Chapter 6 Partial Differential Equations

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14 CHAPTER 6. PARTIAL DIFFERENTIAL EQUATIONS Theorem 6.2.8. Suppose that S is a C 1 hypersurface and a j , b and φ are C 1 real valued functions. Suppose that the vector field V =(a 1 (x, φ(x)), ··· ,a n (x, φ(x))) is not tangent to S at any x ∈ S (this is the noncharacteristic condition). Then there is a unique solution u ∈ C 1 of n∑ a j (x, u)∂ j u = b(x, u) j=1 in a neighborhood Ω of S such that u = φ on S. The proof of this result is constructive. Proof. The uniqueness follows from the discussion above which states that u must be the union of integral curves of A in Ω passing through S ∗ . Now any hypersurface S can be covered by by open sets on which it admits a parametric representation x = g(s), with s ∈ R n−1 . If we solve the problem on each such open set, by uniqueness the local solutions must agree on the overlap of the open sets and hence patch together to give a solution for all of S. It therefore suffices to assume that S is given parametrically by x = g(s) with s ∈ R n−1 . For each s ∈ R n−1 , consider the initial value problem ∂x j ∂t (s, t) =a j(x, y), j=1, ··· ,n, ∂y (s, t) =b(x, y), (6.2.8) ∂t x j (s, 0) = g j (s), y(s, 0) = ϕ(g(s)). Here s is a parameter vector, so we have a system of ODEs in t. By the fundamental existence uniqueness theorem for ODEs, there exists a unique solution (x, y) defined for small t, and (x, y) isaC 1 function of s and t jointly. By the non-characteristic condition (6.2.7) and the inverse mapping theorem, the mapping (s, t) ↦→ (x, t) is invertible on some neighborhood Ω of S, yielding s and t as C 1 functions of x on Ω such that t(x) =0andg(s(x)) = x when x ∈ S. Now set u(x) =y(s(x),t(x)). We have u = ϕ on S, and we claim that u satisfies (6.2.5). By the chain rule and the
6.2. LINEAR AND QUASILINEAR EQUATIONS OF FIRST ORDER 15 fact that ∂s k ∂t = 0, since s k and t are functionally independent, and ∂t ∂t ( n∑ ∂u n∑ n−1 ) ∑ ∂u ∂s k a j = a j + ∂u ∂t ∂x j=1 j ∂s j=1 k ∂x j ∂t ∂x j k=1 This completes the proof. = = = ∑n−1 k=1 ∑n−1 k=1 ∑n−1 k=1 ∂u ∂s k ∂u ∂s k n∑ j=1 n∑ j=1 ∂u ∂s k ∂s k =0+ ∂u ∂t = b. ∂s k a j + ∂u ∂x j ∂t ∂s k ∂x j ∂x j ∂t + ∂u ∂t ∂t + ∂u ∂t ∂t ∂t n∑ j=1 a j ∂t ∂x j n∑ j=1 ∂x j ∂t ∂t ∂x j = 1 we have Example 6.2.9. In R 3 solve the linear initial value problem Lu = x 1 ∂ 1 u+2x 2 ∂ 2 +∂ 3 u =3u with u = ϕ(x 1 ,x 2 ) on the plane x 3 =0. Solvability: We could observe that the solvability condition is satisfied by considering The vector field A(x) =(x 1 , 2x 2 , 1) and the characteristic manifold Char x (L) ={ξ =(ξ 1 ,ξ 2 ,ξ 3 ) ≠0:A(x) · ξ =0}. We note that the initial surface S in this case is x 3 = 0 with constant normal vector ν = (0, 0, 1). In the present case we see that A(x) · ν =1≠0 so the surface is non-characteristic. We could also note that with s =(s 1 ,s 2 ) ∈ R 3−1 and g(s) =(g 1 (s),g 2 (s),g 3 (s))=(s 1 ,s 2 , 0) paramterizing the surface S we have for (6.2.7) ⎛ 1 0 s 1 ⎞ det ⎝0 1 2s 2 ⎠ = ϕ(s 1 ,s 2 ). 0 0 ϕ(s 1 ,s 2 )
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 13: 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

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