TITLE: Inverse trigonometric functions used for difference in angle?
QUESTION [1 upvotes]: I was actually solving a physics question in which I got the equation 3/2 (sin i) = sin r. The graph of (r-i) against i has a constant positive slope till 45 degrees value for i. How do I get this result? The only thing I can get from the equation is that sin r - sin i = sin i/2. I don't have much of knowledge about inverse trigonometric functions, would someone please help me understand this in a simple way?

REPLY [1 votes]: Clearly, the "straight" solution is
$$
{3 \over 2}\sin i = \sin r\quad  \Leftrightarrow \quad r = \arcsin \left( {{3 \over 2}\sin i} \right)\quad  \Leftrightarrow \quad i = \arcsin \left( {{2 \over 3}\sin r} \right)
$$
But, just to catch the behaviour of the solution at lower value of the angles, we can use the Taylor series for $\sin$ to get
$$
\eqalign{
  & {3 \over 2}\sin i = \sin r  \cr
  & {3 \over 2}\left( {i - {{i^3 } \over 6} + O\left( {i^5 } \right)} \right) = r - {{r^3 } \over 6} + O\left( {r^5 } \right)  \cr
  & {3 \over 2}i\left( {1 - {{i^2 } \over 6}} \right) \cong r\left( {1 - {{r^2 } \over 6}} \right)  \cr
  & {3 \over 2}i \cong r \cr}
$$
Of course you can increase the precision of the approximation
by solving the cubic , instead of the last linear approximation.
Otherwise you can use the Taylor series for $\arcsin$, etc.


TITLE: Increasing union of contractible CW complexes
QUESTION [6 upvotes]: Let X be CW complex. I'm trying to prove (using Zorn's lemma) that there is maximal contractible subcomplex. Problem is that I'm not able to show that increasing union of contractible subcomplexes has to be contractible itself.

REPLY [14 votes]: By various standard lemmas, a CW complex $X$ is contractible if and only if every map $u:S^{n-1}\to X$ (for any $n>0$) can be extended over $B^n$.  In this context $u(S^{n-1})$ is compact and therefore (by another standard lemma) contained in some subcomplex with only finitely many cells.  If $X$ is the union of some totally ordered family of subcomplexes $X_\alpha$, it follows that $u(S^{n-1})\subseteq X_\alpha$ for some $\alpha$.  This is enough to prove what you want.


TITLE: Conditional probability of coin tosses
QUESTION [0 upvotes]: Suppose we have the following events:

X = "Number of heads in the first two tosses"
Z = "Number of tosses before first head".

I have to compute $P(Z=1 | X=1)$. I can proceed as follows:
$$ P(Z=1 | X=1) = \frac{P(Z=1 , X=1)}{P(X=1)}.$$
I can compute easily with the Binomial the denominator but I have no idea about the numerator. I know that the result should be $pq$ if the probability of getting head is $p$. Is there a way to visualize this probability?

REPLY [1 votes]: If the number of tosses before the first head is $1$, then in the first two tosses you have necessarily gotten exactly a single head, which means that
$$
(Z = 1)\subseteq (X = 1)
$$
as events. Consequently
$$
P(Z = 1, X = 1) = P(Z = 1)
$$


TITLE: Proving Discreteness of the Harmonic numbers
QUESTION [1 upvotes]: For the set $\{1/n\}$ with $n$ in the positive integers, how do I go about proving that the set is discrete?
I have that the distance between two consecutive terms is greater than zero, but does that finish the proof?

REPLY [1 votes]: If you mean the topological notion of discreteness with respect to the ambient topological space of the real numbers (e.g. here), then what you're saying is sort of the gist of it, but you need an additional line or two. We want to show for each $n$ that there is some inverval around $1/n$ which doesn't contain any other $1/m$ for $m\neq n,m\in\mathbb{N}$. Since $1/1 > 1/2 > 1/3 > \ldots$, it suffices to show that we can find an interval around $1/n$ strictly inside $[1/(n+1),1/(n-1)]$, and now that's easy, as you observed that the distance from $1/n$ to each of the endpoints is positive.


TITLE: Demonstration that vibrating basic particles constitute non-vibrating individuals
QUESTION [2 upvotes]: I am a dilettante in physics; I ask for pardon for my confusion-causing (if any) terminology usage, and also for my imprecise choice of question tags.
I know that basic particles of any individual stopping vibrating contradicts Heisenberg's uncertainty principle, so the particles must whenever vibrate. Then how could basic particles constitute a seemingly non-vibrating individual? Is it a natural limitation of human eyesight or ...? Is there any scientific explanation for it?
A demonstration not using analogy or common sense is seeking after.

REPLY [4 votes]: It is certainly true that as I type this the atoms in my fingers are vibrating, and so are the atoms in my keyboard. Yet I can still type. There are several reasons we don't perceive any vibration.
Firstly, the vibrational frequencies are around $10^{12}$ to $10^{14}$Hz, that's from a trillion times a second to 100 trillion times a second. Since we can't even see the 50Hz refresh on our TV screens it shouldn't be any surprise that at such high frequencies we can't detect any vibration of the atoms in us.
Secondly the vibrations are all in random directions, so on average they balance each other out. On average we aren't vibrating.
Lastly, and this is really the important point, atoms are quantum objects. We tend to visualise a vibrating atom as a little ball flying to and fro on the end of a spring but this isn't a good description of what a vibrating atom looks like. The vibration means the wavefunction of the atom spreads out, so in effect it just becomes a little blurry. Since this blurriness is on a scale of around 0.1 nm it is completely undetectable in everday life.

REPLY [1 votes]: I will use my limited knowledge about this topic to give a best answer I could.
Heisenberg uncertainty principle said that: (according to me)

We cannot at the same time "know" (or measure) the momentum and the position of a particle.

How can we know? How can we define the momentum and the position of a particle? There are many difficulties describing the quantum world by the normal language, because the normal language is created only to describe what we can see, feel directly. Sadly, the quantum world is not as simple as the macro world here. There is no particle (as a ball or a bullet) and there is no wave (as sound wave or water wave) in the micro length. There are just states of physics, which can give us predictions about the measurements we can take.
Then to translate the quantum physics language to more classical language, we must use some analogy of the states to, sometimes, particles and sometimes, waves. The description is therefore not always exactly true or understood in normal sense.

REPLY [1 votes]: You seem to be posing a question about the nature of the Wave-Particle Duality. If you're asking how can a particle be a non "vibrating" particle and a vibrating wave at the same time, then the answer is as follows:
With the Wave-Particle duality, we can view the same subatomic particles that we can apply point particle mechanics to as probability densities. These Probability density functions have wavelike nature. The function is explicitly called the Wavefunction and is typically denoted by $ \psi (\vec{x},t) $ and the probability density is $ |\psi (\vec{x},t)|^2 $. Solutions to the Equations of Motion for quantum potentials ( the Schrödinger Equation ) yield equations of motion analogous to the classical harmonic oscillator.
Even though say, an electron, is a point-like particle that at some point in time has a position with a degree of measurable certainty, we can still view the particle in terms of the derived mathematical construct of wave mechanics. In this manner, one can technically vibrating and not vibrating. It all depends on which framework you chose to utilize in analyzing a system of particles.


TITLE: Primes $p$ and $q$ such that $3p^{q-1}+1$ divides $11^p+17^p$
QUESTION [5 upvotes]: From the Balkan Mathematics Olympiad 2018:

Find all prime numbers $p$ and $q$ such that $3p^{q-1}+1$ divides $11^p+17^p$.

I started using Fermat's Little Theorem, which states that $a^{p-1}-1$ is divisible by $p$ if $p$ is a prime. But I could not find any way to solve the problem.
Any suggestions, please.

REPLY [2 votes]: COMMENT:
$11^p≡11 mod p=k_1 .p + 11$
$17^p≡17 mod p=K_2 .p + 17$
$11^q≡11 mod q=k_3 .q + 11$
$17^p≡17 mod q=K_4 .q + 17$
Summing these relations we get:
$(11^p+17^q)+(11^q+17^p)=56+(k_1+k_2)p+(k_3+k_4)q$
P and q are primes, therefore numbers $(11^p+17^q)$ , $(11^q+17^p)$,  $3.p^{q-1} +1$, $(k_1+k_2)$ and $(k_3+k_4)$ must have common divisors like$ 2, 4, 7, 8, 14, 28 and 56$.
Now $3.p^{q-1}+1$ ≤ $(11^p+17^q)= 56 m$; $m∈N$. This is the condition of question. For example with p=q=3 we get 28. I think better question is: Find all possible divisors of $11^p+17^q$ and $3p^{q-1}+1$


TITLE: Does a well-lit mirror weigh more than an unlit mirror?
QUESTION [22 upvotes]: If you weighed a mirror in a room with no light, and then weighed a mirror in a well lit room so that the mirror reflects light, would the weight be different?

REPLY [0 votes]: While some light energy may be absorbed by the mirror, the only light that has an effect on the image is the light that bounces back off.
For the same reason a ball touching a bat isn't considered as an increase in bat weight, the mirror's weight is not considered as an increase when the light hits the glass.
Further, the light photons that are absorbed (and are not affecting the image appearance) affect nearly every object in similar ways. Thus it is not special to the image either.


TITLE: Why is the matter orbiting a black hole in the shape of a disk?
QUESTION [2 upvotes]: More generally, if a blackhole is a singularity...a point in space...why is matter only circling in a flat disk-like region.  It would seem to me matter would circle it in a sphere surrounding the black hole.
Furthermore...and this also echoes another question asked here at some point...why does the matter circle in one direction?
Being a lay-person in physics, I'm sure there are some basic principles I'm not aware of, but it seems strange to me that the matter around a black hole is a disk spinning in one direction.

REPLY [2 votes]: Well, consider stuff that is far away from the event horizon: if it's far enough away we can just use Newtonian gravity and ask the question there, which is going to be easier.
So the revised question would be: does stuff around large central masses very often form a disk in Newtonian gravity, and if so why?
And the answer is yes, it does: the Solar system is a bunch of stuff orbiting a large central mass, and it's very close to being a disk (with big lumps in, OK).  Saturn's rings are an even better case.  Galaxies are very often disks (but not quite the same thing, as there is no dominating central mass).
So, why?  Well, consider a system where things are orbiting in all sorts of directions. Those things will crash into each other occasionally, and quite often if there are a lot of small things, or gas.  The configurations where things crash into each less often will be more stable, because the interactions which change the configuration happen less often.  Disks are such a configuration.
Perhaps there are more stable configurations than disks?  Well, there's an important constraint on the system: it has to conserve angular momentum.  And disks are pretty much the unique configuration which both conserves angular momentum and minimises the interaction rate.  So you get disks.
So, far from the event horizon, there will be a disk (the solar system would still be a disk if the sun was a black hole).  Closer to the event horizon you need to use GR, but the same considerations turn out to hold, more-or-less.
This argument contains a bunch of hand-waving and omissions: apologies for that.  I think it is right in outline.


text:
\subsection{Crack aligned to the interface}

\subsubsection{Edge Crack aligned to a bimaterial interface in tension} Consider a bimaterial plate with an edge crack $(a/L=$ 0.5$)$ subjected to uniform far field tension $(P=1)$. The geometry, loading and boundary conditions are shown in \fref{fig:crkbimat}. The material is assumed to be isotropic with Young's modulus $E_1/E_2=$ 2 and Poisson's ratio $\nu_1 = \nu_2 = $ 0.3. A state of plane strain is considered in this example. The discontinuities (both the crack and the material interface) are represented independent of the underlying finite element (FE) mesh. From the underlying FE mesh, the scaled boundary region $\Omega^{\rm sbfem}$ is identified by
\begin{itemize}
\item Selecting the nodes whose nodal support is intersected by the discontinuous surface, viz., crack or the material interface. The nodes whose support are intersected by the material discontinuity are treated within the XFEM framework.
\item The elements containing these nodes are then selected. Only the information on the boundary of the domain is retained. The dofs of nodes within the domain are condensed during the solution process.
\end{itemize}
Within this region, i.e, $\Omega^{\rm sbfem}$, the scaled boundary formulation is employed to compute the stiffness matrix. Note that no special numerical integration technique is required. The main advantage of this approach is that it does not require a priori knowledge of the asymptotic fields. \fref{fig:meshdetailsintercrck} shows a typical finite element mesh used for this study. The stiffness matrix of the region $A-B-C-D$ is computed by employing the scaled boundary formulation.
\begin{figure}[htpb]
\centering
\scalebox{0.7}{\input{./Figures/bimater1.pstex_t}}
\caption{Edge crack aligned to a bimaterial inteface in tension: geometry and boundary conditions.}
\label{fig:crkbimat}
\end{figure}

\begin{figure}[htpb]
\centering

\subfigure[Discretization for a bimaterial interface crack]{\includegraphics[scale=0.5]{./Figures/edgcrkmesh1}}
\subfigure[Zoomed in view of the region ABCD]{\includegraphics[scale=0.5]{./Figures/edgcrkmesh2}}
\caption{A typical discretization for a crack aligned to a bimaterial interface crack. The region ABCD is identified  from the FE discretization and then replaced with the SBFEM domain. Note that, the SBFEM domain is constructed from the underlying FE discretization. The dofs of nodes within the SBFEM are suppressed during the solution. In this example, the interface aligns with the element edge. To represent the jump across the crack face, the `circled' nodes are enriched with Heaviside function~\cite{sukumarhuang2004}.}
\label{fig:meshdetailsintercrck}
\end{figure}

Tables \ref{table:mode12meshconve} - \ref{table:mode12layerconve} shows the convergence of the mode I and mode II stress intensity factors with mesh refinement and with number of layers of scaled boundary region in the vicinity of the crack tip. From Table \ref{table:mode12meshconve} it can be seen that with mesh refinement, the numerical SIF approach a constant value. For this study, 3 layers of elements around the crack tip is replaced with the scaled boundary region. The influence of number of layers on the numerical SIFs is demonstrated in Table \ref{table:mode12layerconve} for a structured mesh of 51 $\times$ 102 quadrilateral elements. It is noted that with increasing number of layers around the crack tip, the accuracy of the SIFs improve and approach the values reported in the literature based on the boundary element method~\cite{matsumtotanaka2000} and the XFEM~\cite{liuxiao2004}. It is noted that in~\cite{liuxiao2004}, first 11 terms of the asymptotic solution were retained. From this study, we can conclude that for a given mesh, 4-5 layers of elements should be replaced with the SBFEM domain for reasonable accuracy. However, for more accurate results, more number of layers can be replaced and represented by the SBFEM.

\begin{table}[htpb]
\centering

\caption{Edge Crack aligned to a bimaterial interface in tension: Convergence of the mode I and mode II SIFs with mesh refinement. The number of SBFEM layers around the crack tip is 3.}
\begin{tabular}{llrr}
\hline
Number & Number & \multicolumn{2}{c}{Stress Intensity Factors}\\
\cline{3-4}
of elements & of dofs & $K_{\rm I}$ & $K_{\rm II}$ \\
\hline
11$\times$22 & 459 & 2.7497 & -0.2700 \\
21$\times$42 & 1769 & 2.7630	 & -0.2729 \\
31$\times$62 & 3879 & 2.7666 & -0.2733 \\
41$\times$82 & 6789 & 2.7679	 & -0.2732 \\
51$\times$102 & 10499 & 2.7685 & -0.2730 \\
\hline
\label{table:mode12meshconve}
\end{tabular}
\end{table}

\begin{table}[htpb]
\centering

\caption{Edge Crack aligned to a bimaterial interface in tension: Convergence of the mode I and mode II SIFs with number of layers of scaled boundary region in the vicinity of the crack tip. A structured quadrilateral mesh (51 $\times$ 102 elements) was used.}
\begin{tabular}{lrr}
\hline
Number & \multicolumn{2}{c}{Stress Intensity Factors}\\
\cline{2-3}
of layers & $K_{\rm I}$ & $K_{\rm II}$ \\
\hline
3 & 2.7685 & -0.2730 \\
4 & 2.8280 & -0.2707 \\
5 & 2.8219 & -0.2681 \\
6 & 2.8202 & -0.2674 \\
7 & 2.8181 & -0.2671 \\
8 & 2.8170 & -0.2670 \\
Ref.~\cite{matsumtotanaka2000} & 2.8190 & -0.2680 \\
Ref.~\cite{liuxiao2004} & 2.8440 & -0.2670 \\
\hline
\label{table:mode12layerconve}
\end{tabular}
\end{table}

\paragraph{Crack growth along the interface}
In this case, we assume that the crack grows along the interface. Consider a bimaterial plate with an edge crack subjected to uniform tension (see \fref{fig:crkbimat}). The initial crack length $a/L$ is assumed to be 0.2 and the crack increment at each step is assumed to be 0.2. A structured quadrilateral mesh (51$\times$ 102) with 5 layers of elements replaced by the SBFEM is used for this study. Like the XFEM, as the crack advances, a new region is identified which is intersected by the discontinuous surface and instead of enriching with additional enrichment functions, the SBFEM is employed to compute the stiffness matrix and for further post-processing of the results. \fref{fig:bimatcrkmesh} shows the meshes at the initial step and at intermediate steps. The influence of the crack propagation on the numerical stress intensity factors are shown in \fref{fig:bimatcrkgwth}. It is observed that with increasing crack length $a/L$, the stress intensity factor and the T-stress increases.

\begin{figure}[htpb]
\centering
\subfigure[]{\includegraphics[scale=0.4]{./Figures/mesh1}}
\subfigure[]{\includegraphics[scale=0.4]{./Figures/mesh2}}
\subfigure[]{\includegraphics[scale=0.4]{./Figures/mesh3}}
\subfigure[]{\includegraphics[scale=0.4]{./Figures/mesh4}}
\caption{Edge Crack aligned to a bimaterial interface in tension: crack growth along the interface. Meshes at the initial step and at intermediate steps.}
\label{fig:bimatcrkmesh}
\end{figure}

\begin{figure}[htpb]
\centering

\setlength\figureheight{8cm}
\setlength\figurewidth{10cm}
\subfigure[Stress intensity factor $K_o=\sqrt{K_I^2+K_{II}^2}$]{\scalebox{1}{\input{./Figures/crkgwthSIF.tikz}}}
\subfigure[T-stress $T \sqrt{\pi a}/K_o$]{\scalebox{1}{\input{./Figures/crkgwthTstr.tikz}}}
\caption{Crack growth along the interface: the variation of stress intensity factor and T-stress with the increasing crack length, $a/L$.}
\label{fig:bimatcrkgwth}
\end{figure}

\subsubsection{Bimaterial plate with a center crack in tension} Consider a bimaterial plate with a center crack subjected to uniform far field tension. The geometry, loading and boundary conditions are shown in \fref{fig:crkbimatcenter}. The material is assumed to be isotropic with Young's modulus $E_1$ and $E_2$ and Poisson's ratio $\nu_1$ and $\nu_2$. Due to symmetry, only one half of the plate is considered. A state of plane strain is considered in this example. Based on the observation from the previous example, a structured quadrilateral mesh (51 $\times$ 102 elements) with 5 layers of elements were replaced with the SBFEM in the vicinity of the crack tip. The stress intensity factors $(K_{\rm I}, K_{\rm II})$ and the T-stress are computed from their definitions (see Section \ref{gsiftress}).

\begin{figure}[htpb]
\centering
\scalebox{0.6}{\input{./Figures/bimatinter.pstex_t}}
\caption{Bi-material plate with a center crack in tension: geometry and boundary conditions.}
\label{fig:crkbimatcenter}
\end{figure}

\begin{table}[htpb]
\centering
\renewcommand{\arraystretch}{1.2}
\caption{Bimaterial plate with a center crack in tension: Stress intensity factors and the T-stress, where $K_o = \sqrt{K_I^2 + K_{II}^2}$.}
\begin{tabular}{lrrrrrrr}
\hline
$E_1/E_2$ & \multicolumn{2}{c}{$K_{\rm I}/\sqrt{\pi a}$} & \multicolumn{2}{c}{$K_{\rm II}/\sqrt{\pi a}$} & \multicolumn{3}{c}{$T\sqrt{\pi a}/K_o$} \\
\cline{2-8}
 & Ref.~\cite{song2005} & Present & Ref.~\cite{song2005} & Present  & Ref.~\cite{song2005} & Ref.~\cite{sladeksladek1997} & Present  \\
\hline
1	& 1.189 &	1.1893 & 	0.000 &	0.0000 &	-1.06 & -1.07 & -1.0552 \\
2	& 1.179 &	1.1798	& -0.055 &	-0.0566 & -0.718 & -0.73 & -0.7144 \\
5	& 1.148 &	1.1483	 & -0.104	 & -0.1053 & -0.379 & -0.39 & -0.3770 \\
10 &	1.123 &	1.1237 & 	-0.123 &	-0.1240 &-0.216	 & -0.23 & -0.2147 \\
\hline
\end{tabular}
\label{table:tstressbimaterialinterfacecrk}
\end{table}

Table \ref{table:tstressbimaterialinterfacecrk} presents the stress intensity factors ($K_{\rm I}$ \& $K_{\rm II}$) and the T-stress for various combinations of material property: $E_1/E_2=$ 1, 2, 5 and 10 for a crack length $a/b=$ 0.5. The Poisson's ratio is assumed to be constant with $\nu_1=\nu_2=$ 0.3. When $E_1/E_2 \neq 1$, the order of the singularity is complex and the stress intensity factors are computed from \Eref{eq:GSIFdef-case2-3-1} with the characteristic length $L=2a$. The numerically computed stress intensity factors are normalized with $P\sqrt{\pi a}$. The T-stress is evaluated by using \Eref{eq:TStress-1} to the side of Material 2 (at $\theta= 0^-$) and is normalized with $K_o/\sqrt{\pi a}$. The results from the present formulation are compared with the boundary element technique~\cite{sladeksladek1997} and with the scaled boundary formulation~\cite{song2005} and a very good agreement is observed. \fref{fig:tstrResultsbimatintecrk} shows the singular stress distribution around the crack tip for Young's modulus ratio $E_1/E_2=$ 5 and the T-stress distribution for various combinations of material property. The singular stresses $\bvsig^s$ and the T-stress are evaluated by using \Eref{eq:singularStress_theta0} \&  \Eref{eq:TStress-1}, respectively. It is seen that, as the ratio $E_1/E_2$ increases, the T-stress increases.

\begin{figure}
\centering

\setlength\figureheight{8cm}
\setlength\figurewidth{10cm}
\subfigure[]{\input{./Figures/singstressDis.tikz}}
\subfigure[]{\input{./Figures/centercrackbimat.tikz}}
\caption{Bimaterial plate with a center crack in tension: (a) Singular stress distribution for $E_1/E_2=5$ and (b) the T-stress distribution for various material combinations.}
\label{fig:tstrResultsbimatintecrk}
\end{figure}

\subsubsection{Bimaterial strip with edge loads} In this example, consider a bimaterial strip constrained at the right end while a concentrated load $P_o=$ 1 applied at the left end as shown in \fref{fig:bimstripproblem}. The strip contains two layers of thickness $h_1$ and $h_2$ and is assumed to be in a state of plane strain condition. The length of the bi-material strip is $L=10h_2$ and $h_2=$1 is considered for this study. A structured mesh of 201 $\times$ 51 quadrilateral elements are considered with 10 layers of elements replaced with the scaled boundary domain. The analytical expression for the T-stress is given in~\cite{kimvlassak2006}. Table \ref{table:tstressbimaterialstrip} shows the normalized T-stress computed from the current approach for various combinations of material parameters, where $T_1$ and $T_2$ are the T-stress evaluated at the side of the material 1 (at $\theta=0^+$) and the material 2 (at $\theta=0^-$), respectively  . It can be seen that the results from the present formulation are in excellent agreement with the results available in the literature. It is emphasized that the present formulation does not require a priori knowledge of the asymptotic fields.

\begin{figure}[htpb]
\centering
\subfigure[]{\scalebox{0.7}{\input{./Figures/bimatstrip.pstex_t}}}
\subfigure[]{\includegraphics[scale=0.40]{./Figures/tstrmesh}}
\subfigure[]{\includegraphics[scale=0.40]{./Figures/tstrmesh1}}
\caption{Bi-material strip under edge loads: (a) geometry and boundary conditions; (b) and (c) FE discretization and zoomed in view. The region $A-B-C-D$ is represented by the SBFEM.}
\label{fig:bimstripproblem}
\end{figure}

\begin{table}[htpb]
\centering

\caption{Bimaterial strip with edge loads: material constants and the normalized T-stress. The length of the bimaterial strip, $L=$10$h_2$, the crack length $a=L/2$ and concentrated load, $P_o=$ 1.}
\begin{tabular}{lrrrrrrrrr}
\hline
Case & $E_1/E_2$ & $\nu_1$ & $\nu_2$ & \multicolumn{2}{c}{Ref.~\cite{kimvlassak2006}} & \multicolumn{2}{c}{Ref.~\cite{yuwu2012}} & \multicolumn{2}{c}{Present} \\
\cline{5-10}
   &  &  &  & T$_1^{\rm ref}$/$\sigma_o$ & T$_2^{\rm ref}$/$\sigma_o$ & T$_1^{\rm ref}$/$\sigma_o$ & T$_2^{\rm ref}$/$\sigma_o$ & T$_1^{\rm ref}$/$\sigma_o$ & T$_2^{\rm ref}$/$\sigma_o$ \\
\hline
1 & 7/3 & 1/3 & 1/3 & 0.0709 & 0.0304 & 0.0702 & 0.0301 & 0.07012 & 0.03005\\
 2& 20/9 & 1/4 & 1/8 & 0.0784 & 0.0336 & 0.0773 & 0.0331 & 0.07698 & 0.03299 \\
 3 & 4 & 2/5 & 2/5 & 0.1310 & 0.0328 & 0.1317 & 0.0329 & 0.13150 & 0.03288 \\
 4 & 4 & 1/4 & 1/4 & 0.1424 & 0.0356 & 0.1410 & 0.0353 & 0.14081 & 0.03520 \\
\hline
\end{tabular}
\label{table:tstressbimaterialstrip}
\end{table}

\subsection{Crack terminating at an interface}
In this last example, consider a crack terminating at the bimaterial interface. The crack is assumed to be in material 2 and terminating at the interface between the two materials (see \fref{fig:crackatinter}) and the plate is subjected to a far field tension $P$. Appropriate Dirichlet boundary conditions are enforced to arrest any rigid body motion. A structured quadrilateral mesh $(50 \times 100)$ with 5 layers of element replaced with the SBFEM is used for this study. \fref{fig:tstrResults} shows the stress in the $y-$direction $(\sigma_{yy}$ ahead of the crack tip). The results from the present formulation is compared with the stresses computed using a finite element model with a conforming mesh. \fref{fig:angdistriTcrkatInt} shows the angular stress distribution ahead of the crack tip for $E_1/E_2=$ 100 and the T-stress distribution for various combination of material properties when the crack is perpendicular to the interface, i.e., $\psi=$0$^\circ$. It is seen that, as the ratio $E_2/E_1$ increases, the T-stress increases. The angular distribution of $\sigma_{yy}$ for various crack orientation $\psi$ is shown in \fref{fig:angdistriTcrkatIntvaricrkangle}. When the crack is perpendicular to the interface, the stress distribution is symmetric, whilst in any other case it is unsymmetric. It is observed that the crack orientation strongly influences the stress distribution ahead of the crack tip.

\begin{figure}
\centering
\scalebox{0.6}{\input{./Figures/newcrkatinter.pstex_t}}
\caption{Crack terminating at the material interface: geometry and boundary condition. The crack is assumed to be in material 1 and terminating at the interface between the two materials. $\psi$ is the measured counter clockwise from the positive $x$ axis and determines the inclination of the crack to the interface.}
\label{fig:crackatinter}
\end{figure}

\begin{figure}
\centering

\setlength\figureheight{8cm}
\setlength\figurewidth{10cm}
\input{./Figures/syyaheadcrk.tikz}
\caption{Crack terminating at the material interface: stress $\sigma_{yy}$ ahead of the crack tip (along the path $A-B$, see \fref{fig:crackatinter}). The results are compared with the finite element analysis in which a conforming mesh is generated. The ratio of Young's modulus is $E_1/E_2=$ 1000 with $E_2=$ 1.}
\label{fig:tstrResults}
\end{figure}

\begin{figure}
\centering

\setlength\figureheight{8cm}
\setlength\figurewidth{10cm}
\subfigure[$E_1/E_2=$ 100]{\input{./Figures/strDCrkAtInter.tikz}}
\subfigure[]{\input{./Figures/TstrCrkAtInter.tikz}}
\caption{Angular distribution of singular stresses and T-stress for various combination of material properties for a crack terminating at the material interface.}
\label{fig:angdistriTcrkatInt}
\end{figure}

\begin{figure}
\centering

\setlength\figureheight{8cm}
\setlength\figurewidth{10cm}
\input{./Figures/SigxCrkAtInter.tikz}
\caption{Angular distribution of singular stresses and T-stress for various combination of material properties for a crack terminating at the material interface for various crack orientation. The ratio of Young's modulus is $E_1/E_2=$ 100 with $E_2=$ 1.}
\label{fig:angdistriTcrkatIntvaricrkangle}
\end{figure}

Next we consider the case of a crack deflecting into the material. A crack impinging the interface between two dissimilar materials may advance by either propagating along the interface or deflecting into the material. The competition between these two scenarios have been thoroughly dealt in the literature~\cite{hehutchinson1989,hehutchinson1989a,hutchinsonsuo1992,zhangsuo2007}. The material properties are assumed to be $E_2=100E_1$. We assume that the initial crack is in material 2 and has deflected into the material 1. In this example, material 1 is homogeneous and isotropic and hence the crack growth is governed by the maximum hoop stress criterion, which states that the crack will propagate from its tip in the direction $\theta_c$ where the circumferential stress $\sigma_{\theta \theta}$ is maximum. The critical angle is computed by solving the following equation:
\begin{equation}
K_I \sin \theta_c + K_{II} (3 \cos \theta_c - 1) = 0
\label{eqn:cirstress}
\end{equation}
Solving \Eref{eqn:cirstress} gives the crack propagation angle:
\begin{equation}
\theta_c = 2 \arctan \left[ \frac{-2\left( \frac{K_{II}}{K_I} \right)}{ 1+\sqrt{1 + 8\left( \frac{K_{II}}{K_I} \right)^2}}  \right]
\label{eqn:probdirection}
\end{equation}
The plate is subjected to a far field tension $P$ (see \fref{fig:crackatinter}). A structured quadrilateral mesh (51$\times$102) with 5 layers of element replaced with the SBFEM is employed in this example. \fref{fig:sifvariationcrkgwth} shows the variation of the stress intensity factors (mode I and mode II) with crack length $a/L$. The influence of the initial orientation of the crack is also shown. Since the loading is mode I, we expect the crack to grow in a straight line. For a crack impinging at an angle $\psi$, initially $K_{II}$ is not zero (see \fref{fig:sifvariationcrkgwth}) and hence the crack advances at an angle that is not equal to zero. However, due to the far field tension, as the crack advances, $K_{II}$ becomes zero and the crack propagates in a straight line. If conventional XFEM was employed, a separate set of enrichment functions should be used when the crack tip lies on the interface and these functions are to be replaced with asymptotic fields for homogeneous material when the crack deflects into the material interface. It is noted that in the extended scaled boundary method, no such modification is necessary and moreover the stress intensity factors and the T-stress can be computed from the definitions.

\begin{figure}
\centering

\setlength\figureheight{8cm}
\setlength\figurewidth{10cm}
\input{./Figures/Crkgwthfrminterfacenew.tikz}
\caption{Mode I and II stress intensity factor variation as a function of crack advancement. The crack increment is set in advance and the direction of the crack propagation is computed from \Eref{eqn:probdirection}. }
\label{fig:sifvariationcrkgwth}
\end{figure}

\begin{figure}[htpb]
\centering
\includegraphics[scale=0.6]{./Figures/crackpathcrkatinterface1}
\caption{Crack trajectory for a crack impinging at the interface and deflected into material 1. Two cases are considered: (a) a crack perpendicular to the interface and (b) crack oriented at an angle. The solid black line denotes the discontinuous surface.}
\label{fig:cracktrajectory}
\end{figure}

config: arxiv


TITLE: Uniqueness of Riesz Representation of $C^{*}[a,b]$
QUESTION [0 upvotes]: I have seen this statement: The dual space $C^{*}\left[a,b\right]$ of $C\left[a,b\right]$ is isometrically isomorphic to $BV_{0}\left[a,b\right]$.
where $$BV_{0}\left[a,b\right]=\left\{ \alpha\in BV\left[a,b\right]|\alpha\left(a\right)=0,\alpha\text{ right-continuous on}\left(a,b\right)\right\} $$
Now I want to show:
For $T\in C^{*}\left[a,b\right]$, if there exists a $\alpha\in BV_{0}\left[a,b\right]$ such that $T=T_{\alpha}$. That is, $$T\left(f\right)=T_{\alpha}\left(f\right)=\int_{a}^{b}fd\alpha$$
for all $f\in C\left[a,b\right]$. then $\left\Vert T\right\Vert =T_{a}^{b}\left(\alpha\right)$. And if there exists $\alpha,\beta\in BV_{0}\left[a,b\right]$ such that $T=T_{\alpha}=T_{\beta}$. Then $\alpha=\beta$ on $\left[a,b\right]$.
I have prove that $\left\Vert T_{\alpha}\right\Vert \le T_{a}^{b}\left(\alpha\right)$, but I have trouble proving $\left\Vert T_{\alpha}\right\Vert \geq T_{a}^{b}\left(\alpha\right)$ and the uniqueness, could anyone give some help?

REPLY [0 votes]: If you start with $T \in C[a,b]^*$, then you can extend $T$ to $S \in L^{\infty}[a,b]*
$ in such a way that $\|S\| = \|T\|$. Then $S$ is defined on step functions such as $\chi_{[a,b]}$ and $\chi_{[a,b)}$, etc.. For any positive integer $N$, let
$$
       x_0 = a, \; x_1 = a+\frac{b-a}{N}, x_2=a+2\frac{b-a}{N},\cdots, x_N=b.
$$
For any $f\in C[a,b]$ define $f_N$ to be
$$
          f_N = f(x_0)\chi_{[x_0,x_1)}+f(x_1)\chi_{[x_1,x_2)}+\cdots+f(x_{N-1})\chi_{[x_{N-1},x_N]}.
$$
Then $f_N\rightarrow f$ in $L^{\infty}[a,b]$ as $N\rightarrow\infty$ because $f\in C[a,b]$. Define $\alpha(t)=S(\chi_{[a,t)})$ for $a \le t < b$ and let $\alpha(b)=S(1)$. Then $\alpha$ is of bounded variation and there are unimodular constants $c_n$ such that
$$
       \sum_{n=1}^{N}|\alpha(t_n)-\alpha(t_{n-1})|=\sum_{n=1}^{N-1}c_nS(\chi_{[t_{n-1},t_n)})+c_NS(\chi_{[t_{N-1},t_{N}]}) \\
   = S\left(\sum_{n=1}^{N-1}c_n\chi_{[t_{n-1},t_{n})}+C_N\chi_{[t_{N-1},t_N]}\right)
$$
Therefore,
$$
     \sum_{n=1}^{N}|\alpha(t_n)-\alpha(t_{n-1})| \le \|S\|=\|T\|.
$$
So $\alpha$ is of bounded variation and $V_{a}^{b}(\alpha) \le \|T\|$. Furthermore,
$$
      T(f)=\lim_{N} S(f_N)=\lim_N \sum_{n=1}^{N}f(x_n)\Delta\alpha_n = \int_{a}^{b}f(t)d\alpha(t) \\
         |T(f)| \le \|f\|_{C[a,b]}V_{a}^{b}(\alpha) \\
     \|T\| \le V_{a}^{b}(\alpha).
$$
So $\|T\|=V_a^b(\alpha)$. Modiffying $\alpha$ to be left- or right-continuous will not increase its variation, or the representation of $T(f)=\int_{a}^{b}fd\alpha$. So it still follows that $\|T\|=V_a^b(\tilde{\alpha})$, where $\tilde{\alpha}$ is left- or right- continuous.