Applied Differential Equation
October 22, 2012
Contents
3
4
Second Order Linear Equations
2
3.1
Second Order linear homogeneous equations with constant coefficients . . . . . . . . . .
4
3.2
Solutions of Linear Homogeneous Equations, the Wronskian . . . . . . . . . . . . . . .
5
3.3
Complex Roots of the characteristic Equations . . . . . . . . . . . . . . . . . . . . . . .
9
3.4
Repeated Roots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11
3.5
Nonhomogeneous Equation, Method of undetermined coefficients . . . . . . . . . . . .
13
3.6
Variation of Parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
15
3.7
Mechanical and Electrical Vibrations . . . . . . . . . . . . . . . . . . . . . . . . . . . .
19
3.8
Review of Chapter 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
23
The Laplace Transform
25
4.1
Definition of the Laplace transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
25
4.2
Solution of Initial Value Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
28
4.3
Step functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
31
4.4
Differential Equations with Discontinuous Forcing Functions . . . . . . . . . . . . . . .
36
4.5
Impulse functions – Dirac delta function . . . . . . . . . . . . . . . . . . . . . . . . . .
38
4.6
The Convolution Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
41
4.7
Review of Chapter 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
45
1
Chapter 3
Second Order Linear Equations
Two important areas of application of 2nd order linear equations are in the fields of mechanical and
electrical oscillations.
Example Mechanical system of spring.
l
l+L
KL
m
mg
gravitational force: Fg = mg
m: mass of the ball, g acceleration due to gravity.
Spring force (Hooke’s Law): Fs = −kL
L: displacement. k: spring constant.
Hooke’s Law: Spring force is proportional to the displacement of the spring (L). The direction of the
force is opposite to the direction of the motion.
when the ball is in equilibrium,
mg = kL, ⇒ k =
2
mg
.
L
l
l+L
still
l + L + u(t)
m
still
u(t)
Fs
m
vibrated
Fg = mg
Newton’s Law of motion (neglect damping or resistive force such as air resistance)
Suppose the spring is stretched by a displacement of 0.1L and released later.
• l: original length of the spring.
• L: displacement of the spring caused by gravity in still state.
• u(t): displacement caused by vibration.
• Fg = mg: gravity force.
• Fs = −k(L + u(t)): elastic force.
• Choose the downward direction as ”positive” direction, that is, u(t) is positive, when the spring is
stretched.
• F = ma,
a=
d2 u
= u00 (t)
dt
• mu00 (t) = mg − k(L + u(t)): downward direction is positive, so mg has a positive sign. No matter
what we choose as the positive direction, the elastic force is always opposite to direction of u(t), its
sign is always negative.
mg
: see the second graph, when the spring is still, it is obvious mg = kL.
L
mg
g
• mu00 = mg −
(L + u), u00 = − u
L
L
• mg = kL,
k=
• u(0) = 0.1L,
v = u0 (0) = 0: initial condition.
3

g
00

 u = u
L
•
u(0) = 0.1L

 0
u (0) = 0
Generally speaking, 2nd order DE:
d2 y
dy
= f (x, y, )
2
dx
dx
2nd order linear DE:
3.1
 00
0
 y + p(x)y + q(x)y = g(x)
y(x0 ) = y0
 0
y (x0 ) = y00
Second Order linear homogeneous equations with constant coefficients
ay 00 + by 0 + cy = 0
(a 6= 0)
Example 1: y 00 = y
Possible solutions: ex , e−x , C1 ex , C2 e−x ;
General solution y = C1 ex + C2 e−x .
For ay 00 + by 0 + cy = 0, seek for solution in form y = erx , r to be determined.
(ar2 + br + c)erx = 0
ar2 + br + c = 0 (characteristic equation)
√
−b ± b2 − 4ac
r1,2 =
2a
If b2 − 4ac > 0, then r1 , r2 real and different, then er1 x and er2 x are two different solutions.
y = C1 er1 x + C2 er2 x
C1 and C2 can be determined by y(x0 ) = y0 , y 0 (x0 ) = y00 .
Ex 1. y 00 + 4y 0 + 3y = 0, y(0) = 2, y 0 (0) = −1
r2 + 4r + 3 = 0
(r + 3)(r + 1) = 0
r = −3,
r = −1
⇒y(x) = C1 e−3x + C2 e−x
4

y(0) = C1 e0 + C2 e0 = C1 + C2 = 2





y 0 (x) = −3C1 e−3x − C2 e−x


 y 0 (0) = −3C − C = −1
1
2
1
 ⇒ C1 = − , C2 = 5



2
2



 y(x) = − 1 e−3x + 5 e−x
2
2
Ex 2. Find the general solution of 2y 00 − 3y 0 + y = 0
2r2 − 3r + 1 = 0
(r − 2)(r − 1) = 0
r=2 r=1
⇒y(x) = C1 e2x + C2 ex
Ex 3. y 00 = 9.8y,
y 00 (0) = 0
y(0) = 0.1,
r2 − 9.8 = 0
√
r = ± 9.8
√
y(x) = C1 e
9.8x
+ C2 e−
√
9.8x
0 → x,
0.1 → y, C1 + C2 = 0.1
√
√
√
√
y (x) = 9.8C1 e 9.8x − 9.8C2 e− 9.8x
√
√
0 → x, 0 → y, 9.8C1 − 9.8C2 = 0
0
C1 = C2 = 0.05
√
y(x) = 0.05e
3.2
√
+ 0.05e−
9.8x
Solutions of Linear Homogeneous Equations, the Wronskian
Recall
9.8x
ay 00 + by 0 + cy = 0
r1 , r2 real, different. Then y(x) = C1 er1 x + C2 er2 x
ar2 + br + c = 0
Linear homogeneous equation:
 00
0
 y + p(x)y + q(x)y = 0
y(x0 ) = y0
 0
y (x0 ) = y00
where p(x), q(x) are defined in an interval I = (α, β) and x0 ∈ (α, β).
Theorem 2.1 (Existence and Uniqueness).
y 00 + p(t)y 0 + q(t)y = g(t),
5
y(t0 ) = y0 ,
y 0 (t0 ) = y00
If p(x), q(x) and g(x) are continuous in I : t0 ∈ I, then there is exactly one solution y = φ(x) of the
IVP (Initial Value Problem), and the solution exists throughout the interval I.
Theorem 2.2 (Superposition Principle)
If y1 (x) and y2 (x) are two solutions of
y 00 + p(x)y 0 + q(x)y = 0
Then the linear combination C1 y1 (x) + C2 y2 (x) is also a solution for any values of the constants C1 and
C2 .
Proof:
General Solution: by Crammer’s rule, once y1 and y2 are known, and given initial value y(x0 ) =
y0 , y 0 (x0 ) = y00 , we can determine C1 and C2 as following:
y(x) = C1 y1 (x) + C2 y2 (x),
C1 y1 (x0 ) + C2 y2 (x0 ) = y0
C1 y10 (x0 ) + C2 y20 (x0 ) = y00
y0 y2 (x0 ) 0
y y 0 (x0 ) 0
2
, C2 =
C1 = y1 (x0 ) y2 (x0 ) 0
y (x0 ) y 0 (x0 ) 1
2
y (x ) y2 (x0 )
W [y1 , y2 ](x0 ) = 10 0
y1 (x0 ) y20 (x0 )
y1 (x0 ) y0 0
y (x0 ) y 0 1
0
y1 (x0 ) y2 (x0 )
0
y (x0 ) y 0 (x0 )
1
2
y (x) y2 (x)
W = W [y1 , y2 ](x) = 10
y1 (x) y20 (x)
Wronskian determinant of y1 and y2
Theorem 2.3 (General solution)
If y1 and y2 are two solution of
y 00 + p(x)y 0 + q(x)y = 0
and if there is a point x0 where the wronskian of y1 and y2 is nonzero, i.e.,
y1 (x0 ) y2 (x0 ) ,
W [y1 , y2 ](x0 ) = 0
y1 (x0 ) y20 (x0 ) then the general solution of the DE is
y = C1 y1 (x) + C2 y2 (x)
where C1 and C2 are arbitrary constants. y1 and y2 are said to form a fundamental set of solutions.
6
Example y 00 + 4y 0 + 3y = 0.
r2 + 4r + 3 = 0
(r + 3)(r + 1) = 0
r = −3,
r = −1
Two particular solutions y1 (x) = e−3x and y2 (x) = e−x . At x0 = 0, y1 (x0 ) = e0 = 1, y2 (x0 ) = e0 = 1,
y10 (0) = −3, y20 (0) = −1, So
1
1
= −1 + 3 = 2
W [y1 , y2 ](0) = −3 −1 y = C1 e−3x + C2 e−x
is the fundamental solution.
Def: Linear Independence of functions
Two functions f and g are said to be linearly dependent on interval I if there exist two constants k1 and
k2 , not both zero, such that
k1 f (t) + k2 g(t) = 0, ∀t ∈ I
f and g are said to be linearly independent on I if they are not linearly dependent.
Example 1: f (t) = 3t, g(t) = |t|, (a) I = (0, ∞), (b) I = (−∞, ∞).
k1 · 3t + k2 |t| = 0,
∀t ∈ I
(a)I = (0, ∞):
• k1 · 3t + k2 |t| = k1 · 3t + k2 t = 0
• ⇒ (3k1 + k2 )t = 0 ⇒ k2 = −3k1
So as long as k2 = −3k1 , for example k2 = −3 and k1 = 1, then k1 · 3t + k2 t = 0,
implies f (t) and g(t) are linear dependent on (0, ∞).
k1 · 3t + k2 t = 0, t ∈ [0, ∞)
(b)I = (−∞, ∞) :
k1 · 3t − k2 t = 0, t ∈ (−∞, 0)
• k1 · 3t + k2 t = 0,
t ∈ [0, ∞) ⇒ k2 = −3k1 and k2 = 3k1
• k1 · 3t − k2 t = 0,
t ∈ (−∞, 0) ⇒ k2 = 3k1
• ⇒ k1 = k2 = 0
7
∀t ∈ I, which
which implies f (t) and g(t) are linear independent on (−∞, ∞).
g(t) = t2 ,
Example 2: f (t) = t,
k1 t + k2 t2 = 0,
k1 = −k2 t,
I = (−∞, ∞)
t=−
k1
↔ ∀t ∈ (−∞, ∞)
k2
Linear independence – The Wronkskian
Theorem 3.1 (Abel’s Theorem) If y1 and y2 are two solutions of
y 00 + p(x)y 0 + q(x)y = 0
where p and q are continuous on an open interval I, then the Wronskian W [y1 , y2 ](x) is given by
W [y1 , y2 ](x) = Ce−
R
p(x)dx
where C is a certain constant that depends on y1 and y2 , but not on x. Furthermore, w[y1 , y2 ](x) is either
zero for all x ∈ I (if C = 0) or else is never zero in I (if C 6= 0)
Proof.
[y 00 + p(x)y10 + q(x)y1 ] ·y2 = 0
{z
}
|1
(1)
[y200
(2)
|
+
=0
p(x)y20
{z
=0
+ q(x)y2 ] ·y1 = 0
}
(2) − (1) ⇒ y1 y200 − y2 y100 +p(x) (y1 y20 − y2 y10 ) = 0
|
{z
}
|
{z
}
=W 0
0
0
W = y1 y2 − y2 y1 by definition
W 0 = y10 y20 + y1 y200 − y20 y10 − y2 y100
0
⇒ W + p(x)W = 0,
=W
= y1 y200 − y2 y100
a separable equation
W0
= −p(x), (ln W )0 = −p(x)
Z
R
ln W = − p(x)dx + C, W = Ce− p(x)dx
W
Example Find the Wronskian of x2 y 00 − (x + 2)y 0 + (x + 2)y = 0 without solving this DE.
First, we need to divide the differential equation by the first coefficient of the second derivative term:x2 .
x+2 0 x+2
x+2
This gives us y 00 −
y +
= 0. It turns out p(x) =
.
x2
x2
x2
Z
Z
x+2
x
2
2
dx
=
+ dx = ln x −
x2
x2 x2
x
W [y1 , y2 ](x) = Ce−
R
p(x)dx
2
2
= Celn x− x = Cxe− x
8
Euler’s Formula
Motivation: e3+6i
Taylor series of ex
ex =
eix =
=
∞
X
xn
n=0
∞
X
n=0
∞
X
n=0
n!
(ix)n X i2n x2n X i2n+1 x2n+1
=
+
n!
(2n)!
(2n + 1)!
n=0
n
2n
(−1) x
(2n)!
+i
= cos x + i sin x
∞
X
n=0
(−1)n+1 x2n+1
n=0
(2n + 1)!
Euler’s formula
Example 1: eλ+iµx = eλ eiµx = eλ (cos(µx) + i sin(µx))
e3+6i = e3 (cos(6) + i sin(6))
Property:
3.3
derx
= rerx
dx
Complex Roots of the characteristic Equations
Consider ay 00 + by 0 + cy = 0. y = erx , r to be determined.
Characteristic equation: ar2 + br + c = 0.
If b2 − 4ac < 0, then
p
p
√
−b ± (−1)|b2 − 4ac|
−b ± i |b2 − 4ac|
−b ± b2 − 4ac
=
=
2a
2a
2a
r1 = λ + iµ, r2 = λ − iµ
p
|b2 − 4ac|
−b
λ=
, µ=
2a
2a
y1 (x) = e(λ+iµ)x = eλx (cos µx + i sin µx)
y2 (x) = e(λ−iµ)x = eλx (cos µx − i sin µx)
9
e(λ+iµ)x
e(λ−iµ)x
W [y1 , y2 ](x) = (λ+iµ)x
(λ + iµ)e
(λ − iµ)e(λ−iµ)x
= (λ − iµ)e2λx − (λ + iµ)e2λx
= −2iµe2λx 6= 0
General solution: y(x) = C1 y1 (x) + C2 y2 (x)
y(x) = C1 eλx (cos µx + i sin µx) + C2 eλx (cos µx − i sin µx)
= (C1 + C2 )eλx cos µx + i(C1 − C2 )eλx sin µx
Define: C˜1 = C1 + C2 , C˜2 = i(C1 − C2 ).
r = λ ± iµ
y(x) = C˜1 eλx cos µx + C˜2 eλx sin µx ,
general solution
y˜1 (x) = eλx cos µx
y˜2 (x) = eλx sin µx
Here y˜1 and y˜2 are two real solutions.
Example 1 y 00 = −y
y 00 + y = 0
r2 + 1 = 0
b2 − 4ac = −4 < 0
p
|b2 − 4ac|
b
= 0, µ =
=1
λ=−
2a
2a
y1 (x) = eix = cos x + i sin x, y2 (x) = e−ix = cos x − i sin x
y(x) = C1 eix + C2 e−ix
C˜1 = C1 + C2 , C˜2 = i(C1 − C2 )
ỹ1 (x) = cos x,
ỹ2 = sin x.
Example 2 y 00 + y 0 + 1.25y = 0, y(0) = 3, y 0 (0) = 1
r2 + r + 1.25 = 0
b2 − 4ac = 1 − 5 = −4
p
|4|
b
1
λ=−
=− , µ=
=1
2a
2
2
1
1
y1 = e− 2 x+ix = e− 2 x (cos x + i sin x),
1
y(x) = C1 y1 (x) + C2 y2 (x)
C˜1 = C1 + C2 C˜2 = i(C1 − C2 )
1
ỹ1 (x) = e− 2 x cos x,
1
y2 = e− 2 x−ix = e− 2 x (cos x − i sin x)
1
ỹ2 (x) = e− 2 x sin x.
10
3.4
Repeated Roots
Consider:
ay 00 + by 0 + cy = 0
ar2 + br + c = 0
√
−b + b2 − 4ac
r1,2 =
2a
• b2 − 4ac > 0: r1 and r2 real and different.
• b2 − 4ac < 0: r1 and r1 complex and different.
• b2 − 4ac = 0: r1 = r2 = −
b
real.
2a
In the last case, y1 (x) = e−r1 x = y2 = e−r2 x .
W [y1 , y2 ](x) = 0
General solution?
To find the general solution, we have to find y2 (x) which is different from y1 (x).
Assume that y2 (x) has the form
b
υ(x)y1 (x) = υ(x)er1 x = υ(x)e− 2a x
where υ(x) is to be determined.
b −bx
υe 2a
2a
b
b
b
b
b2
y200 = υ 00 e− 2a x − v 0 e− 2a x + 2 υe− 2a x
a
4a
b
b
b
b
b
b
b
b2
b
a[υ 00 e− 2a x − υ 0 e− 2a x + 2 υe− 2a x ] + b[υ 0 e− 2a x − υe− 2a x ] + cυe− 2a x = 0
a
4a
2a


b
y20 = υ 0 e− 2a x −
b2 

aυ 00 + c −  υ = 0
| {z4a}
=0
00
υ =0
υ(x) = C1 x + C2
y2 (x) = C1 xer1 x + C2 er1 x ,
r1 = −
b
2a
11
y1 (x) = er1 x
y2 (x) = xer1 x
Verify y2 (x) = xer1 x is also a solution.
y20 = er1 x + r1 xer1 x
y200 = r1 er1 x + r1 er1 x + r12 xer1 x = 2r1 er1 x + r12 xer1 x
Check a[2r1 er1 x + r12 xer1 x ] + b[er1 x + r1 xer1 x ] + cxer1 x = 0
2ar1 + ar12 x + b + br1 x + cx
−
b
→ r1 ,
2a
−b +
b2
b2
b2
x+b− x+ x=0X
4a
2a
4a
b
b
General solution: y(x) = C1 e− 2a x + C2 xe− 2a x
Summary General solution of ay 00 + by 0 + cy = 0
ar2 + br + c = 0
√
−b ± b2 − 4ac
r1,2 =
2a
• b2 − 4ac > 0: y(x) = C1 er1 x + C2 er2 x
p
|b2 − 4ac|
b
λ
=
−
,
µ
=
2
• b − 4ac < 0:
2a
2a
y(x) = eλx (C1 cos µx + C2 sin µx)
b
b
• b2 − 4ac = 0: y(x) = C1 e− 2a x + C2 xe− 2a x
Examples:
• y 00 − 2y 0 + 10y = 0
• y 00 = 4y
• 9y 00 + 6y 0 + y = 0
• y 00 + 4y 0 + 2y = 0
• y 00 = −4y
12
3.5
Nonhomogeneous Equation, Method of undetermined coefficients
Homogeneous DE: ay 00 + by 0 + cy = 0
Nonhomogeneous DE: ay 00 + by 0 + cy = g(x)?
Theorem: The general solution of nonhomogeneous equation
y 00 + p(x)y 0 + q(x)y = g(x)
can be written in the form
y = C1 y1 (x) + C2 y2 (x) + Y (x)
where y1 and y2 are a fundamental set of solutions of the corresponding homogeneous equation
y 00 + p(x)y 0 + q(x)y = 0
C1 and C2 are arbitrary constants, and Y is some specific solution of the nonhomogeneous equation, such
that
Y 00 + p(x)Y 0 + q(x)Y = g(x)
Procedure:
• Find y1 and y2
• Find Y (x)
• y = C1 y1 (x) + C2 y2 (x) + Y (x)
How to find Y (x):
The Method of undetermined coefficents
ay 00 + by 0 + cy = g1 (x) + g2 (x)
Splitting ⇒
If Y1 (x) is the solution of ay 00 + by 0 + cy = g1 (x),
Y2 (x) is the solution of ay 00 + by 0 + cy = g2 (x)
then Y1 (x) + Y2 (x) is a solution of ay 00 + by 0 + cy = g1 (x) + g2 (x).
Examples
(a) y 00 − 2y 0 − 3y = 3e2x
• r2 − 2r − 3 = 0,
(r − 3)(r + 1) = 0,
r = −1, 3
13
• y = C1 e−x + C2 e3x + Y (x)
• Guess: Y = Ae2x
• Y 0 = 2Ae2x ,
Y 00 = 4Ae2x
• (4A − 2A − 3)e2x = 3e2x
Y = 3e2x
• A = 3,
• y = C1 e−x + C2 e3x + 3e2x
(b) y 00 + 4y = 5 sin 3x, y(0) = 2, y 0 (0) = −1
• r2 + 4 = 0,
r = ±2i,
λ = 0, µ = 2
• y = C1 cos 2x + C2 sin 2x + Y
• Guess Y = A sin 3x
• Y 0 = 3A cos 3x
• Y 00 = −9A sin 3x
• −9A sin 3x + 4A sin 3x = 5 sin 3x,
, A = −1
• y = y = C1 cos 2x + C2 sin 2x − sin 3x
(c) y 00 − 4y 0 − 12y = sin(2t)
• r2 − 4y − 12 = 0,
−2t
• y = C1 e
r = −2, 6
6t
+ C2 e + Y
• Guess Y = A sin(2t) + B cos(2t)
• Y 0 = 2A cos(2t) − 2B sin(2t)
• Y 00 = −4A sin(2t) − 4B cos(2t)
• (−16A + 8B) sin(2t) − (16B + 8A) cos(2t)
• −16A + 8B = 1,
(16B + 8A) = 0,
A=−
1
,
20
B=
1
40
(d) y 00 − 3y 0 − 4y = −8ex cos 2x
• r2 − 3r − 4 = 0,
−x
• y = C1 e
(r − 4)(r + 1) = 0,
4x
+ C2 e
r = −1, 4
+Y
x
• Guess Y = Ae cos 2x + Bex sin 2x
Y
Y0
Y 00
• ex cos 2x ex cos 2x − 2ex sin 2x −3ex cos 2x − 4ex sin 2x
ex sin 2x ex sin 2x + 2ex cos 2x −3ex sin 2x + 4ex cos 2x
• (−10A − 2B) ex cos 2x + (2A − 10B) ex sin 2x = −8ex cos 2x
|
|
{z
}
{z
}
=−8
15
• A= ,
13
=0
2
B=
13
14
(e) y 00 − 3y 0 − 4y = 2t2 − t + 1
• r2 − 3r − 4 = 0,
−x
• y = C1 e
r = −1, 4
4x
+ C2 e
+Y
2
• Guess Y = At + Bt + C
• Y 0 = 2At + B
• Y 00 = 2A
• 2A − 6At − 3B − 4At2 − 4Bt − 4C = t2 − t − 1

 4A = 1
−(6A + 4B) = −1
•

(2A − 3B − 4C) = −1


 

4
0
0
A
1
•  −6 −4 0   B  =  −1 
2 −3 −4
C
−1
• A = 0.25, B = −0.125, C = 0.4688
3.6
Variation of Parameters
Find a specific solution of the non-homogeneous equation.
Example Find the general solution of y 00 + 4y = sin x
Solution Find y1 and y2 of y 00 + 4y = 0, ⇒
y1 (x) = cos 2x and y2 (x) = sin 2x, y(x) = C1 y1 (x) + C2 y2 (x)(*)
The basic idea in the method of variation of parameters is to replace the constants C1 and C2 in (*) by
functions u1 (x) and u2 (x), resp., and then to determine these functions so that
Y (x) = u1 (x) cos 2x + u2 (x) sin 2x
is a particular solution of the non-homogeneous equation.
Y 0 = u01 cos 2x + u02 sin 2x − 2u1 sin 2x + 2u2 cos 2x
Y 00 = u001 cos 2x + u002 sin 2x − 2u01 sin 2x + 2u02 cos 2x − 2u01 sin 2x
+ 2u02 cos 2x − 4u1 cos 2x − 4u2 sin 2x
= u001 cos 2x + u002 sin 2x − 4u01 sin 2x + 4u02 cos 2x − 4u1 cos 2x − 4u2 sin 2x
Since Y 00 + 4Y = sin x ⇒
[u001 cos 2x + u002 sin 2x − 4u01 sin 2x + 4u02 cos 2x − 4u1 cos 2x − 4u2 sin 2x]
+ 4[u1 cos 2x + u2 sin 2x] = sin x
00
u1 cos 2x + u002 sin 2x − 4u01 sin 2x + 4u02 cos 2x
= sin 2x
15
which is one equation for unknown functions u1 and u2 . To determine them, we need to seek for another
equation or relation.
Let us impose one more restriction u01 cos 2x + u02 sin 2x = 0, then
Y 0 = −2u1 sin 2x + 2u2 cos 2x
Y 00 = −2u01 sin 2x + 2u02 cos 2x − 4u1 cos 2x − 4u2 sin 2x
Since Y 00 + 4Y = sin x ⇒
−2u01 sin 2x + 2u02 cos 2x = sin x
sin 2x 0
u
cos 2x 2
(sin 2x)2
+ 2u02 cos 2x = sin x
2u02
cos 2x
1
u02 = sin x cos 2x
2
u01 = −
Z
1
u2 =
sin x cos 2xdx
2
Z
1
1
=
[sin(x + 2x) − sin(x − 2x)]dx
2
2
1
1
= − cos 3x − cos x + C3
12
4
1
u01 = − sin x sin 2x
2Z
1
u1 = −
sin x sin 2xdx
2
1
1
= − sin x +
sin 3x + C4
4
12
1
1
1
1
Y (x) = [− sin x +
sin 3x] cos 2x + [− cos 3x − cos x] sin 2x
4
12
12
4
y(x) = C1 cos 2x + C2 sin 2x + Y (x)
Summary To solve y 00 + p(x)y 0 + q(x)y = g(x) (*)
Suppose y1 (x) and y2 (x) are solutions of y 00 + p(x)y 0 + q(x)y = 0
Y (x) is a particular solution of (*), in the form of Y (x) = u1 (x)y1 (x) + u2 (x)y2 (x). Take the first
derivative of Y , we have
Y 0 = u01 y1 + u02 y2 + u1 y10 + u2 y20 .
16
Suppose u01 y1 + u02 y2 = 0, then Y 0 = u1 y10 + u2 y20 .
Y 00 = u01 y10 + u02 y20 + u1 y100 + u2 y200
Since Y 00 + pY 0 + qY = g
u01 y10 + u02 y20 + u1 y100 + u2 y200 + p[u1 y10 + u2 y20 ] + [u1 y1 + u2 y2 ] = g
⇒ u01 y10 + u02 y20 + u1 [y100 + py10 + qy1 ] +u2 [y200 + py2 + qy2 ] = g
{z
}
|
{z
}
|
=0
=0
⇒ u01 y10 + u02 y20 = g
u01 y1 + u02 y2 = 0
u01 y10 + u02 y20 = g
0 y2 g y0 y2 g
2
=−
u01 = y1 y2 W [y1 , y2 ](x)
0
y y0 1
2
y1 0 0
y g y1 g
1
=
u02 = y1 y2 W [y1 , y2 ](x)
0
y y0 1
2
Z
Z
gy1
gy2
dx, u2 =
dx
u1 = −
W [y1 , y2 ](x)
W [y1 , y2 ](x)
Z
Z
gy2
gy1
Y (x) = −y1
dx + y2
dx
W [y1 , y2 ](x)
W [y1 , y2 ](x)
By
Z
y = C1 y1 (x) + C2 y2 (x) − y1
gy2
dx + y2
W [y1 , y2 ](x)
Z
gy1
dx
W [y1 , y2 ](x)
Theorem If y1 and y2 are a fundamental set of solutions of
y 00 + p(x)y 0 + q(x)y = 0
then a particular solution of
y 00 + p(x)y 0 + q(x)y = g(x)
is
Z
Y (x) = −y1
gy2
dx + y2
W [y1 , y2 ](x)
Z
gy1
dx
W [y1 , y2 ](x)
and general solution is
y = C1 y1 (x) + C2 y2 (x) + Y (x)
17
Example 1 y 00 + 4y = sin x
y 00 + 4y = 0 gives us y1 = cos 2x, y2 = sin 2x
cos 2x
sin 2x W [y1 , y2 ](x) = =2
−2 sin 2x 2 cos 2x Z
Z
sin 2x sin x
1
1
gy2
dx =
dx = sin x −
sin 3x + C1
W [y1 , y2 ](x)
2
4
12
Z
Z
gy1
cos 2x sin x
1
1
dx =
dx = − cos x −
cos 3x + C2
W [y1 , y2 ](x)
2
4
12
A particular solution is
1
1
1
1
Y (x) = − cos 2x[ sin x −
sin 3x] + sin 2x[− cos x −
cos 3x]
4
12
4
12
General solution is
y(x) = C1 cos 2x + C2 sin 2x + Y (x)
Example 2: y 00 − y 0 − 2y = 2e−x
To find a set fundamental solutions of y 00 − y 0 − 2y = 0,
r2 − r − 2 = 0
(r − 2)(r + 1) = 0
r = 2,
r = −1
y1 = e2x ,
y2 = e−x
−x y1 y2 e2x
e
=
= −ex − 2ex = −3ex
W [y1 , y2 ](x) = 0
y1 y20 2e2x −e−x Z
Z
Z
2e−x · e−x
2
2
gy2
dx =
dx = −
e−3x dx = e−3x
x
W [y1 , y2 ](x)
−3e
3
9
Z
Z
Z
gy1
2e−x · e2x
2
2x
dx =
dx
=
− dx = −
x
W [y1 , y2 ](x)
−3e
3
3
2x
2 2x −x
2
A particular solution is Y = −e2x · e−3x − e−x ·
=−
+
e
9
3
9
3
General solution is y(x) = C1 e2x + C2 e−x + Y (x)
18
3.7
Mechanical and Electrical Vibrations
l
l+L
still
l + L + u(t)
m
still
u(t)
Fs
m
vibrated
Fg = mg
undamped free vibration
m: mass. k: spring constant. g: gravity constant. L: length of spring stretched by mass m.
mg = kL.
mu00 + ku = 0
r
u = A cos
k
t + B sin
m
r
k
t
m
r
= A cos w0 t + B sin w0 t
w0 =
k
m
= R cos(w0 t − δ)
Here A = R cos δ, B = R sin δ or R =
r
w0 =
T =
p
B
A2 + B 2 , tan δ = .
A
k
: in radians per unit time, natural frequency of a vibration.
m
2π
: period of a vibration.
w0
R: amplitude.
19
R
R cos δ I
2π + δ
δ
w0 t
δ: phase or phase angle.
Damped free vibration
air resistance force = γv(t),
r1,2
γ : damped constant
mu00 + γu0 + ku = 0, γ > 0
s
!
p
γ
4km
−γ ± γ 2 − 4mk
=
=
−1 ± 1 − 2
2m
2m
γ
γ 2 − 4km > 0: u = Aer1 t + Ber2 t
γt
γ 2 − 4km = 0: u = (A + Bt)e− 2m
p
4km − γ 2
γ − 4km < 0: u = e
> 0,
(A cos µt + B sin µt) = Re
cos(µt + δ), µ =
2m
p
2π
A
R = A2 + B 2 , tan δ = . µ: quasi frequency, Td =
: quasi-period.
B
µ
2
γ
− 2m
t
γ
− 2m
t
20
Re−γt−δ
R cos δ
δ
Damped vibration with external force
mu00 + γ u0 + ku = F0 cos wt
| {z }
|{z}
damped constant
external periodical force
(a) γ = 0, w 6= w0
u = C1 cos w0 t + C2 sin w0 t +
= R cos(w0 t − δ) +
F0
cos wt
− w2 )
m(w02
F0
cos wt
− w2 )
m(w02
u(0) = 0, u0 (0) = 0 (no initial displacement, no initial velocity)


F0
F0
 R cos δ +
 R=−
=
0
2
2
2
⇒
m(w0 − w )
m(w0 − w2 )


−Rw0 sin δ = 0
δ=0
F0
(cos wt − cos w0 t)
− w2 )
2F0
w0 + w
(w0 − w)t
=
sin
sin
t
2
2
2
2
m(w0 − w )
u=
m(w02
Example 1
u00 + u = 0.5 cos 0.8t
u(0) = 0,
u0 (0) = 0
u = 2.77778 sin 0.1t sin 0.9t
21
Graph of u = 2.77778 sin 0.1t sin 0.9t
2.77778 sin 0.1t
(b) w = w0 u = C1 cos w0 t + C2 sin w0 t +
F0
t sin w0 t
2mw0
t → ∞ u → ∞.
Example 2 u00 + u = 0.5 cos t, u(0) = 0, u0 (0) = 0 then u = 0.25t sin t.
y = 0.25t
This phenomenon is called ”resonance”.
22
3.8
Review of Chapter 3
• Constant coefficients homogeneous equations
b2 − 4ac
>0
<0
ay 00 + by 0 + cy = 0, ar2 + br + c = 0
particular solutions
general solution
r1 x
r2 x
y1 = e , y2 = e
y = C1 er1 x + C2 er2 x
y1 = eλx cos(µx), y2 = eλx sin(µx)
y = eλx [C1 cos(µx) + C2 sin(µx)]
=0
y1 = e− 2a x , y2 = xe− 2a x
b
b
b
y = [C1 + C2 x]e− 2a x
• Constant coefficients non-homogeneous equations
ay 00 + by 0 + cy = g(x)
general solutions:
y = C1 y1 (x) + C2 y2 (x) + Y (x)
where y1 and y2 are a set of fundamental solutions of ay 00 + by 0 + cy = 0, and Y (x) is a particular
solution satisfying
aY 00 + bY 0 + cY = g(x)
– Method 1: undetermined coefficients
a b c and d are known constants, and A, B, C and D are unknown constants to be determined.
g(x)
eax
cos(ax)
sin(ax)
ax3 + bx2 + cx + d
(ax + b)eax
(ax + b) cos(ax)
(ax + b) sin(ax)
sin(ax)ebx
Y (x)
Aeax
A cos(ax) + B sin(ax)
A cos(ax) + B sin(ax)
Ax3 + Bx2 + Cx + D
(Ax + B)eax
(Ax + B) cos(ax) + (Cx + D) sin(ax)
ebx [A sin(ax) + B cos(ax)]
Note: if g(x) is also a solution to the homogeneous equation, then multiply x to Y (x) in the
above table.
– Method 2: Variation of parameters
Z
Z
gy2
gy1
Y (x) = −y1
dx + y2
dx
W [y1 , y2 ](x)
W [y1 , y2 ](x)
• Existence and Uniqueness theorem
y 00 + p(t)y 0 + q(t)y = g(t),
y(t0 ) = y0 ,
y 0 (t0 ) = y00
If p(x), q(x) and g(x) are continuous in I : t0 ∈ I, then there is exactly one solution y = φ(x) of
the IVP (Initial Value Problem), and the solution exists throughout the interval I.
23
• Wronskian
y1 y2 W [y1 , y2 ](x) = 0
y1 y20 Abel’s Theorem If y1 and y2 are two solutions of
y 00 + p(x)y 0 + q(x)y = 0
where p and q are continuous on an open interval I, then the Wronskian W [y1 , y2 ](x) is given by
W [y1 , y2 ](x) = Ce−
R
p(x)dx
• Fundamental set of solutions: If y1 and y2 are two solution of
y 00 + p(x)y 0 + q(x)y = 0
and if there is a point x0 where the wronskian of y1 and y2 is nonzero, i.e.,
y1 (x0 ) y2 (x0 ) ,
W [y1 , y2 ](x0 ) = 0
y1 (x0 ) y20 (x0 ) then the general solution of the DE is
y = C1 y1 (x) + C2 y2 (x)
where C1 and C2 are arbitrary constants. y1 and y2 are said to form a fundamental set of solutions.
• Spring mechanical system
(a) Undamped free vibration: mu00 + ku = 0
(b) Damped free vibration: mu00 + γu0 + ku = 0
(c) Damped free vibration with external periodic force: mu00 + γu0 + ku = F0 cos(wt)
24
Chapter 4
The Laplace Transform
Integral transforms are a class of most useful tools for solving linear differential equations. We are going
to study one of these transforms: the Laplace transform.
4.1
Definition of the Laplace transform
Consider a function f (t) defined for t ≥ 0. Define a new function
Z ∞
L{f (t)} ≡ F (s) =
est f (t)dt
0
Improper integrals:
Z
∞
Z
A
f (t)dt = lim
a
A→∞ a
f (t)dt
divergence, convergence, existence
Piecewise Continuous: A function is said to be piecewise continuous on an interval α ≤ t ≤ β if the
interval can be partitioned by a finite number of points α = t0 < t1 < · · · < tn ≤ β such that
1. f is continuous on each open subinterval ti−1 < t < ti (i = 1, · · · , n).
2. f has a finite limit at the endpoints of each subinterval.
Examples:
• f (t) = tan t,
0 ≤ t ≤ π is not a piecewise continuous function
25
• f (t) =
−1, for t ∈ [−1, 0)
1,
for t ∈ [0, 1]

1, for t ∈ [0, 1)




1



 2 , for t ∈ [1, 2)
···
• f (t) =


1


, for t ∈ [n − 1, n)



 ·n· ·
26
2
1
−2
−1
1
0
2
3
4
5
6
−1
−2
The Laplace transform: if f (t) is a piece-wise continuous function,
Z ∞
e−st f (t)dt
L{f (t)} ≡ F (s) =
0
Examples
∞
Z
• L{1} =
−st
e
0
• L{eat } =
1 −st ∞
1 −∞ 0
1
dt =
e = − [e|{z}
−e ] =
−s
s
s
0
=0
∞
Z
eat · e−st dt =
0
Z
0
Z
• L{sin(at)} =
∞
∞

∞  ∞,
s≤a
1 (a−s)t 1
=
e(a−s)t dt =
e
, s>a

a−s
0
s−a
sin(at)e−st dt =
0
a
a2 + s2
(s > 0)
Property: α, β numbers, f (t), g(t)
L{αf (t) + βf (t)} = αL{f (t)} + βL{g(t)}.
Integration Table
f (t)
af (t)+bg(t)
1
a
F (s) = L{f (t)}
aF (s) + bF (s)
1/s
a/s
t
tn /n!
e−αt
sin(t)
sin(at)
cos(t)
cos(at)
1/s2
1/sn+1 , n is integer
1/(s + α)
1/(1 + s2 )
a/(a2 + s2 )
s/(1 + s2 )
s/(a2 + s2 )
27
4.2
Solution of Initial Value Problems
Theorem 4.2.1 Suppose that |f (t)| ≤ Keat for t ≥ M . Then,
L{f 0 (t)} = sL{f (t)} − f (0). (s > a)
Proof.
∞
Z
0
−st
f (t)e
∞
Z
dt =
e−st df
0
0
= e−st f (t)|∞
0 −
Z
∞
f (t)de−st
Z0 ∞
= [0 − f (0)] + s
f (t)e−st dt
0
= sL{f } − f (0).
Z
∞
00
f (t)e
−st
Z
dt =
0
∞
e−st df 0
0
=e
−st 0
f
(t)|∞
0
0
Z
∞
−
Z0 ∞
= [0 − f (0)] + s
|0
f 0 (t)d−st
f 0 (t)e−st dt
{z
}
=L{f 0 }
= sL{f 0 } − f 0 (0)
= s2 L{f } − sf (0) − f 0 (0).
Generalization:
L{f (n) (t)} = S n L{f (t)} − sn−1 f (0) − sn−2 f 0 (0) − · · · − f (n−1) (0)
L{f 0 (t)} = sL{f (t)} − f (0)
L{f 00 (t)} = s2 L{f (t)} − sf (0) − f 0 (0)
Examples:
1. y 00 − 2y 0 − 2y = 0, y(0) = 2, y 0 (0) = 0
28
Sol:
y 00 (t) − 2y 0 (t) − 2y(t) = 0
0
=0
s
L{y 00 (t)} − 2L{y 0 (t)} − 2L{y(t)} = 0
L{y 00 (t) − 2y 0 (t) − 2y(t)} = L{0} =
Define: Y (s) = L{y(t)}
L{y 0 (t)} = sY (s) − y(0) = sY (s) − 2
L{y 00 (t)} = s2 Y (s) − sy(0) − y 0 (0) = s2 Y (s) − 2s
[s2 · Y (s) − 2s] − 2[s · Y (s) − 2] − 2Y (s) = 0
[s2 − 2s − 2]Y (s) = 2s − 4
2s − 4
Y (s) = 2
s − 2s − 2
The inverse Laplace transform:
y(t) = L−1 {Y (s)}
2s − 4
A
B
√
√
√ +
√
=
[s − (1 + 3)][s − (1 − 3)]
s − (1 + 3) s − (1 − 3)
√
√
(A + B)s − A(1 − 3) − B(1 + 3)
√
√
=
[s − (1 + 3)][s − (1 − 3)]
A + B√
=2
√
A(1 − 3) + B(1 + 3) = 4
1√
1√
2
A
=
4
B
1− 3 1+ 3
A = 0.4226, B = 1.5774
0.4226
1.5774
−1
−1
√
√
y(t) = L
+L
s − (1 + 3)
s − (1 − 3)
√
0.4226
1
−1
−1
√
√
= 0.4226 · L
= 0.4226e(1+ 3)t
L
s − (1 + 3)
s − (1 + 3)
√
1.5774
1
−1
−1
√
√
L
= 1.5774 · L
= 1.5774e(1− 3)t
s − (1 − 3)
s − (1 − 3)
29
method 2:
s2
2s − 4
2s − 4
=
− 2s − 2
(s − 1)2 − 3
2(s − 1) − 2
√
=
(s − 1)2 − ( 3)2
√
s−1
2
3
√
√
=2·
−√ ·
2
2
2
(s − 1) − ( 3)
3 (s − 1) − ( 3)2
s
√
2
s − ( 3)2
↓
s−1
√
(s − 1)2 − ( 3)2
↓
s−1
√
2·
(s − 1)2 − ( 3)2
F (s) =
√
f (t) = cosh( 3t)
√
3
√
2
s − ( 3)2
√↓
3
√
2
(s − 1) − ( 3)2
↓√
2
3
√ ·
√
2
3 (s − 1) − ( 3)2
F (s) =
2. y 00 + y = sin 2t,
y(0) = 2,
↓
√
e · cosh( 3t)
t
↓
√
e · cosh( 3t)
t
√
f (t) = sinh( 3t)
↓
√
et · sinh( 3t)
↓
√
2
√ · et sinh( 3t)
3
y 0 (0) = 1
Let Y (s) = L{y(t)}
L{y 00 + y} = L{sin 2t}
2
L{y 00 } + L{y} = 2
2 + s2
s2 Y − sy(0) − y 0 (0) + Y =
2
22 + s2
2
+ 2s + 1
+ s2
2
2s + 1
+ 2
Y = 2
2
(s + 4)(s + 1) s + 1
2
−1 2s + 1
y = L−1
+
L
(s2 + 4)(s2 + 1)
s2 + 1
(s2 + 1)Y =
22
30
2
A
B
A(s2 + 1) + B(s2 + 4)
=
+
=
(s2 + 4)(s2 + 1)
s2 + 4 s2 + 1
(s2 + 4)(s2 + 1)
A+B =0
⇒
A + 4B = 2
A = −2/3,
B = 2/3
−2/3
2/3
2
−1
−1
=L
+L
L
(s2 + 4)(s2 + 1)
s2 + 2 2
s2 + 1
1
2
2
1
= − · L−1
+ L−1
3
s2 + 2 2
3
s2 + 1
2
1
= − sin(2t) + sin t
3
3
s
1
−1 2s + 1
−1
−1
L
= 2L
+L
s2 + 1
s2 + 1
s2 + 1
= 2 cos t + sin t
−1
2
1
y = − sin(2t) + sin t + 2 cos t + sin t
3
3
3. y (4) − y = 0,
y(0) = 0,
y 0 (0) = 1,
y 00 (0) = 0,
y 000 (0) = 0
Let Y (s) = L{y(t)}
L{y (4) } − L{y} = L{0} = 0
s4 Y − s3 y(0) − s2 y 0 (0) − sy 00 (0) − y 000 (0) − Y = 0
(s4 − 1)Y − s2 = 0
s2
A
B
s2
=
= 2
+
s4 − 1
(s2 − 1)(s2 + 1)
s − 1 s2 + 1
A = B = 1/2
Y =
1/2
1/2
−1
y=L { 2
}+L
s −1
s2 + 1
1 −1
1
1 −1
1
= L
+ L
2
s2 − 1
2
s2 + 1
1
1
= sinh t + sin t
2
2
−1
4.3
Step functions
Definition of the unit step function:
uc (t) =
0, for t ≤ c
1, for t > c
31
or 1 − uc (t)
3
3
y = u(t)
2
uc (t) = u(t − c)
2
1
1
0
0
−1
−1
Hat function

 0, for t ≤ a
c, for a < t ≤ b = c[ua (t) − ub (t)]
f (t) =

0, for t > b
c[ua (t) − ub (t)]
c
a
b
Examples: Express the following functions in terms of unit step function.

 0, for t ≤ 2
1, for 2 < t ≤ 6
• f (t) =

0, for t > 6
Sol: = u2 (t) − u6 (t).
32
c
2
1
−1
0
1
2
3
4
5
6
7
−1

0≤t<1
 1,
−2, 1 ≤ t < 2
• f (t) =

3,
t≥2
sol: f (t) = [u(t) − u1 (t)] + (−2) · [u1 (t) − u2 (t)] + 3 · u2 (t)
5
4
3
2
1
−2
−1
0
1
2
3
4
5
−1
−2
−3

0≤t<1
 t,
t − 1, 1 ≤ t < 2
• f (t) =
 2
t ,
t≥2
Sol: f (t) = t · [u(t) − u1 (t)] + (t − 1) · [u1 (t) − u2 (t)] + t2 · u3 (t).

2,
t < −2


 (t − 1)2 −2 ≤ t < 4
√
• f (t) =

t
4≤t<5


4
5≤t
33
6
Sol: f (t) = 2[1 − u−2 (t)] + (t − 1)2 [u−2 (t) − ut (t)] +
√
t[u4 (t) − u5 (t)] + 4u5 (t)
Translation of a function
• Given b > 0
– f (x − b) is the translation of f (t) to the right by b units.
– f (x + b) is the translation of f (t) to the left by b units.
5
f (x) = x2
y = (x − 2)2
4
3
2
1
−3
−2
−1
0
1
2
3
4
−1
• uc (t)f (t) is the cut-off of f (t) at c.
• uc (t)f (t − b): move f by b units ((a)to the right, if b > 0; (b) to the left if b < 0). And then cut off
f (t − b) at c.
34
6
5
4
f (x) = x2 + 2
u2 (x) · f (x − 2)
3
2
1
−4
−3
−2
−1
0
1
2
3
4
−1
Laplace transforms: Suppose F (s) = L{f (t)}
• L{uc (t)f (t − c)} = e−cs F (s)
• uc (t)f (t − c) = L−1 {e−cs F (s)}
Example 1: Given
π
0≤t<
4
f (t) =
 sin t + cos(t − π/4), t ≥ π
4

 sin t,
find L{f (t)}.
35
5
6
3
2
1
−1
0
1
2
3
4
5
6
7
8
−1
−2
−3
Note that f (t) = sin t + g(t), where

π
 0,
t<
4 = uπ/4 · cos(t − π/4)
g(t) =
 cos(t − π ), t ≥ π
4
4
L{f (t)} = L{sin t} + L{uπ/4 cos(t − π/4)}
πs
= L{sin t} + e− 4 L{cos t}
1
s
= 2
+ e−πs/4 2
s +1
s +1
4.4
Differential Equations with Discontinuous Forcing Functions
Example 1 Find the solution of the differential equation
2y 00 + y 0 + 2y = g(t),
where
g(t) = u5 (t) − u20 (t) =
y(0) = 0,
y 0 (0) = 0
1,
5 ≤ t < 20
0, 0 ≤ t < 5 and t ≥ 20
Soln: Let Y = L{y}
L{2y 00 + y 0 + 2y} = L{g}
2[s2 Y − sy(0) − y 0 (0)] + sY − y(0) + 2Y = L{u5 } − L{u20 }
(2s2 + s + 2)Y = e−5s /s − e−20s /s
1
1
Y = e−5s ·
− e−20s ·
s(2s2 + s + 2)
s(2s2 + s + 2)
36
1
A
Bs + C
= + 2
+ s + 2)
s
2s + s + 2
2
(2A + B)s + (A + C)s + 2A
=
s(2s2 + s + 2)
A = 1/2, B = −1, C = −1/2
−s − 1/2
11
+ 2
=
2 s 2s + s + 2
s(2s2
−s − 1/2
s + 1/2
=− 2
2
2s + s + 2
2(s + 1/2s + 1)
1
s + 1/2
=− · 2
2 s + 1/2s + 1/4 + 3/4
s + 1/4 + 1/4
1
√
=− ·
2 (s + 1/4)2 + ( 3/2)2
1/8
1
s + 1/4
√
√
−
=− ·
2 (s + 1/4)2 + ( 3/2)2 (s + 1/4)2 + ( 3/2)2
11
1
= t
2s
2
√
1
s + 1/4
1
−1
√
L
− ·
= − e−1/4t cos( 3/2t)
2
2
2 (s + 1/4) + ( 3/2)
2
(
)
√
1/8
1
3/2
−1
−1
√
√
L
= √
L
(s + 1/4)2 + ( 3/2)2
8( 3/2)
(s + 1/4)2 + ( 3/2)2
√
1
= √ e−1/4t sin( 3/2t)
4 3
L−1
Example 2
y 00 + 4y = g(t), y(0) = 0, y 0 (0) = 0

 0, 0 ≤ t < 5
(t − 5)/5,
5 ≤ t < 10
g(t) =

1,
t ≥ 10
soln:
g(t) = (t − 5)/5 · [u5 (t) − u10 (t)] + u10 (t)
37
L{y 00 + 4y} = L{g(t)}
1
1
s2 Y − sy(0) − y 0 (0) + 4Y = L{u5 · (t − 5)} − L{u10 · (t − 5)} + L{u10 }
5
5
we can not calculate L{u10 · (t − 5)} directly
u10 · (t − 5) = u10 · (t − 10 + 5) = u10 · (t − 10) + 5u10
e−5s e−10s
−
5s2
5s2
e−5s
e−10s
Y = 2 2
− 2 2
5s (s + 4) 5s (s + 4)
(s2 + 4)Y =
1
A
B
= 2+ 2
s2 (s2 + 4)
s
s +4
(A + B)s2 + 4A
=
s2 (s2 + 4)
A + B = 0, 4A = 1
B = −1/4
A = 1/4,
Let H(s) =
1
5s2 (s2
+ 4)
1
1 −1
L {1/s2 } − L−1 {2/(s2 + 22 )}
20
40
= 1/20t − 1/40 sin(2t)
h(t) = L−1 {H(s)} =
L−1 {e−5s H(s)} = u5 · h(t − 5) = u5 · [1/20(t − 5) − 1/40 sin(2(t − 5))]
L−1 {e−10s H(s)} = u1 0 · h(t − 10) = u1 0 · [1/20(t − 10) − 1/40 sin(2(t − 10))]
4.5
Impulse functions – Dirac delta function
Def: Dirac delta function, or unit impulse function

t 6= 0
 δ(t)
Z ∞ = 0,
δ(t)dt = 1

−∞
Properties
Z
∞
δ(t − t0 )dt = 1
1.
−∞
Z
2.
∞
f (t)δ(t − t0 )dt = f (t0 ) sifting property, or sampling operator.
Z ∞
Z
Z ∞
proof.
f (t)δ(t − t0 )dt =
f (t0 )δ(t − t0 )dt = f (t0 )
δ(t − t0 )dt = f (t0 )
−∞
−∞
−∞
−∞
38
3. L{δ(t − t0 )} = e−st0
Example 1
y 00 + 2y 0 + 2y = δ(t − π),
y(0) = 1, y 0 (0) = 0
Soln: Let Y (s) = L{y(t)}
L{y 00 + 2y 0 + 2y} = L{δ(t − π)}
s2 Y − sy(0) − y 0 (0) + 2[sY − y(0)] + 2Y = e−πs
(s2 + 2s + 2)Y = e−πs + s + 2 (∗)
s+2
1
+
Y = e−πs 2
s + 2s + 2 s2 + 2s + 2
1
1
=
2
s + 2s + 2
(s + 1)2 + 1
1
−1
L
= e−t sin(t)
(s + 1)2 + 1
1
−1
−πs
L
e
= uπ · e−(t−π) sin(t − π)
s2 + 2s + 2
s+2
s+1
1
s+2
=
=
+
2
2
2
s + 2s + 2
(s + 1) + 1
(s + 1) + 1 (s + 1)2 + 1
s+1
1
−1
L
= e−t [cos(t) + sin(t)]
+
(s + 1)2 + 1 (s + 1)2 + 1
y(t) = uπ · e−(t−π) sin(t − π) + e−t [cos(t) + sin(t)]
(s2 + 2s + 2)Y1 = s + 2
Then Y1 (s) is the laplace transform of the
(s2 + 2s + 2)Y2 = e−πs
solution to the homogeneous equation y 00 + 2y 0 + 2y = 0, y(0) = 1, y 0 (0) = 0, and Y2 (s) is the laplace
transform of the solution to y 00 + 2y 0 + 2y = δ(t − π), y(0) = y 0 (0) = 0. And Y = Y1 + Y2 .
Actually, we can rewrite (*) as
Damping system
my 00 + γy 0 + ky = f
m: mass, k: spring constant, γ damping constant, f : external force.
• y 00 + 2y 0 + 2y = 0, y(0) = 1, y 0 (0) = 0. The displacement y(0) = 1 from the equilibrium
position generates the potential energy. Once released, the potential energy and the kinetic energy
are transformed into each other alternatively.
√
√
√
Soln: y = e−t [cos t + sin t] = 2e−t [1/ 2 cos t + 1/ 2 sin t]
√
√
= 2e−t [cos t cos(π/4) + sin t sin(π/4)] = 2e−t cos(t − π/4)
39
1
y=
√
0
y=
2e−t
√
2e−t cos(t − π)
−1
• y 00 + 2y 0 + 2y = δ(t − π), y(0) = y 0 (0) = 0: impulsive force was applied to the system at the
instant moment of time π.
Soln: y = uπ · e−(t−π) sin(t − π)
1
y=e
−t
sin t
y = e−t
0
−1
1
y = e−(t−π) sin(t − π)
0
−1
40
y = e−(t−π)
Graph of y(t)
1
0
−1
Example 2
00
y +y =
10
X
y(0) = 0, y 0 (0) = 0
δ(t − k),
k=0
L{y 00 + y} =
(s2 + 1)Y =
10
X
L{δ(t − k)}
k=0
10
X
e−ks
k=0
1
+1
h(t) = L−1 {H(s)} = sin t
Let H(s) =
s2
L−1 {e−ks H(s)} = u(t − k) sin(t − k)
y(t) =
10
X
time-shifting of sin(t) by k units
u(t − k) sin(t − k)
k=0
4.6
The Convolution Integral
Def: convolution
t
Z
(f ∗ g)(t) =
f (t − τ )g(τ )dτ
0
Properties
1. f ∗ g = g ∗ f (commutative law)
41
Figure 4.1: Graph of y(t)
2. f ∗ (ag1 + bg2 ) = af ∗ g1 + bf ∗ g2 (distributive law)
3. f ∗ (g ∗ h) = (f ∗ g) ∗ h (associative law)
4. f ∗ 0 = 0 ∗ f = 0
5. L{f ∗ g} = F (s)G(s) where F (s) = L{f }, G(s) = L{g}.
Example 1 Find the Laplace transform of the given function
Z
t
(t − τ )2 cos(2τ )dτ
(a) f (t) =
0
Soln: f (t) = t2 ∗ cos(2t), ⇒ F (s) =
Z
(b) f (t) =
t
2
s
·
s3 s2 + 4
e−(t−τ ) sin τ dτ .
0
SolnL f (t) = e−t ∗ sin t, ⇒ F (s) =
1
1
· 2
s−1 s +1
Example 2 Find the inverse laplace transform of
H(s) =
a
s2 (s2 + a2 )
sol:
F (s) = 1/s2 ,
G(s) = a/(s2 + a2 )
f (t) = t,
g(t) = sin(at)
Z t
at − sin(at)
h(t) = (f ∗ g)(t) =
(t − τ ) sin(aτ )dτ =
.
a2
0
42
Integral Equation
1. Volterra integral equations: Find φ(t) of
Z t
φ(t) +
k(t − τ )φ(τ )dτ = f (t)
0
Z
t
k(t − τ )φ(τ )dτ = k ∗ φ
0
L{φ(t) + (k ∗ φ)(t)} = L{f (t)}
Φ(s) + K(s)Φ(s) = F (s)
F (s)
Φ(s) =
1 + K(s)
F (s)
−1
φ(t) = L
1 + K(s)
Example 1 Find φ(t) of
Z
t
(t − τ )φ(τ )dτ = sin(2t)
φ(t) +
0
Z
t
(t − τ )φ(τ )dτ = t ∗ φ(t)
0
L{φ(t) + t ∗ φ(t)} = L{sin(2t)}
1
4
Φ(s) + 2 Φ(s) = 2
s
s +4
2
4s
A
B
Φ(s) = 2
= 2
+
(s + 4)(s2 + 1)
s + 4 s2 + 1
16
4
A= , B=−
3
3
16
1
8
2
8
· 2
= · 2
→ sin(2t)
3 s +4
3 s +4
3
4
1
4
− · 2
→ − sin(t)
3 s +1
3
8
4
y(t) = sin(2t) − sin(t)
3
3
Volterra integral equations find application in demography, the study of viscoelastic materials, and
in insurance mathematics through the renewal equation.
2. Integro-differential equations, find φ(t) of
Z t
0
k(t − τ )φ(τ )dτ = f (t),
φ (t) +
0
43
φ(0) = 0
0
Z
t
k(t − τ )φ(τ )dτ } = L{f (t)}
L{φ (t) +
0
sΦ(s) + φ(0) + K(s)Φ(s) = F (s)
F (s)
Φ(s) =
s + K(s)
Impulse response function: Consider
ay 00 + by 0 + cy = δ(t),
y(0) = y 0 (0) = 0
(i)
L{ay 00 + by 0 + cy} = (as2 + bs + c) Y (s)
|
{z
}
characteristic function
L{δ(t)} = 1
Y (s) =
as2
Let H(s) =
1
+ bs + c
1
. It is known as the transfer function.
as2 + bs + c
The solution to (i), h(t) = L−1 {H(s)} is called the impulse response of the system (i).
Consider the following system with forcing function g(t)
Applying Laplace transform
ay 00 + by 0 + cy = g(t),
y(0) = y 0 (0) = 0
(ii)
(as2 + bs + c)Y (s) = G(s)
1
Y (s) = G(s) · 2
= G(s)H(s)
as + bs + c
y(t) = L−1 {G(s)H(s)} = (g ∗ h)(t)
In the time domain, the transfer function transfers any input g(t) to the output y(t) = (g ∗ h)(t); in the
frequency domain, it transfers the input G(s) to the output Y (s) = G(s)H(s). Once we know the impulse
response function h(t), in the time domain, the solution to the system (ii) with forcing function g(t) is
g ∗ h.
44
4.7
Review of Chapter 4
1. Definition of Laplace transform and the table of common Laplace transform.
Z ∞
est f (t)dt.
L{f } =
0
2. L{f (n) } = sn F (s) − sn−1 f (0) − sn−2 f 0 (0) − · · · − f (n−1) (0)
3. Step function uc (t) = u(t − c): L{u(t)} = 1/s,
L{uc (t)} = e−cs /s
4. Delta function δ(t − c): L{δ(t)} = 1, L{δ(t − c)} = e−cs
Z τ
f (t − τ )g(τ )dτ : L{f ∗ g} = F (s)G(s)
5. Convolution integral (f ∗ g)(t) =
0
6. Transfer function and Impulse response
The Laplace transform of following IVP homogeneous equation
an y (n) + an−1 y (n−1) + · · · a1 y 0 + a0 y = δ(t),
y(0) = y 0 (0) = · · · = y (n−1) (0) = 0
is (an sn + an−1 sn−1 + · · · a1 s + a0 )Y (s) = 1. Here
D(s) = an sn + an−1 sn−1 + · · · a1 s + a0
is the characteristic function of an y (n) + an−1 y (n−1) + · · · a1 y 0 + a0 y
1
H(s) =
is called the transfer function.
an sn + an−1 sn−1 + · · · a1 s + a0
h(t) = L−1 {H(s)} is called the impulse response of the system.
Given h(t), solution to
an y (n) + an−1 y (n−1) + · · · a1 y 0 + a0 y = g(t),
y(0) = y 0 (0) = · · · = y (n−1) (0) = 0
is y(t) = (g ∗ h)(t).
7. Comparison of method of characteristic equation and Laplace transform on constant coefficient
Order
INV
General solution
Step function
Delta function
Nonhomogeneous equation
Characteristic equation
1st, 2nd (so far)
no constraint
yes
no
no
undetermined coefficient
variation of parameter
45
Laplace transform
any order.
initial value at x = 0
no
yes
yes
laplace transform