Chapter 6 Partial Differential Equations

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38 CHAPTER 6. PARTIAL DIFFERENTIAL EQUATIONS Thus the characteristic curves are given by y 2 − x 2 = constant, y 2 + x 2 = constant. We take α = ϕ(x, y) =x 2 + y 2 , β = ψ(x, y) =x 2 − y 2 . We solve for x, y to obtain If we use (6.3.10) we find √ √ α + β α − β x = , y = . 2 2 ˜b =8x 2 y 2 =2(α 2 − β 2 ) ˜d =2(y 2 − x 2 )=−2β ẽ =2(y 2 + x 2 )=2α and also u αβ = 2β 4(α 2 − β 2 ) u 2β α − 4(α 2 − β 2 ) u β = βu a − αu β 2(α 2 − β 2 ) . Example 6.3.6. Suppose that a, b, c, d, e, f in (6.3.4) are constants. Then the characteristics are given by dy dx = b ± √ b 2 − ac ≡ ν ± . a Thus the characteristics are straight lines α = ϕ(x, y) =y − ν + x, β = ψ(x, y) =y − ν − x. In this case we get ã = ˜c = 0 and ˜b = aϕx ψ x + b(ϕ x ψ y + ϕ y ψ x )+cϕ y ψ x = aν + ν − − b(⃗u + + ν − )+c = 2(ac − b2 ) . a
6.3. CHARACTERISTICS AND HIGHER ORDER EQUATIONS 39 For the first order terms we have ˜d = aϕ xx +2bϕ xy + cϕ yy + dϕ x + eϕ y = dν + + e ≡ d ′ , ẽ = aψ xx +2bψ xy + cψ yy + dψ x + eψ y = dν − + e ≡ e ′ . Recalling that there is a 2 timdes the ˜b term and multiplying by K, the transformed equation is u αβ + Kd ′ u α + Ke ′ u β + Kfu = Kg where K = a 4(ac − b 2 ) . We note that a further reduction is always possible in this case. Namely we can remove the first order terms. Let u = e λα+µβ v. Then we have u α = e λα+µβ (λv + v α ) u β = e λα+µβ (µv + v β ) u αα = e λα+µβ (v αα +2λv α + λ 2 v) u ββ = e λα+µβ (v ββ +2µv β + µ 2 v) u αβ = e λα+µβ (v αβ + λv β + µv a + λµv). If we choose µ = −d ′ K, λ = −e ′ K and define f 1 = Kf + K 2 d ′ e ′ , g 1 = e −(λα+µβ) Kg, then the equation can be written in terms of v as v αβ + f 1 v = g 1 . II. Parabolic Case: Assume that D = b 2 − ac = 0. In this case, if a = 0, then b = 0 and we are done since c ≠ 0 (otherwise the equation is not truely second order). Thus we assume that a ≠ 0 and the characteristic equation (6.3.11) factors as ( a v x + b 2 y) a v =0. (6.3.19)
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 31 and 32: 6.3. CHARACTERISTICS AND HIGHER ORD
Page 37: 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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