Let
u
(
x
,
y
)
{\displaystyle u(x,y)}
be a harmonic function on
R
2
{\displaystyle \mathbb {R} ^{2}}
and suppose that
∫
∫
R
2
|
∇
u
|
2
(
x
,
y
)
d
x
d
y
<
∞
.
{\displaystyle \int \int _{\mathbb {R} ^{2}}|\nabla u|^{2}(x,y)\,dx\,dy<\infty .}
Show that
u
{\displaystyle u}
is a constant function.
Let
C
=
∫
∫
R
2
|
∇
u
|
2
(
x
,
y
)
d
x
d
y
<
∞
.
{\displaystyle C=\int \int _{\mathbb {R} ^{2}}|\nabla u|^{2}(x,y)\,dx\,dy<\infty .}
If
u
{\displaystyle u}
is harmonic (i.e.
Δ
u
=
0
{\displaystyle \Delta u=0}
) then so must
v
=
∇
u
{\displaystyle v=\nabla u}
(surely,
Δ
∇
u
=
0
{\displaystyle \Delta \nabla u=0}
). Then since the absolute value as an operator is convex, we have that
|
v
|
=
|
∇
u
|
{\displaystyle |v|=|\nabla u|}
is a subharmonic function on
R
2
{\displaystyle \mathbb {R} ^{2}}
.
Then by the mean value property of subharmonic functions, for any
x
0
∈
R
2
{\displaystyle x_{0}\in \mathbb {R} ^{2}}
we have
|
v
(
x
0
)
|
≤
1
π
r
2
∫
B
(
x
0
,
r
)
|
v
|
d
x
d
y
≤
1
π
r
2
(
∫
B
(
x
0
,
r
)
1
d
x
d
y
)
1
/
2
(
∫
B
(
x
0
,
r
)
|
v
|
2
d
x
d
y
)
1
/
2
≤
C
π
r
2
{\displaystyle {\begin{array}{lll}|v(x_{0})|&\leq &{\frac {1}{\pi r^{2}}}\int _{B(x_{0},r)}|v|\,dx\,dy\\&\leq &{\frac {1}{\pi r^{2}}}\left(\int _{B(x_{0},r)}1\,dx\,dy\right)^{1/2}\left(\int _{B(x_{0},r)}|v|^{2}\,dx\,dy\right)^{1/2}\\&\leq &{\frac {C}{\sqrt {\pi r^{2}}}}\end{array}}}
where the second inequality is due to Cauchy-Schwarz (Hölder) inequality.
This estimate hold for all
r
>
0
{\displaystyle r>0}
. Therefore if we send
r
→
∞
{\displaystyle r\to \infty }
we see that for all
x
0
∈
R
2
,
{\displaystyle x_{0}\in \mathbb {R} ^{2},}
|
v
(
x
0
)
|
=
|
∇
u
(
x
0
)
|
=
0
{\displaystyle |v(x_{0})|=|\nabla u(x_{0})|=0}
which gives us that
u
{\displaystyle u}
is constant.
Let
u
(
x
,
t
)
{\displaystyle u(x,t)}
be a piecewise smooth weak solution of the conservation law
u
t
+
f
(
u
)
x
=
0
,
−
∞
<
x
<
∞
,
t
>
0.
{\displaystyle u_{t}+f(u)_{x}=0,\quad -\infty <x<\infty ,t>0.}
a) Derive the Rankine-Hugoniot conditions at a discontinuity of the solution.
b)Find a piecewise smooth solution to the IVP
u
t
+
(
u
2
+
u
)
x
=
0
,
−
∞
<
x
<
∞
,
t
>
0
{\displaystyle u_{t}+(u^{2}+u)_{x}=0,\quad -\infty <x<\infty ,t>0}
u
(
x
,
0
)
=
{
1
,
x
<
0
−
2
x
>
0.
{\displaystyle u(x,0)=\left\{{\begin{array}{ll}1,&x<0\\-2&x>0.\end{array}}\right.}
When we solve the PDE by methods of characteristics, the characteristic curves can cross, causing a shock, or discontinuity. The task at hand, is to find the curve of discontinuity, call it
C
{\displaystyle C}
.
Multiply the PDE by
v
{\displaystyle v}
, a smooth test function with compact support in
R
×
(
0
,
∞
)
{\displaystyle \mathbb {R} \times (0,\infty )}
. Then by an integration by parts:
0
=
∫
0
∞
∫
−
∞
∞
(
u
t
+
f
(
u
)
x
)
v
d
x
d
t
=
−
∫
0
∞
∫
−
∞
∞
u
v
t
d
x
d
t
+
∫
−
∞
∞
v
u
d
x
|
t
=
0
t
=
∞
−
∫
0
∞
∫
−
∞
∞
f
(
u
)
v
x
d
x
d
t
+
∫
0
∞
f
(
u
)
v
d
t
|
x
=
−
∞
x
=
∞
=
−
∫
0
∞
∫
−
∞
∞
u
v
t
d
x
d
t
−
∫
−
∞
∞
v
u
(
x
,
0
)
d
x
−
∫
0
∞
∫
−
∞
∞
f
(
u
)
v
x
d
x
d
t
.
{\displaystyle {\begin{aligned}0=&\int _{0}^{\infty }\int _{-\infty }^{\infty }(u_{t}+f(u)_{x})v\,dx\,dt\\=&-\int _{0}^{\infty }\int _{-\infty }^{\infty }uv_{t}\,dx\,dt+\left.\int _{-\infty }^{\infty }vu\,dx\right|_{t=0}^{t=\infty }-\int _{0}^{\infty }\int _{-\infty }^{\infty }f(u)v_{x}\,dx\,dt+\left.\int _{0}^{\infty }f(u)v\,dt\right|_{x=-\infty }^{x=\infty }\\=&-\int _{0}^{\infty }\int _{-\infty }^{\infty }uv_{t}\,dx\,dt-\int _{-\infty }^{\infty }vu(x,0)\,dx-\int _{0}^{\infty }\int _{-\infty }^{\infty }f(u)v_{x}\,dx\,dt.\end{aligned}}}
Let
V
l
{\displaystyle V_{l}}
denote the open region in
R
×
(
0
,
∞
)
{\displaystyle \mathbb {R} \times (0,\infty )}
to the left of
C
{\displaystyle C}
and similarly
V
r
{\displaystyle V_{r}}
denotes the region to the right of
C
{\displaystyle C}
. If the support of
v
{\displaystyle v}
lies entirely in either of these two regions, then all of the above boundary terms vanish and we get
0
=
∫
0
∞
∫
−
∞
∞
(
u
t
+
f
(
u
)
x
)
v
d
x
d
t
=
∫
0
∞
∫
−
∞
∞
u
v
t
+
f
(
u
)
v
x
d
x
d
t
.
{\displaystyle 0=\int _{0}^{\infty }\int _{-\infty }^{\infty }(u_{t}+f(u)_{x})v\,dx\,dt=\int _{0}^{\infty }\int _{-\infty }^{\infty }uv_{t}+f(u)v_{x}\,dx\,dt.}
Now suppose the support of
v
{\displaystyle v}
intersects the discontinuity
C
{\displaystyle C}
.
We can calculate
σ
=
F
(
u
l
)
−
F
(
u
r
)
u
l
−
u
r
=
1
2
+
1
−
(
4
−
2
)
1
+
2
=
0
{\displaystyle \sigma ={\frac {F(u_{l})-F(u_{r})}{u_{l}-u_{r}}}={\frac {1^{2}+1-(4-2)}{1+2}}=0}
. Therefore, the shock wave extends vertically from the origin. That is,
u
(
x
,
t
)
=
{
1
x
<
0
−
2
x
>
0
{\displaystyle u(x,t)=\left\{{\begin{aligned}1&x<0\\-2&x>0\end{aligned}}\right.}
Consider the evolution equation with initial data
u
t
t
−
u
x
x
+
u
t
=
(
u
x
t
3
)
x
,
−
∞
<
x
<
∞
,
t
>
0
{\displaystyle u_{tt}-u_{xx}+u_{t}=(u_{xt}^{3})_{x},\quad -\infty <x<\infty ,t>0}
u
(
x
,
0
)
=
f
(
x
)
,
u
t
(
x
,
0
)
=
g
(
x
)
,
u
(
0
,
t
)
=
u
(
1
,
t
)
=
0.
{\displaystyle u(x,0)=f(x),u_{t}(x,0)=g(x),u(0,t)=u(1,t)=0.}
a) What energy quantity is appropriate for this equation? Is it conserved or dissipated?
b) Show that
C
3
{\displaystyle C^{3}}
solutions of this problem are unique.
Consider the energy
E
(
t
)
=
1
2
∫
0
1
u
t
2
+
u
x
2
d
x
{\displaystyle E(t)={\frac {1}{2}}\int _{0}^{1}u_{t}^{2}+u_{x}^{2}\,dx}
. Then
E
˙
(
t
)
=
∫
0
1
u
t
u
t
t
+
u
x
u
x
t
d
x
{\displaystyle {\dot {E}}(t)=\int _{0}^{1}u_{t}u_{tt}+u_{x}u_{xt}\,dx}
. Integrate by parts to get
E
˙
(
t
)
=
∫
0
1
u
t
u
t
t
−
u
x
x
u
t
d
x
+
u
t
u
x
|
0
1
{\displaystyle {\dot {E}}(t)=\int _{0}^{1}u_{t}u_{tt}-u_{xx}u_{t}\,dx+\left.u_{t}u_{x}\right|_{0}^{1}}
.
The boundary terms vanish since
u
(
0
,
t
)
=
0
{\displaystyle u(0,t)=0}
implies
u
t
(
0
,
t
)
=
0
{\displaystyle u_{t}(0,t)=0}
(similarly at
x
=
1
{\displaystyle x=1}
). Then by the original PDE we get
E
˙
(
t
)
=
∫
0
1
u
t
(
(
u
x
t
3
)
x
−
u
t
)
d
x
=
∫
0
1
u
t
(
u
x
t
3
)
x
−
u
t
2
)
d
x
=
∫
0
1
−
u
x
t
4
−
u
t
2
d
x
+
u
x
t
3
u
t
|
0
1
.
{\displaystyle {\begin{aligned}{\dot {E}}(t)&=&\int _{0}^{1}u_{t}((u_{xt}^{3})_{x}-u_{t})\,dx\\&=&\int _{0}^{1}u_{t}(u_{xt}^{3})_{x}-u_{t}^{2})\,dx\\&=&\int _{0}^{1}-u_{xt}^{4}-u_{t}^{2}\,dx+\left.u_{xt}^{3}u_{t}\right|_{0}^{1}.\end{aligned}}}
where the last equality is another integration by parts. The boundary terms vanish again by the same argument. Therefore,
E
˙
(
t
)
<
0
{\displaystyle {\dot {E}}(t)<0}
for all
t
{\displaystyle t}
; that is, energy is dissipated.
Suppose
u
,
v
{\displaystyle u,v}
are two distinct solutions to the system. Then
w
=
u
−
v
{\displaystyle w=u-v}
is a solution to
w
t
t
−
w
x
x
+
w
t
=
(
w
x
t
3
)
x
,
−
∞
<
x
<
∞
,
t
>
0
{\displaystyle w_{tt}-w_{xx}+w_{t}=(w_{xt}^{3})_{x},\quad -\infty <x<\infty ,t>0}
w
(
x
,
0
)
=
0
,
w
t
(
x
,
0
)
=
0
,
w
(
0
,
t
)
=
w
(
1
,
t
)
=
0.
{\displaystyle w(x,0)=0,w_{t}(x,0)=0,w(0,t)=w(1,t)=0.}
This tells us that at
t
=
0
{\displaystyle t=0}
,
w
x
=
w
t
=
0
{\displaystyle w_{x}=w_{t}=0}
. Therefore,
E
(
0
)
=
0
{\displaystyle E(0)=0}
. Since
E
(
t
)
≤
E
(
0
)
{\displaystyle E(t)\leq E(0)}
then
E
(
t
)
=
0
{\displaystyle E(t)=0}
for all
t
{\displaystyle t}
. This implies
w
≡
0
{\displaystyle w\equiv 0}
. That is,
u
=
v
{\displaystyle u=v}
.
Let
Ω
⊆
R
n
{\displaystyle \Omega \subseteq \mathbb {R} ^{n}}
be a bounded open set with smooth boundary
∂
Ω
{\displaystyle \partial \Omega }
. Consider the initial boundary value problem for
u
(
x
,
t
)
{\displaystyle u(x,t)}
:
{
u
t
−
∇
⋅
(
a
(
x
)
∇
u
)
+
b
(
x
)
u
=
q
,
x
∈
Ω
,
t
>
0
u
(
x
,
0
)
=
f
(
x
)
,
x
∈
Ω
u
t
+
∂
u
/
∂
n
+
u
=
0
,
x
∈
∂
Ω
,
t
>
0
{\displaystyle \left\{{\begin{array}{ll}u_{t}-\nabla \cdot (a(x)\nabla u)+b(x)u=q,&x\in \Omega ,t>0\\u(x,0)=f(x),&x\in \Omega \\u_{t}+\partial u/\partial n+u=0,&x\in \partial \Omega ,t>0\end{array}}\right.}
where
∂
u
/
∂
n
{\displaystyle \partial u/\partial n}
is the exterior normal derivative. Assume that
a
,
b
∈
C
1
(
Ω
¯
)
{\displaystyle a,b\in C^{1}({\bar {\Omega }})}
and that
a
(
x
)
≥
0
{\displaystyle a(x)\geq 0}
for
x
∈
Ω
{\displaystyle x\in \Omega }
. Show that smooth solutions of this problem are unique.
Suppose
u
,
v
{\displaystyle u,v}
are two distinct solutions. Then
w
=
u
−
v
{\displaystyle w=u-v}
is a smooth solution to
{
w
t
−
∇
⋅
(
a
(
x
)
∇
w
)
+
b
(
x
)
w
=
0
,
x
∈
Ω
,
t
>
0
w
(
x
,
0
)
=
0
,
x
∈
Ω
w
t
+
∂
w
/
∂
n
+
w
=
0
,
x
∈
∂
Ω
,
t
>
0
{\displaystyle \left\{{\begin{array}{ll}w_{t}-\nabla \cdot (a(x)\nabla w)+b(x)w=0,&x\in \Omega ,t>0\\w(x,0)=0,&x\in \Omega \\w_{t}+\partial w/\partial n+w=0,&x\in \partial \Omega ,t>0\end{array}}\right.}
Consider the energy
E
(
t
)
=
1
2
∫
Ω
w
2
d
x
+
1
2
∫
∂
Ω
a
(
x
)
w
2
d
S
{\displaystyle E(t)={\frac {1}{2}}\int _{\Omega }w^{2}\,dx+{\frac {1}{2}}\int _{\partial \Omega }a(x)w^{2}\,dS}
. It is easy to verify that
E
(
0
)
=
0
{\displaystyle E(0)=0}
. Then
E
˙
(
t
)
=
∫
Ω
w
w
t
d
x
+
∫
∂
Ω
a
(
x
)
w
w
t
d
S
=
∫
Ω
w
∇
⋅
(
a
(
x
)
∇
w
)
−
b
(
x
)
w
2
d
x
+
∫
∂
Ω
a
(
x
)
w
w
t
d
S
=
∫
Ω
−
∇
w
⋅
(
a
(
x
)
∇
w
)
−
b
(
x
)
w
2
d
x
+
∫
∂
Ω
w
a
(
x
)
∂
w
∂
n
d
S
+
∫
∂
Ω
a
(
x
)
w
w
t
d
S
=
∫
Ω
−
a
(
x
)
∇
w
⋅
∇
w
−
b
(
x
)
w
2
d
x
+
∫
∂
Ω
−
a
(
x
)
w
2
−
a
(
x
)
w
w
t
d
S
+
∫
∂
Ω
a
(
x
)
w
w
t
d
S
=
∫
Ω
−
a
(
x
)
∇
w
⋅
∇
w
−
b
(
x
)
w
2
d
x
+
∫
∂
Ω
−
a
(
x
)
w
2
d
S
≤
∫
Ω
−
b
(
x
)
w
2
d
x
≤
‖
b
‖
L
∞
(
Ω
)
∫
Ω
w
2
d
x
{\displaystyle {\begin{aligned}{\dot {E}}(t)=&\int _{\Omega }ww_{t}\,dx+\int _{\partial \Omega }a(x)ww_{t}\,dS\\=&\int _{\Omega }w\nabla \cdot (a(x)\nabla w)-b(x)w^{2}\,dx+\int _{\partial \Omega }a(x)ww_{t}\,dS\\=&\int _{\Omega }-\nabla w\cdot (a(x)\nabla w)-b(x)w^{2}\,dx+\int _{\partial \Omega }wa(x){\frac {\partial w}{\partial n}}\,dS+\int _{\partial \Omega }a(x)ww_{t}\,dS\\=&\int _{\Omega }-a(x)\nabla w\cdot \nabla w-b(x)w^{2}\,dx+\int _{\partial \Omega }-a(x)w^{2}-a(x)ww_{t}\,dS+\int _{\partial \Omega }a(x)ww_{t}\,dS\\=&\int _{\Omega }-a(x)\nabla w\cdot \nabla w-b(x)w^{2}\,dx+\int _{\partial \Omega }-a(x)w^{2}\,dS\\\leq &\int _{\Omega }-b(x)w^{2}\,dx\\\leq &\|b\|_{L^{\infty }(\Omega )}\int _{\Omega }w^{2}\,dx\end{aligned}}}
Therefore
E
˙
(
t
)
≤
‖
b
‖
L
∞
(
Ω
)
E
(
t
)
{\displaystyle {\dot {E}}(t)\leq \|b\|_{L^{\infty }(\Omega )}E(t)}
implies
E
(
t
)
≤
E
(
0
)
e
‖
b
‖
L
∞
(
Ω
)
t
=
0
{\displaystyle E(t)\leq E(0)e^{\|b\|_{L^{\infty }(\Omega )}t}=0}
for all
t
{\displaystyle t}
. Thus,
E
(
t
)
=
0
{\displaystyle E(t)=0}
for all
t
{\displaystyle t}
which implies
w
≡
0
{\displaystyle w\equiv 0}
(
⇒
)
{\displaystyle (\Rightarrow )}
Suppose
u
{\displaystyle u}
minimizes
I
{\displaystyle I}
, i.e.
I
[
u
]
≤
I
[
w
]
∀
w
∈
K
{\displaystyle I[u]\leq I[w]\,\forall w\in K}
. Then for any fixed
w
∈
K
{\displaystyle w\in K}
, if we let
g
(
t
)
=
(
1
−
t
)
u
+
t
v
{\displaystyle g(t)=(1-t)u+tv}
then
I
[
g
(
0
)
]
≤
I
[
g
(
t
)
]
∀
0
≤
t
≤
1
{\displaystyle I[g(0)]\leq I[g(t)]\,\forall 0\leq t\leq 1}
. Let
i
(
t
)
=
I
[
g
(
t
)
]
{\displaystyle i(t)=I[g(t)]}
; then we can say that
i
′
(
0
)
≥
0
{\displaystyle i'(0)\geq 0}
. Now we must compute
i
′
(
0
)
{\displaystyle i'(0)}
. We have
i
(
t
)
=
1
2
∫
Ω
|
(
1
−
t
)
∇
u
+
t
∇
v
|
2
−
f
(
(
1
−
t
)
u
+
t
v
)
d
x
{\displaystyle i(t)={\frac {1}{2}}\int _{\Omega }\left|(1-t)\nabla u+t\nabla v\right|^{2}-f((1-t)u+tv)\,dx}
i
′
(
t
)
=
∫
Ω
−
(
1
−
t
)
|
∇
u
|
2
+
(
1
−
2
t
)
∇
u
⋅
∇
v
+
t
|
∇
v
|
2
−
(
1
−
t
)
f
u
−
t
f
v
d
x
{\displaystyle i'(t)=\int _{\Omega }-(1-t)|\nabla u|^{2}+(1-2t)\nabla u\cdot \nabla v+t|\nabla v|^{2}-(1-t)fu-tfv\,dx}
i
′
(
0
)
=
∫
Ω
−
|
∇
u
|
2
+
∇
u
⋅
∇
v
−
f
u
d
x
{\displaystyle i'(0)=\int _{\Omega }-|\nabla u|^{2}+\nabla u\cdot \nabla v-fu\,dx}
Since we know
i
′
(
0
)
≥
0
{\displaystyle i'(0)\geq 0}
then
∫
Ω
f
(
u
−
v
)
d
x
≥
∫
Ω
|
∇
u
|
2
−
∇
u
⋅
∇
v
=
∫
Ω
∇
u
⋅
∇
(
u
−
v
)
d
x
{\displaystyle \int _{\Omega }f(u-v)\,dx\geq \int _{\Omega }|\nabla u|^{2}-\nabla u\cdot \nabla v=\int _{\Omega }\nabla u\cdot \nabla (u-v)\,dx}
as desired.
(
⇐
)
{\displaystyle (\Leftarrow )}
Conversely suppose
∫
Ω
∇
u
⋅
∇
(
u
−
v
)
d
x
=
∫
Ω
|
∇
|
2
−
∇
u
⋅
∇
v
d
x
≤
∫
Ω
f
(
u
−
v
)
d
x
.
{\displaystyle \int _{\Omega }\nabla u\cdot \nabla (u-v)\,dx=\int _{\Omega }|\nabla |^{2}-\nabla u\cdot \nabla v\,dx\leq \int _{\Omega }f(u-v)\,dx.}
Then
∫
Ω
|
∇
u
|
2
−
f
u
d
x
≤
∫
Ω
∇
u
⋅
∇
v
−
f
v
d
x
≤
∫
Ω
|
∇
u
|
|
∇
v
|
−
f
v
d
x
≤
∫
Ω
1
/
2
|
∇
u
|
2
+
1
/
2
|
∇
v
|
2
−
f
v
d
x
{\displaystyle {\begin{array}{lll}\int _{\Omega }|\nabla u|^{2}-fu\,dx&\leq &\int _{\Omega }\nabla u\cdot \nabla v-fv\,dx\\&\leq &\int _{\Omega }|\nabla u||\nabla v|-fv\,dx\\&\leq &\int _{\Omega }1/2|\nabla u|^{2}+1/2|\nabla v|^{2}-fv\,dx\end{array}}}
Therefore,
1
/
2
∫
Ω
|
∇
u
|
2
−
f
u
d
x
≤
1
/
2
∫
Ω
|
∇
v
|
2
−
f
v
d
x
{\displaystyle 1/2\int _{\Omega }|\nabla u|^{2}-fu\,dx\leq 1/2\int _{\Omega }|\nabla v|^{2}-fv\,dx}
for all
v
∈
K
{\displaystyle v\in K}
.
That is,
I
[
u
]
≤
I
[
v
]
{\displaystyle I[u]\leq I[v]}
for all
v
∈
K
{\displaystyle v\in K}
, as desired.