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EN 202: Problem Set 2

By Douglas R. Lanman

Abstract

Find u(x, y) satisfying ux + uy = u 2, u(x, −x) = x, − ∞ ≤ x ≤ ∞ Sketch the characteristic curves and show the domain of validity of the solution in the (x, y)plane. In this problem we will examine the solution to a nonhomogeneous first-order PDE with constant coefficients. In order to solve this problem we can follow a typical approach: (1) find an explicit form for the characteristics, (2) transform the PDE to an ODE along these characteristics, and (3) conclude by determining the domain of validity of our solution. Consider the vector field v defined as ¯xξ(ξ, η) a(x, y) 1 v = ¯yξ(ξ, η) b(x, y) 1 Similarly, the gradient of u is given by ∇u = ux uy Note that the PDE can be restated as v · ∇u = u 2. Once again, this equation represents a constraint on the component of the gradient of u(x, y) which is parallel to v. To proceed, we want to find a set of characteristics {¯x(ξ, η), ¯y(ξ, η)}, with tracing parameter ξ, such that {¯xξ = a(¯x, ¯y) = 1, ¯yξ = b(¯x, ¯y) = 1} (1) Note that our initial curve Γ is the line y = −x and can be parameterized as ¯x(0, η) = x0(η) = η ¯y(0, η) = y0(η) = −η Integrating Equation 1 with respect to ξ gives ¯x(ξ, η) = ξ + φ(η) and ¯y(ξ, η) = ξ + ψ(η). Recall that, along our initial curve Γ, ξ = 0; as a result, φ(η) = η and ψ(η) = −η. In conclusion, the characteristics are given by ¯x(ξ, η) = ξ + η ¯y(ξ, η) = ξ − η We can invert these coordinates to obtain closed-form expressions for (ξ, η) as follows. ¯ξ(x, y) = x +

Year: 2006
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