Chapter 6 Partial Differential Equations

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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.
6.2. LINEAR AND QUASILINEAR EQUATIONS OF FIRST ORDER 27 Let us demonstrate that u − ϕ(x − c(u)t) = 0 implicitly defines a solution u(x, t) ofthe equation. First, differentiating with respect to x, wehave u x = ϕ ′ (x − c(u)t) [ x − c(u)t ] = x ϕ′ (x − c(u)t) [ 1 − c ′ (u)u x t ] and we can solve for u x u x = ϕ ′ (x − c(u)t) 1+tc ′ (u)ϕ ′ (x − c(u)t) . (6.2.20) Now we differentiate with respect to t, to obtain u t = ϕ ′ (x − c(u)t) [ x − c(u)t ] = t −ϕ′ (x − c(u)t) [ tc ′ (u)u t + c(u) ] and we can solve for u t u t = −c(u)ϕ′ (x − c(u)t) 1+tc ′ (u)ϕ ′ (x − c(u)t) . (6.2.21) So, combining (6.2.20) and (6.2.21), we have u t + c(u)u x = −c(u)ϕ′ (x − c(u)t) 1+tc ′ (u)ϕ ′ (x − c(u)t) + c(u) ϕ ′ (x − c(u)t) 1+tc ′ (u)ϕ ′ (x − c(u)t) = −c(u)ϕ′ (x − c(u)t)+c(u)ϕ ′ (x − c(u)t) =0. 1+tc ′ (u)ϕ ′ (x − c(u)t) and u(x, 0) = ϕ(x − 0) = ϕ(x). t t =1 (ξ,0) x Solution Exists Only for t1 since the characteristics cross beyond that line and the values on u on the intersecting characteristics are different –
Page 1 and 2: Chapter 6 Partial Differential Equa
Page 3 and 4: 6.1. INTRODUCTION 3 In this case we
Page 5 and 6: 6.1. INTRODUCTION 5 Solutions to La
Page 7 and 8: 6.1. INTRODUCTION 7 t u=0 u=1/2 u=1
Page 9 and 10: 6.2. LINEAR AND QUASILINEAR EQUATIO
Page 25: 6.2. LINEAR AND QUASILINEAR EQUATIO
Page 31 and 32: 6.3. CHARACTERISTICS AND HIGHER ORD
Page 53 and 54: Bibliography [1] R. Dautray and J.L

Chapter 6 Partial Differential Equations

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