# One Example about the Relationship between the CD Inequality and CDE’
Inequality

Yijin Gao<EMAIL_ADDRESS>Department of Mathematics,Information School,
Renmin University of China, Beijing 100872, China

###### Abstract.

In this paper,we will give an easy example to satisfy that we can not conclude
CDE’ Inequality just from the CD Inequality.

##  1\. Introduction

The curvature-dimension inequality(CD-inequlaity ) was firstly introduced by
Bakry-Emery as a substitute of the lower Ricci curvature bound of the
underlying space.It was first studies on finite graphs by Lin-Yau[9][11].

Then some special cases $CD(0,\infty)$ are being studid by Liu-
Peyerimhoff.They prove an optimal eigenvalue ratio estimate for finite
weighted graphs satisfying the $CD$ inequality.Also,they show taking Cartesian
products to be an efficient way for construcing new weighted graphs satisfying
$CD(0,\infty)$.[6][7].

In 2015,Paul-Lin-Liu-Yau prove Li-Yau type estimates for bounded and positiv
solutions of the heat equation on graphs,under the assumption of the
curvature-dimension inequality $CDE^{\prime}(0,n)$,which can be consider as a
notion of curvature for graphs.So the relation between $CD$ and $CDE^{\prime}$
is becoming more and more important for the study of the ricci estimate on
finite graphs.[2]

In 2014,2015,Munch introduced a new version of a curvature-dimension
inequality for non-negative curvature.He used this inequlaity to prove a
logarithmic Li-Yau inequality on finite graphs.The new calculus and the new
curvature-dimension inequality coinciden with the common ones.In the case of
graphs,they coincide in a limit.In that case,the new curvature-dimension
inequality gives a more general concept of curvature on graphs.Then he showed
the connection between the $CDE^{\prime}$ and the $CD\psi$ inequality.Also,he
showed that the $CDE^{\prime}$ inequality implies the $CD$ inequality.[3][4]

So can $CD$ inequality implies $CDE^{\prime}$?It is a problem.In this
paper,the author gives a special case to show that the hypothesis is not right
in some special cases.

The paper is organized into three parts:
Chapter $1$ is the introduction of the graph, the Laplacians and CD,CDE’
inequalities on it.
Chapter $2$ introduces some basic conclusions and lemmas in order to get the
main result. These conclusions include some definitions such as local finite
graph, weighted graph and the calculation of the $\Gamma$ operator.
Chapter $3$ is the main conclusion of this thesis.

##  2\. GRAPHS, LAPLACIANS CD INEQUALITIES AND CDE’ INEQUALITIES

Given a graph $G=(V,E)$, for an $x\in V$, if there exists another $y\in V$
that satisfies $(x,y)\in E$, we call them are neighbors, and written as $x\sim
y$. If there exists an $x\in V$ satisfying $(x,x)\in E$, we call it a self-
loop. For a graph $G=(V,E)$.The neighborhood and the degree of a vertx $x\in
V$ are defined,respectively,as $N_{x}=\\{y\in V:xy\in E\\}$ and
$d_{x}=|N_{x}|$.For notational simplicity we work with $\mu_{x}=1/d_{x}$.
Now we will introduce some basic definitions and theorems before we get the
main results.

######  Definition 2.1.

(difference in valuations)For a function $f\in R^{V}$ and two vertices$x,y\in
V$ denote by$f(x,y)=f(y)-f(x)$ the difference in valuations.

######  Definition 2.2.

(locally finite graph) We call a graph G is a locally finite graph if for any
$x\in V$, it satisfies $\\#\\{y\in V|y\sim x\\}<\infty$. Moreover, it is
called connected if there exists a sequence $\\{x_{i}\\}_{i=0}^{n}$
satisfying: $x=x_{0}\sim x_{1}\sim\cdots\sim x_{n}=y$.

######  Definition 2.3.

(Laplacians on locally finite graphs) On a locally finite graph $G=(V,E,\mu)$
the Laplacian has a form as follows:

$\triangle f(x)=\mu_{x}\sum_{y\in N_{x}}(f(y)-f(x)),\quad\forall f\in
C_{0}(V).$

######  Definition 2.4.

(gradient operator $\Gamma$) The operator $\Gamma$ is defined as follows:

$\Gamma(f,g)(x)=\frac{1}{2}(\triangle(fg)-f\triangle g-g\triangle f)(x)$

.

Always we write $\Gamma(f,f)$ as $\Gamma(f)$.

######  Definition 2.5.

(gradient operator $\Gamma_{i}$) The operator $\Gamma_{i}$ is defined as
follows:

$\Gamma_{0}(f,g)=fg$
$\Gamma_{i+1}(f,g)=\frac{1}{2}(\triangle(\Gamma_{i}(f,g))-\Gamma_{i}(f,\triangle
g)-\Gamma_{i}(\triangle f,g))$

.

Also we have
$\Gamma_{2}(f)=\Gamma_{2}(f,f)=\frac{1}{2}\triangle\Gamma(f)-\Gamma(f,\triangle
f)$.

######  Definition 2.6.

($CD(K,n)$ condition) We call a graph satisfies $CD(K,n)$ condition if for any
$x\in V$,we have

$\Gamma_{2}(f)(x)\geq\frac{1}{n}(\triangle f)^{2}(x)+K\Gamma(f)(x).\quad
K\in\mathbb{R}.$

######  Definition 2.7.

($CDE(K,n)$ condition) Let $f:V\rightarrow\mathbb{R}^{+}$ satisfy $f(x)>0$,
$\triangle f(x)<0$. We call a graph satisfies $CDE(x,K,n)$ condition if for
any $x\in V$, we have

$\Gamma_{2}(f)(x)-\Gamma\left(f,\frac{\Gamma(f)}{f}\right)(x)\geqslant\frac{1}{n}(\triangle
f)(x)^{2}+K\Gamma(f)(x).\quad K\in\mathbb{R}.$

######  Definition 2.8.

($CDE^{\prime}(K,n)$ condition) We say that a graph $G$ satisfies the
exponential curvature dimension inequality $CDE(K,n)$ if for any positive
function $f:V\to R^{+}$ such that $\Delta f(x)<0$,we have

$\Gamma_{2}(f)(x)-\Gamma(f,\frac{\Gamma(f)}{f}(x))\geq\frac{1}{n}f(x)^{2}(\Delta
logf)(x)^{2}+k\Gamma(f)(x)$

.

######  Lemma 2.9.

$\Gamma(f)(x)=\frac{1}{2}\mu_{x}\sum_{y\in N_{x}}f(x,y)^{2}.$

Here we define $N_{x}=\\{y\in V:xy\in E\\}$ and $d_{x}=|N_{x}|.$For notational
simplicity we work with $\mu_{x}=\frac{1}{d_{x}}.$

Proof We have

$\left.\begin{aligned}
&\Gamma(f)(x)=\frac{1}{2}\triangle(f^{2})(x)-f(x)(\triangle f)(x)\\\
&=\frac{1}{2}\mu_{x}\sum_{y\in N_{x}}(f^{2})(x,y)-f(x)\mu_{x}\sum_{y\in
N_{x}}f(x,y)\\\ &=\frac{1}{2}\mu_{x}\sum_{y\in
N_{x}}(f(x,y)(f(y)+f(x))-2f(x,y)f(x))\\\ &=\frac{1}{2}\mu_{x}\sum_{y\in
N_{x}}f(x,y)^{2}.\\\ \end{aligned}\right.$

######  Lemma 2.10.

$\Gamma_{2}(f)(x)=\frac{1}{2}(\triangle f)^{2}(x)+\mu_{x}\sum_{y\in
N_{x}}\mu_{y}\sum_{z\in N_{y}}(f(y,z)^{2}-\frac{1}{2}f(x,z)^{2})).$

Proof We have

$\left.\begin{aligned} &\triangle(\Gamma(f))(x)=\mu_{x}\sum_{y\in
N_{x}}\Gamma(f)(x,y)&=\mu_{x}\sum_{y\in N_{x}}\frac{1}{2}\mu_{y}\sum_{z\in
N_{y}}(f(y,z)^{2}-f(x,y)^{2})\\\ \end{aligned}\right.$

and

$\left.\begin{aligned} &\Gamma(f,\triangle
f)(x)=\frac{1}{2}(\triangle(f\cdot\triangle
f)(x)-f(x)\cdot(\triangle^{2}f)(x)-(\triangle f)^{2}(x))\\\
&=-\frac{1}{2}(\triangle f)^{2}(x)+\frac{1}{2}\mu_{x}\sum_{y\in
N_{x}}((f\triangle f)(x,y)-f(x)(\triangle f)(x,y))\\\ &=-\frac{1}{2}(\triangle
f)^{2}(x)+\frac{1}{2}\mu_{x}\sum_{y\in N_{x}}f(x,y)(\triangle f)(y)\\\
&=-\frac{1}{2}(\triangle f)^{2}(x)+\frac{1}{2}\mu_{x}\sum_{y\in
N_{x}}f(x,y)\mu_{y}\sum_{z\in N_{y}}f(y,z)\\\ \end{aligned}\right.$

thus

$\left.\begin{aligned}
&\Gamma_{2}(f)(x)=\frac{1}{2}\triangle(\Gamma(f))(x)-\Gamma(f,\triangle
f)(x)\\\ &=\frac{1}{2}(\triangle f)^{2}(x)+\frac{1}{2}\mu_{x}\sum_{y\in
N_{x}}\mu_{y}\sum_{z\in
N_{y}}(\frac{1}{2}f(y,z)^{2}-\frac{1}{2}f(x,y)^{2}-f(x,y)f(y,z))\\\
&=\frac{1}{2}(\triangle f)^{2}+\frac{1}{2}\mu_{y}\sum_{y\in
N_{x}}\mu_{y}\sum_{z\in N_{y}}(f(y,z)^{2}-\frac{1}{2}f(x,z)^{2})\\\
\end{aligned}\right.$

######  Lemma 2.11.

If $\Delta f(x)<0$ in $x\in V$,$CDE^{\prime}(K,n)$ implies $CDE(K,n).$ Proof
Let $f:V\to R+$ be a positive function for which $\Delta f(x)<0$.Since
$logs\leq s-1$for all positive $s$,we can write

$\Delta logf(x)=\sum_{y\sim x}(logf(y)-logf(x))\leq\sum_{y\sim
x}\frac{f(y)-f(x)}{f(x)}=\frac{\Delta f(x)}{f(x)}<0.$

Hence squaring everything reverses the above inequality and we get

$(\Delta f(x))^{2}\leq f(x)^{2}(\Delta logf(x))^{2},$

and thus $CDE(K,n)$ is satisfied

$\Gamma_{2}(f)(x)\geq\frac{1}{n}f(x)^{2}(\Delta
logf)(x)^{2}+k\Gamma(f)(x)>\frac{1}{n}(\Delta f)(x)^{2}+k\Gamma(f)(x).$

######  Lemma 2.12.

The $CDE^{\prime}$ inequality implies $CD$ inequlaity.
It was the work in [3]

##  3\. BASIC CONCLUSION

######  Remark 3.1.

In this section,we just concern the easiest graph: $x$ is the initial point
and its neighborhood is $y$,also $y$ has another neighborhood $z$.so the graph
consists three points which created from $x$,$x$ and $z$ are not connected.We
give the special graph a name:$EG$.Also,we concern the special case in $CD$
and $CDE^{\prime}$ inequlaity:$CD(0,n)$ and $CDE^{\prime}(0,n)$.

######  Lemma 3.2.

If the $EG$ satisfies $CD(0,m)$,then we have $m\geq 2$.
Proof For the simplicity,we rewrite $f(x)=x$,$f(y)=y$ and $f(z)=z$.Then
according to the lemma before ,we can have the following: $\Delta
f(x)=y-x$,$\Gamma(f)(x)=\frac{1}{2}(y-x)^{2}$. Also,we need to calculate the
$\Gamma_{2}(f)$,

$\left.\begin{aligned}
&\Gamma_{2}(f)(x)=\frac{1}{2}(y-x)^{2}+\frac{1}{4}[(z-y)^{2}-\frac{1}{2}(z-x)^{2}+(x-y)^{2}]\\\
&=\frac{1}{2}(y-x)^{2}+\frac{1}{4}(z,y)^{2}-\frac{1}{8}(z-x)^{2}+\frac{1}{4}(x-y)^{2}\\\
&=\frac{1}{8}[z^{2}+z(2x-4y)+8y^{2}+5x^{2}-12xy]\\\ \end{aligned}\right.$

Absolutely,we will use the knowledge of quadratic function.The minimun of
$z^{2}+z(2x-4y)$ can be obtained when $z=2y-x$.Then we take $z=2y-x$ in the
equality.

$\left.\begin{aligned} &8y^{2}+5x^{2}-12xy+(2y-x)^{2}+(2y-x)(2x-4y)\\\
&=8y^{2}+5x^{2}-12xy+4y^{2}+x^{2}-4xy+4xy-8y^{2}-2x^{2}+4xy\\\
&=4x^{2}+4y^{2}-8xy\\\ \end{aligned}\right.$

So we have the estimate of $\Gamma_{2}(f)$

$\Gamma_{2}(f)\geq\frac{1}{2}(x^{2}+y^{2}-2xy)=\frac{1}{2}(x-y)^{2}\geq\frac{1}{m}(y-x)^{2}$

So we have $m\geq 2$.

######  Theorem 3.3.

If the $EG$ satisfies $CD(0,m)$,we can take the special value in the point
$y$,then it may not be satisfied in $CDE^{\prime}(0,m)$.
Proof Without loss of generality,we assume $x=1$.According the theorem
before,we need to find some special $y$ that make sure the $EG$ does not
satisfy $CDE^{\prime}(0,2)$.
As before,the value of $\Gamma_{2}(f)$ is given

$\Gamma_{2}(f)=\frac{3}{4}(x-y)^{2}+\frac{1}{4}(z,y)^{2}-\frac{1}{8}(z,x)^{2}$

So the next work is to calculate the value of $\Gamma(f,\frac{\Gamma(f)}{f})$

$\left.\begin{aligned}
&\Gamma(f,\frac{\Gamma(f)}{f}=\frac{1}{2}\Delta(\Gamma(f))-\frac{1}{2}f(x)\Delta(\frac{\Gamma(f)}{f(x)})-\frac{1}{2}\Delta(f)\frac{\Gamma(f)}{f(x)}\\\
&=\frac{1}{2}\Delta(\Gamma(f))-\frac{1}{2}\Delta(\frac{\Gamma(f)}{f})-\frac{1}{2}\Delta
f\Gamma(f)\\\ &=I_{1}-I_{2}-I_{3}.\\\ \end{aligned}\right.$

Then we get the value of $I_{1}$.

$\left.\begin{aligned} &I_{1}=\frac{1}{2}(\Gamma(f)(y)-\Gamma(f)(x))\\\
&=\frac{1}{2}{\frac{1}{4}[(y-x)^{2}+(z,y)^{2}]-\frac{1}{2}(y-x)^{2}}\\\
&=\frac{1}{2}[\frac{1}{4}(y-x)^{2}+\frac{1}{4}(y-z)^{2}-\frac{1}{2}(y-x)^{2}]\\\
&=\frac{1}{8}(y-z)^{2}-\frac{1}{8}(y-x)^{2}.\\\ \end{aligned}\right.$

Also the value of $I_{2}$.

$\left.\begin{aligned}
&I_{2}=\frac{1}{2}\frac{\Gamma(f)(y)}{y}-\frac{1}{2}\frac{\Gamma(f)(x)}{x}\\\
&=\frac{1}{2}\frac{1/4[(y-x)^{2}+(z-y)^{2}]}{y}-\frac{1}{4}(y-x)^{2}\\\
\end{aligned}\right.$

Then the value of $I_{3}$.

$\left.\begin{aligned} &I_{3}=\frac{1}{2}\Delta(f)\Gamma(f)(x)\\\
&=\frac{1}{2}(y-x)\frac{1}{2}(y-x)^{2}\\\ &=\frac{1}{4}(y-x)^{3}\\\
\end{aligned}\right.$

We get the following inequality according the definition of
$CDE^{\prime}(m,0)$

$\frac{3}{4}(y-x)^{2}+\frac{1}{4}(y-z)^{2}-\frac{1}{8}(z-x)^{2}-\frac{1}{8}(y-z)^{2}+\frac{1}{8}(y-x)^{2}+\frac{1}{8y}(y-x)^{2}+\frac{1}{8y}(y-z)^{2}\\\
-\frac{1}{4}(y-x)^{2}+\frac{1}{4}(y-x)^{3}\geq\frac{1}{2}(logy)^{2}$

Firstly,we deal with the polynomial with $z$ like the situation before.

$\left.\begin{aligned}
&\frac{1}{8}(y-z)^{2}-\frac{1}{8}(z-x)^{2}+\frac{1}{8y}(y-z)^{2}\\\
&=\frac{1}{8}(y^{2}+z^{2}-2yz-z^{2}-1+2z)+\frac{1}{8y}(y^{2}+z^{2}-2yz)\\\
&=\frac{1}{8}y^{2}-\frac{1}{4}yz+\frac{1}{4}z-\frac{1}{8}+\frac{1}{8}y+\frac{1}{8y}z^{2}-\frac{1}{4}z\\\
&=\frac{1}{8y}z^{2}-\frac{1}{4}yz+\frac{1}{8}y^{2}+\frac{1}{8}y-\frac{1}{8}\\\
\end{aligned}\right.$

Here,we take $z=\frac{\frac{1}{4}y}{\frac{1}{4}\frac{1}{y}}=y^{2}$.Then we get

$\frac{1}{8y}y^{4}-\frac{1}{4}y^{3}+\frac{1}{8}y^{2}+\frac{1}{8}y-\frac{1}{8}=-\frac{1}{8}y^{3}+\frac{1}{8}y^{2}+\frac{1}{8}y-\frac{1}{8}.$

Then the inequality becomes

$\left.\begin{aligned} &The..left..of..the..inequality\\\
&=\frac{1}{4}(y-1)^{3}+\frac{5}{8}(y-1)^{2}+\frac{1}{8y}(y-1)^{2}-\frac{1}{8}y^{3}+\frac{1}{8}y^{2}+\frac{1}{8}y-\frac{1}{8}\\\
&=\frac{1}{4}(y^{3}-3y^{2}+3y-1)+\frac{5}{8}y^{2}+\frac{5}{8}-\frac{10}{8}y-\frac{1}{8}y^{3}+\frac{1}{8}y^{2}+\frac{1}{4}y-\frac{3}{8}+\frac{1}{8y}\\\
&=\frac{1}{8}y^{3}+y(\frac{3}{4}-\frac{10}{8}+\frac{1}{8}+\frac{1}{8})+\frac{1}{8y}\\\
&=\frac{1}{8}y^{3}-\frac{1}{4}y+\frac{1}{8y}\\\ \end{aligned}\right.$

So the inequality becomes:

$y^{3}-2y+\frac{1}{y}\geq 4(logy)^{2}$

. Here we distinguish $y$ into two parts:$y>1$ and $0<y<1$
I:First,we concern the situation $y>1$,because $e^{x}\geq x+1$,so we have
$logy<y-1$
Then we need to prove

$y^{3}-2y+\frac{1}{y}\geq 4(y-1)^{2}$

It equals

$y^{3}-4y^{2}+6y-4+\frac{1}{y}\geq 0$

we set the function $h(y)=y^{3}-4y^{2}+6y-4+\frac{1}{y}.$
The first derivate of $h(y)$ is

$h^{\prime}(y)=3y^{2}-8y+6-\frac{1}{y^{2}}$

The second derivate of $h(y)$ is

$h^{\prime\prime}(y)=6y-8+2\frac{1}{y^{3}}$

Then we use the cauchy inequality to get that

$6y-8+\frac{1}{y^{3}}=2y+2y+2y+2\frac{1}{y^{3}}-8=2(y+y+y+\frac{1}{y^{3}})-8\geq
2*4(yyy\frac{1}{y^{3}})^{(}1/4)-8=0$

So the function $h^{\prime}(y)$ is increasing when $y>1$.

$h^{\prime}(y)>h^{\prime}(1)=3-8+6-1=0$

So the function $h(y)$ is increasing too.

$h(y)\geq h(1)=1-4+6-4+1=0$

. When $y>1$,we can get the $CDE^{\prime}(0,m)$

II:Second,we concern the situation$y<1$ as our discussion.we need to proof the
following

$y^{3}-2y+\frac{1}{y}\geq 4(logf)^{2}$

Actually, it equals that

$y^{3}-2y+\frac{1}{y}\geq 4(log\frac{1}{y})^{2}$

Just as discussed before,then we have to prove

$y^{3}-2y+\frac{1}{y}\geq 4(\frac{1}{y}-1)^{2}$

Here we set
$Q(y)=y^{3}-2y+\frac{1}{y}-4(\frac{1}{y}-1)^{2}=y^{3}-2y+\frac{9}{y}-\frac{4}{y^{2}}-4$.
The next step is to analysis $Q(y)$

$\left.\begin{aligned} &y^{3}-2y+\frac{9}{y}-\frac{4}{y^{2}}-4\\\
&=y^{3}-y^{2}+y^{2}-y-y+1-5+\frac{5}{y}+\frac{4}{y}-\frac{4}{y^{2}}\\\
&=(y-1)(y^{2}+y-1-\frac{5}{y}+\frac{4}{y^{2}})\\\
&=(y-1)(y^{2}-y+2y-2+1-\frac{1}{y}-\frac{4}{y}+\frac{4}{y^{2}}\\\
&=(y-1)^{2}(y+2+\frac{1}{y}-\frac{4}{y^{2}})\\\
&=(y-1)^{2}(y-1+3-\frac{3}{y}+\frac{4}{y}-\frac{4}{y^{2}})\\\
&=(y-1)^{3}(1+\frac{3}{y}+\frac{4}{y^{2}})\\\ \end{aligned}\right.$

Of course $1+\frac{3}{y}+\frac{4}{y^{2}}>0$,but $y-1<0$,so
$(y-1)^{3}(1+\frac{3}{y}+\frac{4}{y^{2}})<0$,which is contrary to what we
need.Clearly,we can set $y=0.1$,then the left is less than the right.So we get
the conclusion that we can not conclude $CDE^{\prime}$ just from $CD$
situation.

## References

  * [1] Chao Gong and Yong Lin, Equivalent Properties for CD Inequalities On Graphs With Unbounded Laplacians, preprint.
  * [2] Paul Horn,Yong Lin,Shuang Liu,Shing-Tung Yau,Volume doubling,Poincare inequality and Gaussian heat kernel estimate for nonnegative curvature graphs,Mathematics,2015
  * [3] Florentin Munch,Remarks on curvature dimension conditions on graphs,Arxiv:1501.05839v1
  * [4] Florentin Munch,Li-Yau inequality on finite graphs via non-linear curvatur dimension conditions,Arxiv:1412.3340v1
  * [5] Yong Lin and Shuang Liu, Equivalent properties of CD inequality on graphs, preprint.
  * [6] David Cushing,Shiping Liu,Norbert Peyerimhoff,Bakry-Emery curvature functions of graphs,Arxiv:1606.01496
  * [7] Shiping Liu,Florentin Munch,Norbert Peyerimhoff,Bakry-Emery curvature and diameter bounds on graphs,Arxiv:1608.07778
  * [8] F. Bauer, P. Horn, Y. Lin, G. Lippner, D. Mangoubi and S.-T. Yau, Li-Yau inequality on graphs, Journal of Differential Geometry,99,359-405, 2015.
  * [9] Y. Lin and S. T. Yau, Ricci curvature and eigenvalue estimate on locally finite graphs. Math. Res. Lett., 17(2):343-356, 2010.
  * [10] B.Hua,Y.Lin,Stochastic completeness for graphs with curvature dimension conditions,preprint,2014
  * [11] Y.lin,L.Lu,S.T.Yau,Ricci curvature of graphs,Tohoku Math.J.63,605-627,2011


## Vertical Bicycle Model of Vehicle with Dry Friction

G.M. Rozenblat1
${}^{1}\leavevmode\nobreak\ $Moscow Automobile and Road Construction State
Technical University
e-mail<EMAIL_ADDRESS>
V.B. Yashin 2
${}^{2}\leavevmode\nobreak\ $Moscow Technical University of Communications and
Informatics
e-mail<EMAIL_ADDRESS>
Abstract. The motion of a vertically positioned bicycle is considered when a
horizontal control force, which maybe both internal and external in relation
to the bicycle, is applied to its pedal. Tangential forces of dryfriction
obeying the Euler – Coulomb law act at points of contact of the wheels with
the horizontal supportplane. The constraint at the points of contact of the
wheels with the support is assumed to be unilateral.The problem of determining
the acceleration of the centre of mass of the bicycle and the realized
motionsof its wheels (with or without slip, and with or without detachment)
with dierent values of the design parameters and control force is solved. This
approach is usefull for control of vehicle and optimization of its dynamical
parameters. Cases of non-uniqueness of motion the Painleve paradox are found.

## Introduction

The movements of bodies by means of change in the positions of internal
(relative to the body) masses occur on exposure to external forces, which
generally are friction forces (resistance forces, hydrodynamic forces,
gravitational forces). The creation of such effective (optimum) external
forces is due to the corresponding synthesis of internal forces on exposure to
which admissible movements of internal masses are realized (Ref. $1,$ p. 344).
Such problems have been examined (see, for example, Refs $2-7$) for the
purposes of creating robotic devices. In some problems, internal forces are
used to create a variable shape of a body moving in liquid, [8–10]. In space
dynamics, internal forces can be used for controlled change in the external
gravitational force, which can move such a body away from the Earth by as
great a distance as desired (the ’graviflight’ effect, see Ref. $11,$ Essay
9).

The movement of a bicycle (and any self-propelled vehicle) also belongs to
this class of problems of the movement of bodies. Here, the internal (control)
force is the force applied to the pedal by the cyclist’s foot. External
forces, which realize the motion of the bicycle, are the gravitational forces
and the forces of reactions at the points of contact of the wheels with the
road.

##  1 The formulation and solution of the two simplest problems of bicycle
motion

Figure 1 depicts a bicycle moving in the vertical position relative to a
stationary right-handed coordinate system Oxyz which is selected in the
following way. The $x$ axis coincides with the horizontal support line along
which the bicycle is moving and is oriented to the right, and the $y$ axis is
directed down the vertical. Thus, the vertical plane coincides with the plane
of the bicycle and with the plane $Oxy.$ The z axis is perpendicular to the
$Oxy$ plane and forms with the $x$ and $y$ axes a right-handed coordinate
system. Thus, the $z$ axis is perpendicular to the plane of the figure and is
directed away from the reader. With this choice of coordinate system, all the
turns, angular velocities, and angular accelerations of the bodies in the
figure that occur in a clockwise direction are considered to be positive. It
is clear that these turns, angular velocities, and angular accelerations, when
viewed from the positive direction of the $z$ axis, are oriented anticlockwise
in this case and are positive, as assumed in theoretical mechanics for three-
dimensional space. Similar properties are also achieved for the moments of
forces: if a force turns the body clockwise about a selected point, from the
point of view of the reader looking at the figure, then the moment of this
force relative to the examined point is positive, and in the opposite case it
is negative. Below, unless stipulated otherwise, we will assume that the
introduced kinematic parameters are positive in the sense indicated above. The
true signs of these quantities are obtained as a result of solving the
problem.

Figure 1:

A bicycle consists of four bodies: _1_ – the sprocket with the pedals; _2 and
3_ – the two wheels; _4_ – the frame to which the bodies 1,2, and 3 are hinged
at points $O_{1},O_{2},$ and $O_{3},$ as shown in Fig. $1.$ It is assumed that
the cyclist moving the bicyle along the horizontal $x$ axis by applying to the
lower pedal $A$ an internal horizontal force $F$ oriented against the
direction of the $x$ axis is also rigidly connected to the frame.

We will introduce the following notation: $v$ and $a$ are the velocity and the
acceleration of the centre of mass of the bicycle along the $x$ axis,
$\omega_{i}$ and $\varepsilon_{i}$ are the angular velocities and
accelerations of the sprocket $(i=1),$ the rear wheel $(i=2),$ and the front
wheel $(i=3),I_{i}(i=1,2,$ 3) is the moment of inertia of the body $i$
relative to the inherent centre of mass, $M=1$ is the total mass of the
bicycle (including the cyclist when seated on the bicycle $,r_{1}$ is the
radius of the sprocket with the pedal, $r_{2}$ is the radius of the sprocket
on the rear wheel, $\mu=r_{2}/r_{1}$ is the gear ratio, $R$ is the external
radius of the wheels 2 and $3,N_{j}$ and $S_{j}(j=2,3)$ are the normal
reaction and the tangential reaction at the point of contact of the wheel $j$
with the road, $b_{j}(j=2,3)$ is the distance from the point of contact of the
wheel $j$ to the centre of mass of the bicycle along the horizontal,
$b=\left(b_{2}+b_{3}\right)/2$ is the half-distance between the points of
attachment of the wheels along the horizontal, $h_{0}$ is the elevation of the
centre of mass of the bicycle (together with the cyclist when seated on the
bicycle) above the road, $F$ is the force applied by the cyclist to the lower
pedal of the sprocket 1 and directed along the horizontal against the positive
direction of the $x$ axis, $L$ is the corresponding arm (i.e., the distance
from the pedal to the horizontal $O_{1}O_{2}$), and $T=T_{1}+T_{2}$ is the
difference in the moduli of the forces of tension of the chain that are
applied to the sprocket 1.

Let us examine the following problems (below it is assumed that there is no
slip of the wheels).

Problem 1. In the standard situation, when the force $F$ applied by the
cyclist is internal relative to the indicated system of four bodies, determine
the acceleration of the centre of mass of the bicycle (together with the
cyclist) along the $x$ axis.

Problem 2. In the non-standard situation, when the force $F$ is external
relative to the indicated system of four bodies (the cyclist has dismounted
from the bicycle and is applying force $F$ to the lower pedal $A$ against the
direction of the $x$ axis or is employing the services of an external aid),
determine the acceleration of the bicycle along the $x$ axis.

In particular, in both cases, if before the application of force $F$ the
bicycle is at rest, then the side to which it will begin to move - in the
direction of force $F$ or in the opposite direction to this force - must be
indicated.

###  1.1 Solutions of the problems

We will use the theorem of change in kinetic energy (Ref. 12, p. 69). Assuming
that the wheels are rolling without slip, and the drive chain of the bicycle
is unstretchable, we will obtain the following kinematic relations:

$\omega_{1}=\mu v/R,\quad\omega_{2}=\omega_{3}=v/R;\quad\mu=r_{2}/r_{1}$ (1.1)

The kinetic energy of the bicycle, with the use of relations (1.1), is
calculated by means of the formula

$T=\frac{1}{2}M_{0}v^{2};\quad M_{0}=1+\frac{I_{1}\mu^{2}+I_{2}+I_{3}}{R^{2}}$
(1.2)

To calculate the power of all internal and external forces, we will consider
that in Problem 1 a non-zero contribution is made by the power of the internal
force $F$, which creates the moment $FL$ applied to the body 1 rotating at
angular velocity $\omega_{1}$. In Problem 2, a non-zero contribution is made
by the external force $F$ applied to the lower pedal of the bicycle $A$ having
the following projection of absolute velocity onto the $x$ axis:

$v_{Ax}=v-\omega_{1}L=v(1-\sigma);\quad\sigma=\mu L/R$ (1.3)

The remaining (both internal and external) forces make a zero contribution
owing to the absence of slip of the wheels, the absence of stretchability of
the drive chain, and the ideal hinges at points $O_{1},O_{2},O_{3}$. Thus, the
power is determined by the expression

$W_{1}=FL\omega_{1}=F\sigma v\text{ for Problem }1,\quad
W_{2}=-FU_{Ax}=-Fv(1-\sigma)\text{ for Problem }2$

We will describe in greater detail how the formula for power $W_{1}$ is
obtained. We will transfer to point $O_{1}$ the force $F$ applied to the pedal
$A$. As a result we obtain the force $F$ applied at point $O_{1}$ and a pair
of forces with moment $FL$. This system of forces is applied to the body 1
rotating about point $O_{1}$ and moving forwards together with this point.
According to Newton’s third law, on the body 4 (on whose side this system of
forces is applied) at point $O_{1}$ is acted upon by a force $F$ and a pair of
forces with moment $FL$. The body 4 does not rotate. and therefore the power
of the latter moment is zero, and the two oppositely directed and applied
forces at point $O_{1}$ together provide zero power. Thus, a non-zero
contribution is made only by the power of the indicated pair of forces applied
to the body 1 rotating at angular velocity $\omega_{1}$, and we arrive at the
expression given above for power $W_{1}$. Then, using the theorem of change in
kinetic energy, we will obtain

$a=\sigma F/M_{0}\text{ for Problem 1}$ (1.4) $a=-(1-\sigma)F/M_{0}\text{ for
Problem 2}$ (1.5)

The parameters $M_{0}$ and $\sigma$ were defined, respectively, by the second
equality of system (1.2) and the second equality of system (1.3). From this it
follows that in problem 1 the bicycle moves along the $x$ axis, against the
force $F$, and in problem 2 with $\sigma\in(0,1)$ (with $\sigma>1$) the
bicycle moves against (along) the $x$ axis, in the direction (against the
direction) of the force $F$. For normal designs of bicycles,
$r_{2}<r_{1}(\mu<1)$, $L<R.$ Then $\sigma\in(0,1),$ and the bicycle moves in
the direction of the external force $F$.

Remark 1. Similar problems for a bicycle (three-wheel) on a qualitative level
were proposed at the All-Union Physics Olympiads in the
$1960\mathrm{\leavevmode\nobreak\ s}$ (see Ref. 13, problems 98 and 106). As
can be seen, the strict and complete solution is fairly complex (for the
student), and the sound solution of Problems 1 and 2 requires physical
intuition.

Remark 2. The excellent book by Arnold 14 contains the essay ’Which force
drives the bicycle forwards?’. The author applies a backward force to the
lower pedal of the bicycle and shows that the bicycle will move forwards. As
can be seen, this outcome contradicts the answer to Problem 2 (with adherence
to the inequality $0<\sigma<1$, as occurs for real bicycle models). The
present author learnt of this discrepancy from V.F. Zhuravlev, and, in part,
it prompted the author to write the present paper. It may be obvious (see Fig.
1) that the force $F$ creating a clockwise moment applied to the sprocket will
rotate the latter likewise in a clockwise direction. The sprocket, in turn,
with the help of the chain, will force the wheels of the bicycle (in the
absence of slippage) to rotate clockwise, which will lead to movement of the
bicycle from left to right, i.e., against the force $F$. However, the moment
from the forces of tension of the chain, which is also applied to the sprocket
with the pedals, must also be taken into account. This moment may be directed
anticlockwise and may be sufficiently great to exceed the moment from the
force $F$ and force the sprocket to rotate anticlockwise. This may lead to
movement of the bicycle from right to left, in the direction of the force $F$.
Such a situation comes about when the force $F$ is external and the parameters
of the bicycle satisfy the condition $\sigma\in(0,1)$. If, however, the force
$F$ is internal, then, as shown above, the given situation can never be
realized, i.e., the bicycle will move against the direction of the force $F$.

##  2 The general problem of controlling the motion of a bicycle in the
vertical plane using an internal or external force $F$ applied to the lower
pedal statement of the problem of determining the accelerations

The force $F$ directed along the horizontal and against the $x$ axis (Fig. 1)
is applied to the pedal (i.e., to the body 1) either as in Problem 1 or as in
Problem 2. Thus, this force is either internal or external. The bicycle at the
initial instant $t=0$ is at rest. It is required to determine the dynamics of
its motion at fairly short times $t>0,$ on the assumption that the wheels are
in contact in a unilateral way with a rough support plane with a Coulomb
friction coefficient $f$.

Remark 1. The use of an internal force $F$ in the way indicated above (as in
Problem 1) is equivalent to applying a corresponding control moment to the
wheel $2,$ which is characteristic of motorcycles, where such a moment is
created by the motor.

Remark 2. It can be assumed that the bicycle has completed uniform
uncontrolled motion without wheel slip, and that the cyclist has then decided
to begin controlled motion in the way indicated above.

Remark 3. Unilateral contact of the wheels with the support plane makes it
possible to consider the well-known trick of motion with one of the wheels
separated from the support. The situation where both wheels are separated is
not ruled out either, but these are different problems.

###  2.1 The principal equations of the kinematics and dynamics of motions of
the bicycle without separation of the wheels

Let the force $F$ be internal (as in Problem 1). We will write three equations
of the dynamics of planar motion of the bicycle, taking into account that the
centre of mass does not move along the vertical:

$a=S_{2}+S_{3}\quad 0=-N_{2}-N_{3}+g$ (2.1)
$I_{1}\varepsilon_{1}+I_{2}\varepsilon_{2}+I_{3}\varepsilon_{3}=N_{2}b_{2}-N_{3}b_{3}-\left(S_{2}+S_{3}\right)h_{0}$
(2.2)

Equations (2.1) are the equations of motion of the centre of mass of the
bicycle along the horizontal and vertical, while Eq. (2.2) is the equation of
change in the angular momentum of the bicycle relative to its intrinsic centre
of mass.

We will describe in greater detail how Eq. (2.2) is obtained. Suppose $v_{k}$
is the velocity of the centre of mass of the body $k(k=1,2,3,4)$ It is clear
that $v_{k}=v,$ where $v$ is the velocity of the centre of mass of the
bicycle, which is directed along the $x$ axis. Let $\mathrm{Cxy}$ be a system
of König coordinates ($C$ is the centre of mass of the bicycle), and let
$m_{k}$ and $y_{k}$ be the mass and vertical coordinate of the centre of mass
of the body $k$ in the $\mathrm{Cxy}$ system. Then, for time derivatives from
the kinetic moments of the bodies relative to their general centre of mass
$\mathrm{C}$, we will obtain

$\dot{K}_{i}=I_{i}\varepsilon_{i}+m_{i}ay_{i},\quad
i=1,2,3;\quad\dot{K}_{4}=m_{4}ay_{4}$

Summing up these expressions and using the equality

$m_{1}y_{1}+m_{2}y_{2}+m_{3}y_{3}+m_{4}y_{4}=0$

we will obtain the left-hand part of Eq. (2.2). The right-hand part is
obtained in the standard way. We will write equations for the changes in the
kinetic moments of the bodies $1,2,$ and 3 separately in relation to their
intrinsic centres of mass, taking into account that forces of tension of the
chain $T_{1}$ and $T_{2}$ are applied to the body 1, and the same forces but
with opposite signs are applied to the body $2.$ We will obtain

$I_{1}\varepsilon_{1}=FL+Tr_{1},\quad
I_{2}\varepsilon_{2}=-S_{2}R-Tr_{2},\quad I_{3}\varepsilon_{3}=-S_{3}R$ (2.3)

Then, we will write the kinematic equation that follows from the condition of
the absence of stretchability of the drive chain between the sprocket 1 and
the sprocket of the wheel 2 :

$\varepsilon_{1}r_{1}=\varepsilon_{2}r_{2}$ (2.4)

The use of Eqs (2.3) makes it possible to write Eq. (2.2) in the form

$N_{3}b_{3}-N_{2}b_{2}+\left(S_{2}+S_{3}\right)h+T\left(r_{1}-r_{2}\right)+FL=0;\quad
h=h_{0}-R$ (2.5)

Excluding $T$ and $\varepsilon_{1}$ from the obtained equations, we will have
the following system of five equations:

$\begin{array}[]{l}a=S_{2}+S_{3},I_{0}\varepsilon_{2}/R=F\sigma-
S_{2},I_{3}\varepsilon_{3}/R=-S_{3}\\\ N_{2}+N_{3}=g,\quad
N_{3}b_{3}-N_{2}b_{2}+\left(S_{2}+S_{3}\right)h+F\sigma
Rk_{2}+S_{2}Rk_{1}=0\end{array}$ (2.6)

where the following notation is introduced:

$I_{0}=I_{1}\mu^{2}+I_{2},\quad k_{1}=I_{1}\mu(\mu-1)/I_{0},\quad
k_{2}=\left(I_{1}\mu+I_{2}\right)/I_{0}$ (2.7)

and parameter $\sigma$ is defined by the last equality in system (1.3). If the
force $F$ is external (as in Problem 2), it is not difficult to show that,
instead of system (2.6), we obtain a system differing from system (2.6) in the
presence of the term $-F$ in the first equation and the term $-F(L+h)$ in the
final equation.

In system $(2.6),S_{j},N_{j},\varepsilon_{j}(j=2,3),$ and $a$ are unknown
quantities to be determined (in both problems, both for an internal and for an
external force $F$).

Thus, for correct and unambiguous solution of system (2.6) or the similar
system in the case of an external force, it is necessary to have two more
equations, which are obtained additionally from physical considerations
following from the proposed possible motion of the bicycle. We will give the
additional equations for different cases of possible motions of this kind:
neither of the wheels 2 and 3 slips:

$a=\varepsilon_{j}R,\quad j=2,3$

the wheel 2 (wheel 3) slips, while the wheel 3 (wheel 2) does not slip:

$a=\varepsilon_{5-j}R,\quad S_{j}=-fN_{j}\operatorname{sign}a_{j},\quad
a_{j}=a-\varepsilon_{j}R,\quad j=2\quad(j=3)$

both the wheels 2 and 3 slip:

$S_{j}=-fN_{j}\operatorname{sign}a_{j},\quad j=2,3$

Before setting out the solutions of the corresponding equations of the
dynamics in the four given cases, we will examine the possibility of static
equilibrium of the bicycle with the indicated application of an internal or
external force $F$ to the pedal and the presence only of frictional forces of
rest at the contact points of the wheels. We will show that for a bicycle
(more accurately, for the combination of rigid bodies making up the bicycle,
with or without a cyclist, depending on the examined problem) such equilibrium
is impossible with any non-zero force $F$ if there are no moments of rolling
friction on its wheels. In fact, if the force $F$ is internal, the equations
of static equilibrium have the form

$\begin{array}[]{l}S_{2}+S_{3}=0,\quad N_{2}+N_{3}-g=0,\quad
N_{2}b_{2}-N_{3}b_{3}-\left(S_{2}+S_{3}\right)h=0\\\ FL+Tr_{1}=0,\quad-
S_{2}R-Tr_{2}=0,\quad-S_{3}R=0\end{array}$ (2.8)

from which it follows that $S_{2}=S_{3}=0,T=0,$ and $F=0$, i.e., equilibrium
is possible only at zero force $F$. If the force $F$ is external, then the
term $-F$ is added to the left-hand part of the first equation of system
$(2.8),$ and the term $F(L+h)$ to the left-hand part of the third equation.
From these equations it follows that $S_{3}=0,S_{2}=F,T=-FL/r_{1},$ and
$T=-FR/r_{2}.$ The last two equations are possible only if $L=Rr_{1}/r_{2}.$
For a real bicycle, $r_{1}>r_{2}$, and the refore $L>R$, i.e., the pedal rests
on the support plane. Thus, static equilibrium is again impossible. Note that
a similar conclusion is also correct when force $F$ is inclined to the
horizontal.

Remark. In their problem book (Ref. 15, p. 81), Gnädig et al. give an
erroneous solution for the case where the force $F$ is external. This error
stems from the $a$ priori assumption that the system of bodies making up the
bicycle is in equilibrium. On this basis, three equations of equilibrium are
examined for the system as a whole as an absolutely rigid body. However, these
equations are the consequence of the indicated assumption and cannot be a
criterion for retention of static equilibrium of the examined system of
bodies. As can be seen, a more detailed analysis taking into account the
specific nature of the examined system of rigid bodies leads to the
impossibility of static equilibrium of the bicycle at any force $F$, either
internal or external.

##  3 Formulating the results for the case of an internal force $F$

In this section, we will formulate the results that are obtained by applying
to the pedal of the bicycle an internal force $F$ (as in Problem 1 from
Section 1) for the four cases indicated above. Everywhere below, unless
stipulated otherwise, it is assumed that

$\mu=1\left(r_{1}=r_{2}\right),\quad b_{2}=b_{3}=b$

Furthermore, to simplify the formulae, it has been assumed that $M=1$ (i.e.,
the total mass of the bicycle with the cyclist is selected as the unit for
measurement of mass). We will introduce the following (already dimensionless)
notation:

$\begin{array}[]{l}M_{0}=1+M_{2}+M_{3},M_{1}=1+M_{2},\quad
M_{2}=\left(I_{1}+I_{2}\right)/R^{2},\quad M_{3}=I_{3}/R^{2}\\\
b_{0}=f\left(h+M_{0}R\right),\quad b_{1}=fh\left(2M_{1}\right),\quad
b_{11}=f\left(h+M_{1}R\right)\\\
\xi_{1}=fg\frac{M_{0}}{2\left(1+M_{3}\right)-b_{0}/b},\quad\xi_{2}=fg\frac{M_{0}}{2M_{3}+b_{0}/b},\quad\xi_{3}=fg\frac{M_{1}b}{b_{11}}\\\
\xi_{4}=fg\frac{\eta}{2\eta-f\left(1+2M_{2}\right)R},\quad\xi_{5}=g\frac{\eta}{\left(1+2M_{3}\right)R},\quad\xi_{6}=g\frac{\eta}{R},\quad\eta=b-fh\\\
a_{21}=fg\frac{2b+fR}{b_{11}}\left(\frac{b}{2b+fR}-\frac{M_{3}bs_{1}}{2M_{3}b+b_{0}}\right),\quad
a_{22}=fg\frac{2b+fR}{2M_{3}b+b_{0}}\left(\frac{b}{2b+fR}+\frac{M_{3}bs_{2}}{b_{11}}\right)\\\
a_{31}=f\frac{gb+F\sigma R}{2\left(1+M_{3}\right)b-fh},\quad
a_{41}=f\frac{F\sigma R}{\eta},\quad
s_{1}=\frac{F\sigma-\xi_{3}}{\xi_{2}-\xi_{3}},\quad
s_{2}=\frac{F\sigma-\xi_{2}}{\xi_{3}-\xi_{2}}\end{array}$ (3.1)

Assertion 1. When the bicycle moves without slip of either wheel, the
following conditions are fulfilled:

$\text{ if }b_{0}<b\text{, then }0<F\sigma<\xi_{1};\text{ if }b\leq
b_{0}\text{, then }0<F\sigma<\xi_{2}\text{.}$

The acceleration $a=a_{11}$ of the centre of mass of the bicycle in this case
is given by formula (1.4).

Assertion 2. To realize motion of the bicycle with slip of the front wheel 3
only (wheel 3 slips forwards), the following conditions are fulfilled (in each
of the examined cases, the corresponding acceleration $a$ of the centre of
mass of the bicycle is given in parentheses):

if $0<b<b_{1},$ then $\xi_{3}<F\sigma<\xi_{2}\left(a=a_{21}>0\right)$

if $b_{1}\leq b<b_{11},$ then $\xi_{2}<F\sigma<\xi_{3}\left(a=a_{22}>0\right)$

if $b_{11}\leq b<b_{0}$, then $\xi_{2}<F\sigma<\xi_{4}\left(a=a_{22}>0\right)$

if $b>b_{0}$, then this motion is impossible.

Assertion 3. When the bicycle moves with slip of the rear wheel 2 only (wheel
2 slips backwards), the following conditions are fulfilled:

$b>b_{0},\xi_{1}<F\sigma<\xi_{5}\left(a=a_{31}>0\right),$

if $b<b_{0},$ tnen this motion is impossible.

Assertion 4. When the bicycle moves with slip of both wheels (the wheel 2 in
this case slips backwards, and the wheel 3 slips forwards), the following
conditions are fulfilled:

if $b_{11}<b<b_{0},$ then $\xi_{4}<F\sigma<\xi_{6}\left(a=a_{41}>0\right)$

if $b_{0}<b<+\infty,$ then $\xi_{5}<F\sigma<\xi_{6}\left(a=a_{41}>0\right)$

if $0<b<b_{11},$ then this motion is impossible.

Substantiation of these assertions is given in Section 4.

The combination of the assertions 1 to 4 leads to the following assertion.

Assertion 5. If $0<b<b_{1}$ and $\xi_{3}<F\sigma<\xi_{2}$, then motion occurs
either without slip of either wheel or with slip of the front wheel
(ambiguity). In all remaining cases, motion is unambiguous in accordance with
the results of the assertions 1 to $4,$ or loss of the unilateral constraint
occurs for the front wheel.

Remark 1. From Assertion 5 it follows that, when certain conditions are
fulfilled, the motion is ambiguous (the Painlevé paradox). The realized motion
of the bicycle depends on the prehistory of its movement. Such issues are
discussed in detail in other papers by the present author. 16,17

Remark 2. The loss (weakening) of the unilateral constraint for the front
wheel of the bicycle occurs at a fairly high force $F$. In this case, the
normal reaction on the front wheel becomes negative. Further motions of the
bicycle with possible separations of the front wheel should be considered in a
separate problem.

Remark 3. If $f\rightarrow\infty$, then only cases of the Assertions 1 and 2
are realized, from which, in part, there follow the results of solving the
Problem 1 from Section 1 for the motion of the bicycle without wheel slip.
However, the ambiguity (i.e., the additional possibility of motion with slip
of the front wheel with certain values of the force $F$ is retained. This was
pointed out to the author by F.L. Chernous’ko.

Remark 4. It is surprising that motions of the bicycle with slip of the front
(driven) wheel are possible. In fact no moments are applied to this wheel.
However, it must be borne in mind that at the centre of the front wheel there
acts a horiz ontal force of reaction from the bicycle frame. It is this force
that may in certain situations cause slip of the front wheel. A problem of
this type for one wheel with a horizontal force applied at its centre is given
in Meshcherskii’s problem book (Ref. 18, problem 35.4). The complete solution
of this problem is available. 19

##  4 Substantiation of the results of Section 3

###  4.1 Proof of Assertion 1

The solution of system of equations (2.6) with the corresponding additional
equations and notation (3.1) leads to the formulae

$\begin{array}[]{l}a=\frac{F\sigma}{M_{0}},\varepsilon_{2}=\varepsilon_{3}=\frac{F\sigma}{M_{0}R}\\\
S_{2}=\frac{F_{0}\left(1+M_{3}\right)}{M_{0}},\quad S_{3}=-\frac{F\sigma
M_{3}}{M_{0}}\\\ N_{2,3}=\frac{gb_{3}\pm F\sigma
d}{2b},d=R+\frac{h}{M_{0}}\end{array}$ (4.1)

For the correct realization of the examined motion, it is necessary to ensure
that the inequalities of the unilateral constraints $N_{2}>0$, $N_{3}>0$ are
observed, and also the Coulomb inequalities $\left|S_{2}/N_{2}\right|<f$ and
$\left|S_{3}/N_{3}\right|<f$ for the frictional forces of rest $S_{2}$ and
$S_{3}$. The use of these inequalities together with relations (4.1) and
notation (3.1) leads to the results of the Assertion $1.$

###  4.2 Proof of Assertion 2

The solution of system of equations (2.6) with the corresponding additional
equations and allowance for notation (3.1) leads to the formulae

$a=\frac{F\sigma-
fN_{3}\delta_{3}}{M_{1}},S_{2}=\frac{F\sigma+fN_{3}\delta_{3}M_{2}}{M_{1}},\delta_{3}=\operatorname{sign}a_{3}$
(4.2)

Then, it is not difficult to estimate that

$a_{3}=\frac{F_{\sigma}}{M_{1}}-fN_{3}\delta_{3}\left(\frac{1}{M_{1}}+\frac{1}{M_{3}}\right)$
(4.3)

from which it follows that the value $l_{3}=-1$ cannot be real ized, as in
this case, under the conditions $N_{3}>0$ and $F>0,$ we obviously obtain
$a_{3}>0,$ which leads to a contradiction when $\mathrm{I}_{3}=signa_{3}$.

Thus, only $\mathrm{l}_{3}=+1,\mathrm{a}_{3}>0$ is possible, i.e., the front
wheel can only slip forwards over the course of motion along the $x$ axis.
Then, we will obtain

$N_{2}=\frac{g\left(b-fd_{0}\right)+F\sigma d_{1}}{q},\quad
N_{3}=\frac{gb-F\sigma d_{1}}{q};\quad q=2b-fd_{0},\quad
d_{0}=\frac{h}{M_{1}},\quad d_{1}=R+d_{0}$ (4.4)

Using the inequality $a_{3}>0$ and the Coulomb inequality for $S_{2}$ and
$\mathrm{N}_{2}$, from formulae (4.2) to (4.4) we will obtain

$N_{2}>0,\quad N_{3}>0,\quad
fN_{3}M_{0}/M_{3}<F\sigma<fN_{2}M_{1}-fN_{3}M_{2}$ (4.5)

$N_{2}$ and $N_{3}$ are defined by formulae (4.4). Then, from relations (4.4)
and (4.5) follow the inequalities

$fM_{0}\frac{gb-F\sigma d_{1}}{M_{3}q}<F\sigma<f\frac{g(b-fh)+D\sigma
d_{1}\left(M_{1}+M_{2}\right)}{q},0<\frac{gb-F\sigma d_{1}}{q}<g$ (4.6)

Hence, when $q>0,$ we obtain

$F\sigma>fg\xi_{2},\quad
F\sigma\left[2(b-fh)-fR\left(M_{1}+M_{2}\right)\right]<fg(b-fh),\frac{g\left(fd_{0}-b\right)}{d_{1}}<F\sigma<\frac{gb}{d_{1}}$
(4.7)

The magnitude of ${\xi}_{2}$ is given by the second equation in the third line
of formulae (3.1). If $\mathrm{q}<0$, then in relations (4.7) all the signs of
the inequalities must be changed to the opposite signs. To solve the obtained
inequalities (4.7), we will introduce, using notation (3.1), the following
functions of parameter $b$:

$F_{k}(b)=\xi_{k},\quad k=1,2,3$ (4.8)

In accordance with inequalities (4.7), these functions limit the upper and
lower values of EST, depending on the value of the parameter
$b\in(0,+\infty)$, and when $b=b_{1}$ they acquire the same value

$F_{0}=fg\frac{h}{2\left(M_{1}R+h\right)}$

An analysis of inequalities (4.7) with the use of functions (4.8) and their
elementary properties leads to the results of the Assertion 2.

###  4.3 Proof of Assertion 3

The solution of system of equations (2.6) with the corresponding additional
equations and notation (3.1) leads to the formula

$a=-fN_{2}\delta_{2}/\left(1+M_{3}\right),\quad\varepsilon_{2}R=\left(F\sigma+fN_{2}\delta_{2}\right)/M_{2},\quad\left(\delta_{2}=\operatorname{sign}a_{2}\right)$

from which we will obtain the equality

$a_{2}=-\frac{F\sigma}{M_{2}}-fN_{2}\delta_{2}\left(\frac{1}{M_{2}}+\frac{1}{1+M_{3}}\right)$
(4.9)

From this, with $\delta_{2}=1$, we arrive at a contradiction, as then, with
$N_{2}>0$ and $F>0$, obviously we will have $a_{2}<0.$ Thus, we have
$\delta_{2}=-1$ and $a_{2}<0$ (the rear wheel 2 may slide only backwards,
against the direction of the $x$ axis), and from the relation (4.9) we will
obtain the inequality

$F\sigma>fN_{2}\frac{M_{0}}{1+M_{3}}$ (4.10)

Furthermore, the Coulomb inequality for the force $S_{3}$ leads to the
relation

$N_{2}M_{3}<N_{3}\left(1+M_{3}\right)$ (4.11)

Then, simple calculations lead to the following formulae for normal reactions:

$N_{2}=\frac{gb+F\sigma R}{2b-f\Delta},\quad N_{3}=\frac{gb-zf\Delta-F\sigma
R}{2b-f\Delta},\Delta=\frac{h}{1+M_{3}}$ (4.12)

For the realization of the examined motion, it is still necessary to add the
inequalities $N_{2}>0$ and $N_{3}>0,$ which, with account taken of formulae
(4.12), leads to the double inequality

$0<\frac{gb+F\sigma R}{2b-f\Delta}<g$ (4.13)

Thus, for the realization of the examined motion, it is necessary to observe
the inequalities $(4.10),(4.11),$ and (4.13) with account taken of the
formulae (4.12). As a result of simple calculations, we obtain the
inequalities of the assertion $3.$

###  4.4 Proof of Assertion 4

The solution of the system of equations (2.6) with the corresponding
additional equations and notation (3.1) leads to the formulae

$\begin{array}[]{l}a_{2}=-\frac{F_{0}}{M_{2}}-fN_{2}\delta_{2}\left(1+\frac{1}{M_{2}}\right)-fN_{3}\delta_{3},\quad
a_{3}=-fN_{2}\delta_{2}-fN_{3}\delta_{3}\left(1+\frac{1}{M_{3}}\right)\\\
\delta_{2}=\operatorname{sign}a_{2},\quad\delta_{3}=\operatorname{sign}a_{3}\end{array}$
(4.14)

Simple analysis of formulae $(4.14).$ with account taken of the inequalities
$\mathrm{N}_{2}>0,\mathrm{\leavevmode\nobreak\ N}_{3}>0,$ and $\mathrm{F}>0,$
leads to the conclusion that only the set $\delta_{2}=-1,\delta_{3}=+1,$ with
which the inequalities $a_{2}<0$ and $a_{3}>0$ should be observed, is
consistent. Thus, here the rear wheel slides backwards, and the front wheel
forwards. Then, from the relations (4.14) we will obtain the inequalities

$F\sigma>fN_{2}\left(1+M_{2}\right)-fN_{3}M_{2},\quad
N_{2}>N_{3}\left(1+\frac{1}{M_{3}}\right)>0$ (4.15)

Simple calculations lead to the following expressions for normal reactions:

$N_{2}=\frac{g\eta+F\sigma R}{2\eta}>0,\quad N_{3}=\frac{g\eta-F\sigma
R}{2\eta}>0$ (4.16)

and inequalities (4.15) acquire the form

$F\sigma>f\frac{g\eta+F\sigma R}{2\eta}+fM_{2}\frac{2F\sigma
R}{2\eta},\frac{2F\sigma R}{2\eta}>\frac{1}{M_{3}}\frac{g\eta-F\sigma
R}{2\eta}$ (4.17)

The parameter $\eta$ is defined by the final equality in the fourth line of
formulae (3.1). Considering the two cases $\eta>0$ and $\eta<0,$ from
inequalities (4.16) and (4.17) we will obtain the results of Assertion $4.$

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