(a) Example 1
As an introduction to the method of characteristics, let us set out to solve the
following simple first-order partial differential equation;
∂f
∂t
þ x
∂f
∂x
¼ 0
ð2:38Þ
with the initial condition;
f x, 0
ð Þ ¼ 2x:
From Eq. (2.38) we note that f is a function of x and t, hence, we write,
f ¼ f x, t
ð Þ
and, therefore,
df ¼
∂f
∂t
dt þ
∂f
∂x
dx,
accordingly,
df
dt
¼
∂f
∂t
þ
dx
dt
∂f
∂x
:
ð2:39Þ
By comparing Eqs. (2.38) and (2.39) we have df/dt ¼ 0 on dx/dt ¼ x, hence,
f ¼ constant
on the curve defined by the equation,
dx
dt
¼ x:
ð2:40Þ
Eq. (2.40) defines a family of curves in the x, t plane, called the characteristics of
Eq. (2.38) and f(x, t) is a constant along each curve (with a different constant in each
case as shown below). Integrating Eq. (2.40) gives
ln x ¼ t þ c,
where c is a constant of integration, hence,
t ¼ ln x À c:
ð2:41Þ
62
2 Waves of Finite Amplitude
As an introduction to the method of characteristics, let us set out to solve the
following simple first-order partial differential equation;
∂f
∂t
þ x
∂f
∂x
¼ 0
ð2:38Þ
with the initial condition;
f x, 0
ð Þ ¼ 2x:
From Eq. (2.38) we note that f is a function of x and t, hence, we write,
f ¼ f x, t
ð Þ
and, therefore,
df ¼
∂f
∂t
dt þ
∂f
∂x
dx,
accordingly,
df
dt
¼
∂f
∂t
þ
dx
dt
∂f
∂x
:
ð2:39Þ
By comparing Eqs. (2.38) and (2.39) we have df/dt ¼ 0 on dx/dt ¼ x, hence,
f ¼ constant
on the curve defined by the equation,
dx
dt
¼ x:
ð2:40Þ
Eq. (2.40) defines a family of curves in the x, t plane, called the characteristics of
Eq. (2.38) and f(x, t) is a constant along each curve (with a different constant in each
case as shown below). Integrating Eq. (2.40) gives
ln x ¼ t þ c,
where c is a constant of integration, hence,
t ¼ ln x À c:
ð2:41Þ
62
2 Waves of Finite Amplitude
