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The Wave Equation
  • One can solve for x and t from eq. 9↑ and substitute in wave equations.
    (10) x = (η + ξ)/(2)     t = (ξ − η)/(2C)
    One can recognize ξ = x − Ct and η = x + Ct as characteristic lines of wave equations.
Step two, calculate derivatives  Recall that the chain rule can be used to calculate derivatives:
  • (11) (∂u)/(∂x) = (∂u)/(∂ξ)(∂ξ)/(∂x) + (∂u)/(∂η)(∂η)/(∂x)
    where
    (12) (∂ξ)/(∂x) = 1
    and
    (13) (∂η)/(∂x) = 1
    so
    (14) (∂u)/(∂x) = (∂u)/(∂ξ) + (∂u)/(∂η)
    If (∂u)/(∂ξ) = p and (∂u)/(∂η) = q then
    (15) (∂u)/(∂x) = p + q
    The second derivative of u with respect to x can be defined as
    (16) (∂2u)/(∂x2) = (∂p)/(∂x) + (∂q)/(∂x)
    (17) (∂p)/(∂x) = (∂p)/(∂ξ) + (∂p)/(∂η)
    and
    (18) (∂q)/(∂x) = (∂q)/(∂ξ) + (∂q)/(∂η)
    Substituting eqs. 17↑ and 18↑ into eq. 16↑ and substituting for p and q, one finally can have:
    (19) (∂2u)/(∂x2) = (∂2u)/(∂ξ2) + 2(∂2u)/(∂ξ∂η) + (∂2u)/(∂η2)
    Similarly
    (20) (∂2u)/(∂t2) = C2(∂2u)/(∂ξ2) − 2C2(∂2u)/(∂ξ∂η) + C2(∂2u)/(∂η2)
Step three, write the wave equation in characteristic coordinates  Substituting these two equations in eq. ↓ leads to
(21) (∂2u)/(∂ξ∂η) = 0
Step four, solve the wave equation in characteristic coordinate  Equation 21↑ can be solved by direct integration:
  • (22) u(ξ, η)  = ⌠⌡⌠⌡(∂2u)/(∂ξ∂η)dξdη      = ⌠⌡(∂u)/(∂ξ)dξ + G(η)      = F(ξ) + G(η)
    Finally,
    (23) u(ξ, η) = F(ξ) + G(η) = F(x − Ct) + G(x + Ct)
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