Fourier Series and Partial Differential Equations Lecture Notes

Chapter 4. The wave equation 42
Figure 14
Again, the sum
y(x,t) := F(x−ct)+G(x+ct), (4.84)
is a solution of the wave equation. It will now be shown that every solution of the wave
equation must be of the form (4.84).
To verify this introduce new independent variables
and seek a solution y(x,t) = Y(ξ,η). Then
∂y
∂t
∂y ∂Y
=
∂x ∂ξ
ξ := x−ct, η := x+ct, (4.85)
+ ∂Y
∂η ,
= −c∂Y
∂ξ +c∂Y
∂η ,
and substitution into the wave equation gives
∂ 2 y
∂x 2 = ∂2 Y
∂ξ 2 +2 ∂2 Y
∂2y ∂t2 = c2∂2 Y
∂ξ2 −2c2 ∂2Y ∂ 2 Y
∂ξ 2 +2 ∂2 Y
∂ξ∂η + ∂2 Y
∂η 2 = ∂2 Y
∂ξ 2 −2 ∂2 Y
∂ξ∂η + ∂2Y , (4.86)
∂η2 ∂ξ∂η +c2∂2 Y
, (4.87)
∂η2 ∂ξ∂η + ∂2Y . (4.88)
∂η2 Hence in the new variables the wave equation transforms to the equation
i.e.
∂ 2 Y
∂ξ∂η
= 0, (4.89)
∂ ∂Y
= 0. (4.90)
∂ξ ∂η
Thus ∂Y/∂η is independent of ξ and is a function of η only, say G ′ (η), i.e.
and so
∂Y
∂η = G′ (η), (4.91)
∂
[Y −G(η)] = 0. (4.92)
∂η
Thus Y −G(η) is a function of ξ only, say F(ξ), and therefore
and
Y −G(η) = F(ξ), (4.93)
Y = F(ξ)+G(η) =⇒ y(x,t) = F(x−ct)+G(x+ct). (4.94)
Further use of this conclusion will be made later.