MAE 200B Homework #3 Solutions
University of California, Irvine
Winter 2005
Problem 1 (Haberman 5.3.2):
Consider this equation:
ρ
∂2u
∂2u
∂u
=
T
+
αu
+
β
0
∂t2
∂x2
∂t
a)
The term αu describes a force that is proportional to displacement, such as a spring or
friction force. For such restoring forces α < 0.
describes an aerodynamic damping term. This damping should reduce the
The term β ∂u
∂t
velocity so β < 0
b)
u(x, t) = h(t)φ(x)
Now we can use separation of variables to solve the equation.
ρ(x)h00 φ = T0 φ00 h + α(x)hφ + β(x)h0 φ
ρ(x)h00 φ − β(x)h0 φ = T0 φ00 h + α(x)hφ
hφ
h00
h0
φ00
ρ(x) − β(x) = T0 + α(x)
h
h
φ
h00 β(x) h0
T0 φ00
1
−
=
+
α(x)
h
ρ(x) h
ρ(x) φ
ρ(x)
For separation of variables to work the left side must only be a function of t and the right
side must only be a function of x. Therefore for this to hold true β(x)
= c where c = constant
ρ(x)
or β(x) = cρ(x).
1
c)
If we look at the spatial equation:
T0 φ00
1
+
α(x) = −λ2
ρ(x) φ
ρ(x)
T0 φ00 + α(x)φ = −λ2 ρ(x)φ
T0 φ00 + α(x) + λ2 ρ(x) φ = 0
"
#
d
dφ
T0
+ α(x) + λ2 ρ(x) φ = 0
dx
dx
Now we can put this in Strum-Liouville form:
"
#
d
dφ
p(x)
+ q(x) + λ2 σ(x) φ = 0
dx
dx
Therefore, p(x) = T0 , q(x) = α(x), and σ(x) = ρ(x)
Now we can solve the time equation with β(x) = cρ(x) :
h00 β(x) h0
−
= −λ2
h
ρ(x) h
h00
h0
− c = −λ2
h
h
h00 − ch0 + λ2 h = 0
The roots for this equation come out to be:
r1,2
Case 1)
c2
4
c
= ±
2
s
c2
− λ2
4
− λ2 > 0
The solution to the time equation is:
h(t) = Cn e
Case 2)
c2
4
c
+
2
q
c2
−λ2
4
− λ2 = 0
2
t
+ Dn e
c
−
2
q
c2
−λ2
4
t
c
c
h(t) = Cn e( 2 )t + Dn te( 2 )t
Case 3)
c2
4
− λ2 < 0


s
s

c2
c2
h(t) = Cn e cos  λ2 − t + sin  λ2 − t
4
4
c
t
2
Note c < 0 because β < 0
3
Problem 2 (Haberman 5.3.9):
Consider this equation and boundary conditions:
x2
d2 φ
dφ
+ λφ = 0
+x
2
dx
dx
φ(1) = 0
φ(b) = 0
a)
If we divide through by
1
x
we can put this equation into Strum-Liouville form:
1 2 d2 φ
dφ
x
+
x
+ λφ = 0
x
dx2
dx
"
x
#
d2 φ dφ 1
+
+ λφ = 0
dx2 dx x
Notice that:
d
dφ
d2 φ dφ
x
=x 2 +
dx dx
dx
dx
"
#
Therefore:
"
#
dφ
1
d
x
+ λφ = 0
dx dx
x
is in S.L. form where p(x) = x, q(x) = 0, and σ(x) =
1
x
b)
We can solve this problem using the Raleigh quotient
λ=
−pφ dφ
|b
dx a
+
Rb
p
a
Rb
a
dφ 2
dx
− qφ dx
φ2 σdx
In our case a = 1, b = b, p(x) = x, q(x) = 0, and σ(x) = x1 . Therefore:
λ=
−pφ dφ
|b +
dx 1
Rb
Rb
1
x
1 2
1 x φ dx
4
dφ 2
dx
dx
Since φ(1) = φ(b) = 0,
dφ 2
dx
1
dx
Rb 1
2
1 x φ dx
Rb
λ=
x
For 1 ≤ x ≤ b,
Z
1
b
dφ
x
dx
!2
dx ≥ 0
and
b
Z
1
1 2
φ dx > 0
x
Thus λ ≥ 0
c)
When solving the equidimentional or Cauchy-Euler equation we look for solutions in the
form of φ = xr . Substituting this solution into our problem we get,
x2 r(r − 1)xr−2 + xrxr−1 + λxr = 0
[r(r − 1) + r + λ] xr = 0
r(r − 1) + r + λ = 0
r2 − r + r + λ = 0
r2 = −λ
√
r = ±i λ
This gives us a the solution,
√
φ = x±i
λ
Since,
√
x±i
λ
√
±i λ
= eln(x
√
φ = e±i
5
λln(x)
)
The general solution is then,
φ = Ccos
√
λln(x) + Bsin
√
λln(x)
Applying the first BC:
φ(1) = 0 = Ccos
√
√
λln(1) + Bsin
λln(1)
C=0
Applying the second BC:
φ(b) = 0 = Bsin
√
λln(b)
√
λln(b) = nπ
nπ
λn =
ln(b)
q
λn =
nπ
ln(b)
!2
n = 1, 2, 3, ...
Therefore,
φ(x) = Bsin
q
λn ln(x)
d)
!
nπ
φn (x) = sin
ln(x)
ln(b)
!
mπ
φm (x) = sin
ln(x)
ln(b)
By definition,
σ(x) =
< φn , φm >=
Z
1
x
b
a
φn (x)φm (x)σ(x)dx
6
< φn , φm >=
=
Z
1
b
b
Z
1
!
!
mπ
1
nπ
ln(x) sin
ln(x)
dx
sin
ln(b)
ln(b)
x
"
!
!#
11
π
π
cos
(n − m)ln(x) − cos
(n + m)ln(x)
2x
ln(b)
ln(b)
dx
If m 6= n and we let ln(x) = u and x1 dx = du then:
< φn , φm >=
Z
ln(b)
0

"
!
1
π
π
cos
(n − m)u − cos
(n + m)u
2
ln(b)
ln(b)
1 ln(b)  sin
=
2 π
π
(n
ln(b)
− m)u
−
n−m
=
sin
π
(n
ln(b)
+ m)u
!#
du
ln(b)

n+m
0
1 ln(b)
[0 + 0] = 0
2 π
If m = n,
< φn , φm >=
Z
ln(b)
0
"
!#
1
π
1 + cos
2nu
2
ln(b)
du
1 ln(b)
1
1
[sin (2πn) − sin (0)] = ln(b)
= ln(b) − 0 +
2
2 2πn
2
Therefore
if n 6= m
if n = m
< φn , φm >= 0
1
< φn , φm >= ln(b)
2
e)
!
nπ
φn = sin
ln(x) = sin(nzπ)
ln(b)
z=
1<x<b
ln(x)
ln(b)
As x goes from 1 to b, nz goes from 0 to n. Therefore φn = sin(nzπ) has the following zeros
for x in (1, b)
7
k
n
zk =
k
k
xk = e n ln(b) = b n ,
k = 1, 2, 3, ..., n − 1
Problem 3:
Consider the periodic Strum-Liouville problem:
X 00 (x) = −ω 2 X(x)
−L<x<L
X(−L) = X(L)
X 0 (−L) = X 0 (L)
a)
Plugging in the given solutions into the above equation yields:
φn = e−iωn x
ωn =
nπ
L
φ0n = −iωn e−iωn x
φ00n = −ωn2 e−iωn x
φ00n = −ωn2 φn
ok
φn = eiωn x
φ0n = iωn eiωn x
φ00n = −ωn2 eiωn x
φ00n = −ωn2 φn
ok
These solutions can be related to sin and cos by:
e−iβ = cos(β) − isin(β)
eiβ = cos(β) + isin(β),
Therefore,
cos(β) =
1 iβ
e + e−iβ ,
2
sin(β) =
β = ωn x
8
1 iβ
e − e−iβ
2
b)
By definition, for
φn = e−iωn x
φm = e−iωm x
< φn , φm >=
< φn , φm >=
−L
L
Z
L
Z
φn φ∗m dx
=
L
Z
e−iωn x eiωm x dx
−L
e−i(ωm −ωn )x dx =
Z
−L
(
< φn , φm >=
L
π
e−i L (m−n)x dx
−L
2L
m=n
m 6= n)
π
L
ei L (m−n)x |−L
L
i(m−n)π
(
< φn , φm >=
2L m = n
0 m=
6 n)
For the same m,
<
−
φ+
m , φm
>=
Z
L
π
π
im L
x im L
x
e
e
dx =
Z
L
2π
ei L x
−L
−L
−
< φ+
m , φm >=
L i 2π x L
e L |−L = 0
i2π
Therefore, these eigenfunctions are orthogonal.
c)
We can find the coefficients using the definition of the inner product.
f (x) =
+∞
X
nπ
Cn ei L x
n=−∞
Therefore,
nπ
nπ
< f (x), ei L x >
1
1 ZL
i nπ
x
Cn =
=
< f (x), e L >=
f (x)e−i L x dx
i nπ
x
2
2L
2L −L
||e L ||
9
d)
If f (x) is real, then
nπ
< f (x), e−i L x >
Cn =
2L
By plugging in the value of n into the equation generated above we can show that
C−n
(−n)π
(n)π
1 ZL
1 ZL
−i L x
f (x)e
f (x)ei L x dx
=
dx =
2L −L
2L −L
∗
C−n
(n)π
1 ZL
=
f (x)e−i L x dx = Cn
2L −L
10
Problem 4:
The governing equation is the 1D String Equation with a forcing function:
2
∂2u
2∂ u
=
c
+ Asin(ωt)
∂t2
∂x2
(0 < x < L)
BC:
u(0, t) = 0
ux (L, t) = 0
IC:
ut (x, 0) = 0
u(x, 0) = f (x)
(
u(x, 0) = f (x) =
3h
x
L
h
(0 < x < L/3)
(L/3 < x < L)
First we will solve this problem assuming that ω 6= ωn = ckn . We need to solve this
problem using the method of Eigenfunction Expansion. We start by looking at the related
homogeneous problem which we have solved before.
2
∂2v
2∂ v
=c
∂t2
∂x2
(0 < x < L)
Using the method of separation of variables we can find the eigenfunctions of this related
problem:
φn (x) = sin(kn x)
kn =
nπ
(2n − 1)π
=
2L
2L
This now can give us a general solution in terms of an eigenfunction expansion,
u(x, t) =
∞
X
an (t)φn (x)
n=1
Now we can expand the non-homogeneous part of the equation in terms of our eigenfunctions,
11
g(x, t) =
∞
X
An (t)φn (x) = Asin(ωt)
n=1
Since this is a periodic forcing with no spacial dependence we can easily find the coefficients
for the expansion.
RL
An (t) =
An (t) =
0
2ZL
g(x, t)φn (x)dx
=
Asin(ωt)φn (x)dx
RL
2
L 0
0 φn (x)dx
Z L
2
2
2A
Asin(ωt)
sin(kn x)dx =
Asin(ωt)(1 − cos(kn L)) =
sin(ωt)
L
Lkn
Lkn
0
Our goal now is to find the coefficient’s an (t) that will satisfy our equation. To find these
coefficients we can plug our general solutions back into the original equation.
Recall,
2
∂2u
2∂ u
=c
+ Asin(ωt)
∂t2
∂x2
and
u(x, t) =
∞
X
an (t)φn (x)
n=1
f (x, t) =
∞
X
An (t)φn (x) = Asin(ωt)
n=1
φ00n (x) = k 2 φn (x)
Differentiating our solution term by term yields,
∞
∞
X
∂2u X
∂2
=
a
(t)φ
(x)
=
a00n (t)φn (x)
n
2 n
∂t2
∂t
n=1
n=1
c2
∞
∞
∞
2
X
X
∂2u X
2∂ u
2
00
=
c
a
(t)φ
(x)
=
−c
a
(t)φ
(x)
=
c2 kn2 an (t)φn (x)
n
n
n
n
∂x2 n=1 ∂x2
n=1
n=1
Therefore,
∞ h
X
i
a00n (t)φn (x) = −c2 kn2 an (t)φn (x) + An (t)φn (x)
n=1
12
This reduces to a second order ODE in time,
∞ h
X
i
a00n (t) = −c2 kn2 an (t) + An (t)
n=1
Since we know An (t) we can rewrite this equation,
a00n (t) + c2 kn2 an (t) =
Bn =
2A
sin(ωt) = Bn sin(ωt)
Lkn
2A
4A
=
Lkn
(2n − 1)π
Recall that this differential equation can be solved by finding the homogeneous solution and
the particular solution. The particular solution in this case can be found using the method
of undetermined coefficients.
The homogeneous equation is an ODE which we have done many times. The solution is,
a00n,h (t) + c2 kn2 an,h (t) = 0
a00n,h (t) = −c2 kn2 an,h (t)
an,h (t) = Cn cos(ckn t) + Dn sin(ckn t)
For the particular solution we can guess a solution in the form of an,p (t) = En sin(ωt)
a00n,p (t) + c2 kn2 an,p (t) = Bn sin(ωt)
−ω 2 En sin(ωt) + c2 kn2 En sin(ωt) = Bn sin(ωt)
En (−ω 2 + c2 kn2 ) = Bn
En =
(−ω 2
Bn
+ c2 kn2 )
Therefore the solution for the temporal coefficients are,
an (t) = an,h (t) + an,p (t) = Cn cos(ckn t) + Dn sin(ckn t) +
Bn
sin(ωt)
(−ω 2 + c2 kn2 )
That makes the whole solution so far,
u(x, t) =
∞
X
n=1
"
#
Bn
Cn cos(ckn t) + Dn sin(ckn t) +
sin(ωt) sin(kn x)
(−ω 2 + c2 kn2 )
13
Now we can apply the the first IC,
ut (x, t) = 0 =
∞
X
n=1
"
#
Bn
ωcos(ωt) cos(kn x)
−ckn Cn sin(ckn t) + ckn Dn cos(ckn t) +
2
(−ω + c2 kn2 )
ut (x, 0) = 0 =
∞
X
n=1
"
#
Bn
Dn ckn +
ω sin(kn x)
(−ω 2 + c2 kn2 )
Therefore,
Dn =
(ω 2
Bn
ω
− c2 kn2 )ckn
Now we can apply the second IC to our new solution,
u(x, t) =
∞
X
n=1
"
#
Bn
Bn
Cn cos(ckn t) + 2 2
ωsin(ck
t)
+
sin(ωt) sin(kn x)
n
(c kn + ω 2 )ckn
(ω 2 − c2 kn2 )
u(x, 0) =
∞
X
Cn sin(kn x) =
∞
X
an (0)sin(kn x) = f (x)
n=1
n=1
As you can see we arrive at the same equation as we did in Homework #2 when trying
to determine the coefficients for the Fourier Series. Notice that these coefficients are the
temporal coefficients at the initial time.
an (0) = an,h (0) + an,p (0) = Cn cos(ckn (0)) + Dn sin(ckn (0)) +
Bn
sin(ω(0)) = Cn
(−ω 2 + c2 kn2 )
Therefore,
2ZL
Cn =
f (x)sin(kn x)dx
L 0
thus,
Cn =
2 Z L/3 3h
2ZL
xsin(kn x)dx +
hsin(kn x)dx
L 0
L
L L/3
Now we can integrate by parts or use a software package to find the solution for the
coefficients.
Cn =
(2n − 1)π
24h
sin(
)
2
((2n − 1)π)
6
14
So our final solution is,
u(x, t) =
∞
X
n=1
"
#
Bn
Bn
ωsin(ckn t) +
sin(ωt) sin(kn x)
Cn cos(ckn t) + 2
2
2
2
(ω − c kn )ckn
(−ω + c2 kn2 )
Note that this solution only works if ω 6= ckn for all n.
If we now look at the case when ω = ω2 = ck2 we see a problem in our particular solution.
In this case we need to come up with a different form of the solution for this part of the
problem.
For the particular solution if we guess a solution in the form of an,p (t) = En sin(ω2 t)
a00n,p (t) + c2 kn2 an,p (t) = Bn sin(ω2 t)
−ω22 En sin(ω2 t) − c2 kn2 En sin(ω2 t) = Bn sin(ω2 t)
En (−ω22 + c2 kn2 ) = Bn
En =
(−ω22
Bn
+ c2 kn2 )
When n = 2, En will not be defined.
Therefore we look for a different particular solution in the form of an,p (t) = En tsin(ω2 t) +
Fn tcos(ω2 t),
a0n,p (t) = En sin(ω2 t) + ω2 En tcos(ω2 t) + Fn cos(ω2 t) − Fn ω2 tsin(ω2 t)
a00n,p (t) = ω2 En cos(ω2 t)+ω2 En cos(ω2 t)−En ω22 tsin(ω2 t)−Fn ω2 sin(ω2 t)−Fn ω2 sin(ω2 t)−Fn ω22 tcos(ω2 t)
a00n,p (t) = 2ω2 En cos(ω2 t) − En ω22 tsin(ω2 t) − Fn 2ω2 sin(ω2 t) − Fn ω22 tcos(ω2 t)
a00n,p (t) + c2 kn2 an,p (t) = Bn sin(ω2 t)
2ω2 En cos(ω2 t)−En ω22 tsin(ω2 t)−Fn 2ω2 sin(ω2 t)−Fn ω22 tcos(ω2 t)−c2 kn2 (En tsin(ω2 t)+Fn tcos(ω2 t))
= Bn sin(ω2 t)
Now we can separate out the sin and cos terms to solve for the coefficients.
En ω22 tsin(ω2 t) − Fn 2ω2 sin(ω2 t) − c2 kn2 En tsin(ω2 t) = Bn sin(ω2 t)
En (ω22 t − c2 kn2 ) − Fn 2ω2 = Bn
15
2ω2 En cos(ω2 t) − Fn ω22 tcos(ω2 t) − c2 kn2 Fn tcos(ω2 t) = 0
2ω2 En − Fn (ω22 t − c2 kn2 t) = 0
Now when n = 2,
−F2 2ω2 = B2
2ω2 E2 = 0
B2
B2
, so our particular solution is, a2,p (t) = − 2ω
tcos(ω2 t)
Therefore E2 = 0 and F2 = − 2ω
2
2
So our solution for this resonance case is,
B2
u(x, t) = C2 cos(ck2 t) + D2 sin(ck2 t) −
tcos(ω2 t) sin(k2 x)+
2ω2
∞
X
"
n=1,n6=2
#
Bn
sin(ωt) sin(kn x)
Cn cos(ckn t) + Dn sin(ckn t) +
2
(−ω + c2 kn2 )
Now we can apply the ICs,
B2
B2
ut (x, t) = 0 = −ck2 C2 sin(ck2 t) + ck2 D2 cos(ck2 t) −
cos(ω2 t) +
tsin(ω2 t) sin(k2 x)+
2ω2
2
∞
X
"
n=1,n6=2
#
Bn
−ckn Cn sin(ckn t) + ckn Dn cos(ckn t) +
ωcos(ωt) cos(kn x)
2
(−ω + c2 kn2 )
∞
X
B2
Bn
ut (x, 0) = 0 = ck2 D2 −
sin(k2 x) +
Dn ckn +
ω sin(kn x)
2
2ω2
(−ω + c2 kn2 )
n=1,n6=2
"
#
∞
X
B2
Bn
− ck2 D2 sin(k2 x) =
Dn ckn +
ω sin(kn x)
2
2ω2
(−ω + c2 kn2 )
n=1,n6=2
"
#
Now we can multiply both sides by sin(kn x),
2 Z L B2
Bn
ω
=
− ck2 D2 sin(k2 x)sin(kn x)
(−ω 2 + c2 kn2 )
L 0 2ω2
Dn ckn +
For n 6= 2
Dn ckn +
(−ω 2
Bn
ω=0
+ c2 kn2 )
16
Dn = −
Bn
ω
(−ω 2 + c2 kn2 )ckn
For n = 2
L
B2
− ck2 D2
=0
2ω2
2
B2
= ck2 D2
2ω2
B2
B2
= D2 =
2ω2 ck2
2ω22
Now we can apply the second IC,
!
B2
B2
tcos(ω2 t) sin(k2 x)+
u(x, t) = C2 cos(ck2 t) + 2 sin(ck2 t) −
2ω2
2ω2
∞
X
"
n=1,n6=2
#
Bn
Bn
Cn cos(ckn t) −
ωsin(ck
t)
+
sin(ωt) sin(kn x)
n
(−ω 2 + c2 kn2 )ckn
(−ω 2 + c2 kn2 )
∞
X
u(x, 0) = C2 sin(k2 x) +
Cn sin(kn x) =
∞
X
Cn sin(kn x) = f (x)
n=1
n=1,n6=2
As you can see this solution will give you all the coefficients from a Fourier Sine Series
expansion. Therefore,
Cn =
Cn =
2ZL
f (x)sin(kn x)dx
L 0
24h
(2n − 1)π
sin(
)
2
((2n − 1)π)
6
So our final solution is,
!
B2
B2
u(x, t) = C2 cos(ck2 t) + 2 sin(ck2 t) −
tcos(ω2 t) sin(k2 x)+
2ω2
2ω2
∞
X
n=1,n6=2
"
#
Bn
Bn
ωsin(ckn t) +
sin(ωt) sin(kn x)
Cn cos(ckn t) + −
2
2
2
2
(−ω + c kn )ckn
(−ω + c2 kn2 )
17
Now we can make the appropriate plots of the solutions for each of the cases.
Case ω 6= ωn
The first plots show each of the first 5 terms of the Fourier Series and the associated sum
.
for at 5 different times. For these plots, ω = 2πc
L
18
This next plot shows the time dependent solution at the center of the domain for each of the
first five terms and the sum of those terms. Notice how the solution looks slightly different
from the previous homework solution due to the effect of the forcing function.
19
These last plots are a 3D plot of the domain over the time of one period. If you would like
to view the movie of these plots please refer to the Course Files on E3.
20
Case ω = ω2
The first plots show each of the first 5 terms of the Fourier Series and the associated sum
for at 5 different times.
This next plot shows the time dependent solution at the center of the domain for each of
the first five terms and the sum of those terms. Notice that at resonance we have a growing
term that is unbounded.
21
22
These last plots are a 3D plot of the domain over the time of one period. If you would like
to view the movie of these plots please refer to the Course Files on E3.
23
Problem 5:
The governing equation is the 1D String Equation with a forcing function:
2
∂2u
∂u
2∂ u
=
c
+
b
+ Asin(ωt)
∂t2
∂t
∂x2
(0 < x < L)
BC:
u(0, t) = 0
ux (L, t) = 0
IC:
ut (x, 0) = 0
u(x, 0) = f (x)
(
u(x, 0) = f (x) =
3h
x
L
h
(0 < x < L/3)
(L/3 < x < L)
First we will solve this problem assuming that ω 6= ωn = ckn . We need to solve this
problem using the method of Eigenfunction Expansion. We start by looking at the related
homogeneous problem which we have solved before.
2
∂v
∂2v
2∂ v
+b
=c
∂t2
∂t
∂x2
(0 < x < L)
Using the method of separation of variables we can find the eigenfunctions of this related
problem:
φn (x) = sin(kn x)
kn =
nπ
(2n − 1)π
=
2L
2L
This now can give us a general solution in terms of an eigenfunction expansion,
u(x, t) =
∞
X
an (t)φn (x)
n=1
Now we can expand the non-homogeneous part of the equation in terms of our eigenfunctions,
24
g(x, t) =
∞
X
an (t)φn (x) = Asin(ωt)
n=1
Since this is a periodic forcing with no spacial dependence we can easily find the coefficients
for the expansion.
Z L
2
2A
an (t) = Asin(ωt)
sin(kn x)dx =
sin(ωt)
L
Lkn
0
Our goal now is to find the coefficient’s an (t) that will satisfy our equation. To find these
coefficients we can plug our general solutions back into the original equation.
Recall,
2
∂u
∂2u
2∂ u
+
b
=
c
+ Asin(ωt)
∂t2
∂t
∂x2
and
u(x, t) =
∞
X
an (t)φn (x)
n=1
f (x, t) =
∞
X
an (t)φn (x) = Asin(ωt)
n=1
φ00n (x) = k 2 φn (x)
Differentiating our solution term by term yields,
∞
∞
X
∂2u X
∂2
=
an (t)φn (x) =
a00n (t)φn (x)
2
2
∂t
n=1 ∂t
n=1
∞
∞
X
∂u X
∂
=
an (t)φn (x) =
a0n (t)φn (x)
∂t
∂t
n=1
n=1
c2
∞
∞
∞
2
X
X
∂2u X
2∂ u
2
00
=
c
a
(t)φ
(x)
=
−c
a
(t)φ
(x)
=
c2 kn2 an (t)φn (x)
n
n
n
n
∂x2 n=1 ∂x2
n=1
n=1
Therefore,
∞ h
X
i
a00n (t)φn (x) + ba0n (t)φn (x) = −c2 kn2 an (t)φn (x) + an (t)φn (x)
n=1
25
This reduces to a second order ODE in time,
∞ h
X
i
a00n (t) + ba0n (t) = −c2 kn2 an (t) + an (t)
n=1
Since we know Ωn (t) we can rewrite this equation,
a00n (t) + ba0n (t) + c2 kn2 an (t) =
Bn =
2A
sin(ωt) = Bn sin(ωt)
Lkn
2A
4A
=
kn
(2n − 1)π
Recall that this differential equation can be solved by finding the homogeneous solution and
the particular solution. The particular solution in this case can be found using the method
of undetermined coefficients.
The homogeneous equation is an ODE which we have solved in the previous homework
assignment. Recall that the solutions depend on the value of b,
a00n,h (t) + ba0n (t) + c2 kn2 an,h (t) = 0
Now if
ωn = ckn
ζ=
b
2ωn
This equation has two roots
q
q
b
r1,2 = −ζωn ± ωn ζ 2 − 1 = − ± ωn ζ 2 − 1
2
Case #1 (Under-damped mode): ζ < 1 or b < 2ωn =
r1,2
2n−1
πc
L
q
b
= − ± iωn 1 − ζ 2
2
Therefore, the general solution is:
b
an,h (t) = e− 2 t [Cn cos(βn t) + Dn sin(βn t)]
26
√
Where βn = ωn 1 − ζ 2
Case #2 (Critically damped mode): ζ = 1 or b = 2ωn =
r1 = r2 =
2n−1
πc
L
−b
2
Therefore, the general solution is:
an,h (t) = Cn e
−b
t
2
+ Dn te
Case #3 (Over-damped mode): ζ > 1 or b > 2ωn =
−b
t
2
2n−1
πc
L
q
b
r1,2 = − ± ωn ζ 2 − 1
2
Therefore, the general solution is:
b
√
an,h (t) = Cn e(− 2 +ωn
ζ 2 −1)t
b
+ Dn e(− 2 −ωn
√
ζ 2 −1)t
an,h (t) = Cn cos(ckn t) + Dn sin(ckn t)
For the particular solution we can guess a solution in the form of an,p (t) = En sin(ωt) +
Fn cos(ωt). Note that since there is a damping term the form of our particular solution is
different.
a00n,p (t) + ba0n (t) + c2 kn2 an,p (t) = Bn sin(ωt)
−ω 2 (En sin(ωt)+Fn cos(ωt))+b(En ωcos(ωt)−Fn ωsin(ωt))+c2 kn2 (En sin(ωt)+Fn cos(ωt)) = Bn sin(ωt)
Like in the previous problem, we separate the sin(ωt) and cos(ωt) terms,
−ω 2 En − bFn ω + ωn2 En = Bn
En (ωn2 − ω 2 ) − bFn ω = Bn
Fn (ωn2 − ω 2 ) + bEn ω = 0
En =
ω 2 − ωn2
Fn
bω
27
En = Bn
ω 2 − c2 kn2
(bω)2 − (ω 2 − c2 kn2 )2
Fn = Bn
bω
(bω)2 − (ω 2 − c2 kn2 )2
Therefore the particular solution for the temporal equation takes this form,
an,p (t) = Bn
ω 2 − ωn2
bω
sin(ωt) + Bn 2
cos(ωt)
2
2
2
2
bω − (ω − ωn )
bω − (ω 2 − ωn2 )2
Note that this solution works even when ω = ωn = ckn
an,p (t) = Bn
bωn
ωn2 − ωn2
sin(ωn t) + Bn
cos(ωn t)
2
2
2
2
2
(bωn ) − (ωn − ωn )
(bωn ) − (ωn2 − ωn2 )2
an,p (t) =
Bn
cos(ωn t)
ωn
This solution also conforms with theory because the resonance phenomena is 90 degrees out
of phase with the forcing function.
Now we can solve for the initial conditions for each of the cases.
, Case #1 will be true for n = 1, 2, 3, ... (all modes are under1) Consider b < 2ω1 = πc
L
damped) thus the solution is:
b
un (x, t) = e− 2 t (Cn cos(βn t) + Dn sin(βn t)) sin(kn x)+
"
ω 2 − ωn2
bω
Bn
sin(ωt) + Bn
cos(ωt) sin(kn x)
2
2
2
2
2
(bω) − (ω − ωn )
(bω) − (ω 2 − ωn2 )2
#
For simplicity we will define,
Pn = Bn
Qn = Bn
ω 2 − ωn2
(bω)2 − (ω 2 − ωn2 )2
(bω)2
bω
− (ω 2 − ωn2 )2
Therefore,
u(x, t) =
∞ h
X
b
i
e− 2 t (Cn cos(βn t) + Dn sin(βn t)) + Pn sin(ωt) + Qn cos(ωt) sin(kn x)
n=1
28
Now we apply the first IC:
ut (x, t) =
∞
X
− 2b t
e
n=1
"
#
b
− (Cn cos(βn t) + Dn sin(βn t)) + (−Cn βn sin(βn t) + Dn βn cos(βn t)) sin(kn x)+
2
[Pn ωcos(ωt) + Qn ωsin(ωt)] sin(kn x)
ut (x, 0) =
∞
X
n=1
"
#
b
− (Cn ) + (Dn βn ) + Pn ω sin(kn x) = 0
2
Because of the uniqueness of the Fourier Series
"
#
b
− (Cn ) + (Dn βn ) + Pn ω = 0
2
Dn =
b
Pn ω
(Cn ) −
2βn
βn
Therefore,
u(x, t) =
∞
X
"
− 2b t
e
n=1
!
!
#
Pn ω
b
(Cn ) −
sin(βn t) + Pn sin(ωt) + Qn cos(ωt) sin(kn x)
Cn cos(βn t) +
2βn
βn
Now we can apply the second IC:
u(x, 0) =
∞
X
n=1
"
− 2b (0)
e
!
!
#
b
Pn ω
Cn cos(βn (0)) +
(Cn ) −
sin(βn (0)) + Pn sin(ω(0)) + Qn cos(ω(0)) sin(
2βn
βn
u(x, 0) =
∞
X
(Cn + Qn ) sin(kn x) = f (x)
n=1
Using superposition we see that the coefficients can be found using the Fourier series.
Cn + Qn =
2ZL
f (x)sin(kn x)dx
L 0
2 Z L/3 3h
2ZL
Cn =
hsin(kn x)dx + Qn
xsin(kn x)dx +
L 0
L
L L/3
Again the coefficients can be solved using integration by parts or using a software program.
Cn =
24h
(2n − 1)π
sin(
) + Qn
2
((2n − 1)π)
6
29
Therefore the final solution for this case is:
u(x, t) =
∞
X
"
− 2b t
e
n=1
!
!
#
b
Pn ω
Cn cos(βn t) +
(Cn ) −
sin(βn t) + Pn sin(ωt) + Qn cos(ωt) sin(kn x)
2βn
βn
2) Consider b = 2ω1 = πc
< 2ωn n = 2, 3, .... In this case mode 1 is critically damped and
L
all other modes are under-damped.
u1 (x, t) = C1 e
−b
t
2
+ D1 te
−b
t
2
+ P1 sin(ωt) + Q1 cos(ωt) sin(k1 x)
b
un (x, t) = e− 2 t (Cn cos(βn t) + Dn sin(βn t) + Pn sin(ωt) + Qn cos(ωt)) sin(kn x)
n = 2, 3, ...
Therefore the general solution is:
u(x, t) = C1 e
∞
X
−b
t
2
+ D1 te
−b
t
2
+ P1 sin(ωt) + Q1 cos(ωt) sin(k1 x)+
b
e− 2 t [(Cn cos(βn t) + Dn sin(βn t)) + Pn sin(ωt) + Qn cos(ωt)] sin(kn x)
n=2
First we can find the coefficients when n = 1.
u(x, t) = C1 e
−b
t
2
+ D1 te
−b
t
2
+ P1 sin(ωt) + Q1 cos(ωt) sin(k1 x)
Applying the first B.C.
!
ut (x, t) =
−b
−b
b
−b −b
C1 e 2 t + D1 e 2 t + D1 t e 2 t + P1 ωcos(ωt) − Q1 ωsin(ωt) sin(k1 x)
2
2
"
#
b
ut (x, 0) = C1 + D1 + P1 ω sin(k1 x) = 0
2
−b
C1 + D1 + P1 ω = 0
2
b
D1 = C1 − P1 ω
2
Therefore,
u(x, t) = C1 e
∞
X
−b
t
2
!
!
−b
b P1
+
−
ω te 2 t + P1 sin(ωt) + Q1 cos(ωt) sin(k1 x)+
2 C1
b
e− 2 t [(Cn cos(βn t) + Dn sin(βn t)) + Pn sin(ωt) + Qn cos(ωt)] sin(kn x)
n=2
30
For n = 2, 3, .... we can apply the first IC as in the precious problem and find that,
Dn =
Pn ω
b
(Cn ) −
2βn
βn
So we can rewrite our solution as:
u(x, t) = C1 e
∞
X
"
− 2b t
Cn e
n=2
−b
t
2
!
!
−b
Q1
−b P1
P1
−
ω te 2 t +
sin(ωt) +
cos(ωt) sin(k1 x)+
+
2
C1
C1
C1
!
!
#
b
Pn ω
cos(βn t) +
−
sin(βn t) + Pn sin(ωt) + Qn cos(ωt) sin(kn x)
2βn Cn βn
Now we can apply the second IC:
u(x, 0) = (C1 + Q1 ) sin(k1 x) +
∞
X
(Cn + Qn ) sin(kn x) = f (x)
n=2
Using superposition we see that the coefficients can be found using the Fourier series.
2ZL
Cn + Qn =
f (x)sin(kn x)dx
L 0
Cn =
2 Z L/3 3h
2ZL
hsin(kn x)dx − Qn
xsin(kn x)dx +
L 0
L
L L/3
Again the coefficients can be solved using integration by parts or using a software program.
Cn = (1 + Qn )
24h
(2n − 1)π
sin(
) − Qn
2
((2n − 1)π)
6
Therefore the final solution for this case is:
u(x, t) = C1 e
∞
X
n=2
"
− 2b t
Cn e
−b
t
2
!
!
−b
−b P1
+
−
ω te 2 t + P1 sin(ωt) + Q1 cos(ωt) sin(k1 x)+
2
C1
!
!
#
b
Pn ω
cos(βn t) +
−
sin(βn t) + Pn sin(ωt) + Qn cos(ωt) sin(kn x)
2βn Cn βn
3) Consider b = 2πc
= 4ω1 . We know that ω1 = 2πc
, ω2 = 3 2πc
= 3ω1 , ω3 = 5 2πc
= 5ω1 .
L
L
L
L
Thus 2ω1 < b < 2ω2 < .... Therefore the first mode is over-damped and the rest are still
under-damped.
31
√
b
u1 (x, t) = C1 e(− 2 +ω1
ζ 2 −1)t
b
+ D1 e(− 2 −ω1
√
ζ 2 −1)t
+ P1 sin(ωt) + Q1 cos(ωt) sin(k1 x)
b
h
i
un (x, t) = e− 2 t (Cn cos(βn t) + Dn sin(βn t)) + Pn sin(ωt) + Qn cos(ωt) sin(kn x)
n = 2, 3, ...
Therefore the general solution is:
(− 2b +ω1
√
u(x, t) = C1 e
∞ h
X
ζ 2 −1)t
(− 2b −ω1
√
+ D1 e
ζ 2 −1)t
+ P1 sin(ωt) + Q1 cos(ωt) sin(k1 x)+
b
i
e− 2 t (Cn cos(βn t) + Dn sin(βn t)) + Pn sin(ωt) + Qn cos(ωt) sin(kn x)
n=2
First we can find the coefficients when n = 1.
b
√
u(x, t) = C1 e(− 2 +ω1
ζ 2 −1)t
b
+ D1 e(− 2 −ω1
√
ζ 2 −1)t
+ P1 sin(ωt) + Q1 cos(ωt) sin(k1 x)
Applying the first B.C.
"
ut (x, t) =
√
√
q
q
b
b
b
b
2
2
− + ω1 ζ 2 − 1 C1 e(− 2 +ω1 ζ −1)t + − − ω1 ζ 2 − 1 D1 e(− 2 −ω1 ζ −1)t sin(k1 x)+
2
2
!
!
#
+ (P1 ωcos(ωt) − Q1 ωsin(ωt)) sin(k1 x)
"
!
!
#
q
q
b
b
2
− + ω1 ζ − 1 C1 + − − ω1 ζ 2 − 1 D1 + P1 ω sin(k1 x) = 0
2
2
ut (x, 0) =
!
!
q
q
b
b
− + ω1 ζ 2 − 1 C1 + − − ω1 ζ 2 − 1 D1 + P1 ω = 0
2
2
√
− b + ω1 ζ 2 − 1
√
D1 = − 2b
C1 − P1 ω
− 2 − ω1 ζ 2 − 1
Therefore,
(− 2b +ω1
u(x, t) = C1 e
√
ζ 2 −1)t
√
!
!
√
− 2b + ω1 ζ 2 − 1 P1
(− 2b −ω1 ζ 2 −1)t
√
−
−
ω e
sin(k1 x)+
− 2b − ω1 ζ 2 − 1 C1
+ (P1 sin(ωt) + Q1 cos(ωt)) sink1 x
∞
X
h
b
i
e− 2 t (Cn cos(βn t) + Dn sin(βn t)) + Pn sin(ωt) + Qn cos(ωt) sin(kn x)
n=2
For n = 2, 3, ....
32
This was the same as in the previous case, therefore:
Dn =
Pn ω
b
(Cn ) −
2βn
βn
So we can rewrite our solution as:
(− 2b +ω1
√
u(x, t) = C1 e
ζ 2 −1)t
√
!
!
√
− 2b + ω1 ζ 2 − 1 P1
(− 2b −ω1 ζ 2 −1)t
√
−
−
ω e
sin(k1 x)+
− 2b − ω1 ζ 2 − 1 C1
+ (P1 sin(ωt) + Q1 cos(ωt)) sink1 x
∞
X
"
− 2b t
Cn e
n=2
!
!
#
b
Pn ω
−
cos(βn t) +
sin(βn t) + Pn sin(ωt) + Qn cos(ωt) sin(kn x)
2βn Cn βn
Now we can apply the second IC:
u(x, 0) = C1
√
!
!
∞
X
− 2b + ω1 ζ 2 − 1 P1
Q1
√
(Cn + Qn ) sin(kn x)
−
sin(k
x)
+
1−
ω
+
1
C1
− 2b − ω1 ζ 2 − 1 C1
n=2
The solution for the coefficients deliver the same integral as in the previous cases with the
addition that the first coefficient has a over-damped factor in front of it. So for the coefficients
we find that:
1
C1 = −
1−
Cn =
− 2b +ω1
− 2b −ω1
π
24h
√
sin( ) − Q1
2
2
6
√ζ −1 π
ζ 2 −1
24h
(2n − 1)π
sin(
) − Qn
2
((2n − 1)π)
6
n = 2, 3, ...
Therefore the final solution for this case is:
(− 2b +ω1
u(x, t) = C1 e
√
ζ 2 −1)t
√
!
!
√
− 2b + ω1 ζ 2 − 1 P1
(− 2b −ω1 ζ 2 −1)t
√
−
−
ω e
sin(k1 x)+
− 2b − ω1 ζ 2 − 1 C1
+ (P1 sin(ωt) + Q1 cos(ωt)) sink1 x
∞
X
n=2
"
− 2b t
Cn e
!
!
#
b
Pn ω
cos(βn t) +
−
sin(βn t) + Pn sin(ωt) + Qn cos(ωt) sin(kn x)
2βn Cn βn
We can now examine the plots for each of these cases and compare it to the case of no
damping in Problem #4.
33
1):
The first case is known as an under-damped case. The first plots show each of the first 5
terms of the Fourier Series and the associated sum for at 5 different times.
34
This next plot shows the time dependent solution at the center of the domain for each of the
first five terms and the sum of those terms. Notice how the solution looks slightly different
from the previous homework solution due to the affect of the forcing function.
35
These last plots are a 3D plot of the domain over the time of one period. If you would like
to view the movie of these plots please refer to the Course Files on E3.
36
2)
This case is known as an critically damped case. The first plots show each of the first 5
terms of the Fourier Series and the associated sum for at 5 different times.
This next plot shows the time dependent solution at the center of the domain for each of
the first five terms and the sum of those terms.
37
38
These last plots are a 3D plot of the domain over the time of one period. If you would like
to view the movie of these plots please refer to the Course Files on E3.
You can see the oscillatory behavior if you look at a zoomed in graph later in time.
39
3)
The last case is known as an over-damped solution.
This next plot shows the time dependent solution at the center of the domain for each of
the first five terms and the sum of those terms.
40
41
These last plots are a 3D plot of the domain over the time of one period. If you would like
to view the movie of these plots please refer to the Course Files on E3.
42
After a long time the transient solutions will die out leaving behind only the forced reponse.
u(x, t) =
X
(Pn cos(ωt) + Qn sin(ωt))sin(kn t)
Therefore the amplitude of oscillation is
Rn (ω) =
q
Pn2 + Q2n
The following plot shows Rn for various values of n.
43