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SOLUTION OF QUIZ 4 MINGFENG ZHAO March 14, 2013 1. [5 Points] Let f : R2 → R be differentiable. Prove that f (x, y) − f (a, b) = (x − a)fx (c, y) + (y − b)fy (a, d) for some numbers c between a and x and d between b and y. Proof. For any x, y, a, b ∈ R and fix them, by the mean value theorem, there exists some c ∈ (a, x) and d ∈ (b, y) such that f (x, y) − f (a, y) = fx (c, y)(x − a), and f (a, y) − f (a, b) = fy (a, d)(y − b). Hence we get f (x, y) − f (a, b) = f (x, y) − f (a, y) + f (a, y) − f (a, b) = (x − a)fx (c, y) + (y − b)fy (a, d). 2. [5 Points] Let u(x1 , · · · , xn , t) : Rn+1 → R be C 2 . We say that u satisfies the n-dimensional wave equation of speed c if ∂2u = c2 ∂t2 ∂2u ∂2u + ··· + 2 ∂x1 ∂x2n , where t is the time variable, x1 , · · · , xn are the spatial variables, and the constant c is the propagation speed of the wave. This equation appears naturally in physics. For example, it describes water waves, sound waves, and light waves. 1 2 MINGFENG ZHAO While it is not easy to find solutions to the wave equation in general, the solution to the onedimensional wave equation has a simpler form. For any C 2 functions f, g : R → R, we define u(x, t) = f (x − ct) + g(x + ct). Verify that u satisfies the one-dimensional wave equation of speed c, i.e., ∂2u ∂2u = c2 2 . 2 ∂t ∂x Proof. By the chain rule, we know that ∂u (x, t) ∂t = f 0 (x − ct)(−c) + g 0 (x + ct)c = −cf 0 (x − ct) + cg 0 (x + ct) ∂2u (x, t) ∂t2 = −cf 00 (x − ct)(−c) + cg 00 (x + ct)c = c2 [f 00 (x − ct) + g 00 (x + tc)] ∂u (x, t) ∂x ∂2u (x, t) ∂x2 = f 0 (x − ct) + g 0 (x + ct) = f 00 (x − ct) + g 00 (x − ct). Hence we know that 2 ∂2u 2∂ u (x, t) − c (x, t) ∂t2 ∂x2 = c2 [f 00 (x − ct) + g 00 (x + tc)] − c2 [f 00 (x − ct) + g 00 (x + tc)] = 0. Therefore, u satisfies the one-dimensional wave equation of speed c. Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT 06269-3009 E-mail address: [email protected]