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: mingfeng.zhao@uconn.edu