Ordinary Differential Equations Higher-order ODEs Problem Set 5 (Version 1.0)

Transcription

Ordinary Differential Equations
Higher-order ODEs
Problem Set 5
(Version 1.0)
Wayne Hacker
Copyright ©Wayne Hacker 2007, 2014. All rights reserved.
higher-order odes, problem set 5
Copyright ©Wayne Hacker 2007. All rights reserved. 1
Contents
1 Higher-order ODEs
1.1
2
Definitions, theorems, and terminology for higher-order ODEs . . . . . .
2
1.1.1
Determining when functions are dependent or independent . . . .
2
1.1.2
Verifying solutions to higher-order equations . . . . . . . . . . . .
4
1.2
Differential operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5
1.3
Second-order linear homogeneous ODEs . . . . . . . . . . . . . . . . . .
7
1.4
Reduction of order for second-order ODEs . . . . . . . . . . . . . . . . .
8
1.5
Higher-order linear homogeneous ODEs . . . . . . . . . . . . . . . . . . .
9
1.6
Cauchy-Euler equations
. . . . . . . . . . . . . . . . . . . . . . . . . . .
10
1.7
The method of undetermined coefficients . . . . . . . . . . . . . . . . . .
11
1.8
Variation of parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . .
12
1.9
Applications: harmonic oscillating systems . . . . . . . . . . . . . . . . .
13
1.9.1
Harmonic motion w/o friction and external forcing . . . . . . . .
13
1.9.2
Damped harmonic motion w/o external forcing . . . . . . . . . .
14
1.9.3
Forced harmonic motion w/o damping . . . . . . . . . . . . . . .
16
1
1.1
1.1.1
Higher-order ODEs
Definitions, theorems, and terminology for higher-order ODEs
Determining when functions are dependent or independent
For the following problems determine whether the functions given in the form: {f1 (x), f2 (x)}
or {f1 (x), f2 (x), f3 (x)} are independent or dependent on the entire real line (−∞, ∞).
Problem 1. Let f1 (x) = x and f2 (x) = |x|.
Problem 2. Let f1 (x) = 1, f2 (x) = x, and f3 (x) = x2 .
Problem 3. Let f1 (x) = 1, f2 (x) = cos2 x, and f3 (x) = sin2 x.
Problem 4. Let f1 (x) = x, f2 (x) = 2x, and f3 (x) = x + 3.
Problem 5. Let f1 (x) = 1, f2 (x) = cos(2x), and f3 (x) = cos2 x.
Problem 6. Let f1 (x) = ex , f2 (x) = e−x , and f3 (x) = cosh x.
For the following problems use the Wronskian to determine whether the functions given
in the form: {f1 (x), f2 (x)} or {f1 (x), f2 (x), f3 (x)} are independent or dependent on the
given interval.
Problem 7. Let f1 (x) = 1, f2 (x) = x, and f3 (x) = x2 .
Problem 8. Let f1 (x) = x1/2 and f2 (x) = x on (0, ∞).
Problem 9. Let f1 (x) = cos x and f2 (x) = sin x on (−∞, ∞).
Problem 10. Let f1 (x) = eαx and f2 (x) = xeαx on (−∞, ∞).
Problem 11. (i) Show graphically that f1 (x) = x2 and f2 (x) = x|x| are linearly independent on (−∞, ∞).
(ii) Even though |x| is not differentiable, you can still compute the Wronskian on the
intervals (0, ∞) and (−∞, 0). What result does this give? Does it contradict the result
in part (i)?
Problem 12. [When are a function and its derivative independent]
Let f (x) be a smooth function of x. It is often advantageous to know when a function
and its derivative are independent. Show that f (x) and f 0 (x) are independent so long as
f is not a solution of the differential equation
f f 00 − (f 0 )2 = 0 .
1.1.2
Verifying solutions to higher-order equations
Problem 13. Verify that y = c1 + c2 cos x + c3 sin x is a three-parameter family of
solutions of the ODE:
y 000 + y 0 = 0 on the interval (−∞, ∞).
Then find a member of the family satisfying the initial conditions: y(π) = 0, y 0 (π) = 2,
and y 00 (π) = −1.
Problem 14. Find the largest possible interval about x = 0 for which theorem 5.1
guarantees a unique solution to the following IVP:
y 00 + (tan x) y = ex ,
y(0) = 1,
y 0 (0) = 0 .
Problem 15. Determine the values of λ for which the family of curves
y = c1 cos(λx) + c2 sin(λx)
form a nontrivial solution to the boundary-value problem (BVP):
y 00 + λy = 0 y(0) = 0,
y(π) = 0 .
Such a problem where only certain parameters will satisfy boundary conditions is known
as an eigenvalue problem.
Problem 16. For the ODE x2 y 00 + xy 0 + y = 0, show that {cos(ln x), sin(ln x)} is a
fundamental solution set on the interval (−∞, ∞).
Problem 17. [a nonlinear equation]
a) Verify that y1 (x) = 1 and y2 (x) = ln x are solutions of the nonlinear differential
equation
y 00 + (y 0 )2 = 0 .
(b) Do the solutions satisfy the superposition principle? In particular, is y1 + y2 also a
solution to the ODE?
1.2
Differential operators
dy
+ 8y = x + 3 in the form L(y) = f (x),
dx
d
where L is a differential operator with constant coefficients and D = dx
.
Solution: All we have to do is change notation:
Problem 1. Rewrite the differential equation 4
4Dy + 8y = x + 3
factor from the right
−−−−−−−−−−−−−−−→
compress notation
−−−−−−−−−−−−−→
(4D + 8) y = x + 3
L(y) = 1 − sin x,
where L = 4D + 8
Problem 2. Rewrite the differential equation y (4) − 2y 00 + y = e−3x + e2x in the form
d
.
L(y) = f (x), where L is a differential operator with constant coefficients and D = dx
Solution: All we have to do is change notation:
D4 y − 2D2 y + y = e−3x + e2x
factor from the right
−−−−−−−−−−−−−−−→ (D4 − 2D2 + 1)y = e−3x + e2x
compress notation
−−−−−−−−−−−−−→
L(y) = e−3x + e2x ,
where L = D4 − 2D2 + 1
Problem 3. Factor the differential operator 2D2 − 3D − 2.
Solution: 2D2 − 3D − 2 = (2D + 1)(D − 2)
Problem 4. Factor the differential operator D3 + 4D.
Solution: D3 + 4D = D(D2 + 4). This won’t factor any farther over the reals.
Problem 5. Factor the differential operator D3 + 4D2 + 3D.
Solution: D3 + 4D2 + 3D = D(D2 + 4D + 3) = D(D + 1)(D + 3).
Problem 6. Factor the differential operator D4 − 8D2 + 16.
Solution:
D4 − 8D2 + 16 = (D2 )2 − 2 · 4 · (D2 ) + 42
= (D2 − 4)2
= ((D − 2)(D + 2))2
= (D − 2)2 (D + 2)2
Problem 7. Show that D4 annihilates y = 10x3 − 2x.
Solution:
D4 (10x3 − 2x) = D3 (D(10x3 ) − 2Dx)
= D3 (30x2 − 2)
= D2 (30Dx2 − 2D(1))
= D2 ((30 · 2x − 0)
= 60D2 x
= 60 · 2D(1)
= 120 · 0 = 0
Problem 8. Show that (D − 2)(D + 5) annihilates y = 4e2x .
Solution:
(D − 2)(D + 5)(4e2x ) = 4(D + 5)(D − 2)e2x
(We’ve used the fact that differential operators with constant coefficients commute)
= 4(D + 5)(De2x − 2e2x )
= 4(D + 5)(2e2x − 2e2x )
= 4(D + 5)(0) = 0
Remember that since Dn+1 annihilates xn for every n ∈ {0, 1, 2, . . .}, Dn+1 must likewise
annihilate any nth-degree polynomial Pn (x):
D
n+1
Pn (x) = D
n+1
n
X
k=0
k
ak x =
n
X
k=0
ak D
n+1 k
x =
n
X
ak · 0 = 0
k=0
Problem 9. Find a differential operator that annihilates the given function x3 (1 − 5x).
Solution: Write: x3 (1 − 5x) = x3 − 5x4 . This is a 4th-degree polynomial, so D5
annihilates it.
Problem 10. Find a differential operator that annihilates the given function 8x−sin x+
10 cos(5x).
Solution: Look at this term by term:
D2 (8x) = 0
(D2 + 1) sin x = − sin x + sin x = 0
(D2 + 52 ) cos(5x) = −52 cos(5x) + 52 cos(5x) = 0
The operators D2 + 1 and D2 + 52 don’t factor over the real numbers. Thus:
(D2 + 52 )(D2 + 1)D2
annihilates 8x − sin x + 10 cos(5x)
and is the operator of lowest possible order that does so: if we removed any of the “prime”
factors, then the operator would no longer annihilate the function.
Problem 11. Find linearly independent functions that are annihilated by the given
operator D2 + 4D.
Solution: First, we’ll factor the operator:
D2 + 4D = D(D + 4)
This is a second-order operator, so we’ll look for two independent solutions: one that is
annihilated by D, the other by D + 4.
Dy1 = 0 ⇒
(D + 4)y2 = 0
y1 = constant = 1
⇒ y2 = e−4x
1 and e−4x are linearly independent.
Problem 12. Find linearly independent functions that are annihilated by the given
operator D2 (D − 5)(D − 7).
Solution: This operator is already factored into prime factors. Each of these factors will
annihilate one of the independent functions {y1 , y2 , y3 , y4 }. The functions y1 and y2 are
independent functions that are both annihilated by D2 ; we’ll refer to them collectively
as y1,2 .
D2 y1,2 = 0
(D − 5)y3 = 0
(D − 7)y4 = 0
⇒
⇒
⇒
y1,2 = ax + b
y3 = e5x
y4 = e7x
⇒
y1 = 1;
y2 = x
We have a 4th-order operator, so we have 4 independent solutions: {1, x, e5x , e7x }
1.3
Second-order linear homogeneous ODEs
Problem 1. Find the general solution to the ODE y 00 − y = 0.
Problem 2. Find the general solution to the ODE y 00 + y = 0.
Problem 3. Find the general solution to the ODE y 00 − y 0 − 12y = 0.
Problem 4. Find the general solution to the ODE y 00 − 2y 0 + 5y = 0.
1.4
Reduction of order for second-order ODEs
Problem 1. Given that y1 (x) = xe−x is a solution to the ODE y 00 + 2y 0 + y = 0,
find a second linearly independent solution to the ODE (do not use the guess method:
y2 = emx ).
Problem 2. Consider the ODE y 00 +2y 0 +y = 0. Since this is a constant-coefficient linear
homogeneous ODE the standard approach is to seek a solution of the form: y = emx .
The resulting characteristic equation is
m2 + 2m + 1 = (m + 1)2 = 0
⇒
m = 1 (a double root).
However, this method only gives one solution. One short cut is to differentiate w.r.t.
the parameter. However, a more straight-forward method, interns of understanding, is
to use the method of reduction of order to find the second solution. Find the second
independent solution using this method and verify its independence.
Problem 3. Given that y1 (x) = x sin(ln x) is a solution to the ODE x2 y 00 − xy 0 + 2y = 0,
find a second linearly independent solution to the ODE.
Problem 4. Given that y1 (x) = 1 is a solution to the homogeneous component of the
ODE y 00 +y 0 = 1, find a second linearly independent solution to the ODE and a particular
solution. Write down the general solution.
1.5
Higher-order linear homogeneous ODEs
Problem 1. Solve for the general solution to the ODE
y 000 + 3y 00 + 2y 0 = 0 .
y 000 + 6y 00 + 11y 0 + 6y = 0 .
y 000 + 2y 00 − y 0 − 2y = 0 .
y (4) − y 000 + y 00 − y 0 = 0 .
1.6
Cauchy-Euler equations
Problem 1. Solve the IVP
x2 y 00 − 3xy 0 + 4y = 0,
y(1) = 5,
y 0 (1) = 3.
x2 y 00 − 4xy 0 + 6y = 0,
y(−2) = 8,
y 0 (−2) = 0
by transforming the equation onto the interval (0, ∞).
Problem 3. Find the general solution to the inhomogeneous ODE
x2 y 00 − 4xy 0 + 6y = 2 ln x
by using the substitution x = et , or equivalently, t = ln x.
1.7
The method of undetermined coefficients
Note: When solving inhomogeneous equations always solve for the homogeneous solution
first, since the relation between the fundamental set to the homogeneous equation and
the form of the inhomogeneous term determines what method we us to seek a particular
solution to the inhomogeneous equation. If the inhomogeneous term is not independent
from the fundamental set, then you must modify your particular solution appropriately.
Problem 1. Solve for the general solution to the inhomogeneous ODE
ay 00 + by = c ,
where a, b ∈ R+ and c ∈ R.
y 00 + y 0 − 6y = 2x .
y 00 + 4y = x2 − 3 sin(2x) .
1.8
Variation of parameters
Problem 1. Find the general solution to the linear inhomogeneous equation
y 00 + 2y 0 + y = e−x ln x on (0, ∞).
3
1
2 00
0
2
x y + xy + x −
y = x 2 on (0, ∞).
4
x2 y 00 + xy 0 + y = sec(ln x) on (0, ∞).
1.9
1.9.1
Applications: harmonic oscillating systems
Harmonic motion w/o friction and external forcing
Problem 1. A 20-kg mass is attached to one end of a hanging spring. The other end
of the spring is attached to the ceiling of your physics lab. The frequency of the simpleharmonic motion is observed to be 2/π, what is the spring constant? You may ignore
the mass of the spring compared to the 20-kg mass.
Problem 2. When a 400 newton weight is attached to a massless spring, the spring is
stretched by 2 meters. A mass of 50 kg is attached to the end of the spring and released
from the equilibrium position with an downward velocity 10 m/s. If we take the x-axis
in the downward direction, what is the equation of motion?
Problem 3. Use the alternative form of the solution x(t) = A sin (ωt + φ) to the ODE
x¨ + ω 2 x = 0 to show that the time interval between two successive maxima of x(t) is
2π/ω.
1.9.2
Damped harmonic motion w/o external forcing
We can add damping to a mass-spring system by immersing it in a viscous fluid. If we
assume that the attached mass m is much greater than the mass of the spring mspring and
the resistance is proportional to the velocity of the mass, then the governing equation
with the initial conditions are of the form:
x¨ + 2λx˙ + ω 2 x = 0,
x(0) = x0 ,
x(0)
˙
= v0 ,
where λ is the constant of proportionality for the frictional term. If we let x(t) = emt ,
then the characteristic equation is m2 + 2λm + ω 2 = 0 with roots:
√
m = −λ ± λ2 − ω 2 .
Fact: The solution to the IVP in the case of underdamped motion (i.e., λ2 − ω 2 < 0) is
of the form:
√
ω 2 − λ2 t + φ .
x(t) = Ae−λt sin
This is known as the alternative form. The quasi-period is defined to be the roots of the
equation, which can be shown to be
2π
.
Tperiod = √
2
ω − λ2
x¨ + 2x˙ + 10x = 0,
x(0) = −2,
x(0)
˙
= 0.
Express the solution in the alternative form.
Problem 5. For the case of underdamped
motion, show that the ratio between any two
2πλ
consecutive maxima is exp √ω2 −λ2 .
Problem 6. In the case of over-damped motion (i.e., λ2 −ω 2 > 0), show that the solution
can be expressed in the form
√
x(t) = Ae−λt sinh
ω 2 − λ2 t + φ .
1.9.3
Forced harmonic motion w/o damping
Problem 7. Solve the following initial value problem analytically
x¨ + ω 2 x = AF sin(ΩF t),
x(0) = 0,
x(0)
˙
= v0 ,
where ω, ΩF , and AF are all positive constants with ω 6= ΩF (no resonance).
Problem 8. Graph the solution to the initial value problem for v0 = ±1, ±2 over the
time interval [0, 4π].
x¨ + x = sin(2t),
x(0) = 0,
x(0)
˙
= v0 .
(i) How does the sign of v0 affect the solution?
(ii) How does the magnitude of v0 affect the solution?
(iii) Compare you numerical results here to the previous solution.
Problem 9. (near-resonance) In the notes it was shown that the solution to the IVP
x¨ + ω 2 x = AF sin(ΩF t),
x(0) = 0,
x(0)
˙
= 0,
with ω 6= ΩF was given by
AF
x(t) =
ω
AF
x(t) = 2
ω
"
ΩF sin(ω t) − ω sin(ΩF t)
Ω2F − ω 2
,
which can be recast as
ΩF
ω
sin(ω t) − sin(ΩF t)
ΩF 2
−1
ω
#
.
(1.1)
We can study the case of near resonance by letting ΩF = (1 + )ω in the above solution
and examine the solution for small (i.e., 0 < 1).
Graph the solution for AF = 1, ω = 1, and = 1/10, 1/100, 1/1000 over the time interval
[0, 10π]. What trend do you notice in the solutions x(t; ) as → 0+ ? Can you justify
observation by using a Taylor expansion of the solution x(t; ) about = 0 with t fixed?
Problem 10. (A variable frequency) Solve the initial value problem
x¨ + x = sin(ΩF (t) · t),
x(0) = 0, x(0)
˙
=0
√
over the time interval [0, 40π] for ΩF (t) = 1/ t. Since the sine function has a variable
argument, you will need to solve the equation numerically using either Mathematica or
Matlab. Also graph the forcing function sin(ΩF (t) · t) on the same graph. Is there any
striking features that you notice between the response curve (i.e., the solution curve) and
the forcing curve?

Ordinary Differential Equations Higher-order ODEs Problem Set 5 (Version 1.0)

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