TITLE: Periodic Automorphism Towers
QUESTION [24 upvotes]: In Scott's classic textbook on Group Theory, he asks:
Suppose that $G$ is a finite group. Is the sequence of isomorphism types
of the groups $Aut^{(n)}(G)$ for $n \in \mathbb{N}$ eventually periodic?
Here $Aut^{(2)}(G) = Aut(Aut(G))$ etc. Equivalently, is the sequence
$|Aut^{(n)}(G)|$ always bounded above?
It apparently remains opens whether the sequence of automorphism types of $Aut^{(n)}(G)$ is in fact always eventually constant. (A wonderful theorem of Wielandt says that if $G$ is a finite centerless group, then the sequence is eventually constant.) So I would like to ask:
Does there exists a finite group such that $Aut(G) \not \cong G$ but
$Aut^{(n)}(G) \cong G$ for some $n \geq 2$?
Edit: Joel has pointed out that my question is perhaps even open for infinite groups. This sounds like an interesting question which doesn't seem amenable to the standard tricks.

REPLY [8 votes]: As a follow up to my previous answer, I have computed all the automorphism series for groups of order up to 24 within the limits of what I could do with GAP. I will share some of what I found, throughout I will use the notation $(n,d)$ for the group of order $n$ with GAP library id $d$. For large groups not contained in the library, I will use $<r,n>$ to denote a group with minimum generating set of size $r$ and of order $n$, if I know only a bound for $r$, I will use $<\leq r,n>$ to denote that fact. We will say that a group $G$ stabilizes (at $k$) if $Aut^{(k-1)}(G) \not\simeq Aut^{(k)}(G) \simeq Aut^{(k+1)}(G)$. We will call a group $G$ stable if $G \simeq Aut(G)$.
I have verified that all groups of order up to 24 stabilize except for the following groups:
$(16,5)$, $(16,6)$, $(16,7)$, $(16,8)$, $(16,9)$, $(16,10)$, $(16,11)$, $(24,4)$, $(24,6)$, $(24,7)$, $(24,9)$, and $(24,10)$.
Of note is that many of these groups have the same automorphism group, hence their series is identical from $k=1$ onwards. Specifically, $(16,5)$, $(16,6)$, $(16,8)$, $(24,9)$, and $(24,10)$ all have $(16,11)$ as their automorphism group. The groups $(16,7)$, $(16,9)$ have the same automorphism group and so do the groups $(24,4)$, $(24,6)$, and $(24,7)$.
The following is a list of the groups I know to be stable, it is complete only up to order 24, and I give a Structure Description of the ones of order up to 24:
$(1,1) \simeq \mathbb{Z}_1$, $(6,1) \simeq S_3$, $(8,3) \simeq D_8$, $(12,4) \simeq D_{12}$, $(20,3) \simeq \mathbb{Z}_5 \rtimes \mathbb{Z}_4$, $(24,12) \simeq S_4$.
These are still in the library: $(40,12)$, $(42,1)$, $(48,48)$, $(54,6)$, $(110,1)$, $(144,183)$, $(336,208)$, $(384,5678)$, $(432,734)$, $(1152,157849)$.
These are too big to be in the library: $<2,40320> \simeq S_8$, $<4,442368> \simeq Aut^{(6)}((16,3))$.
These results give me two (general) ideas on how to attack this problem: One, analyze those groups for which I haven't been able to determine stabilization to see if I can find anything they have in common and use it show the conjecture is false. Two, Analyze the stable groups to see what causes them to be stable and use that knowledge to (somehow) show that every group must eventually stabilize. Both will likely require a detailed analysis of how $Aut(G)$ arises from $G$, to see how $G$ controls the properties of $Aut(G)$.
Edit:
I now have a complete list of stable groups of order up to 511, their GAP structure descriptions already reveal some very interesting patterns:
$(1,1)\simeq\mathbb{Z}_{1}$
$(6,1)\simeq S_{3}$
$(8,3)\simeq D_{8}$
$(12,4)\simeq D_{12}$
$(20,3)\simeq\mathbb{Z}_{5}\rtimes\mathbb{Z}_{4}$
$(24,12)\simeq S_{4}$
$(40,12)\simeq\mathbb{Z}_{2}\times(\mathbb{Z}_{5}\rtimes\mathbb{Z}_{4})$
$(42,1)\simeq(\mathbb{Z}_{7}\rtimes\mathbb{Z}_{3})\rtimes\mathbb{Z}_{2}$
$(48,48)\simeq\mathbb{Z}_{2}\times S_{4}$
$(54,6)\simeq(\mathbb{Z}_{9}\rtimes\mathbb{Z}_{3})\rtimes\mathbb{Z}_{2}$
$(84,7)\simeq\mathbb{Z}_{2}\times((\mathbb{Z}_{7}\rtimes\mathbb{Z}_{3})\rtimes\mathbb{Z}_{2})$
$(108,26)\simeq\mathbb{Z}_{2}\times((\mathbb{Z}_{9}\rtimes\mathbb{Z}_{3})\rtimes\mathbb{Z}_{2})$
$(110,1)\simeq(\mathbb{Z}_{11}\rtimes\mathbb{Z}_{5})\rtimes\mathbb{Z}_{2}$
$(120,34)\simeq S_{5}$
$(120,36)\simeq S_{3}\times(\mathbb{Z}_{5}\rtimes\mathbb{Z}_{4})$
$(144,182)\simeq((\mathbb{Z}_{3}\times\mathbb{Z}_{3})\rtimes\mathbb{Z}_{8})\rtimes\mathbb{Z}_{2}$
$(144,183)\simeq S_{3}\times S_{4}$
$(156,7)\simeq(\mathbb{Z}_{13}\rtimes\mathbb{Z}_{4})\rtimes\mathbb{Z}_{3}$
$(168,43)\simeq((\mathbb{Z}_{2}\times\mathbb{Z}_{2}\times\mathbb{Z}_{2})\rtimes\mathbb{Z}_{7})\rtimes\mathbb{Z}_{3}$
$(216,90)\simeq(((\mathbb{Z}_{2}\times\mathbb{Z}_{2})\rtimes\mathbb{Z}_{9})\rtimes\mathbb{Z}_{3})\rtimes\mathbb{Z}_{2}$
$(220,7)\simeq\mathbb{Z}_{2}\times((\mathbb{Z}_{11}\rtimes\mathbb{Z}_{5})\rtimes\mathbb{Z}_{2})$
$(240,189)\simeq\mathbb{Z}_{2}\times S_{5}$
$(252,26)\simeq S_{3}\times(\mathbb{Z}_{7}\rtimes\mathbb{Z}_{3})\rtimes\mathbb{Z}_{2}$
$(272,50)\simeq\mathbb{Z}_{17}\rtimes\mathbb{Z}_{16}$
$(312,45)\simeq\mathbb{Z}_{2}\times(\mathbb{Z}_{13}\rtimes\mathbb{Z}_{4})\rtimes\mathbb{Z}_{3}$
$(320,1635)\simeq((\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_2)\rtimes\mathbb{Z}_5)\rtimes\mathbb{Z}_4$
$(324,118)\simeq S_{3}\times(\mathbb{Z}_9\rtimes\mathbb{Z}_3)\rtimes\mathbb{Z}_2)$
$(336,208)\simeq PSL(3,2)\rtimes\mathbb{Z}_2$
$(342,7)\simeq (\mathbb{Z}_{19}\rtimes\mathbb{Z}_{9})\rtimes\mathbb{Z}_2$
$(384,5677)\simeq((((\mathbb{Z}_{4}\times\mathbb{Z}_{4})\rtimes\mathbb{Z}_{3})\rtimes\mathbb{Z}_{2})\rtimes\mathbb{Z}_{2})\rtimes\mathbb{Z}_{2}$
$(384,5678)\simeq((((\mathbb{Z}_{2}\times\mathbb{Z}_{2}\times\mathbb{Z}_{2}\times\mathbb{Z}_{2})\rtimes\mathbb{Z}_{3})\rtimes\mathbb{Z}_{2})\rtimes\mathbb{Z}_{2})\rtimes\mathbb{Z}_{2}$
$(432,520)\simeq(((\mathbb{Z}_{3}\times\mathbb{Z}_{3})\rtimes\mathbb{Z}_{3})\rtimes Q_{8})\rtimes\mathbb{Z}_{2}$
$(432,523)\simeq(((\mathbb{Z}_{6}\times\mathbb{Z}_{6})\rtimes\mathbb{Z}_{3})\rtimes\mathbb{Z}_{2})\rtimes\mathbb{Z}_{2}$
$(432,533)\simeq\mathbb{Z}_{2}\times((((\mathbb{Z}_{2}\times\mathbb{Z}_{2})\rtimes\mathbb{Z}_{9})\rtimes\mathbb{Z}_{3})\rtimes\mathbb{Z}_{2})$
$(432,734)\simeq(((\mathbb{Z}_{3}\times\mathbb{Z}_{3})\rtimes Q_{8})\rtimes\mathbb{Z}_{3})\rtimes\mathbb{Z}_{2}$
$(480,1189)\simeq(\mathbb{Z}_{5}\rtimes\mathbb{Z}_{4})\times S_{4}$
$(486,31)\simeq(\mathbb{Z}_{27}\rtimes\mathbb{Z}_{9})\rtimes\mathbb{Z}_{2}$
$(500,18)\simeq(\mathbb{Z}_{25}\rtimes\mathbb{Z}_{5})\rtimes\mathbb{Z}_{4}$
$(506,1)\simeq(\mathbb{Z}_{23}\rtimes\mathbb{Z}_{11})\rtimes\mathbb{Z}_{2}$


TITLE: Isometry from Manhattan plane to Euclidean plane?
QUESTION [3 upvotes]: Does there exist an isometry from a Manhattan plane $A$ to a Euclidean plane $B$?
I.e. a function $\varphi:A \to B$ that suffices $\|\varphi(a)\|_B = \|a\|_A$ for all $a \in A$, where  $\| \cdot \|_A$ is the Manhattan norm and $\| \cdot \|_B$ the Euclidean norm.

REPLY [0 votes]: I actually had the same question, and the answer is that there is no isometry, but you can find a relation between both norms. It takes to add a second variable to the norm, that is the number $N(a)$ of minimal paths to cover  $a$ on the manhattan plane.
$||a||_B = \phi(||a||_A,N(a))$ or
$||a||_A = \psi(||a||_B,N(a))$
You can verify easily on a few examples that it works, and if you compute the number of paths, you will find out that as long as you are just after the euclidean norm. If you compute the number of paths, you'll find explicit forms for $\phi$ and $\psi$.


TITLE: Vectors in 3 dimensions
QUESTION [1 upvotes]: If $a$ is a vector that makes equal angles with ${\mathbf i},{\mathbf j},{\mathbf k}$ and has magnitude $3$, then find the angle of $a$ with either of these unit vectors?
Wouldn't the answer simply be $120$ degrees? However, the answer is given us $\cos^{-1} (1/\sqrt{3})$.

REPLY [2 votes]: Let $a=(x,y,z)$ We know that $a\cdot (1,0,0)=a\cdot (0,1,0)=a\cdot (0,0,1)$ and so $x=y=z$ so $a=\lambda(1,1,1)$ for some $\lambda\in\mathbb{R}$. Given that $\|a\|=3$ then we have $\sqrt{3\lambda^2}=3$ so $\lambda=\sqrt{3}$.
So, $a\cdot (1,0,0)=\sqrt{3}$ and using the dot product formula, this means that $$\sqrt{3}=\|a\| \|(1,0,0)\|\cos\theta=(3)(1)\cos\theta$$ so $\cos\theta=\frac{\sqrt{3}}{3}=\frac{1}{\sqrt{3}}$. Finally then $\theta=\cos^{-1}\frac{1}{\sqrt{3}}$.


TITLE: matrix isomorphism apparent contradiction - where is the bug?
QUESTION [4 upvotes]: Let $M_2(\mathbb R)$ be the algebra of 2x2 matrices over $\mathbb R$.
Let $M^{op}_2$ be the same algebra, but with product reverted (that is,
$A\cdot B \ \hbox{(in $M^{op}$)}= B\cdot A \ \hbox{(in $M$)}$.
By the Skolem-Noether theorem (if I'm not wrong), every isomorphism of $M$ into $M^{op}$ is
of the form $f(M) = U^{-1}MU$, where $U$ is an invertible matrix of $M_2$ (depending only on $f$).
Now, matrix transposition is such an isomorphism. But I think that if there were exist a matrix $U$ such that $M^T= U^{-1}MU$ for all $M$, this would be known (a joke, this is impossible of course).
So, the theorem of Skolem-Noether is false!
where is the bug in my thinking ?
Edit: the Skolem-Noether theorem:
Let $R$, $S$ be finite dimensional algebras, $R$ simple and $S$ central simple. If $f, g :\ R → S$ are homomorphisms then there is an element $s ∈ S$ such that, for all $r ∈ R$, $g(r) = s^{−1} f (r)s$.

REPLY [3 votes]: Your statment cannot possibly be true, since $M\mapsto U^{-1}MU$ is not a ring homomorphism from $M$ onto $M^{\text{op}}$; it does not reverse the order.


TITLE: variation problem of constrained area and minimized distance
QUESTION [0 upvotes]: $$c=\int_{x_1}^{x_2}f_{gr}(x)\;dx$$
The integral is a time-like curve between $x_1$ and $x_2$ and at imagine fgf(x1) is a lower left corner of the rectangle and fgf(x2) is the upper right corner and $x_2-x_1$ is the length of the base of the rectangle. The geodesic is the shortest length curve parametrized by this on the condition that the area = c. The solution isn't always a strait line but could be some curved function. How do I find the function ? I know this is a variational problem involving a length integral and an area integral.
Do I need to find the 2x2 metric tensor for the 2-d parameterization curve and solve a pair of differential equations like here ?

REPLY [1 votes]: You can incorporate the area constraint with the Lagrange multiplier method; for example, the book of John Oprea Differential geometry and its applications has an accessible treatment of such variational problems.
However I recommend to look for a curve in parametric form $x=f(t), y=g(t)$ rather than in the graph form $y=f(x)$. The reason is as follows. The free part of the boundary (the curve you are looking for) wants to have constant curvature. Indeed, if a curve has different curvature $\kappa_1>\kappa_2$ at two points $p_1,p_2$, then we can push a little at the point $p_1$ to decrease the curvature, and push from the opposite side at $p_2$ to increase it. This perturbation will decrease the length without changing the area. A planar curve of constant curvature is a circular arc. So, the boundary wants to be a circular arc with given endpoints, bounding the given area. When such an arc is a graph $y=f(x)$, it is the solution. But in some cases (depending on your parameters) the extremal circular arc is not a graph, because it intersects some vertical line (either $x=x_1$  or $x=x_2$) twice. If you insist on the graph form of the solution, you create an obstacle problem, which makes the problem substantially more difficult.
On the geometric grounds (without doing the computations) I expect that when the obstacle comes into effect, there will not be a minimizer satisfying the given boundary conditions. Instead, a minimizing sequence will converge to a function which violates one or both boundary conditions, and whose graph is a circular arc tangent to the effective obstacle(s). For a simple example, take $x_1=-1, x_2=1$, prescribe the boundary conditions $f(-1)=f(1)=0$ and demand the area to be $10+\pi/2$. The minimizing functions will converge to $f(x)=5+\sqrt{1-x^2}$, which fails the boundary conditions. To recover the existence of minimizer, you can relax the problem by adding the length of vertical segments to the length of the proper graph: that is, the new variational problem is to minimize $|f(x_1)-y_1|+|f(x_2)-y_2|+\int_{x_1}^{x_2}\sqrt{1+f\,'(x)^2}\,dx$ subject to $\int_{x_1}^{x_2} f(x)\,dx=c$, and no boundary conditions. Still, an additional complication arises when the prescribed area $c$ is so small that $f$ wants to be negative.


TITLE: Bremermann's limit vs Planck frequency
QUESTION [0 upvotes]: Bremermann's limit, as maximum possible computation power or CPU total computing frequency, is known to be on the order $10^{50}~\text{Hz}/\text{kg}$.
Why max computation frequency for unit mass can exceed Plank frequency, which is on the order of $10^{43} ~\text{Hz}$ and how it is related to it?

REPLY [0 votes]: Maybe I have not expressed my question in a good way. The core thing which I was interested in was,- Can a Bremermann's limit be calculated using only Plank units or at least approximated by it ?
Thanks to @user253751, I can now understand that. Usually maximum possible bit-rate is achieved by equating quanta energy to $mc^2$ :
$$mc^2 = hf $$,
then maximum bit-rate possible per unit mass is :
$$R_b = \frac fm = \frac {c^2}{h}$$
which is $1.3 \times 10^{\,50} ~\text{bps}/\text{kg}$, here "bps" means bits-per-second. Now, technically bit-rate has dimensions of $\text{Hz/kg}$, so one can approximate maximum bit-rate using Plank frequency and Plank mass :
$$ R_b = \frac{f_{_P}}{m_{_P}} $$
Using these we get $R_b$ also on the order $10^{50}$ bps/kg ! So conclusion is that bit-rate maximum limit can be quite accurately modeled using Plank units only.


TITLE: Is this a good explanation of the Higgs mechanism?
QUESTION [3 upvotes]: Possible Duplicate:
Is there an accepted analogy/conceptual aid for the Higgs field?

A video I watched explains the Higgs mechanism as follows.
Take massless particles. These can only ever travel at the speed of light, but they can bounce off things. Now take a particle with mass. Imagine that it can also only travel at the speed of light, but it interacts with the (very dense) Higgs field all the time, by constantly bouncing back and forth off the field. Thus it appears to travel slower than the speed of light.
Would you say this is a reasonable explanation of the mechanism? Does the described process happen in any real sense, or is this one of those interpretations where we can't really pick one over the other because they are mathematically equivalent?
For example, if the particles were really travelling at the speed of light but constantly bouncing off the field, wouldn't GR predict that they would never decay as observed in any frame of reference?

REPLY [5 votes]: The best non-technical explanation I've seen of this is Matt Strassler's blog article, though even this is still fairly technical, so let me see if I can interpret it a bit.
The key point is that an electron is not a particle. In Quantum Field Theory it's described as an excitation in the electron quantum field. This excitation will propagate in spacetime, and from the way it propagates e.g. the propagation velocity and how much momentum it carries, we can calculate it's mass. And in fact if the electron field has no interaction with the Higgs field the propagation speed would be $c$ so the electron would be massless.
The Higgs field is another quantum field, and the Higgs field and electron field interact. That means you cannot just write an electron just as an excitation of the electron field, but instead it has to be written as an excitation of both the electron and Higgs fields together. Because the interaction is relatively small we can write the excitation as a slightly perturbed electron field excitation, that is we write it as an excitation of the electron field plus a bit of the Higgs field. If we now calculate how this excitation propagates we find it travels at less than the speed of light i.e. the excitation of the combined fields has a mass. The amount of mass is proportional to the strength of the interaction between the electron and Higgs fields.
The idea of an electron bouncing off Higgs bosons is more easily visualised, but it's a very crude analogy, and like all analogies when pushed too far it ceases to be helpful. That's why you're (quite reasonably!) confused about the time dilation the electrons would suffer when moving at light speed in between collisions. It's not correct to think of electrons as intermittently bouncing off Higgs bosons. The electron field and Higgs field mix so the electrons cease to be pure electrons.
This applies to all fermions, i.e. electrons, muons, taus and quarks (and I think neutrinos though I'm not sure about this). The vector bosons like the W and Z particles get their mass through a different interaction. Photons and gluinos remain massless.


text:
\begin{document}
\title[Spectrum for Isentropic Gas Dynamics with Capillarity]{On the Shock Wave Spectrum for Isentropic Gas Dynamics with Capillarity}
\author{Jeffrey Humpherys}
\address{Department of Mathematics, Brigham Young University, Provo, UT 84602}
\email{jeffh at math.byu.edu}

\subjclass{35Q30, 35Q35}
\date{\today}

\keywords{gas dynamics, shock wave stability, capillarity}
\thanks{The author thanks Kevin Zumbrun for useful conversations throughout this project.  This work was supported in part by the National Science Foundation, award number DMS-0607721.}

\begin{abstract}
We consider the stability problem for shock layers in Slemrod's model of an isentropic gas with capillarity.  We show that these traveling waves are monotone in the weak capillarity case, and become highly oscillatory as the capillarity strength increases.  Using a spectral energy estimate we prove that small-amplitude monotone shocks are spectrally stable.  We also show, through the use of a novel spectral energy estimate, that monotone shocks have no unstable real spectrum regardless of amplitude; this implies that any instabilities of these monotone traveling waves, if they exist, must occur through a Hopf-like bifurcation, where one or more conjugate pairs of eigenvalues cross the imaginary axis.  We then conduct a systematic numerical Evans function study, which shows that monotone and mildly oscillatory profiles in an adiabatic gas are spectrally stable for moderate values of shock and capillarity strengths.  In particular, we show that the transition from monotone to non-monotone profiles does not appear to trigger any instabilities.
\end{abstract}
\maketitle
\thispagestyle{empty}

\section{Introduction}
\label{int}
We consider Slemrod's model \cite{HaSl,Sl.1,Sl.2} for a one-dimensional isentropic gas with capillarity
\begin{equation}
\begin{split}
\label{slemrod}
v_{t}-u_{x} &=0, \\
u_{t}+p(v)_{x} &= \left(\frac{u_x}{v}\right)_x - d v_{xxx},
\end{split}
\end{equation}
where physically, $v$ is the specific volume, $u$ is the velocity in Lagrangian coordinates, $p(v)$ is the pressure law for an ideal gas, that is $p'(v) < 0$ and $p''(v) > 0$, and the coefficient, $d \geq 0$, accounting for capillarity strength, is constant.  This is model is a generalization of the compressible isentropic Navier-Stokes equations, or the $p$-system with semi-parabolic (or real) viscosity,
\begin{equation}
\begin{split}
\label{int:navierstokes}
v_{t}-u_{x} &=0, \\
u_{t}+p(v)_{x} &= \left(\frac{u_x}{v}\right)_x.
\end{split}
\end{equation}
It has recently been shown that viscous shock wave solutions of \eqref{int:navierstokes} are spectrally stable for all amplitudes in the case of an adiabatic gas law $p(v) = v^{-\gamma}$, $\gamma\in[1,3]$; see \cite{BHRZ,HLZ}.  We remark that this result, together with Mascia \& Zumbrun's work  \cite{MZ.2, MZ.3} implies that viscous shocks are asymptotically orbitally stable (hereafter called nonlinearly stable).   In this paper, we make the first step toward generalizing this work to Slemrod's model by showing that monotone and mildly oscillatory smooth shock profiles of small to moderate amplitude are likewise spectrally stable.

More generally, we are interested in understanding the degree to which the analytic methods used to study shock wave stability in viscous conservation laws extend to viscous-dispersive systems.  We view Slemrod's model as an important test case as it is physically realistic and yet captures some of the essential mathematical hurdles found in more extensive models of compressible fluid flow.  In particular, Slemrod's model is symmetrizable and genuinely coupled, having only semi-parabolic diffusion; see \cite{Hu.2} for details.

A few notable results in the study of shock wave stability for viscous conservation laws include the works of Kawashima \cite{Ka.1,Ka.2,KS}, who proved that genuinely coupled symmetrizable systems has a stable essential spectrum, the works of Goodman and others \cite{Go.1,MN, KMN, HuZ.1}, who proved small-amplitude spectral stability for viscous shocks through the use of cleverly chosen weighted energy estimates, and the works of Zumbrun and collaborators \cite{ZH,MZ.1,MZ.2, MZ.3, Z.2}, who proved that spectral stability implies nonlinear stability for viscous shocks in conservation laws for both strictly parabolic and semi-parabolic viscosities.  The missing piece in this overall program is to determine whether moderate- and large-amplitude viscous shocks are spectrally stable.  Very recently, however, spectral stability for large-amplitude shocks for \eqref{int:navierstokes} was proven in the case of an adiabatic gas \cite{HLZ}, and spectral stability was numerically demonstrated for the intermediate range through an extensive Evans function study \cite{BHRZ}.  There is some hope that this overall strategy will extend to more general systems of viscous conservation laws and perhaps even viscous-dispersive models.

We remark that Kawashima's admissibility results, mentioned above, were recently extended to viscous-dispersive (and higher-order) systems \cite{Hu.2}.  Also, Howard \& Zumbrun showed that spectral stability implies nonlinear stability for scalar viscous-dispersive conservation laws \cite{HZ}.   However, the remaining pieces of the general program for viscous-dispersive systems, described above, are still open.

This paper is organized as follows:  In Section \ref{prelim}, we set the stage by first proving the existence of shock profiles for \eqref{slemrod} through the use of a Lyapunov function argument.  Then using geometric singular perturbation theory, we show that small-amplitude shock profiles converge to the zero-capillarity case and are thus monotone.  Following that, we use a qualitative ODE argument to show that our profiles are monotone for weak capillarity yet become highly oscillatory as the capillarity strength $d$ increases.   We then provide a short estimate on the derivative bounds of the profile, which are used later in the stability analysis. Finally, we formulate the integrated eigenvalue problem, which makes the stability problem more amenable to analysis; see for example \cite{Go.1,ZH}.  In Section  \ref{spec}, we generalize the work of Matsumura \& Nishihara \cite{MN} and Barker, Humpherys, Rudd, \& Zumbrun \cite{BHRZ} by using a spectral energy estimate to prove that small-amplitude monotone shocks of \eqref{slemrod} are spectrally stable.  In Section \ref{noreal}, we further extend the results in \cite{BHRZ} and offer a short and novel proof that monotone shocks have no unstable real spectrum regardless of amplitude.  This restricts the class of admissible bifurcations for monotone profiles to those of Hopf-type, where one or more conjugate pairs of eigenvalues cross the imaginary axis.  The approach used here is different than many energy methods in that we use a spectral energy estimate that does not appear to have a time-asymptotic equivalent, whereas most energy estimates can be performed in either domain.  In Section \ref{hfb}, we extend the spectral bounds in \cite{BHRZ} to \eqref{slemrod} by proving that high-frequency instabilities cannot occur for adiabatic monotone profiles of any amplitude for $d\leq 1/3$.  Finally in Section \ref{evans}, we carry out a systematic numerical Evans function study showing that adiabatic monotone and mildly oscillatory profiles are spectrally stable for moderate shock and capillarity strengths.

We remark that highly oscillatory profiles in the scalar KDV-Burgers model were shown by Pego, Smerka, \& Weinstein \cite{PSW} to be unstable in certain cases.  Thus for some, perhaps extreme, parameters, one can reasonably expect instabilities to occur in our system as well.  It is challenging, however, with current technology to explore these extreme cases numerically.  We plan on exploring this in the future.

\section{Preliminaries}
\label{prelim}

In this section, we derive the profile ODE and provide a convenient scaling for our analysis.  We prove the existence of shock profiles for \eqref{slemrod} through the use of a Lyapunov function argument.  Then using geometric singular perturbation theory, we show that small-amplitude shock profiles converge to the zero-capillarity case and are thus monotone.  Through a qualitative ODE argument, we then show that profiles are monotone for weak capillarity yet become oscillatory as the capillarity strength $d$ increases beyond the transition point $d_*$.  We then provide a short estimate on the derivative bounds of the profile, which will be used later in the stability analysis. Finally, we formulate the spectral stability problem and change to integrated coordinates making it more amenable to analysis; see for example \cite{Go.1,ZH}.

\subsection{Shock Profiles}
\label{scalingsection}
By a shock layer (or shock profile) of \eqref{slemrod}, we mean a traveling wave solution
\begin{equation*}
\begin{split}
v(x,t)&=\hat{v}(x-s t),\\
u(x,t)&=\hat{u}(x-s t),
\end{split}
\end{equation*}
with asymptotically constant end-states $(\hat{v},\hat{u})(\pm \infty) =(v_{\pm},u_{\pm})$. Rather by translating $x \rightarrow x-s t$, we can instead consider  stationary solutions of
\[
\begin{split}
v_{t} - s v_{x} - u_{x} &= 0,\\
u_{t} - s u_{x} + p(v)_{x} &= \left(\frac{u_x}{v}\right)_x - d v_{xxx}.
\end{split}
\]
Under the rescaling
$(x,t,u) \rightarrow (-sx, s^2 t, -u/s)$, our system takes the form
\begin{equation}
\begin{split}
\label{trav:cap2}
v_t + v_x - u_x &= 0,\\
u_t + u_x + a  p(v)_{x} &= \left(\frac{u_x}{v}\right)_x - d v_{xxx},
\end{split}
\end{equation}
where $a = 1/s^2$.  Thus, the shock profiles of \eqref{slemrod} are solutions of the ordinary differential equation
\begin{align*}
v' - u' &= 0,\\
u' + a  p(v)' &= \left(\frac{u'}{v}\right)' - d v''',
\end{align*}
subject to the boundary conditions $(v,u)(\pm \infty) =(v_{\pm},u_{\pm})$. This simplifies to
\begin{equation*}
v' + a  p(v)' = \left(\frac{v'}{v}\right)' - d v'''.
\end{equation*}
By integrating from $-\infty$ to $x$, we get our profile equation,
\begin{equation}
\label{profile}
v-v_{-} + a  (p(v)-p(v_{-})) = \frac{v'}{v} - d v'' ,
\end{equation}
where $a$ is found by setting $x=+\infty$, thus yielding the Rankine-Hugoniot condition
\begin{equation}
\label{RH}
a = -\frac{v_+ - v_-}{p(v_+) - p(v_-)}.
\end{equation}
Without loss of generality, we will assume that $0 < v_+ < v_-$.   We remark that small-amplitude shocks  occur when $v_+$ is close to $v_-$ and large-amplitude shocks arise to when $v_+$ nears zero.

\begin{remark}
\label{cap:rmk1}
In the absence of capillarity, that is when $d=0$, the profile equation \eqref{profile} is of first order, and thus has a monotone solution.  As we will show, small values of $d$ likewise yield monotone profiles whereas large values of $d$ produce oscillatory profiles.  We make this precise below.
\end{remark}

\subsection{Adiabatic Gas}
\label{adiabatic}
Although much of the analysis in this paper holds for ideal gases, that is when $p'(v)<0$ and $p''(v)>0$, our numerical study focuses on the special case of an adiabatic gas law,
\begin{equation}
\label{gammalaw}
p(v) = v^{-\gamma}, \gamma\geq 1,
\end{equation}
together with the rescaling
\[
(x,t,v,u,a,d) \rightarrow (\varepsilon x, \varepsilon t, v/\varepsilon, u/\varepsilon,a \varepsilon^{-\gamma-1},\varepsilon^2 d),
\]
where $\varepsilon$ is chosen so that $v_-=1$;  see \cite{BHRZ,HLZ} for more details.  This choice simplifies our analysis in Section \ref{hfb} and also gives the Mach number $M$ the simplifying form $M=1/\sqrt{\gamma a}$.

\subsection{Existence}
We prove existence of profiles by the following Lyapunov function argument.

By writing \eqref{profile} as a first order system, we get

\begin{subequations}
\label{firstorder}
\begin{align}
v' &= w, \label{firstorder1}\\
w' &= \frac{1}{d} \left[ \frac{w - \phi(v) }{v} \right], \label{firstorder2}
\end{align}
\end{subequations}

where
\begin{equation}
\label{phi}
\phi(v)= v(v-v_{-} + a  (p(v)-p(v_{-})).
\end{equation}
The zero-diffusion case is conservative and has a corresponding Hamiltonian that provides us with the needed Lyapunov function.  Specifically, let
\begin{equation}
E(v,w) = \frac{1}{2} w^2 - \frac{1}{d} \int^{v_{-}}_{v} \frac{\phi(\tilde v)}{\tilde v} d\tilde v.
\end{equation}
Since $\phi(v) < 0$ on $(v_+,v_-)$, then $E(v,w)$ is non-negative for $v \in [v_{+},v_{-}]$.  It follows that
\begin{equation}
\frac{d}{dx} E(v(x),w(x))= \nabla E \cdot
(v',w')^T=\frac{w^2}{d v} > 0.
\end{equation}
Hence with diffusion, bounded (homoclinic) orbits at $(v_+,0)$ are pulled into the minimum $(v_{-},0)$ of $E(v,w)$ as $x\rightarrow -\infty$.  Thus there exists a connecting orbit from $v_+$ to $v_-$.

\subsection{The Small-Amplitude Limit}
\label{smallamp}

We now show that small-amplitude shocks of \eqref{slemrod} are monotone and follow the same asymptotic limits as the $d=0$ case presented in \cite{MP,P,BHRZ,HLZ}.  We accomplish this by rescaling and showing, via geometric singular perturbation theory \cite{F,GS,J}, that the profile converges, in the small-amplitude shock limit, to the (monotone) non-dispersive case.  Thus, monotonicity of small-amplitude shocks of \eqref{slemrod} is implied by the monotonicity of the non-dispersive case, as mentioned above.

\begin{lemma}
Small-amplitude shocks of \eqref{slemrod} are monotone for any fixed $d$.
\end{lemma}

\begin{proof}
We scale according to the amplitude $\varepsilon = v_{-}-v_{+}$.  Let $\bar{v} = (v-v_{0})/\varepsilon$ and $\bar{x}  = \varepsilon x$, where $v_{0}=v_{-}-\varepsilon \bar{v}_-$.  This frame is chosen so that the end-states of the profile are fixed at $\bar{v}_{\pm} = \mp 1/2$.  Additionally, we expand the pressure term $p(v)$ and the viscosity term $v^{-1}$ about $v_{-}$. Hence \eqref{profile} becomes
\begin{equation}
\label{jxexp}
\begin{split}
&\varepsilon (\bar{v}-\bar{v}_{-}) \left( 1+ a  p(v_{-}) \right) + \varepsilon^2
\frac{a p''(v_{-})}{2}(\bar{v}-\bar{v}_{-})^2 +
\mathcal{O}(\varepsilon^3)(\bar{v}-\bar{v}_{-})^3 \\
& \qquad = \varepsilon^2 \frac{\bar{v}'}{\bar{v}_-} + \mathcal{O}(\varepsilon^3)
(\bar{v}-\bar{v}_{-}) \bar{v}' + \varepsilon^3 d \bar{v}''.
\end{split}
\end{equation}
By expanding the Rankine-Hugoniot equality, $\varepsilon = a   (p(v_{+}) - p(v_{-}))$, about $v_{-}$, we obtain
\begin{equation}
\label{jxexp1}
1+ a  p'(v_{-}) = \frac{a p''(v_{-})}{2}\varepsilon + \mathcal{O}(\varepsilon^2).
\end{equation}
Substituting \eqref{jxexp1} into \eqref{jxexp} and simplifying gives (recall that $\bar{v}_{-} = 1/2$)
\begin{equation}
\frac{a p''(v_{-})}{2}(\bar{v}^2 - \frac{1}{4}) + \varepsilon R(\bar{v},\bar{v}') = \frac{\bar{v}'}{\bar{v}_-} +
\varepsilon^3 d \bar{v}'' \label{jxexp2}.
\end{equation}
where $R(\bar{v},\bar{v}') = \mathcal{O}(1)$.  Thus, in the $\varepsilon=0$ limit,
\eqref{jxexp2} becomes
\begin{equation}
\bar{v}' = \frac{a p''(v_{-})v_-}{2}(\bar{v}^2 - \frac{1}{4})
\label{jxexp3},
\end{equation}
which is essentially the same reduction obtained for the viscous Burgers equation.  Note that the capillarity term vanishes as well, and thus the reduction is the same as the zero-capillarity $(d=0)$ case.

The slow dynamics of \eqref{jxexp2} take the form
\begin{subequations}\label{jxslow}
\begin{align}
\bar{v}' &= \bar{w}, \label{jxslow:1}\\
\varepsilon \bar{w}' &= \frac{1}{d} \left[ \frac{a p''(v_{-})}{2}(\bar{v}^2 - \frac{1}{4}) +
\varepsilon R(\bar{v},\bar{v}')- \frac{\bar{w}}{\bar{v}_-} \right]. \label{jxslow:2}
\end{align}
\end{subequations}
The fast dynamics, obtained by rescaling $x \rightarrow x/\varepsilon$, take the form
\begin{subequations}\label{jxfast}
\begin{align}
\bar{v}' &= \varepsilon \bar{w}, \label{jxfast:1}\\
\bar{w}' &= \frac{1}{d} \left[ \frac{a p''(v_{-})}{2}(\bar{v}^2 - \frac{1}{4}) + \varepsilon R(\bar{v},\bar{v}') - \frac{\bar{w}}{\bar{v}_-} \right]. \label{jxfast:2}
\end{align}
\end{subequations}
We can see from the slow dynamics that solutions will remain on the parabola defined by
\begin{equation*}
\bar{w} = \frac{a p''(v_{-}) v_-}{2}(\bar{v}^2-\frac{1}{4}).\label{jxexp5}
\end{equation*}
In addition, we can see from the fast dynamics that any jumps will be vertical, that is, $v=$ constant.  Since there are no vertical branches, no jumps occur and thus it follows that small-amplitude shocks approach the solutions for \eqref{jxexp3}.

Hence, for sufficiently small amplitudes, the profiles are monotone.
\end{proof}

\begin{remark}
In the original scale, small-amplitude profiles of \eqref{int:navierstokes} have the asymptotic properties $|\bV_x|=\CalO(\varepsilon^2)$ and $|\bV_{xx}| = |\bV_x| \CalO(\varepsilon)$, where $\varepsilon = v_--v_+$ is the amplitude; see \cite{MP,P}.  From the above argument, these asymptotic properties hold with our scaling in \eqref{trav:cap2} as well.  It is also straightforward to establish these asymptotic properties directly; see for example Theorem \ref{asympthm} below.
\end{remark}

\begin{figure}[p]
\begin{center}
$\begin{array}{cc}
\includegraphics[width=6cm]{figures/cap1_profile} & \includegraphics[width=6cm]{figures/cap1_phase} \\
\multicolumn{2}{c}{\mbox{\bf (a)}}\\
\includegraphics[width=6cm]{figures/cap2_profile} & \includegraphics[width=6cm]{figures/cap2_phase} \\
\multicolumn{2}{c}{\mbox{\bf (b)}}\\
\includegraphics[width=6cm]{figures/cap3_profile} & \includegraphics[width=6cm]{figures/cap3_phase} \\
\multicolumn{2}{c}{\mbox{\bf (c)}}
\end{array}$
\end{center}
\caption{Images of the profiles and their derivatives (left) and corresponding phase portraits (right)  for an adiabatic monatomic gas ($\gamma=5/3$) with $v_+=0.1$ and $d$ varying (note $d_*\approx 0.259$).  We demonstrate $(a)$ monotone profiles with $d=0.2$, $(b)$ non-monotone profiles which are mildly oscillatory with $d=2$, and $(c)$ non-monotone profiles that are highly oscillatory with $d=200$.
}
\label{profiles}
\end{figure}

\subsection{Classification of Profiles}

We show that smooth shock profiles are monotone for small values of $d$ and transition to highly oscillatory fronts when $d$ gets large; see Figure \ref{profiles} for illustrative examples.  The transition point between monotone and non-monotone profiles is found to be
\begin{equation}
\label{dstar}
d_* = \frac{1}{4v_-^2(1+a p'(v_-))},
\end{equation}
and in the case of an adiabatic gas with $v_-=1$, see Section \ref{adiabatic}, this becomes
\begin{equation}
\label{dstar2}
d_* = \frac{1}{4(1-a\gamma)} =  \frac{M^2}{4(M^2-1)} ,
\end{equation}
where $M$ is the Mach number.  In particular as the amplitude approaches zero, we have that $1+a p'(v_-) \rightarrow 0$, see \eqref{jxexp1}, thus making all profiles monotone regardless of $d$; this is consistent with the results in Section \ref{smallamp}.  In the large-amplitude limit, we have that $a\rightarrow 0$ and thus $d_*\rightarrow 1/(4 v_-^2)$.  Hence, for values of $d$ less than $1/(4 v_-^2)$, all profiles are monotone, regardless of amplitude, and for $d\geq 1/(4 v_-^2)$ a transition from monotone to non-monotone occurs for moderate to large amplitude fronts.  We have the following:

\begin{theorem}
Shock profiles of \eqref{slemrod} are monotone iff $0\leq d\leq d_*$.
\end{theorem}

\begin{proof}
By a geometric singular perturbation argument very similar to the one in Section \ref{smallamp}, we know that profiles are monotone for sufficiently small values of $d$.  When $d=d_*$, we can show that the local behavior near the fixed point $(v_-,0)$ transitions from that of an unstable node to an unstable spiral, which is clearly non-monotone.  Hence, it suffices to show that the profile does not lose monotonicity until $d$ passes through $d_*$.  By linearizing \eqref{firstorder} about $v_-$, we get the system
\begin{equation}
\label{linearode}
\begin{pmatrix}v\\w\end{pmatrix}' = \begin{pmatrix}0 & 1\\\frac{-(1+a p'(v_-))}{d} & \frac{1}{d v_-}\end{pmatrix} \begin{pmatrix}v\\w\end{pmatrix}.
\end{equation}
If monotonicity is lost before $d$ gets to $d_*$, then for some $d_0<d_*$ the phase curve connects to $v_-$ vertically.  This would require the vector field near $v_-$ to admit a vector in the $w$ direction.  However, since
\[
\begin{pmatrix}0 & 1\\\frac{-(1+a p'(v_-))}{d_0} & \frac{1}{v_-}\end{pmatrix} \begin{pmatrix}0\\-w\end{pmatrix} = \begin{pmatrix}-w\\-\frac{w}{d_0 v_-}\end{pmatrix},
\]
this cannot happen.  Hence the profiles are monotone whenever $d<d_*$.
\end{proof}

\begin{figure}[ht]
\begin{center}
\includegraphics[width=12cm]{figures/cap_vx}
\end{center}
\caption{Two phase portraits for an adiabatic monatomic gas ($\gamma=5/3$) for $v_+=0.1$.  The dark line corresponds to a non-monotone profile with $d=5$ and the dotted line is the zero-capilarity profile $w=\phi(v)$.  Note that the dotted curve intersects the dark one at its minimum.  Hence to bound the derivative of the non-monotone profile, we need only bound the derivative of the monotone profile.}
\label{vx}
\end{figure}

\subsection{Bounds on $|\bV_x|$}

W now provide bounds on $\bV_x$ that are used later in our analysis.  We show that $|\bV_x| \leq \varepsilon^2/4$, where $\varepsilon=v_--v_+$ is the amplitude of the profile.  This bound holds regardless of capillarity strength, and is important for our analysis in Section \ref{hfb}.  The idea behind the proof follows from Figure \ref{vx}, where we see that the maximum value of $|\bV_x|$ occurs at the point where the profile intersects the zero-capillarity profile.  Thus we need only find a bound on the zero-capillarity profile.

\begin{theorem}
\label{asympthm}
Shock profiles of \eqref{slemrod} satisfy $|\bV_x| \leq \varepsilon^2/4$, where $\varepsilon=|v_--v_+|$.

\end{theorem}

\begin{proof}
Consider the phase portrait of the profile.  Let $v_0$ denote the point that maximizes $|\bV_x|$.  This occurs when $w'=0$ in \eqref{firstorder2}, or in other words, when $w = \phi(v)$, which is the zero-capillarity profile.  Hence, the maximum point for $|\bV_x|$ coincides with the zero-capillarity curve, which we can show is bounded above by $\varepsilon^2/4$.  This follows easily since
\[
\sup_{x\in\mathbb{R}} |\bV_x| = \sup_{v\in[v_+,v_-]} |\phi(v)| < \sup_{v\in[v_+,v_-]} |\bV(\bV-v_-)| \leq \frac{|v_--v_+|^2}{4} = \frac{\varepsilon^2}{4}.
\]

\end{proof}

\begin{remark}
In the $\varepsilon\rightarrow 0$ limit we can likewise show that $|\bV_{xx}|=|\bV_x|\mathcal{O}(\varepsilon)$.
\end{remark}

\subsection{Stability problem}

By linearizing \eqref{trav:cap2} about the profile $(\hat{v},\hat{u})$, we get the eigenvalue problem
\begin{equation}
\label{jx6}
\begin{split}
&\lambda v + v' - u' =0,\\
&\lambda u + u' - (f(\bV) v)' = \left(\frac{u'}{\bV}\right)' - d v''',
\end{split}
\end{equation}
where $f(\bV) = -a p'(\hat{v})-\bV_x/\bV^2$.  We say that a shock profile of \eqref{slemrod} is spectrally stable if the linearized system \eqref{jx6}  has no spectra in the closed deleted right half-plane given by $P = \{\R(\lambda) \geq 0\}\setminus\{0\}$, that is, there are no growth or oscillatory modes.  To show that the essential spectrum is stable, we linearize \eqref{trav:cap2} about the endstates $(v_\pm,u_\pm)$ and show that the resulting constant-coefficient system is stable; see \cite{He}.  This was done for general viscous-dispersive and higher-order systems in \cite{Hu.2}.  Thus it suffices to show that the point spectrum is also stable.  However, since traveling wave profiles always have a zero-eigenvalue due to translational invariance, it is often difficult to get good uniform bounds in energy estimates.  Hence, we use the standard technique of transforming into integrated coordinates; see \cite{Go.1, ZH, BHRZ}.  This goes as follows:

Suppose that $(v,u)$ is an eigenfunction of \eqref{jx6} with eigenvalue $\lambda\ne 0$.  Then
\[
\tilde{u}(x) = \int_{-\infty}^x u(z) dz , \quad
\tilde{v}(x) = \int_{-\infty}^x v(z) dz,
\]
and their derivatives decay exponentially as $x \rightarrow \infty$; see \cite{ZH}. Thus, by substituting and then integrating, $(\tilde{u},\tilde{v})$ satisfies (suppressing the tilde)
\begin{subequations}\label{ep}
\begin{align}
&\lambda v + v' - u' =0, \label{ep:1}\\
&\lambda u + u' - f(\bV) v' = \frac{u''}{\bV} - d v'''\label{ep:2}
\end{align}
\end{subequations}
This new eigenvalue problem is important because its point spectrum differs from that of \eqref{jx6} only at $\lambda=0$.  It follows that spectral stability of \eqref{jx6} is implied by spectral stability of \eqref{ep}.  Hence, we will use \eqref{ep} instead of \eqref{jx6} in the remainder of our stability analysis.

\section{Small-Amplitude Spectral Stability}
\label{spec}

In this section we show that small-amplitude smooth shock profiles are spectrally stable.  This work generalizes the energy methods in \cite{MN,BHRZ} to the case of an isentropic gas with capillarity.

\begin{theorem}
Small-amplitude shocks of \eqref{slemrod} are spectrally stable.
\end{theorem}

\begin{proof}

Suppose that $\R\lambda\geq 0$.  Recall that small-amplitude profiles are monotone with $\bV_x<0$ and thus also satisfy $f(\bV) > 0$ and $f'(\bV)<0$.  By multiplying \eqref{ep:2} by the conjugate $\bar{u}/f(\bV)$ and integrating in $x$ from $-\infty$ to $\infty$, we have
\[
\ip \frac{\lambda u \bar{u}}{f(\bV)} + \ip \frac{u' \bar{u}}{f(\bV)} -  \ip v' \bar{u} = \ip
\frac{u''\bar{u}}{\bV f(\bV)} - \ip \frac{d v''' \bar{u}}{f(\bV)}.
\]
Integrating the last three terms by parts and appropriately using \eqref{ep:1} to substitute for $u'$ in the third term gives us
\begin{align*}
&\ip \frac{\lambda |u|^2}{f(\bV)} + \ip \left[ \frac{1}{f(\bV)} + \left(\frac{1}{\bV f(\bV)}\right)' \right] u' \bar{u} + \ip v (\overline{\lambda v + v'}) + \ip \frac{|u'|^2}{\bV f(\bV)}\\
&\quad= d \ip \frac{1}{f(\bV)} v'' \bar{u}' + d\ip \left(\frac{1}{f(\bV)}\right)' v'' \bar{u}.
\end{align*}

We take the real part and appropriately integrate by parts:
\begin{align*}
&\R(\lambda)\ip \left[ \frac{|u|^2}{f(\bV)}+|v|^2 \right] + \ip g(\bV) |u|^2 + \ip \frac{|u'|^2}{\bV f(\bV)} \\
&\qquad= d\:\R\left[\ip \frac{1}{f(\bV)} v'' \bar{u}'+ \ip \left(\frac{1}{f(\bV)}\right)'v'' \bar{u}\right],
\end{align*}
where
\begin{equation}
\label{g}
g(\bV)= - \frac{1}{2} \left[ \left(\frac{1}{f(\bV)}\right)' + \left(\frac{1}{\bV f(\bV)}\right)'' \right].
\end{equation}
Thus, by integrating the last two terms by parts and further simplifying, for $\lambda \geq 0$, we have
\begin{equation}
\begin{split}
&  \ip g(\bV) |u|^2  + \ip \frac{|u'|^2}{\bV f(\bV)} - \frac{d}{2}\ip \left(\frac{1}{f(\bV)}\right)' |v'|^2 \\
&\qquad \leq - d\:\R\left[ 2\ip \left(\frac{1}{f(\bV)}\right)'v' \bar{u}' + \ip \left( \frac{1}{f(\bV)}\right)'' v' \bar{u}\right].
\end{split}
\end{equation}
We note that since $d \geq 0$ and $\bV_x <0$, then all the terms on the left-hand side are non-negative.  Moreover, since $|\bV_x|=\CalO(\varepsilon^2)$ and $|\bV_{xx}| = |\bV_x| \CalO(\varepsilon)$, it follows that the right-hand side of the above equation is bounded above by
\begin{align*}
- 2 d \ip \left(\frac{1}{f(\bV)}\right)' |v'| |u'| + C d \ip \varepsilon |\bV_x| |v'||u|.
\end{align*}
Thus, by Young's inequality, we have
\begin{align*}
& \ip g(\bV) |u|^2  + \ip \frac{|u'|^2}{\bV f(\bV)} - \frac{d}{2}\ip \left(\frac{1}{f(\bV)}\right)' |v'|^2 \notag \\
&\qquad < - 2 d \ip \left(\frac{1}{f(\bV)}\right)' \left[ \frac{|v'|^2}{4 \eta_1} +  \eta_1 |u'|^2 \right] + C \ip \varepsilon |\bV_x| \left[ \frac{|v'|^2}{4 \eta_2} +  \eta_2 |u|^2 \right].
\end{align*}
We can see that for $\eta_1 >1$ and $\eta_2,\varepsilon$ sufficiently small, the left side dominates the right side, which is a contradiction.
\end{proof}

\section{Monotone large-amplitude shocks}
\label{noreal}

In this section, we show that monotone profiles have no unstable real spectrum.  Our proof follows from a novel energy estimate that generalizes that of \cite{BHRZ} to a general ideal gas law and the addition of a capillarity term.  This restricts the class of admissible bifurcations for monotone profiles to those of Hopf-type, where one or more conjugate pairs of eigenvalues cross the imaginary axis.

\begin{theorem}
Monotone shocks of \eqref{slemrod} have no unstable real spectrum.
\end{theorem}
\begin{proof}
Suppose that $\lambda\in [0,\infty)$.  Since profiles are monotone, we have that $\bV_x < 0$. We multiply \eqref{ep:2} by the conjugate $\bar{v}$ and integrate in $x$ from $-\infty$ to $\infty$.  This gives
\[
\ip \lambda u \bar{v} + \ip u' \bar{v} -  \ip f(\bV) v' \bar{v} = \ip \frac{u''\bar{v}}{\bV} - d \ip v''' \bar{v}.
\]
Notice that on the real line, $\bar{\lambda}=\lambda$.  Thus, we have
\[
\ip \bar{\lambda}u \bar{v} + \ip u' \bar{v} - \ip f(\bV) v' \bar{v} = \ip \frac{u''\bar{v}}{\bV} + d \ip v'' \bar{v}'.
\]
Using \eqref{ep:1} to substitute for $\overline{\lambda v}$ in the first term and for $u''$ in the last term, we get
\[
\ip u (\bar{u}'-\bar{v}') + \ip u' \bar{v} - \ip f(\bV) v' \bar{v} = \ip \frac{(\lambda v' + v'')\bar{v}}{\bV} + d \ip v'' \bar{v}'.
\]
Separating terms and simplifying gives
\[
\ip u \bar{u}' + 2 \ip u' \bar{v} -\ip f(\bV) v' \bar{v} = \lambda  \ip \frac{v'\bar{v}}{\bV} + \ip \frac{v'' \bar{v}}{\bV} + d \ip v'' \bar{v}'.
\]
We further simplify by substituting for $u'$ in the second term and integrating  the last terms by parts to give,
\[
\ip u \bar{u}' + 2 \ip (\lambda v + v') \bar{v} - \ip \left( f(\bV) + \frac{\bV_x}{\bV^2} + \frac{\lambda}{\bV} \right) v' \bar{v} + \ip  \frac{|v'|^2}{\bV} = d \ip v'' \bar{v}',
\]
which yields
\[
\ip u \bar{u}' + 2 \lambda \ip |v|^2 + \ip \left(2 +a p'(\bV) - \frac{\lambda}{\bV}\right) v' \bar{v} + \ip \frac{|v'|^2}{\bV} = d \ip v'' \bar{v}'.
\]
By taking the real part (recall that $\lambda \in [0,\infty)$), we arrive at
\[
\lambda \ip \left(2 - \frac{\bV_x}{2\bV^2} \right) |v|^2 - \frac{a}{2} \ip  p''(\bV) \bV_x |v|^2 + \ip \frac{|v'|^2}{\bV} = 0.
\]
This is a contradiction.  Thus, there are no positive real eigenvalues for monotone shock layers in \eqref{slemrod} .
\end{proof}

\section{High-frequency bounds}
\label{hfb}

In this section, we prove high-frequency spectral bounds for monotone large-amplitude smooth shock profiles.  This provides a ceiling as to how far along both the imaginary and real axes that one must explore for point spectra when doing Evans function computations.  Indeed to check for roots of the Evans function in the unstable half-plane, say using the argument principle, one needs only compute within these bounds.  If no roots are found therein, then we have a numerical verification of spectral stability.  We remark that in this section and the next, we depart from the generality of an ideal gas, and restrict ourselves to the adiabatic case; see Section \ref{adiabatic}.  We remark, however, that we could have carried out our analysis for an ideal gas as long as $v_-=1$, which we can achieve by rescaling.  We have the following lemmata:

\begin{lemma}
The following identity holds for $\varepsilon_1,\varepsilon_2,\theta>0$ and $\R \lambda \geq 0$:
\begin{align}
(\R(\lambda) &+ |\I (\lambda)|)  \ip \bV |u|^2 + (1-\varepsilon_1 - \varepsilon_2) \ip |u'|^2\notag\\
& \leq \left[ \frac{1}{4\varepsilon_1} + \frac{C}{2 \theta}\right] \int \bV |u|^2  + d^2 \int \left[\frac{1}{4} + \frac{1}{2\varepsilon_2}\right] |v''|^2 + \theta \int f(\bV) |v'|^2  \label{id1},
\end{align}
where $C = \sup |f(\bV)\bV|$.
\end{lemma}

\begin{proof}
We multiply \eqref{ep:2} by $\bV {\bar u}$ and integrate along $x$ from $-\infty$ to $\infty$.  This yields
\[
\lambda \ip \bV |u|^2 + \ip \bV u'\bar{u} + \ip |u'|^2 = \ip f(\bV) \bV v'\bar{u} + d\int \bV_x v'' \bar{u} + d\int \bV v'' \bar{u}' .
\]
Taking the real and imaginary parts, adding them together, and noting that $|\R(z)| + |\I(z)| \leq \sqrt{2}|z|$, yields
\begin{align*}
&(\R(\lambda) + |\I (\lambda)|)  \ip \bV |u|^2 -\frac{1}{2}\ip \bV_x|u|^2 + \ip |u'|^2\\
& \leq \ip \bV |u||u'| + \sqrt{2} \ip f(\bV)\bV|v'||u| + \sqrt{2} d \left[ \ip |\bV_x| |v''| |u| + \ip \bV |v''| |u'|\right] \\
& \leq \varepsilon_1 \ip \bV |u'|^2 + \frac{1}{4\varepsilon_1}\ip \bV |u|^2 +\theta \ip f(\bV) |v'|^2 + \frac{1}{2\theta}\ip f(\bV) \bV^2|u|^2 \\
&\qquad + \frac{1}{2} \ip |\bV_x| |u|^2 + d^2 \int |\bV_x| |v''|^2 + \varepsilon_2 \ip \bV |u'|^2 + \frac{d^2}{2 \varepsilon_2} \ip \bV |v''|^2\\
& < (\varepsilon_1+\varepsilon_2) \ip |u'|^2 + \left[ \frac{1}{4\varepsilon_1} + \frac{C}{2\theta}\right] \ip \bV |u|^2 +\theta \ip f(\bV) |v'|^2 \\
&\qquad + \frac{1}{2} \ip |\bV_x| |u|^2 + d^2 \int \left[\frac{1}{4} + \frac{1}{2\varepsilon_2}  \right]|v''|^2.
\end{align*}
Rearranging terms yields \eqref{id1}.
\end{proof}

\begin{lemma}
\label{kawashima}
The following identity holds for $\R \lambda \geq 0$:
\begin{equation}
\label{id2}
\ip |u'|^2 \geq \frac{1}{2} \ip \left[ f(\bV) - p'(\bV) \right] |v'|^2 + d \ip |v''|^2.
\end{equation}
\end{lemma}

\begin{proof}
We multiply \eqref{ep:2} by ${\bar v'}$ and integrate along $x$ from $-\infty$ to $\infty$.  This yields
\[
\lambda \ip u\bar{v}' + \ip u'\bar{v}' - \ip f(\bV) |v'|^2 = \ip \frac{1}{\bV}u''\bar{v}' - d\ip v'''\bar{v}'.
\]
Using \eqref{ep:1} on the right-hand side, integrating by parts, and taking the real part gives
\[
\R \left[ \lambda \ip u\bar{v}' + \ip u'\bar{v}'\right] = \ip \left[ f(\bV) + \frac{\bV_x}{2 \bV^2} \right] |v'|^2 + \R(\lambda)\ip \frac{|v'|^2}{\bV} + d\ip|v''|^2.
\]
In our domain of interest, this yields
\begin{equation}
\label{id3_1}
\R \left[ \lambda \ip u\bar{v}' + \ip u'\bar{v}'\right] \geq \frac{1}{2} \ip \left[f(\bV) - p'(\bV) \right] |v'|^2  + d\ip|v''|^2
\end{equation}
Now we manipulate the left-hand side.  Note that
\begin{align*}
\lambda \ip u\bar{v}' + \ip u'\bar{v}' &= (\lambda+\bar{\lambda}) \ip u\bar{v}' - \ip u(\bar{\lambda}\bar{v}' + \bar{v}'')\\
&= -2\R(\lambda) \ip u' \bar{v} - \ip u \bar{u}''\\
&= -2\R(\lambda) \ip (\lambda v + v') \bar{v} + \ip |u'|^2.
\end{align*}
Hence, by taking the real part we get
\[
\R \left[ \lambda \ip u\bar{v}' + \ip u'\bar{v}'\right] = \ip |u'|^2 - 2\R(\lambda)^2 \ip |v|^2.
\]
This combines with \eqref{id3_1} to give \eqref{id2}.
\end{proof}

Now we prove our high-frequency bounds.

\begin{theorem}
\label{thmhfb}
For a monotone profile with $0 \leq d \leq 1/3$, any eigenvalue $\lambda$ of \eqref{ep} with nonnegative real part satisfies
\begin{equation}
\label{hfbounds}
\R(\lambda) + |\I(\lambda)| \leq 3 + \frac{12 C}{5},
\end{equation}
where $C = \sup|f(\bV)\bV|$.
\end{theorem}

\begin{proof}
Combining \eqref{id1} and \eqref{id2}, we have
\begin{align*}
(\R(\lambda) &+ |\I (\lambda)|)  \ip \bV |u|^2 + (1-\varepsilon_1 - \varepsilon_2) \left[ \frac{1}{2}\ip f(\bV)|v'|^2 + d \ip |v''|^2 \right] \\
& \leq \left[ \frac{1}{4\varepsilon_1} + \frac{C}{2 \theta}\right] \int \bV |u|^2  + d^2 \int \left[\frac{1}{4} + \frac{1}{2\varepsilon_2}\right] |v''|^2 + \theta \int f(\bV) |v'|^2.
\end{align*}
Setting $\theta = (1-\varepsilon_1 - \varepsilon_2)/2$ yields
\begin{align*}
(\R(\lambda) &+ |\I (\lambda)|)  \ip \bV |u|^2 + (1-\varepsilon_1 - \varepsilon_2) d \ip |v''|^2 \\
& \leq \left[ \frac{1}{4\varepsilon_1} + \frac{C}{1-\varepsilon_1 - \varepsilon_2}\right] \int \bV |u|^2  + d^2 \int \left[\frac{1}{4} + \frac{1}{2\varepsilon_2}\right] |v''|^2.
\end{align*}
Hence for  $0 \leq d \leq 1/3$, choose $\varepsilon_1 = 1/12$ and $\varepsilon_2 = 1/2$ to get \eqref{hfbounds}.
\end{proof}

\begin{remark}
For an adiabatic gas, $p(\bV) = v^{-\gamma}$, $\gamma\geq 1$, we can show that $C \leq \gamma$; see \cite{BHRZ}.  Thus in the range $\gamma\in[1,3]$ we can safely bound the unstable spectrum with a half circle of radius 12.  This compactifies the region of admissible unstable spectrum, thus allowing us to numerically compute winding numbers of the Evans function and determine whether shock layers are spectrally stable.
\end{remark}

\section{Evans function computation}
\label{evans}

In this section, we numerically compute the Evans function to determine whether any unstable eigenvalues exist in our system.  The Evans function $D(\lambda)$ is analytic to the right of the essential spectrum and is defined as the Wronskian of decaying solutions of \eqref{ep}; see \cite{AGJ}.  In a spirit similar to the characteristic polynomial, we have that $D(\lambda)=0$ if and only if $\lambda$ is an eigenvalue of the linearized operator \eqref{ep}.  While the Evans function is generally too complex to compute analytically, it can readily be computed numerically; see \cite{HuZ.2} and references within.

Since the Evans function is analytic in the region of interest, we can numerically compute its winding number in the right-half plane.  This allows us to systematically locate roots (and hence unstable eigenvalues) within.  As a result, spectral stability can be determined, and in the case of instability, one can produce bifurcation diagrams to illustrate and observe its onset.  This approach was first used by Evans and Feroe \cite{EF} and has been applied to various systems since; see for example \cite{PSW,AS,Br.2,BDG}.

\subsection{Numerical Setup}

We begin by writing \eqref{ep} as a first-order system $W' = A(x,\lambda) W$, where
\begin{equation}
\label{evans_ode}
A(x,\lambda) = \begin{pmatrix}0 & \lambda & 1 & 0\\0 & 0 & 1 & 0\\0 & 0 & 0 & 1\\ \lambda/d&\lambda/d& h/d & -(d\bV)^{-1}\end{pmatrix},\quad W = \begin{pmatrix} u\\v\\v'\\v''\end{pmatrix},
\end{equation}
and $h = h(\bV,\lambda) := 1+a p'(\bV)+\bV_x/\bV^2-\lambda/\bV$.  Note that eigenvalues of \eqref{ep} correspond to nontrivial solutions of $W(x)$ for which the boundary conditions $W(\pm\infty)=0$ are satisfied.  We remark that since $\bV$ is asymptotically constant in $x$, then so is $A(x,\lambda)$.  Thus at each end-state, we have the constant-coefficient system
\begin{equation}
\label{apm}
W' = A_\pm(\lambda) W, \qquad A_\pm(\lambda) := \lim_{x\rightarrow\pm\infty}A(x,\lambda).
\end{equation}
Hence solutions that satisfy the needed boundary condition must emerge from the 2-dimensional unstable manifold $W_1^-(x) \wedge W_2^-(x)$ at $x=-\infty$ and also the 2-dimensional stable manifold $W_3^+(x) \wedge W_4^+(x)$ at $x=\infty$.  In other words, eigenvalues of \eqref{ep} correspond to the values of $\lambda$ for which these two manifolds intersect, or more precisely, when $D(\lambda) = 0$, where
\[
D(\lambda) := (W_1^- \wedge W_2^- \wedge W_3^+ \wedge W_4^+)_{\mid x=0} = \det(W_1^- W_2^- W_3^+ W_4^+)_{\mid x=0}.
\]

We cannot naively produce the stable and unstable manifolds numerically.  Indeed with two exponential growth and decay modes, problems with stiffness arise.  Hence, we use the compound-matrix method to analytically track the stable and unstable manifolds; see \cite{AlBr,BDG, Br.1, Br.2,HuZ.2}.  Specifically we lift $A(x,\lambda)$ into the exterior-product space $\Lambda^2(\mathbb{C}^4) \approx \mathbb{C}^6$ to get
\[
A^{(2)}(x,\lambda) = \begin{pmatrix}
0&1&0&-1&0&0\\
0&0&1&\lambda&0&0\\
\lambda/d &h/d&-(d\bV)^{-1}&0&\lambda&1\\
0&0&0&0&1&0\\
-\lambda/d&0&0&h/d&-(d\bV)^{-1}&1\\
0&-\lambda/d&0&-\lambda/d&0 & -(d\bV)^{-1}.
\end{pmatrix}
\]
We then consider single trajectories $W_\pm(x)$ of the ``lifted'' problem
\[
W' = A^{(2)}(x,\lambda) W
\]
on each side corresponding to the simple dominant growth and decay modes at the left and right end states, respectively.  These trajectories correspond to the 2-forms  $W_1^-(x) \wedge W_2^-(x)$ and $W_3^+(x) \wedge W_4^+(x)$, and can be effectively wedged together when they meet at zero; see \cite{AlBr} for an excellent overview of this method.

As an alternative, we consider the adjoint formulation of the Evans function \cite{PW,BSZ}.  Specifically, we integrate the trajectory $\widetilde{W}_+$ along the largest growth mode of the adjoint ODE
\begin{equation}
\label{adjode}
\widetilde{W}' = -A^{(2)}(x,\lambda)^* \widetilde{W}.
\end{equation}
starting at $x=\infty$.  We then define the (adjoint) Evans function to be $D_+(\lambda): = (\widetilde{W}_+\cdot  W_-)_{\mid x=0}$.  Note that $\widetilde{W}_+$ corresponds to the orthogonal complement of the 2-form $W_3^+(x) \wedge W_4^+(x)$ and so orthogonality of $\widetilde{W}_+$ and $W_-$ corresponds to intersection of the stable and unstable manifolds.

To further improve the numerical efficiency and accuracy of the shooting scheme, we rescale $W$ and $\widetilde{W}$ to remove exponential growth/decay at infinity, and thus eliminate potential problems with stiffness.  Specifically, we let $W(x) = e^{\mu^- x} V(x)$, where $\mu^-$ is the largest growth rate of the unstable manifold at $x=-\infty$, and we solve instead $V'(x) = (A^{(2)}(x,\lambda)-\mu^- I)V(x)$.  We initialize $V(x)$ at $x=-\infty$ as eigenvector $r^-$ of $A^{(2)}_-(\lambda)$ corresponding to $\mu^-$.  Similarly, it is straightforward to rescale and initialize $\widetilde{W}(x)$ at $x=\infty$.  This method is known to have excellent accuracy \cite{Br.1,Br.2,BrZ,BDG,HuZ.2,BHRZ};  in addition, the adaptive refinement gives automatic error control.  Finally, in order to maintain analyticity, the initial eigenvectors $r^-(\lambda)$ are chosen analytically using Kato's method; see \cite[pg. 99]{Kato} and also \cite{BrZ,BDG,HSZ}.

\begin{figure}[t]
\begin{center}
$\begin{array}{cc}
\includegraphics[width=6cm]{figures//typical1} & \includegraphics[width=6cm]{figures//typical2} \\
\mbox{\bf (a)} & \mbox{\bf (b)}\\
\includegraphics[width=6cm]{figures//typical3} & \includegraphics[width=6cm]{figures//typical4} \\
\mbox{\bf (c)} & \mbox{\bf (d)}\\
\includegraphics[width=6cm]{figures//typical5} & \includegraphics[width=6cm]{figures//typical6} \\
\mbox{\bf (e)} & \mbox{\bf (f)}\\
\end{array}$
\end{center}
\caption{Evans function output for semi-circular contour of radius 12 with $d=0.45$ and (a) $v_+=0.65$, (b) $v_+=0.45$, (c) $v_+=0.35$, (d) $v_+=0.25$, (e) $v_+=0.20$, and (f) $v_+=0.15$.  Although the contours wrap around the origin as the shock strength increases, they clearly have winding number zero, thus demonstrating spectral stability.}
\label{runs}
\end{figure}

\subsection{Numerical Experiments}

We truncate the domain to a sufficiently large interval $[L_-,L_+]$ in order to do numerical computation.  Some care needs to be taken, however, to make sure that we go out far enough to produce good results.  Our experiments, described below, were primarily conducted using $L_\pm=\pm 25$, but for weaker shocks we had to go out as far as $L_\pm=\pm 50$.  For highly oscillatory profiles, very large values of $L_\pm$ are needed because the (under-damped) decay rate can be small.  To compute the profile, we used Matlab's {\tt bvp4c} routine, which is an adaptive Lobatto quadrature scheme.

\begin{figure}[t]
\begin{center}
\includegraphics[width=12cm]{figures/several}
\end{center}
\caption{Evans function output of a semi-circular contour with $d=0.75$ and $v_+\in [0.20,0.80]$.  As the shock strength increases, the contours get closer to the origin and begin to wrap around it.  In the small shock limit, the contour drifts away from the origin and gets smaller.}
\label{several}
\end{figure}

\begin{figure}[t]
\begin{center}
\includegraphics[width=12cm]{figures/several2}
\end{center}
\caption{Evans function output of a semi-circular contour with $v_+=0.25$ and $d\in [0.15,0.80]$. As the $d$ decreases, the contours get larger and more spread out. }
\label{several2}
\end{figure}

\begin{figure}[t]
\begin{center}
\includegraphics[width=12cm]{figures/param}
\end{center}
\caption{Dots correspond to runs with parameters $(v_+,d)$.  The upward increasing curve corresponds to the critical value $d_*$ between monotone and oscillatory shock profiles.}
\label{param}
\end{figure}

Our experiments were carried out uniformly on the range
\[
(v_+, d)\in [0.10,0.80]\times [0.05,0.80],
\]
with $\gamma=1.4$ (diatomic gas).  In terms of Mach number, this corresponds roughly to $1.15 \leq M \leq 5$, which covers the supersonic range and goes into the hypersonic regime.  Indeed $M\approx 5$ may even go beyond the physical range of the model.  For each $(v_+, d)$ on our grid, we computed the Evans function along a semi-circular contour in the right-half plane of radius $12$ centered at the origin.  Recall that for $d\leq 1/3$, this contains the admissible region of unstable spectrum from our high-frequency bounds.  The ODE calculations for individual values of $\lambda$ were carried out using Matlab's {\tt ode45} routine, which is the adaptive 4th-order Runge-Kutta-Fehlberg method (RKF45).  Typical runs involved between $100$ and $700$ mesh points, with error tolerance set to {\tt AbsTol = 1e-6} and {\tt RelTol = 1e-8}.  Values of $\lambda$ were varied on the semi-circular contour with $70$ points in the first quadrant, 40 on the arc and 30 along the imaginary axis, and then reflected along the real axis due to the conjugate symmetry of the Evans function, that is, $\overline{D(\lambda)}=D(\overline{\lambda})$.

In Figure \ref{runs}, we see a typical run for increasing $v_+$.  Notice that the contour wraps around the origin as the shock strength increases.  Thus it is difficult to conclude stability in the strong shock limit; this is a topic for future consideration.  Note that the graph gets farther away from the origin as the shock strength decreases, thus strongly suggesting stability in the small-amplitude limit.  In Figure \ref{several}, we see this effect more clearly.  In Figure \ref{several2}, we hold the shock strength fixed and vary $d$.  As $d$ approaches zero, we see the contour getting larger and more spread out.  Otherwise output does not seem to vary much in $d$, at least in our region of interest.

The actual parameter values computed were
\[
(v_+,d) \in \{0.10,0.15,\ldots,0.80\} \times \{ 0.05,0.10,\ldots,0.80\};
\]
see Figure \ref{param}.  In total $240$ runs were conducted, all of which had winding number zero.  This effectively demonstrates spectral stability for monotone and nearly monotone profiles with $d\leq 1/3$ and strongly suggests spectral stability elsewhere in our region of study. Indeed the output is strikingly similar throughout.  Nonetheless, for $d>>1$ our profile becomes highly oscillatory and so it is not unreasonable to expect an instability to occur in the extreme.  This is a good direction for future work.

\section{Discussion and Open Problems}

We note that \eqref{evans_ode} blows up as $v_+\rightarrow 0$, and moreover the eigenvalues get far apart, thus  causing extreme stiffness.  Hence we have numerical difficulties for strong shocks, e.g, $M>>5$.  Difficulties also arise for both large and small values of $d$.  In particular the profile becomes highly oscillatory and numerically intractable for very large values of $d$, and as $d\rightarrow 0$ we likewise have that \eqref{evans_ode} blows up.  Nonetheless, we may be able to demonstrate stability as $d\rightarrow 0$ analytically as a singular limit of the $d=0$ case, which is stable; see \cite{HLZ}.

Slemrod's model is an ideal system for further investigation.  Not only is it physically relevant, and in some sense a canonical viscous-dispersive system, but it also pushes the boundaries of current numerical methods.  While this model has nice features such as monotone profiles, it also has highly complex and numerically taxing obstacles such as highly oscillatory profiles and large spectral separation between modes of \eqref{evans_ode} in the extreme parameter regime.  We intend to study this model further.

\def\ocirc#1{\ifmmode\setbox0=\hbox{$#1$}\dimen0=\ht0 \advance\dimen0
  by1pt\rlap{\hbox to\wd0{\hss\raise\dimen0
  \hbox{\hskip.2em$\scriptscriptstyle\circ$}\hss}}#1\else {\accent"17 #1}\fi}
  \def\cprime{$'$}

config: arxiv


TITLE: Show $\int_{0}^{1}(\int_{0}^{x}g(t)dt)^2 dx\leq\frac{1}{2}\int_0^1(1-x^2)(g(x))^2 dx$ for any $g(x)$ continuous
QUESTION [0 upvotes]: Prove (or disprove) that
$$\int_{0}^{1}\left(\int_0^x g(t)\ dt\right)^2dx\leq\frac{1}{2}\int_0^1 (1-x^2)(g(x))^2 dx$$
for any $g(x)$ continuous on $[0,1]$.
I have verified the cases of $g(x)$ being monomials like $x^k (k\in \mathbb{N})$ and found the equality holds iff $k=0$, namely, $g(x)\equiv 1$.

REPLY [1 votes]: \begin{align*}
\int_{0}^{1}\left(\int_{0}^{x}g(t)dt\right)^{2}dx&=\int_{0}^{1}\left(\int_{0}^{1}\chi_{0\leq t\leq x}g(t)dt\right)^{2}dx\\
&\leq\int_{0}^{1}\left(\left(\int_{0}^{1}\chi_{0\leq t\leq x}^{2}dt\right)^{1/2}\left(\int_{0}^{1}(\chi_{0\leq t\leq x}g(t))^{2}dt\right)^{1/2}\right)^{2}dx\\
&=\int_{0}^{1}\left(\int_{0}^{x}dt\right)\left(\int_{0}^{x}(g(t))^{2}dt\right)dx\\
&=\int_{0}^{1}x\left(\int_{0}^{x}(g(t))^{2}dt\right)dx\\
&=\int_{0}^{1}\int_{0}^{x}x(g(t))^{2}dtdx\\
&=\int_{0}^{1}\int_{t}^{1}x(g(t))^{2}dxdt\\
&=\int_{0}^{1}(g(t))^{2}\int_{t}^{1}xdxdt\\
&=\int_{0}^{1}(g(t))^{2}\dfrac{1}{2}(1-t^{2})dt\\
&=\dfrac{1}{2}\int_{0}^{1}(1-x^{2})(g(x))^{2}dx.
\end{align*}


TITLE: Natural numbers verifying $P(n) = n^2 - 42n + 440$, where $P(n)$ is the product of the digits
QUESTION [4 upvotes]: Let $P(n)$ be the product of the digits of the number $n$, with $n \in  \mathbb{N}$.
What is the product of all the natural numbers $n$ that verify the equation $P(n) = n^2 - 42n + 440$?

I factorized the polynomial to get $P(n) = (n - 20)(n - 22)$. Then I observed that $n = 20$ is a solution because $P(20) = 0$.
But at this point I got stuck, so I wrote an Haskell program that would calculate the solutions for me. After letting it run for a while I checked the output, which was: $[18, 20, 24]$. So apparently the answer is $18 \cdot 20 \cdot 24 = 8640$.
What is the correct, mathematical way to solve this?

REPLY [2 votes]: Let $n = \sum_{i=0}^{k-1} d_i \cdot 10^i$  be a number with $k$ digits and $d_i \in \{0,...,9\}$. Then
$$P(n) := \prod_{i=0}^{k-1} d_i$$
So the question is what the solutions are for the equation
\begin{align}
P(n) &= n^2 - 42n + 440\\
\Leftrightarrow \prod_{i=0}^{k-1} d_i &= \left ((\sum_{i=0}^{k-1} d_i \cdot 10^i)-20 \right ) \cdot \left ((\sum_{i=0}^{k-1} d_i \cdot 10^i)-22 \right)\\
\end{align}
Restrictions for solutions

We know that every solution is a natural number (hence > 0).
As every digit is at most $9$, the left hand side of the equation is at most $9^k = 9^{\lfloor \log n \rfloor + 1} = 9 \cdot 9^{\lfloor \log n \rfloor} < 9 \cdot 9^{\log n} = 9 \cdot n^{\log{9}} < 9 \cdot n$.
It is easy to show that $\exists n_0 \in \mathbb{N}: \forall n > n_0: n^2- 42n+440 > 9 n$ is true:

\begin{align}
n^2- 42n+440 &\geq 9n\\
\Leftrightarrow n^2- 51n+440 &\geq 0\\
\Leftrightarrow (n-11) \cdot (n-40) &\geq 0
\end{align}

So for $n > n_0 = 40$ we know that there cannot be a solution.

Check candidates
Use the following Python script to check the remaining 40 candidates:
#!/usr/bin/env python

def P(n):
    prod = 1
    for digit in str(n):
        prod *= int(digit)
    return prod

def check(n):
    return P(n) == n**2 - 42*n + 440

if __name__ == '__main__':
    for i in range(41):
        if check(i):
            print(i)

Answers

18
20
24


TITLE: Semicircle Question
QUESTION [5 upvotes]: I need help with the question in the image. I just need someone to help by pointing me in the right direction. I don't want a full solution. I want to try to work out this question myself but I just need someone to direct me.

I was thinking if the question had something to do with joining the three points in the circle where the circle is touching the semicircles?
Or perhaps this question has something to do with similar triangles? I know that an angle subtended by an arc inside a semicircle is 90 degrees.

REPLY [1 votes]: Let the center of small semi-circle be C, point on the AB where circle touches touches is D, p denotes center of the circle, E is the point where circle touches bigger semi-circle. Le the circle touch the smaller semi-circle at T. Let the radius of the circle be $r$, then
$$OP=2-r, OD=\sqrt{(2-r)^2-r^2}=\sqrt{4-4r}, CD=\sqrt{(1+r)^2-r^2}=\sqrt{2r+1}.$$
Finally,  $$CD-OD=1 \implies \sqrt{1+2r}-\sqrt{4-4r}=1 \implies r=8/9.$$


TITLE: Is "being decomposable" preserved under taking a subspace?
QUESTION [1 upvotes]: Let $V$ be a vector space over some field, and $W \le V$ a vector subspace. Let $1<k<\dim V$ be an integer.

Suppose $\omega \in \bigwedge^k W$ is decomposable as an element in $\bigwedge^k V$. Is it decomposable as an element in $\bigwedge^k W$?

REPLY [1 votes]: The answer is positive. Let $w_i$ be a basis for $W$. Write $\omega=\sum_{1 \le i_1 <\dots<i_k \le \dim W)}  w_{i_1} \wedge \ldots \wedge w_{i_k}$. Let $v \in V \setminus W$. Since $w_i,v$ are linearly independent, we can complete them into a basis of $V$. This implies that $\omega \wedge v=\sum_{1 \le i_1 <\dots<i_k \le \dim W)}  w_{i_1} \wedge \ldots \wedge w_{i_k} \wedge v \neq 0$.
By assumption, $\omega \in \bigwedge^k W$ is decomposable in $\bigwedge^k V$, i.e. $\omega=v_1 \wedge \dots \wedge v_k$, for some $v_i \in V$. Since $v_1 \wedge \dots \wedge v_k \wedge v=\omega \wedge v \neq 0$, we deduce that $v \notin \{v_1,\dots,v_k\}$. Since $v \in V \setminus W$ was arbitrary, we have proved that $V \setminus W=W^c \subseteq \{v_1,\dots,v_k\}^c$, so $ \{v_1,\dots,v_k\} \subseteq W$, so $\omega=v_1 \wedge \dots \wedge v_k$ is decomposable in $\bigwedge^k W$.
An alternative proof:
$\omega$ is decomposable in $\bigwedge^k V$, if and only if $V_{\omega}=\{ v \in V \, | \, v \wedge \omega=0\}$ is $k$-dimensioal.
Now, $v \in V \setminus W \Rightarrow v \wedge \omega \neq 0 \Rightarrow v \notin V_{\omega}$, i.e. $W^c \subseteq V_{\omega}^c$. Thus,
 so $V_{\omega} \subseteq W$, hence
$$ W_{\omega}=V_{\omega} \cap W=V_{\omega}$$
is also $k$-dimensional, so $\omega$ is decomposable in $\bigwedge^k W$.


TITLE: How prove this limit $\lim_{n\to\infty}a_{n}=0$
QUESTION [0 upvotes]: Question:
let $f(x)\gt0(x\geqslant0)$ is increasing function, let three sequence
$\{a_{n}\},\{b_{n}\},\{c_{n}\}$, such $a_{n}\geqslant 0,b_{n}\geqslant 0,c_{n}\geqslant 0$ and
$$a_{n+1}\leqslant a_{n}-b_{n}f(a_{n})+c_{n}a_{n},\sum_{b=1}^{\infty}b_{n}=\infty,\sum_{n=1}^{\infty}c_{n}\lt\infty$$
show that:
$$\lim_{n\to\infty}a_{n}=0$$
since
$$a_{n+1}-a_{n}\leqslant -b_{n}f(a_{n})+c_{n}a_{n}$$
so
$$a_{n}-a_{1}\leqslant -(b_{1}f(a_{1})+b_{2}f(a_{2})+\cdots+b_{n-1}f(a_{n-1}))+c_{1}a_{1}+c_{2}a_{2}+\cdots+c_{n-1}a_{n-1}$$
Then I am stuck , it is said this is hard analysis problem.and I want know this problem how prove it,Thank you

REPLY [1 votes]: Set $x_n=a_n/d_n$, $1\le d_n=(1+c_1)...(1+c_{n-1})\le d_*<\infty$. Then
$$x_{n+1}\le x_n-b_n\frac{f(d_nx_n)}{d_{n+1}}\le x_n-b_n\frac{f(x_n)}{d_*}.$$
This reduced form should still be sufficient to exclude convergence towards any positive limit.

For the first step notice that rearranging the given inequality reads as
$$a_{n+1}\le (1+c_n)a_n-b_nf(a_n)$$
So dividing by $d_{n+1}$ gives $\frac{(1+c_n)a_n}{d_{n+1}}=\frac{a_n}{d_n}=x_n$. Note that $d_n$ is increasing and bounded by
$$d_n\le e^{c_1+\dots+c_{n-1}}\le\exp\left(\sum_{k=1}^\infty c_n\right)<\infty,$$
so there exists a limit $d_*=\lim_{n\to\infty}d_n=\sup_{n\to\infty}d_n$.


TITLE: Task Selection Problem
QUESTION [0 upvotes]: I am trying to distribute 15 tasks to two people. Each task can only be assigned to one person and each person has a time budget.
I want to express this problem as a linear program (ultimately in the standard linear form) so that I can write a computer program to solve it.
\begin{array}{|c|c|c|c|c|}
\hline
Workers& Task 1 & Task2 & Task3 & ... & Time budget \\ \hline
 Jack& 0.5& 0.25& 0.25 & & 9\\ \hline
 John&  0.75& 1.0 &0.75 & & 8\\ \hline
\end{array}
I can find plenty of examples how to do it without a budget constraint. Without it, it is an Assignment Problem that can be solved with The Hungarian Method.

REPLY [0 votes]: Hints: Let binary decision variable $x_{i,j}$ indicate whether person $i$ is assigned task $j$. Now you need a constraint that each task gets assigned to exactly one person, as well as a time budget constraint for each person.


TITLE: Proof: $ A - (B - C) \subseteq (A - B) - C$
QUESTION [0 upvotes]: Question:
Prove or disprove the following statements:
For all sets $A, B, C$:

a) $A - (B - C) \subseteq (A - B) - C$
b) $(A - B) - C \subseteq A - (B - C)$
c) If $A - (B - C) \subseteq (A - B) - C$, then $A \cap C= \varnothing$

My attempt to answer:
a) This is false.
b) This is true.
c) This is true.
Are these answers I correct?
Proof: Working on it.

REPLY [0 votes]: $(A \setminus B) \setminus C$ is all the elements of $A$ that are not in either B or C (removed one after the other). Intuitively, you would expect it to contain fewer elements that the set that contains all the elements of $A$ save for just the ones that are in $B$ but not $C$.
To see this rigorously:
$A \setminus (B \setminus C) = A \cap (B \setminus C)^ c = A \cap (B \cap C^c)^c = A \cap (B^c \cup C) = (A \cap B^c) \cup (A \cap C) = (A \setminus B) \cup (A \cap C) \supseteq (A \setminus B) \supseteq (A \setminus B) \setminus C$.
Naturally, for equality to hold, you need the second term in the set union to be empty, i.e. $A \cap C = \emptyset$. This ensures the second set inequality becomes an equality too.


TITLE: What is the name for the topology where every point forms an open set
QUESTION [5 upvotes]: I remember there is some name for a    special topology on a set, such that    every subset is open i.e. every point    forms an open set, but I cannot    recall it.
Please tell me its name or say I am    wrong.
Thanks!

REPLY [7 votes]: I believe it is called the Discrete Topology.


TITLE: Finding the $n$-th derivatives of $x^n \ln x$ and $\frac{\ln x}{x}$.
QUESTION [8 upvotes]: How can I prove the following identities:
$$ \left( x^n \ln x \right)^{(n)}=  n! \left(\ln x + 1 + \frac{1}{2} +\cdots +\frac{1}{n} \right), \quad x>0, \quad n\ge 1, \tag{a}$$
$$ \left( \frac{\ln x}{x} \right)^{(n)}= (-1)^n \ n! \ x^{-n-1}  \left( \ln x - 1 - \frac{1}{2} - \cdots -\frac{1}{n} \right), \quad x>0,\quad n\ge 1, \tag{b}$$
Here $(\cdot )^{(n)}$ indicates the $n$-th derivative with respect to $x$.
I don't know where to start!

REPLY [5 votes]: To finish off Yuval's argument, we try an inductive proof on
$$\sum_{k=1}^n \binom{n}{k}\frac{(-1)^k}{k}=-\sum_{k=1}^n \frac1{k}=-H_n$$
The equality is easily verified for $n=1$. We now try to simplify
$$\sum_{k=1}^{n+1} \binom{n+1}{k}\frac{(-1)^k}{k}-\sum_{k=1}^n \binom{n}{k}\frac{(-1)^k}{k}$$
which goes a bit like this:
$$\begin{align*}
&\frac{(-1)^{n+1}}{n+1}+\sum_{k=1}^n \binom{n+1}{k}\frac{(-1)^k}{k}-\sum_{k=1}^n \binom{n}{k}\frac{(-1)^k}{k}\\
&\frac{(-1)^{n+1}}{n+1}+\sum_{k=1}^n \left(\binom{n+1}{k}-\binom{n}{k}\right)\frac{(-1)^k}{k}\\
&\frac{(-1)^{n+1}}{n+1}+\frac1{n+1}\sum_{k=1}^n (-1)^k\binom{n+1}{k}\\
&\frac{(-1)^{n+1}}{n+1}+\frac1{n+1}\left(-1-(-1)^{n+1}+\sum_{k=0}^{n+1} (1)^{n-k+1}(-1)^k\binom{n+1}{k}\right)\\
&\frac{(-1)^{n+1}}{n+1}+\frac1{n+1}\left(-1-(-1)^{n+1}+(1-1)^{n+1}\right)=-\frac1{n+1}\\
\end{align*}$$
and since we also have $-H_{n+1}-(-H_n)=-\dfrac1{n+1}$, the equality is established.

Here's an explicit application of Leibniz's product rule for the second problem, which uses the identity $\dfrac{\mathrm d^n}{\mathrm dx^n}\dfrac1{x}=\dfrac{(-1)^n n!}{x^{n+1}}$:
$$\begin{align*}
\frac{\mathrm d^n}{\mathrm dx^n}\frac{\ln\,x}{x}&=\frac{(-1)^n n!}{x^{n+1}}\ln\,x+\sum_{k=1}^n\binom{n}{k}\left(\frac{\mathrm d^k}{\mathrm dx^k}\ln\,x\right)\left(\frac{\mathrm d^{n-k}}{\mathrm dx^{n-k}}\frac1{x}\right)\\
&=\frac{(-1)^n n!}{x^{n+1}}\ln\,x+\sum_{k=1}^n\frac{n!}{\color{blue}{k!}\color{magenta}{(n-k)!}}\left(\frac{\color{red}{(-1)^{k-1}}\color{blue}{(k-1)!}}{\color{orange}{x^k}}\right)\left(\frac{\color{red}{(-1)^{n-k}}\color{magenta}{(n-k)!}}{\color{orange}{x^{n-k+1}}}\right)\\
&=\frac{(-1)^n n!}{x^{n+1}}\ln\,x+\frac{\color{red}{(-1)^{n-1}} n!}{\color{orange}{x^{n+1}}}\sum_{k=1}^n\frac1{\color{blue} k}=\frac{(-1)^n n!}{x^{n+1}}\left(\ln\,x-\sum_{k=1}^n\frac1{k}\right)
\end{align*}$$


TITLE: Idea of using etale site
QUESTION [14 upvotes]: I have just read an article which mentions that, when Grothendieck considered using etale morphism, he did borrow the idea from Riemann that multivalued function on an open subset of complex plane should live in Riemann surface covering it instead of the open subset itself.
The question is: how these things are related? Any detailed explanation is very welcome and appreciated. More generally, what are the good properties etale morphism has which make it essential in solving Weil conjecture.I understand that Grothendieck tried to search for a cohomology theory in algebraic geometry which is similar to classical cohomology theory on manifold theory but  I dont know what obstructions are there preventing a cohomology theory in AG behaves similar to one in manifold for which etale site was introduced to overcome them.

REPLY [6 votes]: With regards to the Weil conjectures:
I believe it was known to Weil that, to prove most of the Weil conjectures, a cohomology theory for varieties with coefficients in a field of characteristic zero would just have to satisfy a few properties that are familiar from singular cohomology:

The Lefschetz fixed point formula, applied to the endomorphism Frob_p, and finiteness of Betti numbers immediately give rationality of the zeta function.
Poincare duality, suitably twisted to account for the fact that Frob_p should act on $H^{2n} (X)$ for $X$ an $n$-dimensional smooth projective variety by multiplication by $p^n$, gives the functional equation.
The Riemann hypothesis is the only one I'm not sure about - Deligne's proof of this involves many clever steps that had no antecedent in singular cohomology.
The fact that Betti numbers are constant in smooth proper fibrations suffices for the Betti number conjecture, and a comparison theorem to singular cohomology over the complex numbers, (although Weil may not have had the full picture of an arithmetic family of varieties as being the same as a geometric family as clearly as Grothendieck did)

So why does etale cohomology have these features? I think experience shows that any cohomology theory we define has these features as long as it satisfies some very basic tests - off the top of my head, I can't think of any cohomology theory that has the right Betti numbers for a curve of genus $g$ that doesn't have all these properties.
$\ell$-adic etale cohomology avoids the trivial failure of all the cohomology theories you can construct with just the Zariski topology - either cohomology of the constant sheaf $\mathbb Z$ in the Zariski topology, or cohomology of some coherent sheaf, probably powers of the cotangent bundle to mimic Dolbeaut cohomology in characteristic zero. The first one has the wrong Betti numbers, and the second one has coefficient field the base field, so has coefficients of characteristic $p$ in characteristic $p$, and so cannot prove the Weil conjectures.


TITLE: Homological dimension in algebraic geometry
QUESTION [0 upvotes]: 1-Do we have some relation between the projective dimension of $k[X]$ as $K[x_{1},..x_{n}]$-module and the Krull dimension of the affine variety $X$ of $A^{n}$.
2- If we have to affine subvarieties $X$,$Y$ of $A^{n}$. Do we have some relation between
$pdk[X]$ and $pdk[Y]$?
Thanks!

REPLY [3 votes]: 1) Assuming for simplicity that $X$ is integral, what you can say is $\ \mathrm{pd}(K[X])\geq n-\dim(X)$. After localizing at the (prime) ideal of $X$, this is an easy case of the Auslander-Buchsbaum theorem. And, of course, $\ \mathrm{pd}(K[X])\leq n$.
2) I just don't understand the question. If $X$ and $Y$ are arbitrary subvarieties, how can you expect any relations between the projective dimensions?


TITLE: Borel Measures: Single-Valued
QUESTION [0 upvotes]: Given the complex plane $\mathbb{C}$.
Consider the Dirac measure:
$$\mu_\lambda(A):=\chi_A(\lambda)$$
Then it attains only zero and one:
$$\mu_\lambda(A)=0,1$$
Are there any other such measures?

REPLY [2 votes]: Same proof, less technical.  Suppose $\mu$ is a measure taking exactly values $0,1$.  We claim $\mu$ is a dirac measure.
Let $U$ be the union of all open sets of measure zero.  Since $\mathbb C$ is 2nd countable, $U$ is the union of coutably many open sets of measure zero, so $\mu(U) = 0$.  Let $F$ be the complement of $U$.  Then $F$ has measure $1$.  I claim that $F$ is a singleton.  If not, there are two points $x \ne y$ in $F$, and thus there are disjoint open neighborhoods $U_x, U_y$ of them.  Since $U_x$ is not a subset of $U$, $\mu(U_x) = 1$.  Similarly $\mu(U_y)=1$.  So in fact $\mu(\mathbb C) \ge 2$ a contradiction.  Thus $F$ is a singleton, $\mu(F)=1$, $\mu(\mathbb C \setminus F) = 0$.
So I guess the point of doing another proof is: you don't need the exact structure of $\mathbb C$, merely that it is a 2nd countable Hausdorff space.


TITLE: Question about proof that every f.g. projective module over a local ring is free.
QUESTION [2 upvotes]: I'm reading the proof here.
I'm at the line where they say
$$
\psi\pi(f)=\psi(f+FR)=\varphi(f)+PR.$$
Since $\psi\pi$ is surjective, it should follow that $\{\varphi(f)+PR:f\in F\}=P/PR$. I don't understand the notation $\mathrm{im}(\varphi)+PR=P$. I assume $\mathrm{im}(\varphi)+PR=\{\varphi(f)+PR:f\in F\}$? If that's a subset of $P/PR$, how can it equal $P$?

REPLY [3 votes]: (This should just be a comment, but too many of these questions are left without posted answers...)
The notation $\text{im}(\varphi) + PR$ refers to the sum of the two submodules $\text{im}(\varphi)$ and $PR$ (these are subsets of $P$, not $P/PR$). Explicitly,
$$\text{im}(\varphi) + PR = \{\varphi(f) + \sum_{\text{finite}} pr \mid f \in F, p \in P, r \in R \}$$
The reason that $\text{im}(\varphi) + PR = P$, is the equality $\{\varphi(f)+PR:f\in F\}=P/PR$, which says that $\{\varphi(f)\}$ forms a complete set of coset representatives of $PR$ in $P$, so every element of $P$ is a sum of some $\varphi(f)$ with an element of $PR$.