Let
U
⊂
R
n
{\displaystyle U\subset \mathbb {R} ^{n}}
be a bounded open set with smooth boundary
∂
U
{\displaystyle \partial U}
. Let
x
↦
P
(
x
)
:
U
¯
→
R
n
×
n
{\displaystyle x\mapsto P(x):{\bar {U}}\to \mathbb {R} ^{n\times n}}
be a smooth family of symmetric
n
×
n
{\displaystyle n\times n}
real matrices that are uniformly positive definite. Let
c
≥
0
{\displaystyle c\geq 0}
and
f
(
x
)
{\displaystyle f(x)}
be smooth functions on
U
¯
{\displaystyle {\bar {U}}}
. Define the functional
I
[
u
]
=
∫
U
(
1
2
⟨
∇
u
,
P
(
x
)
∇
u
⟩
+
1
2
c
(
x
)
u
2
−
f
(
x
)
u
)
d
x
{\displaystyle I[u]=\int _{U}\left({\frac {1}{2}}\langle \nabla u,P(x)\nabla u\rangle +{\frac {1}{2}}c(x)u^{2}-f(x)u\right)\,dx}
where
⟨
⋅
,
⋅
⟩
{\displaystyle \langle \cdot ,\cdot \rangle }
is the scalar product on
R
n
{\displaystyle \mathbb {R} ^{n}}
. Suppose that
u
∈
C
2
(
U
)
∩
C
0
(
U
¯
)
{\displaystyle u\in C^{2}(U)\cap C^{0}({\bar {U}})}
is a minimizer of this functional subject to the Dirichlet condition
u
=
g
{\displaystyle u=g}
on
∂
U
{\displaystyle \partial U}
, with
g
{\displaystyle g}
continuous.
a) Show that
u
{\displaystyle u}
satisfies the variational equation
∫
U
⟨
∇
u
,
P
(
x
)
∇
v
⟩
+
c
(
x
)
u
v
=
∫
U
f
v
{\displaystyle \int _{U}\langle \nabla u,P(x)\nabla v\rangle +c(x)uv=\int _{U}fv}
for any
v
∈
V
=
{
v
∈
Lip
(
U
)
:
v
|
∂
U
=
0
}
{\displaystyle v\in V=\{v\in \operatorname {Lip} (U):\left.v\right|_{\partial U}=0\}}
.
b) What is the PDE satisfied by
u
{\displaystyle u}
?
c) Suppose
f
,
g
≥
0
{\displaystyle f,g\geq 0}
. Show that
v
=
min
(
u
,
0
)
{\displaystyle v=\min(u,0)}
is an admissible test function, and use this to conclude that
u
≥
0
{\displaystyle u\geq 0}
. (Hint: You may use the fact
∇
v
=
χ
{
u
<
0
}
∇
u
{\displaystyle \nabla v=\chi _{\{u<0\}}\nabla u}
a.e.)
d) Show that there can be only one minimizer of
I
[
u
]
{\displaystyle I[u]}
or, equivalently, only one solution
u
∈
C
2
(
U
)
∩
C
0
(
U
¯
)
{\displaystyle u\in C^{2}(U)\cap C^{0}({\bar {U}})}
of the corresponding PDE.
For
v
∈
V
{\displaystyle v\in V}
, define
ϕ
(
t
)
=
I
[
u
+
t
v
]
{\displaystyle \phi (t)=I[u+tv]}
. Then since
u
{\displaystyle u}
minimizes the functional,
ϕ
′
(
0
)
=
0
{\displaystyle \phi '(0)=0}
. We can calculate
ϕ
′
{\displaystyle \phi '}
(by exploiting the symmetry of
P
{\displaystyle P}
):
ϕ
′
(
t
)
=
d
d
t
∫
U
1
2
⟨
∇
(
u
+
t
v
)
,
P
(
x
)
∇
(
u
+
t
v
)
⟩
+
1
2
c
(
x
)
(
u
+
t
v
)
2
−
f
(
x
)
(
u
+
t
v
)
d
x
=
d
d
t
∫
U
1
2
⟨
∇
u
,
P
∇
u
⟩
+
t
⟨
∇
u
,
P
∇
v
⟩
+
1
2
t
2
⟨
∇
v
,
P
∇
v
⟩
+
1
2
c
(
x
)
(
u
+
2
t
u
v
+
t
2
v
2
)
−
f
(
x
)
(
u
+
t
v
)
d
x
=
∫
U
⟨
∇
u
,
P
(
x
)
∇
v
⟩
+
t
⟨
∇
v
,
P
(
x
)
∇
v
⟩
+
c
(
x
)
(
u
v
+
t
v
2
)
−
f
(
x
)
v
d
x
{\displaystyle {\begin{aligned}\phi '(t)=&{\frac {d}{dt}}\int _{U}{\frac {1}{2}}\langle \nabla (u+tv),P(x)\nabla (u+tv)\rangle +{\frac {1}{2}}c(x)(u+tv)^{2}-f(x)(u+tv)\,dx\\=&{\frac {d}{dt}}\int _{U}{\frac {1}{2}}\langle \nabla u,P\nabla u\rangle +t\langle \nabla u,P\nabla v\rangle +{\frac {1}{2}}t^{2}\langle \nabla v,P\nabla v\rangle +{\frac {1}{2}}c(x)(u+2tuv+t^{2}v^{2})-f(x)(u+tv)\,dx\\=&\int _{U}\langle \nabla u,P(x)\nabla v\rangle +t\langle \nabla v,P(x)\nabla v\rangle +c(x)(uv+tv^{2})-f(x)v\,dx\end{aligned}}}
And so
ϕ
′
(
0
)
=
∫
U
⟨
∇
u
,
P
(
x
)
∇
v
⟩
+
c
(
x
)
u
v
−
f
(
x
)
v
d
x
=
0
{\displaystyle \phi '(0)=\int _{U}\langle \nabla u,P(x)\nabla v\rangle +c(x)uv-f(x)v\,dx=0}
which proves the result.
We have
∫
U
f
v
=
∫
U
⟨
∇
u
,
P
(
x
)
∇
v
⟩
+
c
(
x
)
u
v
=
∫
U
P
(
x
)
∇
u
⋅
∇
v
+
c
(
x
)
u
v
=
−
∫
U
div
(
P
(
x
)
∇
u
)
v
+
∫
∂
U
P
(
x
)
∇
u
v
+
∫
U
c
(
x
)
u
v
{\displaystyle {\begin{aligned}\int _{U}fv=&\int _{U}\langle \nabla u,P(x)\nabla v\rangle +c(x)uv\\=&\int _{U}P(x)\nabla u\cdot \nabla v+c(x)uv\\=&-\int _{U}\operatorname {div} (P(x)\nabla u)v+\int _{\partial U}P(x)\nabla uv+\int _{U}c(x)uv\end{aligned}}}
The boundary terms vanish since
v
∈
V
{\displaystyle v\in V}
and we've obtained a weak form of the PDE. Thus,
u
{\displaystyle u}
is a solution to the following PDE:
{
−
div
(
P
(
x
)
∇
u
)
+
c
(
x
)
u
=
f
in
U
u
=
g
on
∂
U
.
{\displaystyle \left\{{\begin{array}{rl}-\operatorname {div} (P(x)\nabla u)+c(x)u=f&{\text{ in }}U\\u=g&{\text{ on }}\partial U.\end{array}}\right.}
First we need to show that
v
=
min
(
u
,
0
)
∈
V
{\displaystyle v=\min(u,0)\in V}
. Firstly, on
∂
U
{\displaystyle \partial U}
,
v
=
min
(
g
,
0
)
=
0
{\displaystyle v=\min(g,0)=0}
since we've assumed
g
≥
0
{\displaystyle g\geq 0}
. Secondly,
u
{\displaystyle u}
, hence
v
{\displaystyle v}
, must be Lipschitz continuous since
u
∈
C
2
(
U
)
{\displaystyle u\in C^{2}(U)}
and
U
{\displaystyle U}
is a bounded domain in
R
n
{\displaystyle \mathbb {R} ^{n}}
(i.e.
∇
u
∈
C
1
(
U
)
{\displaystyle \nabla u\in C^{1}(U)}
) and so
|
∇
u
|
{\displaystyle |\nabla u|}
must achieve a (finite) maximum in
U
{\displaystyle U}
, hence the derivative is bounded, hence
u
{\displaystyle u}
is Lipschitz. Therefore,
v
∈
V
{\displaystyle v\in V}
.
This gives
∫
U
f
v
=
∫
U
⟨
∇
u
,
P
(
x
)
∇
v
⟩
+
c
(
x
)
u
v
=
∫
{
u
<
0
}
⟨
∇
u
,
P
(
x
)
∇
u
⟩
+
c
(
x
)
u
2
{\displaystyle \int _{U}fv=\int _{U}\langle \nabla u,P(x)\nabla v\rangle +c(x)uv=\int _{\{u<0\}}\langle \nabla u,P(x)\nabla u\rangle +c(x)u^{2}}
.
But notice that since
P
(
x
)
{\displaystyle P(x)}
is uniformly positive definite, then
∫
{
u
<
0
}
⟨
∇
u
,
P
(
x
)
∇
u
⟩
=
∫
{
u
<
0
}
(
∇
u
)
T
P
(
x
)
(
∇
u
)
≥
0.
{\displaystyle \int _{\{u<0\}}\langle \nabla u,P(x)\nabla u\rangle =\int _{\{u<0\}}(\nabla u)^{T}P(x)(\nabla u)\geq 0.}
Therefore, we have
0
≤
∫
{
u
<
0
}
(
∇
u
)
T
P
(
x
)
(
∇
u
)
+
c
u
2
=
∫
{
u
<
0
}
f
u
<
0
{\displaystyle 0\leq \int _{\{u<0\}}(\nabla u)^{T}P(x)(\nabla u)+cu^{2}=\int _{\{u<0\}}fu<0}
a contradiction, unless
m
(
{
u
<
0
}
)
=
0
{\displaystyle m(\{u<0\})=0}
, i.e.
u
≥
0
{\displaystyle u\geq 0}
a.e.
Suppose
u
1
,
u
2
{\displaystyle u_{1},u_{2}}
are two distinct such solution. Let
w
=
u
1
−
u
2
{\displaystyle w=u_{1}-u_{2}}
. Then
w
{\displaystyle w}
is Lipschitz (since
u
1
,
u
2
{\displaystyle u_{1},u_{2}}
must both be) and
w
=
0
{\displaystyle w=0}
on
∂
U
{\displaystyle \partial U}
. Therefore, the variational equation gives
∫
U
⟨
∇
w
,
P
(
x
)
∇
w
⟩
+
c
(
x
)
w
2
=
0
{\displaystyle \int _{U}\langle \nabla w,P(x)\nabla w\rangle +c(x)w^{2}=0}
. Since
P
(
x
)
{\displaystyle P(x)}
is positive definite, this gives
∫
U
c
w
2
≤
0
{\displaystyle \int _{U}cw^{2}\leq 0}
, a contradiction unless
w
=
0
{\displaystyle w=0}
a.e.