The theorem 3.2.6 on complex solutions
This section §3.3
Sample I (unstable oscillation)
Sample II: Ex 22 (Dampened Oscillation)
Sample III (Stable oscillation)
Sample IV: Ex 24 (A Little More)
III: Second Order DE
§3.3 Complex roots of the CE
Satya Mandal, KU
Satya Mandal, KU
III: Second Order DE §3.3 Complex roots of the CE
The theorem 3.2.6 on complex solutions
This section §3.3
Sample I (unstable oscillation)
Sample II: Ex 22 (Dampened Oscillation)
Sample III (Stable oscillation)
Sample IV: Ex 24 (A Little More)
Homogeneous LSODEs
◮
Recall a Homogeneous LSODEs has forms:
L(y ) = y ” + p(t)y ′ + q(t)y = 0
◮
where p(t), q(t), g (t) are functions of t.
Then, (1) can also be written as:
L(y ) = P(t)y ” + Q(t)y ′ + R(t)y = 0
◮
(1)
(2)
where P(t), Q(t), R(t), G (t) are functions of t.
The Trivial Solution: For any homogeneous equation (1,
2), y = 0 is a solution.
Satya Mandal, KU
III: Second Order DE §3.3 Complex roots of the CE
The theorem 3.2.6 on complex solutions
This section §3.3
Sample I (unstable oscillation)
Sample II: Ex 22 (Dampened Oscillation)
Sample III (Stable oscillation)
Sample IV: Ex 24 (A Little More)
Theorem 3.2.6 (page 153) on complex solutions
Theorem 3.2.6: Consider the homogeneous equation (1):
L(y ) = y ” + p(t)y ′ + q(t)y = 0
(1).
As always p(t), q(t) are real valued functions.
◮ Let y = ϕ(t) = u(t) + iv (t) is a complex solutions of the
equation (1), where u is the real part and v is the
imaginary part of y .
Then, both u, v are solutions the same equation (1).
Proof: Use linearlity.
◮
Satya Mandal, KU
III: Second Order DE §3.3 Complex roots of the CE
The theorem 3.2.6 on complex solutions
This section §3.3
Sample I (unstable oscillation)
Sample II: Ex 22 (Dampened Oscillation)
Sample III (Stable oscillation)
Sample IV: Ex 24 (A Little More)
Complex roots of the CE
◮
In this section (as was in §3.1), we consider Homogeneous
LSODEs, with constant coefficients:
L(y ) = ay ” + by ′ + cy = 0
◮
where a, b, c ∈ R are real numbers.
Recall the CE of (3) is:
ar 2 + br + c = 0
◮
(3)
(4)
Unlike §3.1, in this section, roots of the CE will be
complex. This will be the case, when b2 − 4ac < 0.
Satya Mandal, KU
III: Second Order DE §3.3 Complex roots of the CE
The theorem 3.2.6 on complex solutions
This section §3.3
Sample I (unstable oscillation)
Sample II: Ex 22 (Dampened Oscillation)
Sample III (Stable oscillation)
Sample IV: Ex 24 (A Little More)
Continued
◮
So, characteristic equation(4) has two conjugate complex
roots:
r1 = λ + iµ,
r2 = λ − iµ
◮
As in §3.1 (3) has two solutions:
y1 (t) = e r1 t = exp[(λ + iµ)t] = e λt e iµt
y2 (t) = e r2 t = exp](λ − iµ)t] = e λt e −iµt
◮
(5)
However, (5) involves complex exponentiation e iµt , e −iµt .
Satya Mandal, KU
III: Second Order DE §3.3 Complex roots of the CE
The theorem 3.2.6 on complex solutions
This section §3.3
Sample I (unstable oscillation)
Sample II: Ex 22 (Dampened Oscillation)
Sample III (Stable oscillation)
Sample IV: Ex 24 (A Little More)
Complex Exponentiation
◮
◮
For real numbers θ, we define e iθ = cos θ + i sin θ.
For complex numbers z = ρ + iθ define
e z :=e ρ+iθ := e ρ e iθ = e ρ (cos θ + i sin θ)
◮
All the rules of exponentiation that you are familiar with
works, with this definition of e z . In particular
e z+w = e z e w
◮
for all z, w ∈ C.
The author gives a motivation for this definition of e z .
Satya Mandal, KU
III: Second Order DE §3.3 Complex roots of the CE
The theorem 3.2.6 on complex solutions
This section §3.3
Sample I (unstable oscillation)
Sample II: Ex 22 (Dampened Oscillation)
Sample III (Stable oscillation)
Sample IV: Ex 24 (A Little More)
Solution of (3)
◮
◮
So the solution (5) of (3) reduces to:
y1 (t) = e λt (cos µt + i sin µt)
y2 (t) = e λt (cos µt − i sin µt)
(6)
By Theorem 3.2.6 both the real and complex parts (of y1
or y2 ) are solution of (3). We get two real solutions:
u(t) = e λt cos µt
(7)
v (t) = e λt sin µt
Satya Mandal, KU
III: Second Order DE §3.3 Complex roots of the CE
The theorem 3.2.6 on complex solutions
This section §3.3
Sample I (unstable oscillation)
Sample II: Ex 22 (Dampened Oscillation)
Sample III (Stable oscillation)
Sample IV: Ex 24 (A Little More)
Continued
◮
Wronskian of u, u is
u(t) v (t)
W (u, v )(t) = ′
u (t) v ′ (t)
λt
λt
e
cos
µt
e
sin
µt
= λt
λt
λt
λt
λe cos µt − e µ sin µt λe sin µt + e µ cos µt cos
µt
sin
µt
= µe 2λt
= e 2λt λ cos µt − µ sin µt λ sin µt + µ cos µt Satya Mandal, KU
III: Second Order DE §3.3 Complex roots of the CE
The theorem 3.2.6 on complex solutions
This section §3.3
Sample I (unstable oscillation)
Sample II: Ex 22 (Dampened Oscillation)
Sample III (Stable oscillation)
Sample IV: Ex 24 (A Little More)
Continued
◮
◮
◮
So, (since µ 6= 0), Wronskian W (u, v )(t) = µe 2λt 6= 0.
So, u, v form a fundamental set of solutions of (3).
So, the general (real) solution of (3) has the form
y = c1 u(t) + c2 v (t) = c1 e λt cos µt + c2 e λt sin µt
◮
(8)
where c1 , c2 are arbitrary constants.
We can write the same as
y = e λt (c1 cos µt + c2 sin µt)
Satya Mandal, KU
(9)
III: Second Order DE §3.3 Complex roots of the CE
The theorem 3.2.6 on complex solutions
This section §3.3
Sample I (unstable oscillation)
Sample II: Ex 22 (Dampened Oscillation)
Sample III (Stable oscillation)
Sample IV: Ex 24 (A Little More)
Behavior of the solution
If the CE (4) has complex roots or µ 6= 0, then, the solution
(9) has two factors:
◮ The exponential factor:
E (t) = e λt
◮
Depending on the sign of λ this part will "blow up" to
∞ or "decay" to the x−axis (horizontal asymptote).
The periodic factor.
Φ(t) = c1 cos µt + c2 sin µt
Satya Mandal, KU
with periodicity=
2π
µ
III: Second Order DE §3.3 Complex roots of the CE
The theorem 3.2.6 on complex solutions
This section §3.3
Sample I (unstable oscillation)
Sample II: Ex 22 (Dampened Oscillation)
Sample III (Stable oscillation)
Sample IV: Ex 24 (A Little More)
◮
Further,
−(|c1 | + |c2 |) ≤ Φ(t) ≤ (|c1 | + |c2 |)
◮
The graph of Φ(t) contributes to a steady oscillation.
The behavior of the solution y (as in (9)) will be a
combination of (1) the exponential rise/decay due to
E (t) and (2) the periodic oscillation due to Φ(t).
Satya Mandal, KU
III: Second Order DE §3.3 Complex roots of the CE
The theorem 3.2.6 on complex solutions
This section §3.3
Sample I (unstable oscillation)
Sample II: Ex 22 (Dampened Oscillation)
Sample III (Stable oscillation)
Sample IV: Ex 24 (A Little More)
Continued
So, the nature of the solutions is summarized as follows:
◮ If λ = 0 then the solution would be a steady oscillation.
◮ If λ > 0, it will be unsteady oscillation.
◮ If λ < 0 the oscillation will stabilize with time.
Satya Mandal, KU
III: Second Order DE §3.3 Complex roots of the CE
The theorem 3.2.6 on complex solutions
This section §3.3
Sample I (unstable oscillation)
Sample II: Ex 22 (Dampened Oscillation)
Sample III (Stable oscillation)
Sample IV: Ex 24 (A Little More)
Sample I
Consider the IVP:

 y ” − 2y ′ + 5y = 0
y (π/8) = 0
 ′
y (π/8) = e π/8
◮
◮
◮
Solve the problem
Sketch the graph
Describe the nature of the solution, as t → ∞
Satya Mandal, KU
III: Second Order DE §3.3 Complex roots of the CE
The theorem 3.2.6 on complex solutions
This section §3.3
Sample I (unstable oscillation)
Sample II: Ex 22 (Dampened Oscillation)
Sample III (Stable oscillation)
Sample IV: Ex 24 (A Little More)
Solution: Sample I
◮
◮
◮
The CE: r 2 − 2r + 5 = 0
Roots of the CE: r1 = 1 + 2i, r2 = 1 − 2i.
By solution (9), the general solution
y = e λt (c1 cos µt + c2 sin µt) = e t (c1 cos 2t + c2 sin 2t)
◮
We can already answer (3) that the solution will be an
unsteady oscillation; because the exponential part is e t
Satya Mandal, KU
III: Second Order DE §3.3 Complex roots of the CE
The theorem 3.2.6 on complex solutions
This section §3.3
Sample I (unstable oscillation)
Sample II: Ex 22 (Dampened Oscillation)
Sample III (Stable oscillation)
Sample IV: Ex 24 (A Little More)
Continued
◮
Before we use the initial values, compute
y ′ = e t (c1 cos 2t + c2 sin 2t) + e t (−2c1 sin 2t + 2c2 cos 2t)
◮
Initial value conditions:

 y (π/8) = e π/8 √c1 + √c2 = 0
2 2
c
c
′
π/8
π/8
 y (π/8) = e
√1 + √2
√1 +
− 2c
+e
2
2
2
c1 + c2 = 0
−c1 + c2 =
√1
2
Satya Mandal, KU
=⇒
(
2c
√2
2
c1 = − 2√1 2
c2 = 2√1 2
= e π/8
III: Second Order DE §3.3 Complex roots of the CE
The theorem 3.2.6 on complex solutions
This section §3.3
Sample I (unstable oscillation)
Sample II: Ex 22 (Dampened Oscillation)
Sample III (Stable oscillation)
Sample IV: Ex 24 (A Little More)
Continued
◮
So, the solution is
1
1
t
y = e − √ cos 2t + √ sin 2t
2 2
2 2
◮
Repeat: y = y (t) has an unsteady/unstable oscillation.
Satya Mandal, KU
III: Second Order DE §3.3 Complex roots of the CE
The theorem 3.2.6 on complex solutions
This section §3.3
Sample I (unstable oscillation)
Sample II: Ex 22 (Dampened Oscillation)
Sample III (Stable oscillation)
Sample IV: Ex 24 (A Little More)
Graph of y = y (t):
Sample I: Unstable Oscillation
50
40
30
20
10
0
−10
0
0.5
1
1.5
2
Satya Mandal, KU
2.5
3
3.5
4
4.5
5
III: Second Order DE §3.3 Complex roots of the CE
The theorem 3.2.6 on complex solutions
This section §3.3
Sample I (unstable oscillation)
Sample II: Ex 22 (Dampened Oscillation)
Sample III (Stable oscillation)
Sample IV: Ex 24 (A Little More)
On the Matlab Graph
◮
◮
It took some trial and error to get a good graph.
Following commands were used to get this graph:
◮
◮
◮
t=[0:.01:5];
y=exp(t).*(sin(2*t)-cos(2*t))/(2.*sqrt(2));
plot(t,y), title(’Sample I: Unstable Oscillation’)
Satya Mandal, KU
III: Second Order DE §3.3 Complex roots of the CE
The theorem 3.2.6 on complex solutions
This section §3.3
Sample I (unstable oscillation)
Sample II: Ex 22 (Dampened Oscillation)
Sample III (Stable oscillation)
Sample IV: Ex 24 (A Little More)
Sample II: Ex 22 (Dampened Oscillation)
Consider the IVP:

 y ” + 2y ′ + 2y = 0
y (π/4) = 2
 ′
y (π/4) = −2
◮
◮
◮
Solve the problem
Sketch the graph
Describe the nature of the solution, as t → ∞
Satya Mandal, KU
III: Second Order DE §3.3 Complex roots of the CE
The theorem 3.2.6 on complex solutions
This section §3.3
Sample I (unstable oscillation)
Sample II: Ex 22 (Dampened Oscillation)
Sample III (Stable oscillation)
Sample IV: Ex 24 (A Little More)
Solution: Ex 22
◮
◮
◮
The CE: r 2 + 2r + 2 = 0
Roots of the CE: r1 = −1 + i, r2 = −1 − i.
By solution (9), the general solution
y = e λt (c1 cos µt + c2 sin µt) = e −t (c1 cos t + c2 sin t)
◮
We can already answer (3) that the solution will be an
stabilized/dampened/ decaying oscillation; because the
exponential part is e −t
Satya Mandal, KU
III: Second Order DE §3.3 Complex roots of the CE
The theorem 3.2.6 on complex solutions
This section §3.3
Sample I (unstable oscillation)
Sample II: Ex 22 (Dampened Oscillation)
Sample III (Stable oscillation)
Sample IV: Ex 24 (A Little More)
Continued
◮
Before we use the initial values, compute
y ′ = −e −t (c1 cos t + c2 sin t) + e −t (−c1 sin t + c2 cos t)
◮
Initial value conditions:

 y (π/4) = e −π/4 √c1 + √c2 = 2
2
2
c2
c
−π/4
′
−π/4
1
 y (π/4) = −e
√ + √
− √c12 +
+
e
2
2
(
√
c1 + c2 = 2 2e π/4
−2 + e −π/4 − √c12 +
c2
√
2
Satya Mandal, KU
= −2
=⇒
c2
√
2
= −2
√
c1 + c2 = 2 2e π/4
c1 = c2
III: Second Order DE §3.3 Complex roots of the CE
The theorem 3.2.6 on complex solutions
This section §3.3
Sample I (unstable oscillation)
Sample II: Ex 22 (Dampened Oscillation)
Sample III (Stable oscillation)
Sample IV: Ex 24 (A Little More)
Continued
◮
◮
√
So, c1 = c2 = 2e π/4
So, the solution is
√
√
y = e −t
2e π/4 cost + 2e π/4 sin t
=
◮
√ −t+π/4
2e
(cos t + sin t)
Repeat: the y = y (t) has stabilized/dampened/ decaying
oscillation.
Satya Mandal, KU
III: Second Order DE §3.3 Complex roots of the CE
The theorem 3.2.6 on complex solutions
This section §3.3
Sample I (unstable oscillation)
Sample II: Ex 22 (Dampened Oscillation)
Sample III (Stable oscillation)
Sample IV: Ex 24 (A Little More)
Graph of y = y (t): The exponential part E (t) = e −t+π/4
dampens (flattens) the graphs very quickly.
Sample II: Dampened Oscillation
3.5
3
2.5
2
1.5
1
0.5
0
−0.5
0
1
2
Satya Mandal, KU
3
4
5
6
7
III: Second Order DE §3.3 Complex roots of the CE
The theorem 3.2.6 on complex solutions
This section §3.3
Sample I (unstable oscillation)
Sample II: Ex 22 (Dampened Oscillation)
Sample III (Stable oscillation)
Sample IV: Ex 24 (A Little More)
On the Matlab Graph
◮
◮
It took some trial and error to get a good graph.
Following commands were used to get this graph:
◮
◮
◮
t=[0:.01:7];
y=sqrt(2).*exp(-t+(pi/4)).*(cos(t)+sin(t));
plot(t,y), title(’Sample II: Dampened Oscillation’)
Satya Mandal, KU
III: Second Order DE §3.3 Complex roots of the CE
The theorem 3.2.6 on complex solutions
This section §3.3
Sample I (unstable oscillation)
Sample II: Ex 22 (Dampened Oscillation)
Sample III (Stable oscillation)
Sample IV: Ex 24 (A Little More)
Sample III
Consider the IVP:
◮
◮
◮

 y ” + 9y = 0
y (0) = 0
 ′
y (0) = 1
Solve the problem
Sketch the graph
Describe the nature of the solution, as t → ∞
Satya Mandal, KU
III: Second Order DE §3.3 Complex roots of the CE
The theorem 3.2.6 on complex solutions
This section §3.3
Sample I (unstable oscillation)
Sample II: Ex 22 (Dampened Oscillation)
Sample III (Stable oscillation)
Sample IV: Ex 24 (A Little More)
Solution: Sample III (Stable oscillation)
◮
◮
◮
The CE: r 2 + 9 = 0
Roots of the CE: r1 = 3i, r2 = −3i.
By solution (9), the general solution
y = e λt (c1 cos µt + c2 sin µt) = c1 cos 3t + c2 sin 3t
◮
We can already answer (3) that the solution will be a
STABLE oscillation; because there is no exponential part.
Satya Mandal, KU
III: Second Order DE §3.3 Complex roots of the CE
The theorem 3.2.6 on complex solutions
This section §3.3
Sample I (unstable oscillation)
Sample II: Ex 22 (Dampened Oscillation)
Sample III (Stable oscillation)
Sample IV: Ex 24 (A Little More)
Continued
◮
Before we use the initial values, compute
y ′ = −3c1 sin 3t + 3c2 cos 3t
◮
Initial value conditions:
y (0) = 3c2 = 0
y ′ (0) = 3c1 = 1
Satya Mandal, KU
=⇒
c1 = 13
c2 = 0
III: Second Order DE §3.3 Complex roots of the CE
The theorem 3.2.6 on complex solutions
This section §3.3
Sample I (unstable oscillation)
Sample II: Ex 22 (Dampened Oscillation)
Sample III (Stable oscillation)
Sample IV: Ex 24 (A Little More)
Continued
◮
So, the solution is
1
sin 3t
3
Repeat: y = y (t) has an STABLE oscillation.
y=
◮
Satya Mandal, KU
III: Second Order DE §3.3 Complex roots of the CE
The theorem 3.2.6 on complex solutions
This section §3.3
Sample I (unstable oscillation)
Sample II: Ex 22 (Dampened Oscillation)
Sample III (Stable oscillation)
Sample IV: Ex 24 (A Little More)
Graph of y = y (t):
Sample III: Stable Oscillation
0.4
0.3
0.2
0.1
0
−0.1
−0.2
−0.3
−0.4
0
1
2
3
4
Satya Mandal, KU
5
6
7
8
9
10
III: Second Order DE §3.3 Complex roots of the CE
The theorem 3.2.6 on complex solutions
This section §3.3
Sample I (unstable oscillation)
Sample II: Ex 22 (Dampened Oscillation)
Sample III (Stable oscillation)
Sample IV: Ex 24 (A Little More)
On the Matlab Graph
◮
◮
It took some trial and error to get a good graph.
Following commands were used to get this graph:
◮
◮
◮
t=[0:.01:10];
y=sin(3*t)/3;
plot(t,y), title(’Sample III: Stable Oscillation’)
Satya Mandal, KU
III: Second Order DE §3.3 Complex roots of the CE
The theorem 3.2.6 on complex solutions
This section §3.3
Sample I (unstable oscillation)
Sample II: Ex 22 (Dampened Oscillation)
Sample III (Stable oscillation)
Sample IV: Ex 24 (A Little More)
Sample IV : Ex 24 (A Little More)
Consider the IVP:

 3y ” − y ′ + 2y = 0
y (0) = 2
 ′
y (0) = 0
◮
◮
◮
Solve the problem
Sketch the graph
For t > 0, find first time at which |y (t)| = 10
Satya Mandal, KU
III: Second Order DE §3.3 Complex roots of the CE
The theorem 3.2.6 on complex solutions
This section §3.3
Sample I (unstable oscillation)
Sample II: Ex 22 (Dampened Oscillation)
Sample III (Stable oscillation)
Sample IV: Ex 24 (A Little More)
Solution: Sample IV: Ex 24 (A Little more)
◮
◮
◮
The CE: 3r 2 − r + 2 = 0
√
Roots of the CE: r1 = 16 + 623 i, r2 =
By solution (9), the general solution
1
6
−
√
23
i.
6
y = e λt (c1 cos µt + c2 sin µt)
=e
◮
t
6
c1 cos
√
23t
+ c2 sin
6
√
23t
6
!
We can see the solution will be an unsteady oscillation;
t
because the exponential part is e 6
Satya Mandal, KU
III: Second Order DE §3.3 Complex roots of the CE
The theorem 3.2.6 on complex solutions
This section §3.3
Sample I (unstable oscillation)
Sample II: Ex 22 (Dampened Oscillation)
Sample III (Stable oscillation)
Sample IV: Ex 24 (A Little More)
Continued
◮
Before we use the initial values, compute
!
√
√
t
1
23t
23t
y ′ = e 6 c1 cos
+ c2 sin
6
6
6
+e
◮
t
6
√
23
−
c1 sin
6
√
!
√
√
23t
23
23t
+
c2 cos
6
6
6
Initial value conditions:
y (0) = c1 = 2√
y ′ (0) = 16 2 + 623 c2 = 0
Satya Mandal, KU
=⇒
c1 = 2
c2 = − √223
III: Second Order DE §3.3 Complex roots of the CE
The theorem 3.2.6 on complex solutions
This section §3.3
Sample I (unstable oscillation)
Sample II: Ex 22 (Dampened Oscillation)
Sample III (Stable oscillation)
Sample IV: Ex 24 (A Little More)
Continued
◮
So, the solution is
y =e
=e
◮
◮
t
6
t
6
c1 cos
2 cos
√
√
23t
+ c2 sin
6
23t
2
− √ sin
6
23
√
23t
6
√
23t
6
!
!
Repeat: y = y (t) has an unsteady/unstable oscillation.
(3): It looks from the graph in the next frame:
approximately, at T = −17 it crosses −10.
Satya Mandal, KU
III: Second Order DE §3.3 Complex roots of the CE
The theorem 3.2.6 on complex solutions
This section §3.3
Sample I (unstable oscillation)
Sample II: Ex 22 (Dampened Oscillation)
Sample III (Stable oscillation)
Sample IV: Ex 24 (A Little More)
Graph of y = y (t):
Sample IV: an unstable Oscillation.
30
20
10
0
−10
−20
−30
−40
−50
−60
0
2
4
6
8
Satya Mandal, KU
10
12
14
16
18
20
III: Second Order DE §3.3 Complex roots of the CE
The theorem 3.2.6 on complex solutions
This section §3.3
Sample I (unstable oscillation)
Sample II: Ex 22 (Dampened Oscillation)
Sample III (Stable oscillation)
Sample IV: Ex 24 (A Little More)
On the Matlab Graph
◮
◮
It took some trial and error to get a good graph.
Following commands were used to get this graph:
◮
◮
◮
t=[0:.01:20];
y=exp(t/6).*(2.*cos(sqrt(23)*t/6)2.*sin(sqrt(23)*t/6)/sqrt(23));
plot(t,y), title(’Sample IV: an unstable Oscillation.’)
Satya Mandal, KU
III: Second Order DE §3.3 Complex roots of the CE
The theorem 3.2.6 on complex solutions
This section §3.3
Sample I (unstable oscillation)
Sample II: Ex 22 (Dampened Oscillation)
Sample III (Stable oscillation)
Sample IV: Ex 24 (A Little More)
§3.3 Assignments and Homework
◮
◮
◮
Read Example 1, 2,3 (the method of solution)
Suggested Problems: Exercise §3.3 (page 164) Exercise
7-16; 17-22; 23, 24
Homework: §3.3 See Homework site!
Satya Mandal, KU
III: Second Order DE §3.3 Complex roots of the CE