•
If
A
is an
n
×
n
matrix, then
e
A
=
∑
∞
n
=0
A
n
n
!
. The significance of this
is that
x
(
t
) =
e
At
x
0
is the solution to the system of equations
x
0
=
Ax
with initial condition
x
(0) =
x
0
. For example, if
A
=
0

1
1
0
, then
e
At
=
cos
t

sin
t
sin
t
cos
t
. If
X
(
t
) is a “fundamental matrix” (a matrix
whose columns are linearly independent solutions to
x
0
=
Ax
), then
e
At
=
X
(
t
)
X
(0)

1
.
(Matrix exponentials were not on the syllabus,
but I explained this in class because I think it clarifies things a lot.)
•
To solve an inhomogeneous system
x
0
=
Ax
+
f
, choose a fundamental
matrix
X
(
t
) and write
x
(
t
) =
X
(
t
)
v
(
t
). Then the equation to be solved
is equivalent to
v
0
(
t
) =
X
(
t
)

1
f
. This can be integrated to obtain
v
(
t
),
and then multiplied by
X
(
t
) to give
x
(
t
). The answer will contain
n
integration constants, which correspond to adding a particular solution
to
x
0
=
Ax
+
f
to a general solution to the homogeneous equation
x
0
=
Ax
.
Differential equations chapter 10.
•
The Fourier series of a periodic function
f
with period 2
π
is given by
f
(
x
) =
a
0
2
+
∞
X
n
=1
(
a
n
cos(
nx
) +
b
n
sin(
nx
))
where
a
n
=
1
π
Z
π

π
f
(
x
) cos(
nx
)
dx
and
b
n
=
1
π
Z
π

π
f
(
x
) sin(
nx
)
dx.
5
If
f
is an odd function then
a
n
= 0 for all
n
; and if
f
is an even
function then
b
n
= 0 for all
n
. If
f
and
f
0
are piecewise continuous
then the Fourier series for
f
(
x
) converges to the average of the left and
right limits of
f
at
x
.
•
The Fourier sine series of a function
f
on the interval [0
, π
] is
f
(
x
)
∼
∞
X
n
=1
b
n
sin(
nx
)
where
b
n
=
2
π
Z
π
0
f
(
x
) sin(
nx
)
dx.
The Fourier cosine series of
f
is
f
(
x
)
∼
a
0
2
+
∞
X
n
=1
a
n
cos(
nx
)
where
a
n
=
2
π
Z
π
0
f
(
x
) cos(
nx
)
dx.
•
How to find some solutions to the heat equation by separation of vari
ables.
•
The solution to the heat equation
∂u
∂t
=
β
∂
2
u
∂x
2
for
x
∈
[0
, π
] and
t
∈
R
with boundary conditions
u
(0
, t
) =
u
(
π, t
) = 0 and initial condition
u
(
x,
0) =
f
(
x
) is given by
u
(
x, t
)
∼
∞
X
n
=1
b
n
e

βn
2
t
sin(
nx
)
where the constants
b
n
are the coefficients of the Fourier sine series of
f
. Here we ignore convergence issues.
•
Similarly, if we change the boundary condition above to
∂u
∂x
(0
, t
) =
∂u
∂x
(
π, t
) = 0, then the solution is
u
(
x, t
)
∼
a
0
2
+
∞
X
n
=1
a
n
e

βn
2
t
cos(
nx
)
where the constants
a
n
are the coefficients of the Fourier cosine series
of
f
.
6
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 Spring '08
 Chorin
 Math, Differential Equations, Linear Algebra, Algebra, Equations, LINEAR ALGEBRA Chapter