p1
Example. Find the solution of the differential equation
2y 00 + y 0 + 2y = g(t),
where
(
g(t) = u5 (t) − u20 (t) =
1, 5 ≤ t ≤ 20,
0, 0 ≤ t < 5 and t ≥ 20.
Assume that the initial conditions are
y(0) = 0,
y 0 (0) = 0.
p2
This problem governs the charge on the capacitor(電容器) in a
simple electric circut with a unit voltage pulse(電壓脈衝) for
5 ≤ t < 20. Alternatively, y may represent response of a
damped(阻尼) oscillator(振盪器) subject to the applied force(受所
施加的力) g(t).
Recall,
L{f (n) (t)}
=sn L{f (t)} − s(n−1) f (0) − s(n−2) f 0 (0) − · · ·
− sf (n−2) (0) − f (n−1) (0).
The Laplace transform of this equation
p3
2s2 Y (s) − sy(0) − 2y 0 (0) + sY (s) − y(0) + 2Y (s)
=L{u5 (t)} − L{u20 (t)}
=
e−5s − e−20s
,
s
initial values y(0) = 0, y 0 (0) = 0 and solve for Y (s), we obtain
Y (s) =
e−5s − e−20s
= (e−5s − e−20s )H(s)
s(2s2 + s + 2)
1
where H(s) = s(2s2 +s+2)
. Let y = φ(t) is a solution of the initial
value problem, then
p4
φ(t) =L−1 {Y (s)} = L−1 {(e−5s − e−20s )H(s)}
=u5 (t)h(t − 5) − u20 (t)h(t − 20)
where h(t) = L−1 {H(s)}.
H(s) =
1
a
bs + c
= + 2
.
s(2s2 + s + 2)
s 2s + s + 2
Determine the coefficients, we find that a = 1/2, b = −1, and
c = −1/2. Thus
H(s) =
1/2 (−1)s + (−1/2)
1/2 1
(s + 1/4) + 1/4
+
=
− ·
.
2
s
2s + s + 2
s
2 (s + 1/4)2 + 15/16
p5
Recall, L{1} = 1s ; s > 0, and L{eat sin bt} = (s−a)b2 +b2 ;
s−a
and L{eat cos bt} = (s−a)
s > a. We have
2 +b2 ,
h(t) =
s > a,
i
√
√
√
1 1 h −t/4
−
e
cos( 15t/4) + ( 15/15)e−t/4 sin( 15t/4) .
2 2
the graph of y = φ(t).
p6
The solution consists of three distinct parts:
(1) For 0 < t < 5;the differential equation
2y 00 + y 0 + 2y = 0;
y(0) = 0,
y 0 (0) = 0
(2) For 5 < t < 20 the differential equation
2y 00 + y 0 + 2y = 1; ;
y(5) = 0,
y 0 (5) = 0
(3) For t > 20;the differential equation
2y 00 + y 0 + 2y = 0;
y(20) ≈ 0.50162,
y 0 (20) ≈ 0.01125
p7
Exercise.
One can also show by direct computation from the equation
φ(t) = u5 (t)h(t − 5) − u20 (t)h(t − 20)
where
h(t) =
i
√
√
√
1 1 h −t/4
−
e
cos( 15t/4) + ( 15/15)e−t/4 sin( 15t/4)
2 2
that φ(t) and φ0 (t) are continuous at t = 5 and t = 20. However,
lim φ00 (t) = 0,
t→5−
and
1
lim φ00 (t) = .
2
t→5+
Consequently, φ00 (t) has a jump of 1/2 at t = 5. In a similar way,
we can show that φ00 (t) has a jump of −1/2 at t = 20.
p8
Remark. The general second order linear equation
y 00 + p(t)y 0 + q(t)y = g(t),
where p and q are continuous on some interval α < t < β, but g is
only piecewise continuous there. If φ(t) is a solution, then φ and
φ0 are continuous on α < t < β, but φ00 (t) has jump discontinuities
at the same points as g.
Proof. Exercise.
p9-0
Remark. The function δ defined to have the properties
δ(t) = 0,
and
Z
t 6= 0;
∞
δ(t)dt = 1
−∞
is called an idealized unit impulse function.
p9
Impulse function
phenomena of an impulsive nature(衝動性):
for example, voltages or forces of large magnitude that act over
very short time intervals.
Such problems often lead to differential equations of the form
ay 00 + by 0 + cy = g(t),
where g(t) is large during a short interval t0 − τ < t < t0 + τ and
is otherwise zero (g(t) = 0 for each t ∈
/ (t0 − τ, t0 + τ )). Thus, the
integral I(τ ), defined by
Z t0 +τ
I(τ ) =
g(t)dt,
t0 −τ
p10
then we have
Z
t0 +τ
I(τ ) =
Z
∞
g(t)dt =
t0 −τ
g(t)dt
−∞
since g(t) = 0 for each t ∈ (−∞, t0 − τ ] ∪ [t0 + τ, ∞).
Note: I(τ ) is a measure of the strength of the forcing function.
(1) In a mechanical system, where g(t) is a force, I(τ ) is the total
impulse of the force g(t) over time interval (t0 − τ, t0 + τ ).
(2) if y is the current in an electric circut and g(t) is the time
derivative of the voltage, then I(τ ) represents the total
voltage impressed on the circuit during the interval
(t0 − τ, t0 + τ ).
p11
t0 = 0 and g(t) is given by
(
g(t) = dτ (t) =
1
2τ ,
0,
−τ < t < τ,
t ≤ −τ or
t ≥ τ,
where τ is a small positive constant. In this case,
Z τ
1
I(τ ) =
dt = 1
−τ 2τ
independent of the value of τ , as long as τ 6= 0. to act over
shorter and shorter time interval, that is τ → 0, then we obtain
lim dτ (t) = 0,
τ →0
t 6= 0.
and since I(τ ) = 1 as τ 6= 0, so
lim I(τ ) = 1.
τ →0
p12
Remark. The function δ defined to have the properties
δ(t) = 0,
and
Z
t 6= 0;
∞
δ(t)dt = 1
−∞
is called an idealized unit impulse function. It is usually called
the Dirac delta function. A unit impulse at an arbitrary point
t = t0 is given by δ(t − t0 )
δ(t − t0 ) = 0,
and
Z
t 6= t0 ;
∞
δ(t − t0 )dt = 1
−∞
p13
Let δ(t) be defined by
δ(t) = lim dτ (t)
τ →0
where
(
dτ (t) =
1
2τ ,
0,
−τ < t < τ,
t ≤ −τ or
t ≥ τ.
We will assume that t0 > 0 and define L{δ(t − t0 )} by the equation
L{δ(t − t0 )} = lim L{dτ (t − t0 )}.
τ →0
p14
Since when τ → 0, then we have τ < t0 and t0 − τ > 0. and
dτ (t − t0 ) 6= 0 in the interval (t0 − τ, t0 + τ ). Then
Z ∞
L{dτ (t − t0 )} =
e−st dτ (t − t0 )dt
0
Z t0 +τ
=
e−st dτ (t − t0 )dt
t0 −τ
Substituting for dτ (t), we obtain
Z t0 +τ
1
1 −st t0 +τ
L{dτ (t − t0 )} =
e−st dt = −
e 2τ t0 −τ
2sτ
t0 −τ
1 −st0 sτ
=
e
(e − e−sτ )
2sτ
p15
or
L{dτ (t − t0 )} =
sinhsτ −st0
e
.
sτ
By L’Hospital’s rule,
sinhsτ
scoshsτ
= lim
= 1.
τ →0
τ →0
sτ
s
lim
Thus, we find
L{δ(t − t0 )} = e−st0 .
When t0 → 0, we have
L{δ(t)} = lim e−st0 = 1.
t0 →0
p16
In a similar way, any continuous function f , we have
Z ∞
Z ∞
δ(t − t0 )f (t)dt = lim
dτ (t − t0 )f (t)dt.
τ →0 −∞
−∞
By mean value theorem for integral, we have
Z t0 +τ
Z ∞
1
1
· 2τ · f (t∗ ),
dτ (t − t0 )f (t)dt =
f (t)dt =
2τ
2τ
−∞
t0 −τ
where t0 − τ < t∗ < t0 + τ . Hence t∗ → t0 as τ → 0, we obtain
Z ∞
δ(t − t0 )f (t)dt = f (t0 )
−∞
p17
Example. Find the solution of the initial value problem
2y 00 + y 0 + 2y = δ(t − 5),
y(0) = 0,
Hint: L{δ(t − 5)} = e−5s and L{eat sin bt} =
Sol.
y 0 (0) = 0.
b
(s−a)2 +b2
L{2y 00 + y 0 + 2y} = L{δ(t − 5)}
⇒2s2 Y (s) − 2sy(0) − 2y 0 (0) + sY (s) − y(0) + 2Y (s) = e−5s
⇒(2s2 + s + 2)Y (s) = e−5s
Thus,
Y (s) =
e−5s
e−5s
1
=
.
2
2s + s + 2
2 (s + 1/4)2 + 15/16
p18
and
−1
L
√
1
15
4 −t/4
{
sin
}= √ e
t.
2
(s + 1/4) + 15/16
4
15
Hence, we obtain
−1
y=L
or
√
15
2
−(t−5)/4
(t − 5),
sin
{Y (s)} = √ u5 (t)e
4
15
(
y=
0,
√2 e−(t−5)/4 sin
15
√
15
4 (t
t < 5,
− 5), t ≥ 5.
p19
HW
(a) By the method of variation of parameters, show that the
solution of the initial value problem
y 00 + 2y 0 + 2y = f (t);
is
Z
y=
t
y(0) = 0, y 0 (0) = 0
e−(t−τ ) f (τ ) sin(t − τ )dτ.
0
(b) Show that if f (t) = δ(t − τ ), then the solution of part (a)
reduces to
y = uπ (t)e−(t−π) sin(t − π).
p20
(c) Use a Laplace transform to solve the given initial value
problem with f (t) = δ(t − τ ) and confirm that the solution
agrees with result of part (b).