Advanced book on Mathematics Olympiad

556 Real Analysis

∫π

0

f′′(x)(nsinx−sinnx)dx=

π

2

(na 1 −n^2 an).

Therefore,

0 ≤

∫π

0

(f (x)−f′′(x))(nsinx−sinnx)dx=

π

2

(na 1 −an−na 1 +n^2 an)

=

π

2

(n^2 − 1 )an.

This implies thatan ≥ 0, forn ≥ 2. We deduce thatan =0 forn ≥ 2, and so
f(x)=a 1 sinx, for anyx∈[ 0 ,π].
(S. Radulescu, M. R ̆ adulescu, ̆ Teoreme ̧si Probleme de Analiza Matematic ̆ ̆a(Theorems
and Problems in Mathematical Analysis), Editura Didactica ̧ ̆si Pedagogic ̆a, Bucharest,
1982).

498.This is an exercise in the product and chain rules. We compute

∂v

∂t

(x, t)=

∂

∂t

(

t−

(^12)
e−
x 42 t
u(xt−^1 ,−t−^1 )

)

=−

1

2

t−

(^32)
e−
x 42 t
v(x, t)+
x^2 t−
(^52)
4
e−
x 42 t
v(x, t)−xt−
(^52)
e−
x 42 t∂u
∂x
(xt−^1 ,−t−^1 )
+t−
(^52)
e−
x 42 t∂u
∂t
(xt−^1 ,−t−^1 ),
then
∂v
∂x
(x, t)=t−
(^12)
e−
x 42 t
(−

1

2

t−^1 x)u(xt−^1 ,−t−^1 )+t−

(^32)
e−
x 42 t∂u
∂x
(xt−^1 ,−t−^1 )
and
∂^2 v
∂x^2
(x, t)=

1

4

x^2 t−

(^52)
e−
x 42 t
v(x, t)−

1

2

t−

(^32)
e−
x 42 t
v(x, t)−

1

2

xt−

(^52)
e−
x 42 t∂u
∂x
(xt−^1 ,−t−^1 )

−

1

2

xt−

(^52)
e−
x 42 t∂u
∂x
(xt−^1 ,−t−^1 )+t−
(^52)
e−
x 42 t∂^2 u
∂x^2
(xt−^1 ,−t−^1 ).
Comparing the two formulas and using the fact that∂u∂t = ∂
(^2) u
∂t^2 , we obtain the desired
equality.
Remark.The equation
∂u
∂t

=

∂^2 u

∂x^2

is called the heat equation. It describes how heat spreads through a long, thin metal bar.