is no

damping, . To make a connection to our previous discussions let us go through

this same argument with an explicitly complex Ansatz . The

complex analogue of Eq. (14.21) is

2

0

2

2

1,2 02

0

0

.2 4

i ta ib c z e

a ib c

b c bi i

a a a

(14.24)

In this case the complementary solution is written in the familiar form

0 0 0 0

0 0 0 0 0

Re cos

cos cos sin sin .

t i t t

c

t

x t z e e z e t

z e t t

(14.25)

where, as usual, , is the phase of the complex constant . Note, in particular, that

as long as 0 is real (and the exponential is complex), the operation of taking the

Real part includes the second solution with the 0 exponent and we don’t need to

explicitly include it to find the general solution. On the other hand, if 0 is

imaginary (and the exponentials are real) as in Eq. (14.23), we do need to include

both exponentials. In any case, we still have 2 (real) constants to be determined by

the initial conditions. We are just writing the same result in a different notation. We

can make the identifications

1 2

1 2 ,t t

cx t x e x e

1 2,x x 0 , 0x x

1,2 1,2

0b

0Re i tx t z e

0 0z


1 2

2

1 20 2

1 0 2 0

,2 2

,4 2

, .

b

a

c bi

a a

i i

(14.26)

In either notation the explicit form of the solution will depend on the values of the

(physics dependent) constants . So let us list the various possibilities.

- with the damping dominant, the are both real and negative,

while the “frequency” 0 is imaginary and smaller in magnitude than .

This is the “over-damped” case with the complementary solution exhibiting

only exponentially damped behavior as in Eq. (14.23).

2

1 2 02,

4 2

b c b

a a a - the critically damped case where the

complementary solution looks like 1 2

t

cx t x t x e (the reader is

encouraged to check that this is a solution for any value of the arbitrary

constants).

2 2

1,2 02 2,

4 2 4

b c b c bi i

a a a a a - complex conjugates, the under-

damped, oscillatory case with 0 0 0cost

cx t d e t , where the

constants to fit the initial conditions are now 0 0,d . Note that

2 2 2

0 0,c a , where is the natural frequency of the system

with no damping.

As a first look at the inhomogeneous problem with a “driving term”, consider the

general form . We base the general solution on the

(complementary) solution of the homogeneous problem plus a(ny) particular solution

of the inhomogeneous equation, . Thus our goal is now to

find particular solutions for the various possible forms of . Note these particular

, ,a b c

2

24

b c

a a

1,2

ax bx cx F t

c px t x t x t

F t


solutions are not unique, we can always add any amount of the solution of the

homogenous problem. So we consider some familiar forms for the driving function:

, a power series in t. Here we use the method of undetermined

coefficients to write the particular solution as a corresponding power series,

, and solve for the in terms of the and by equating

the coefficients of the powers of t on the two sides of the initial equation. For

example,

, ;

, ;

, ;

etc.

tF t e , an exponential, where 1 2, . With the Ansatz t

px t be we

find easily that 1 2

t

px t e a . Note that this works also if

i , i.e., a sinusoidal driving function, Re i tF t e . In this case the

maximum response occurs when c a , i.e., when you drive the system

near its natural frequency. To see this we note that the particular solution for a

sinusoidal driving function (in complex notation, 0

i t

pz t z e ) is given by

1 2 0 0

2 2 2 2 2

0

1

2 222 2 2 2

2 2

2, tan .

4

i t i t

p

i t i t

i t i

e ez t

a i i a i i

e e

a i a i

e

a

The general form of this solution (or more correctly its square) is called a

0

Nn

n

n

F t t

0

Nn

p n

n

x t b t

nb n , ,a b c

0N 0 0b c

1N 2

1 1 0 0 1,b c b c b c

2N 2 2 3

2 2 1 1 2 0 0 2 1 2, 2 , 2 2b c b c b c b c ac b c b c


Lorentzian shape and it arises so commonly that web pages are dedicated to it.

From this result we see that the driving force and the response are generally “out

of phase”, i.e., 0 .

The other question we want to answer is when is the magnitude of the response

largest? To answer this accurately we need to be careful because there are really

two questions. If we mean for a fixed system, i.e., fixed values of 0, , we ask

what value of the driving frequency yields the largest response, i.e., we want

to find when 0pz . In this case the response is largest, i.e., the

magnitude of the denominator is minimum, for 2 2 2 2 2

02 .

However, we could also ask what value of should we “tune” the system to in

order to obtain the largest response from a given driving frequency (this is

how radio tuners work). Here we want 0pz and the answer is . In

both cases the large response is labeled a resonance.

tF t e , but with 1 (or

2 ). We must proceed carefully as we did

for the degenerate homogeneous case. The trick again is to add an extra power

of t and try the Ansatz t

px t bt e . Then we find 1

1 2

t

px t te a .

tF t e , but with 1 2 . We must proceed even more carefully and

try 2 t

px t bt e . Then we find 12 2t

px t t e a . The reader is encouraged

to check this result.

0

Nt n

n

n

F t e t

, where we combine our previous results. We can write a

particular solution in terms of undetermined coefficients, 0

Nt n

p n

n

x t e b t

, and

solve for the coefficients term by term.

The underlying idea is this last example is the every important concept of linear

superposition, which is always applicable to a linear equation. If the right-hand-side

of the inhomogeneous equation is a sum of terms, we find the particular solution as a

sum of particular solutions for each of the individual driving terms and sum them up.

In the next lectures we will look at periodic driving terms in more detail, i.e., use

Fourier series techniques.

Lecture 14: Introduction to (Review of) Differential Equations [See Chapter 8,courses.washington.edu/ph227814/228/ellis/Phys_228_09... · 2014-07-29 · Physics 228 Lecture 14 1 Winter

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