9Partial differential equations (PDEs)
IA Differential Equations
9.2 Second-order wave equation
We consider equations in the following form:
∂
2
y
∂t
2
= c
2
∂
2
y
∂x
2
.
This is the second-order wave equation, and is often known as the “hyperbolic
equation” because the form resembles that of a hyperbola (which has the form
x
2
− b
2
y
2
= 0). However, the differential equation has no connections to
hyperbolae whatsoever.
This equation models an actual wave in one dimension. Consider a horizontal
string, along the
x
axis. We let
y
(
x, t
) be the vertical displacement of the string
at the point x at time t.
y
Suppose that
ρ
(
x
) is the mass per unit length of a string. Then the restoring
force on the string
ma
=
ρ
∂
2
y
∂t
2
is proportional to the second derivative
∂
2
y
∂x
2
. So
we obtain this wave equation.
(Why is it proportional to the second derivative? It certainly cannot be
proportional to
y
, because we get no force if we just move the whole string
upwards. It also cannot be proportional to
∂y/∂x
: if we have a straight slope,
then the force pulling upwards is the same as the force pulling downwards, and
we should have no force. We have a force only if the string is curved, and
curvature is measured by the second derivative)
To solve the equation, suppose that c is constant. Then we can write
∂
2
y
∂t
2
− c
2
∂
2
y
∂x
2
= 0
∂
∂t
+ c
∂
∂x
∂
∂t
− c
∂
∂x
y = 0
If
y
=
f
(
x
+
ct
), then the first operator differentiates it to give a constant (as in
the first-order wave equation). Then applying the second operator differentiates
it to 0. So y = f(x + ct) is a solution.
Since the operators are commutative,
y
=
f
(
x − ct
) is also a solution. Since
the equation is linear, the general solution is
y = f(x + ct) + g(x − ct).
This shows that the solution composes of superpositions of waves travelling to
the left and waves travelling to the right.
We can show that this is indeed the most general solution by substituting
ξ
=
x
+
ct
and
η
=
x −ct
. We can show, using the chain rule, that
y
tt
−c
2
y
xx
≡
−4c
2
y
ηξ
= 0. Integrating twice gives y = f(ξ) + g(η).
How many boundary conditions do we need to have a unique solution? In
ODEs, we simply count the order of the equation. In PDEs, we have to count
over all variables. In this case, we need 2 boundary conditions and 2 initial
conditions. For example, we can have:
– Initial conditions: at t = 0,
y =
1
1 + x
2
∂y
∂t
= 0
– Boundary conditions: y → 0 as x → ±∞.
We know that the solution has the form
y = f(x + ct) + g(x − ct).
The first initial condition give
f(x) + g(x) =
1
1 + x
2
(1)
The second initial condition gives
∂y
∂t
= cf
0
(x) − cg
0
(x) = 0 (2)
From (2), we know that
f
0
=
g
0
. So
f
and
g
differ by a constant. wlog, we can
assume that they are indeed equal, since if we had, say,
f
=
g
+ 2, then we can
let
y
= (
f
(
x
+
ct
) + 1) + (
g
(
x − ct
) + 1) instead, and the new
f
and
g
are equal.
From (1), we must have
f(x) = g(x) =
1
2(1 + x
2
)
So, overall,
y =
1
2
1
1 + (x + ct)
2
+
1
1 + (x − ct)
2
Where we substituted x for x + ct and x − ct in f and g respectively.