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From: Mark Blome <mblome@au...>  20050110 21:10:25

Hi Michael, I am using libmesh to solve for the electric potential of a current source within a 3d cubic domain. The calculations are used for a geoelectrical forward solver. On the top side of the domain, I apply a Neuman boundary condition (du/dn=0) and on the other sides I use Dirichlet boundary conditions (u=0). The second term div( (sigmasigma0) grad Un) is added to remove the singularity arising due to the point current source. When I discretize the second term with Fe(i) +=  (sigmasigma0) * JxW[qp] * (gradVn[qp] * dphi[i][qp]) (of course, as John noticed, I forgot to mention the surface integral term, but in my case it is zero anyway cause sigma=sigma0 at the outer boundary. ) I get wrong results: The resulting potential U is the potential that does not contain the singular potential Un=1/(2*pi) *1/r anymore. For the case of a homogenous cube with a high conducting cube centered inside, the introduction of the discretization above results in a strange dipole field within the high conducting cube, which physically doesnt make any sense. Thanks for the help so far, Mark Am Montag 10 Januar 2005 19:28 schrieb Michael Schindler: > Hi Mark, > > On 10.01.05, Mark Blome wrote: > > Hi everybody, > > > > I am trying to descretize a laplace equation which contains a second > > term. In my opinion the problem should be very easy, but I become more > > and more confused the more I think about it. The laplace equation reads: > > div(sigma grad U) + div( (sigmasigma0) grad Un) =0 > > where Un is a potential for which the equation is known ( Un=j0/(2 * PI * > > sigma0) * 1/r), hence the term has to go to the right hand side. > > > > For the term div(sigma grad U)=0 I simply get matrix terms: > > Ke(i,j) +=  sigma * JxW[qp]*(dphi[i][qp]*dphi[j][qp]); > > But what to add for div( (sigmasigma0) grad Un)? > > The FEM method is to find a weak solution with a testfunction phi > > int div(sigma grad U) phi = int div( (sigmasigma0) grad Un) phi > > Depending on sigma you can use the Galerkinansatz where phi is the > same function you approximate U with. If sigma is constant, omitting > the surface terms. > >  sigma int grad(U)*grad(phi) =  (sigmasigma0) int grad(Un)*grad(phi) > > This is what you tried first. > > > I tried: > > Fe(i) +=  (sigmasigma0) * JxW[qp] * (gradVn[qp] * dphi[i][qp]) > > ; where gradVn[qp] simply is grad(Un) analytically calculated at the > > quadrature points. But it doesnt work like this. > > What actually does not work? What boundary conditions do you use? > > Michael. 
From: Mark Blome <mblome@au...>  20050110 21:10:25

Hi Michael, I am using libmesh to solve for the electric potential of a current source within a 3d cubic domain. The calculations are used for a geoelectrical forward solver. On the top side of the domain, I apply a Neuman boundary condition (du/dn=0) and on the other sides I use Dirichlet boundary conditions (u=0). The second term div( (sigmasigma0) grad Un) is added to remove the singularity arising due to the point current source. When I discretize the second term with Fe(i) +=  (sigmasigma0) * JxW[qp] * (gradVn[qp] * dphi[i][qp]) (of course, as John noticed, I forgot to mention the surface integral term, but in my case it is zero anyway cause sigma=sigma0 at the outer boundary. ) I get wrong results: The resulting potential U is the potential that does not contain the singular potential Un=1/(2*pi) *1/r anymore. For the case of a homogenous cube with a high conducting cube centered inside, the introduction of the discretization above results in a strange dipole field within the high conducting cube, which physically doesnt make any sense. Thanks for the help so far, Mark Am Montag 10 Januar 2005 19:28 schrieb Michael Schindler: > Hi Mark, > > On 10.01.05, Mark Blome wrote: > > Hi everybody, > > > > I am trying to descretize a laplace equation which contains a second > > term. In my opinion the problem should be very easy, but I become more > > and more confused the more I think about it. The laplace equation reads: > > div(sigma grad U) + div( (sigmasigma0) grad Un) =0 > > where Un is a potential for which the equation is known ( Un=j0/(2 * PI * > > sigma0) * 1/r), hence the term has to go to the right hand side. > > > > For the term div(sigma grad U)=0 I simply get matrix terms: > > Ke(i,j) +=  sigma * JxW[qp]*(dphi[i][qp]*dphi[j][qp]); > > But what to add for div( (sigmasigma0) grad Un)? > > The FEM method is to find a weak solution with a testfunction phi > > int div(sigma grad U) phi = int div( (sigmasigma0) grad Un) phi > > Depending on sigma you can use the Galerkinansatz where phi is the > same function you approximate U with. If sigma is constant, omitting > the surface terms. > >  sigma int grad(U)*grad(phi) =  (sigmasigma0) int grad(Un)*grad(phi) > > This is what you tried first. > > > I tried: > > Fe(i) +=  (sigmasigma0) * JxW[qp] * (gradVn[qp] * dphi[i][qp]) > > ; where gradVn[qp] simply is grad(Un) analytically calculated at the > > quadrature points. But it doesnt work like this. > > What actually does not work? What boundary conditions do you use? > > Michael. 
From: Michael Schindler <mschindler@us...>  20050110 18:28:34

Hi Mark, On 10.01.05, Mark Blome wrote: > > Hi everybody, > > I am trying to descretize a laplace equation which contains a second term. In > my opinion the problem should be very easy, but I become more and more > confused the more I think about it. The laplace equation reads: > div(sigma grad U) + div( (sigmasigma0) grad Un) =0 > where Un is a potential for which the equation is known ( Un=j0/(2 * PI * > sigma0) * 1/r), hence the term has to go to the right hand side. > > For the term div(sigma grad U)=0 I simply get matrix terms: > Ke(i,j) +=  sigma * JxW[qp]*(dphi[i][qp]*dphi[j][qp]); > But what to add for div( (sigmasigma0) grad Un)? The FEM method is to find a weak solution with a testfunction phi int div(sigma grad U) phi = int div( (sigmasigma0) grad Un) phi Depending on sigma you can use the Galerkinansatz where phi is the same function you approximate U with. If sigma is constant, omitting the surface terms.  sigma int grad(U)*grad(phi) =  (sigmasigma0) int grad(Un)*grad(phi) This is what you tried first. > I tried: > Fe(i) +=  (sigmasigma0) * JxW[qp] * (gradVn[qp] * dphi[i][qp]) ; > where gradVn[qp] simply is grad(Un) analytically calculated at the quadrature > points. But it doesnt work like this. What actually does not work? What boundary conditions do you use? Michael.  "A mathematician is a device for turning coffee into theorems" Paul Erdös. 
From: John Peterson <peterson@cf...>  20050110 15:02:26

Mark Blome writes: > > Hi everybody, > > I am trying to descretize a laplace equation which contains a second term. In > my opinion the problem should be very easy, but I become more and more > confused the more I think about it. The laplace equation reads: > div(sigma grad U) + div( (sigmasigma0) grad Un) =0 > where Un is a potential for which the equation is known ( Un=j0/(2 * PI * > sigma0) * 1/r), hence the term has to go to the right hand side. > For the term div(sigma grad U)=0 I simply get matrix terms: > Ke(i,j) +=  sigma * JxW[qp]*(dphi[i][qp]*dphi[j][qp]); > But what to add for div( (sigmasigma0) grad Un)? > I tried: > Fe(i) +=  (sigmasigma0) * JxW[qp] * (gradVn[qp] * dphi[i][qp]) ; > where gradVn[qp] simply is grad(Un) analytically calculated at the quadrature > points. But it doesnt work like this. Hi. This is pretty close! The second term in your equation above, div( (sigmasigma0) grad Un) generates two terms when you multiply by a test function: \int_{Omega} div ( (sigmasigma0) grad(Un) ) * phi = + \int_{dOmega} (sigmasigma0) ( grad(Un) * n ) * phi dS (Term 1)  \int_{Omega} (sigmasigma0) grad(Un) * grad(phi) dx (Term 2) So you will have to code something pretty much like the first part, with the associated boundary term, but just put it on the rhs. John 
From: Mark Blome <mblome@au...>  20050110 11:05:32

Hi everybody, I am trying to descretize a laplace equation which contains a second term. In my opinion the problem should be very easy, but I become more and more confused the more I think about it. The laplace equation reads: div(sigma grad U) + div( (sigmasigma0) grad Un) =0 where Un is a potential for which the equation is known ( Un=j0/(2 * PI * sigma0) * 1/r), hence the term has to go to the right hand side. For the term div(sigma grad U)=0 I simply get matrix terms: Ke(i,j) +=  sigma * JxW[qp]*(dphi[i][qp]*dphi[j][qp]); But what to add for div( (sigmasigma0) grad Un)? I tried: Fe(i) +=  (sigmasigma0) * JxW[qp] * (gradVn[qp] * dphi[i][qp]) ; where gradVn[qp] simply is grad(Un) analytically calculated at the quadrature points. But it doesnt work like this. Then I tried to express Un using the form functions, ie Un=sum ( Un_0 * phi_i ), which would result to Fe(i) =  (sigmasigma0) * JxW[qp] * ( (dphi[i][qp] * Vn[qp] + gradVn[qp] *phi[i][qp]) * dphi[i][qp] ) That doesnt work as well. I also tried Fe(i) = (sigmasigma0) * JxW[qp] * (dphi[i][qp] * dphi[i][qp]) * Vn[i]; which doesnt make so much sense to me and as I expected doesnt work eather. I think this is really an easy thing, but my thoughts seem to go in a completely wrong direction. I would be very happy for any comments on this, Mark 