TITLE: Fourier multipliers and transference on cyclic groups
QUESTION [3 upvotes]: It seems to be a commonplace in harmonic analysis that if some operator (say, Fourier multiplier) is bounded on $L^p(\mathbb{R}^n)$ then by transference the similar operator is also bounded on $L^p(\mathbb{T}^n)$ (the most common example of application of this general principle is Hilbert transform for $n=1$). There exist certain general theorems which allow us to transfer the results in the setting of $\mathbb{R}^n$ to the setting of $\mathbb{T}^n$, and vice versa (for example, such theorems may be found in Grafakos, "Classical Fourier Analysis").
Now, my question is -- can we do the same for the cyclic group $\mathbb{Z}_n$ (with the standard Haar measure on it)? Say, for arbitrary number $m<n$ we may define the operator $\widehat{Tf}=\chi_{[0,m]}\hat{f}$ on $L^p(\mathbb{Z}_n)$. It seems to be not difficult to see directly that such operator is bounded and its norm does not depend on $n$ and $m$ but can we somehow derive it from the fact that Hilbert transform is bounded on $L^p(\mathbb{R})$ and $L^p(\mathbb{T})$? The intuition here is that for large $n$ the group $\mathbb{Z}_n$ should resemble $\mathbb{Z}$ (or $\mathbb{T}$) but I do not know how to do a rigorous proof of such "transference".
After that, one may go further and prove the "Littlewood--Paley theorem for $\mathbb{Z}_n$" (uniformly in $n$) and maybe certain multiplier theorems. Did anybody address such questions (with or without transference)? Or maybe these questions are trivial for some reason?

REPLY [1 votes]: Ok, I found the answer myself, so I'll post it here. It turned out that the results for multipliers should be transferred not from $L^p(\mathbb{T})$ but from $\ell^p(\mathbb{Z})$. That is, the boundedness of operators that figure in the question follows from the boundedness of the following operator on $\ell^p(\mathbb{Z})$: we take a sequence in $\ell^p(\mathbb{Z})$, consider its Fourier transform (which is a function on $\mathbb{T}$), multiply it by characteristic function of an arc and take the Fourier coefficients of the resulting function. The boundedness of such operator on $\ell^p(\mathbb{Z})$ is of course known --- it can be found in the book by Edwards and Gaudry "Littlewood--Paley and multiplier theory" (1977).
As for this transference, it is known, too (and not very difficult): it is written in an article "Transference methods in analysis" by Coifman and Weiss (in a more general context; to be more specific, Theorem 3.15 and Corollary 3.16 can be applied here).


TITLE: Integral equation $u(t)=f(t)+a\int_0^t u(s)ds\quad t\geq 0$
QUESTION [5 upvotes]: Let $a\in\mathbb R$ and $f\colon [0,1]\to\mathbb R$ a continuous function. Solve the integral equation
$$u(t)=f(t)+a\int_0^t u(s)\mathrm ds,\quad t\geq 0$$ and find an explicit formula for the solution.       Thank you

REPLY [4 votes]: One approach is to solve this when $f$ is differentiable, to find a formula for $u$ involving only $f$ (this can be done using integration by parts to get rid of $f'$), and to show that this formula also yields the unique solution when $f'$ does not exist.
Another approach is to consider the function $v=u-f$ and to note that
$$
v(t)=g(t)+a\int_0^tv(s)\mathrm ds,\qquad g(t)=a\int_0^tf.
$$
The function $g$ is $C^1$ hence one can solve this new integral equation the usual way, by differentiatig it. One gets $v'=g'+av$, hence $(\mathrm e^{-at}v(t))'=\mathrm e^{-at}g'(t)$, that is,
$$
v(t)=\mathrm e^{at}\left(v(0)+\int_0^t\mathrm e^{-as}g'(s)\mathrm ds\right).
$$
Since $v(0)=0$ and $g'=af$, this shows that
$$
u(t)=f(t)+v(t)=f(t)+a\mathrm e^{at}\int_0^t\mathrm e^{-as}f(s)\mathrm ds.
$$
Sanity check: The first approach described in this post yields the same formula.


TITLE: Surfaces and Planes in Three-Dimensional Spaces
QUESTION [1 upvotes]: A roof will be constructed to fit the surface given by the function:
$$ z = f(x,y) = \frac 52 + {1\over 200} (9x^2 - 4y^2) \tag{1} $$
On the domain $[-6, 6] \times [-6, 6]$, with $x$, $y$, and $z$ all measured in metres. Nine timber joists will be cut, and positioned such that they lie on cross sections of the surface. Shadecloth will be laid on top of the joists to produce a curved roof effect (See figure).

Recall that a vertical plane has the form:
$$ ax + by = c \tag{2} $$
and that, generally, cross-sections of a surface $z = f(x,y)$ are the lines of intersections with vertical planes. Usually, the vertical planes are either taken as $x$ is constant, or $y$ is constant. For the surface, $z$, this gives the cross-sections:
$$f(x,c) = \frac52 + {1\over200}(9x^2 - 4c^2)$$
$$ f(c,y) = \frac52 + {1\over200}(9c^2 - 4y^2)$$
which are parabolas. Since the joists are cut from straight pieces of timber, these are not helpful.

a) Which values of $a$, $b$, and $c$ can be chosen in equation $(2)$ such that the cross-sections of $(1)$ are straight lines in three-dimensional space? Give the general equation of the cross section in the form:
$$\mathbf r(\lambda) = \mathbf r_0 +\lambda \mathbf v$$

b) Each joist $J_k$, $k = 1, \dots, 9$ is to have one end attached to the position $P_k$ and the other end attached to the position $Q_k$ in three-dimensional space. Specifically, the $P_k$ are given by:
$$ \begin{align}
P_1 &= (-6, 3, f(-6, 3)) \\
P_2 &= (-6, 0, f(-6, 0)) \\
P_3 &= (-6, -3, f(-6, -3)) \\
P_4 &= (-6, -6, f(-6, -6)) \\
P_5 &= (-4, -6, f(-4, -6)) \\
P_6 &= (-2, -6, f(-2, -6)) \\
P_7 &= (0, -6, f(0, -6)) \\
P_8 &= (2, -6, f(2, -6)) \\
P_9 &= (4, -6, f(4, -6))
\end{align}
$$
Determine the other endpoints $Q_k$, such that each joist $J_k$ lies on the surface $f(x,y)$ and the $x$ and $y$ coordinates of each $Q_k$ lie on the boundary of the domain $[-6, 6] \times [-6, 6]$.

c) Calculate the length of each joist. (Give the answer correct to two decimal places).

REPLY [1 votes]: I'm going to give a hint for the first part to get you started. You should know that it's not in good form too ask so many questions in the same post. This forum does not like posts that are copied homework questions. Here's a quick guide on how to ask a good question

Our function has the general form
$$ f(x,y) = Ax^2 - By^2 + C $$
where $A,B,C$ are known, positive constants
Since a vertical plane is only given in terms of $x$ and $y$, it's a good idea to put one in terms, of the other, so we can rewrite $(2)$ as
$y = px + q $
where $p = -a/b$, $q = c/b$
Plugging this in gives
$$ z = f(x,px+q) = C + Ax^2 - B(px+q)^2 = (A-Bp^2)x^2 - 2Bpqx + (C-Bq^2) $$
You want this cross-section to be linear, which means the $x^2$ coefficients must be zero, therefore we have
$$ A - Bp^2 = 0 $$
This gives you a value for $p$, which is the "slope" of the vertical plane. There are no constraints on $q$
As for the parametric form, notice that everything is already in terms of $x$, so one possible parametrization is
$(x,y,z) = (\lambda,p\lambda+q,f(\lambda,p\lambda+q))$
If you have one specific point on this cross section, it's easy to find the value of $q$ (since $(2)$ is always valid and you already know what $p$ is). There are actually two solutions passing the same point, since you have two different values for $p$.


text:
\section{Conclusion}
\label{sec:conclusion}
Interest in machine learning for scientific applications has intensified in recent
years, in part due to advances in algorithms, data storage, and computational
analysis capabilities \citep{baker2019workshop,stevens2020ai}.  Fundamental challenges still remain when developing
and using a machine learning model for scientific applications which must satisfy physical
principles.  Enforcing such principles as constraints helps ensure the behavior of the
model is consistent with prior physical knowledge when queried in an extrapolatory
region. In other words, in addition to supplementing limited or expensive scientific data,
constraints help improve the generalizability of the model in ways that simply increasing dataset size may not.
Many approaches have been developed to perform ``physics-informed machine learning.''
In this survey, we have focused on constraint implementation in Gaussian processes,
which are popular as a machine-learned metamodels or emulators for a computational simulation.

Our survey focused on several important classes of constraints for Gaussian processes.
These included positivity or bound constraints on the Gaussian processes in Section \ref{sec:bound_constraints}.
When positivity constraints are applied to the derivatives of a Gaussian process, they lead to monotonicity and  convexity constraints as in Section \ref{sec:monotonicity_constraints}. This is a special example of regression with a linear transformation of a Gaussian process, which is the basis of Gaussian processes constrained by linear differential equations reviewed in Section \ref{sec:pde_constraints}. We discuss boundary value constrained Gaussian processes in Section \ref{sec:boundary_constraints}.
Throughout, we see that constraints can be enforced in an implicit way through data that satisfies the constraint, by construction of a tailored sample space, by derivation of a constrained covariance kernel, or by modifying the output or likelihood of the Gaussian process. The constraints may be enforced in a ``global sense'', at a finite set of ``virtual'' or ``auxiliary'' points, or only in an approximate sense. We have pointed to these aspects as key features distinguishing the constraints in this survey.

Constraints introduce new practical challenges into the GPR framework. These include: the analytical construction of sample spaces, transformations, or covariance kernels that inherently provide constraints; the sampling of truncated multivariate normals or intractable posterior distributions that arise when using non-Gaussian likelihoods; increased data and covariance matrix size when enforcing constraints with ``virtual'' data that leads to expanded ``four-block'' covariance; calculation of eigenvalues/eigenfunctions in bounded domains with complex geometry; the placement of virtual points or construction of spline grids in higher dimensions; and maximum likelihood training (optimization) of the hyperparameters of constrained Gaussian processes. Numerical issues are the focus of Section \ref{sec:computation_considerations}. In that section, we have also reviewed established numerical strategies for accelerating GPR. Some of these techniques have been applied to constrained Gaussian processes in the literature, while others have not. In general, the adaptation of computational strategies to constrained GPR is a relatively new field, and best practices have not yet been established. Moreover, while several codebases have been developed for constrained GPR, such \texttt{GPStuff} for non-Gaussian likelihoods \citep{vanhatalo2013gpstuff} and the \texttt{lineqGPR} package \citep{lopez2018lineqgpr} for the spline approach including constrained MLE, constraints have not made their way into the most widely used production codes for GPR. Furthering these computational aspects of constrained GPR remains a promising area for future work.

The field of constrained Gaussian processes has made significant advances over the past decade, and we expect significant development to continue. The purpose of this survey, while non-exhaustive, has been to catalog and investigate some of the more common approaches, guide the practitioner to identify which strategies are most appropriate for his or her needs, and point out the new computational challenges of constrained Gaussian processes and how they can be approached.

config: arxiv


TITLE: Using method of variation to find a general solution - a problem.
QUESTION [1 upvotes]: The general solution is of the form $$y_g = u_1y_1+u_2y_2$$
where $y_1$ and $y_2$ are solutions of the homogenous equation and $u_1$ and $u_2$ are functions that we want to calculate where
$$u_1' = \frac{-y_2g}{W(y_1,y_2)} \ u_2' = \frac{y_1 g}{W(y_1,y_2)}$$
where $W(y_1,y_2)$ is the wronskian. Then solve $$y''-7y'+10y = g(t)$$
Now I found out that $y_1= e^{5t} \ y_2=e^{2t}$ by solving the corresponding homogenous equation, then $W(y_1,y_2)=-3e^{7t}$
Now here g is arbitrary, so I ended up with:
$$u_1 =  \frac{1}{3} \int -e^{-5t}g(t)dt + c_1 \\ u_2= \frac{-1}{3} \int e^{-2t}g(t)dt + c_2$$
Then I ended up with :
$$y_g = c_1e^{5t}+c_2e^{2t}+\frac{e^{5t}}{3} \int e^{-5t}g(t)dt - \frac{e^{2t}}{3} \int e^{-2t}g(t)dt$$
I believe (checked a lot) I didn't make a calculation mistake. Answer sheet says the answer is :
$$y_c = c_1e^{2t}+c_2{e^{5t}} + \int (e^{5(t-s)}- e^{2(t-s)}) g(s) ds$$
I couldn't understand the part with integration. Isn't the Wronskian obvious? Why the variable s? I couldn't understand anything..
Thanks for your help.

REPLY [0 votes]: Note that the $t$ inside the integral is a dummy variable, and thus can be replaced by any letter we desire.  One motivation for doing it in this case is so as not to confuse it with the $t$ variable outside of the integral.  I will say that technically there should be limits on the integrals ($0$ to $t$) when writing it this way. So your answer could be rewritten as $$ y_g = c_1e^{5t}+c_2e^{2t}+\frac{e^{5t}}{3} \int_0^t e^{-5s}g(s)ds - \frac{e^{2t}}{3} \int_0^t e^{-2s}g(s)ds $$ and then we can bring the exponentials with the $t$ variables inside the integral (because there will not be any incorrect canceling) and combine the integrals: $$ y_g = c_1e^{2t}+c_2{e^{5t}} + \frac{1}{3}\int_0^t (e^{5(t-s)}- e^{2(t-s)}) g(s) ds $$
I am not sure why the answer has the $\frac{1}{3}$ disappearing. In general, I would say that your answer, or the way you wrote your answer are not wrong, just different.


TITLE: How can I bound this equation?
QUESTION [3 upvotes]: This is a problem I stuck in Berkeley Problems in Mathematics, fall 1983:
Let $x(t)=(x_{1}(t)...x_{n}(t))$ be a differentiable function from $\mathbb{R}$ to $\mathbb{R}^{n}$. It satisfies a differential equation of the form $$x'(t)=f(x(t))$$
where $f:\mathbb{R}^{n}\rightarrow \mathbb{R}$ is a continuous function. Assuming that $f$ satisfies the condition $$\langle f(y),y\rangle\le |y|^{2}$$ derive an inequality showing that $|x(t)|$ grows at most exponentially.
I am thinking about decompoising $f(y)$ into $y^{T}$ and $y$ directions, but this does not allow me to integrate $x'(t)$ from $t_{0}$ to $t$. If I can show $|f(y)|\le K|y|$ then the inequality would be immediate; but this is false as $f(y)$ can be orthogonal to $y$ at every point $y=x(t)$ and have arbitrarily large norm. So I do not know how to solve this in an elegant way. I can solve it in the most radical case that $x'(t)\cdot x(t)=0\forall t$ and $x'(t)=K(t)x(t)$, but I do not know how to deal with arbitrarily curves.

REPLY [3 votes]: Let $V(t) = \frac{1}{2} \|x(t)\|^2$. Then  $\dot{V}(t) = \langle x(t), \dot{x}(t) \rangle = \langle x(t), f(x(t)) \rangle \leq \|x(t)\|^2 = 2 V(t)$.
Now consider $\eta(t) = e^{-2t}V(t)$; we have $\dot{\eta}(t) \leq 0$, from which we get $\eta(t) \leq \eta(0)$, which is equivalent to $V(t) \leq V(0) e^{2t}$. Multiplying by $2$ and taking square roots gives $\|x(t)\| \leq \|x(0)\| e^t$.


TITLE: Moving wedge and pulley system
QUESTION [1 upvotes]: A wedge of mass $M$ rests on a rough horizontal surface with coefficient of static friction $\mu$. The face of the wedge is a smooth plane inclined at an angle $ \alpha$ to the horizontal. A mass  $m_1$ hangs from a light string which passes over a smooth peg at the upper end of the wedge and attaches to a mass $m_2$ which slides without friction on the face of the wedge.
A) Find the accelerations of $ m_1, m_2 $ and the tension in the string when $ \mu $ is very large.
B)Find the smallest coefficient of friction such that the wedge will remain at rest.
Attempts:
No specification on whether or not any of the masses $m_i$ are bigger than the others, so let us assume that $m_1$ goes down and $m_2$ goes up the slope. Then A) is fine. (a minus error in the final answer for A) is introduced if I assume the other orientation)
As for B) if we consider the wedge+masses as a system, then to conserve linear momentum if the masses move as above, then the wedge must move rightwards.  So the frictional force acts leftwards until the wedge slips.  The only other force acting on the wedge is via the normal contact force from $m_2$.  A component of this force acts to accelerate the wedge horizontally.  Let's write this relative to some fixed inertial frame of reference with $x$ rightwards. Then the above recast into symbols gives $$-\mu (M+m_1+m_2)g - m_2 g \cos \alpha \sin \alpha = 0 $$ at the point of slip of the wedge.  However, solving this for $\mu$ gives a negative result.  Where did I go wrong?

REPLY [1 votes]: Part A
$\mu$ large implies that the wedge does not move. Therefore the two forces on the small masses along their direction of movement are:
$$F_1 = m_1 g, \quad F_2 = m_2 g \sin(\alpha).$$
They pull in different directions on the pulley, so the net acceleration (with respect to $m_1$ going down) is:
$$F = F_1 - F_2 = g [m_1 - m_2 \sin(\alpha)].$$
Then the acceleration of both masses is:
$$ a = \frac Fm = \frac{g}{m_1 + m_2}[m_1 - m_2 \sin(\alpha)].$$
You should not be able to introduce a minus error just by choosing a convention. I assume that you just did not flip all the minus signs.
Part B
The wedge has to be at rest, so the force of friction has to compensate the tangential forces exactly. This means
$$ \mu F_\text{down} \overset != F_\text{tangential}. $$
I now construct the forces that act on the masses:

The tensions will change with the acceleration:
$$ T_1 = m_1[g-a], \quad T_2 = m_2[g\sin(\alpha) + a].$$
To digest this, set $a =0$, the static case. Then the tensions are just the forces that gravity exerts on the masses. The extreme case, where $m_1$ is free falling is $a = g$. Then $T_1$ has to be zero, since the object is free falling.
The mass $m_2$ exerts a normal force onto the wedge that will push it to the left. The tension $T_2$ will drag on the pulley which will then push on the wedge to the right. The force to the left is given by
$$ m_2 g \cos(\alpha)\sin(\alpha)$$
whereas the force to the right (just the $x$-component) is given by
$$ m_2[g \sin(\alpha) + a] \cos(\alpha).$$
In the static case $a = 0$, those two forces are equal and do not make the wedge move. In the dynamic case, they try to move the wedge. This is what you meant by conservation of linear momentum, just written down with the forces.
The net force to the right (assuming $a > 0$ then is
$$ m_2 a \cos(\alpha) = F_\text{tangential}.$$
Now the forces that push the wedge downward and give it a frictional force. The mass $M$ will drag downward. Then $T_1$ will push the pulley down, so does the $y$-component of $T_2$. The last thing is the $y$-component of the normal force. The last two mostly add up to just $m_2 g$, there is still a $m_2 a \sin(\alpha)$ left from the dynamics in $T_2$:
$$\underbrace{m_1[g-a]}_{T_1} + \underbrace{m_2[g\sin(\alpha) + a]\sin(\alpha)}_{{T_2}_y} + m_2 g \cos(\alpha)^2$$
Together, this is:
$$F_\text{down} = Mg + m_1[g-a] + m_2[g + a\sin(\alpha)].$$
Now set them equal, plug in $a$ and solve for $\mu$. You might want to omit all the terms with $a$ here, since that makes it really complicated. If $M$ is large, you can do that without a large error.


TITLE: How to find the PDF of one random variable when the PDF of another random variable and the relationship between the two random variables are known?
QUESTION [8 upvotes]: The probability density function of random variable X is given as  $ f_x(x) = \lambda e^ {-\lambda x} , x \ge 0. $ A new random variable $ Y = e^ {-\lambda X}$ is formed. Find the PDF of Y.

REPLY [4 votes]: There is a nice "change of variables" formula for doing this.  If $g$ is a monotonic function, then
$$
f_Y(y)=\left\vert\frac{d}{dy}(g^{-1}(y))\right\vert f_X(g^{-1}(y))
$$
So, let $g(x)=e^{-\lambda x}$; this is a monotonic function, so the formula applies.  $g^{-1}(y)=-\ln(y)/\lambda$, and so
$$
\left\vert\frac{d}{dy}g^{-1}(y)\right\vert=\frac{1}{\lambda y}
$$
putting the pieces together, we get:
$$
f_Y(y)=\frac{1}{\lambda y}\lambda \exp[-\lambda g^{-1}(y)]=\frac{1}{y}\exp[\ln(y)]=1
$$
Notice that since $x\geq 0$, $0<y\leq 1$ (assuming $\lambda>0$) so this distribution makes sense.


TITLE: Density of a distribution given by a Gaussian copula and a set of marginals
QUESTION [3 upvotes]: Suppose the distribution of an $n$-dimensional random vector $X$ is characterized by a Gaussian copula $C_R$ with correlation matrix $R$ and a set of marginal $\{(F_{X_i}, f_{X_i})\}_{i=1}^n$ (pairs of a distribution function and a density function). I would like to derive the probability density of $X$. Wikipedia says that the density of the copula is as follows:
$$
f_{C_R}(u)
= \frac{1}{\sqrt{\det{R}}}\exp\left(-\frac{1}{2}
\begin{pmatrix}\Phi^{-1}(u_1)\\ \vdots \\ \Phi^{-1}(u_n)\end{pmatrix}^T \cdot
\left(R^{-1}-I\right) \cdot
\begin{pmatrix}\Phi^{-1}(u_1)\\ \vdots \\ \Phi^{-1}(u_n)\end{pmatrix}
\right)
$$
Given this information, I decided to apply a change-of-variables transformation and, thereby, express the density of $X$ in terms of the density of $U \sim C_R$. I arrived at the following formula:
$$
f_X(x) = \left| \prod_{i=1}^n f_{X_i}(x_i) \right| \, \, f_{C_R}\left(
\begin{pmatrix}F_{X_1}(x_1)\\ \vdots \\ F_{X_n}(x_n)\end{pmatrix}
\right)
$$
The coefficient before $f_{C_R}$ is the absolute value of the determinant of the Jacobian of the inverse of the transformation from $U$ to $X$, which is just an application of $\{F^{-1}_{X_i}\}$ to the corresponding elements of $U$.
Since I do not have a proper background, I am not entirely sure should that all the steps that I have taken are legitimate. I, therefore, would be grateful if somebody could confirm that the above computations are correct.
I also wonder why $f_{C_R}$ differs so much from the density of a centered multivariate Gaussian vector:
$$
f_Y(y) =
\frac{1}{\sqrt{(2\pi)^n \det{\Sigma}}}
\exp\left(-\frac{1}{2}y^\mathrm{T}\Sigma^{-1}y
\right).
$$
How did $-I$ in $f_{C_R}$ come about? And why does not $f_{C_R}$ have $\sqrt{(2\pi)^n}$? It would be nice if somebody could explain this.
Thank you!
Regards,
Ivan

REPLY [2 votes]: The Gaussian copula (with correlation matrix $R$):
$$
C_R(u)=\Phi_n(\Phi^{-1}(u_1),\dots,\Phi^{-1}(u_n);R).
$$
Then, letting $\zeta:=(\Phi^{-1}(u_1),\dots,\Phi^{-1}(u_n))^{\top}$,
$$
c_R(u)=\frac{\phi_n(\Phi^{-1}(u_1),\dots,\Phi^{-1}(u_n);R)}{\phi(\Phi^{-1}(u_1))\cdots \phi(\Phi^{-1}(u_1))} \\=(2\pi)^{-n/2}\lvert R \rvert^{-1/2}\frac{-\exp(\zeta^{\top}R^{-1}\zeta/2)}{\prod_{i=1}^n (2\pi)^{-1/2}\exp(-\zeta_i^2/2)}=\lvert R \rvert^{-1/2}\frac{\exp(-\zeta^{\top}R^{-1}\zeta/2)}{\exp(-\zeta^{\top}I_n\zeta/2)}.
$$

Here, $\Phi_n(\cdot;R),\phi_n(\cdot;R)$ are zero-mean multivariate normal CDF and PDF with covariance matrix $R:n\times n$.