Chapter 6 Partial Differential Equations

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34 CHAPTER 6. PARTIAL DIFFERENTIAL EQUATIONS 2. D =0implies Parabolic 3. D0 implies ã = ˜c =0 2. D =0implies ã = ˜b =0 3. D 0. If a = c = 0 we are done so we assume without loss of generality that a ≠ 0. We seek α = ϕ(x, y) and β = ψ(x, y) so that ã and ˜c are zero. We first note that the calculations (6.3.10) suggest that we consider an expression of the form av 2 x +2bv x v y + cv 2 y = 0 (6.3.11)
6.3. CHARACTERISTICS AND HIGHER ORDER EQUATIONS 35 where v could represent either ϕ or ψ. We note that (6.3.11) can be factored into [ ( √ ) ][ ( √ ) ] −b + b2 − ac −b − b2 − ac a v x − v y v x − v y . (6.3.12) a a Thus from (6.3.12) we seek ϕ and ψ so that, for example, [ ( √ ) ] −b + b2 − ac ϕ x − ϕ y =0, (6.3.13) a and [ ψ x − ( √ ) ] −b − b2 − ac ψ y =0. (6.3.14) a With these choices (6.3.11) is satisfied with v given by ϕ and ψ. In addition we must impose a noncharacteristic solvability condition ∣ ∂(ϕ, ψ) ∣∣∣∣ ∂(x, y) = ϕ x ψ x ϕ y ψ y ∣ ∣∣∣∣ ( √ ) −b + b2 − ac = ϕ y ψ y − a ( −b − √ b2 − ac a ) ϕ y ψ y = 2 a √ b2 − ac ϕ y ψ y ≠0. (6.3.15) A solution of (6.3.13) is found by solving dx dt =1, ( √ ) dy −b + dt = − b2 − ac , a dϕ dt =0. For this problem we take the initial conditions x 0 (s) =x 0 ,y 0 (s) =y 0 + s, ϕ 0 (s) =s. With this choice we find (in our earlier notation we used a and b which are not the same as those occuring in the present problem) ∣ ∣∣∣∣∣∣ a x ′ 1 0 0(s) ∣b y 0(s) ′ ∣ = ( √ ) b − b2 − ac ≠0. (6.3.16) 1 a ∣
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 33: 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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