9Partial differential equations (PDEs)
IA Differential Equations
9.1 First-order wave equation
Consider the equation of the form
∂y
∂t
= c
∂y
∂x
,
with
c
a constant and
y
a function of
x
and
t
. This is known as the (first-order)
wave equation. We will later see that solutions correspond to waves travelling in
one direction.
We write this as
∂y
∂t
− c
∂y
∂x
= 0.
Recall that along a path x = x(t) so that y = y(x(t), t),
dy
dt
=
∂y
∂x
dx
dt
+
∂y
∂t
=
dx
dt
∂y
∂x
+ c
∂y
∂x
by the chain rule. Now we choose a path along which
dx
dt
= −c. (1)
Along such paths,
dy
dt
= 0 (2)
So we have replaced the original partial differential equations with a pair of
ordinary differential equations.
Each path that satisfies (1) can be described by
x
=
x
0
− ct
, where
x
0
is a
constant. We can write this as x
0
= x + ct.
From (2), along each of these paths,
y
is constant. So suppose for each
x
0
, the
value of
y
along the path
x
0
=
x
+
ct
is given by
f
(
x
0
), where
f
is an arbitrary
function. Then the solution to the wave equation is
y = f(x + ct),
By differentiating this directly, we can easily check that every function of this
form is a solution to the wave equation.
The contours of y look rather boring.
x
t
O
contours of y
Note that as we move up the time axis, we are simply taking the
t
= 0 solution
and translating it to the left.
The paths we’ve identified are called the “characteristics” of the wave equation.
In this particular example,
y
is constant along the characteristics (because the
equation is unforced).
We usually have initial conditions e.g.
y(x, 0) = x
2
− 3
Since we know that y = f(x + ct) and f(x) = x
2
− 3, y must be given by
y = (x + ct)
2
− 3.
We can plot the xy curve for different values of t to obtain this:
x
y
We see that each solution is just a translation of the t = 0 version.
We can also solve forced equations, such as
∂y
∂t
+ 5
∂y
∂x
= e
−t
, y(x, 0) = e
−x
2
.
Along each path
x
0
=
x −
5
t
, we have
dy
dt
=
e
−t
. So
y
=
f
(
x
0
)
− e
−t
for some
function f.
Using the boundary conditions provided, At
t
= 0,
y
=
f
(
x
0
)
−
1 and
x
=
x
0
.
So f(x
0
) − 1 = e
−x
2
0
, i.e. f(x
0
) = 1 + e
−x
2
0
. So
y = 1 + e
−(x−5t)
2
− e
−t
.