Math 2280-001
Wed Apr 15
Laplace transform, continued.
, Today and Friday we will continue discussing Chapter 7 Laplace transform material. We will see how
this material applies to our favorite linear differential equations, with interesting new forcing functions for
mechanical (or electrical) oscillation problems.
Exercise 1: correction to Monday's notes: On Monday we checked the table entry
L ea t f t s = L f t s K a = F s K a
for a 2 = (Exercise 6). But then in Exercise 7 I had us using the formula for a 2 C instead of a 2 =.
This can be justified but the proof is not easy. However, the particular transform we needed for Exercise
8, namely
1
L t ea t s =
2
sKa
for a = a C i k can be shown directly from the definition of Laplace transform and integration by parts.
Check this. (Bonus: This exercise is a lot like your first homework exercise this week.)
, Then complete Monday's notes, Exercises 8-10. As a follow-up to Exercise 9, where w = w0 , solve
Exercise 2) Use Laplace transforms to write down the solution to
2
x## t C w0 x t = F0 sin w t
x 0 = x0
x# 0 = v0
with w s w0 . Hint: There's a short cut on the partial fractions in this problem.
Partial fractions set-up practice:
Exercise 3a) What is the form of the partial fractions decomposition for
K356 C 45 s K 100 s2 K 4 s5 K 9 s4 C 39 s3 C s6
X s =
.
2
s K 3 3 s C 1 2 C 4 s2 C 4
3b) What is x t = LK1 X s
t , using the constants from a?
3c) Have Maple compute the precise partial fractions decomposition.
3d) Have Maple compute the inverse Laplace transform directly.
> X d s/
K356 C 45$s K 100$s2 K 4 $s5 K 9$s4 C 39$s3 C s6
s K 3 3$
> convert X s , parfrac, s ;
> with inttrans ;
> invlaplace X s , s, t ;
>
sC1
2
C 4 $ s2 C 4
2
;
A harder table entry to understand is the following one - go through this computation and see why it seems
reasonable, even though there's one step that we don't completely justify. The table entry is
tf t
KF# s
Here's how we get it:
F s =L f t
d
d
0 ds F s = ds
s d
N
f t eKs t dt
0
N
N
f t eKs t dt =
0
0
d
Ks t
f t e
dt
ds
d
through the integral sign that needs more justification. We know that the
ds
derivative of a sum is the sum of the derivatives, and the integral is a limit of Riemann sums, so this step
does at least seem reasonable. The rest is straightforward:
It's this step of passing
N
0
d
Ks t dt =
ds f t e
N
f t Kt eKs t dt =KL t f t
s
□.
0
Exercise 3)
a) Deduce then for n 2 ;
tn f t
K1
n
F
n
s
b) Use L t f t s =KF# s and part a to show the following. Note the first transform is the one we
checked in Exercise 1, and the generalization is the one we checked illegally in Monday's Exercise 7.
1
L t ea t s =
,a2C
2
sKa
n!
L tn ea t s =
,a2C .
nC1
sKa
New flavor Laplace table entries for today and Friday:
f t % CeM t
f t with
Fs d
N
f t eKs t dt
for s O M
comments
0
u tKa
eKa s
s
unit step function
f tKa u tKa
eKa sF s
d tKa
eKa s
t
f t g t K t dt
for turning components on and
off at t = a .
more complicated on/off
unit impulse/delta "function"
convolution integrals to invert
Laplace transform products
Fs Gs
0
Recall from our discussion on Monday that the unit step function with jump at t = 0 is defined to be
0, t ! 0
u t =
.
1, t R 0
Its graph is shown below. Notice that this function is called the "Heaviside" function in Maple, after the
person who popularized it (among a lot of other accomplishments) and not because it's heavy on one side.
http://en.wikipedia.org/wiki/Oliver_Heaviside
> with plots :
plot Heaviside t , t =K3 ..3, color = green, title = `graph of unit step function` ;
graph of unit step function
1
K3
K2
K1
0
1
2
3
Technically the vertical line should not be in the graph above - a more precise picture would have a solid
point at 0, 1 and a hollow circle at 0, 0 , for the graph of u t . In terms of Laplace transform integral
definition it doesn't actually matter what we define u 0 to be.
Then
u tKa =
0, t K a ! 0; i.e. t ! a
1, t K a R 0; i.e. t R a
and has graph that is a horizontal translation by a to the right, of the original graph, e.g. for a = 2:
> plot Heaviside t K 2 , t =K1 ..5, color = green, title = `graph of u(t-2)` ;
graph of u(t-2)
1
K1
1
2
3
4
5
Exercise 4) Verify the table entries below. We discussed the first one on Monday. It also follows from
the second entry, which is the one we will check today.
u tKa
unit step function
f tKa u tKa
Exercise 5)
eKa s
s
eKa sF s
for turning components on and
off at t = a .
more complicated on/off
Exercise 5a) Explain why the description above leads to the differential equation initial value problem for
x t
x## t C x t = .2 cos t 1 K u t K 10 p
x 0 =0
x# 0 = 0
5b) Find x t . Show that after the parent stops pushing, the child is oscillating with an amplitude of
exactly p meters (in our linearized model).
Alternate approach via Chapter 3 is doable, but involves more work:
step 1) solve
x## t C x t = .2 cos t
x 0 =0
x# 0 = 0
for 0 % t % 10 p .
step 2) Then solve
y## t C y t = 0
y 0 = x 10 p
y# 0 = x# 10 p
and set x t = y t K 10 for t O 10 .
Pictures for the swing:
> plot1 d plot .1$t$sin t , t = 0 ..10$Pi, color = black :
plot2 d plot Pi$sin t , t = 10$Pi ..20$Pi, color = black :
plot3 d plot Pi, t = 10$Pi ..20$Pi, color = black, linestyle = 2 :
plot4 d plot KPi, t = 10$Pi ..20$Pi, color = black, linestyle = 2 :
plot5 d plot .1$t, t = 0 ..10$Pi, color = black, linestyle = 2 :
plot6 d plot K.1$t, t = 0 ..10$Pi, color = black, linestyle = 2 :
display plot1, plot2, plot3, plot4, plot5, plot6 , title = `adventures at the swingset` ;
adventures at the swingset
3
0
K3
2 p 4 p 6 p 8 p 10 p 12 p14 p16 p18 p20 p
t
Laplace transform convolution
For Laplace transforms the convolution of f and g is defined to be
t
f)g t d
f t g t K t dt .
0
Exercise 6) Show f)g t = g)f t by doing a change of variables in the convolution integral.
t
0
f t g t K t dt =
t
g t f t K t dt
convolution integrals to invert
Laplace transform products
Fs Gs
0
Exercise 7) Verify that the convolution integral table entry is correct, for
f t = sin t
Fs =
g t = cos t
Gs =
f)g t
Hint: you might use the trig identity sin t 2 =
t
>
0
>
sin t $cos t K t dt;
1
s2 C 1
s
s2 C 1
Fs Gs =
1 K cos 2 t
2
.
s
s2 C 1
2
.
Why convolutions are so important in applications:
Consider a mechanical or electrical forced oscillation problem for x t , and the particular solution that
begins at rest:
a x##C b x#C c x = f t
x 0 =0
x# 0 = 0 .
Then in Laplace land, this equation is equivalent to
a s2 X s C b s X s C c X s = F s
0 X s a s2 C b s C c = F s
1
0X s =F s $ 2
dF s W s .
a s Cb sCc
LK1 W s
t = w t is called the weight function for the physical system, and because of the
convolution table entry the solution is given by
t
x t = f)w t = w)f t =
w t f t K t dt .
0
This idea generalizes to much more complicated mechanical and circuit systems, and is how engineers
experiment mathematically with how proposed configurations will respond to various input forcing
functions, once they figure out the weight function for their system. Or, they may have a certain
characteristics of the response function x t that they would like to get, no matter the forcing function, and
they design the weight function to get it. We'll study some interesting forcing functions on Friday, using
the convolution method.
f t , with
f t
% CeM t
c f t Cc f t
1 1
2 2
N
f t eKs t dt for s O M
Fs d
0
Y
verified
c F s Cc F s
A
1
s
1
A
1 1
2 2
1
t
(s O 0
A
s2
2
t2
A
s3
n!
tn, n 2 ;
A
sn C 1
1
ea t
(s O R a
sKa
s
cos k t
sin k t
cosh k t
sinh k t
ea tcos k t
ea tsin k t
f
n
t
f t dt
0
tf t
t2 f t
tn f t , n 2 Z
f t
t
t cos k t
1
2 k t sin k t
A
(s O 0
A
k
s2 C k2
s
s2 K k2
k
s2 K k2
sKa
(s O k
(s O k
2 C k2
(s O a
A
sOa
A
k
2 C k2
sKa
A
F sKa
ea t f t
f# t
f ## t
t , n2;
(s O 0
s2 C k2
sKa
A
s F s Kf 0
s K s f 0 K f# 0
sn F s K sn K 1f 0 K...Kf n K 1 0
Fs
s
s2F
A
A
A
A
KF# s
F## s
K1 n F n s
N
F s ds
s
s2 K k2
s2 C k2
s
s2 C k2
2
2
A
A
1
1
2 k3
sin k t K k t cos k t
s2 C k2
t ea t
1
tn e a t, n 2 Z
sKa
n!
sKa
t
f)g t =
f t g t K t dt
2
2
nC1
Fs Gs
0
u tKa f tKa
Laplace transform table
eKa s F s
A
A