MATH 254
Lectures 16-21.
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2nd order, linear, constant coefficient equations with time dependent forcing
Method of undetermined coefficients
Use of SYSTEMS
Forced vibrations
Applications to mass-spring systems and to RLC circuits
Notion of resonance
Notion of beating
Method of variation of parameters.
1
Consider ode’s of the form
L(y) =
d2 y
dy
+ p(t)
+ q(t)y = g(t)
dt2
dt
If there is a coefficient in front of
d2 y
dt2 ,
(6.1)
divide it through the equation.
Let p(t), q(t), g(t), be continuous over some interval I. The general solution of
(6.1) in I is the sum of:
• the general solution of
L(y) =
d2 y
dy
+ p(t)
+ q(t)y = 0
dt2
dt
(6.2)
• any specific or particular solution of (6.1)
The former is often called the complementary solution; the latter is the particular
solution. Let Y1 (t), Y2 (t) be any two solutions of (6.1). Then (easy to prove)
Y1 (t) − Y2 (t) satisfies (6.2). Therefore, we may write the difference between any
two solutions of (6.1) as Ay1 (t) + By2 (t), where y1 (t) and y2 (t) are two linearly
independent solutions of (6.2).
Now let Y (t) be the general solution of (6.1). This means that Y (t) satisfies (6.1)
for all t ∈ I and satisfies two arbitrarily assigned initial conditions
Y (t0 ) = Y0 , Y 0 (t0 ) = Y00
at some point t0 ∈ I.
Let Y2 (t) be any particular solution of (6.1). It does not have to satisfy the initial
conditions Y (t0 ) = Y0 , Y 0 (t0 ) = Y00 . Then we may write
Y (t) = Y2 (t) + Ay1 (t) + By2 (t)
and we choose A, B to satisfy
Y0 = Y2 (t0 ) + Ay1 (t0 ) + By2 (t0 )
Y00 = Y20 (t0 ) + Ay10 (t0 ) + By20 (t0 )
Since y1 (t) and y2 (t) are linearly independent on I, these two linear algebraic
equations can be solved to give unique values of A and B. (Recall: y1 , y2 l.i.⇒
W (y1 , y2 ) = (y1 y20 − y2 y10 6= 0.) In what is to come, we write yp (t) for Y2 (t), the
particular solution of (6.1). We often denote the complementary solution, (6.2), by
yc (t).
Example:
1
1
1
y 00 + y 0 − 2 y = ,
t
t
t
y(1) = 0, y 0 (1) = 10.
2
Check that, t, 1t are l.i. solutions of y 00 + 1t y − t12 y = 0 on both the intervals (−∞, 0)
and (0, ∞) excluding t = 0.
Try the particular solution
yp = A1 t + A2 , A1 , A2 constant.
1
1
1
0 + A1 − 2 (A1 t + A2 ) = .
t
t
t
Equate powers of t: this doesn’t work. Try the particular solution
yp = A1 t loge t + A2 t + A3
y 0 = A1 loge t + A1 + A2
yp00 =
A1
t
+ 1t (A1 loge t + A1 + A2 ) −
A1
t
1
2
t (A1 t loge
t + A2 t + A3 ) = 1t .
⇒ A1 = 12 , A2 arbitrary, A3 = 0. yp = 21 t loge t + 3t is a particular solution; the
general solution is y = 12 t loge t + (A + 3)t + Bt . So
y(1) = 0 = A + 3 + B
y 0 (1) = 10 = 21 loge 1 +
1
2
+A+3
⇒
A+3=
19
19
t, B = −
2
2
and
1
19
19
y(t) + t loge t + t − .
2
2
2t
Note that the solution becomes discontinuous at t = 0 where p(t), q(t), and g(t) are
all discontinuous.
Exercise:
y 00 + 5y 0 + 4y = et , y(0) = 0, y 0 (0) = 1/10.
Step 1: Find the general solution of y 00 + 5y 0 + 4y = 0.
Step 2: Try yp = A1 et , substitute into yp00 + 5yp0 + 4yp = et and determine A1 .
4 −t
1 −4t
1 t
(Answer: y(t) = − 30
e + 30
e
+ 10
e .)
Exercise:
y 00 + y = 1, y(0) = 0, y 0 (0) = 1.
(Answer: y(t) = 1 + sin t.)
How to find yc (t) and yp (t)
In general, this is difficult and in most cases approximations (for example, expanding the solution in series form, numerical approximations of solutions, representations of solutions in integral form) must be made.
But, for a special and important class of cases, we can calculate yc (t) and yp (t)
explicitly. These are the cases where p(t) and q(t) are constants and g(t) ∈ C,
where C is the class of functions which are sums of terms which are of the form
tα eβt cos γt or tα eβt sin γt,
where α is a nonnegative integer and β, γ are real.
3
The Method of Undetermined Coefficients
y 00 + py 0 + qy = g(t).
(6.3)
Step 1: Determine two l.i. solutions y1 (t), y2 (t) of
y 00 + py 0 + qy = 0.
(6.4)
We know that y1 and y2 are in the class C.
Step 2: Derive the auxiliary set of l.i. functions {g(t)} obtained from g(t) by
repeated differentiation. This set is finite because g(t) is in the class C.
Example:
g(t) = t2 ⇒ {g(t)} = (t2 , t, 1).
g(t) = t sin t ⇒ {g(t)} = (t sin t, t cos t, sin t, cos t)
because g(t) = t sin t; g 0 (t) = sin t + t cos t; g 00 (t) = 2 cos t − t sin t; g 000 (t) = −3 sin t −
t cos t, and so on. All derivatives can be written as a linear combination of members
of {g(t)}.
Step 3: Find out if any member of the auxiliary set solves the homogeneous
equation (6.4). If not, do not modify the auxiliary set. If one does, multiply the
auxiliary set by tm , where m is the lowest integer such that no member of the new
auxiliary set solves (6.4). This is the modified auxiliary set {g(t)}.
Step 4: Seek yp (t) in the form of a linear combination of members of the auxiliary
(or, if it has been modified) the modified auxiliary set. Substitute this expression
into (6.3) and determine the unknown coefficients by using the linear independence
of the members of the auxiliary set.
Remark: If g(t) = g1 (t) + g2 (t) (e.g. g(t) = t2 + t sin t), determine yp (t) for g1 (t)
(i)
and g2 (t) separately and add. If yp (t) satisfies y 00 + py 0 + qy = gi (t) for i = 1, 2,
(2)
(1)
then yp = yp + yp satisfies y 00 + py 0 + qy = g1 + g2 .
Suppose that
{g(t)} = (φ1 (t), φ2 (t), . . . , φr (t)).
Then let
yp (t) = A1 φ1 (t) + . . . + Ar φr (t).
Substitute into (6.3) and equate coefficients of all the linearly independent functions.
We will discuss why these will be enough information to determine the r constants
A1 , . . . , Ar . You might give some thought to this and try to construct a proof that
this must be so.
4
But, first, you should do lots of examples.
Worked Examples (see HW on NSS pg. 211, #’s 1,7,9,13,29)
The best way to learn the method is to do lots of examples
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
(ix)
(x)
y 00 − 2y 0 − 3y = 3e2t ,
y 00 − 2y 0 − 3y = −3te−t ,
y 00 + 9y = t2 e3t + 6,
y 00 + ω02 y = cos(ωt),
y 00 + ω02 y = cos(ω0 t),
1
y 00 + py 0 + qy = m
F cos(ωt),
00
3
y =t ,
y 00 + y 0 = te−t ,
y 00 + y = t sin t,
y 00 + 2y 0 + 5y = 4e−t cos(2t),
y(0) = 0, y 0 (0) = 1.
y(0) = y 0 (0) = 0.
y(0) = y 0 (0) = 0.
ω 6= ω0 , y(0) = y 0 (0) = 0.
y(0) = y 0 (0) = 0.
p2 < 4q, y(0) = α, y 0 (0) = β.
y(0) = y 0 (0) = 1.
y(0) = y 0 (0) = 0.
y(0) = y 0 (0) = 0.
y(0) = 1, y 0 (0) = 0.
Example: (i) y 00 − 2y 0 − 3y = 3e2t , y(0) = 0, y 0 (0) = 1. First find yc . . .
Step 1: y = ert , r2 − 2r − 3 = 0 ⇒ r = 3, −1, so yc = Ae3t + Be−t .
Now find yp . . .
Step 2: Repeated differentiation of g(t) = 3e2t simply gives multiples of e2t , so
{g(t)} = (e2t ).
Step 3: The auxiliary set does not need modification.
Step 4: yp = A1 e2t ; yp00 − 2yp0 − 3yp = 3e2t .
So, (4A1 − 4A1 − 3A1 )e2t = 3e2t ⇒ A1 = −1.
Remark: You could have added any solution of y 00 − 2y 0 − 3y = 0 to yp and this
term would disappear because y100 + py10 + qy1 = 0.
The general solution is y(t) = Ae3t + Be−t − e2t . Using the initial conditions:
y(0) = 0 = A + B − 1 and y 0 (0) = 1 = 3A − B − 2 give A = 1, B = 0, so we have
the particular solution y = e3t − e2t .
Example: (ii) y 00 − 2y 0 − 3y = −3te−t , y(0) = y 0 (0) = 0.
Step 1: yc = Ae3t + Be−t .
Step 2: g(t) = −3te−t ; {g(t)} = (te−t , e−t ).
5
Step 3: The modified auxiliary set is {g(t)} = (t2 e−t , te−t ).
Remark: If you had sought solutions in the form yp = A1 te−t + A2 e−t , A2 would
disappear because (e−t )00 − 2(e−t )0 − 3(e−t ) = 0.
Step 4: Let
yp
=
A1 t2 e−t + A2 te−t ,
yp0
=
−A1 t2 e−t + (2A1 − A2 )te−t + A2 e−t ,
yp00
=
A1 t2 e−t + (−4A1 t + A2 t)e−t + (2A1 − 2A2 )e−t .
Then
yp00 − 2yp0 − 3yp = −3te−t
⇒ t2 e−t (A1 + 2A1 − 3A1 ) + te−t (−4A1 + A2 − 4A1 + 2A2 − 3A2 )
+e−t (2A1 − 2A2 − 2A2 ) = −3te−t ∀t.
Now t2 e−t , te−t , e−t are l.i. Therefore, the coefficients of each of these expressions
must vanish, so A1 + 2A1 − 3A1 = 0, this is true automatically.
−8A1
2A1 − 4A2
= −3 ⇒ A1 = 3/8
= 0 ⇒ A2 = 3/16.
y(t) = Ae3t + Be−t + 38 t2 e−t +
y(0) = 0 = A + B
y 0 (0) = 0 = 3A − B +
3
16
3
−t
16 te ,
3
⇒ A = −B = − 64
.
There is a unique solution:
y(t) = −
3 3t
3
3
3
e + e−t + t2 e−t + te−t .
64
64
8
16
Example: (iii) Do yourself.
y(t)
¡
1 2
18 t
−
1
27 t
¢
1
e3t + 32 ,
¡ 1 2162 1
¢ 3t
1
= A cos 3t + B sin 3t + 18 t − 27 t + 162
e
1
2
109
y(0) = 0 ⇒ 0 = A + 162 + 3 ⇒ A = − 162 .
1
1
y 0 (0) = 0 ⇒ 0 = 3B − 54
⇒ B = 162
.
yp (t) =
6
+
+ 32 ,
Example: (iv)
yc = A cos ω0 t + B sin ω0 t,
yp = A1 cos ωt + A2 sin ωt.
1
,
ω02 −ω 2
yp00 + ω02 yp = cos ωt ⇒ A1 =
1
ω02 −ω 2
y= A cos ω0 t + B sin ω0 t +
0
y = (0) = y (0) = 0 ⇒ A =
so y(t) =
A2 = 0.
cos ωt,
1
− ω2 −ω
2,B
0
=0
cos ωt − cos ω0 t
t sin ω0 t
, ⇒ lim y(t) =
(Use L’Hôpital’s Rule).
ω→ω0
ω02 − ω 2
2ω0
Example: (v) y 00 + ω02 y = cos ω0 t, y(0) = y 0 (0) = 0.
Step 1: Solve y 00 + ω02 y = 0.
y = ert ⇒ r2 + ω02 = 0 ⇒ r = ±iω0 , yc = A cos ω0 t + B sin ω0 t.
Step 2: g(t) = cos ω0 t ⇒ {g(t)} = (cos ω0 t, sin ω0 t).
Step 3: The modified auxiliary set is {g(t)} = (t cos ω0 t, t sin ω0 t).
Step 4: Let yp = A1 t cos ω0 t + A2 t sin ω0 t, so that
yp0 = A1 cos ω0 t − A1 ω0 t sin ω0 t + A2 sin ω0 t + A2 ω0 t cos ω0 t,
yp00 = −2A1 ω0 sin ω0 t − A1 ω02 cos ω0 t + 2A2 ω0 cos ω0 t − A2 ω02 sin ω0 t.
yp00 = ω02 yp = cos ω0 t ⇒ −2A1 ω0 sin ω0 t + 2A2 ω0 cos ω0 t = cos ω0 t.
But sin ω0 t and cos ω0 t are linearly independent, so A1 = 0, A2 =
1
2ω0 .
1
t sin ω0 t
2ω0
y(0) = 0 ⇒ A = 0; y 0 (0) = 0 ⇒ B = 0.
y(t) = A cos ω0 t + B sin ω0 t +
Therefore, y(t) =
1
2ω0 t sin ω0 t.
(Compare the previous exercise.)
Examine: (vi) This is a very important example. Work through it carefully. y 00 +
1
py 0 + qy = m
F cos(ωt), p2 < 4q, y(0) = α, y 0 (0) = β.
Step 1: The solution of the homogeneous equation y 00 + py 0 + qy = 0 for p2 < 4q is
Ã
!
r
r
p2
p2
−p
t
yc = e 2 A cos q − t + B sin q − t .
4
4
Step 2: g(t) =
1
mF
cos ωt, so {g(t)} = (cos ωt, sin ωt).
7
Step 3: The auxiliary set does not need to be changed.
Step 4: Let yp = A1 cos ωt + A2 sin ωt, ⇒ yp0 = −A1 ω sin ωt + A2 ω cos ωt,
⇒ yp00 = −A1 ω 2 cos ωt − A2 ω 2 sin ωt.
Then yp00 + pyp0 + qyp =
F
m
cos ωt, so
2
(−A1 ω + A1 q) cos ωt + A2 pω cos ωt − A1 pω sin ωt + (A2 q − A2 ω 2 ) sin ωt =
2
⇒ A1 (q − ω ) + pωA2 =
F
m
cos ωt,
F
m
⇒ −A1 pω + A2 (q − ω 2 ) = 0
⇒ A1 =
q
F
ω −ω
q
mω ( ω
−ω)2 +p2 , A2
µ
q
−pt/2
p2
4 t
y(t) = e
A cos q −
³ qm
(
−mω) cos ωt
+ Fω ( qmω−mω)2 +(mp)2 +
ω
=
p
F
q
mω ( ω
−ω)2 +p2 .
¶
q
+ B sin
p2
4 t
q−
mp sin ωt
´
2
2
( qm
ω −mω) +(mp)
.
A and B are determined from the two equations
α=A+
p
β =− A+
2
r
q−
mq
F
ω − mω
,
ω ( mq
−
mω)2 + (mp)2
ω
p2
F
mpω
B+
.
2 + (mp)2
4
ω ( mq
−
mω)
ω
p
For long times, p2 t → ∞ so e− 2 t ¿ 1,
y(t) → yp (t) =
F 1
cos(ωt − φ),
ω Z(ω)
(6.5)
where
1
1
= p mq
, cos φ =
Z(ω)
( ω − mω)2 + (mp)2
mq
ω
− mω
mp
, sin φ =
.
Z(ω)
Z(ω)
(6.6)
Interpretation of (vi) and its solutions (6.5) and (6.6)
The response of the damped oscillator system to periodic (here sinusoidal) forcing
is described by the equation
µ ¶µ
¶
F
1
yp (t) =
cos(ωt − φ),
ω
Z(ω)
where
1
Z(ω)
is the amplification factor and φ is the phase lag.
8
Draw the graph of
1
Z(ω)
against ω
for m = q = 1 and p = 1, 0.2. We will study this example closely.
Example:
y 00 = t3 , y(0) = y 0 (0) = 1.
5
t
By straightforward integration, y = 20
+ At + B. y(0) = 1 ⇒ B = 1; y 0 (0) = 1 ⇒
t5
A = 1, so we have y(t) = 1 + t + 20 by the method of undetermined coefficients.
Step 1: The homogeneous equation is y 00 = 0. This gives y = ert with r2 = 0
which has repeated root 0, so
yc = (At + B)e0t = At + B.
Step 2: g(t) = t3 , so {g(t)} = (t3 , t2 , t, 1).
Step 3: The modified auxiliary set is {g(t)} = (t5 , t4 , t3 , t2 ). (We cannot just
multiply by t because the element t would satisfy y 00 = 0.)
9
Step 4: Let
yp
=
A1 t5 + A2 t4 + A3 t3 + A4 t2
yp0
=
5A1 t4 + 4A2 t3 + 3A3 t2 + 2A4 t
yp00
=
⇒ A1
=
20A1 t3 + 12A2 t2 + 6A3 t + 2A4 = t3
1
, A2 = A3 = A4 = 0.
20
Hence we have y(t) = At + B +
1 5
20 t ,
as we found above by direct integration.
Exercise:
y 00 + y = te−t , y(0) = y 0 (0) = 0.
(Answer: y(t) = − 21 cos t + 12 (t + 1)e−t ).
Exercise:
y 00 + y = t sin t, y(0) = y 0 (0) = 0.
(Answer: y(t) = − 41 t2 cos t + 14 t sin t.)
Exercise:
y 00 + 2y 0 + 5y = 4e−t cos 2t, y(0) = 1, y 0 (0) = 0.
(Answer: y(t) = e−t cos 2t + te−t sin 2t.)
Exercise:
Solve
d2 y
dx2
+ λy = sin πx, y(0) = 0, y(1) = 0..
For what values of λ is there a unique solution?
For what values of λ is there no solution?
For what values of λ is there an infinite number of solutions?
Message: BVP’s are fundamentally different from IVP’s!
10
Forced vibrations. The response of a weakly damped mass-spring system
to periodic forcing.
Models.
1. Mass-spring system under the influences of gravity, friction and external forcing.
d2 x
b dx
k
F
+
+ x=
cos ωt
dt2
m dt
m
m
Mass m (kgs.),
friction force = −b dx
dt ,
tension or spring force = −kx,
external force = F cos ωt.
Note: Gravity disappears from equation!
2. Mass-spring with damping and external forcing.
m
d2 x
dx
+b
+ kx = F cos ωt
dt2
dt
d2 x
b dx kx
F
+
+
=
cos ωt.
dt2
m dt
m
m
11
3. Pendulum.
Newton’s second law in angular direction
d dθ
dθ
= −β
− mg sin θ + F (t).
m l
dt dt
dt
dθ
Where l dθ
dt is the velocity of the bob in angular direction, −β dt is friction, −mg sin θ
is gravity and F (t) is the applied external force in angular direction. In the direction
along the pendulum, the tension in the rod which is absorbed at the hinge O is
balanced by the gravity and centrifugal forces. That is T = mg cos θ + ml(dθ/dt)2 .
d2 θ
β dθ g
F (t)
+
+ sin θ =
.
(6.7)
dt2
ml dt
l
ml
(6.7) is second order, nonlinear and nonautonomous if F depends on t. For small
oscillations, however, sin θ is well approximated by θ and then (6.7) is linear.
4. Electrical Circuits. (See Lectures 7-8-9).
I have already told you how to derive these equations. You may also consult NSS
§5.5. For now we will just deal with the equation. Consider the RLC circuit pictured
to which the emf (voltage) V (t) is applied. Then the charge Q(t) satisfies
and the current i(t) =
dQ
dt
d2 Q
dQ
1
+R
+ Q = V (t)
2
dt
dt
C
satisfies
L
(6.8)
d2 i R di
1
1 dV
+
+
i=
.
2
dt
L dt LC
L dt
(6.9)
L - inductance, R - resistance, C - capacitance.
12
If V (t) = V0 cos ωt, then the RHS of (6.8) is
V0
cos ωt.
L
Remark: Because we have already used q as a coefficient in (6.3), we will use cap
q or Q to denote the charge on the capacitor of the RLC circuit.
Each of these models is described by a linear, second order, constant coefficient,
nonhomogeneous ode.
d2 x
F0
dx
+ qx =
cos ωt
(6.10)
+p
dt2
dt
m
where the external force is periodic with frequency ω.
In the mass-spring system p = b/m, q = k/m, F0 is the magnitude of the force and
m is the mass.
In the electric circuit, p = R/L, q = 1/LC, Fm0 = V0 /L. Recall from exercise (vi)
that for p2 < 4q the general solution of (6.10) is:
a.
x(t)
=
xc (t) + xp (t) where
b. xc (t)
=
µ
q
e−pt/2 A cos q −
c.
=
F0
1
ω Z(ω)
xp (t)
d. Z(ω) =
e.
cos φ
=
p2
4 t
p2
4 t
(6.11)
2
2
( mq
ω − mω) + (mp) =
Z(ω)
q−
cos(ωt − φ)
p
mq
ω −mω
¶
q
+ sin
=
k/ω−mω
Z(ω) , sin φ
q
( ωk − mω)2 + b2
=
mp
Z(ω)
=
b
Z(ω)
Today we are going to look at this solution, study it, investigate how it changes as
we change the parameters p, q, and ω; and learn from it.
Observations:
(1) The general solution consists of two parts, the complementary function
xc (t) in which the unknown constants A and B are determined by the
initial conditions, and the particular solution xp (t).
(2) As t → ∞, xc (t) decays to zero on a time scale proportional to (p/2)−1 .
It is called the transient part of the solution because, while very important
for small t and for matching the initial conditions, it eventually dies away.
(3) As t → ∞, x(t) → xp (t). The long term response of the mass-spring /
pendulum / electric circuit to periodic forcing is given by
xp (t) =
F0 1
cos(ωt − φ).
ω Z(ω)
13
(6.12)
For the electric circuit, V = V0 cos ωt, and
Qp (t) =
Qp (t) = √
ω
cos φ = √
(4)
(5)
(6)
(7)
V0
ωZ(ω)
cos(ωt − φ)
V0
1
( ωC
−ωL)2 +R2
1
ωC −ωL
1
( ωC −ωL)2 +R2
cos(ωt − φ)
, sin φ = √
(6.13)
R
.
1
( ωC
−ωL)2 +R2
In observation 9 below, we discuss the long term current of the RLC
circuit.
We are interested in those cases when the damping is small; that is not
√
√
only is p2 < 4q but p ¿ 2 q i.e. (p/2) ¿ q or that the decay rate p/2
p
of the transient is a lot less than its natural frequency q − p2 /4 which is
√
almost equal to q.
In these cases, the natural oscillations of the spring xc (t) has frequency
r
p2
√
q−
≈ q
4
and is weakly damped, whereas the forced oscillation xp (t) has the same
frequency ω as the external force.
It is natural to ask: When is the response to the forcing maximum? Clearly
this occurs for the maximum value of
1
1
= p mq
Z(ω)
( ω − mω)2 + (mp)2
√
which occurs when ω = ± q, namely when the forcing frequency is the
same as the natural frequency of the oscillator with zero damping.
√
The strength of the response when ω 2 = q is 1/Z(ω = q) = 1/mp,
which is very large when mp is very small.
Recall from exercises (iv) (as ω0 → ω) and (v), the response of the un2
damped oscillator ddt2x + qx = 0 to the external forcing (F0 /m) cos ωt when
q → ω 2 was
F0
d2 x
+ qx =
cos ωt
dt2
m
giving
F0 t
x(t) = A cos ωt + B sin ωt +
·
sin ωt;
m 2ω
namely x(t) grows like t.
The effect of damping, no matter how weak is to arrest this unbounded
behavior and give rise to a bounded response xp (t) with an amplitude which
is proportional to the inverse of the damping.
We call this strong response of an oscillator which is brought into contact
with another oscillator of the same frequency by the name resonance.
Observe that Z(ω) has the dimensions or units of friction or resistance. To
see this look at
1
1
.
(Z −1 )max =
√ =
Z( q)
mp
For mass spring systems 1/mp = 1/b where the friction force was b dx
dt . For
electric circuits,
1
1
1
= R =
mp
R
LL
14
where the resistance of the circuit is R and the voltage drop due to resistance
is Ri (Ohm’s Law).
(8) Let us examine response when the circuit or mass-spring system is tuned
i.e. when q is chosen to be ω 2 . Then cos φ = 0, sin φ = 1, Z = mp = b so
that φ = π/2 and
F0 1
sin ωt
xp (t) =
ω b
or
dxp
b
= F0 cos ωt.
dt
So in a forced mass-spring system which is tuned so that the ‘natural fre√
quency’ q of the undamped, unforced system is chosen to be the forcing
2
frequency ω, the mass-spring system behaves as if the ddt2x and qx terms are
ignored (they exactly cancel each other out) and the remaining force is the
instantaneous friction and the external force.
d2 x
dx
m 2 +b
+ kx = F0 cos ωt.
dt
dt
When k is tuned so that k/m = ω 2 , then the forced response
xp =
F0 1
sin ωt
ω b
‘balances’
d2 xp
+ kxp ;
dt2
i.e. it makes it zero, leaving friction to balance the external field,
dxp
b
= F0 cos ωt
dt
(9) Note that if the long term charge Qp (t) is given by (6.13), then the long
dQ
term current ip = dtp is given by
m
ip =
−V0
sin(ωt − ϕ).
Z(ω)
At resonance, namely when ω =
− sin( π2 − ωt) = − cos ωt,
√1 ,
LC
ϕ =
(6.14)
π
2
and since sin(ωt −
π
2)
=
V0
cos ωt
(6.15)
R
This is very interesting because it is just like Ohm’s law Rip = V (t) for
a time dependent voltage. Look at equation (6.8) at the resonance where
1
ω = √LC
;
ip =
d2 Q
dQ
1
+R
+ Q = V0 cos ωt,
(6.16)
2
dt
dt
C
On resonance, the first and third terms on the LHS balance, i.e. cancel,
leaving
dQ
R
= Ri = V0 cos ωt,
(6.17)
dt
So that, at each instant of time, the ohmic resistance of the circuit balances
the external voltage. Since Z(ω), which has the units of the resistance R, is
like a resistance with terms added to account for the lack of tuning. Z(ω)
L
15
1
is called the impedance of the RLC circuit, R the resistance and ωC
− ωL
the reactance.
(10) Recall again the graph of 1/Z(ω) which is the amplitude of the response of
the weakly damped oscillator to external forcing.
For p very small,
(
0,
for q 6= ω 2
1
≈
1
2
Z(ω)
mp , for q = ω
The response of the oscillator as a function of ω (or if you fix ω and vary
q, as a function of q) is very sensitive to ω the smaller p is. For p very, very
√
small, if ω = + q + σp, (σ = 1, 2, 3), the response amplitude decays very
quickly as σ is increased. The width of the response is inversely proportional
to the height. The latter is called the ‘Q’ of the circuit or oscillator.
(11) Analogy between mechanical and electrical systems
Mechanical mass-spring
damping and forcing
2
Electrical RLC circuit
in series
2
m ddt2x + b dx
dt + kx = F (t)
dQ
L ddtQ
2 + R dt +
Displacement x(t)
Velocity V (t) = dx
dt
Mass m
friction/damping b
spring constant k
External force F (t)
charge Q(t)
current i(t) = dQ
dt
inductance L
resistance R
(capacitance)−1 C1
External voltage V (t)
16
1
CQ
= V (t)
(12) Beating between oscillators. The notion of amplitude modulation.
Let us return to exercise (iv)
d2 x
+ ω02 x = cos ωt
dt2
dx
x(0) =
(0) = 0.
dt
Solution:
x(t) =
(6.18)
cos ωt − cos ω0 t
.
ω02 − ω 2
Let us imagine that ω is close to ω0 and write,
ω0 +ω
− ω02−ω ,
2
ω0 = ω02+ω − ω02−ω
ω=
and then rewrite x(t) as
x(t) =
Now if ω0 ≈ ω so that
2 sin ω02−ω t
ω0 + ω
sin
t.
2
2
ω0 − ω
2
(6.19)
¯
¯
¯ ω0 − ω ¯
ω0 + ω
¯
¯
¯ 2 ¯¿
2
the factor sin ω02+ω in x(t) changes much faster than the factor sin ω02−ω . We see
this when we use SYSTEMS to plot x(t) with ω0 = 1 and ω = 0.9.
In fact, the response of the oscillator to the forcing frequency which is almost in
resonance is an amplitude modulation of the average frequency ω02+ω .
µ
x(t) =
=
For
ω0 −ω
2 t
Slowly varying
amplitude
¶
¢
¡
× sin ω02+ω t
We call the ‘fast’ frequency
the carrier frequency.
×
small,
sin
so
ω −ω
2 sin 02 t
ω02 −ω 2
ω0 − ω
ω0 − ω
t≈
t
2
2
x(t) ≈
t
ω0 + ω
sin
t.
ω0 + ω
2
There is rapid resonant growth and the amplitude grows like t. However by the time
ω0 −ω
π
2 t ≈ 2 (assume ω0 > ω), the fact that the frequencies of the natural oscillation
17
and the forcing are slightly different has registered. The amplitude reaches its
2
maximum ω2 −ω
2.
0
For π2 < ω02−ω t < π, the phase of the forcing is such that it is opposing the motion
x(t) and so x(t) decreases back to zero at
ω0 − ω
t = π.
2
Then the cycle repeats. This phenomenon is know as beating. We will talk about
the beating of oscillators during the lecture.
I will also tell you about the phenomenon of synchronization in which a nonlinear
oscillator (imagine the restoring force to be proportional to k(x − γx3 )) can adjust
its amplitude so as to tune its frequency (which now depends on its amplitude)
to match that of the forcing frequency. This phenomenon was first observed by
Swiss clockmakers who noticed that clocks on the same wall (which would act as a
coupling between the clocks) would synchronize; i.e. keep exactly the same time.
Method of Variation of Parameters
Recall the method of how to find particular solutions yp (t) of
d2 y
dy
+ p(t)
+ q(t)y = g(t)
(6.20)
dt2
dt
where p(t) and q(t) were constant and g(t) was a linear combination of terms each
of which had the form of a product of a polynomial, an exponential and a sinusoidal
factor. Namely, g(t) was a linear combination of terms of the form
L[y] =
tα eβt cos γt or tα eβt sin γt.
(6.21)
The method of undetermined coefficients was very useful because it was reasonably
easy to implement. But what happens when p(t), q(t) are not constants of g(t) is
not of the required form?
Here I am going to show you a very general method which may be used if you
happen to know two l.i. solutions y1 (t), y2 (t) of the homogeneous equation
d2 y
dy
+ p(t)
+ q(t)y = 0.
(6.22)
2
dt
dt
In fact, all you need to know is one solution y1 (t) of (6.23) because by reduction of
order (lectures 11-16) you can find the second.
L[y] =
I will first apply the method to the first order equation
dy
+ p(t)y = g(t).
dt
Let y1 (t) satisfy
dy
dt
+ p(t)y = 0. Then set yp (t) = y1 (t)v(t) and find
y1
dv
dy1
+v
+ p(t)y1 v = g(t)
dt
dt
18
or
dv
1
=
g(t).
dt
y1 (t)
Thus
Z
t
g(s)
ds
y1 (s)
or
Z t
g(s)
yp (t) = y1 (t)
ds
y1 (s)
³ R
´
t
is a particular solution. Note that y1 (t) = exp − p(t0 )dt0 so that the particular
solution found here agrees exactly with what you would have found by using an
integrating factor.
v=
Let y1 (t), y2 (t) be two l.i. solutions of (6.22). Set
yp (t) = y1 (t)v1 (t) + y2 (t)v2 (t),
(6.23)
yp0 = y1 v10 + y2 v20 + y10 v1 + y20 v2 .
(6.24)
and differentiating find
Now since we introduced two unknown functions v1 (t), v2 (t) in place of one unknown yp (t), we are at liberty to choose one relation between v1 and v2 for our
convenience. We will choose
(6.25)
y1 v10 + y2 v20 = 0
so that now
yp0 = y10 v1 + y20 v2 .
(6.26)
Differentiate to find
yp00 = y100 v1 + y200 v2 + y10 v10 + y20 v20 .
(6.27)
Substitute (6.23), (6.26), (6.27), in (6.23) to find
y100 v1 + y200 v2 + y10 v10 + y20 v20 + p(t)(y10 v1 + y20 v2 ) + q(t)(y1 v1 + y2 v2 ) = g(t).
But yi00 + pyi0 + qyi = 0, (i = 1, 2), so that
y10 v10 + y20 v20 = g(t).
Now, (6.25) and (6.28) are two equations for the derivatives
known functions v1 (t), v2 (t). We find,
−g(t)y2 (t)
,
W (y1 , y2 )
(6.28)
v10
and
v20
g(t)y1 (t)
,
W (y1 , y2 )
´
³ R
t
where W (y1 , y2 ) = y1 y20 − y10 y2 = constant exp − p(t0 )dt0 . Thus
Rt
Rt
v1 (t) = − g(s) Wy(y2 (s)
ds,
v2 (t) = g(s) Wy(y1 (s)
ds,
1 ,y2 )
1 ,y2 )
R
Rt
t
ds + y2 (t) = g(s) Wy(y1 (s)
ds.
yp (t) = −y1 (t) g(s) Wy(y2 (s)
1 ,y2 )
1 ,y2 )
v10 (t) =
v20 (t) =
Example: Find the general solution of
t2 y 00 − t(t + 2)y 0 + (t + 2)y = 2t3 , t > 0.
19
of the un(6.29)
(6.30)
Writing this in the form (6.20)
t+2 0 t+2
y 00 −
y + 2 y = 2t, t > 0.
t
t
From an earlier lecture, we know that
y1 (t) = t, y2 (t) = tet .
Set
yp (t) = tv1 + (tet )v2
yp (t)0 = v1 + (t + 1)et v2 + tv10 + (tet )v20 .
Choose
tv10 + (tet )v20 = 0,
(6.31)
so that now
yp0 = v1 + (t + 1)et v2 .
Then
Substitute
yp , yp0 ,
yp00
yp00 = v10 + (t + 2)et v2 + (t + 1)et v20 .
in equation;
t+2
t+2
(v1 + (t + 1)et v2 ) + 2 (tv1 + tet v2 ) = 2t.
t
t
Check that the coefficients of v1 ((t + 2)/2 + t(t + 2)/t2 ) and v2 ((t + 2)et − (t + 1)(t +
2)(et )/t + t(t + 2)(et )/t2 ) are zero. We find
v10 + (t + 2)et v2 + (t + 1)et v20 −
v10 + (t + 1)et v20 = 2t.
(6.32)
Solving (6.31), (6.32), to find
0
tet
2t (t + 1)et
v10 =
= −2,
t
tet
1 (t + 1)et
v20 =
t
1
t
1
0
2t
= 2e−t
tet
(t + 1)et
v1 = −2t + c1 ,
v2 = −2et + c2 .
yp (t) = t(−2t + c1 ) + tet (−2e−t + c2 ) = −2t2 + (c1 − 2)t + c2 (tet ).
The last two terms are solutions of the homogeneous equation and will automatically be included in the complementary functions yc (t), so we can take
yp (t) = −2t2 .
Check: t2 [−4] − t(t + 2)[−4t] + (t + 2)[−2t2 ] = 2t3 .
General solution:
y(t) = yc (t) + yp (t) = At + Atet + (−2t2 ).
Exercises: (Some of these may be done using the method of undetermined coefficients.)
(i) y 00 + y = cos ωt, ω 2 6= 1.
(ii) y 00 + y = cos t.
20
(iii) y 00 + y = tan t, 0 < t < π/2.
(iv) t2 y 00 + ty 0 − y = t2 et , t > 0.
21
TRIAL EXAM #2. To help you prepare for Exam #2.
Work in groups.
Should take 4-5 hours.
Q.1. Linear independence (l.i) and linear dependence (l.d) of functions on intervals. In what follows, I will give you a set of functions and an interval I.
You have to say whether they are l.i or l.d on that interval.
Function set
Interval
(i)
1, t, t2 , t|t|
(−1, 1)
(ii)
1, t, t2 , t|t|
(0, 1)
(iii)
sin t, t sin t
(0, π)
(iv)
e2t , te2t , t2 e2t
(−∞, ∞)
(v)
1, et , 5 + 2et , tet
(−∞, ∞)
(vi)
5 + 2et , et , tet
(−∞, ∞)
(vii)
t3 , t2 , t2 |t|
(−1, 1)
(viii)
t3 , t2 , t2 |t|
(0, 1)
(ix)
sin t, cos t, sin 2t
(−∞, ∞)
l.i or l.d
On (i), set
k1 + k2 t + k3 t2 + k4 t|t| = 0
22
for all t ∈ (−1, 1)
t = 0 ⇒ k1 = 0
Choose
t = − 12 ⇒ − k22 +
Choose
1
2
t=
t=
∴
1
4
⇒ − k22 +
⇒
k2
4
+
k3
16
k3
4
−
k4
4
=0
Add ⇒
k3
4
+
k4
4
+
k4
16
= 0 ⇒ k2 = 0.
=0
k3 = 0
k4 = −2k2
µ
¶
∴ k4 = −2k2
k3 = 0
k1 = k2 = k3 = k4 = 0
¯
¯
W (1, t, t , t|t|)¯¯
2

1 t
0 1
= det 
0 0
0 0
t positive

t2 t|t| = t2
2t 2|t| = 2t
 = 0.

2
2
0
0
So W = 0 for t > 0
W = 0 for t < 0
W not defined at t = 0.
We can draw no conclusion from W test in this case.
Q2. The ode,
dx
x
(*) an dn
− ∞ < t < ∞ has associated with it the
dtn + · · · + a1 dt + a0
characteristic eqn.
(**) an rn + · · · + a1 r + a0 = 0. Below, I give the roots of (**). You give the
general solution of (*).
23
Roots of **
General soln. of (*)
n = 5; 1, i, −i, i, −i
x(t) = A1 et + A2 cos t + A3 sin t
+A4 t cos t + A5 t sin t
n = 7; 0, 0, 1, −1 + 2i, −1 + 2i,
−1 − 2i, −1 − 2i
n = 4; 0, 0, 0, 0
n = 10; 0, 0, 0, 1, 1, 2,
−1 + i, −1 − i,
−1 + i, −1 − i.
Q3. Solve the following initial value problems
(i)
d2 x
dx
dx
+2
+ 5x = 0, x(0) = 1, (0) = 0.
dt2
dt
dt
d2 x(0)
dt2
(ii)
d4 x
dt4
+ 2 ddt2x + x = 0, x(0) = 1, dx(0)
dt =
2
(iii)
d2 x
dt2
dx
+ 5 dx
dt + 4x = 0, x(0) = 1, dt (0) = 0.
(iv)
d2 x
dt2
dx
+ 4 dx
dt + 4x = 0, x(0) = 1, dt (0) = 0.
(v)
d2 x
dt2
dx
+ 2 dx
dt + 4x = 0, x(0) = 1, dt (0) = 0.
(vi)
d5 x
dt5
−
(vii)
d2 x
dt2
dx(0)
+ 0.2 dx
dt + 9.01x = 0, x(0) = 1, dt = 0.
(viii)
d2 x
dt2
+ 0.2 dx
dt + 9.01x = cos 3t, x(0) =
(ix)
d2 x
dt2
− x = tet , x(0) = 1, dx
dt (0) = 0.
(x)
d4 x
dt4
− 2 ddt2x + x = t2 et , x(0) = 1, dx(0)
dt =
(xi)
d3 x
dt3
− 3 ddt2x + 2 dx
dt = 5, x(0) =
d3 x
dt3
d3 x(0)
dt3
= 0.
s
x(0)
= 0, x(0) = 1, d dt
= 0, s = 1, 2, 3, 4.
s
dx(0)
dt
2
2
=
dx(0)
dt
24
=
= 0.
d2 x(0)
dt2
d2 x(0)
dt2
=
= 0.
d3 x(0)
dt3
= 0.
Q.4a. A very light spring (its extension under its own right is negligible) hangs
from a hinge (Fig A). A weight of .1 kg is added to the pan (Fig B) and the spring
stretches by 20cms. The pan is stretched a further 4 cms. (Fig C) and then is
released with zero velocity. If g, gravity, is taken to be 10m/sec2 and if the friction
force experienced by the spring is numerically equal to .2 dx
dt in Newtons, find the
equation which x(t) satisfies. What are the initial conditions x(0), dx
dt (0)? Express
the solution x(t) = Ceat cos(bt − ϕ) and determine C, a, b and ϕ.
A
o
o
o
o
o
- - -4- - - - -
B
C
o
o
o
o
o
o
o
o
o
o
o
- o- - o
o
o 20 cms o
o
o
o
o
- - - -4
-r - - - - - -oo - ↓ 4 cms 4r x(0)
-------mg
Q4b. If in addition, the spring experiences an external force numerically
equal to
√
.1 cos ωt in Newtons, find x(t) for ω 2 = 50, ω 2 = 49, ω 2 = 25 i.e. ω 50, 7, 5.
In particular, write
x(t) = Ceat cos(bt − ϕ1 ) + D cos(ωt − ϕ2 )
and determine C, a, b, ϕ1 , D, ϕ2 .
You will find a < 0. Therefore as t → ∞, x(t) → D cos(ωt − ϕ2 ).
What is the value of ϕ2 when ω =
√
50?
Q5. An LRC circuit with inductance L = .1 henry, R = .2 ohms and C = 0.2
farads is set in motion by a voltage of .1 cos ωt applied at time t = 0.
Given the initial q(t) and current i(t) =
√
t > 0 when ω = 50.
dq
dt
are zero, determine q(t) and i(t) for
Q6. Write down the general solution to
m
q
where Z =
d2 x
dx
F0
+b
+ kx =
sin(ωt − ϕ)
dt2
dt
Z
( ωk − mω)2 + b2
What happens when k is chosen k/m = ω 2 . Show then that b dx
dt = F0 cos ωt.
Q7. Consider the ode
for p = 5, 4 and 2.
d2 x
dx
+p
+ 4x = 0
2
dt
dt
25
(a) Write the ode as a system of 2 first order ode’s
dx
=y
dt
dy
=
?
dt
(b) For p = 5, write the solutions as
µ
¶
µ ¶
µ ¶
x(t)
¤ −t
¤ −4t
=A
e +B
e
y(t)
¤
¤
Fill-in squares
Given A 6= 0, draw the trajectory x(t), y(t) in the x,y plane and in particular note the direction along which it approaches (0, 0)
(c) For p = 4, write the general solution
µ
¶
µ ¶
µ ¶
x(t)
¤
¤
=A
¤+B
¤.
y(t)
¤
¤
Fill-in squares
(d) For p = 2, write the solution in the form
x(t) = Ceat cos(bt − ϕ)
y(t) =
and draw the trajectory x(t), y(t) in the x,y plane.
Use SLOPES or SYSTEMS if you wish.
26
MATH 254 Mock exam #2 (≈ 1 14 hours).
Name
(1) Given y1 (x) = x satisfies
(∗)
1
d2 y
1 dy
+ 2 y = 0 for x > 0
=
2
dx
x dx x
find a 2nd l.i. soln. y2 (x) =
.
Solve (∗) with initial condition y(1) = 0,
dy
dx (1)
= 1.
y(x) =
Hint: you should find y(e2 ) = 2e2 .
(2) Solve the following initial value problems.
d2 y
dy
dy
x
(a) dx
2 − 2 dx + y = e , y(0) = 0, dx (0) = 0
y(x) =
(b)
d2 y
dx2
dy
+ y = cos x, y(0) = 0, dx
(0) = 0
y(x) =
(c)
d2 y
dx2
dy
dy
+ 5 dx
+ 4y = 0, y(0) = 1, dx
(0) = 0.
y(x) =
(d)
d2 x
dt2
1
+ 2 dx
dt + 2x = cos 2t, x(0) = − 10 ,
Write x(t) = A cos(2t − ϕ).
Find A, cos ϕ, sin ϕ.
A=
27
dx
dt (0)
=
2
5
cos ϕ =
sin ϕ =
x(t) =
(3) An LRC circuit, shown below, is acted upon by an emf of V (t) = V0 cos ωt
for t > 0.
A
ffffff
B
L
∼i
E
C
R
∧∨∨∨∨∨
D
VA − VE = V (t), t > 0
0, t < 0.
1
For L = .1 henry, R = .2 ohms, C = 2 farads, V0 = 10
volts and ω = 2
rad/sec, write down the ode for q(t), the charge in Coulombs on the capacitor. (Do not solve the equation). Given the initial charge is zero and the
initial current is zero, what are q(0) and dq
dt (0)?
ode
q(0) =
dq
dt (0)
=
28