Chapter 6 Partial Differential Equations

28 CHAPTER 6. PARTIAL DIFFERENTIAL EQUATIONS
thus, as a function, u is not well defined. More specifically, recall that our solution is defined
by
u(x, t) =ϕ(ξ) where x = ϕ(ξ)t + ξ.
The characteristics are described by
⎧
⎨ 2t + ξ, ξ < 0
x = (2 − ξ)t + ξ, 0 ≤ ξ ≤ 1 .
⎩
t + ξ, ξ > 1
We can compute that the characteristics intersect at
((2 − ξ)t + ξ) ∣ ∣
ξ=0
=(t + ξ) ∣ ∣
ξ=1
,
or 2t =1+t, i.e., t =1. Fort1)
t
t =1
x = t +1
x =2t
u =2
(ξ,0)
u =1
x
Solution
If 0 ≤ ξ ≤ 1 the characteristic passing through (ξ,0) is x =(2− ξ)t + ξ which implies
ξ =(x − 2t)/(1 − t) and so
( ) x − 2t
u(x, t) =2− = 2 − x , 2t ≤ x ≤ t +1, t < 1.
1 − t 1 − t
u
t =1
u =2
t
u =1
x