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

dφ

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).