Math 124A – September 30, 2010
Viktor Grigoryan
2
Firstorder linear equations
Last time we saw how some simple PDEs can be reduced to ODEs, and subsequently solved using ODE
methods. For example, the equation
u
x
= 0
(1)
has “constant in
x
” as its general solution, and hence
u
depends only on
y
, thus
u
(
x, y
) =
f
(
y
) is the
general solution, with
f
an arbitrary function of a single variable. Today we will see that any linear
first order PDE can be reduced to an ordinary differential equation, which will then allow as to tackle
it with already familiar methods from ODEs.
Let us start with a simple example. Consider the following constant coefficient PDE
au
x
+
bu
y
= 0
.
(2)
Here
a
and
b
are constants, such that
a
2
+
b
2
6
= 0, i.e. at least one of the coefficients is nonzero (otherwise
this would not be a differential equation). Using the inner (scalar or dot) product in
R
2
, we can rewrite
the left hand side of (2) as
(
a, b
)
·
(
u
x
, u
y
) = 0
,
or
(
a, b
)
· ∇
u
= 0
.
Denoting the vector (
a, b
) =
v
, we see that the left hand side of the above equation is exactly
D
v
u
(
x, y
),
the directional derivative of
u
in the direction of the vector
v
. Thus the solution to (2) must be constant
in the direction of the vector
v
=
a
i
+
b
j
.
v
=
H
a
,
b
L

5
5
x

5
5
y
Figure 1: Characteristic lines
bx

ay
=
c
.
Ξ
Η
H
a
,
b
L
H

b
,
a
L
H
x
,
y
L
Ξ
=
H
x
,
y
L
H
a
,
b
L
Η
=
H
x
,
y
L
H

b
,
a
L

10

5
5
x

5
5
y
Figure 2: Change of coordinates.
The lines parallel to the vector
v
have the equation
bx

ay
=
c,
(3)
since the vector (
b,

a
) is orthogonal to
v
, and as such is a normal vector to the lines parallel to
v
. In
equation (3)
c
is an arbitrary constant, which uniquely determines the particular line in this family of
parallel lines, called
characteristic lines
for the equation (2).
As we saw above,
u
(
x, y
) is constant in the direction of
v
, hence also along the lines (3). The line
containing the point (
x, y
) is determined by
c
=
bx

ay
, thus
u
will depend only on
bx

ay
, that is
u
(
x, y
) =
f
(
bx

ay
)
,
(4)
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
where
f
is an arbitrary function. One can then check that this is the correct solution by plugging it
into the equation. Indeed,
a∂
x
f
(
bx

ay
) +
b∂
y
f
(
bx

ay
) =
abf
0
(
bx

ay
)

baf
0
(
bx

ay
) = 0
.
The geometric viewpoint that we used to arrive at the solution is akin to solving equation (1) simply
by recognizing that a function with a vanishing derivative must be constant. However one can approach
equation (2) from another perspective, by trying to reduce it to an ODE.
2.1
The method of characteristics
To have an ODE, we need to eliminate one of the partial derivatives in the equation. But we know that
the directional derivative vanishes in the direction of the vector (
a, b
). Let us then make a change of the
coordinate system to one that has its “
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '08
 Ponce
 Differential Equations, Linear Equations, Equations, Partial Differential Equations, Trigraph, Elementary algebra, Partial differential equation, Coordinate system, ξ

Click to edit the document details