TITLE: Understanding log equations $\log^2(x) + \frac{1}{2}\log(x^2)-2 = 0$ and $\log_2(x)+\log_2(x+2)-2 = 0$
QUESTION [0 upvotes]: I want to understand how these equations have been solved:
Here's the first equation:
$$\log^2(x) + \frac{1}{2}\log(x^2)-2 = 0$$
$$= \log^2(x)+\log(x)-2=0 \text{ (What is this rule called here?)}$$
$$\Rightarrow u \equiv \log(x): u^2 + u-2 = 0 \text{ (Substitution)}$$
$$\Rightarrow u^2 + u - 2 = (u+2)(u-1) \Rightarrow u_1 = -2, u_2 = 1 $$
$$x_1 =e^{-2}, x_2=e^1 \text{ (How do we know that we have to use e here?)}$$
Here's the second equation:
$$\log_2(x)+\log_2(x+2)-2 = 0$$
$$=\log_2(x^2+2x)-2= 0 \text{ (Can we just multiply it even though it is a sum? } x \cdot (x+2))$$
$$\Leftrightarrow \log_2(x^2+2x) = 2$$
$$\Leftrightarrow x^2 +2x = 2^2 = 4 \text{ }$$
$$\Rightarrow x_{1,2} = \frac{1}{2} \big(-2 \pm \sqrt{4-4\cdot(-8)}\big) = -1 \pm \sqrt{1+8}$$
$$\Rightarrow x_1=2, x_2 = -4$$

REPLY [0 votes]: The questions's been answered in the comments
$$\log_ab^x \Rightarrow x \log b$$
$$\log_axy \Rightarrow \log_a x + \log_a y$$
and an error on the r.h.s


TITLE: finding speed given acceleration and displacement but without velocity and time
QUESTION [0 upvotes]: Your task is to correctly answer the following question: Assuming the spinner starts from rest, what will its speed be (in cm/s)    after travelling 250 cm?
the acceleration is 1.57cm/s/s but how are you supposed to solve this without velocity and time using algebra?

REPLY [0 votes]: Using kinematics:
$$v_f^2=v_i^2+2ad$$
$$v_f=\sqrt{(1.57)(2)(250)}$$
$$v_f=28 \,cm/s$$


TITLE: Variance and mean of balls in bins limited capacity
QUESTION [3 upvotes]: Let there be $m$ indistinguishable balls, $k$ bins, $C$ capacity.
Let $X_j$ denote the total balls in bin $j$.
I've seen ways to calculate the total number of combinations, but I'm not sure how to go about calculating the mean and variance of $X_j$.
It is understood that the ball is always thrown into one of the empty bins with uniform probability. edit: nonfull bins not empty bins sorry.

REPLY [1 votes]: This seems computationally very difficult to me.  You might do better with simulation.
It $C\ge m$ then the constraint has no effect, and we are just looking for  the mean and variance of the number of balls in bin $j.$  This has a binomial distribution, and the answer is well known.
When ${m\over k}\leq C< M,$ the situation is much more complicated, because the result depends on the sequence in which the balls were thrown into the bins.  Before any bin fills up, a ball has probability $1/k$ of landing in bin $j.$  After some other ball fills up, it has probability $1/(k-1)$ of landing in bin $j$.  The situation is even more complicated if it's possible for more than one ball to fill up.
Consider the case $k=3, j=1.$  Let $P(x_1,x_2,x_3)$ be the probablity that at some point there are $x_i$ balls in bin $i$ for $i=1,2,3.$  To compute the expectation, we have to sum up terms of the form $aP(a,b,c)$ where $a+b+c=m.$  If $\max\{a,b,c\}<C,$ we have  simply $$P(a,b,c)={m\choose a,b,c}3^{-n}\tag{1}$$  But suppose $c=C.$  Then $$P(a,b,c)=P(a,b,C)=\frac12P(a-1,b,C)+\frac12P(a,b-1,C)+\frac13P(a,b,C-1)\tag{2}$$ taking account of all bins the last ball might have been thrown into.
The last term on the right-hand side of $(2)$ can be computed directly from $(1),$ but the first two have to be computed recursively from $(2)$.  It won't take many bins, or many balls, before this computation becomes unwieldy, if the capacity constraint is effective.
There are so many possible sequences of throws that memoization is unlikely to be effective, so far as I can see.  I've been tying to think of ways to simplify the calculations, but so far, I don't have a glimmer.


TITLE: Does the formula $f(x,y)=(x^2-y)yx^{-4}\mathbf 1_{0<y<x^2}$ define a continuous function?
QUESTION [1 upvotes]: Let $f : \mathbb{R}^2 \to \mathbb{R}$ be the function such that
  $$
f(x, y) = \begin{cases}
\frac{(x^2 - y)y}{x^4}, & 0 < y < x^2\\
0,                      & \text{elsewhere}
\end{cases}
$$
  Is $f$ continuous?

My attempt:
I tried to use the characterizations of continuity by sequence. So I took the sequence $x_k = \frac{2}{k}$ and $y_k = \frac{1}{k^2}$ and so I had $(x_k, y_k) \rightarrow (0, 0)$ but $f(x_k, y_k) = \frac{3}{16}$ wich converges trivially to $\frac{3}{16}$. And this value is different from $f(0, 0) = 0$. So $f$ is not continuous at $(0,0)$ but what about the rest of the point? And also I don't know how to manipulate the condition $0 < y < x^2$ ir order to prove or disprove continuity.
Any hint or answer would be really appreciated.

REPLY [1 votes]: We have that for $x\neq 0$ $f(x,y)$ is continuous that is
$$\lim_{y\to (x^2)^-}\frac{(x^2 - y)y}{x^4} =0 \quad \lim_{y\to 0^+}\frac{(x^2 - y)y}{x^4} =0$$
and for $(x,y)\to (0,0)$ by $x=t$ and $y=t^2-t$ with $t\to 0$
$$\frac{(x^2 - y)y}{x^4}=\frac{(t^2-t^2+t)(t^2-t)}{t^4}=\frac{t-1}{t^2}\to -\infty$$
therefore the function is continuous in $\mathbb{R^2}\setminus\{(0,0)\}$.


text:
\section{Problem formulation}
\label{sec:probform}

\subsection{Governing equations}
\label{sec:goveq}
\noindent On the micro scale, the fluid velocity $\mathbf{u}$ and pressure $p$ of a flow can be computed by solving Stokes equations. Here, an infinite expanse of fluid, $\Omega_0$, with viscosity $\mu_0$ is considered. The fluid contains $n$ droplets, denoted by $\Omega_k$, for $k = 1, \hdots , n$, each with viscosity $\mu_k$. The interface between droplet $\Omega_k$ and bulk fluid $\Omega_0$ is denoted $\Gamma_k$, see Figure~\ref{fig:form_domain} for an example domain. The Stokes equations take the form
\begin{align}
	\begin{split}
		\mu_0 \Delta \mathbf{u}_0 = \nabla p_0,& \text{ for } \mathbf{x}\in\Omega_0,  \\
		\mu_k \Delta \mathbf{u}_k = \nabla p_k,& \text{ for } \mathbf{x}\in\Omega_k, \; k=1,\hdots,n.
	\end{split}
	\label{eq:form_stokeseq}
\end{align}
For droplets, $\mu_k \neq 0$ and the velocity is continuous over the interfaces. Also, the solution fulfils the normal stress balance over the interfaces,
\begin{align}
	-\left(p_0 - p_k\right)\mathbf{n} + 2\left(\mu_0\mathbf{e}_0 - \mu_k\mathbf{e}_k\right)\cdot\mathbf{n} = - \sigma\kappa\mathbf{n} + \nabla_s\sigma \text{ on } \Gamma,
	\label{eq:form_normalstress}
\end{align}
where $\Gamma = \bigcup_{k=1}^n \Gamma_k$. This condition states that the jump in normal stress over an interface is proportional to the curvature and Marangoni force. Furthermore, the kinematic interface condition states that the normal velocity of the interface, $\frac{d\mathbf{x}}{dt}\cdot\mathbf{n}$, is equal to the normal fluid velocity,
\begin{align}
	\dfrac{d\mathbf{x}}{dt}\cdot\mathbf{n} = \mathbf{u}\cdot\mathbf{n}.
	\label{eq:form_kincond}
\end{align}
Here, $\mathbf{n}$ is the inward facing normal, $\sigma$ surface tension coefficient, $\kappa = \nabla_s\cdot\mathbf{n}$ curvature and $\mathbf{e}$ the rate of strain tensor. Note that this curvature is negative for a circular drop. The term $\nabla_s\,\sigma$, where $\nabla_s$ is the surface gradient operator, gives the tangential stress (Marangoni force) which is the result of a non-uniform surface tension. The interfaces are discretised clockwise by $s\in[0,L_k(t)]$, where $L_k(t)$ is the length of the interface $\Gamma_k$ at time $t$.

The limit when $\mu_k = 0$ corresponds to the study of inviscid bubbles. The continuity condition of velocity across the interface can then be disregarded, as there is no velocity inside the bubbles. The kinematic condition \eqref{eq:form_kincond} is still necessary, as is the normal stress balance \eqref{eq:form_normalstress} modified to contain only pressure and strain tensor of the bulk fluid.

\begin{figure}[h!]
	\centering
	\includegraphics[width=0.5\textwidth]{graphics/domain.png}
	\caption{Example of domain consisting of three droplets.}
	\label{fig:form_domain}
\end{figure}

As the fluid domain $\Omega_0$ is unbounded, an additional condition at infinity is needed,
\begin{align*}
	\mathbf{u} \rightarrow \mathbf{u}_\infty, \text{ as } |\mathbf{x}|\rightarrow \infty.
\end{align*}
Here it is typical to impose a linear far-field flow, where
\begin{align}
	\mathbf{u}_\infty = \begin{pmatrix} Q_0 & B_0+\frac{G_0}{2} \\
										B_0-\frac{G_0}{2} & -Q_0
						\end{pmatrix} \cdot \mathbf{x}.
	\label{eq:form_farfield}
\end{align}
The two imposed linear flows considered in this paper are: \textit{extensional flow}, where $B_0=G_0=0$ and \textit{shear flow}, where $Q_0=0$ and $G_0=2B$.

The surfactants considered here are insoluble, i.e. they exist only on the interfaces of the droplets. Their concentration, $\rho(s,t)$, is described by the convection-diffusion equation \cite{stone}
\begin{align}
	\frac{D\rho}{Dt} + \rho\left(\nabla_s \cdot\mathbf{u}\right) = \frac{1}{D_\Gamma}\nabla_s^2\, \rho, \text{ for } \mathbf{x}\in\Gamma.
	\label{eq:form_surfacteq}
\end{align}
Here, $D_\Gamma$ is the diffusion coefficient along the interface and $\frac{D}{Dt}$ the material derivative. As the surfactants are insoluble, their mass on each interface is conserved,
\begin{align}
	\frac{d}{dt}\int_{\Gamma_k(t)} \rho(t)dS = 0, \quad k=1,\hdots,n.
	\label{eq:form_conservmass}
\end{align}

The surfactant concentration and the surface tension at an interface are coupled through an equation of state. Two common equations of states are the Langmuir equation of state,
\begin{align}
	\sigma(\rho) = \sigma_0 + RT\log\left(\rho_\infty-\rho\right)
	\label{eq:form_eqstate_langmuir}
\end{align}
and  the linearised equation of state
\begin{align}
	\sigma(\rho) = \sigma_0 - RT\rho,
	\label{eq:form_eqstate_linear}
\end{align}
as described in \cite{Pawar1996}. Here,  $R$ is the universal gas constant, $T$ the temperature and $\sigma_0$ the surface tension coefficient of a clean interface. Furthermore, $\rho_\infty$ is the maximum monolayer packing concentration of surfactants on $\Gamma(t)$. Generally, the Langmuir equation of state is considered more accurate. The linear equation of state is mostly used for problems with low saturation levels. It is also commonly used in many validation cases, which is the reason it is included here. To switch between the two is trivial.

\subsection{Nondimensionalisation}
\noindent The above equations are nondimensionalised using a characteristic length $r_0$, defined as the radius of an initial droplet. The characteristic velocity is defined as the ratio $\frac{\sigma_0}{\mu_0}$, where $\mu_0$ is the viscosity of the bulk and $\sigma_0$ the surface tension coefficient of a clean interface. Furthermore the characteristic pressure is defined as $\frac{\sigma_0}{r_0}$. For the remainder of this paper, all physical quantities are considered in their non-dimensional form.

Rewriting the equations, \eqref{eq:form_stokeseq} becomes
\begin{align}
	\begin{split}
		\Delta\mathbf{u}_0 = \nabla p_0,& \text{ for } \mathbf{x}\in\Omega_0, \\
		\lambda_k \Delta\mathbf{u}_k = \nabla p_k,& \text{ for } \mathbf{x}\in\Omega_k, \; k=1,\,\hdots,\,n,
	\end{split}
	\label{eq:form_stokesnondim}
\end{align}
and \eqref{eq:form_normalstress}
\begin{align}
	-\left(p_0-p_k\right)\mathbf{n} + 2\left(\mathbf{e}_0-\lambda_k\mathbf{e}_k\right) \cdot \mathbf{n} = -\sigma\kappa\mathbf{n} + \nabla_s\sigma, \text{ on } \Gamma.
	\label{eq:form_normalstressnondim}
\end{align}
Here, $\lambda_k = \frac{\mu_k}{\mu_0}$ is the viscosity ratio between droplet $k$ and bulk fluid. Also the imposed far-field flow is nondimensionalised, such that
\[
	\left(Q,\;G,\,B \right)~=~\frac{r_0\mu_0}{\sigma_0}\left(Q_0,\;G_0,\;B_0\right).
\]
The nondimensionalisation of the surfactant concentration depends on which equation of state is used:
\paragraph{Case 1} In the case of the Langmuir equation of state, it is natural to nondimensionalise $\rho$ with the maximum monolayer packing concentration $\rho_\infty$, and \eqref{eq:form_eqstate_langmuir} reads
\begin{align}
	\sigma = 1 + E\log(1-\rho),
	\label{eq:form_eqstate_langmuir_nd}
\end{align}
where $E=\frac{RT\rho_\infty}{\sigma_0}$ is the so-called elasticity number.

\paragraph{Case 2} When it comes to the linearised equation of state, it is instead common to nondimensionalise with the initial surfactant concentration, $\rho_0$. Then, \eqref{eq:form_eqstate_linear} becomes
\begin{align}
	\sigma = 1 - E\rho,
	\label{eq:form_eqstate_linear_nd}
\end{align}
with $E=\frac{RT\rho_0}{\sigma_0}$.
\\ \\
In both cases, \eqref{eq:form_surfacteq} becomes
\begin{align}
	\frac{D\rho}{Dt} + \rho\left(\nabla_s \cdot\mathbf{u}\right) = \frac{1}{Pe_\Gamma}\nabla_s^2\, \rho, \text{ for } \mathbf{x}\in\Gamma.
	\label{eq:form_surfacteq_nondim}
\end{align}
where $Pe_\Gamma=\frac{\sigma_0 r_0}{\mu_0 D_\Gamma}$ is the Peclet number.
\\ \\
In summary, to evolve the deforming interfaces of surfactant-covered droplets in time, the following coupled system needs to be solved for $\mathbf{x}\in\Gamma(t)$:
\begin{align}
	\begin{split}
		\frac{d\mathbf{x}}{dt} &= \mathbf{u}(\mathbf{x},\rho,t), \\
		\frac{D\rho}{Dt} &= -\rho\left(\nabla_s \cdot\mathbf{u}\right) + \frac{1}{Pe_\Gamma}\nabla_s^2\, \rho.
	\end{split}
	\label{eq:form_system}
\end{align}
The solution to this system will be explained in steps. First, the boundary integral equation (BIE) formulation to compute $\mathbf{u}$ is described below in \S\ref{sec:bieform}.  In \S\ref{sec:methodBIE} a numerical method to solve the BIE accurately is explained. Then, \S\ref{sec:nummethod} explains the numerical solution of the surfactant concentration, the coupling of the system and its evolution in time.

config: arxiv