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.