Partial Differential Equations in Action: Complements and Exercises (UNITEXT)

Partial Differential Equations in Action: Complements and Exercises (UNITEXT)

Language: English

Pages: 431

ISBN: 331915415X

Format: PDF / Kindle (mobi) / ePub


This textbook presents problems and exercises at various levels of difficulty in the following areas: Classical Methods in PDEs (diffusion, waves, transport, potential equations); Basic Functional Analysis and Distribution Theory; Variational Formulation of Elliptic Problems; and Weak Formulation for Parabolic Problems and for the Wave Equation. Thanks to the broad variety of exercises with complete solutions, it can be used in all basic and advanced PDE courses.

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constant. Recalling that L .u t / .x; / D U .x; / u .x; 0/ D U .x; / ; we deduce that U solves7 U Uxx D 0; U .0; / D 1; 7 Set D D 1 for simplicity. 0

C1 X 0 ur .r;  / D ˛ 0 .r/ C ˛ 0n .r/ cos n C ˇ 0n .r/ sin n ; nD1 urr .r;  / D ˛ 000 .r/ C C1 X ˛ 00n .r/ cos n C ˇ 00n .r/ sin n ; nD1 u .r;  / D C1 X n2 Œ˛ n .r/ cos n C ˇ n .r/ sin n : nD1 Substituting into equation (2.20) we get ˛ 000 .r/ C1 ²Ä ˛ 0 .r/ X C C 0 r nD1 1 n2 ˛ 00n .r/ C ˛ 0n .r/ ˛ n .r/ cos n C r r2 Ä ³ 1 0 n2 00 C ˇ n .r/ C ˇ n .r/ ˇ .r/ sin n D 0 r r2 n from which we obtain the family of problems, for n 0: 8 1 ˆ <˛ 000 .r/ C ˛ 00 .r/ D 0 r a A ˆ :˛ 0

Poisson kernel is harmonic, so is u. Now for y > 0, 1 Z R .x 1 y ds D s/2 C y 2 Thus ju .x; y/j Ä 1 Z R Ä Â Ã 1 y x arctan dx D x2 C y2 y Z R .x C1 1 D 1: (2.28) y jg .x/j ds Ä sup jgj R s/2 C y 2 and therefore u is bounded. Let now .x; y/ ! .x0 ; 0/; given " > 0, we take jg .s/ for js g .x0 /j < " x0 j < ı " . Then ju .x; y/ 1 Z y jg .s/ g .x0 /j ds x/2 C y 2 R .s Z Z 1 1 ds C ds : D ¹js x0 j<ı " º ¹js x0 j ı " º „ ƒ‚ … „ ƒ‚ … g .x0 /j D I1 I2 From (2.28) we infer I1 < " .

for t D log 3, any s. Hence we expect to encounter some problems near p Á p ÁÁ p X s; log 3 ; Y s; log 3 D .0; 3=3/: At any rate, as e t D y, the first equation in (3.20) implies à  1 3y D x; s 2 2y 3.2 Solved Problems 183 i.e. 2xy , t D log y: 3y 2 1 Substituting into the third equation of (3.20) we find  à 3y 2 C 1 2xy 3y 1 C Dx 2 : u.x; y/ D 2 3y 1 2 2y 3y 1 sD As expected, the solution is defined only when p 3 3 (recall that the initial datum is given on y D 1). y> Problem 3.2.14

h2 = D 1? b) Find the explicit expression of p . c) Show that in the limit the particle reaches L in finite time with probability 1. Observe, in particular, that for t > 0 Z L p .x; t / dx < 1: 1 a b [18, Chap. 2, Sect. 4]. The probability that the particle will reach x at time t . 28 1 Diffusion Solution. a) This being a symmetric walk, the limit density p solves 1 pxx D 0 2 pt for x < L and t > 0; as the particle starts from the origin, we have p .x; 0/ D ı.x/ on . 1; L/. To understand

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