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CAAM 336
·
DIFFERENTIAL EQUATIONS
Problem Set 8
·
Solutions
Posted Thursday 21 October 2010. Due Wednesday 27 October 2010, 5pm.
1. [50 points: 18 points for (a); 12 points for (b); 10 points each for (c) and (d)]
Consider the following three matrices:
(i)
A
=
±
0 1
1 0
²
(ii)
A
=
±

50
49
49

50
²
(iii)
A
=
±
0
1

1 0
²
.
(a) For each of the matrices (i)–(iii), compute the matrix exponential
e
t
A
.
(You may use
eig
for the eigenvalues and eigenvectors, but please construct the matrix exponential
‘by hand’ (not with
expm
). For diagonalizable
A
=
VΛV

1
, recall the formula
e
t
A
=
V
e
t
Λ
V

1
.)
(b) Use your answers in part (a) to explain the behavior of
x
0
(
t
) =
Ax
(
t
) as
t
→ ∞
given that
x
(0) = [2
,
0]
T
(e.g., exponential growth, exponential decay, or neither) for each of the three
matrices (i)–(iii).
(c) For the matrix (ii), describe how large one can choose the time step Δ
t
so that the forward Euler
method applied to
x
0
(
t
) =
Ax
(
t
),
x
k
+1
=
x
k
+ Δ
t
Ax
k
,
will produce a solution that qualitatively matches the behavior of the true solution (i.e., the
approximations
x
k
should grow, decay, or remain of the same size as the true solution does).
Answer the same question for the backward Euler method
x
k
+1
=
x
k
+ Δ
t
Ax
k
+1
.
(d) For the matrix in (iii), describe how the forward Euler method behaves
for all
Δ
t
as
k
→ ∞
for
x
(0) = [1
,
1]
T
. Now describe how the backward Euler method must behave as
k
→ ∞
for the
same matrix and initial condition.
Solution.
[GRADERS: it is acceptable for students to use MATLAB to compute eigendecompositions, but they
should not simply use the
expm
command. In particular, only give half credit if students computed
e
t
A
for a ﬁxed value of
t
. The correct answer should depend on the variable
t
.]
(a) We compute the matrix exponentials for each matrix in turn.
(i) Note that det(
λ
I

A
) =
λ
2

1 = (
λ
+ 1)(
λ

1) and hence the eigenvalues of
A
are
λ
1
=

1
and
λ
2
= 1. The corresponding (normalized) orthogonal eigenvectors are
u
1
=
√
2
2
±
1

1
²
,
u
2
=
√
2
2
±
1
1
²
.
As
A
is symmetric, if we set
U
= [
u
1
u
2
] and
Λ
= diag(
λ
1
,λ
2
), we have
A
=
UΛU
*
and
e
t
A
=
U
e
t
Λ
U
*
=
U
±
e

t
0
0
e
t
²
U
*
.
Multiplying this out gives
e
t
A
=
1
2
±
e
t
+
e

t
e
t

e

t
e
t

e

t
e
t
+
e

t
²
.
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View Full Document(ii) If we denote the matrix in part (a) as
A
1
, then we ﬁnd that the
A
in part (b) can be written
as
A
=

50
I
+ 49
A
1
, from which it follows (a slight modiﬁcation of problem 1 on the ﬁrst
exam) that
A
has eigenvalues
λ
1
=

99 and
λ
2
=

1 with the same eigenvectors as in
part (a):
u
1
=
√
2
2
±
1

1
²
,
u
2
=
√
2
2
±
1
1
²
.
Again
A
is symmetric, and we have that
e
tA
=
U
e
t
Λ
U
*
=
1
2
±
e

t
+
e

99
t
e

t

e

99
t
e

t

e

99
t
e

t
+
e

99
t
²
.
(iii) [GRADERS: please be a bit more lenient with this problem, as this
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 Fall '09
 Tompson

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