As we mentioned in the Introduction, we will consider only special discontinuities in the coefficient
f
and we now make this precise. The function
f
:
Ω
→
R
will be piecewise Lipschitz continuous across a Lipschitz hypersurface according to the following definition.
Technically crucial although straightforward in what follows is the next remark.
Proof.
Since (20) is invariant by an increasing change of the dependent variable, it is not restrictive to assume that
u
and
v
are bounded by taking
tanh
u
,
tanh
v
instead of
u
,
v
.
Assume by contradiction that there exists a point
(
x
o
,
t
o
)
∈
Ω
∞
such that
u
x
o
,
t
o

v
x
o
,
t
o
=
A
>
0
. Fix any
α
∈
R
,
0
<
α
<
A
/
t
o
and let
x
^
,
t
^
be a maximum point of
(21)
Φ
x
,
t
≔
u
x
,
t

v
x
,
t

α
t
,
x
,
t
∈
Ω
¯
∞
.
Clearly the maximum of
Φ
exists since
Ω
is bounded,
u

v
is upper semicontinuous,
Φ
(
x
o
,
t
o
)
=
A

α
t
o
>
0
,
Φ
≤
0
on
∂
Ω
∞
, and
Φ
(
x
,
t
)
≤
0
for large
t
since
u
,
v
are bounded. Notice that
(
x
o
,
t
o
)
∉
∂
Ω
∞
and
max
Ω
¯
×
0
,
+
∞
Φ
≥
Φ
x
o
,
t
o
=
A

α
t
o
=
2
γ
>
0
; therefore every maximum point of
Φ
cannot be on
∂
Ω
∞
.
We now define
(22)
ω
ε
x
,
y
,
t
,
s
≔
u
x
,
t

v
y
,
s

γ
4
x

y
ε
+
η
4

γ
2
t

s
ε
2

α
t
,
where
η
=
η

=

η
+
is as in Remark 3, and consider
(
x
ε
,
y
ε
,
t
ε
,
s
ε
)
∈
Ω
∞
×
Ω
∞
such that
(23)
ω
ε
x
ε
,
y
ε
,
t
ε
,
s
ε
=
max
ω
ε
x
,
y
,
t
,
s
:
x
,
y
,
t
,
s
∈
Ω
¯
∞
×
Ω
¯
∞
.
As above the maximum of
ω
ε
exists and it is positive since
(24)
ω
ε
x
^
,
x
^
,
t
^
,
t
^
=
Φ
x
^
,
t
^

γ
4
η
4
≥
2
γ

γ
4
≥
γ
>
0
.
Moreover from
ω
ε
(
x
ε
,
y
ε
,
t
ε
,
s
ε
)
≥
ω
ε
(
x
^
,
x
^
,
t
^
,
t
^
)
>
0
and the boundedness of
u
,
v
we obtain that
(25)
x
ε

y
ε
ε
+
η
,
t
ε

s
ε
ε
are uniformly bounded in
ε
>
0
since we have assumed that
u
and
v
are bounded, and we get that
x
ε

y
ε
,
t
ε

s
ε
→
0
if
ε
→
0
. Therefore
(
x
ε
,
t
ε
)
,
(
y
ε
,
s
ε
)
stay in a bounded region uniformly in
ε
>
0
and
(26)
lim
ε
→
0
+
x
ε
,
y
ε
,
t
ε
,
s
ε
=
x
¯
,
x
¯
,
t
¯
,
t
¯
∈
Ω
∞
¯
×
Ω
∞
¯
.
Using the upper semicontinuity of
u
and the continuity of
v
we compute
(27)
Φ
x
¯
,
t
¯
=
u
x
¯
,
t
¯

v
x
¯
,
t
¯

α
t
¯
≥
limsup
ε
→
0
+
u
x
ε
,
t
ε

v
y
ε
,
s
ε

α
t
ε
≥
limsup
ε
→
0
+
ω
ε
x
ε
,
y
ε
,
t
ε
,
s
ε
≥
lim
ε
→
0
+
ω
ε
x
^
,
x
^
+
ε
η
,
t
^
,
t
^
=
lim
ε
→
0
+
u
x
^
,
t
^

v
x
^
+
ε
η
,
t
^

α
t
^
=
Φ
x
^
,
t
^
≥
Φ
x
¯
,
t
¯
.
From here we obtain that
(
x
¯
,
t
¯
)
is a maximum point for
Φ
, so
(
x
¯
,
t
¯
)
∈
Ω
∞
and moreover
(28)
lim
ε
→
0
+
u
x
ε
,
t
ε
=
u
x
¯
,
t
¯
.
Expanding again
(29)
ω
ε
x
¯
,
x
¯
+
ε
η
,
t
¯
,
t
¯
≤
ω
ε
x
ε
,
y
ε
,
t
ε
,
s
ε
.
By the continuity of
v
and (28) we improve previous estimates as
(30)
lim
ε
→
0
+
x
ε

y
ε
ε
+
η
=
0
,
lim
ε
→
0
+
t
ε

s
ε
ε
=
0
.
Hence for
ε
sufficiently small
(31)
x
ε

y
ε
+
ε
η
≤
c
ε
,
where
c
appears in Remark 3. In particular (31) shows that if
x
¯
∈
Γ
and
x
ε
∈
Ω

∪
Γ
, then
y
ε
∈
Ω

by Remark 3 which is something that we keep in mind for later. Up to now we never used the equation. Now consider two different cases.
Case 1. Suppose that there exists a sequence
ε
k
→
0
+
such that
y
ε
k
=
x
ε
k
+
ε
k
η
for all
k
. We omit from now on the subindex
k
. Since
ω
ε
x
,
y
ε
,
t
,
s
ε
≤
ω
ε
x
ε
,
y
ε
,
t
ε
,
s
ε
for any
(
x
,
t
)
∈
Ω
∞
we get
(32)
max
Ω
∞
u

φ
+
=
u
x
ε
,
t
ε

φ
+
x
ε
,
t
ε
,
where
(33)
φ
+
x
,
t
=
γ
4
x

y
ε
ε
+
η
4
+
γ
2
t

s
ε
ε
2
+
α
t
.
Similarly we see that
(34)
min
Ω
∞
v

φ

=
v
y
ε
,
s
ε

φ

y
ε
,
s
ε
,
with
(35)
φ

y
,
s
=

γ
4
x
ε

y
ε
+
η
4

γ
2
t
ε

s
ε
2
.
Since we have assumed that
x
ε
=
y
ε
+
ε
η
we have
D
φ
+
x
ε
,
t
ε
=
D
2
φ
+
x
ε
,
t
ε
=
0
and
D
φ

y
ε
,
s
ε
=
D
2
φ

y
ε
,
s
ε
=
0
. And so, using the fact that
u
is a viscosity subsolution and
v
a supersolution of (20) we obtain by computing derivatives
(36)
α
+
γ
ε
2
t
ε

s
ε
≤
0
≤
γ
ε
2
t
ε

s
ε
,
since here
F
∗
(
y
ε
,
s
ε
,
0,0
)
=
F
∗
(
x
ε
,
t
ε
,
0,0
)
=
0
. This contradicts
α
>
0
.
Case 2. We now suppose that
x
ε
≠
y
ε
+
ε
η
for
ε
small enough. We set
ξ
=
(
x
,
t
)
,
ζ
=
(
s
,
y
)
, and
(37)
φ
x
,
t
,
y
,
s
=
γ
4
x

y
ε
+
η
4
+
γ
2
t

s
ε
2
.
We now use classical matter. Observe that since
(
x
,
t
,
y
,
s
)
↦
u
(
x
,
t
)

v
(
y
,
s
)

α
t

φ
(
x
,
t
,
y
,
s
)
takes its maximum over
Ω
∞
¯
×
Ω
∞
¯
at
(
x
ε
,
t
ε
,
y
ε
,
s
ε
)
, we see that
(38)
D
ξ
φ
ξ
ε
,
ζ
ε
D
ζ
φ
ξ
ε
,
ζ
ε
,
A
∈
J
2
,
+
u
ξ
ε

v
ζ
ε

α
t
ε
,
the superjet of the indicated function, where
(39)
A
=
D
ξ
ξ
2
φ
ξ
ε
,
ζ
ε
D
ξ
ζ
2
φ
ξ
ε
,
ζ
ε
D
ζ
ξ
2
φ
ξ
ε
,
ζ
ε
D
ζ
ζ
2
φ
ξ
ε
,
ζ
ε
.
Now we apply the theorem on sums (see [1]) to find that for every
λ
>
0
there exist two matrices
X
,
Y
∈
S
n
such that
(40)
α
+
φ
t
ξ
ε
,
ζ
ε
,
D
x
φ
ξ
ε
,
ζ
ε
,
X
∈
P
¯
2
,
+
u
x
ε
,
t
ε

φ
s
ξ
ε
,
ζ
ε
,

D
y
φ
ξ
ε
,
ζ
ε
,
Y
∈
P
¯
2
,

v
y
ε
,
s
ε
the closed parabolic super and subjets, and
(41)

1
λ
+
A
0
I
≤
X
O
O

Y
≤
A
0
+
λ
A
0
2
,
where
(42)
A
0
=
D
x
x
2
φ
x
ε
,
t
ε
,
y
ε
,
s
ε
D
x
y
2
φ
x
ε
,
t
ε
,
y
ε
,
s
ε
D
y
x
2
φ
x
ε
,
t
ε
,
y
ε
,
s
ε
D
y
y
2
φ
x
ε
,
t
ε
,
y
ε
,
s
ε
.
Using again the fact that
u
and
v
are, respectively, sub and supersolution of (20) and that
D
x
φ
(
x
ε
,
t
ε
,
y
ε
,
s
ε
)
=

D
y
φ
(
x
ε
,
t
ε
,
y
ε
,
s
ε
)
≠
0
, we have by the equation of
u
(43)
α
+
γ
t
ε

s
ε
+
f
∗
x
ε
D
x
φ
x
ε
,
t
ε
,
y
ε
,
s
ε
+
F
D
x
φ
x
ε
,
t
ε
,
y
ε
,
s
ε
,
X
≤
0
and by the equation of
v
(44)
γ
t
ε

s
ε
+
f
∗
y
ε

D
y
φ
x
ε
,
t
ε
,
y
ε
,
s
ε
+
F

D
y
φ
x
ε
,
t
ε
,
y
ε
,
s
ε
,
Y
≥
0
.
Adding these two inequalities we obtain
(45)
α
≤
f
∗
y
ε

f
∗
x
ε
D
x
φ
x
ε
,
t
ε
,
y
ε
,
s
ε
+
F

D
y
φ
,
Y

F
D
x
φ
,
X
.
Since
(46)
D
x
x
2
φ
x
ε
,
t
ε
,
y
ε
,
s
ε
=
D
y
y
2
φ
x
ε
,
t
ε
,
y
ε
,
s
ε
=

D
x
y
2
φ
x
ε
,
t
ε
,
y
ε
,
s
ε
=

D
y
x
2
φ
x
ε
,
t
ε
,
y
ε
,
s
ε
multiplying the second matrix inequality in (41) twice by vectors of the form
(
ξ
,
ξ
)
∈
R
n
+
1
×
R
n
+
1
implies
X
≤
Y
. Therefore since
F
is elliptic,
(47)
F

D
y
φ
x
ε
,
t
ε
,
y
ε
,
s
ε
,
Y
=
F
D
x
φ
x
ε
,
t
ε
,
y
ε
,
s
ε
,
Y
≤
F
D
x
φ
x
ε
,
t
ε
,
y
ε
,
s
ε
,
X
and the inequality in (45) simply becomes
(48)
α
≤
f
∗
y
ε

f
∗
x
ε
D
x
φ
x
ε
,
t
ε
=
γ
ε
f
∗
y
ε

f
∗
x
ε
x
ε

y
ε
ε
+
η
3
.
Now we analyze the right hand side of (48) as
ε
→
0
. If
x
¯
∈
Γ
and
x
ε
∈
Ω

∪
Γ
then (31) implies that also
y
ε
∈
Ω

, for all
ε
sufficiently small as we already observed. Inequality (48) and Lipschitz continuity of
f
in the region
Ω

therefore yield
(49)
α
≤
γ
L
f
x
ε

y
ε
ε
x
ε

y
ε
ε
+
η
3
,
where
L
f
is a Lipschitz constant for
f
in
Ω

. Notice that the same estimate holds true if
x
¯
∉
Γ
as
f
is Lipschitz continuous in a neighborhood of
x
¯
. Finally, letting
ε
→
0
+
and using (30) we get a contradiction in these cases since the right hand side of (49) tends to
0
as
ε
→
0
+
.
We are left with the case
x
¯
∈
Γ
and
x
ε
∈
Ω
+
along a subsequence. We have two further subcases: either for
ε
small
y
ε
∈
Ω
+
∪
Γ
and we proceed similarly as above by the Lipschitz continuity of
f
in the region
Ω
+
or for
y
ε
∈
Ω

on a subsequence. In the latter situation we observe that
(50)
lim
ε
→
0
+
f
y
ε

f
x
ε
=
f
∗
x
¯

f
∗
x
¯
<
0
and again we obtain a contradiction in (48) for
ε
small ending the proof.
An immediate consequence of the previous result is the uniqueness of solutions of (1).