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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