Chapter 6 Partial Differential Equations

26 CHAPTER 6. PARTIAL DIFFERENTIAL EQUATIONS
which implies that x = C 1 t + C 2 . Now since u is constant on a characteristic, then on a
characteristic, say the characteristic passing through (ξ,0),
dx
dt
= c(u(x(t),t)) = c(u(ξ,0)) = c(ϕ(ξ)).
Thus we see that the slope of the characteristics depend on c(u) and the initial data. The
equation for the characteristic passing through (ξ,0) is given by
and the solution on this line is given by
where x = c ( ϕ(ξ) ) t + ξ.
x = c ( ϕ(ξ) ) t + ξ,
u(x, t) =ϕ(ξ) =ϕ(x − c(ϕ(ξ))t)
t
(x, t)
(ξ,0)
x
Characteristic through (ξ,0)
Example 6.2.18. One particularly famous example of a hyperbolic conservation law which
is often used as a one dimensional model for the Navier-Stokes equations is the Burgers’
equation given by
Let us consider (6.2.18) subject to initial conditions
u t + uu x =0, x ∈ R, t>0. (6.2.18)
⎧
⎨ 2, x < 0
u(x, 0) = ϕ(x) = 2 − x, 0 ≤ x ≤ 1 . (6.2.19)
⎩
1, x > 1
For x1 the slope is 1.