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1

1

$begingroup$

Let $Dsubset mathbbR^3$. Let $D$ be a connected subset of $mathbbR^3$. Show that if there is a solution of the system of equations
beginequation
Delta u=f text in D, fracdudn=gtext on boundary of D,
endequation
then $int_D f dV=int_textboundary of D g ds$.

My partial answer:

Assume that $int_D f dVneqint_textboundary of D g dS$ and $u$ is the solution of the system of equation, then

beginequation
int_D Delta u dV= int_D f dVneqint_textboundary of D g dS=
int_textboundary of D fracdudn dS.
endequation
This contradicts Green's first identity.

Please let me know that idea of my answer is correct. Is it possible to prove this question without using a contradiction.

ordinary-differential-equations multivariable-calculus pde

share|cite|improve this question

edited Mar 10 at 19:46

dmtri

1,6172521

asked May 11 '14 at 12:57

beginnerbeginner

401935

$endgroup$

add a comment | 

1

1

$begingroup$

Let $Dsubset mathbbR^3$. Let $D$ be a connected subset of $mathbbR^3$. Show that if there is a solution of the system of equations
beginequation
Delta u=f text in D, fracdudn=gtext on boundary of D,
endequation
then $int_D f dV=int_textboundary of D g ds$.

My partial answer:

Assume that $int_D f dVneqint_textboundary of D g dS$ and $u$ is the solution of the system of equation, then

beginequation
int_D Delta u dV= int_D f dVneqint_textboundary of D g dS=
int_textboundary of D fracdudn dS.
endequation
This contradicts Green's first identity.

Please let me know that idea of my answer is correct. Is it possible to prove this question without using a contradiction.

ordinary-differential-equations multivariable-calculus pde

share|cite|improve this question

edited Mar 10 at 19:46

dmtri

1,6172521

asked May 11 '14 at 12:57

beginnerbeginner

401935

$endgroup$

add a comment | 

$begingroup$

Let $Dsubset mathbbR^3$. Let $D$ be a connected subset of $mathbbR^3$. Show that if there is a solution of the system of equations
beginequation
Delta u=f text in D, fracdudn=gtext on boundary of D,
endequation
then $int_D f dV=int_textboundary of D g ds$.

My partial answer:

Assume that $int_D f dVneqint_textboundary of D g dS$ and $u$ is the solution of the system of equation, then

beginequation
int_D Delta u dV= int_D f dVneqint_textboundary of D g dS=
int_textboundary of D fracdudn dS.
endequation
This contradicts Green's first identity.

Please let me know that idea of my answer is correct. Is it possible to prove this question without using a contradiction.

ordinary-differential-equations multivariable-calculus pde

share|cite|improve this question

edited Mar 10 at 19:46

dmtri

1,6172521

asked May 11 '14 at 12:57

beginnerbeginner

401935

$endgroup$

share|cite|improve this question

edited Mar 10 at 19:46

dmtri

1,6172521

asked May 11 '14 at 12:57

beginnerbeginner

401935

share|cite|improve this question

edited Mar 10 at 19:46

dmtri

1,6172521

asked May 11 '14 at 12:57

beginnerbeginner

401935

edited Mar 10 at 19:46

dmtri

1,6172521

edited Mar 10 at 19:46

dmtri

1,6172521

asked May 11 '14 at 12:57

beginnerbeginner

401935

asked May 11 '14 at 12:57

beginnerbeginner

401935

1 Answer
1

active

oldest

votes

2

$begingroup$

The solution is correct. It can be phrased in a more direct way, if your statement is made less "negative". Instead of saying "there does not exist a solution unless equality holds", we could say "if there exists a solution, then equality holds". Then the proof is direct: let $u$ be a solution then use Green's identity and finally conclude that the equality holds.

In any case, yours is not really a proof by contradiction (reductio ad absurdum), but rather a proof of contrapositive statement.

share|cite|improve this answer

edited Mar 10 at 19:43

dmtri

1,6172521

answered May 11 '14 at 18:36

user147263

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @dmtri: Your edit (to the Question above) largely obviates the Accepted Answer here. Normally I would reject it on that basis. But (as the Answer points out) rephrasing the Question really does improve its clarity, and neither the OP of the Question or the Answer is currently active, so I think it is worthwhile in this restricted case.
  $endgroup$
  – hardmath
  Mar 10 at 20:04
  
  

  
  
  
  

add a comment | 

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2

$begingroup$

The solution is correct. It can be phrased in a more direct way, if your statement is made less "negative". Instead of saying "there does not exist a solution unless equality holds", we could say "if there exists a solution, then equality holds". Then the proof is direct: let $u$ be a solution then use Green's identity and finally conclude that the equality holds.

In any case, yours is not really a proof by contradiction (reductio ad absurdum), but rather a proof of contrapositive statement.

share|cite|improve this answer

edited Mar 10 at 19:43

dmtri

1,6172521

answered May 11 '14 at 18:36

user147263

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @dmtri: Your edit (to the Question above) largely obviates the Accepted Answer here. Normally I would reject it on that basis. But (as the Answer points out) rephrasing the Question really does improve its clarity, and neither the OP of the Question or the Answer is currently active, so I think it is worthwhile in this restricted case.
  $endgroup$
  – hardmath
  Mar 10 at 20:04
  
  

  
  
  
  

add a comment | 

2

$begingroup$

The solution is correct. It can be phrased in a more direct way, if your statement is made less "negative". Instead of saying "there does not exist a solution unless equality holds", we could say "if there exists a solution, then equality holds". Then the proof is direct: let $u$ be a solution then use Green's identity and finally conclude that the equality holds.

In any case, yours is not really a proof by contradiction (reductio ad absurdum), but rather a proof of contrapositive statement.

share|cite|improve this answer

edited Mar 10 at 19:43

dmtri

1,6172521

answered May 11 '14 at 18:36

user147263

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @dmtri: Your edit (to the Question above) largely obviates the Accepted Answer here. Normally I would reject it on that basis. But (as the Answer points out) rephrasing the Question really does improve its clarity, and neither the OP of the Question or the Answer is currently active, so I think it is worthwhile in this restricted case.
  $endgroup$
  – hardmath
  Mar 10 at 20:04
  
  

  
  
  
  

add a comment | 

$begingroup$

The solution is correct. It can be phrased in a more direct way, if your statement is made less "negative". Instead of saying "there does not exist a solution unless equality holds", we could say "if there exists a solution, then equality holds". Then the proof is direct: let $u$ be a solution then use Green's identity and finally conclude that the equality holds.

In any case, yours is not really a proof by contradiction (reductio ad absurdum), but rather a proof of contrapositive statement.

share|cite|improve this answer

edited Mar 10 at 19:43

dmtri

1,6172521

answered May 11 '14 at 18:36

user147263

$endgroup$

share|cite|improve this answer

edited Mar 10 at 19:43

dmtri

1,6172521

answered May 11 '14 at 18:36

user147263

edited Mar 10 at 19:43

dmtri

1,6172521

edited Mar 10 at 19:43

dmtri

1,6172521