Hello Jean,
Thank you very much for your derivation. It does help clear up many
questions that I had. I greatly appreciate it!
The strong form that I will be using are simplified versions of the form
that you derived in the description.
- delta V = \rho^{f} / [\epsilon_{0} \epsilon_{r}]
becomes
- delta V = \rho / [\epsilon_{0}]
when f = 1 and when the material is free space (eplison_r = 1 in air).
I am starting off with a more simpler simulation to learn the ropes better.
For now, I am assuming that the charge density will be zero (\rho). The
strong form then becomes:
- delta V = 0
Later, I plan on expanding this for the full delta V equation (- delta V =
\rho^{f} / [\epsilon_{0} \epsilon_{r}]). Since I am just starting out, I
would like to keep things "simple".
I was originally thinking to obtain the electric field (*e* = -grad V)
separately. First, i would solve the simulation for V using the equation -
delta V = 0. Then solve for the electric field. But, I do not know how I
can do this in deal.ii. In fact, I wasn't even sure if this was possible.
Am I able to solve for the electric field using the results I obtained when
I run the solver for the voltage? If so, would you be able to point me to
some resources on how I can implement this using deal.ii?
Again, at the time, I didn't think this was possible. So, I decided to
treat the equations as 2 separate equations and solve for them at the same
time. Which lead me to where I am at currently.
Right now, I am not concerned about solving the electric displacement.
Unless, it is necessary for me to solve this.
You have also confirmed my idea that solving for
- delta V = 0
div E = 0
will not work since they are the same equation.
Basically, from the description that you provided, I should solve for the
voltage first and then solve for the electric field using the results I
obtained from the first simulation. If this is the case, would you be able
to point to me some resources that I can use in order to implement this in
deal.ii? Concerning the continuity equations, should I check my results
against them to make sure the results are valid?
Thank you
On Wednesday, September 5, 2018 at 3:17:14 AM UTC-4, Jean-Paul Pelteret
wrote:
>
> Dear Philip,
>
> It looks to me like you’re mixing things up a little bit. In all of the
> equations that you presented there is a mixture of governing equations,
> assumed relationships and the consequence of combining them. From what I
> can tell you’re actually trying to use the same equation more than once
> which is problematic. (I hope that I have clearly understood what you’re
> trying to do, otherwise what follows is a bit of a waste of space and time…)
>
> I think that if you work from the irrefutable, then there should be little
> confusion. So let me see if I can help by sketching out electrostatics from
> the basics, as I know it:
>
> The starting point is Maxwell’s equations. Let’s immediately simplify
> things by saying that we already assume a quasi-static state — this means
> that the electric and magnetic fields are now decoupled. We then get
> (1) div *d* = \rho^{f} [Gauss’ law]
> (2) curl *e* = *0* [Faraday’s law]
> on B. Here *d* is the electric displacement vector, *e* is the electric
> field vector, and \rho^{f} is the free charge density.
>
> There is a further fundamental constitutive law that links *d *and *e,*
> namely
> *(*3) *d* = \epsilon_{0} \epsilon_{r} *e*
> where \epsilon_{0} is the electric permittivity of free space and
> \epsilon_{r} is the relative permittivity of the material. Here I have
> assumed that this material is linearly polarisable, so that is that the
> polarisation is in the same direction as the electric field (that’s where
> the scaling factor \epsilon_{r} comes from).
>
> There are also the continuity conditions for Maxwells equations, which for
> this simplified scenario would be
> (4) *n* x [[*e*]] = *0*
> (5) *n* . [[*d*]] = 0 (assuming no surface charges)
> on dB. Here *n* is a surface normal and [[ ]] denotes the jump of a
> quantity, “.” the dot product and “x” the cross product.
>
> So, going back to Maxwell’s equations, you now need to choose which
> variable is going to be the primary variable, and which one follows from it
> (so use (3) in (1) or (2)). Since in general you wish to include source
> terms \rho^{f}, it is appropriate to choose the electric field to be the
> primary variable. what comes next is that you substitute (3) into (1)
> thereby eliminating the electric displacement from the equation:
> (6) div *[*\epsilon_{0} \epsilon_{r} *e*] = \rho^{f}
> —>
> (7) div *[**e*] = \rho^{f} / [\epsilon_{0} \epsilon_{r}]
>
> So now (7) represents an amended form of (1) that needs to be solved, but
> we still need to satisfy (2). So what we do is to exploit the identity
> (a) curl (grad (s)) = *0 *
> for all arbitrary scalars s. So we postulate the existence of an electric
> scalar potential field V that is linked to the electric field by
> (8) *e* = -grad V
> Using the identity (a), you can see that this satisfies (2):
> (b) curl *e* = curl (- grad V) == *0*
>
> Now we can put (8) into (7), knowing that if we solve
> (9) div *[*-grad V] = \rho^{f} / [\epsilon_{0} \epsilon_{r}]
> —>
> (10) - delta V = \rho^{f} / [\epsilon_{0} \epsilon_{r}]
> for V then we automatically satisfy (2). Remember that (10) is nothing
> other that (1) in disguise. So this actually satisfies both of Maxwell’s
> electrostatic governing equations at the same time (since we solve for the
> electric scalar potential).
>
> Now, about those continuity conditions… So when you discretise the field V
> using continuous finite elements, then (4) is automatically satisfied
> because the continuity of the solution ensures that there is no jump in the
> solution V, and therefore no tangential jump in grad V. You can see this
> from
> (11) *0 *= *n* x [[*e*]] =(8)= *n* x [[-grad V]] = *n* x grad [[V]]
> Finally, (5) is satisfied automatically through this same assumption
> (12) 0 = *n* . [[*d*]] =(3)= *n* . [[\epsilon_{0} \epsilon_{r} *e*]]
> =(8)= \epsilon_{0} \epsilon_{r} *n* . [[-grad V]] = -\epsilon_{0}
> \epsilon_{r} *n* . grad[[V]]
>
> So, in summary, the strong form of the governing equation to be
> implemented is (10), with the solution for V leading to the electric field
> by (8) and subsequently the electric displacement by (3). The continuity
> conditions (4,5) are satisfied if the solution for V is continuous.
>
> Does this make sense, and does it help clear up things?
>
> Best regards,
> Jean-Paul
>
> On 05 Sep 2018, at 04:59, phillip mobley <[email protected]
> <javascript:>> wrote:
>
> Hello Wolfgang,
> Thank you for your reply. What you described above is the first approach
> that I took. There are some differences but overall, it is the same thing.
> I am not familiar with the form:
>
> div E - Delta V = 0
>
> but, I was able to figure out how you came this. Instead, I kept the
> equations separate and applied the technique described in lecture 19. For
> reference, the equations that I am referring to are:
>
> -grad E = 0
>
> delta V = 0
>
> where E is the electric field vector.
>
> I saw a new approach that I could take which I will describe here. Coming
> back to the original equations:
>
> delta V = 0
>
> -grad V = E
>
> And then applying lecture 19 technique:
>
> (delta phi_V, V) - (grad phi_V, V) = (E, phi_V)
>
> However, I run into the problem which I believe you briefly touched upon
> on your answer which is I am multiplying by a single test function; phi_V.
> So I am also thinking that this is the wrong approach.
>
> There is a third equation that I have not touched yet which is the curl E
> = 0. If I were to account for this equation, then this would become the
> third approach to the problem. Thus the "complete" set of equations would be
>
> delta V = 0
>
> -grad V = E
>
> curl E = 0
>
> If curl E = 0 is taken into account, then I would be working with the form:
>
> (delta phi_V, V) - (grad phi_V, V) + (curl phi_E, E) = (phi_V, E)
>
> Now, it would seem that i am working with two test functions, phi_V and
> phi_E. So it would appear that this form is more appropriate to work with.
> Is this the direction that I should be taking or is there another approach
> that I should consider?
>
> On Tuesday, September 4, 2018 at 5:25:43 PM UTC-4, Wolfgang Bangerth wrote:
>>
>>
>> > So far, I have found that there are two paths that I can take to get my
>> > bilinear form. I have already attempted one path and I ran into a
>> compile
>> > error with Deal.II. So, I am thinking that this bilinear form may not
>> be
>> > appropriate.
>>
>> That's not necessarily the right conclusion. Just because you chose the
>> wrong
>> syntax (= compiler error) does not mean that you chose the logically
>> wrong
>> approach.
>>
>>
>> > But I wanted to turn to the community here to see if my bilinear
>> > form has an issue or if my code has an issue.
>> >
>> > For the first path, since V and E are my unknowns, I wanted both V and
>> E to be
>> > on the left-hand side of their PDEs. With a bit of substitution, the
>> equations
>> > that I would be solving turned into this:
>>
>> I don't see your inlined equations in my mail program, but assuming that
>> you
>> are talking about the equation
>>
>> (div E, phi_E) - (grad phi_V, grad V) = 0
>>
>> then this is not correct. If you start out from the equation
>> div E - Delta V = 0
>> then you will want to multiply this (scalar) equation with a test
>> function and
>> integrate by parts where necessary. But it's *one* test function, so you
>> need
>> to end up multiplying *both* terms with the same test function phi_V, not
>> different test functions for the two terms.
>>
>> That said, here are two questions:
>> * In essence, you want to solve the Laplace equation for V, but derive a
>> mixed
>> form, which with your variables is usually stated as
>> E + grad V = 0
>> whereas you choose
>> div E + Delta V = 0
>> This raises the question of why you want to do it this way? You are
>> requiring more differentiability than necessary, negating the
>> advantages of
>> the mixed form.
>>
>> * If you chose to go with the formulation you have, I suspect that you
>> want to
>> integrate by parts:
>> - (E, grad phi_V) - (grad phi_V, grad V) = 0
>> so that all derivatives are on the V test functions, not on E. Any
>> good
>> reasons not to do so? I'm not sure that your formulation is uniquely
>> solvable at all.
>>
>> Best
>> W.
>>
>> --
>> ------------------------------------------------------------------------
>> Wolfgang Bangerth email: [email protected]
>> www: http://www.math.colostate.edu/~bangerth/
>>
>>
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