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Lagrangian multipliers can be used to mini-/maximize a multivariable function $f()$ subject to one or multiple constraints. Dedner et al. used the technique of generalized Lagrangian multipliers (GLM) to minimize the divergence of the magnetic field $\vec{B}(x,y,z)$ in the ideal MHD equations.

They transformed the equations \begin{equation} \begin{array}{rcl} \displaystyle \nabla\cdot\vec{B} &=& 0 \,,\\ \displaystyle \frac{\partial}{\partial t}\vec{B} &=& \nabla\times\left(\vec{u}\times\vec{B}\right)\, , \end{array}\tag{1} \end{equation} into the form \begin{equation} \begin{array}{rcl} \displaystyle \nabla\cdot\vec{B} + \mathcal{D}(\psi) &=& 0 \,,\\ \displaystyle \frac{\partial}{\partial t}\vec{B} + \nabla \psi &=& \nabla\times\left(\vec{u}\times\vec{B}\right)\, . \end{array}\tag{2} \end{equation}

Suitable choices for the differential operator $\mathcal{D}(\psi)$ seemingly lead to a scheme that removes the divergence in the field $\vec{B}$, e.g. \begin{equation}\tag{3} \mathcal{D}(\psi) = -\frac{1}{c^2}\frac{\partial}{\partial t}\psi\, , \end{equation} with another newly introduced parameter $c$.

My question is specifically not about the involved physics, but the mathematical background (why is it allowed to do this?). I'd really love to be able to understand the underlying math to get a better understanding of what is happening here. However, I have never seen the concept of GLM being used like this. It is entirely unclear to me how you get from (1) to (2) and also why you would have to introduce yet another variable "$c$" in (3).

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    $\begingroup$ Is it the same $\vec B$ in (1) and (2)? And what divergence is removed when $\vec B$ is already divergence free in (1)? $\endgroup$ – md2perpe Aug 3 '17 at 15:41
  • $\begingroup$ Yes, it is the same $\vec{B}$. If $\vec{B}$ is already divergence free, then the first equation of (2) will be $0 = \frac{\partial}{\partial t}\psi$ and hence $\psi$ will stay constant in time. Assuming $\psi$ is initialized with $\psi_0 = 0$, then its gradient, $\nabla \psi$, will vanish and (2) will in fact be identical to (1). $\endgroup$ – MrD Aug 3 '17 at 19:05
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    $\begingroup$ Could you show their calculations? I'm not going to buy the paper. $\endgroup$ – md2perpe Aug 3 '17 at 21:46
  • $\begingroup$ @md2perpe I can only understand this. You may find it here. See "Relevant publications" -> "Hyperbolic Divergence Cleaning for the MHD Equations" $\endgroup$ – MrD Aug 4 '17 at 10:32
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I don't think equation (1) is equivalent to (2). What they do in that paper is introducing equation (2), which has an additional unknown, the function $\psi$. The advantage in doing so is that equation (2) maybe simpler to solve or simpler to study numerically. The reason why they consider the parameter $c$ in the hyperbolic choice (3) is that when $c\to\infty$ equation (2) converges to equation (1), and so solutions of (3) should converge to solutions of (1). This is what I think they mean with the sentence "We try to choose $\mathcal{D}$ and the initial and boundary conditions for $\psi$ in such a way that a numerical approximation to (4), (5) is a good approximation to the original system (1c), (2)". If you look at the reference [35] on Proposition 2 in the last page, they consider a corrector $\mathcal{D}$ of the form $\varepsilon \partial_t p$ and then prove that as $\varepsilon\to 0$ solutions of the $\varepsilon$ problem converge to solutions of the original problem.

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  • $\begingroup$ So you would agree to the statement that this is not a real application of the method of Lagrangian multipliers, but they only name it like this as (a) they introduce a new field and (b) they are so-to-say looking for an extremum (minimal $\nabla\cdot\vec{B}$)? $\endgroup$ – MrD Aug 6 '17 at 11:26
  • $\begingroup$ Well, yes and no. They explained the name in the introduction. If you take a PDE and add a constraint, say, $\int_\Omega B\,dx=c$, then by the Lagrange multiplier theorem you need to modify your PDE by adding a constant to it. So this is a generalization, since instead of adding a constant, you are adding a function to the PDE. $\endgroup$ – Gio67 Aug 6 '17 at 11:35

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