EN 202 : Problem Set 4
نویسنده
چکیده
For each of the following functions u 0 (x), construct solutions to the initial value problem (1 + ku)u x + u y = 0, y ≥ 0 u(x, 0) = u 0 (x) is a given function for −∞ ≤ x ≤ ∞ Draw the characteristics in the (x, y)-plane. If you find the wave-breaking phenomenon arising, determine the time and place of the initial break appearance and find a solution containing a shockwave. For this first problem, consider u 0 (x), for k ≥ 0 and ε > 0, given by u 0 (x) = 1, x ≥ ε x ε , 0 < x < ε 0, x ≤ 0 As discussed in class on 2/17/06, this problem focuses on the solution of a quasilinear first-order PDE. Note that our initial curve Γ is the line y = 0. As a result, we can parameterize it as ¯ x(0, η) = x 0 (η) = η ¯ y(0, η) = y 0 (η) = 0 In addition, we define the vector field v(ξ, η) as v(ξ, η) = ¯ x ξ (ξ, η) ¯ y ξ (ξ, η) = 1 + kU (ξ, η) 1 As usual, the PDE becomes an ODE along the characteristics such that v·∇u = 0. Applying the chain rule, we find U ξ (ξ, η) = 0. We can integrate this ODE with respect to ξ to determine U (ξ, η) = φ(η). Notice that, along Γ, U (0, η) = u 0 (η). Applying this initial condition, we have U (ξ, η) = u 0 (η), for − ∞ ≤ η ≤ ∞ (1) Substituting this result into our expression for v(ξ, η) we find v(ξ, η) = ¯ x ξ (ξ, η) ¯ y ξ (ξ, η) = 1 + ku 0 (η) 1 (2) At this point, we require an explicit form of the characteristics. First, consider the solution to the ODE ¯ y ξ (ξ, η) = 1. Integrating with respect to ξ, we find ¯ y = ξ + φ(η). Applying the initial value on Γ, where y 0 (η) = 0 and ξ = 0, we find φ(η) = 0. Next, consider the solution to the ODE ¯ x ξ (ξ, η) = 1+ku 0 (η). Integrating with respect to ξ, we find ¯ x = [1+ku 0 (η)]ξ+ψ(η).
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تاریخ انتشار 2006