higher order equations

7/31/2019 Higher Order Equations

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Higher order equations Second order linear equations

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4 Higher order equations

4.1 Second order linear equationsThe general second order linear ordinary differential equation may be written as

( ) ( ) ( ) ( )2

2

d y dy R t P t Q t y F t

dt dt + + = .

This is linear in y (and its derivatives). If F = 0 the equation is homogeneous ; if F 0 the equationis inhomogeneous.

Assume that R(t ) does not vanish, then can rewrite as2

2

d y dy p qy f

dt dt

+ + =

where p(t ) = P / R, q(t ) = Q/ R and f (t ) = F / R.

End of Lecture 12

4.1.1 R EDUCTION TO FIRST ORDER SYSTEM By defining u1 = y and u2 = dy/dt we may rewrite the second order equation as a system of firstorder equations:

12 1

duu g

dt = ,

22

2 12

d y dy du p qy pu qu f

dt dt dt + + = + + =

2 2 1 2du

pu qu f g dt

= + .

Define the vectors u = (u1, u2) and g = ( g 1, g 2), then

d dt

=u g .

To be able to solve this, we also require initial or boundary conditions. As we have two first order equations, we require two initial/boundary conditions. Note that the coupling allows them to both beon the same variable and/or at different times. For example, we could have u1(0) and u2(0), or u1(0)and u1(T ), or u1(0) and u2(T ).

As with the first order linear equations, we shall see that the linearity allows us to combine the particular integral solution of the inhomogeneous equations, with the complementary functionsolution of the homogeneous equation. Here, however, there will be two complementary functions,reflecting the second-order nature of the equation.

4.2 Equations with constant coefficients

Suppose p and q are independent of t .


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Higher order equations Equations with constant coefficients

91

An example of this is a mass m on a spring (spring constant k ), optionally subject to a viscousfriction (drag, 2 ):

This gives for unit mass and external forcing f

2 y y ky f + + =## #

Note: In lectures I used k as the spring constant in this bit (as above), but later in 4.3 I switch tousing k 2 for the spring constant. For completeness, I have included in these notes the material bothas lectured (normal text), and how it would have been if I had used k 2 throughout for the springconstant (text with violet background).

22 y y k y f + + =## #

4.2.1 HOMOGENEOUS EQUATION For f = 0 (the unforced or freely oscillating case)

2 0 y y ky + + =## # (*)22 0 y y k y + + =## #

Recalling that t t d

e edt

= and2

22

t t d e edt

= suggests that y = A e t has the correct functional

form to satisfy (*).

Substituting A ( 2 + 2 + k )e t = 0, A ( 2 + 2 + k )e t = 0

which must hold t . Hence either A = 0 or

2 + 2 + k = 0 (**)

2 + 2 + k 2 = 0

with arbitrary A. This is the characteristic equation.

Solving 2 k = 2 2k = .

There are two possible solutions to the characteristic equation because this is a second order equation. This leads to two solutions to the homogeneous equation the complementary functions .Any linear combination of these will also be a solution, hence

( ) ( )

( )

2 2

2 2

k t k t

k t k t

y Ae Be

e Ae Be

+

= +

= +.

( ) ( )

( )

2 2 2 2

2 2 2 2

k t k t

k t k t

y Ae Be

e Ae Be

+

= +

= +.

This linear combination is called the principle of superposition and applies to all linear equations.


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Clearly the two solutions coincide when 2 = k 2 = k 2. We shall return to this situation later.

When 2 < k the square root is imaginary. Since we have not restricted the constants A and B to bereal, we may then write

( ) ( )( )2 2 cos sint y e A k t B k t = + .

( ) ( )( )2 2 2 2 cos sint y e A k t B k t = + .4.2.2 PRINCIPLE OF SUPERPOSITION Define the second order linear differential operator L as

( ) ( )2

2

d d L p x q x

dx dx

+ +

,

then L[u] Lu = u + p u + q u . This provides us with a compact notation for the differentialequation.

Suppose L[u1] = 0 and L[u2] = 0, then the principle of superposition states that L[ Au1 + Bu2] = 0 for arbitrary constants A, B. Proof is by straight forward substitution.

4.2.3 I NITIAL VALUE PROBLEMS The two boundary conditions are applied at the same (initial) value of x.

Suppose u = Au1 + Bu2 where L[u1] = 0 and L[u2] = 0 and u1 and u2 are linearly independent ( i.e. they are the complementary function of L[u] = 0).

If we require u(0) = u0 and u (0) = v0, we therefore have the linear system A u1(0) + B u2(0) = u0

A u 1(0) + B u 2(0) = v0.

This has a nontrivial solution provided the determinant

( ) ( )( ) ( )

1 2

1 2

0 0

0 0

u u

u u

does not vanish. If the determinant does vanish, then u1 and u2 are not (linearly) independent.

4.2.4 LINEAR INDEPENDENCE Two functions u1( x) and u2( x) are linearly dependent in an interval I if constants k 1, k 2 0 suchthat

k 1 u1( x) + k 2 u2( x) = 0

x I . If they are not linearly dependent, then they are linearly independent .

Theorem:

Suppose u1( x) and u2( x) are differentiable functions in the interval I . Then if the determinant

( )( ) ( )( ) ( )

1 21 2

1 2,

u x u xW u u u x u x=


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does not vanish at any x = x0 in I , then u1 and u2 are linearly independent in I . Further, if u1 and u2 are linearly dependent in I then W = 0 x I . This determinant is known as the Wronskian .

Proof: Suppose there exits k 1, k 2 such that k 1 u1( x) + k 2 u2( x) = 0 in I , then we also havek 1 u1 ( x) + k 2 u2 ( x) = 0 and substitution into the determinant W proves the result.

Note that we have proven that if the functions are linearly dependent, then W = 0. We have notshown that W = 0 implies the functions are linearly dependent.

Note there is a similarity here with linear independence in vector algebra.

x , x ( )

dx /dx = x 1; dx /dx = x 1

( )1 1

1 1

1 1 x xW x x x x

+ +

= = =

which is nonzero (except at x = 0)

cos kx, sin kx (k 0)

( )2 2cos sin cos sin 0sin cos

kx kxW k kx kx k

k kx k kx= = + =

e x, e x ( )

( )

( )( )1 1

0

x x x x

x x

e e

W e ee e

+ += = = .

Warning: the term linear is often omitted from discussions about linear independence.

4.2.5 ABEL S THEOREM

If u1 and u2 satisfy L[u] u + p( x) u + q( x) u = 0, where p and q are continuous in interval x I ,then the Wronskian W (u1, u2) satisfies

( )0

p x dxW W e

=

where W 0 is a constant.

Consequently, when p remains finite, either W = 0 x in I , or W never vanishes in I .

Proof:

u2 (u1 + p u1 + q u1) = 0u1 (u2 + p u2 + q u2) = 0

Subtracting u1 u2 u2 u1 + p(u1 u2 u2 u1 ) = 0

0dW

pW dx

+ = since W = u1 u2 u2 u1

( )0 p x dx

W W e

=


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Note: this result carries over to higher order equations. For a nth order equation the Wronskian is thedeterminant of the nn matrix

( )( ) ( ) ( )

1 2

1 2

1 2

1 1 11 2

, , ,

n

n

n

n n nn

u u u

u u uW u u u

u u u

=

!

!!

% % & %

!

.

4.2.6 EXISTENCE AND UNIQUENESS THEOREMS We saw before that, given a set of initial conditions, we could uniquely determine the solution tothe homogeneous equation from two linearly independent solutions. Indeed, the solution is uniqueregardless of the choice of u1, u2 satisfying L[u] = 0, and two linearly independent u1, u2 alwaysexist. [This statement shall not be proven in this course.]

Consequently, if W (u1,u2) 0, thenu = A u1 + B u2

is the most general solution of L[u] = 0.

End of Lecture 13

4.2.7 HOMOGENEOUS EXAMPLES

y + 9 y = 0 subject to y(0) = 0, y (0) = 6

Characteristic equation 2 + 9 = 0

= 3i complex sinesoidalGeneral solution y = A cos 3 x + B sin 3 x A, B = const

Derivative y = 3 A sin 3 x + 3 B cos 3 x

At y(0) = 0 A = 0

y (0) = 6 3 B = 6 B = 2.

Hence y = 2 sin 3 x.

y + 4 y + y = 0 subject to y(0) = 1, y(1) = 0

Characteristic equation 2 + 4 + 1 = 0

2 4 1 2 3 = =

General solution ( )2 3 3t t t y e Ae Be = + y(0) = 1 A + B = 1

y(1) = 1 2 3 2 3 0 Ae Be + + =

Solving2 3

1

1 A

e

=

,2 3

2 3 1

e B

e=


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Hence3 2 3 3

2

2 3 1

t t t e e e y e

e

+=

.

y + 2 y + y = 0

Characteristic equation 2 + 2 + 1 = ( + 1) 2 = 0

= 1 double rootBut we need two complementary functions!

Consider y = ( A + Bt ) et

y = ( A B + Bt ) et , y = ( A 2 B + Bt ) et

Substitute ( A 2 B + Bt ) et 2( A B + Bt ) et + ( A + Bt ) et = 0.This is true for any A, B.

4.2.8 R

EPEATED ROOTS

Consider the equation

( ) ( )2

2 2 0d y dy

ydt dt

+ + + + =

The characteristic equation of this

2 + (2 + ) + ( + ) = ( + )( + + ) = 0

gives = and = ( + ), and so the general solution

y = A e t + B e ( + )t = ( A + B e t ) e t .

Now as 0 the two roots tend towards a repeated root and e t 1 t , thus the general solutiontends towards

y = [( A + B) B t ] e t = [C + Dt ] e t .

Note that as A, B are arbitrary at this point, we can have 0 but keep B finite. This means thatthe next term in the Taylor Series expansion of e t (proportional to 2) will vanish.

Another way of looking at the characteristic equation is that in general we may rewrite thedifferential equation in the form

( )2

2 0d y dy d d y ydt dt dt dt

+ + + = + + =

where = and = are the roots of the characteristic equation. The complementary functionse t and e t can then be seen individually to be solutions of the respective first-order equations

0d

ydt

+ =

and 0d

ydt

+ =

.

The inverses of these first-order operators are therefore give1

0 t d

y Aedt

= + = and1

0 t d

y Bedt

= + = . A, B constants


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Hence we may interpret the second order equation as the solution of 1

0 t d d

y Aedt dt

+ = + = . (*)

As seen in 2, the solution of this is the complementary function B e t plus the particular integral

t t A y e Ae

B

= =

,

thus the general solution y = A e t + B e t .

When = , (*) becomes1

0 t d d

y Aedt dt

+ = + =

which (as we saw in 2.1.2) yields a particular integral of the form t e t , and thus the generalsolution y = ( A + Bt ) e t .

By considering the Wronskian of e t and t e t we can show that the two complementary functionsare linearly independent.

4.2.9 PARTICULAR INTEGRAL As with the first order equation in 2.1.2, the inhomogeneous equation, L[u] = f , introduces the needfor a particular integral .

Suppose for linearly independent u1, u2

L[u1] = 0, L[u2] = 0 complementary functions

and is any solution of L[ ] = f principal integral

then u = A u1 + B u2 + general solution

and L[u] = L[u1] + L[u2] + L[ ] = 0 + 0 + f = f .

Moreover, u is in fact the most general solution of L[u] = f .

This may be demonstrated by supposing that is another solution of L[u] = f (i.e. L[ ] = f ). Hence

L[ ] = L[ ] L[ ] = f f = 0,

i.e. satisfies the homogeneous equation, so we must be able to write

= C u1 + D u2

so = + C u1 + D u2.

Trial solution

As with the first order inhomogeneous equation, we can pose trial solutions for the particular integral of the second order inhomogeneous equation.

y + y = 1

The homogeneous equation gives rise to the characteristic equation 2 + 1 = 0, yielding = i and complementary functions sin t and cos t .Consider the trial solution y = a + bt + ct 2, and substitute:


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2c + a + bt + ct 2 = 1

As this must hold for all t , we have the particular integral is y = 1, and the general solution is

y = A sin t + B cos t + 1.

Clearly, introducing any other powers of t to the trial solution would demonstrate they had zerocoefficients.

y 3 y y = t 2 + 2

The characteristic equation 2 3 1 = 0 gives = (3 13)/2 and exponentialcomplementary functions.

Try y = a + bt + ct 2, and substitute

2c 3(b + 2ct ) (a + bt + ct 2) = (2c a 3b) + ( 6c b)t + (c)t 2 = t 2 + 2Equating coefficients for different powers of t :

2c a 3b = 2 t 0

6c + b = 0 t 1

c = 1 t 2 Solving simultaneously

c = 1,

b = 6c = 6

a = 2c 3b 2 = 2 18 2 = 22and the general solution

y = Ae(3+ 13)t + Be(313) t t 2 + 6t 22. y + 2 y + y = sin t

Characteristic equation 2 + 2 + 1 = 0 = 1 (twice).

Trial solution y = a sin t + b cos t and substitute

2(a sin t + b cos t ) + 2 (a cos t b sin t ) + a sin t + b cos t = sin t Equating terms

a 2 2b + a = (1 2)a 2b = 1

b 2

+ 2 a + b = 2 a + (1 2)b = 0

Simultaneous solution gives

( )

2

22

1

1a

=+

,( )

22

2

1b

=+

and the general solution

( )( ) ( )

2

2 22 2

1 2sin cos

1 1

t y A Bt e t t

= + + + +

.

y + y = cos t


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Higher order equations Applications

98

Characteristic equation 2 + 1 = 0 = i Complementary functions sin t and cos t , so trial function cannot simply be a sin t + b cos t .Recall a similar problem in 2.1.2, which suggests we try y = a t sin t + b t cos t . Substituting

(2a cos t a t sin t 2b sin t b t cos t ) + a t sin t + b t cos t = cos t Equating terms

at 2b + at = 2b = 02a b t + bt = 1

a = , b = 0

y = A sin t + B cos t + t sin t .

y + 2 y + y = et

The characteristic equation yields a double root = 1, suggesting the solution to the

homogeneous problem is y = ( A + Bt ) et

. Hence the trial solution for the particular integral mustcontain more than just a et and b t et . Try ( a + bt + ct 2)et :

(a 2b+2c + (b4c)t + ct 2)et + 2( a+b + (b+2c)t ct 2)et + (a + bt + ct 2)et = et Equating coefficients

a2b+2c 2a+2b + a = 2c = 1

b4c 2b+4c + b = 0 = 0

c 2c + c = 0 = 0 Note that the last two do not tell us anything: this is no surprise, since we know that a and b are

arbitrary as they represent the homogeneous solution. c =

Thus the general solution y = ( A + Bt + t 2) et .

4.3 Applications

4.3.1 FREE OSCILLATORS As noted in 4.2, the behaviour of a mass on a spring may be modelled by a second-order linear ordinary differential equation.


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If y is the location of the mass, then its acceleration is a = d 2 y/dt 2, and we assume the force imparted by the spring is Ky, where K is the spring constant (force per unit extension). We shall also assumethat the mass is sitting on a lubricating film ( e.g. oil) or attached to a dashpot so that the friction is

proportional to the speed (this will be covered in more detail in the Dynamics course next term),giving drag force D = 2 J dy /dt , say. Following Newton, F = ma, where F is the net force, m themass and a the acceleration, hence

2

22dy d y

Ky J mdt dt

= .

If k 2 = K /m and = J /m, then2

22 2 0

d y dyk y

dt dt + + = .

WARNING: My notation here is not quite consistent with 4.2 where I had k rather than k 2 in frontof the y term.

We get an equation of the same form in a variety of situations, from clock pendulum to car suspension and washing machines.

The characteristic equation

2 + 2 + k 2 = 0 (**)

has solutions 2 2k = .

Undamped oscillationWhen = 0, then = ik , and the homogeneous problem has oscillatory solutions

y = A cos kt + B sin kt

with frequency k rad/s.

End of Lecture 14

Damped oscillation

When < k , the homogeneous problem has oscillatory solutions that decay with time

y = e t ( A cos t + B sin t ),


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where 2 = k 2 2. Note that the damping has reduced the frequency.

0.0 1.0 2.0 3.0 4.0 5.0

t/2pi

-1.0

-0.5

0.0

0.5

1.0

y

Critical damping The decay rate of the oscillation increases as increases and the frequency decreases until = k , atwhich point the system is said to be critically damped (oscillations have zero frequency).

When = k the characteristic equation gives a double root = = k and the general solution is

y = ( A + Bt ) e t .

Note: Car suspension is roughly critically damped. Too little damping will cause the car to bounce ,while too much damping will limit the suspensions ability to absorb shocks.

Over damping

When > k the system is over damped and there are no oscillations.

4.3.2 FORCED OSCILLATOR In the previous section we considered free oscillations, but of course something like a car or awashing machine has a continuous forcing. Suppose this forcing is sinusoidal in nature, then

22

2 2 sind y dy

k y t dt dt

+ + = .

This has the same set of homogeneous solutions as before, plus a particular integral. Try

y = a sin t + b cos t , 2(a sin t + b cos t ) + 2 (a cos t b sin t ) + k 2(a sin t + b cos t ) = sin t


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(k 2 2)a 2 b = 1

2 a + (k 2 2)b = 0

( ) ( )

2 2

2 22 2 2 2 2 2 2 2

2sin cos

4 4

k y t t

k k

= + +

Undamped solution

Noting that the particular integral is the steady (large time) solution, we can see that for = 0 thesolution tends towards the steady oscillation

2 2

1sin y t

k

=

,

which is singular if the forcing frequency matches the frequency of free oscillations, k . Thissituation is referred to as resonance.

0.0 1.0 2.0 3.0 4.0 5.0

omega/k

0.0

2.0

4.0

6.0

8.0

10.0

y

Note that as the forcing frequency comes closer and closer to the resonant frequency, the amplitudeof the particular integral increases dramatically.

Damped oscillator Damping prevents the amplitude from growing without bound as resonance is approached.


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0.0 1.0 2.0 3.0 4.0 5.0

omega/k

0.0

2.0

4.0

6.0

8.0

10.0

| y |

4.3.3 R ADIOACTIVE SEQUENCES

Suppose A and B are radioactive elements with decay constants and , respectively, and C isstable. The decay process is

A B + b,

B C + c.We shall not concern ourselves with the particles b, c emitted. Let X (t ), Y (t ) and Z (t ) be the

abundances of A, B and C , with initial values X 0, Y 0 and Z 0, respectively.dX

X dt

= , (1)

dY X Y

dt = , (2)

dZ Y

dt = (3)

Note that although Z depends on Y (and thence X ), neither X nor Y depend on Z , and so we can

consider the first two equations in isolation.

Differentiating (2)2

22

d Y dX dY dY X

dt dt dt dt = = (using dX X

dt = )

and eliminating X using (2)2

2

d Y dY dY Y

dt dt dt

= +

( )2

2 0d Y dY

Y dt dt

+ + + = .

Characteristic equation 2 + ( + ) + = 0

Y = H e t + K e t .


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At t = 0, Y = Y 0 H + K = Y 0.

What is the other condition we need to tie down H and K ?

As we know X 0, then we can determine dX /dt at t = 0, and so also know that

0 00t

dY X Y dt = = ,

Since t t dY

He Kedt

= , then H K = X 0 Y 0

( ) K = X 0 + ( )Y 0

0 0 K Y X

=

, 0 H X

=

so ( )00 t t t X Y Y e e e

= +

. (*)

Of course, we could have solved this in a different way, since X does not depend on Y :

dX X

dt = X = X 0 e

t

so 0t dY Y X e

dt + =

which we know gives the complementary function Y = Y 0 e t and we can seek a particular integral

of the form Y = a e t

to recover (*) once the initial condition is imposed. Note that if = then, as we have seen before, we have a degenerate case where we need tointroduce a solution of the form a t e t to the trial solution (or to the solution of the second order system).

The second approach is easier in this case, but the first approach, using an equation and itsderivative to help eliminate a variable, is often useful.

Solving a differential equation is only half the story: we also need to understand the solution. It isinstructive to rewrite the solution as

( )( )00 1 t t t X Y Y e e e = +

.

The first term is simply the decay of the quantity of B initially in the system.

The second term embodies the balance between A being converted to B, and B being converted toC .

Suppose ' , then B evolves faster than A, and for large time we have

0 t X X Y e

=


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Higher order equations Reduction of order

104

and we have Y move towards a quasi-equilibrium with X . [For short time we may consider X asapproximately constant.] Overall, there is a rapid initial response towards this quasi-equilibrium,followed by a slow run-down on timescale 1/ .

Suppose $ , then initial A will be converted to B faster than B can decay, thus

Y (Y 0 + X 0) e t .

Suppose = . As we have seen before, this will introduce terms of the form t e t and we find

Y = (Y 0 + X 0 t ) e t .

To determine Z , the quantity of C , we can solve

( )0 0 t dZ

Y Y X t e

dt

= = + Z (0) = Z 0

to find

( ) ( )0 00 1 1 1t t Y X Z Z e t e = + + .

4.4 Reduction of orderWe have seen that there is a relationship between the differential equation (with constantcoefficients) and the polynomial characteristic equation.

For a general polynomial equation of degree n, P n( ) = 0, then if we know one solution = 1, say,we can divide P n( ) by 1 exactly, and thus form a new polynomial P n-1( ) of degree n1.For linear ordinary differential equations we can undertake a similar process, as was illustrated in4.2.9. Suppose Ln[ y] = 0 is a linear ode of order n, then we may (at least in principle) divide theequation by a known solution to obtain a ode of order n1, i.e. Ln-1[ y ] = 0. If we can solve this for

y , then we can simply integrate the solution to find y. Note that we need not restrict attention toequations with constant coefficients.

Consider

L[ y] y + p( x) y + q( x) y = 0.Suppose u satisfies L[u] = 0 (hence u is a solution to the homogeneous equation). We then define anew function

v = u

v = u + u

v = u + 2 u + u

L[v] = (u + pu + qu) + u + 2 u + p u = 0.

Now the first term in brackets vanishes, so we may rewrite this as

2u

pu

=

.


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Higher order equations Generalisation to higher orders

105

Integrating ln 2ln u p dt = 2 p dxu e =

hence ( ) 2 x

p dx

u x e dx

=

so that ( ) ( ) ( ) ( )2 x

p dx

v u x x u x u x e dx

= = is a solution. y 5 y + 6 y = 0.

Characteristic equation 2 5 + 6 = ( 2)( 3)= 0 = 2, 3

Taking the e2 x solution,5

2 4 4 5

x

dx p dx x x x xu e e e e e e

= = = = ,

and integrating = e x

so v = u = e x

e2 x

= e3 x

is another solution.

End of Lecture 15

y + 2 k y + k 2 y = 0

Characteristic equation gives double root = k .

Take u = ekx as one solution, then2

2 2 2 2 0 1

x

k dx p dx kx kx kxu e e e e e e

= = = = = .

Integrating = x so the other complementary function is u = x ekx, and the general solution is

y = ( A + B t ) ekx

4.5 Generalisation to higher ordersThe techniques we have developed for second order odes may be applied to higher order equations.Suppose we have the nth order equation

L[ y] an y(n) + an-1 y(n1) + + a 1 y + a 0 y = 0 a i = const

This gives the polynomial characteristic equationa n n + a n-1 n

1 + + a1 + a0 y = 0

with roots (potentially complex) = 1, 2, n and complementary functions 1 2, , , n x x xe e e

being individual solutions of first order equations of the form y 1 y = 0.

Suppose one of the roots, k is repeated m times. From the discussion on repeated roots in 4.2.8, we might predict that this will produce a series of complementary functions of the form

( )2 10 1 2 1 k xmm A A x A x A x e + + + +! .Recall integrating factors from 2.3.2 that for y + qy = 0 then


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Higher order equations More on forcing

106

( )dy d I qy Iydx dx

+ = or ( )1d d q y Iy

dx I dx + =

,

where I is the integrating factor

q dx

I e+

= . Now for our repeated root k we have

0m

k k k k d d d d

y ydx dx dx dx

= =

! .

Noting that ( )1k d d

y Iydx I dx

=

, then

( )

( ) ( )

( )

( )

1

2 2 2

2

3 3

3

1

1 1 1

1

10

m m

k k

m m

k k

m

k

m

m

d d d y Iy

dx dx I dx

d d d d d I Iy Iy

dx I dx I dx dx I dx

d d Iy

dx I dx

d Iy

I dx

= = =

=

= =

Hence, integrating this mth order equation,

Iy = A0 + A1 x + A2 x2

+ + Am-1 xm1

,and for y k y = 0, the integrating factor is k x I e = , so

( )2 10 1 2 1 k xmm y A A x A x A x e = + + + +! ,as expected.

4.6 More on forcing

4.6.1 ELECTRICAL CIRCUITS

Here we consider electrical circuits that contain resistors, capacitors and inductors, connected by perfect conductors (wires; of course, real copper wire is not a perfect conductor).

Ohms law

The resistor is the most basic component of a circuit (other than a wire). The resistance R relates thevoltage V across the resistor to the current I that passes through it.


18/45


107

V = I R.

Current is measured in amps (A; coulombs per second; one coulomb contains 6.25 1018 electrons)and, by convention, flows from positive to negative. [In fact, the electrons flow from negative to

positive.] Voltage, sometimes referred to as potential , is measured in volts (V), and resistance inohms ( ). Resistors in series, with no intermediate connections, must each have the same current through

them, so the effective resistance is additive.

31 2

1 2 3 1 2 3

V V V V I R R R R R R

= = = =+ +

Resistors in parallel share the same voltage, with each contributing to the current.

1 2 31 2 3

V V V I I I I

R R R= + + = + + .

V

I

R3

I 1 I 2 I 3

R2 R1

I R1 R2 R3

V 1 V 2 V 3

V

V

I

R


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108

Capacitors

Capacitors (sometimes referred to as condensers ) consist of two conductors separated by aninsulator. The conductors accumulate a charge (electrons) in response to the application of a

potential (voltage) across them. The charge is created or dissipated by a current flowing into or out

of the capacitor, but no current passes from one conductor to the other.

The voltage across the capacitor lags behind the current:

dV I dt C = .

The capacitance, C is measured in Farads, representing charge Q (coulombs) per volt.

Inductors

Inductors typically consist of a coiled conductor that creates a magnetic field. The magnetic field isgenerated by the current passing through the coil, but a changing magnetic field in turn induces avoltage in the coil.

The current lags behind the voltage:

dI V L

dt = ,

and the inductance L is measured in Henry.

Simple circuit

Consider a simple circuit excited by an oscillating voltage:

V

I L

V

I C


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109

Suppose the circuit is excited by the voltage E = E 0 ei t [ E 0 is complex and we assume the actualvoltage is the real part of E 0 e

i t , thus the complex amplitude E 0 also contains the phaseinformation.]

Conservation of charge implies here that each of the components must also see the same current I .

( )2 1d I

V V dt C

=

3 2dI

V V Ldt

= ( )2

3 2 2

d d I V V L

dt dt =

V 4 V 3 = RI ( )4 3d dI

V V Rdt dt

=

Adding the contributions for the individual components, and noting that V 4 V 1 must be equal to E :

( )2

4 1 02 1 i t d d I dI V V L R I i E edt dt dt C = + + = ,

a second order linear ordinary differential equation with constant coefficients. Rearranging2

02

1 i t i E d I R dI I edt L dt LC L

+ + = .

We can see that the resistance R leads to damping, and that the frequency of the undamped free

oscillation is1

LC . More generally, the characteristic equation has solutions

21

2 2 R R L L LC

= ,

For the particular integral, try the form I = a ei t (a complex):

2 0i E R aa ia L LC L

+ + =

L

RC

E V 1

V 2 V 3

V 4

I


21/45


110

( )( )

( ) ( )

0 02

2

20

2 22

1 1

1

1

iE i CE a

R L LC i RC i LC L

E C RC i LC

LC RC

= = + +

+ =

+

The particular integral a ei t has the same form as we saw previously for the resonance of a mass ona spring,

( ) ( )2 2

2 22 2 2 2 2 2 2 2

2sin cos

4 4

k y t t

k k

= + +

except that the forcing has an additional factor of in it since it is the differential of E rather than E itself that is important in driving the circuit.

The modulus of the complex amplitude

( ) ( )( )( ) ( )

2 220

2 22

1

1

E C LC RC a

LC RC

=

+

1 10 6 2 10 6 3 10 6 4 10 6Frequency

24

68

1012

14

Amplitude a

Amplitude with frequency for L = 1 H, C = 1 F and R = 0.01, 0.1 and 1 .

The resonance peak is damped by the presence of the resistive element in the circuit.


22/45


111

0.00002 0.00004 0.00006 0.00008 0.0001Time t

-0.15

-0.1

-0.05

0.05

0.1

Current I

L = 10 H, C = 1 F, R = 10 , = 1 MHz.

Typical tuning circuit

This is the basic idea of the tuning circuit in a radio or television, however, the amplitude of theresonant signal is strongly affected by the resistance, R. A typical tuning circuit is a little morecomplex than the previous example. This may be illustrated by

Analysis is similar, but with

I R = V 4 V 3 ( )4 3d dI

V V Rdt dt

=

I = I 1 + I 2

( ) 13 1 I d

V V dt C

=

22 1

dI V V L

dt = ( )

22

2 1 2

d I d V V L

dt dt =

I 2 r = V 3 V 2 ( ) 23 2 dI d V V r dt dt =

L

R

C

E

V 1 V 3

V 4

I

r

I 2

I 1

V 2


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112

Eliminating V 2 23 1 2dI

V V L rI dt

= +

Noting that inductive and capacitative branches see the same voltage difference2

1 2 22

I d I dI L r C dt dt = +

Equating V 4 V 1 with E 24 1 2 0i t dI V V L rI RI E e

dt = + + =

Noting I = I 1 + I 2

( ) ( )

( ) ( )

22 2 2 2

2 1 2 2 2

22 2

2 02i t

dI dI d I dI L rI R I I L R r I RC L r

dt dt dt dt

d I dI LRC L rRC R r I E e

dt dt

+ + + = + + + +

= + + + + =

2

02 222

11 i t

E d I dI L r L r I e

dt RC dt C R RC + + + + =

Similar to before, but with

L R r

RC +

1 11

r C C R

+

00

iE E RC

.

The characteristic equation

2 1 1 0 L r

L r RC C R

+ + + + =

has solutions

21 1 4

12 2

L L L r r r

L RC L RC C R

= + + + .

Suppose R and r 0, then

0

1 1 42 2

1 12

12

L Lr

L RC L C

Lr i

L RC LC

Lr i

L RC

+ = +

== +

where 0 = ( LC ) is the oscillation in the L C part of the circuit, with little leakage into the resetof the circuit as R is large. The first term is small and represents the slow exponential decay, both


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113

due to the small resistance in series with the inductor (causing dissipation in the L C network) anddue to the small leakage through R.

For particular integral, again try a solution of the form I 2 = a ei t :

2 01 1 E L r

a L ia r a RC C R RC

+ + + + =

( )

2 0

2 20

2 201

22 2 2 21

11

11

E RCL L r a a L ia r a

r r RC C R i R RC L

E i RCL

= + + + + = + + +=

+

where 12 = 02 (1 + r / R) and = r / L + 1/ RC . The functional form of the dependence on is, of course, the same as before.

When r $ ( L/C ) and R ' ( L/C ) then $ 0 = 1/( LC ), and r / R $ 1, so 1 and the circuitwill oscillate at a resonant frequency close to 1/ ( LC ).

End of Lecture 16

4.6.2 CAR SUSPENSION Consider again our idealised car suspension

Let d be the equilibrium height of the car above the road surface, and d + be the instantaneousheight above the mean road level, so is the departure form the equilibrium location.

Excess upward force on car due to spring S = ( - z ).

Upward force on car due to dashpot ( )2 D J z = # # .Assuming wheel has no mass and the tyre is rigid,

M

d +

z Actual roadMean road level


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114

( ) ( )2 0 M J z z + + =## # # .[We also assume the wheel remains in contact with the road.] Let = J / M and 2 = / M , then

2 22 2 z z + + = +## # # .

As noted earlier in 4.3.2, an ideal car is critically damped, so = . [A real car typically has > to allow for future wear of the dashpot.]

Mounting a curb

Suppose at time t = 0 the car goes over a step in the road from z = 0 for t < 0 to z = h for t > 0.

If the car is travelling at a constant speed U , there is an equivalence between position and timethrough x = U t . Road surface may be described as

z = h H (t )

where H (t ) is the Heaviside step function (see 1.11). The vertical speed of this surface is therefore

( ) z h t =#

where (t ) is the Dirac delta function. Note that we are assuming here that car wheel movesvertically upward as it changes level, whereas the trajectory will be more gradual for a typicalwheel with a diameter much larger than the step height.

The equation therefore becomes

( ) ( )2 22 2 h t hH t + + = +## # .

Clearly, for t < 0 nothing happens and = 0. As car goes over curb, adjusts continuously: thespring supplies a finite force, although the dashpot provides an infinite one because z # is infinite

but # is finite.

For t > 0, we therefore have2 22 h + + =## # ,

with the initial condition (0) = 0. But need a second one, on #, say.

To get # , integrate equations over instant of crossing the step,

0

0

0dt +

= ,0

0

0dt +

= # because # is finite

h


26/45


27/45


116

2 4 6 8 10sigma t

0.2

0.4

0.6

0.8

1

xi h

If the car had been underdamped, < , then going over the curb or over a speed ramp will set uposcillations.

Suppose = , then for a curb taken at speed we have for, t > 0,2 2h + + =## # ,

giving the solution to the characteristic equation

( )2

2 1 32 4 2

i

= =

and the general solution

12

3 3cos sin

2 2t A t B t e

= + .

The initial condition (0) = 0 gives A = 0, while12

1 3 33 cos sin

2 2 2t B t t e

= #

must give #(0) = 2 kh = h, so23

B h= and the solution is

12

2 3sin

23t t e

h

=


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117

2 4 6 8 10sigma t

-1.5

-1

-0.5

0.5

1

1.5xi h

Response to an impulse with critical damping (upper) and half critical damping (lower).

Normal road surface

A normal road surface will have undulations over a variety of wave lengths. However, we canunderstand much by considering a road surface with simple sinusoidal undulations of constant wavelength and frequency. [The idea of superposition allows us to simply add the response for differentsurfaces. This links in with the idea of Fourier decomposition you will come across later in thetripos.]

Suppose z = h cos x = h cos Ut where U is the speed of the car and 2 / and h are the wavelength and amplitude of the undulationsof the road surface. For a car with critically damped suspension we therefore have

( )( )2 22 2 Re 2 i t z z h i e + + = + = +## # # where = U . For a car moving steadily, the initial transients will dye out and only the particular integral will remain. Trying a solution of the form a ei t leads to

( )2 22 2a ia a h i + + = +

( )( ) ( )( )22 22 2

2h i h ia

i i

+ += = + +

( )

( )

2 2 2 22

22 2 2 2

4

4

ha

+=

+


29/45

Higher order equations Non-constant coefficients

118

2 4 6 8 10

0.2

0.4

0.6

0.8

1

h

As , a 2 h/ .Driving very slowly means you go up and down over every bump. Driving fast minimises thevertical motion of the car (so long as your suspension does not bottom out ). The performance isgenerally worse at lower speeds.

Of course we need to worry about the tyres on a real car: both add to the suspension and damping(just as well for those of us who ride bikes without suspension as the flexing of the frame providesvery little damping).

4.7 Non-constant coefficientsConsider x2 u + (1 ) x u + u = 0 for , constant.Try a solution of the form u = x . Substitution leads to

[ ( 1) + (1 ) + ] x = 0.For a non-trivial solution, the expression in square brackets must vanish. This is the indicial equation . Rearranging this for

2 ( + ) + = ( )( ) = 0

shows that u = x and u = x are independent solutions to the differential equation (at least provided ).More generally, for any equation of the form

a 0 y + a 1 x y + a 2 x2 y + a 3 x

3 y + = 0 (*)

is said to be homogeneous in x and possesses solutions of the form y = x . Moreover, the values of are solutions to a polynomial equation, the indicial equation , analogous to the characteristicequation for equations with constant coefficients.

We may consider this differential equation in a different way by introducing a new independent

variable , say, such thatd d

xdx d

= . Thus if we rewrite (*) in the form

2 3

0 1 2 3 0d d d A y A x y A x y A x ydx dx dx + + + + =

!


30/45

Higher order equations Singular points

119

then2 3

0 1 2 32 3 0dy d y d y

A y A A Ad d d

+ + + + =!

the complementary functions for which are of the form y = e .

Noting thatd d

x dx d = dx

xd = = ln x, then y = e

= e ln x

= x

as before.

From this relationship between equations with constant coefficients and equations that arehomogeneous in x, we know how to handle cases such as repeated roots. For equations withconstant coefficients we found the solution to the homogeneous equation to be of the form

y = ( A + Bt + Ct 2 + ) e t .

Correspondingly, for equations that are homogeneous in x with repeated roots of the indicialequation, we will have solutions of the form

y = ( A + B ln x + C (ln x)2 +) x .

We shall return to this later.

End of Lecture 17

4.8 Singular pointsFor an equation of the form

( ) ( ) ( ) ( )2

2

d y dy R t P t Q t y F t

dt dt + + = .

singular points of the equation occur when R(t ) vanishes. More generally, a singular point occurs at points where the coefficient in front of the highest derivative vanishes.

Consider again x2 u + (1 ) x u + u = 0 for , constant.which we have seen gives the indicial equation

2 ( + ) + = ( )( ) = 0

and complementary functions u = x and u = x that are independent solutions to the differentialequation (at least provided ).Recall the Wronskian

( ) 11 1 x xW x x x

+

= = + ,

which vanishes at x = 0 if + > 1. Writing the equation as

2

10u u u

x x + + = ,

then we can see that this is consistent with Abels theorem ( 4.2.5) :

( ) ( )1

1 1 ln0 0 0

10

dx p dx x xW W e W e W e

W x

+

= = =

=


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120

where p = (1 )/ x. Hence W vanishes at x = 0 if + > 1 and p diverges at x = 0.

If and are not both positive integers, then at least one of x , x is not well behaved at x = 0 in thesense that some of its derivatives are unbounded and a Taylor Series does not exist for it. For

example, if < 0, then x

diverges, whereas if = 3/2, then the second derivative diverges at x = 0.Something special must happen at the singular point. In the neighbourhood of this the coefficient of the highest derivative is very small, so there is a possibility of either at least one of the solutions

behaving badly (such as u to keep x2 u finite to balance u if u 0 at x = 0), or if R( x) y does become small, the equation behaves as though it is an equation of lower order. In the latter case, the linearly independent solutions that existed away from the singular point may no longer belinearly independent near the singular point, as is shown by the Wronskian vanishing at the singular

point.

4.8.1 SERIES SOLUTION NEAR SINGULAR POINT

To establish what happens near a singular point we shall use a series solution. However, unlike the previous examples, we shall pose

0

nn

n

y x a x

== an 0

for an equation of the form

R( x) y + P ( x) y + Q( x) y = 0

in which P, Q and R are continuous and differentiable. Our choice of the form of the series solutionis motivated by the discussion in 4.4 where, given a solution u to L[ y] = 0, we sought a second

solution of the form v = u. Here we know that there is a solution of the form x

and we will nowseek a second solution.

As we will be equating coefficients for different powers of x, we will also need Taylor Seriesexpansions of P ( x), Q( x) and R( x) about the singular point. Our requirement that P , Q and R arecontinuous and differentiable means that this is possible (or at least can be made possible, e.g. if Q = x2 / ( x-a ) then multiply through by xa if analysing the solution about x = a). Hence, assumingthe singular point is x = 0,

P = P 0 + P 1 x + P 2 x2 + ,

Q = Q0 + Q1 x + Q2 x2

+ , R = R0 + R1 x + R2 x2 + .

L[ y] x2 y + 3 x y + (1+ x) y = 0 Note that P , Q and R are already power series in x, but the equation is not homogeneous in x.

Pose0

nn

n

y x a x

== ( ) 1

0

nn

n

dyn a x

dx

== + and ( )( )

22

20

1 nnn

d y x n n a x

dx

== + +

Substituting


32/45


121

( )( ) ( ) ( )

( )( ) ( ) ( )

2 2 1

0

0

1 3 1

1 3 1 0

n n nn n n

n

nn n n

n

x n n a x x n a x x a x

x x n n a n a x a

=

=

+ + + + + +

= + + + + + + =

Rearranging

( )( )( ) 10

2 1 0n n nn

x x n n a a

=

+ + + + + = a -1 0

Since xn+ are linearly independent n, then square brackets must vanish n.

For n = 0 [ ( +2)+1] an = ( +1) 2 a n = 0 since a -1 0

= 1

and for n > 0

( )( )

1

2 1

nn

aa

n n

=

+ + + +

Noting that this is only valid when = 1, we have

( )( )1 1

21 1 1n n

n

a aa

n n n = =

+ +

[Note that the n2 in the demoninator ensures rapid convergence when x = O(1)].

a0 arbitrary (from boundary conditions)

a1 = a0 a

2= a

0/22 = a

0

( )0 0

3 22 22 3 3!

a aa

= =

( )( )

02

0

1!

nn

n

a y

x n

==

We now have one solution (complementary function) for the equation, but since the equation issecond order there must be a second solution.

There is more than one approach, but we shall use what we know about reduction of order in4.4. We proceed by rewriting the equation in the form

y + 3 x1 y + (1+ x) x2 y = 0.

Let u = x1 (1 x + x2/4 ) be the known solution, then y = v u is another solution, where

2 p dxu e =

and p = 3/ x. Hence


33/45


122

( )

22 3ln 2 3 3

22

1

11

41

1 2 2

x xu e u x x

x x

x x x

= = = +

= + + = + +

!

! !

using binomial expansion

Integrating = ln x + 2 x +

Thus y = u ln x + x power series [Could equivalently write y = u (ln x + power series ).]

L[ y] x2 y + x y + ( x 2) y = 0

Pose0

nn

n

y x a x

== ( ) 1

0

nn

n

dyn a x

dx

== + and ( )( )

22

20

1 nnn

d y x n n a x

dx

== + +

and substitute

( )( ) ( )

( )( ) ( )

( )

2 2 1 2

0

2 1

0

2 21

0

1

1

n n n nn n n n

n

n nn n

n

nn n

n

x n n a x x n a x a x xa x

x n n n a x a x

x n a a x

=

+

=

=

+ + + + +

= + + + + +

= + +

a -1 = 0

For n = 0 get indicial equation 2 = 2 = .

n > 0( )

12 2

nn

aan

=+

Have two sets of solutions. For = + ,

( )1

2n

n

aa

n n =

+

If is integer, then a 0 = arbitrary from boundary conditions

01

1 2

aa

=+ ,

( ) ( )( )

( )0

2 0

2 !11 2 2 2 2 2! 2 2 !

aa a

= =

+ + + ,

( )( ) 0

2 !

! 2 !na a

n n

=

+

and( )

( )0

0

2 !

! 2 !n

n

y a xn n

+

==

+

[Note: Factorials can be generalised for non-integer .]


34/45


123

For the second solution to the indicial equation we had = . Hence

( ) ( )1 12 2 2

n nn

a aa

n nn = =

+ .

But if 2 is integer, then the denominator vanishes at n = 2 and a 2 =

Thus we can see asimple series expansion does not work!

4.8.2 FROBENIUS

Not examinable!

There is a way around this: the method of Frobenius . However, this is not an examinable part of this course. The idea is that if = 1 is a repeated root of the indicial equation giving a solution

y = u = u( x; ) at = 1 that satisfies L[u] = 0, then L[u] must also be stationary with respect to

at = 1, so we also have1

0

u

L =

= .

Recall earlier that we found solutions of the form x and x ln x when we had a repeated root. We

could have predicted this from the above discussion as ln lnln ln x x x e x e x x

= = = .

4.8.3 LEGENDRE S EQUATION Consider

(1 x2) y 2 x y + ( + 1) y = 0

for 1 x 1. This represents axisymmetric oscillations of a sphere when x = cos ( 0 1).Clearly there are singular points at x = 1 (corresponding to = 0, , the poles). We are interestedin the solution between the singularities.

To study the equation near the singular points it is convenient to introduce = x + 1, whichtransforms the equation to

(2 ) y + 2(1 ) y + ( +1) y = 0

and pose0

nn

n

y a

== a 0 0.

End of Lecture 18

Substitute into the equation

( )( )( ) ( )( ) ( )

( )( ) ( )( ) ( ) ( ) ( )

( ) ( )( ) ( )

2 1

0

1 10

2

10

2 1 2 1 1

2 1 1 2 1 2 1

2 1 1 1

n n nn n n

n

nn n n n n

n

n

n nn

n n a n a a

n n a n n a n a n a a

n a n n a

=

++ +

=

+

+=

+ + + + + +

= + + + + + + + + + + +

= + + + + + + +


35/45


124

with a -1 = 0. Setting n = 1 gives the indicial equation

2 2 a0 = 0,

giving = 0 (double root).

The recurrence relation

( ) ( )( )

1 2

1 1

2 1n n

n na a

n

+

+ +=

+.

However, as n , we have a n+1/a n and so the series posed for y will diverge (at least for sufficiently large x) unless it terminates. We thus need to be a non-negative integer so that seriesgives a +1 = 0. The solution y is therefore a polynomial of degree .

4.8.4 IRREGULAR SINGULAR POINTS

Not examinable! Not all singular points are this simple. Here we are looking at a regular singular point ; solutions inthe neighbourhood of a regular singular point never diverge faster than a power of x. Singularitiesthat diverge faster than a power of x are essential singularities and occur in the neighbourhood of anirregular singular point .

x3 y + x y + y = 0

Pose0

nn

n

y x a x

== an 0

( )( ) ( )

( )( ) ( )( )

3 2 1

0

10

1

1 2 0

n n nn n n

n

nn n

n

x n n a x x n a x a x

n n a n a x

=

+

=

+ + + + +

= + + + + + =

a -1 = 0

n = 0 + = 0

so = / . Although this is a second order equation, we appear to have only one root and theindicial equation is linear rather than quadratic

( )( )( ) 1 1

1 21 2

n n n

n nn n

a a an n

+ + = =+ + +

Assuming / is not a positive integer, then anan-1 ~ n as n , so series diverges rapidlyif it does not terminate. Hence, whenever x 0 the solution diverges; it has a zero radius of convergence and the series solution is not valid anywhere. (This does not necessarily mean thereis no valid solution to the equation.)

Could try solutions of a different form in x3 y + x y + y = 0. Here try

y = e y = e and y = ( + 2) e

x2 ( + 2) e + x e + e = 0


36/45

Higher order equations Particular integrals

125

hence + 2 + x2 + x3 = 0.

The second and third terms suggest = / x2 as an appropriate form. Substitute

2 / x3 +( )/ x4 + / x3 = 0

which is dominated by the second term as x 0, hence ~ / x, and y = e / x as x 0.

4.8.5 CONDITIONS FOR SINGULAR POINT TO BE REGULAR For the equation

R( x) y + P ( x) y + Q( x) y = 0

singular points occur at x = x0 where R( x) = 0. For this singular point to be regular, we require both

( ) ( )( )0 0lim

x x

P x x x

R x

and ( ) ( )( )02

0lim x xQ x

x x R x

to be finite. This means that R/ P and R/Q must not tend to zero faster than they do in an equationthat is homogeneous in x at that singular point.

Thus in xr y + A x p y + B xq y = 0

we must have r p 1 and r q 2for the singular point to be regular at x = 0.

4.9 Particular integrals

As we have seen, the general solution to an inhomogeneous equation is the sum of thecomplementary function(s) (solutions to the homogeneous problem) and a particular integral thatsatisfies the inhomogeneous problem. Although the particular integral is not unique, such solutionsdiffer only by some linear combination of the complementary functions.

4.9.1 EQUATIONS THAT ARE HOMOGENEOUS IN x

Consider x2 u + (1 ) x u + u = 1 for , constant.

As we saw in 4.7, solutions have the form x , leading to the indicial equation

2 ( + ) + = ( )( ) = 0

which reveals that u = x and u = x are independent solutions to the differential equation (at least provided ).For the particular integral, as with the equations with constant coefficients, we try a solution of theform u = a + bx + cx2. Substituting

x2(2c) + (1 ) x(b+2cx) + (a+bx+cx2) = 1 ,

hence c = 0, b = 0 and a = 1/ . Hence the general solution is

1u Ax Bx

= + + .


37/45


126

In general, a polynomial right-hand side will require a polynomial form for the particular integral.The technique we have used here and previously is to pose a form with unknown parameters,substitute and then solve to find the unknown parameters.

Consider x2 u 2 x u + 2 u = x + x2,

the indicial equation suggests the complementary functions x and x2

. As with equations withconstant coefficients, we would then expect to have other terms in the principal integral. We mightthink of trying a + bx + cx2 + dx3 + ex4. Substituting

x2(2c+6dx+12 ex2) 2 x(b+2cx+3dx2+4ex3) + 2( a+bx+cx2+dx3+ex4) = x + x2.

Inspection shows that this requires a = 0, d = 0, e = 0, but cannot be satisfied for any values of b or c.

Recall that for y + 2 y + y = et in 4.2.9 we had complementary functions of the form ( A+ Bt )et and required a trial function of the form at 2 et . We can get some guidance here by recalling that for a double root in the indicial equation we needed to introduce a solution of the form ( A + B ln x) x ,

so we could try u = ax ln x + bx2

ln x. Now u = a ln x + a + 2bx ln x + bx,

u = a/ x + 2b ln x + 2b + b

Substituting

(ax + 2bx2 ln x +3bx2) 2(ax ln x + 2 bx2 ln x + ax + bx2) + 2( ax ln x + bx2 ln x) = x + x2 Equating different terms

x: a 2a = 1 a = 1

x2: 2b 2b + 2 b = 1 b =

x ln x 2a + 2a = 0 OK

x2 ln x 2b 4b + 2b = 0 OK hence the particular integral is

u = x ln x + x2 ln x and the general solution

u = Ax + Bx2 x ln x + x2 ln x.If the indicial equation had a double route, and the right-hand side had contained one of the

complementary equations, then we would have had to introduce terms of the form (ln x)2

x

, etc. This is no surprise given the equivalence between equations with constant coefficients and thosethat are homogeneous in the independent variable.

4.9.2 METHOD OF VARIATION OF PARAMETERS What do we do if our guess at a suitable form for the particular integral fails?

Consider the general second order linear ordinary differential equation

L[ y] y + py + qy = f ,where p, q and f are all functions of the independent variable x.


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127

Two known complementary functions

Let y = u1 and y = u2 are complementary functions solving the homogeneous problem ( L[u1] = 0, L[u2] = 0). We could then try solutions of the form

y = (t ) u1 + (t ) u2.

How do we choose (t ) and (t )? Noting that

y = u1 + u2 + u1 + u2

we make and (which are, at present, arbitrary) look like constants by requiring

u1 + u2 = 0

The second derivative is then

y = u1 + u2 + u1 + u2 .

Substituting into the ode,

y + py + qy = ( u1 + u2 + u1 + u2 ) + p( u1 + u2 ) + q( u1 + u2) = f

noting that (u1 + p u1 + q u1) = 0 and (u2 + p u2 + q u2) = 0, then we require

u1 + u2 = f .

Hence we have two simultaneous equations for and :

u1 + u2 = 0

u1 + u2 = f .

(u1 u2 u1 u2 ) = u2 f

2

1 2 1 2

u f dxu u u u

=

and 1

1 2 1 2

u f dx

u u u u = .

Of course the denominator here is just the Wronskian for this to make sense we must have u1 andu2 linearly independent.

Car driving up curb ( 4.6.2) : ( ) ( )2 22 2 h t hH t + + = +## #

Suspension critically damped, characteristic equation has double root and the complementaryfunctions are u1 = e t , u2 = t e

t . The Wronskian is

2 21

1t t t W e e

t

= =

,

and noting ( ) ( ) ( ) ( )0t

f s s ds f H t

= and ( ) ( ) ( ) ( )0

t t

f s H s ds H t f s ds

=


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128

then

( ) ( )( )

( ) ( )

( )( ) ( )

22

2

00 0

2

1 1

t s

s

t t t s s s

t

u f sedt h s H s ds

W e

hH t se ds hH t se e ds

h t e H t

= = +

= = = +

and

( ) ( )( )

( ) ( )( )

( )( )

12

0

2

2 2 1

1

t s

s

t s t

t

u f edt h s H s ds

W e

hH t e ds hH t e

hH t e

= = +

= + = + = +

Hence, the particular integral is( ) ( )( ) ( )( )( )( )

( ) ( )( )

1 2 1 1 1

1

1 1

t t t t

t t

t

u u hH t t e e hH t e te

hH t t e te t

hH t t e

= + = + + +

= + +

=

.

The general solution

= h H (t ) (1 (1 t )e t ) + ( A + Bt )e t ,

needs to be matched by initial conditions that = 0 for t = , hence A = B = 0 and the generalsolution is simply

= h H (t ) (1 (1 t )e t )which, apart from the presence of the step function, is the same as we found earlier for t > 0 in4.6.2.

End of Lecture 19

4.9.3 O NLY ONE COMPLEMENTARY FUNCTION KNOWN Consider the inhomogeneous equation

L[ y] = f . If you only know one solution, y = u, say, to the homogeneous equation L[ y] = 0, could search out asecond solution using the approach we used earlier ( 4.4) to find a second solution of the form

y = u. However, if we also need to find the particular integral, it might be better to combine thetwo steps.

Set y = u,

and seek such that L[ y] y + py + qy = f . Substituting,

u + 2 u + u + p( u + u) + q u = f .

Noting that (u + pu + qu) = 0, then2 u + u + p u = f ,


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Higher order equations Coupled systems

129

2u f

pu u

+ + =

,

which is a first order linear equation for for which we have already discussed solutiontechniques. Ultimately, is a double integral; the constants of integration lead to the general

solution.Car driving up curb, again: ( ) ( )2 22 2 h t hH t + + = +## #

Suppose we know one complementary function, u = e t ,

= u = e t ,

require( )

( ) ( )( )

2 2 2

2t

u p

u f

he t H t u

+ + = + + =

= = +

( ) ( )( ) ( ) ( )

( )( ) ( )( )0

2 2

2 1 1

t t t

t t

h e t H t dt h H t H t e dt b

hH t e b hH t e b

= + = + +

= + + = + +

( )( ) ( ) ( )

( ) ( )0

1 1

11

t t s

t

hH t e b dt hH t e ds bt c

hH t t e bt c

= + + = + + +

= + + +

so again ( )( ) ( ) ( )( ) ( )1 1 1t t t tu e hH t t e bt c hH t t e bt c e = = + + + = + + ,with the arbitrary constants of integration b and c disappearing as = 0 for t < 0.

4.10 Coupled systems

4.10.1 CONVERTING TO A COUPLED SYSTEM As we saw in 4.1.1, we may reduce a second order linear differential equation into a pair of coupled first order equations. For

y + py + qy = f ,

define z = y , then

y = z ,

z = f pz qy.We can do a similar thing for any order of linear ode: an nth order equation may be converted into asystem of n first order equations.

4.10.2 CONVERTING TO A SINGLE EQUATION

Conversely, any pair of coupled equations, such as


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130

y = ay + bz + f ,

z = cy + dz + g ,

can be converted into a second order equation for y, say, by rewriting the first equation as

y ay f z b

=

and substituting into the second

( ) ( ) ( )2 y ay a y f b y ay f b d

cy y ay f g b b

= + +

b ab b f

y a d y ad bc a y f df g b b b

+ + + + = + .

Obviously, we could do the same thing for z to give an equivalent second order equation for thatinstead.

4.10.3 SOLVING COUPLED SYSTEMS Consider

y + 4 y 24 z = 4 e x

z y + 2 z = e x.

Method 1

Use one of the equations to eliminate one variable, ending up with a second order system.

y = z + 2 z e x

y = z + 2 z e x

Substitute z + 2 z e x + 4( z + 2 z e x) 24 z = 4e x

z + 6 z 16 z = 9 e x

Characteristic equation gives = 8, 2, so complementary functions e8 x and e2 x. For particular integral try z = a e x

a + 6a 16a = 9 a = 1

so z = A e8 x + B e2 x e x.

and z = 8 A e8 x + 2 B e2 x e x

so y = z + 2 z e x

= -8 A e-8 x + 2 B e2 x e x + 2( A e8 x + B e2 x e x) e x

= 6 A e8 x + 4 B e2 x 4 e x.This is a simple, but cumbersome approach, particularly for equations where the coefficients arefunctions of x. It can, however, make finding the particular integrals more complex than necessary.

Method 2

Add a linear combination of the two equations. Here, add times the second equation to the first:


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131

1 y + 4 y 24 z = 4 e x

( z y + 2 z ) = e x

add ( y + z ) + (4 ) y + (24 + 2 ) z = (4+ )e x Rearrange

( ) ( ) ( )24 24 44

x y z y z e

++ + + = + .

So far has been arbitrary. We now choose such that the term in square brackets is equal to y + z . Thus

24 24

+=

2 2 24 = 0

which has solutions = 4, 6.For = 4:

u = y 4 z ,

and u + 8u = 0

u = A e8 x.

For = 6:

v = y + 6 z ,

and v 2v = 10 e x

v = B e2 x 10 e x.

We now have a pair of solutions for u and v. These are known as the normal modes of the system,each having just one term in the complementary function.

u: y 4 z = A e8 x

v: y + 6 z = B e2 x 10 e x

Solve simultaneously for y and z :

10 z = Ae8 x + Be2 x 10e x

z = ( A/10) e8 x + ( B/10) e2 x e x and y = (3 A/5)e8 x + (2 A/5) e2 x 4 x.

Note: The constants A and B are not the same as with the previous method.

Method 3: Matrix method

This is really a generalisation of the second method.

Write the system of first order linear equations in matrix form:

w Mw = b .For our present system


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132

y

z

=

w ,4 24

1 2

= M , and

4 x

x

e

e

=

b .

For the complementary functions, try w = w0 e x,

w0 e x

Mw 0 e x

= 0, Mw 0 = w0

and are the eigenvalues of M , with w0 the corresponding eigenvectors. Of course, the eigenvaluesmust satisfy

|M I | = 0.For the present example,

( )( ) ( )( )24 24

4 2 24 6 16 8 21 2

= = + = +

,

hence = 8, 2, as before. The corresponding eigenvectors are

01

6

1

=

w , 024

1

=

w

and the solution to homogeneous equation

8 2 8 201 02

6 4

1 1 x x x x y A e B e A e B e

z = + = +

w w .

For the particular integral, try w = a e x,

ae x Ma e x = b

1 1

2 2

4 24 5 24 4

1 2 1 3 1

a a

a a

= = I

a1 = 4, a2 = 1,giving the general solution

8 26 4 4

1 1 1 x x x y A e B e e

z = + +

.

This method is readily generalised to systems comprising a larger number of coupled equations. Itcan also be generalised to handle more complex right-hand sides.

4.10.4 COUPLED HIGHER ORDER SYSTEMS For a system of higher order equations, can either reduce to a larger system of first order equations,or extend the previous ideas.

y y 10 z = K cos 3 x,

z + 5 y + 14 z = cos 3 x,

subject to y, y , z , z = 0 at x = 0.

Using the second method, multiplying the second equation by and adding to the first


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Higher order equations Second order difference equations

133

( ) ( ) ( )10 141 5 cos31 5

y z y z K x

++ + + + = + .

Selecting so that term in square brackets has same form as first term

10 141 5

+= +

5 2 15 + 10 = 0

= 1, 2.

For = 1, have

u + 4u = ( K 1) cos 3 x

u = A cos 2 x + B sin 2 x (( K 1)/5) cos 3 x

and for = 2,

v +9v = ( K 2) cos 3 x

v = C cos 3 x + D sin 3 x + (( K 2)/6) x sin 3 x

Initial conditions give u, u , v, v = 0 at x = 0, so

u = (( K 1)/5) (cos 2 x cos 3 x),

v = (( K 2)/6) x sin 3 x which may be solved simultaneously for y and z :

( )( ) ( )2 11 cos 2 cos3 2 sin 35 6

y K x x K x x= ,

( )( ) ( )1 11 cos 2 cos3 2 sin 35 6

z K x x K x x= + .

4.11 Second order difference equations

Homogeneous equation

Consider the difference equation

yn+2 + yn+1 + yn = 0.

Noting the similarity to the second order ode y + qy + ry = 0 which has solutions of the form y = e x, try a solution of the form yn = an = e n ( = ln a):

(a2 + a + )a n = 0.

For a non-trivial solution,

( )21 42a = .Like the second order ode, the second order linear homogeneous difference equation has twosolutions.


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Higher order equations Second order difference equations

As with odes, if there are repeated roots, ( e.g. if 2 = 4 in the above), then try y = n a n for thesecond solution:

(n+2)a n+2 + (n+1)an+1 + (n/4) 2an = 0

(n+2)a 2 + (n+1) a + (n/4) 2 = 0

22 2

2 2

1 1 1 1 12 2 4 2 4 2

1 1 12 1 2

2 2 2 21 1 12 2

n n na

n n n

nn n n n

n nn

n

+ + = + + + += + + + +

+ = +

and so both a n and n a n are solutions when a = ./2.

Particular integral Consider yn+2 + yn+1 + yn = c.

Again there is an analogy with the ode y + qy + ry = f .

Try yn = A (1 + + ) A = c

A = c/(1 + + ).

If 1 + + = 0, then inspection will show that yn = const is a solution to the homogeneousequation. We can try instead yn = A n. If we have a repeated root, then would need to use yn = A n2,etc.

Reduction of order

In the same way as we can convert a second order linear ode into a pair of coupled first order linear odes, we can convert a second order difference equation into a system of coupled first order difference equations.

Consider yn+2 + yn+1 + yn = f n.

Let + = and = , then

yn+2 + yn+1 + ( yn+1 + yn) = f n.

Defining z n = yn+1 + yn, then we have the system yn+1 + yn = z n,

z n+1 + z n = f n.

End of Lecture 20

higher order equations

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