@AnonSubmitter85 I am not sure, since Philip Roe seemed to have other questions related to Hermite interpolation, finite elements, von Neumann analysis and so on, but since I am not sure I answered only the question I could make sense of. However based on the comments I am assuming that it is something related to Hermite interpolation elements and their approximation qualities for numerical solutions of PDEs, and specifically some kind of Fourier analysis of said error. Without more details I can't say anything though, and even with more details I may still be unable to.

@lightxbulb. You are correct(ish) There are surprising differences of expectation between apparently similar fields. In signal processing errors may already have occurred in measuring the signal. In numerical PDEs the data is fine, but errors accumulate as the solution proceeds. In computer graphics we often require an aesthetically pleasing surface. In PDEs we often deal with non-smooth behavior. Hermite methods are almost unused in hyperbolic PDEs but I think they may have merit. Before getting in too deeply I wanted to find what was known from other fields. Your guesses were helpful.

@PhilipRoe Hermite elements should be perfectly applicable to hyperbloic PDEs. For example onsider as a model problem the wave equation $\partial_{tt} u = c^2\Delta u$. Compared to Lagrange elements Hermite elements allow you to prescribe derivative information at vertices: useful to implement boundary conditions involving $\partial_{n} u = g$ or to prescribe derivative constraints within the domain. In 2D this generalizes in many different ways: see Hermite, Thomas-Raviart, Nedelec, Morley, Bell, and Argyris elements. The latter three can be used for PDEs with $\Delta^2$.

@PhilipRoe As far as Shannon-like sampling theorems go - I don't think those are very useful for FEM solutions of PDEs. The latter cares more about the smoothness of your solution space (that's why one typically studies error convergence in terms of the polynomial degree), while Shannon is concerned with when one can reproduce a continuous function exactly using a sinc interpolant, or trigonometric polynomials in the periodic case. So Shannon-like stories may be useful in FEM analysis if you e.g. use trigonometric polynomials for your finite-dimensional subspace.

@lightxbulb Thank you for these remarks. You are speaking largely from the viewpoint of second-order PDEs. I prefer first-order PDEs as more descriptive of the physics. Currently people are looking at high-order FD and FE methods that are very accurate at low frequencies. In practice, breakdown occurs at high frequencies and I think that Hermite methods may have some advantage there. I am looking at the model advection problem $\partial_tu+a\partial_xu=0$ by interpolating in a Hermite spline that incorporates p derivatives. I have used MAPLE to find the dispersion relationship.

@PhilipRoe Transport equations are trickier I agree. In FDM they require upwinding/downwinding schemes. I haven't tackled those in an FEM setting, but at least in 1D with linear Lagrange elements on a regular grid it should result in essentially the same thing as for FDM if you implement it correctly. It's just about in which direction you discretise $\partial_x$ so that it can actually "see" what must be propagated. I don't think that Hermite elements should change much w.r.t. the above logic - you'll just have more degrees of freedom.

@lightxbulb I have to solve a polynomial of order p+1. For p=0,1,2,3 I find accuracy 2p+1. I can choose roots of the polynomial that collectively provide plausible behavior up to frequencies $(p+1)\pi$, but my gut tells me that this is too good to be true. I was very interested in the Papoulis result, which I was not aware of, and gives theoretical support to my tentative findings. My student already has some remarkable results with simplex elements in 2D, and is planning 3D, but I feel a need to also find better foundations in 1D

@lightxbulb The conventional wisdom is that because hyperbolic problems have discontinuous solutions, they are best tackled with discontinuous elements, but this becomes a source of difficulty in more than 1D. The Hermitian advantage is that the richer information allows minimal stencil size, and "captured" discontinuities are actually very compact with only slight overshooting,

What do you mean by you have to solve for a polynomial of order p+1? You mean for prescribing the elements or what?

Don't you just spatially discretise d_x in d_t u + ad_x u = 0 with FEM and just integrate over time? Are you talking about the matrix assembly process?

I have seen a lot of people use FVM for hyperbolic problems, because it models the fluxes naturally, but as mentioned I am not really familiar with the literature. You can probably ask a specific question on the computation science stackexchange - there are many people that are much more knowledgeable than me there, e.g. Wolfgang Bangerth is on there.

In fact I am one of the FVM experts myself! I am devising a new methods called Active Flux that started as an offshoot of FVM but is developing connections to FEM. Probably best to think of it as FDM with no doctrinal attachments.

The Hermitian aspect is a recent thought. The simplest version is this. At irregularly spaced points $x_j$ I have a function and its first $p$ derivatives. I think of these as a vector $\vec{u}$ of length $p+1$ whose evolution is governed by the system $\partial_t\vec{u}+aI\partial_x\vec{u}=0$ with the Jacobian matrix $aI$.

In each interval $j,j+1$ I form the Hermite spline using $2(p+1)$ pieces of information to define a polynomial of order $2p+1$. I interpolate in this polynomial to the foot of the characteristic $dx/dt=a$ to find $\vec{u}$ at $j+1$ at the next time level.

so you solve a transport equation for the p+1 derivatives of the function?

(counting from zero)

Correct.

is this what you mean by finding the roots?

getting the polynomial in terms of the 2(p+1) pieces of information?

Again correct. I think you are ahead of me!

you can probably just use the Hermite basis functions if you're interpolating the derivatives

they have them on wikipedia for cubic hermite, but it's not too hard to derive them for higher orders too

I am guessing it's not too hard to analytically integrate them either

so your solution space is essentially a spline of order 2p+1

the smoothness depends on how you prescribe the junctions

if you have a small enough problem you can also compute the spectral solution explicitly by using the eigendecomposition of ID_x, i.e. u(t) = exp(-taID_x)u(0)

Yes that is what I do. I assume that $\vec{u}_j^n=G-tothe-n\exp(ij\theta}\vec{U}$ and need the eigenvalues and eigenvectors of the $(p+1)\times(p+1)$ matrix G. THis is where the polynomial come in.

So were you referring to the characteristic polynomial then?

Maple has been able to do this for $p=0, 1, 2, 3$ Maybe my earlier post makes more sense now. I actually think that large $p$ is not practical but as a springboard into higher dimensions I want to understand 1D better.

So you are referring to the characteristic polynomial?

Yes, The matrix $G$ has elements that are polynomial in the Courant number.

Afaik eigendecomposition of discrete derivative operators is not nice. The svd is easy enough to compute (on a regular grid), but the eigendecomposition is a mess. I would just suggest using a numerical method to compute the eigendecomposition.

In practice on larger problems one would just use timestepping methods as you mentioned. Have you compared this to e.g. Lagrange polynomials?

Both Hermite and Lagrange would span the same polynomial space afaik if you match the degrees, but the latter are probably more unwieldy for prescribing derivatives. So I guess you picked Hermite because of that, which makes perfect sense.

Compared with Lagrange it is MUCH better! What interests me is whether the ability to compute high frequencies is really as good as the naive dispersion analysis suggests. If it is, this could be a gateway into turbulence.

Since both Lagrange and Hermite give you the same polynomial, I wouldn't expect one to be intrinsically better though - you're projection over the same Galerkin space, only the basis is different, no?

E.g. if I have 2 derivatives and 2 values at the endpoints then I get a cubic polynomial. But I can get a cubic polynomial by also prescribing 2 values at the endpoints and 2 values somewhere on the inside

Then since both result in a cubic polynomial the spanned space should be exactly the same unless I am missing something

The difference is the size of the stencil. Lagrange requires either I use points outside the interval (ie outside the true domain of dependence) or crowded inside the interval (which reduces the timestep) I am not (yet) intersted in the derivative for its own sake, just as a device to find the values.

aha, makes sense

so it's the same polynomial space, but you get a different space discretisation, I think I am starting to get it

There seems to be a bit of a minefield associated with finding which polynomial root is

I talked with a guy that studies spectral decomposition of Toeplitz matrices recently, and he was pretty pessimistic about odd derivative operators

the one you want, and what the others may mean. I can pick roots that tell a nice story, but it may not be true.

we're still talking about the characteristic polynomial of the derivative matrix right?

yes

what do you mean by pick a root you want?

you need all roots for the eigendecomposition, or do you just use a reduced version?

here's the page of the person studying symbols of toeplitz matrices: 2pi.se/publications

I should get an eigenvalue $g(\theta)$ equal to 1,0 when $\theta=0$ remaining <1 for all theta, but I expect it to taper off like $\theta^k$ for some $k$. If I plot all roots I can trace through them a path that does what I expect. If I just have a quadratic and alwayswant to take one sign or the other.

I don't know what you mean

At higher order the "plausible" root hops from branch to branch but I dont know if this means anything. I will start experiments soon, but hoped for a bit more understanding before I begin.

What do you mean by plausible root? Are you not solving this numerically?

$g(theta)$ is the complex amplification after one timestep. Its amplitude should be 1.0

and its phase should be $-CFL\theta$. Only one root starts off like this., but that root does not continue as expected. The expected trajectory is picked up by another root.

I can't say anything since this seems pretty involved and specific. I also have to get to sleep so I have to go. It was nice talking to you!

Yes I have enjoyed this too, despite our linguistic separation! THank you for the Papoulis reference. It was what I was hoping for. Sleep well!