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Existence and uniqueness for parabolic equations with Robin BCs

Poisson's equation with Robin boundary conditionsVariational formulation of Robin boundary value problem for Poisson equation in finite element methodsRegularity when Dirichlet conditions are posed on the interior of a domain.Weak formulation of a system of biharmonic pdesShowing coercivity of the bilinear form associated with a robin boundary value problemExistence of time derivative in the Galerkin equation of parabolic PDEsProof of Weak Maximum Principle for Parabolic EquationsImplementing boundary conditions for the Biharmonic equation using $C^1$ elements.Why 1/2 in the energy functional formula for weak solutionNumerical Analysis and Differential equations book recommendations focusing on the given topics.Motivation for the definition of weak solutions to parabolic equations of second order

2

$begingroup$

I'm following Evans's book for PDEs, and the existence and uniqueness for parabolic problems is analised for Dirichlet BCs. I'm trying to analise this for Robin BCs, but in the weak formulation a term involving a boundary integral appears. Consequently, it is not posible to apply the Theorem 3 (section 7.1.2 c).

I suppose that I'm not the first one facing this problem. Is there any approach that I should follow?

Thanks in advance.

pde

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edited yesterday

kim_8

asked yesterday

kim_8kim_8

133

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2

$begingroup$

I'm following Evans's book for PDEs, and the existence and uniqueness for parabolic problems is analised for Dirichlet BCs. I'm trying to analise this for Robin BCs, but in the weak formulation a term involving a boundary integral appears. Consequently, it is not posible to apply the Theorem 3 (section 7.1.2 c).

I suppose that I'm not the first one facing this problem. Is there any approach that I should follow?

Thanks in advance.

pde

share|cite|improve this question

edited yesterday

kim_8

asked yesterday

kim_8kim_8

133

New contributor

kim_8 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

I'm following Evans's book for PDEs, and the existence and uniqueness for parabolic problems is analised for Dirichlet BCs. I'm trying to analise this for Robin BCs, but in the weak formulation a term involving a boundary integral appears. Consequently, it is not posible to apply the Theorem 3 (section 7.1.2 c).

I suppose that I'm not the first one facing this problem. Is there any approach that I should follow?

Thanks in advance.

pde

share|cite|improve this question

edited yesterday

kim_8

asked yesterday

kim_8kim_8

133

New contributor

kim_8 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited yesterday

kim_8

asked yesterday

kim_8kim_8

133

New contributor

kim_8 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited yesterday

kim_8

asked yesterday

kim_8kim_8

133

New contributor

kim_8 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked yesterday

kim_8kim_8

133

New contributor

kim_8 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked yesterday

kim_8kim_8

133

1 Answer
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1

$begingroup$

I believe you will essentially have to reprove 7.1.2 a-c with this new boundary condition. If you let $frac{partial u}{partial n}=alpha(x,t)u$ on $partial Utimes[0,T]$, then you still get the weak formulation $(u',v) + B[u,v,t] = (f,v)$, but there are 2 big differences.

1. Your space is now $H^1(U)$, not $H_0^1(U)$, since functions can take on any value on $partial U$ (in the trace sense). This means that you have to be careful using Poincare's Inequality depending on what version you are used to seeing, as some forms of it only apply to function that are 0 on the boundary.




2. Your bilinear form is now $$B[u,v,t] = int_Usum_{i=1,j=1}^Na_{ij}(x,t)u_{x_i}v_{x_i}+sum_{i=1}^Nb_i(x,t)u_{x_i}v+c(x,t)uv dx+int_{partial U}alpha(x,t)uv dS.$$

If you look at the proofs of 7.1.2 b and c, they require bounds on $B$ that were derived in the section on elliptic equations for Dirichlet BCs (6.2.2). If you can prove similar bounds using this new bilinear form, you should be able to plug them into the proofs for 7.1.2 a-c and things will work out.

I haven't proved this for myself, but my intuition is that if you let $alphain:L^{infty}(partial U_T)$, then you can probably bound that surface integral by the $L^2$ norms of the functions and its contribution to the bound will just result in a larger constant $gamma$, but there is sure to be some subtleties in relating the boundary values of these functions and their norms and will definitely involve the constant from the trace operator somewhere.

I didn't provide a full answer but I hope this helped you understand what steps you can take and what tools to use.

edit: There is also a discussion of a similar question here: Poisson's equation with Robin boundary conditions

share|cite|improve this answer

edited yesterday

answered yesterday

whpowell96whpowell96

48819

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you so much for you answer. So I see that if I use the proof in the link that you have provided, the proofs from Evans's book would be valid also in my case. Is that correct?
  $endgroup$
  – kim_8
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I didn't look super carefully at the answer I linked to see if it was complete or not, but I believe that if follow 7.1.2 a-c, the only properties specific to the operator/BC are the bounds they use so once they are established (as I believe the linked thread is doing), everything is the same
  $endgroup$
  – whpowell96
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you again for your answer. And one last question. To obtain some energy estimates, I suppose that if I want a bound for integrals like $$int_{partial Omega} (Tr u)^2 mathrm{d}x=||Tr u||_{L^2(partial Omega)}^2$$ I can use the trace theorem and conclude that $$int_{partial Omega} (Tr u)^2 mathrm{d}x=||Tr u||_{L^2(partial Omega)}^2 leq int_{Omega} u^2 mathrm{d}x=||u||_{L^2(Omega)}^2$$ Am I right?
  $endgroup$
  – kim_8
  20 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  IIRC the trace theorem says that the last norm should be the $H^1(Omega)$ norm and there should be a proportionality constant but other than that I think it's fine
  $endgroup$
  – whpowell96
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, I forgot that.
  $endgroup$
  – kim_8
  11 hours ago
  

  
  
  
  

add a comment | 

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1

$begingroup$

I believe you will essentially have to reprove 7.1.2 a-c with this new boundary condition. If you let $frac{partial u}{partial n}=alpha(x,t)u$ on $partial Utimes[0,T]$, then you still get the weak formulation $(u',v) + B[u,v,t] = (f,v)$, but there are 2 big differences.

1. Your space is now $H^1(U)$, not $H_0^1(U)$, since functions can take on any value on $partial U$ (in the trace sense). This means that you have to be careful using Poincare's Inequality depending on what version you are used to seeing, as some forms of it only apply to function that are 0 on the boundary.




2. Your bilinear form is now $$B[u,v,t] = int_Usum_{i=1,j=1}^Na_{ij}(x,t)u_{x_i}v_{x_i}+sum_{i=1}^Nb_i(x,t)u_{x_i}v+c(x,t)uv dx+int_{partial U}alpha(x,t)uv dS.$$

If you look at the proofs of 7.1.2 b and c, they require bounds on $B$ that were derived in the section on elliptic equations for Dirichlet BCs (6.2.2). If you can prove similar bounds using this new bilinear form, you should be able to plug them into the proofs for 7.1.2 a-c and things will work out.

I haven't proved this for myself, but my intuition is that if you let $alphain:L^{infty}(partial U_T)$, then you can probably bound that surface integral by the $L^2$ norms of the functions and its contribution to the bound will just result in a larger constant $gamma$, but there is sure to be some subtleties in relating the boundary values of these functions and their norms and will definitely involve the constant from the trace operator somewhere.

I didn't provide a full answer but I hope this helped you understand what steps you can take and what tools to use.

edit: There is also a discussion of a similar question here: Poisson's equation with Robin boundary conditions

share|cite|improve this answer

edited yesterday

answered yesterday

whpowell96whpowell96

48819

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you so much for you answer. So I see that if I use the proof in the link that you have provided, the proofs from Evans's book would be valid also in my case. Is that correct?
  $endgroup$
  – kim_8
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I didn't look super carefully at the answer I linked to see if it was complete or not, but I believe that if follow 7.1.2 a-c, the only properties specific to the operator/BC are the bounds they use so once they are established (as I believe the linked thread is doing), everything is the same
  $endgroup$
  – whpowell96
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you again for your answer. And one last question. To obtain some energy estimates, I suppose that if I want a bound for integrals like $$int_{partial Omega} (Tr u)^2 mathrm{d}x=||Tr u||_{L^2(partial Omega)}^2$$ I can use the trace theorem and conclude that $$int_{partial Omega} (Tr u)^2 mathrm{d}x=||Tr u||_{L^2(partial Omega)}^2 leq int_{Omega} u^2 mathrm{d}x=||u||_{L^2(Omega)}^2$$ Am I right?
  $endgroup$
  – kim_8
  20 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  IIRC the trace theorem says that the last norm should be the $H^1(Omega)$ norm and there should be a proportionality constant but other than that I think it's fine
  $endgroup$
  – whpowell96
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, I forgot that.
  $endgroup$
  – kim_8
  11 hours ago
  

  
  
  
  

add a comment | 

1

$begingroup$

I believe you will essentially have to reprove 7.1.2 a-c with this new boundary condition. If you let $frac{partial u}{partial n}=alpha(x,t)u$ on $partial Utimes[0,T]$, then you still get the weak formulation $(u',v) + B[u,v,t] = (f,v)$, but there are 2 big differences.

1. Your space is now $H^1(U)$, not $H_0^1(U)$, since functions can take on any value on $partial U$ (in the trace sense). This means that you have to be careful using Poincare's Inequality depending on what version you are used to seeing, as some forms of it only apply to function that are 0 on the boundary.




2. Your bilinear form is now $$B[u,v,t] = int_Usum_{i=1,j=1}^Na_{ij}(x,t)u_{x_i}v_{x_i}+sum_{i=1}^Nb_i(x,t)u_{x_i}v+c(x,t)uv dx+int_{partial U}alpha(x,t)uv dS.$$

If you look at the proofs of 7.1.2 b and c, they require bounds on $B$ that were derived in the section on elliptic equations for Dirichlet BCs (6.2.2). If you can prove similar bounds using this new bilinear form, you should be able to plug them into the proofs for 7.1.2 a-c and things will work out.

I haven't proved this for myself, but my intuition is that if you let $alphain:L^{infty}(partial U_T)$, then you can probably bound that surface integral by the $L^2$ norms of the functions and its contribution to the bound will just result in a larger constant $gamma$, but there is sure to be some subtleties in relating the boundary values of these functions and their norms and will definitely involve the constant from the trace operator somewhere.

I didn't provide a full answer but I hope this helped you understand what steps you can take and what tools to use.

edit: There is also a discussion of a similar question here: Poisson's equation with Robin boundary conditions

share|cite|improve this answer

edited yesterday

answered yesterday

whpowell96whpowell96

48819

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you so much for you answer. So I see that if I use the proof in the link that you have provided, the proofs from Evans's book would be valid also in my case. Is that correct?
  $endgroup$
  – kim_8
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I didn't look super carefully at the answer I linked to see if it was complete or not, but I believe that if follow 7.1.2 a-c, the only properties specific to the operator/BC are the bounds they use so once they are established (as I believe the linked thread is doing), everything is the same
  $endgroup$
  – whpowell96
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you again for your answer. And one last question. To obtain some energy estimates, I suppose that if I want a bound for integrals like $$int_{partial Omega} (Tr u)^2 mathrm{d}x=||Tr u||_{L^2(partial Omega)}^2$$ I can use the trace theorem and conclude that $$int_{partial Omega} (Tr u)^2 mathrm{d}x=||Tr u||_{L^2(partial Omega)}^2 leq int_{Omega} u^2 mathrm{d}x=||u||_{L^2(Omega)}^2$$ Am I right?
  $endgroup$
  – kim_8
  20 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  IIRC the trace theorem says that the last norm should be the $H^1(Omega)$ norm and there should be a proportionality constant but other than that I think it's fine
  $endgroup$
  – whpowell96
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, I forgot that.
  $endgroup$
  – kim_8
  11 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

I believe you will essentially have to reprove 7.1.2 a-c with this new boundary condition. If you let $frac{partial u}{partial n}=alpha(x,t)u$ on $partial Utimes[0,T]$, then you still get the weak formulation $(u',v) + B[u,v,t] = (f,v)$, but there are 2 big differences.

1. Your space is now $H^1(U)$, not $H_0^1(U)$, since functions can take on any value on $partial U$ (in the trace sense). This means that you have to be careful using Poincare's Inequality depending on what version you are used to seeing, as some forms of it only apply to function that are 0 on the boundary.




2. Your bilinear form is now $$B[u,v,t] = int_Usum_{i=1,j=1}^Na_{ij}(x,t)u_{x_i}v_{x_i}+sum_{i=1}^Nb_i(x,t)u_{x_i}v+c(x,t)uv dx+int_{partial U}alpha(x,t)uv dS.$$

If you look at the proofs of 7.1.2 b and c, they require bounds on $B$ that were derived in the section on elliptic equations for Dirichlet BCs (6.2.2). If you can prove similar bounds using this new bilinear form, you should be able to plug them into the proofs for 7.1.2 a-c and things will work out.

I haven't proved this for myself, but my intuition is that if you let $alphain:L^{infty}(partial U_T)$, then you can probably bound that surface integral by the $L^2$ norms of the functions and its contribution to the bound will just result in a larger constant $gamma$, but there is sure to be some subtleties in relating the boundary values of these functions and their norms and will definitely involve the constant from the trace operator somewhere.

I didn't provide a full answer but I hope this helped you understand what steps you can take and what tools to use.

edit: There is also a discussion of a similar question here: Poisson's equation with Robin boundary conditions

share|cite|improve this answer

edited yesterday

answered yesterday

whpowell96whpowell96

48819

$endgroup$

share|cite|improve this answer

edited yesterday

answered yesterday

whpowell96whpowell96

48819

answered yesterday

whpowell96whpowell96

48819

answered yesterday

whpowell96whpowell96

48819