74 2. FOUR IMPORTANT LINEAR PDE Consequently formula (25) becomes (26) u(x, t) = 1 2 ∂t t2 B(x,t) g(y) (t2 |y x|2)1/2 dy + t2 2 B(x,t) h(y) (t2 |y x|2)1/2 dy. But t2 B(x,t) g(y) (t2 |y x|2)1/2 dy = t− B(0,1) g(x + tz) (1 |z|2)1/2 dz, and so ∂t t2 B(x,t) g(y) (t2 |y x|2)1/2 dy = B(0,1) g(x + tz) (1 |z|2)1/2 dz + t− B(0,1) Dg(x + tz) · z (1 |z|2)1/2 dz = t− B(x,t) g(y) (t2 |y x|2)1/2 dy + t− B(x,t) Dg(y) · (y x) (t2 |y x|2)1/2 dy. Hence we can rewrite (26) and obtain the relation (27) u(x, t) = 1 2 B(x,t) tg(y) + t2h(y) + tDg(y) · (y x) (t2 |y x|2)1/2 dy for x R2, t 0. This is Poisson’s formula for the solution of the initial- value problem (11) in two dimensions. The trick of solving the problem for n = 3 first and then dropping to n = 2 is the method of descent. d. Solution for odd n. In this subsection we solve the Euler–Poisson– Darboux PDE for odd n 3. We first record some technical facts. LEMMA 2 (Some useful identities). Let φ : R R be Ck+1. Then for k = 1, 2,... (i) d2 dr2 ( 1 r d dr )k−1 ( r2k−1φ(r) ) = ( 1 r d dr )k r2k dr (r) , (ii) ( 1 r d dr )k−1 ( r2k−1φ(r) ) = ∑k−1 j=0 βj krj+1 dj φ drj (r), where the constants βj k (j = 0,...,k 1) are independent of φ. Furthermore, (iii) β0 k = 1 · 3 · 5 · · · (2k 1).
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