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I work in a field (elasticity reconstruction) that frequently uses standard Finite Element methods to solve the first order PDE, where we are given u and solve for mu and lambda:

$$ (\mu(u_{i,j} + u_{j,i}))_{,j} + (\lambda~div~\mathbf{u})_{,i} = -\rho \omega^2\mathbf{u}_{i} $$

known as the Navier-Lame equation. Here we try to solve the "inverse" case (given u solve for mu and lambda) which is a first-order PDE (while the equation is more commonly solved for u which is a second order PDE).

The results overall have been a bit underwelming. Over the past year I was able to speak briefly about this problem to four quite prominent applied mathematicians. Their consensus was that standard FEM, while appropriate for elliptic PDEs of second order, is an inappropriate and unstable mathematical basis for a first order PDE, while better solutions might be Finite Volume or discontinuous Galerkin.

All I remember of the discussions is that first order PDEs have characteristic curves and can handle jumps, while finite elements, like second order PDEs, are smooth in C1 throughout.

I cannot find any supporting information for this viewpoint in a text. Can anyone explain why standard FEM is a poor choice for a first order PDE? Or is this not right?

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    $\begingroup$ Although I am not an expert in first-order PDE's, I often point people to Ern, Guermond. Theory and Practice of Finite Elements (2004) chapter 5 which is called First order PDEs. They discuss some difficulties that finite element methods encounter when approximating first-order problems. They also discuss how these difficulties have been fixed in some cases. $\endgroup$ – knl Sep 14 '17 at 10:50
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    $\begingroup$ It should be pointed out that "finite element methods" covers a very broad class of methods, of which the standard piecewise linear $H^1$ conforming approach for second order elliptic PDEs is only one (albeit very well known) example -- there are others that are specifically developed for first-order PDEs. Besides the excellent reference that @knl mentioned, I'd recommend the book by Di Pietro and Ern, Discontinuous Galerkin Methods. $\endgroup$ – Christian Clason Sep 14 '17 at 11:46
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    $\begingroup$ Also, I'd be amiss if I didn't point out that this is not at all the way to solve inverse problems -- it's horribly unstable and will blow up in your face. Please, google "parameter identification" and "(Tikhonov) regularization". (Briefly, you want to treat this as a least squares problem, not an equation.) $\endgroup$ – Christian Clason Sep 14 '17 at 11:48
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    $\begingroup$ @ChristianClason I think you should turn your comment into an answer, arguably that's more the source of OP's difficulty than how to discretize first-order PDE. $\endgroup$ – Daniel Shapero Sep 14 '17 at 16:13
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    $\begingroup$ I edited the title, since you mention that discontinuous Galerkin methods (which are finite element methods!) were recommended to you for this problem, and also to indicate that the issues involved are not necessarily generic to all first-order PDEs. $\endgroup$ – David Ketcheson Sep 15 '17 at 5:47
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The issue with finite elements in the current context is not that you have a first order differential equation, but with the kind of first order equation you have. In general, finite element methods can be made to perform quite well for first order problems such as advection equations. But that's beside the point here.

To illustrate the issue, think of the simpler problem of identifying the coefficient $a(x)$ from the equation $$ -\nabla \cdot (a(x) \nabla u(x)) = f(x) $$ assuming that you know $f(x)$ and $u(x)$. It is true that you can write this as a first order equation for the unknown $a(x)$: $$ \beta \cdot \nabla a + \gamma a = f(x) $$ where $\beta(x) = -\nabla u(x)$ looks a lot like an advection direction, and $\gamma(x)=-\Delta u(x)$ like a reaction coefficient. So this looks like an advection-reaction equation.

But the solution of this problem is not well posed unless (i) $|\beta|\ge \beta_\text{min}>0$, i.e., the "wind speed" is always bounded away from zero, (ii) unless each point $x$ in the domain is connected by a "streamline" of the vector field $\beta$ to the boundary, and (iii) you have boundary values for $a(x)$ at all inflow boundary parts, i.e., where $\beta\cdot n<0$.

But, in general cases, none of these three conditions are satisfied. You can easily see this if you consider a circular domain and you have $f=1$ in the outer half of the circle and $f=-1$ in the inner half (and let's assume $a=1$ everywhere). Then your solution $u(x)$ will be rotationally symmetric, and from the boundary first rise up and the decrease again -- i.e., you'll have a circular ring. In this case, each point outside the rim of the hill is connected to the boundary by a streamline, but not on the inside of the hill. Furthermore, $\nabla u=0$ along the rim of the circular hill.

In situations like this, the first-order PDE is simply not a well posed problem for $a(x)$. No method will be able to yield a satisfactory solution in such cases, and that includes the finite element method.

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  • $\begingroup$ I suppose there is no unique solution to the inverse problem. But one can add an artificial time derivative $a_t + \beta\cdot\nabla a + \gamma a = f$ and then solve this to steady state. You can specify $a$ at inflow boundaries where $\beta \cdot n < 0$. $\endgroup$ – cpraveen Sep 16 '17 at 3:53
  • $\begingroup$ @PraveenChandrashekar -- the time derivative now also requires you to pose initial conditions; I'm not sure that's going to make things easier. As for specifying boundary conditions: the question is of course what you are going to specify. There is no information in the problem that would enable you to choose one choice over another. $\endgroup$ – Wolfgang Bangerth Sep 16 '17 at 19:06
  • $\begingroup$ I'll add that adding the time derivative does not resolve the issue at points where $\beta=0$. At those points, the steady state solution of the time dependent problem is either zero or $\infty$, depending on the sign of $\gamma$, assuming for a moment that $f=0$ at these points. What this shows is that you've lost the ability to control the solution at these points. $\endgroup$ – Wolfgang Bangerth Sep 16 '17 at 19:08

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