If $u$ is harmonic and vanishes on the boundary, then $uequiv 0$.

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Let's assume $u$ is harmonic on $D=x^2+y^2leq1$. If $u=0$ on $partial D$ then $uequiv 0$ on $D$.

$textbfProof$:
Let's use Green's theorem and set $P=-ufracpartial upartial y$ and $Q=ufracpartial upartial x$.

Then

$$1. fracpartial Qpartial x-fracpartial Ppartial y
=left(fracpartial upartial xright)^2+
uleft(fracpartial upartial xright)^2+
left(fracpartial upartial yright)^2+
uleft(fracpartial upartial yright)^2
=left(fracpartial upartial xright)^2+
left(fracpartial upartial yright)^2.
$$

By Green's theorem,

$$
0=int_partial D -ufracpartial upartial ydx+ufracpartial upartial xdy=int int_D
left(
left(fracpartial upartial xright)^2+
left(fracpartial upartial yright)^2right)dxdy.
$$

Because $uin C^2$, we get

$$
left(fracpartial upartial xright)^2 +
left(fracpartial upartial yright)^2=0.
$$

So $uequiv c$ on $D$ but on $partial D$ we have $u=0$. So $uequiv 0$.

What have we done on the first move for $1$.?

calculus analysis

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edited Jul 29 at 16:26

Mee Seong Im

2,5591517

asked Jul 29 at 15:02

newhere

742310

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You can also use the strong max/min principle of Harmonic Functions, which states that non-constant Harmonic Functions attain their max/min only on the boundary. Since the value on the boundary is identically 0, we see it's max and min are 0, so it must be constant, e.g. $u equiv 0$
  – Raymond Chu
  Jul 29 at 17:06
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

1

Let's assume $u$ is harmonic on $D=x^2+y^2leq1$. If $u=0$ on $partial D$ then $uequiv 0$ on $D$.

$textbfProof$:
Let's use Green's theorem and set $P=-ufracpartial upartial y$ and $Q=ufracpartial upartial x$.

Then

$$1. fracpartial Qpartial x-fracpartial Ppartial y
=left(fracpartial upartial xright)^2+
uleft(fracpartial upartial xright)^2+
left(fracpartial upartial yright)^2+
uleft(fracpartial upartial yright)^2
=left(fracpartial upartial xright)^2+
left(fracpartial upartial yright)^2.
$$

By Green's theorem,

$$
0=int_partial D -ufracpartial upartial ydx+ufracpartial upartial xdy=int int_D
left(
left(fracpartial upartial xright)^2+
left(fracpartial upartial yright)^2right)dxdy.
$$

Because $uin C^2$, we get

$$
left(fracpartial upartial xright)^2 +
left(fracpartial upartial yright)^2=0.
$$

So $uequiv c$ on $D$ but on $partial D$ we have $u=0$. So $uequiv 0$.

What have we done on the first move for $1$.?

calculus analysis

share|cite|improve this question

edited Jul 29 at 16:26

Mee Seong Im

2,5591517

asked Jul 29 at 15:02

newhere

742310

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You can also use the strong max/min principle of Harmonic Functions, which states that non-constant Harmonic Functions attain their max/min only on the boundary. Since the value on the boundary is identically 0, we see it's max and min are 0, so it must be constant, e.g. $u equiv 0$
  – Raymond Chu
  Jul 29 at 17:06
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

1

up vote
1
down vote

favorite

1

1

Let's assume $u$ is harmonic on $D=x^2+y^2leq1$. If $u=0$ on $partial D$ then $uequiv 0$ on $D$.

$textbfProof$:
Let's use Green's theorem and set $P=-ufracpartial upartial y$ and $Q=ufracpartial upartial x$.

Then

$$1. fracpartial Qpartial x-fracpartial Ppartial y
=left(fracpartial upartial xright)^2+
uleft(fracpartial upartial xright)^2+
left(fracpartial upartial yright)^2+
uleft(fracpartial upartial yright)^2
=left(fracpartial upartial xright)^2+
left(fracpartial upartial yright)^2.
$$

By Green's theorem,

$$
0=int_partial D -ufracpartial upartial ydx+ufracpartial upartial xdy=int int_D
left(
left(fracpartial upartial xright)^2+
left(fracpartial upartial yright)^2right)dxdy.
$$

Because $uin C^2$, we get

$$
left(fracpartial upartial xright)^2 +
left(fracpartial upartial yright)^2=0.
$$

So $uequiv c$ on $D$ but on $partial D$ we have $u=0$. So $uequiv 0$.

What have we done on the first move for $1$.?

calculus analysis

share|cite|improve this question

edited Jul 29 at 16:26

Mee Seong Im

2,5591517

asked Jul 29 at 15:02

newhere

742310

Let's assume $u$ is harmonic on $D=x^2+y^2leq1$. If $u=0$ on $partial D$ then $uequiv 0$ on $D$.

$textbfProof$:
Let's use Green's theorem and set $P=-ufracpartial upartial y$ and $Q=ufracpartial upartial x$.

Then

$$1. fracpartial Qpartial x-fracpartial Ppartial y
=left(fracpartial upartial xright)^2+
uleft(fracpartial upartial xright)^2+
left(fracpartial upartial yright)^2+
uleft(fracpartial upartial yright)^2
=left(fracpartial upartial xright)^2+
left(fracpartial upartial yright)^2.
$$

By Green's theorem,

$$
0=int_partial D -ufracpartial upartial ydx+ufracpartial upartial xdy=int int_D
left(
left(fracpartial upartial xright)^2+
left(fracpartial upartial yright)^2right)dxdy.
$$

Because $uin C^2$, we get

$$
left(fracpartial upartial xright)^2 +
left(fracpartial upartial yright)^2=0.
$$

So $uequiv c$ on $D$ but on $partial D$ we have $u=0$. So $uequiv 0$.

What have we done on the first move for $1$.?

calculus analysis

share|cite|improve this question

edited Jul 29 at 16:26

Mee Seong Im

2,5591517

asked Jul 29 at 15:02

newhere

742310

share|cite|improve this question

share|cite|improve this question

edited Jul 29 at 16:26

Mee Seong Im

2,5591517

edited Jul 29 at 16:26

Mee Seong Im

2,5591517

edited Jul 29 at 16:26

Mee Seong Im

2,5591517

2,5591517

asked Jul 29 at 15:02

newhere

742310

asked Jul 29 at 15:02

newhere

742310

asked Jul 29 at 15:02

newhere

742310

742310

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You can also use the strong max/min principle of Harmonic Functions, which states that non-constant Harmonic Functions attain their max/min only on the boundary. Since the value on the boundary is identically 0, we see it's max and min are 0, so it must be constant, e.g. $u equiv 0$
  – Raymond Chu
  Jul 29 at 17:06
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You can also use the strong max/min principle of Harmonic Functions, which states that non-constant Harmonic Functions attain their max/min only on the boundary. Since the value on the boundary is identically 0, we see it's max and min are 0, so it must be constant, e.g. $u equiv 0$
  – Raymond Chu
  Jul 29 at 17:06
  

  
  
  
  

You can also use the strong max/min principle of Harmonic Functions, which states that non-constant Harmonic Functions attain their max/min only on the boundary. Since the value on the boundary is identically 0, we see it's max and min are 0, so it must be constant, e.g. $u equiv 0$
– Raymond Chu
Jul 29 at 17:06

You can also use the strong max/min principle of Harmonic Functions, which states that non-constant Harmonic Functions attain their max/min only on the boundary. Since the value on the boundary is identically 0, we see it's max and min are 0, so it must be constant, e.g. $u equiv 0$
– Raymond Chu
Jul 29 at 17:06

add a comment | 

2 Answers
2

active

oldest

votes

up vote
1
down vote



accepted

In the second line of your proof, it should be:

beginalign*
fracpartial Qpartial x-fracpartial Ppartial y
&=left(fracpartial upartial xright)^2+
ufracpartial^2 upartial x^2+
left(fracpartial upartial yright)^2+
u fracpartial^2 upartial y^2 \
&= left(fracpartial upartial xright)^2
+ left(fracpartial upartial yright)^2
+ u
underbraceleft( fracpartial^2 upartial x^2
+ fracpartial^2 upartial y^2right)_0 \
&=left(fracpartial upartial xright)^2+
left(fracpartial upartial yright)^2
endalign*
since $u$ is harmonic on $D$, i.e., $u$ must satisfy Laplace's equation.

share|cite|improve this answer

answered Jul 29 at 16:45

Mee Seong Im

2,5591517

add a comment | 

up vote
1
down vote

We have used the product rule; e.g.

$$
partial_x Q = partial (u partial_x u) = (partial_x u)^2 + u partial_x^2 u.
$$

(Your notation seems a bit off at that point.)

share|cite|improve this answer

answered Jul 29 at 15:30

AlgebraicsAnonymous

66611

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Is the sentence true only on a circle?
  – newhere
  Jul 29 at 15:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  In fact, I now recall that using probabilistic methods, one can show uniqueness on arbitrary open and bounded sets. @newhere
  – AlgebraicsAnonymous
  Jul 29 at 15:39
  
  

  
  
  
  

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

In the second line of your proof, it should be:

beginalign*
fracpartial Qpartial x-fracpartial Ppartial y
&=left(fracpartial upartial xright)^2+
ufracpartial^2 upartial x^2+
left(fracpartial upartial yright)^2+
u fracpartial^2 upartial y^2 \
&= left(fracpartial upartial xright)^2
+ left(fracpartial upartial yright)^2
+ u
underbraceleft( fracpartial^2 upartial x^2
+ fracpartial^2 upartial y^2right)_0 \
&=left(fracpartial upartial xright)^2+
left(fracpartial upartial yright)^2
endalign*
since $u$ is harmonic on $D$, i.e., $u$ must satisfy Laplace's equation.

share|cite|improve this answer

answered Jul 29 at 16:45

Mee Seong Im

2,5591517

add a comment | 

up vote
1
down vote



accepted

In the second line of your proof, it should be:

beginalign*
fracpartial Qpartial x-fracpartial Ppartial y
&=left(fracpartial upartial xright)^2+
ufracpartial^2 upartial x^2+
left(fracpartial upartial yright)^2+
u fracpartial^2 upartial y^2 \
&= left(fracpartial upartial xright)^2
+ left(fracpartial upartial yright)^2
+ u
underbraceleft( fracpartial^2 upartial x^2
+ fracpartial^2 upartial y^2right)_0 \
&=left(fracpartial upartial xright)^2+
left(fracpartial upartial yright)^2
endalign*
since $u$ is harmonic on $D$, i.e., $u$ must satisfy Laplace's equation.

share|cite|improve this answer

answered Jul 29 at 16:45

Mee Seong Im

2,5591517

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

In the second line of your proof, it should be:

beginalign*
fracpartial Qpartial x-fracpartial Ppartial y
&=left(fracpartial upartial xright)^2+
ufracpartial^2 upartial x^2+
left(fracpartial upartial yright)^2+
u fracpartial^2 upartial y^2 \
&= left(fracpartial upartial xright)^2
+ left(fracpartial upartial yright)^2
+ u
underbraceleft( fracpartial^2 upartial x^2
+ fracpartial^2 upartial y^2right)_0 \
&=left(fracpartial upartial xright)^2+
left(fracpartial upartial yright)^2
endalign*
since $u$ is harmonic on $D$, i.e., $u$ must satisfy Laplace's equation.

share|cite|improve this answer

answered Jul 29 at 16:45

Mee Seong Im

2,5591517

In the second line of your proof, it should be:

beginalign*
fracpartial Qpartial x-fracpartial Ppartial y
&=left(fracpartial upartial xright)^2+
ufracpartial^2 upartial x^2+
left(fracpartial upartial yright)^2+
u fracpartial^2 upartial y^2 \
&= left(fracpartial upartial xright)^2
+ left(fracpartial upartial yright)^2
+ u
underbraceleft( fracpartial^2 upartial x^2
+ fracpartial^2 upartial y^2right)_0 \
&=left(fracpartial upartial xright)^2+
left(fracpartial upartial yright)^2
endalign*
since $u$ is harmonic on $D$, i.e., $u$ must satisfy Laplace's equation.

share|cite|improve this answer

answered Jul 29 at 16:45

Mee Seong Im

2,5591517

share|cite|improve this answer

share|cite|improve this answer

answered Jul 29 at 16:45

Mee Seong Im

2,5591517

answered Jul 29 at 16:45

Mee Seong Im

2,5591517

answered Jul 29 at 16:45

Mee Seong Im

2,5591517

2,5591517

add a comment | 

add a comment | 

up vote
1
down vote

We have used the product rule; e.g.

$$
partial_x Q = partial (u partial_x u) = (partial_x u)^2 + u partial_x^2 u.
$$

(Your notation seems a bit off at that point.)

share|cite|improve this answer

answered Jul 29 at 15:30

AlgebraicsAnonymous

66611

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Is the sentence true only on a circle?
  – newhere
  Jul 29 at 15:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  In fact, I now recall that using probabilistic methods, one can show uniqueness on arbitrary open and bounded sets. @newhere
  – AlgebraicsAnonymous
  Jul 29 at 15:39
  
  

  
  
  
  

add a comment | 

up vote
1
down vote

We have used the product rule; e.g.

$$
partial_x Q = partial (u partial_x u) = (partial_x u)^2 + u partial_x^2 u.
$$

(Your notation seems a bit off at that point.)

share|cite|improve this answer

answered Jul 29 at 15:30

AlgebraicsAnonymous

66611

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Is the sentence true only on a circle?
  – newhere
  Jul 29 at 15:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  In fact, I now recall that using probabilistic methods, one can show uniqueness on arbitrary open and bounded sets. @newhere
  – AlgebraicsAnonymous
  Jul 29 at 15:39
  
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

We have used the product rule; e.g.

$$
partial_x Q = partial (u partial_x u) = (partial_x u)^2 + u partial_x^2 u.
$$

(Your notation seems a bit off at that point.)

share|cite|improve this answer

answered Jul 29 at 15:30

AlgebraicsAnonymous

66611

We have used the product rule; e.g.

$$
partial_x Q = partial (u partial_x u) = (partial_x u)^2 + u partial_x^2 u.
$$

(Your notation seems a bit off at that point.)

share|cite|improve this answer

answered Jul 29 at 15:30

AlgebraicsAnonymous

66611

share|cite|improve this answer

share|cite|improve this answer

answered Jul 29 at 15:30

AlgebraicsAnonymous

66611

answered Jul 29 at 15:30

AlgebraicsAnonymous

66611

answered Jul 29 at 15:30

AlgebraicsAnonymous

66611

66611

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Is the sentence true only on a circle?
  – newhere
  Jul 29 at 15:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  In fact, I now recall that using probabilistic methods, one can show uniqueness on arbitrary open and bounded sets. @newhere
  – AlgebraicsAnonymous
  Jul 29 at 15:39
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Is the sentence true only on a circle?
  – newhere
  Jul 29 at 15:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  In fact, I now recall that using probabilistic methods, one can show uniqueness on arbitrary open and bounded sets. @newhere
  – AlgebraicsAnonymous
  Jul 29 at 15:39
  
  

  
  
  
  

Is the sentence true only on a circle?
– newhere
Jul 29 at 15:38

Is the sentence true only on a circle?
– newhere
Jul 29 at 15:38

In fact, I now recall that using probabilistic methods, one can show uniqueness on arbitrary open and bounded sets. @newhere
– AlgebraicsAnonymous
Jul 29 at 15:39

In fact, I now recall that using probabilistic methods, one can show uniqueness on arbitrary open and bounded sets. @newhere
– AlgebraicsAnonymous
Jul 29 at 15:39

add a comment | 

 

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