Chapter 6 Partial Differential Equations

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40 CHAPTER 6. PARTIAL DIFFERENTIAL EQUATIONS Thus we need to find a solution ϕ(x, y) of (6.3.19) and then select ψ(x, y) so that ∣ ∂(ϕ, ψ) ∣∣∣∣ ∂(x, y) = ϕ x ψ x ϕ y ψ y ∣ ∣∣∣∣ = ϕ x ψ y − ψ x ϕ y = − b a ϕ yψ y − ψ x ϕ y = −ϕ y (ψ x + b ) a ψ y ≠0. We may, for example, take ψ = x and prescribe inital conditions for (6.3.18) so that ϕ y ≠ 0. With these choices, ã = 0 (since ϕ satisfies (6.3.18)) and from (6.3.10) we find ˜b = aϕx ψ x + b(ϕ x ψ y + ϕ y ψ x )+cϕ y ψ y ( = a − b ) [( a ϕ y ψ x + b − b ) ] a ϕ y ψ y + ϕ y ψ x + cϕ y ψ x [ ] = ϕ y −bψ x − b2 a ψ y + bψ x + cψ y [ ] = −ϕ y ψ y − b2 a + c =0. Thus with α = ϕ(x, y), β = ψ(x, y), (6.3.4) becomes ˜cu ββ + ˜du α + ẽu β + ˜fu = ˜g. Furthermore, since b 2 = ac, and with the choice ψ = x we have ( ˜c = aψx 2 +2bψ x ψ y + cψy 2 = a ψ x + b 2 y) a ψ = a ≠0. Therefore we can divide by ˜c = a to get the desired form. If ϕ(x, y) is a solution of (6.3.19) and ϕ(x, y) = constant, then it follows from (6.3.18) that the characteristics could be obtained from dy dx = b a . Example 6.3.7. Let us transform the equation u xx − 2xu xy + x 2 u yy − 2u y =0
6.3. CHARACTERISTICS AND HIGHER ORDER EQUATIONS 41 into canonical form. Here we have b 2 −ac = x 2 −x 2 = 0, so the equation is of parabolic type and the characteristics are found from or Take dy dx = b a = −x, y = − x2 2 + c. ϕ(x, y) =y + x2 2 , ψ(x, y) =x. Then ˜d = aϕ xx +2bϕ xy + cϕ yy + dϕ x + eϕ y = (1)(1)+0+0+0+(−2)(1) = −1, ẽ = aψ xx +2bψ xy + cψ yy + dψ x + eψ y =0, and so u ββ − u α =0. III. Elliptic Case: Assume that D = b 2 − ac < 0. For this case we choose to proceed in a purely formal fashion. This problem turns out to be rather messy and lengthy to carry out in detail. For a detailed treatment we refer to the text by Garabedian [6]. In particular, we will assume that a, b and c are analytic functions near the point in question (x 0 ,y 0 ). This can be avoided but not without great difficulty. The characteristic equation (6.3.18) factors in this case into a ( v x − [ −b + i √ ac − b 2 a ] v y )( v x − [ √ ] ) −b − i ac − b 2 v y =0. a We seek a solution of the characteristic equation solving, for example, ( [ √ ] ) −b + i ac − b 2 v x − v y =0. a Let us suppose that we find z(x, y) =ϕ(x, y)+iψ(x, y) so that Note that in the present case we must have ac < 0 which means that a and c have the same sign and are not zero. We seek a holomorphic function ϕ = ϕ 1 + iϕ 2 such that aϕ x + (b + i √ ) ac − b 2 ϕ y =0,
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 39: 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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