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By V.G. Shervatov

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The method of images also solves the Neumann boundary value problem in a half-space using an even mirror reflection in x1 = 0. It shows that for the Neumann condition, the reflection coefficient is equal to 1. 4. i. 7) 81uIx1=O = 0. ii. 8), then the even extension of u to R1+d is a smooth even solution of u = 0. 1. 8) t/n > 0, 2n+1 8x1 I xi=0 = 0. 1. 4. 2. Prove uniqueness of solutions by the energy method. Hint. Use the local energy identity. 3. Verify the assertion concerning the reflection coefficient by following the examples above.

Simple Examples of Propagation 22 There are conservations of all orders. 4) ue(0, x) = y(x) eixl/E 'y E n Hs(Rd) . 3). When e is small, the initial value is an envelope or profile y multiplied by a rapidly oscillating exponential. 3) with g = 0 and u(0, ) = F(y(x) eixl/e) = '(C - ei/e) yields u = u+ + u_ with 21 u' (t, x) 'Y( - ei/e) (2x)d/2 ei(xC 1 tICI) < Analyze u+. The other term is analogous. For ease of reading, the subscript plus is omitted. Introduce ei/e, = ei + e( so , and e f e-itlel+ECI/E d(.

Haar's inequal- ity applied to u - v implies that u - v = 0. 1. Simple Examples of Propagation 10 The proof also yields finite speed of propagation of signals. Define Amin(t, x) and A.. (t, x) to the smallest and largest eigenvalues of A(t, x). Then the functions A are uniformly Lipschitzean on [0, T] x R. 9) shows that the diagonal elements cj (t, x) of the right-hand side are the eigenvalues of A. Their expression by the left-hand side shows that their partial derivatives of first order (in fact any order) are bounded on [0, T] x R.

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Hyperbolic Functions by V.G. Shervatov


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