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Solvability of Neumann problem

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1

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$begingroup$

Let $D$ be the domain in $mathbbR^2$ such that $x^2 + y^2 <4$ , and $alpha, beta$ and $ gamma $ be real constants. For which values of these three constants there is no solution for:$$Delta u =0 , (x,y)in D$$
$$partial_n u= alpha x^2 + beta y + gamma, (x,y) in partial D$$
How to solve the equation for the values of these three constants which the solution exists?

Thanks.

pde

share|cite|improve this question

asked 2 days ago

Math LoverMath Lover

938

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add a comment | 

1

1

$begingroup$

Let $D$ be the domain in $mathbbR^2$ such that $x^2 + y^2 <4$ , and $alpha, beta$ and $ gamma $ be real constants. For which values of these three constants there is no solution for:$$Delta u =0 , (x,y)in D$$
$$partial_n u= alpha x^2 + beta y + gamma, (x,y) in partial D$$
How to solve the equation for the values of these three constants which the solution exists?

Thanks.

pde

share|cite|improve this question

asked 2 days ago

Math LoverMath Lover

938

$endgroup$

add a comment | 

$begingroup$

Let $D$ be the domain in $mathbbR^2$ such that $x^2 + y^2 <4$ , and $alpha, beta$ and $ gamma $ be real constants. For which values of these three constants there is no solution for:$$Delta u =0 , (x,y)in D$$
$$partial_n u= alpha x^2 + beta y + gamma, (x,y) in partial D$$
How to solve the equation for the values of these three constants which the solution exists?

Thanks.

pde

share|cite|improve this question

asked 2 days ago

Math LoverMath Lover

938

$endgroup$

share|cite|improve this question

asked 2 days ago

Math LoverMath Lover

938

share|cite|improve this question

asked 2 days ago

Math LoverMath Lover

938

asked 2 days ago

Math LoverMath Lover

938

asked 2 days ago

Math LoverMath Lover

938

2 Answers
2

active

oldest

votes

2

$begingroup$

In polar coordinates $(x,y) = (rcosphi,rsinphi)$ the general solution (which is finite at $r=0$) is

$$ u(r,phi) = A_0 + sum_n=1^infty r^n big[A_ncos(nphi) + B_nsin(nphi)big] $$

where $D = 0<r < 2$. The gradient is

$$ u_r(r,phi) = alpha r^2cos^2phi + beta rsin(phi) + gamma = fracalpha r^22big[cos(2phi) + 1big] + beta r sin(phi) + gamma $$

Evaluated at the boundary:

$$ u_r(2,phi) = 2alphacos(2phi) + 2beta sin(phi) + (2alpha + gamma) $$

You also have, from the general solution

$$ u_r(2,phi) = sum_n=1^infty n2^n-1big[A_ncos(nphi) + B_nsin(phi)big] $$

Comparing coefficients gives

begincases
2alpha + gamma = 0 \
B_1 = 2beta \
4A_2 = 2alpha
endcases

The first condition is required for the solution to exist.

share|cite|improve this answer

answered 2 days ago

DylanDylan

13.8k31027

$endgroup$

add a comment | 

0

$begingroup$

The Neumann problem has a solution iff $int_Sfdsigma=0$ , where $S$, is the circle with center the origin and radius $2$ and $f(x,y)=alpha x^2 + beta y + gamma$ . A parametrization for this circle is $(2cost, 2sint)$. Now if we take the above mentioned line - integral we get the desired condition: $2a+gamma=0$.

Edit The proposition above is

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

dmtridmtri

1,6082521

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is $f$? ---
  $endgroup$
  – Dylan
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Dylan, $f(x,y)=alpha x^2 + beta y + gamma$
  $endgroup$
  – dmtri
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It would help to specify that in your answer. Also mention which theorem this came from. This is a good answer but needs some more context.
  $endgroup$
  – Dylan
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok, thanks. See please "Introduction to PDE with Applications", by Zachamanoglou and Toe. It is just a ref in page 259 , line 14. Potential Theory is used here.
  $endgroup$
  – dmtri
  2 days ago
  

  
  
  
  

add a comment | 

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2

$begingroup$

In polar coordinates $(x,y) = (rcosphi,rsinphi)$ the general solution (which is finite at $r=0$) is

$$ u(r,phi) = A_0 + sum_n=1^infty r^n big[A_ncos(nphi) + B_nsin(nphi)big] $$

where $D = 0<r < 2$. The gradient is

$$ u_r(r,phi) = alpha r^2cos^2phi + beta rsin(phi) + gamma = fracalpha r^22big[cos(2phi) + 1big] + beta r sin(phi) + gamma $$

Evaluated at the boundary:

$$ u_r(2,phi) = 2alphacos(2phi) + 2beta sin(phi) + (2alpha + gamma) $$

You also have, from the general solution

$$ u_r(2,phi) = sum_n=1^infty n2^n-1big[A_ncos(nphi) + B_nsin(phi)big] $$

Comparing coefficients gives

begincases
2alpha + gamma = 0 \
B_1 = 2beta \
4A_2 = 2alpha
endcases

The first condition is required for the solution to exist.

share|cite|improve this answer

answered 2 days ago

DylanDylan

13.8k31027

$endgroup$

add a comment | 

2

$begingroup$

In polar coordinates $(x,y) = (rcosphi,rsinphi)$ the general solution (which is finite at $r=0$) is

$$ u(r,phi) = A_0 + sum_n=1^infty r^n big[A_ncos(nphi) + B_nsin(nphi)big] $$

where $D = 0<r < 2$. The gradient is

$$ u_r(r,phi) = alpha r^2cos^2phi + beta rsin(phi) + gamma = fracalpha r^22big[cos(2phi) + 1big] + beta r sin(phi) + gamma $$

Evaluated at the boundary:

$$ u_r(2,phi) = 2alphacos(2phi) + 2beta sin(phi) + (2alpha + gamma) $$

You also have, from the general solution

$$ u_r(2,phi) = sum_n=1^infty n2^n-1big[A_ncos(nphi) + B_nsin(phi)big] $$

Comparing coefficients gives

begincases
2alpha + gamma = 0 \
B_1 = 2beta \
4A_2 = 2alpha
endcases

The first condition is required for the solution to exist.

share|cite|improve this answer

answered 2 days ago

DylanDylan

13.8k31027

$endgroup$

add a comment | 

$begingroup$

In polar coordinates $(x,y) = (rcosphi,rsinphi)$ the general solution (which is finite at $r=0$) is

$$ u(r,phi) = A_0 + sum_n=1^infty r^n big[A_ncos(nphi) + B_nsin(nphi)big] $$

where $D = 0<r < 2$. The gradient is

$$ u_r(r,phi) = alpha r^2cos^2phi + beta rsin(phi) + gamma = fracalpha r^22big[cos(2phi) + 1big] + beta r sin(phi) + gamma $$

Evaluated at the boundary:

$$ u_r(2,phi) = 2alphacos(2phi) + 2beta sin(phi) + (2alpha + gamma) $$

You also have, from the general solution

$$ u_r(2,phi) = sum_n=1^infty n2^n-1big[A_ncos(nphi) + B_nsin(phi)big] $$

Comparing coefficients gives

begincases
2alpha + gamma = 0 \
B_1 = 2beta \
4A_2 = 2alpha
endcases

The first condition is required for the solution to exist.

share|cite|improve this answer

answered 2 days ago

DylanDylan

13.8k31027

$endgroup$

share|cite|improve this answer

answered 2 days ago

DylanDylan

13.8k31027

answered 2 days ago

DylanDylan

13.8k31027

answered 2 days ago

DylanDylan

13.8k31027

0

$begingroup$

The Neumann problem has a solution iff $int_Sfdsigma=0$ , where $S$, is the circle with center the origin and radius $2$ and $f(x,y)=alpha x^2 + beta y + gamma$ . A parametrization for this circle is $(2cost, 2sint)$. Now if we take the above mentioned line - integral we get the desired condition: $2a+gamma=0$.

Edit The proposition above is

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

dmtridmtri

1,6082521

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is $f$? ---
  $endgroup$
  – Dylan
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Dylan, $f(x,y)=alpha x^2 + beta y + gamma$
  $endgroup$
  – dmtri
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It would help to specify that in your answer. Also mention which theorem this came from. This is a good answer but needs some more context.
  $endgroup$
  – Dylan
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok, thanks. See please "Introduction to PDE with Applications", by Zachamanoglou and Toe. It is just a ref in page 259 , line 14. Potential Theory is used here.
  $endgroup$
  – dmtri
  2 days ago
  

  
  
  
  

add a comment | 

0

$begingroup$

The Neumann problem has a solution iff $int_Sfdsigma=0$ , where $S$, is the circle with center the origin and radius $2$ and $f(x,y)=alpha x^2 + beta y + gamma$ . A parametrization for this circle is $(2cost, 2sint)$. Now if we take the above mentioned line - integral we get the desired condition: $2a+gamma=0$.

Edit The proposition above is

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

dmtridmtri

1,6082521

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is $f$? ---
  $endgroup$
  – Dylan
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Dylan, $f(x,y)=alpha x^2 + beta y + gamma$
  $endgroup$
  – dmtri
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It would help to specify that in your answer. Also mention which theorem this came from. This is a good answer but needs some more context.
  $endgroup$
  – Dylan
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok, thanks. See please "Introduction to PDE with Applications", by Zachamanoglou and Toe. It is just a ref in page 259 , line 14. Potential Theory is used here.
  $endgroup$
  – dmtri
  2 days ago
  

  
  
  
  

add a comment | 

$begingroup$

The Neumann problem has a solution iff $int_Sfdsigma=0$ , where $S$, is the circle with center the origin and radius $2$ and $f(x,y)=alpha x^2 + beta y + gamma$ . A parametrization for this circle is $(2cost, 2sint)$. Now if we take the above mentioned line - integral we get the desired condition: $2a+gamma=0$.

Edit The proposition above is

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

dmtridmtri

1,6082521

$endgroup$

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

dmtridmtri

1,6082521

answered 2 days ago

dmtridmtri

1,6082521

answered 2 days ago

dmtridmtri

1,6082521