MATH 220: SECOND ORDER CONSTANT COEFFICIENT PDE
ANDRAS VASY
We consider second order constant coefficient scalar linear PDEs on
R
n
. These
have the form
L
u
=
f,
L
=
n
summationdisplay
i,j
=1
a
ij
∂
x
i
∂
x
j
+
n
summationdisplay
i
=1
b
i
∂
x
i
+
c
where
a
ij
,
b
i
and
c
are (complex) constants, and
f
is given. As a general principle,
the leading, i.e. second order, terms are the most important in analyzing the PDE.
The second order terms then take the form
L
=
n
summationdisplay
i,j
=1
a
ij
∂
x
i
∂
x
j
;
here we are interested in the case when the
a
ij
are real constants. Since
∂
x
i
∂
x
j
=
∂
x
j
∂
x
i
, we might as well assume that
a
ij
=
a
ji
(otherwise replace both by
1
2
(
a
ij
+
a
ji
), which does not change
L
), i.e. that
A
is symmetric. We rewrite
L
in a different
form. With
∇
x
=
∂
x
1
∂
x
2
...
∂
x
n
and
A
=
a
11
... a
1
n
...
a
n
1
... a
nn
,
we can write
L
=
∇
†
x
A
∇
x
,
where
∇
†
x
is the transpose of
∇
x
:
∇
†
x
=
bracketleftbig
∂
x
1
... ∂
x
n
bracketrightbig
.
Technically, there is no need for the
x
in the subscripts, and one should write e.g.
∇
†
=
bracketleftbig
∂
1
... ∂
n
bracketrightbig
. However, the subscripts are very useful below as they enable
us to compress (and abuse) the notation in the change of variables formula.
We would like to bring it to a simpler form; the simpler form we wish for may
depend on the intended use. As a first step we would like to have
L
brought into a
diagonal form, i.e.
A
made diagonal. We can achieve this by changing coordinates.
Namely, if we let
ξ
= Ξ(
x
), i.e.
ξ
j
= Ξ
j
(
x
1
,...,x
n
),
j
= 1
,...,n
, with inverse
function
X
, i.e.
x
=
X
(
ξ
), meaning that
X
◦
Ξ and Ξ
◦
X
are both the identity
maps (both
X
and Ξ are assumed to be
C
1
), then by the chain rule for any
C
1
function
u
we have
∂
x
k
(
u
◦
X
) =
n
summationdisplay
ℓ
=1
∂
Ξ
ℓ
∂x
k
(
∂
ξ
ℓ
u
)
◦
X.
1
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2
ANDRAS VASY
We can rewrite this in a matrix form:
(1)
∂
x
1
...
∂
x
n
=
∂
Ξ
1
∂
x
1
...
∂
Ξ
n
∂x
1
...
∂
Ξ
1
∂
xn
...
∂
Ξ
n
∂x
n
∂
ξ
1
...
∂
ξ
n
,
so in particular if
ξ
is a linear function of
x
, i.e.
ξ
=
Bx
,
B
=
bracketleftbig
b
ij
bracketrightbig
n
i,j
=1
, so
ξ
i
=
∑
n
j
=1
B
ij
x
j
, then
(2)
∇
x
=
B
†
∇
ξ
.
Thus,
L
=
∇
†
x
A
∇
x
=
∇
†
ξ
BAB
†
∇
ξ
.
Hence, in order to make
L
be given by a diagonal matrix in the
ξ
coordinates, we
need that
BAB
†
is diagonal. Recalling that
A
is symmetric, we proceed as follows.
Any diagonal matrix, such as
A
, can be diagonalized by conjugating it by an
orthogonal matrix, i.e. there is an orthogonal matrix
O
(recall that orthogonal
means that
OO
†
=
O
†
O
= Id, i.e.
O
†
=
O
−
1
) and a diagonal matrix Λ such that
A
=
O
†
Λ
O,
where the diagonal entries
λ
j
of Λ are the eigenvalues of
A
and the rows of
O
,
i.e. the columns of
O
†
, are corresponding (unit length) eigenvectors
e
j
.
(Thus,
the orthogonality of
O
is a consequence of the orthogonality of the
e
j
.) Choosing
B
=
O
, we deduce that
BAB
†
=
OO
†
Λ
OO
†
= Λ
is indeed diagonal.
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 Fall '10
 Vasy
 Partial differential equation, Diagonal matrix, initial conditions, Hyperbolic partial differential equation, diagonal entries

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