54 2. FOUR IMPORTANT LINEAR PDE
Also, let us introduce the useful function
(21) ψ := −
n
2
log(−4πs) +
|y|2
4s
+ n log r
and observe ψ = 0 on ∂E(r) − {(0, 0)}, since Φ(y, −s) =
r−n
on ∂E(r). We
utilize (21) to write
B =
1
rn+1
E(r)
4us
n
i=1
yiψyi dyds
= −
1
rn+1
E(r)
4nusψ + 4
n
i=1
usyi yiψ dyds;
there is no boundary term since ψ = 0 on ∂E(r) − {(0, 0)}. Integrating by
parts with respect to s, we discover
B =
1
rn+1
E(r)
−4nusψ + 4
n
i=1
uyi yiψs dyds
=
1
rn+1
E(r)
−4nusψ + 4
n
i=1
uyi yi −
n
2s
−
|y|2
4s2
dyds
=
1
rn+1
E(r)
−4nusψ −
2n
s
n
i=1
uyi yi dyds − A.
Consequently, since u solves the heat equation,
φ (r) = A + B
=
1
rn+1
E(r)
−4nΔuψ −
2n
s
n
i=1
uyi yi dyds
=
n
i=1
1
rn+1
E(r)
4nuyi ψyi −
2n
s
uyi yi dyds
= 0, according to (21).
Thus φ is constant, and therefore
φ(r) = lim
t→0
φ(t) = u(0, 0)
(
lim
t→0
1
tn
E(t)
|y|2
s2
dyds
)
= 4u(0, 0),
as
1
tn
E(t)
|y|2
s2
dyds =
E(1)
|y|2
s2
dyds = 4.
We omit the details of this last computation.
