Mathematics, Sustainability, and a Bridge to Decision Sup.docx

Mathematics, Sustainability, and a Bridge to Decision Support
Author(s): Mary Lou Zeeman
Source: The College Mathematics Journal, Vol. 44, No. 5
(November 2013), pp. 346-349
Published by: Mathematical Association of America
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GUEST EDITORIAL
Mathematics, Sustainability, and a Bridge to
Decision Support
Mary Lou Zeeman
Mary Lou Zeeman ([email protected]) is the Wells
Johnson Professor of Mathematics at Bowdoin College.
She received her Ph.D. at Berkeley, worked at U. T. San
Antonio for 15 years, and has held postdocs at the IMA and
MIT, as well as visiting positions at Michigan and Cornell.
Her research interests include dynamical systems,
population dynamics and fisheries, neuroscience,
endocrinology, and climate science. Zeeman is also
involved in several interdisciplinary initiatives focused on
the health of the planet. She co-directs the Mathematics
and Climate Research Network, which links researchers
across the U.S. and beyond to develop the mathematics
needed to better understand the earth’s climate
(http://www.mathclimate.org). She helped found the
Institute for Computational Sustainability based at Cornell
University, and she is on the organizational team of the
Mathematics of Planet Earth 2013 initiative.
The Mathematics of Planet Earth. Scientific societies,
universities, and organi-
zations around the world have come together this year to focus
attention on the Math-
ematics of Planet Earth (MPE 2013). The MPE web portal [6]
lists research programs,
curriculum materials, public events, hosts a daily blog, and

more. The initiative con-
tinues beyond 2013, so please keep sending events and ideas to
the portal. This issue
of The College Mathematics Journal, part of the MPE initiative,
has articles that il-
lustrate all four MPE themes: a planet to discover; a planet
supporting life; a planet
organized by humans; and a planet at risk. All four are essential
for understanding our
changing climate and addressing our pressing sustainability
challenges. We are used to
the idea that mathematically-rich subjects such as statistics,
economics, engineering,
and climate science have a role to play in supporting decision-
making at the front lines
of sustainability challenges. What about those of us in
mathematics departments? We,
too, have a lot to contribute.
What is decision support? The core idea, according to the
National Research
Council [7], is “making scientific knowledge useful for
practical decision making.”
What might that entail for mathematicians? We could, for
example, help translate sci-
entific results into non-technical language—especially when
uncertainty is involved.
Or we could provide data, analysis tools, and software. I’ve
deliberately phrased these
suggestions to emphasize a professional service component of
decision support. Ser-
vice, of course, can be rewarding in its own right, but the reader
of this JOURNAL
may wonder how decision support is tied to mathematics
research or the mathematics

http://dx.doi.org/10.4169/college.math.j.44.5.346
346 „ THE MATHEMATICAL ASSOCIATION OF AMERICA
2017 18:37:17 UTC
http://www.mathclimate.org
curriculum. Let’s explore this by analogy with the application
of mathematics to other
disciplines. Mathematical biology is my example; you may
substitute your favorite
application.
Analogy with mathematical biology. There is a well-established
bridge between
the mathematics and biology communities. As interdisciplinary
researchers, we can
position ourselves wherever we feel most comfortable on the
bridge. Some of us like
to work at the biology end of the bridge, immersing ourselves in
lab experiences and
collaborating directly with experimentalists to understand what
can be measured, what
can be controlled, and what are sources of uncertainty. From
there, we collabora-
tively develop models and design experiments to test
hypotheses, tease apart systems,
and unravel biological mysteries. Others prefer to work in the
middle of the bridge,
abstracting the ideas and analyzing the structures that recur in
the models. For ex-
ample, there are thriving mathematical biology communities

that study coupled oscil-
lators, excitability, and bifurcation, generalizing those ideas to
networks, and explor-
ing the delicate balance nature walks between stability and
adaptability in a stochastic
world. Finally, others prefer to work at the mathematical end of
the bridge, proving
theorems that unify the consequences of these structures, and
developing methods for
their analysis. Over a career, some enjoy moving back and forth
across the bridge.
An essential feature of the bridge is the two-way nature of the
interactions. Both
disciplines are enriched with new insights, new questions, and
new ways of looking at
old questions.
A mathematics-decision support bridge. We are beginning to
build a similar
bridge between mathematics and decision stakeholder
communities. The term ‘stake-
holder’ includes all who care about a decision, particularly
those who make it and those
impacted by it. For example, regarding a decision of where to
build a wind farm and
what kind of turbines to use, stakeholders include private
investors, the energy com-
pany, local residents, town councils, land trusts, field
naturalists, and others. Those
of us who enjoy working at the stakeholder end of the bridge
immerse ourselves in
the decision question, collaborating directly with stakeholders
to understand the dif-
ferent components of the system, what can be measured, what
can be controlled, what
creates uncertainty, what is valued, and what are the associated

costs, economic and
otherwise. (See [12] for a description of how to bring
stakeholder communities to-
gether to describe a complex system). From there, we
collaboratively develop models,
design computer experiments to test scenarios, tease apart
systems, and identify new
decision options. In the early stages of bridge building, it may
feel like isolated deci-
sions are separated by their specific details. But, just as in
mathematical biology, from
multiple case studies common structures emerge, generating
abstractions to explore at
the middle of the bridge and stimulating new mathematics all
along the bridge. One
of my favorite emerging structures is that of resilience. Systems
with inherent positive
feedback often exhibit multiple stable states. For example,
rangelands can be grassy
or become wooded or desertified, depending on patterns of
grazing, fires, and precip-
itation [12]. The resilience of a state measures how much
perturbation a system can
withstand without transitioning to a qualitatively different state.
Estimating resilience,
therefore, involves understanding the interplay between between
deterministic basins
of attraction and the stochastic characteristics of noise and
disruptive events in a grad-
ually changing environment [10, 11, 12]. More examples of
mathematical structures
common to decision questions are showcased in the MPE daily
blog, and education
and workshop pages [6].
VOL. 44, NO. 5, NOVEMBER 2013 THE COLLEGE

MATHEMATICS JOURNAL 347
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Strengthening the bridge. Enrichment for the communities at
both ends of the
mathematics-biology bridge derives from healthy overlap
between individuals along
the bridge. To increase this overlap, we have a range of
opportunities for individuals
at all levels to broaden their reach along the bridge. In
mathematical biology, we
have curricula for undergraduate and graduate students, funding
agencies that spon-
sor immersion experience programs for researchers, and
interdisciplinary societies,
conferences, and institutes. Let’s do the same thing for the
mathematics-decision
support bridge. Let’s apply our collective, intelligent minds to
create opportunities
for individuals to broaden their reach. In the language of the
National Research
Council, let’s figure out how the mathematics community can
help “create the condi-
tions for the production of decision-relevant information and for
its appropriate use”
[7, p. 34].
Getting involved. We all have something to contribute to this
effort, if we choose.
We may step onto the bridge ourselves, or we may empower
others to do so. One

way to step onto the bridge is to seek out a sustainability
organization and learn what
questions it grapples with. At the local level, for example, this
could involve joining a
sustainability solutions seminar series. To connect with decision
makers at the state or
federal level, Angus King, a former Governor of Maine and
current U.S. Senator, rec-
ommends talking to to a “staff member for energy and the
environment”. King points
out, “Decision-makers are often in search of compelling data as
a basis for public pol-
icy and effectively presented mathematical data can have a
significant influence on
policy outcomes.” (Personal communication, 2013).
How do we empower others? Straightforward mechanisms
include supporting in-
terdisciplinary seminars, guest speakers, cross-disciplinary
conference travel, and so
on. Those of us who are senior are in a position to facilitate
discussions about how to
evaluate and reward colleagues for deeply intellectual decision-
support work, whose
tangible products are not research publications. This is
especially important for tenure
decisions. At Bowdoin, the scholarship criterion for tenure
candidates is “professional
distinction recognized by members of their guild outside the
College,” [1], a criterion
that can certainly encompass the production of decision-relevant
information and its
appropriate communication.
In the classroom. We must also create opportunities for our
students to increase

their reach along the math-decision support bridge.
Sustainability questions are highly
multidisciplinary, blending mathematics with science, social
science, and economics.
None of us can be expert in all these disciplines, so it takes
cross-disciplinary team-
work and courage to attack the questions. There are many ways
to broaden the cur-
riculum to help students develop those skills. In doing so, we
also broaden the appeal
of mathematics to students. There are excellent examples in the
articles in this issue,
and more examples on the MPE education and curriculum
materials pages [6]. They
range from one-day modules for core math classes [2] to
annotated reading lists by
climate experts around which to design seminar courses [5]. The
Multidisiplinary Sus-
tainability Education project at Ithaca College is a model for
enriching mathematics
and science education by using sustainability questions as an
organizing principle for
linking existing courses from several departments [4, 8]. I
experienced a quantum leap
in my own appreciation for cross-disciplinary teams when I
helped develop and run
Cornell’s State of the Planet course, which had the theme:
Whatever your talent, what-
ever your passion, you can use them to help the planet [3, 9].
348 „ THE MATHEMATICAL ASSOCIATION OF AMERICA
2017 18:37:17 UTC

Into the future. I hope that some of these suggestions resonate
with you, and that
you will share your own examples and suggestions through the
MPE blog, curriculum
pages, articles in this JOURNAL, and elsewhere.
Our students are young. They are inheriting the planet as it is.
Their enthusiasm
for finding solutions to sustainability challenges is palpable,
bringing new energy and
creativity into the mathematics classroom. Harness this energy
and empower them. Be
honest about the state of the planet, but don’t get trapped in
gloom and doom. Teach
material that is about solutions more than it is about problems.
Acknowledgment. I am grateful to Marty Anderies, Steve
Cantrell, Chris Cosner, Michael
Henle, David McCobb, Elan Shapiro, and members of the
Mathematics and Climate Research
Network for helpful conversations, and to Bowdoin College and
NSF for their support (DMS-
0940243 and CCF-0832788).
References
1. Bowdoin College, Faculty handbook (2012); available at
http://www.bowdoin.edu/
academic-affairs/forms-policies/policies/pdf/12-
13FacultyHandbook.pdf.
2. Center for Discrete Mathematics and Theoretical Computer
Science, DIMACS sustainability modules for
undergraduate mathematics classes (2013); available at

http://dimacs.rutgers.edu/MPE.
3. T. Eisner, L. E. Fletcher, J. G. Hamilton, D. P. McCobb, and
M. L. Zeeman, Empower your students: Bring
a State of the Planet course to your school, Mathematics
Awareness Month theme essay (2009 and 2013);
available at http://www.mathaware.org/mam/2013/essays.
4. J. Hamilton, M. Rogers, T. Pfaff, and A. Erkan,
Multidisciplinary collaborations in the traditional classroom:
Wrestling with global climate change to improve science
education, Transformations: The Journal of Inclu-
sive Scholarship and Pedagogy 21 (2010) 89–98.
5. Mathematics and Climate Research Network, MCRN
annotated reading lists (2013); available at http:
//www.mathclimate.org/education/annotated-reading-lists.
6. Mathematics of Planet Earth, MPE 2013 web portal; available
at http://mpe2013.org.
7. Panel on Strategies and Methods for Climate-Related
Decision Support; National Research Council, Inform-
ing Decisions in a Changing Climate, The National Academies
Press, Washington DC, 2009; available at
http://www.nap.edu/catalog.php?record_id=12626.
8. T. Pfaff, A. Erkan, J. Hamilton, and M. Rogers, Ithaca
College multidisciplinary sustainability education
project, (2010); available at http://www.ithaca.edu/mse.
9. K. L. Rypien, J. Anderson, J. Andras, R. W. Clark, G. A.
Gerrish, J. T. Mandel, M. L. Nydam, and D. K.
Riskin, Students unite to create State of the Planet course,
Nature 44 (2007) 775.

10. M. Scheffer, Critical Transitions in Nature and Society,
Princeton University Press, Princeton NJ, 2009.
11. B. Walker and D. Salt, Resilience Thinking: Sustaining
Ecosystems and People in a Changing World, Island
Press, Washington DC, 2006.
12. , Resilience Practice: Building Capacity to Absorb
Disturbance and Maintain Function, Island Press,
Washington DC, 2012.
Joseph Fourier on Global Temperature
In his “Mémoire sure les températures du globe terreste et des
espace planétaires”
(see also p. 363), Fourier advances a principle that might be
considered the theme
of this issue,
In the present writing I have set myself another goal, that of
calling atten-
tion to one of the greatest objects of Natural Philosophy, and to
set forth an
overview of the general conclusions. I have hoped that the
geometers will
not see these researches only as a question of calculation, but
that they will
also consider the importance of the subject.
VOL. 44, NO. 5, NOVEMBER 2013 THE COLLEGE
MATHEMATICS JOURNAL 349
2017 18:37:17 UTC

http://www.bowdoin.edu/academic-affairs/forms-
policies/policies/pdf/12-13FacultyHandbook.pdf
http://www.bowdoin.edu/academic-affairs/forms-
policies/policies/pdf/12-13FacultyHandbook.pdf
http://dimacs.rutgers.edu/MPE
http://www.mathaware.org/mam/2013/essays
http://www.mathclimate.org/education/annotated-reading-lists
http://www.mathclimate.org/education/annotated-reading-lists
http://mpe2013.org
http://www.nap.edu/catalog.php?record_id=12626
http://www.ithaca.edu/mseThe Mathematics of Planet
Earth.What is decision support?Analogy with mathematical
biology.A mathematics-decision support bridge.Strengthening
the bridge.Getting involved.In the classroom.Into the
future.Joseph Fourier on Global Temperature
LECTURES
In Module 2, we'll discuss three aspects of organizational
change. The first is your role as a change agent. You will learn
about how the skills, knowledge, experience, attitude, and
behavior of the change agent affect the change process. The
personality and image of the change agent plays a vital role in
forming relationships with people within the organization. The
nature of this relationship often contributes to the success or
failure of the attempted change.
The second aspect is problem diagnosis. You will learn the
importance of correctly identifying problems within the
organization. A quote that is often repeated among OD change
agents is, "Correct diagnosis of the problem is half the problem
solved."
The third aspect is resistance to change. You as a change agent
should necessarily expect a reasonable degree of resistance to
change. People are usually fixed in their ways and find it

difficult to cope with change. We'll discuss ways in which you
can manage resistance to change initiatives.
The Change Agent (1 of 3)
Before we proceed further in the course you need to understand
your role as a change agent. A change agent initiates change,
espouses its need, and works toward its implementation. The
change agent may be an outsider who focuses primarily on
managing organizational change or an employee of the
organization who visualizes the need for change.
Whether internal or external the change agent brings with
him/her a package of skills and personality, which affects how
the change process is implemented and the relationships of
people involved in the change process.
Change Agent—External versus Internal
In layman's terms your role as a change agent is to handle
change—regardless of whether you are from within or outside
the organization. The status of the change agent influences
critical issues such as low productivity and high employee
turnover in the change process.
An external change agent comes with fresh and unbiased
perspectives, with new and creative ideas. The external agent is
usually someone of high repute with considerable success in
change management and may already have the respect of the
people within the organization.
Unlike an external change agent, an internal change agent may
not have an unbiased stance due to increased level of
involvement in the change initiative. This does not however
mean that the insider is less successful. Often the external
change agent is brought into the change process only when
things are out of hand. Alternatively the internal change agent

has the advantage of recognizing problem symptoms as they
arise.
The internal agent has an already established relationship with
the employees of the organization and may find it easier to
gather information. The insider also knows who to contact to
get specific information. The external change agent does not
have this advantage and has to work on getting to know
employees and gain their trust.
The trust levels may reverse when there is a great divide
between the management and employees. The internal change
agent may be viewed as an individual who is biased towards
senior management. Employees may fear repercussions for
themselves if the confidentiality of the information they share
with the change agent is not maintained. In these situations the
external change agent usually has an advantage if he/she can
convince employees that confidentiality will be maintained.
Another key challenge is the availability of the change agent
after the change process is complete. The external change agent
disengages from the organization, which may need to handle
any post-implementation issues that may arise. However the
internal change agent is available to work with the organization
on post-implementation issues.
You learned that internal and external change agents bring some
inherent advantages and disadvantages with them. However
their position in the organization is not the only factor that
influences the consulting process. Consulting styles also play a
very important role.
Change Agent—The Consulting Style
Supervisors are often classified as task-oriented or relationship-
oriented. Task-oriented supervisors focus on performance and
achievement, and relationship-oriented supervisors try to build

and maintain relationships with employees. On a similar vein
change agents, both internal and external, can be classified on
the basis of the degree of task- and relationship-orientation. The
pathfinder is high in both relationship- and task-orientation, but
the stabilizer is low in both. The cheerleader is high in
relationship-orientation, but the analyzer is high in task-
orientation. The persuader tries to find a middle ground to
achieve a balance between task- and relationship-orientation.
Research shows that each consulting style can be associated
with specific skills that are handy during the consulting process.
Being a change agent you'll find it extremely useful to identify
your own consulting style. You also need to be aware that you
should not be limited by your consulting style. Instead of
bracketing yourself into one category and identifying your
current skills, you may wish to identify critical skills that you
need to build on.
It is important to reiterate that internal and external change
agents, regardless of their consulting style, should build a
productive relationship with the employees for the change
initiative to be successful.
Forming the Agent-Client Relationship (1 of 3)
If you are an internal change agent, you'll likely have
relationships in place even before the change process is
initiated. As an external change agent your relationship begins
when the client contacts you with a potential problem. You can
decline from entering into this relationship if you want.
The Contract
After the initial meeting and probably a site visit, the first step
is to chart out a written contract. This contract should define the
scope of the assignment, the remuneration, time lines, list of
resources, responsibilities of the client versus the change agent,
expected results, operating rules, arbitration rules, and so on.

Contracts are necessary but they can also be restricting if they
are not flexible. Therefore it is a good idea to include
contingency plans in the contract.
If you are an internal change agent you may not be able to
define an all-inclusive contract with internal clients. An
alternative is to develop a written proposal and obtain necessary
approval from the concerned authority.
A contract is extremely useful from a legal perspective, but its
contribution to the success of the change initiative is limited.
The success of the initiative depends on the interpersonal
relationship between the client and you as the change agent.
The Agent-Client Relationship
The agent-client relationship begins even before the drafting of
the contract. The contract or proposal provides broad guidelines
for the entire consulting process and sets the stage for the
agent-client relationship. The relationship between you as a
change agent and client progresses along with the consulting
assignment.
As a change agent you should be aware of what the client
expects from you. Similarly as a client you should be able to set
expectations for the change agent. It is important to engage in
impression management. Both the change agent as well as the
client needs to gauge the other party in initial meetings. They
should use their intuition to assess the nature of the other
person and their commitment to the proposed change initiative.
Impression management plays a critical role in the change
initiative. If you are a change agent you need to impress upon
the client your skills and capabilities. In doing so you may be
tempted to talk about successful previous assignments. Keep in
mind however not to reveal the identities of earlier clients or
share confidential information. This may cost you the

assignment because you break the trust of your earlier clients
and indicate to your prospective client that you cannot be
trusted with company information. Instead you can show
reference letters if the client requests.
If you are a change agent it is also important to assess the
client's commitment to change and check if the client is
interested merely in superficial changes. On the other side as a
client in an organization seeking a change agent's services,
you'll need to assess the change agent's degree of commitment.
You may want to determine if the change agent will treat your
organization as just another project or is really concerned about
the success of your organization.
A change initiative will be successful if the relationship
between the change agent and the client is based on trust,
understanding, and similarity in personal values. While the
relationship is cordial please bear in mind that too much
closeness between the change agent and client might not be in
the best of interests. For the external change agent it may be
difficult to disengage from the organization at the end of the
assignment.
Finally the change agent should establish a network within the
organization. You learned from the systems view of
organizations that a change initiative involves various parts of
the organization either directly or indirectly. Therefore
relationship- and trust-building should extend to the entire
organization. As a change agent you will then be able to ensure
a greater degree of participation and involvement from
employees.
When you encourage greater participation from employees
you'll observe that you are able to correctly identify problems
and gather sufficient information to gauge employees' resistance

to change.
Diagnosing the Problem (1 of 3)
As a student of organizational change management and a future
change agent, you need to understand the role of problem
diagnosis in the process of change management. Diagnosing the
problem correctly is fundamental to implementing a change
process.
The Symptom versus the Problem
When you go to a doctor with a problem, the doctor first puts
you at ease, asks a series of questions about your symptoms,
and tries to diagnose the problem. For example, you have a
headache, fever, and nausea. The doctor may suggest a pain
reliever to ease your pain but this solution is temporary. The
doctor then tries to identify your problem based on your
symptoms and narrows down the diagnosis to flu, stress, or
bacterial infection. The doctor tries to diagnose the problem to
prescribe the appropriate medication.
Note that the doctor analogy mirrors the change agent who
attempts to diagnose the problem in the organization. Being a
change agent you are the doctor and the client is the patient.
There are two mistakes that change agents can make in the
problem diagnosis stage. They may treat only the symptoms and
provide quick-fix solutions without addressing the root cause of
the problem. For example, you cannot cure a bacterial infection
by prescribing a pain reliever. Second, change agents may not
diagnose the problem correctly. For example, prescribing an
antibiotic to treat flu may actually cause serious harm to a
patient.
Here is a story of how a change agent realized the difference
between symptoms and the actual underlying problem. The
change agent was called in to help a company solve the problem
of high manpower turnover. The change agent learned about

various methods to increase retention in organizations such as
raising pay and making tasks more challenging. Armed with
several ideas, the change agent presented her list of suggestions
to the management. To her surprise she found that the
organization had already tried many of the techniques and had
failed.
Out of curiosity she asked to examine company records and was
given the permission to do so. She discovered that the turnover
was high in one department, but the figures for other
departments were at par with industry figures. Closer
examination revealed that the manager of the department in
question was domineering and hard to work with. Employees
did not like working in this atmosphere and either quit the
company or sought transfers out of the department.
In this case high turnover was only the symptom of the problem.
The actual problem was one of interpersonal issues. Based on
this information the change agent was able to provide
suggestions to resolve the issues.
It is not easy to correctly diagnose problems. Similar to the
doctor who received relevant training, the OD change agent
needs to learn about tools and concepts related to the diagnostic
process.
Using Diagnostic Tools
Several diagnostic tools are available for managers and change
agents to diagnose problems. These tools are:
Performance-gap analysis
Force-field model
Systems model
Six-box model
Cultural climate surveys

Each of these tools has advantages and disadvantages. None of
them are perfect and despite various claims, none can be
classified as the best tool for diagnosing all problems. You'll be
able to perform better as a change agent if you understand the
advantages and disadvantages of using each tool, and you may
often use more than one tool.
In addition correct diagnosis is not possible without gathering
relevant data from the client organization.
Data Collection
In the doctor analogy the process of data collection is similar to
the doctor asking the patient what the symptoms are. There are
several ways of collecting data from an organization. Archival
data can be retrieved from company records. Focus groups and
interviews provide rich personal data. Surveys can be used to
reach out to larger audiences. The type of data being collected
determines the appropriateness of the method to be used. Some
methods of data collection assure anonymity more than others.
For example, it may be harder to identify an employee from a
survey than from personal interviews.
You should remember that trust and interpersonal relationships
play important roles in effective data collection. Data collection
is also easier if support from top management is evident and if
the change agent does not appear biased towards the
management.
Data that is gathered is essential to diagnose the problem, re-
evaluate the problem if necessary, determine appropriate
measures of intervention to correct the problem, and assess
organizational resistance to change.
Resistance to Change (1 of 2)
Organizational consultants and change agents caution against
resistance to change. But do people really abhor change?

Change is not bad or difficult. After all who wouldn't like to
buy a new car or who wouldn't like to replace their old
wardrobe? The problem with change arises when it is forced.
Often while initiating change the change agent and management
do not involve the employees of the organization who are
usually the people most directly affected by the proposed
change. They are often involved in the process only during the
change implementation phase. No wonder then that they resist
the change!
Fear of the unknown is another reason for resistance to change.
The results from the change initiative are uncertain, and there is
a fear of negative repercussions. This fear can be mitigated if
there is a support mechanism as a buffer.
It is reasonable to expect resistance to change. Yet you as a
change agent should also expect that there may be some who
favor and welcome the change initiative. When this happens you
can try to gather support from this group of people to sell the
change initiative to the rest of the organization.
Resistance to Change (2 of 2)
Here are some tips to reduce resistance to change:
Use a participatory approach to change.
Involve employees in diagnosing the problem.
Welcome and encourage suggestions from employees.
Obtain employee buy-in to the change initiative.
Involve employees in the change implementation plan.
Provide employees information about the pros and cons of the
change initiative.
Encourage the organization to provide a support system to
answer questions and queries from the employees.

Do not make false promises.
Summary
This week you learned about three aspects of organizational
change.
The first aspect was your role as a change agent. You learned
how the skills, knowledge, experience, attitude, and behavior of
the change agent affect the change process. In addition you
learned that the personality and image of the change agent plays
a vital role in forming relationships with people within the
organization.
The second aspect of organizational change that you learned
this week was problem diagnosis. You learned the importance of
correctly diagnosing the problem within the organization.
Resistance to change is the third aspect that you learned about
this week. You learned that as a change agent you should expect
a reasonable degree of resistance to change. People are usually
fixed in their ways and find it difficult to cope with change.
You also learned a few useful tips to manage resistance to
change.
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Lancret helices
Alexandre F. da Fonseca, C. P. Malta ∗
Instituto de F́ ısica, Universidade de São Paulo, USP

Rua do Matão, Travessa R 187, Cidade Universitária, 05508-
900, São Paulo,
Brazil
Abstract
Helical configurations of inhomogeneous symmetric rods with
non-constant bend-
ing and twisting stiffness are studied within the framework of
the Kirchhoff rod
model. From the static Kirchhoff equations, we obtain a set of
differential equations
for the curvature and torsion of the centerline of the rod and the
Lancret’s theorem
is used to find helical solutions. We obtain a free standing
helical solution for an
inhomogeneous rod whose curvature and torsion depend on the
form of variation of
the bending coefficient along the rod. These results are obtained
for inhomogeneous
rods without intrinsic curvature, and for a particular case of
intrinsic curvature.
Key words:
Kirchhoff rod model, inhomogeneous rod, Lancret’s theorem,
tendrils of climbing

plants
PACS: 46.70.Hg, 87.15.La, 02.40.Hw
∗ C. P. Malta
Email address: [email protected] (C. P. Malta).
Preprint submitted to Elsevier Science 2 February 2008
http://arXiv.org/abs/physics/0507105v1
1 Introduction
Helical filaments are tridimensional structures commonly found
in Nature.
They can be seen in microscopic systems, as biomolecules [1],
bacterial fibers [2]
and nanosprings [3], and in macroscopic ones, as ropes, strings
and climbing
plants [4,5,6]. Usually, the axis of all these objects is modeled
as a circular
helix, i. e. a 3D-space curve whose mathematical geometric
properties, namely
the curvature, kF , and the torsion, τF , are constant [7,8,9].
This kind of helical
structure has been shown to be a static solution of the Kirchhoff
rod model [7].

The Kirchhoff rod model [10,11] has been proved to be a good
framework to
study the statics [7,12,13] and dynamics [29] of long, thin and
inextensible
elastic rods. Applications of the Kirchhoff model range from
Biology [1,14,5]
to Engineering [15] and, recently, to Nanoscience [16]. In most
cases, the rod or
filament is considered as being homogeneous, but the case of
nonhomogeneous
rods have also been considered in the literature. It has been
shown that non-
homogeneous Kirchhoff rods may present spatial chaos [17,18].
In the case of
planar rods, Domokos and collaborators have provided some
rigorous results
for non-uniform elasticae [19] and for constrained Euler
buckling [20,21]. Devi-
ations of the helical structure of rods due to periodic variation
of the Young’s
modulus were verified numerically by da Fonseca, Malta and de
Aguiar [22].
Nonhomogeneous rods subject to given boundary conditions
were studied by

da Fonseca and de Aguiar in [23]. The effects of a
nonhomogeneous mass
distribution in the dynamics of unstable closed rods have been
analyzed by
Fonseca and de Aguiar [24]. Goriely and McMillen [25] studied
the dynamics
of cracking whips [26] and Kashimoto and Shiraishi [27]
studied twisting waves
in inhomogeneous rods.
2
The stability analysis of helical structures is of great
importance in the study
of the elastic behavior of filamentary systems and has been
performed both
experimentally [28] and theoretically [29,30,31]. It has been
also shown that
the type of instability in twisted rods strongly depends on the
anisotropy of
the cross section [32,33,34].
Here, we consider a rod with nonhomogeneous bending and
twisting coef-
ficients varying along its arclength s, B(s) and C(s),

respectively. We are
concerned with the following question: is there any helical
solution for the sta-
tionary Kirchhoff equations in the case of an inhomogeneous
rod ? The answer
is ‘yes’ and it will be shown that the helical solution for an
inhomogeneous
rod with varying bending coefficient cannot be the well known
circular helix,
for which the curvature, kF , and torsion, τF , are constant. To
this purpose,
we shall derive a set of differential equations for the curvature
and the torsion
of the centerline of an inhomogeneous rod and then apply the
condition that
a space curve must satisfy to be helical: the Lancret’s theorem.
We shall ob-
tain the simplest helical solutions satisfying the Lancret’s
theorem and show
that they are free standing helices, i.e., helices that are not
subjected to axial
forces [30]. A resulting helical structure different from the
circular helix, from
now on, will be called a Lancret helix.

According to the fundamental theorem for space curves [9], the
curvature
kF (s), and the torsion, τF (s), completely determine a space
curve, but for
its position in space. We shall show that the kF (s) and τF (s) of
a Lancret he-
lix depend directly on the bending coefficient, B(s), an expected
result since
the centerline of the rod does not depend on the twisting
coefficient (see for
example, Neukirch and Henderson [13]).
3
Some motivations for this work are related to defects [35] and
distortions [36]
in biological molecules. These defects and distortions could be
modeled as
inhomogeneities along a continuous elastic rod.
In Sec. II we review the general definition of a space curve, the
Frenet basis
and the so-called Lancret’s theorem. In Sec. III we present the
static Kirchhoff

equations for an intrinsically straight rod with varying stiffness,
and derive
the differential equations for the curvature and torsion of the
rod. In Sec.
IV we use the Lancret’s theorem for obtaining helical solutions
of the static
Kirchhoff equations and we show that they cannot be circular
helices if the
bending coefficient is not constant. As illustration, we compare
a homogeneous
rod with two simple cases of inhomogeneous rods: (i) linear and
(ii) periodic
bending coefficient varying along the rod. The circular helix
has a well known
relation of the curvature and torsion with the radius and pitch of
the helix. In
Sec. V we define a function involving all these variables in such
a way that for
the circular helix its value is identically null. We have verified,
numerically,
that this function approaches zero for the inhomogeneous cases
considered
here. In Sec. VI we analyse the cases of null torsion (straight
and planar

rods). Since helical solutions of intrinsically straight rods are
not dynamically
stable [30], in Sec. VII we consider a rod with a given helical
intrinsic curvature
and we obtain, for this case, a helical solution of the static
Kirchhoff equations
similar to that of an intrinsically straight inhomogeneous rod. In
Sec. VIII we
summarize the main results.
4
2 Curves in space
A curve in space can be considered as a path of a particle in
motion. The
rectangular coordinates (x, y, z) of the point on a curve can be
expressed as
function of a parameter u inside a given interval:
x = x(u) , y = y(u) , z = z(u) , u1 ≤ u ≤ u2 . (1)
We define the vector x(u) ≡ (x(u), y(u), z(u)). If u is the time,
x(u) represents
the trajectory of a particle.
2.1 The Frenet frame and the Frenet-Serret equations

The vector tangent to the space curve at a given point P is
simply dx/du. It
is possible to show [9] that if the arclength s of the space curve
is considered
as its parameter, the tangent vector at a given point P of the
curve x(s) is
a unitary vector. So, using the arclength s to parametrize the
curve, we shall
denote by t its tangent vector
t =
dx
ds
, (2)
‖t‖ = 1. The tangent vector t points in the direction of increasing
s.
The plane defined by the points P1, P2 and P3 on the curve,
with P2 and P3
approaching P1, is called the osculating plane of the curve at P1
[9]. Given a
point P on the curve, the principal normal at P is the line, in the
osculating
plane at P , that is perpendicular to the tangent vector at P . The
normal

vector n is the unit vector associated to the principal normal (its
sense may
5
be chosen arbitrarily, provided it is continuous along the space
curve).
From t.t = 1, differentiating with respect to s (indicated by a
prime) it follows
that:
t.t′ = 0 , (3)
so that t and t′ are orthogonal. It is possible to show that t′ lies
in the
osculating plane, consequently t′ is in the direction of n. This
allows us to
write
t
′ = kF n , (4)
kF being called the curvature of the space curve at P .
The curvature measures the rate of change of the tangent vector
when moving
along the curve. In order to measure the rate of change of the
osculating plane,

we introduce the vector normal to this plane at P : the binormal
unit vector
b. At a point P on the curve, b is defined in such a way that
b = t × n . (5)
The frame {n, b, t} can be taken as a new frame of reference
and forms the
moving trihedron of the curve. It is commonly called the Frenet
frame. The rate
of change of the osculating plane is expressed by the vector b′.
It is possible
to show that b′ is anti-parallel to the unit vector n [9]. So we
can write
b
′ = −τF n , (6)
τF being called the torsion of the space curve at P .
6
The rate of variation of n [9] can be obtained straightforwardly.
It is given by
n
′ = −kF t + τF b . (7)
The set of differential equations for {t, n, b} is

t′ = kF n ,
n′ = −kF t + τF b ,
b′ = −τF n ,
(8)
and are known as the formulas of Frenet or the Serret-Frenet
equations [9].
2.2 The Fundamental theorem of space curves
A space curve parametrized by its arclength s is defined by a
vectorial func-
tion x(s). The form of x(s) depends on the choice of the
coordinate system.
Nevertheless, there exists a form of characterization of a space
curve given
by a relation that is independent of the coordinates. This
relation gives the
natural equation for the curve.
kF (s) gives the natural equation in the case of planar curves.
Indeed, if ϕ is
the angle between the tangent vector of the planar curve and the
x-axis of
the coordinate system, it is possible to show that kF = dϕ/ds.
Since cos(ϕ) =

dx/ds and sin(ϕ) = dy/ds, knowing kF (s), then ϕ(s), x(s), and
y(s) of the
planar curve can be obtained immediatly:
ϕ(s) =
s
∫
s0
kF (s) ds, x(s) =
s
∫
s0
cos ϕ(s)ds, y(s) =
s
∫
s0
sin ϕ(s)ds. (9)
7
In the case of non-planar curves, if we have two single valued
continuous func-
tions kF (s) and τF (s), s > 0, then there exists one and only one

space curve,
determined but for its position in space, for which s is the
arclength, kF (s)
the curvature, and τF (s) the torsion. It is the Fundamental
theorem for space
curves [9]. The functions kF (s) and τF (s) provide the natural
equations of the
space curve.
2.3 Curves of constant slope: the Lancret’s theorem
A space curve x(s) is a helix if the lines tangent to x make a
constant angle
with a fixed direction in space (the helical axis) [8,9]. Denoting
by a the unit
vector of this direction, a helix satisfies
t.a = cos α = constant . (10)
Differentiating Eq. (10) with respect to s gives a.n = 0.
Therefore a lies in
the plane determined by the vectors t and b:
a = t cos α + b sin α . (11)
Differentiating Eq. (11) with respect to s, gives
0 = (kF cos α − τF sin α)n ,

or
kF
τF
= tan α = constant. (12)
8
This result says that for curves of constant slope the ratio of
curvature over
torsion is constant. Conversely, given a regular curve for which
the equation
(12) is satisfied, it is possible to find [9] a constant angle α
such that
n (kF cos α − τF sin α) = 0 ,
d
ds
(t cos α + b sin α) = 0 ,
implying that the vector a = t cos α + b sin α is the unit vector
along the axis.
Moreover, a.t = cos α=constant, so that the curve has constant
slope. This
result can be expressed as:
A necessary and suficient condition for a space curve to be a

curve of constant
slope (a helix) is that the ratio of curvature over torsion be
constant. It is the
well known Lancret’s theorem, dated of 1802 and first proved
by B. de Saint
Venant [9,37].
If a helical curve x(s) is projected onto the plane perpendicular
to a, the
vector x1(s) representing this projection is given by
x1(s) = x − (x.a)a . (13)
It is possible to show [9] that the curvature k1 of the projected
curve is given
by:
k1(s) =
kF (s)
sin2 α
. (14)
The shape of the planar curve obtained by projecting a helical
curve onto the
plane perpendicular to its axis is used to characterize it. For
example, the well
known circular helix projects a circle onto the plane
perpendicular to its axis.

9
The spherical helix projects an arc of an epicycloid onto a plane
perpendicular
to its axis [9]. The logarithmic spiral is the projection of a
helical curve called
conical helix [9].
3 The static Kirchhoff equations
The statics and dynamics of long and thin elastic rods are
governed by the
Kirchhoff rod model. In this model, the rod is divided in
segments of infinites-
imal thickness to which the Newton’s second law for the linear
and angular
momentum are applied. We derive a set of partial differential
equations for
the averaged forces and torques on each cross section and for a
triad of vectors
describing the shape of the rod. The set of PDE are completed
with a linear
constitutive relation between torque and twist.
The central axis of the rod, hereafter called centerline, is

represented by a
space curve x parametrized by the arclength s. A Frenet frame is
defined for
this space curve as described in the previous section. For a
physical filament
the use of a local basis, {d1, d2, d3}, to describe the rod has the
advantage of
taking into account the twist deformation of the filament. This
local basis is
defined such that d3 is the vector tangent to the centerline of
the rod (d3 = t),
and d1 and d2 lie on the cross section plane. The local basis is
related to the
Frenet frame {n, b, t} through
(d1 d2 d3) = (n b t)

cos ξ − sin ξ 0
sin ξ cos ξ 0
0 0 1

, (15)
10
where the angle ξ is the amount of twisting of the local basis
with respect to
t.
In this paper, we are concerned with equilibrium solutions of
the Kirchhoff
model, so our study departs from the static Kirchhoff equations
[38]. In scaled
variables, for intrinsically straight isotropic rods, these
equations are:
F
′ = 0 , (16)

M
′ = F × d3 , (17)
M = B(s) k1 d1 + B(s) k2 d2 + C(s) k3 d3 , (18)
the vectors F and M being the resultant force, and
corresponding moment
with respect to the centerline of the rod, respectively, at a given
cross section.
As in the previous section, s is the arclength of the rod and the
prime ′ denotes
differentiation with respect to s. ki are the components of the
twist vector,
k, that controls the variations of the director basis along the rod
through the
relation
d
′
i
= k × di , i = 1, 2, 3 . (19)
k1 and k2 are related to the curvature of the centerline of the
rod (kF =
√
k21 + k
2
2) and k3 is the twist density. B(s) and C(s) are the bending and

twisting coefficients of the rod, respectively. In the case of
macroscopic fila-
ments the bending and twisting coefficients can be related to the
cross section
radius and the Young’s and shear moduli of the rod. Writing the
force F in
the director basis,
F = f1d1 + f2d2 + f3d3 , (20)
the equations (16–18) give the following differential equations
for the compo-
11
nents of the force and twist vector:
f ′1 − f2 k3 + f3 k2 = 0 , (21)
f ′2 + f1 k3 − f3 k1 = 0 , (22)
f ′3 − f1 k2 + f2 k1 = 0 , (23)
(B(s) k1)
′ + (C(s) − B(s)) k2 k3 − f2 = 0 , (24)
(B(s) k2)
′ − (C(s) − B(s)) k1 k3 + f1 = 0 , (25)
(C(s) k3)

′ = 0 . (26)
The equation (26) shows that the component M3 = C(s) k3 of
the moment
in the director basis (also called torsional moment), is constant
along the
rod, consequently the twist density k3 is inversely proportional
to the twisting
coefficient C(s)
k3(s) =
M3
C(s)
. (27)
In order to look for helical solutions of the Eqs. (21–26) the
components of
the twist vector k are expressed as follows:
k1 = kF (s) sin ξ , (28)
k2 = kF (s) cos ξ , (29)
k3 = ξ
′ + τF (s) , (30)
where kF (s) and τF (s) are the curvature and torsion,
respectively, of the space
curve that defines the centerline of the rod and ξ is given by Eq.
(15). If the

rod is homogeneous, the helical solution has constant kF and τF
, and ξ
′ is
proved to be null [5].
Substituting Eqs. (28–30) in Eqs. (21–26), extracting f1 and f2
from Eqs. (25)
and (24), respectively, differentiating them with respect to s,
and substituting
in Eqs. (21), (22) and (23), gives the following set of nonlinear
differential
equations:
12
[M3 kF (s) − B(s) kF (s) τF (s)]
′ − (B(s) kF (s))
′ τF (s) = 0 , (31)
(B(s) kF (s))
′′ + kF (s) τF (s)[M3 − B(s) τF (s)] − f3(s) kF (s) = 0 , (32)
(B(s) kF (s))
′ kF (s) + f
′
3(s) = 0 . (33)

Appendix A presents the details of the derivation of Eqs. (31–
33).
The Eqs. (31–33) for the curvature, kF , and torsion, τF do not
depend on the
twisting coefficient, C(s). Therefore, the centerline of an
inhomogeneous rod
does not depend on the twisting coefficient like in the case of
homogeneous
rods (see, for example, Eqs. (13) and (14) of Ref. [13]).
Langer and Singer [39] have obtained a set of first-order
ordinary differential
equations for the curvature and torsion of the centerline of a
homogeneous
rod that contains terms proportional to k2
F
and τ 2
F
. The Eqs. (31–33) have the
advantage of involving only terms linear in kF and τF .
4 Helical solutions of inhomogeneous rods
In order to find helical solutions for the static Kirchhoff
equations, we apply
the Lancret’s theorem to the general equations (31–33). We first

rewrite the
Lancret’s theorem in the form:
kF (s) = β τF (s) , (34)
with β 6= 0. From Eq. (12),
β ≡ tan α = Constant . (35)
13
Substituting Eq. (34) in Eq. (31) we obtain
τ ′
F
(M3 − B τF ) − 2 τF (B τF )
′ = 0 . (36)
Substituting Eq. (34) in Eq. (32) and extracting f3, we obtain
f3 =
(B τF )
′′
τF
+ τF (M3 − B τF ) . (37)
Differentiating f3 with respect to s and substituting in Eq. (33)
we obtain the
following differential equation for τF :

(B τF )
′′′
τF
−
(B τF )
′′τ ′
F
τ 2
F
+ (β2 + 1) τF (B τF )
′ = 0 , (38)
where the Eq. (36) was used to simplify the above equation.
One immediate
solution for this differential equation is
(B τF )
′ = 0 , (39)
that substituted in Eq. (36) gives
τ ′
F
(M3 − BτF ) = 0 . (40)
For non-constant τF , the Eq. (40) gives the following solution
for τF :

τF (s) =
M3
B(s)
. (41)
Substituting the Eqs. (39) and (41) in Eq. (37) we obtain that
f3(s) = 0 . (42)
14
Substituting Eq. (41) in (34) we obtain:
kF (s) = β
M3
B(s)
. (43)
Substituting Eq. (15) in Eq. (20), the force F becomes
F = (f1 cos ξ − f2 sin ξ) n + (f1 sin ξ + f2 cos ξ) b + f3 t , (44)
where {n, b, t} is the Frenet basis. Using the Eqs. (A.7) and
(A.8) for f1 and
f2 (Appendix A), we obtain
F = −(B kF )
′
n + kF [M3 − B τF ] b + f3 t , (45)
where f3, in the inhomogeneous case, must satisfy the Eq. (33).

Substituting the Eqs. (41–43) in the Eq. (45), and using Eq.
(35), it follows
F = 0. Therefore, the helical solutions satisfying (39) are free
standing.
Now, we prove that a circular helix cannot be a solution of the
static Kirchhoff
equations for a rod with varying bending stiffness. If a helix is
circular, k′
F
= 0
and τ ′
F
= 0, and from Eq. (31) we obtain:
2 kF τF B
′ = 0 . (46)
Since B′(s) 6= 0, Eq. (46) will be satisfied only if kF = 0 and/or
τF = 0.
Therefore, it is not possible to have a circular helix as a
solution for a rod
with varying bending coefficient.
The solutions for the curvature kF , Eq. (43), and the torsion τF
, Eq. (41), can
be used to obtain the unit vectors of the Frenet frame (by

integration of the
Eqs. (8)). From Eqs. (43), (35) and (11), we can obtain α and a
once kF (0),
15
M3 and B(s) are given. By choosing the z-direction of the fixed
cartesian basis
as the direction of the unit vector a, we can integrate t in order
to obtain the
three-dimensional configuration of the centerline of the rod.
Figure 1 displays the helical solution of the static Kirchhoff
equations for rods
with bending coefficients given by
Fig 1a: Ba(s) = 1 , (47)
Fig 1b: Bb(s) = 1 + 0.007 s , (48)
Fig 1c: Bc(s) = 1 + 0.1 sin(0.04s + 2) . (49)
The case of constant bending (47) produces the well known
circular helix
displayed in Fig. 1a. Figs. 1b–1c show that non-constant
bending coefficients
(Eqs. (48–49)) do not produce a circular helix.

The helical solutions displayed in Fig. 1 satisfy the Lancret’s
theorem, Eq. (12).
The tridimensional helical configurations displayed in Fig. 1
were obtained by
integrating the Frenet-Serret equations (8) using the following
initial condi-
tions for the Frenet frame: t(s = 0) = (0, sin α, cos α), n(s = 0) =
(−1, 0, 0)
and b(s = 0) = (0, − cos α, sin α). This choice ensures that the z-
axis is par-
allel to the direction of the helical axis, vector a. The centerline
of the helical
rod is a space curve x(s) = (x(s), y(s), z(s)) that is obtained by
integration
of the tangent vector t(s). We have taken the helical axis as the
z-axis and
placed the initial position of the rod at x(0) = 1/k1(0), y(0) = 0
and z(0) = 0
(in scaled units), where k1(0) is the curvature of the planar
curve at s = 0
obtained by projecting the space curve onto the plane
perpendicular to the
helical axis (Eq. (14)). From Eq. (14) we have
k1(0) =

kF (0)
sin2 α
. (50)
16
Using the Eq. (35), it follows that
sin2 α =
β2
1 + β2
. (51)
From Eq. (43), setting s = 0, we get
β =
kF (0) B(0)
M3
. (52)
Substituting Eqs. (51) and (52) in Eq. (50), we obtain:
x(0) =
1
k1(0)
=
kF (0) B
2(0)

M 23 + k
2
F
(0)B2(0)
. (53)
kF (0) and M3 are free parameters that have been chosen so that
the helical
solutions displayed in Fig. 1 have the same angle α. The
parameters kF (0) =
0.24 and M3 = 0.05 give x(0) ≃ 4 for the helical solutions
displayed in Figs.
1a and 1b, and the parameters kF (0) = 0.22 and M3 = 0.05 give
x(0) ≃ 4.36
for the helical solution displayed in the Fig. 1c.
For short the projection of the space curve onto the plane
perpendicular to the
helical axis will be called projected curve. As mentioned in Sec.
II, the circle
is the projected curve of the most common type of helix, the
circular helix.
Fig. 2 displays the projected curves related to the helical
solutions displayed in
Fig. 1. Fig. 2a shows that the helical solution of the
inhomogeneous rod with

constant bending coefficient projects a circle onto the plane
perpendicular to
the helical axis.
If required, the natural equations for the projected curves
displayed in Fig. 2
are easily obtained, for instance, by substitution of the solution
(Eq. (43)) for
the curvature kF (s) of the helical rod into Eq. (14). The natural
equation of
17
the projected curve is given by its curvature,
k1(s) =
βM3
B(s)
sin−2 α , (54)
where β and sin−2 α can be obtained by Eqs. (51) and (52).
Then, in the
Eq.(54), setting B(s) = Bi(s), i = a, b, c, as given in Eqs. (47–
49), produces the
natural equation for the corresponding projected curve
displayed in Fig. 2. The
helical rod displayed in Fig. 1b is a conical helix since the

radius of curvature
of its projected curve (inverse of k1(s)) is a logarithmic spiral
(1/k1(s) is a
linear function of s [9]).
From Eqs. (28–30), (27) and (41) we obtain the variation of the
angle ξ between
the local basis, di, i = 1, 2, 3, and the Frenet frame, {n, b, t}:
ξ′ = k3(s) − τF (s) =
(
M3
C(s)
−
M3
B(s)
)
= M3
B(s) − C(s)
B(s) C(s)
. (55)
Eq. (55) shows that ξ′ 6= 0 for the general case of B(s) 6= C(s),
i. e. helical
filaments corresponding to inhomogeneous rods are not

twistless. The circular
helix is a helicoidal solution for the centerline of an
inhomogeneous rod having
constant bending coefficient. We emphasize that the
inhomogeneous rod is not
twistless in contrast with the homogeneous case where it has
been proved that
ξ′ = 0 [5].
A homogeneous rod has B(s) and C(s) constant so that k3 =
Constant (from
Eq. (26)). Since ξ′ has been proved to be null for a helical
solution of a ho-
mogeneous rod (see reference [5]), Eq. (30) shows that the
torsion τF must
be a constant. In order to satisfy the Lancret’s theorem (Eq.
(12)) the curva-
ture kF of the helical solution must also be a constant.
Therefore, the only
type of helical solution for a homogeneous rod is the circular
helix, while an
18
inhomogeneous rod may present other types of helical

structures.
5 Radius and Pitch of the helical solution
The radius R of a helix is defined as being the distance of the
space curve to
its axis. The pitch P of a helix is defined as the height of one
helical turn, i.e.,
the distance along the helical axis of the initial and final points
of one helical
turn.
For a circular helix, R, P, kF and τF are constant, and it is easy
to prove that
λ =
(
√
R2 + P2/(4π2)
)
−1
=
√
k2
F
+ τ 2
F

. (56)
For other types of helix, it constitutes a very hard problem in
differential
geometry to obtain the relation between the curvature kF and
the torsion τF
with the radius R and the pitch P. We have seen in Sec. II that
the definitions
of curvature and torsion involve the calculation of the modulus
of the tangent
and normal vectors derivative with respect to the arclength of
the rod. We also
saw that the Frenet-Serret differential equations for the Frenet
frame depend
on the curvature and torsion. The difficulty of integration of the
Frenet-Serret
equations for the general case where kF (s) and τF (s) are
general functions of
s poses the problem of finding an analytical solution for the
centerline of the
rod, thus the difficulty of relating non constant curvature and
torsion with non
constant radius and pitch. Due to this difficulty we shall test the
possibility

of generalizing the relation (56) to the present inhomogeneous
case. In order
to do so, from the equation (56) we define:
gλ(s) ≡
(
√
R2(s) + P2(s)/(4π2)
)
−1
−
√
k2
F
(s) + τ 2
F
(s) , (57)
19
where R(s) and P(s) are the radius and the pitch of the helical
structure as
function of s. In the case of a circular helix, from Eq. (56),
gλ(s) = 0 for all s.
Since the z-axis is defined as being the axis of the helical

solution we can
calculate the radius R(s) through: R(s) =
√
x2(s) + y2(s), where x(s) and
y(s) are the x and y components of the vector position of the
centerline of the
helical rod.
The pitch of the helix is the difference between the z-coordinate
of the initial
and final positions of one helical turn. A helical turn can be
defined such that
the projection of the vector position of the spatial curve along
the xy-plane
(vector x1 of Eq. (13)), rotates of 2π around the z-axis.
Fig. 3 shows gλ(s) for the free standing helix of Fig. 1b. We see
that gλ(s)
oscillates, its maximum amplitude being smaller than 0.006. For
the helical
shape displayed in Fig. 1c we found that the maximum value of
gλ(s) is smaller
than 0.008 (data not shown). While for a circular helix gλ = 0,
for the free
standing helices displayed in Fig. 1b and Fig. 1c the function gλ

oscillates
around zero with small amplitude.
The small amplitude of these oscillations suggests that the
relations R(s) ≃
kF (s)[
√
k2
F
(s) + τ 2
F
(s)]−1 and P(s) ≃ 2πτF (s)[
√
k2
F
(s) + τ 2
F
(s)]−1, valid for cir-
cular helices, could be used to derive approximate functions for
the radius and
the pitch of different types of helical structures, but the
oscillatory behavior
indicates that these relations are not simple functions of the
geometric features
of the helix.

20
6 Straight and planar inhomogeneous rods
Straight rods (kF = 0), and planar rods (kF 6= 0), have null
torsion (τF = 0),
and constitute particular cases of helices. In both cases there is
at least one
direction in space that makes a constant angle α = π/2 with the
vector tangent
to the rod centerline.
The straight inhomogeneous rod is a solution of the static
Kirchhoff equations
that has non-constant twist density (Eq. (27)), in contrast with
the homoge-
neous case for which the twist density is constant.
The twisted planar ring (kF = Constant) is a solution of the
static Kirchhoff
equations only if the bending coefficient can be written in the
form:
B(s) = A0 cos(kF s) + B0 sin(kF s) + CI /k
2
F

, (58)
with A0, B0 and CI constant. If kF is function of s (instead of
being a constant)
there exist no solutions for Eqs. (21–26). So, the existence of a
planar solution
related to the general form of the components of the twist vector
given by
equations (28–30) requires kF = Constant.
7 Helical structure with intrinsic curvature
The helical shape displayed in Fig. 1b resembles that exhibited
by the tendrils
of some climbing plants. In these plants the younger parts have
smaller cross
section diameter, giving rise to non-constant bending
coefficient. The main
difference between the solution displayed in Fig. 1b and the
tendrils of climbing
plants is that the solution in Fig. 1b was obtained for an
intrinsically straight
21
rod while the tendrils have intrinsic curvature [5].

The tendrils of climbing plants are stable structures while the
helical solution
displayed in Fig. 1b is not stable because the rod is intrinsically
straight [30].
We shall show that a rod with intrinsic curvature may have a
static solution
of the Kirchhoff equations similar to that displayed in Fig. 1b.
The intrinsic curvature of a rod is introduced in the Kirchhoff
model through
the components of the twist vector, k(0), in the unstressed
configuration of the
rod as
k
(0)
1 = k
(0)
F
(s) sin ξ , (59)
k
(0)
2 = k
(0)
F
(s) cos ξ , (60)

k
(0)
3 = ξ
′ + τ
(0)
F
(s) , (61)
where k
(0)
F
(s) and τ
(0)
F
(s) are the curvature and torsion of the space curve that
represents the axis of the rod in its unstressed configuration,
simply called
intrinsic curvature of the rod. We consider that the unstressed
configuration
of the axis of the rod forms a helical space curve with the
intrinsic curvature
satisfying
B(s) k
(0)
F

(s) = K0 , (62)
B(s) τ
(0)
F
(s) = T0 , (63)
where K0 and T0 are constant and B(s) is the bending
coefficient of the rod.
The linear constitutive relation (Eq. (18)) becomes
M = B(s)(k1 − k
(0)
1 )d1 + B(s)(k2 − k
(0)
2 )d2 + C(s)(k3 − k
(0)
3 )d3 , (64)
where C(s) is the twisting coefficient of the rod. The static
Kirchhoff equations
for this case, Eqs. (16), (17) and Eq. (64), are given by
22
f ′1 − f2 k3 + f3 k2 = 0 , (65)
f ′2 + f1 k3 − f3 k1 = 0 , (66)

f ′3 − f1 k2 + f2 k1 = 0 , (67)
(B(s)(k1 − k
(0)
1 ))
′ − B(s)(k2 − k
(0)
2 )k3 + C(s)(k3 − k
(0)
3 ) − f2 = 0 , (68)
(B(s)(k2 − k
(0)
2 ))
′ + B(s)(k1 − k
(0)
1 )k3 − C(s)(k3 − k
(0)
3 ) + f1 = 0 , (69)
(C(s)(k3 − k
(0)
3 ))
′ + B(s)(k
(0)
1 k2 − k
(0)
2 k1) = 0 . (70)
The components of the twist vector are expressed as:

k1 = kF (s) sin χ , (71)
k2 = kF (s) cos χ , (72)
k3 = χ
′ + τF (s) . (73)
In order to obtain the simplest solution for the static Kirchhoff
equations Eqs.
(65–70) with the intrinsic curvature given by Eqs. (59–61) and
(62–63) we shall
look for a solution such that χ = ξ in Eqs. (71–73). This solution
preserves
the intrinsic twist density of the helical structure. In this case,
the Eq. (70)
becomes simply [C(s)(τF − τ
(0)
F
)]′ = 0 or
M3 = C(s)(τF − τ
(0)
F
) = Constant , (74)
and we obtain the following differential equations for the
curvature kF (s), and
the torsion τF (s), of the rod:

[M3 kF − B τF (kF − k
(0)
F
)]′ − [B(kF − k
(0)
F
)]′ τF = 0 ,
[B(kF − k
(0)
F
)]′′ + τF [M3 kF − B τF (kF − k
(0)
F
)] − f3 kF = 0 ,
[B(kF − k
(0)
F
)]′ kF + f
′
3 = 0 ,
(75)
where we have omitted the dependence on s to simplify the
notation. In order
to obtain a helical solution of these equations we apply the

Lancret’s theorem,
23
Eq. (12), to the Eqs. (75). We obtain the following results:
f3(s) = 0 , (76)
[B(s)(kF (s) − k
(0)
F
(s))]′ = 0 ⇒ kF (s) − k
(0)
F
(s) =
K
B(s)
, (77)
[B(s)(τF (s) − τ
(0)
F
(s))]′ = 0 ⇒ τF (s) − τ
(0)
F
(s) =
T
B(s)

, (78)
where K and T are integration constants. From Eqs. (74) and Eq.
(78) we
obtain
T =
B(s)
C(s)
M3 , (79)
so that the ratio B(s)/C(s) has to be constant.
From Eqs. (77), (78), (62), (63) and (12) we have
kF (s)
τF (s)
=
K + K0
T + T0
= tan α . (80)
From Eqs. (68), (69), (77) and (78) it follows that
f1 = 0 ,
f2 = 0 .
(81)
Therefore, the helical solution given by Eqs. (76–78) (obtained

imposing χ =
ξ) is a free standing helix (F = (f1, f2, f3) = 0).
It follows from Eqs. (76–78), (62) and (63) that the solutions
for the curvature
kF (s), and the torsion τF (s), of the rod with helical intrinsic
curvature are
similar to those of intrinsically straight rods, Eqs. (41–43).
Therefore, rods
with intrinsic curvature and a non-constant bending coefficient
given by Eq.
24
(48) (Eq. (49)) can have a three-dimensional configuration
similar to that
displayed in Fig. 1b (Fig. 1c).
8 Conclusions
The existence of helical configurations for a rod with non-
constant stiffness has
been investigated within the framework of the Kirchhoff rod
model. Climbing
and spiralling solutions of planar rods have been studied by
Holmes et. al. [40].

Here, we have shown that helical spiralling three-dimensional
structures are
possible solutions of the static Kirchhoff equations for an
inhomogeneous rod.
From the static Kirchhoff equations, we derived the set of
differential equations
(31–33) for the curvature and the torsion of the centerline of a
rod whose
bending coefficient is a function of the arclength s. We have
shown that the
circular helix is the type of helical solution obtained when B(s)
is constant,
independently of the rod being homogeneous or inhomogeneous.
Though the differential equations for the curvature and torsion
are general,
we have obtained only the simplest helical solutions (Eqs. (39)
and (41–43)),
obtained when the Lancret’s theorem is applied to the
differential equations.
We show that these solutions are free standing and that the
curvature and
torsion depend directly on the form of variation of the bending
coefficient.
Figures 1b and 1c are examples of helical solutions of

inhomogeneneous rods
whose bending coefficients are given by Eqs. (48) and (49). The
helical struc-
ture displayed in Fig. 1b is a conical helix since the projected
curve onto the
plane perpendicular to the helical axis is a logarithmic spiral, i.
e., 1/k1(s) is
a linear function of s [9].
25
In the particular case of an inhomogeneous rod with the
intrinsic curvature
defined by Eqs. (59–61) and (62–63), with B(s)/C(s) constant,
we also ob-
tain the helical solutions displayed in Figs. 1b and 1c. The
tendrils of some
climbing plants present a three-dimensional structure similar to
that displayed
in Fig. 1b. In these plants, the cross-section diameter of the
tendrils varies
along them, giving rise to non-constant bending coefficient, and
the differen-
tial growth of the tendrils produces intrinsic curvature [5]. The

bending and
twisting coefficients of a continuous filament with circular
cross-section are
proportional to its moment of inertia I. It implies that B(s)/C(s)
is constant
for an inhomogeneous rod. Therefore, the tendrils of climbing
plants can be
well described by the Kirchhoff model for an inhomogeneous
rod with a linear
variation of the bending stiffness.
Acknowledgements
This work was partially supported by the Brazilian agencies
FAPESP, CNPq
and CAPES. The authors would like to thank Prof. Manfredo do
Carmo for
valuable informations about the Lancret’s theorem.
A Appendix: The differential equations for the curvature and
tor-
sion
Here, we shall derive the Eqs. (31–33). Substitution of Eqs.
(28–30) into Eqs.
(21–26) gives:

f ′1 − f2 (ξ
′ + τF ) + f3 kF cos ξ = 0 , (A.1)
f ′2 + f1 (ξ
′ + τF ) − f3 kF sin ξ = 0 , (A.2)
26
f ′3 − f1 kF cos ξ + f2 kF sin ξ = 0 , (A.3)
(B(s) kF sin ξ)
′ + (C(s) − B(s)) kF cos ξ (ξ
′ + τF ) − f2 = 0 , (A.4)
(B(s) kF cos ξ)
′ − (C(s) − B(s)) kF sin ξ (ξ
′ + τF ) + f1 = 0 , (A.5)
(C(s) (ξ′ + τF ))
′ = 0 . (A.6)
First, we extract f1 and f2 from Eqs. (A.5) and (A.4),
respectively:
f1 = −(B(s) kF )
′ cos ξ + [M3 kF − B(s) kF τF ] sin ξ , (A.7)
f2 = (B(s) kF )
′ sin ξ + [M3 kF − B(s) kF τF ] cos ξ , (A.8)
where M3 = C(s) (ξ
′ + τF ) is the torsional moment of the rod that is con-

stant by Eq. (A.6). Differentiating f1 and f2 with respect to s,
substituting in
Eqs. (A.1) and (A.2), respectively, and using Eqs. (A.7) and
(A.8), gives the
following equations:
{−(B(s) kF )
′′ − τF [M3 kF − B(s) kF τF ] + f3 kF } cos ξ +
{[M3 kF − B(s) kF τF ]
′ − τF (B(s) kF )
′} sin ξ = 0 ,
(A.9)
{(B(s) kF )
′′ + τF [M3 kF − B(s) kF τF ] − f3 kF } sin ξ +
{[M3 kF − B(s) kF τF ]
′ − τF (B(s) kF )
′} cos ξ = 0 .
(A.10)
Multiplying Eq. (A.9) (Eq. (A.10)) by sin ξ (cos ξ) and then
adding the re-
sulting equations, we obtain the Eq. (31) for the curvature and
torsion:
[M3 kF − B kF τF ]

′ − (B kF )
′τF = 0 . (A.11)
Multiplying Eq. (A.9) (Eq. (A.10)) by − cos ξ (+ sin ξ) and then
adding the
resulting equations, we obtain the Eq. (32):
(B kF )
′′ + kF τF (M3 − BτF ) − f3 kF = 0 . (A.12)
27
Finally, the Eq. (33) is obtained by substituting Eqs. (A.7) and
(A.8) in Eq.
(A.3):
(B kF )
′ + f ′3 = 0 . (A.13)
References
[1] T. Schlick, Curr. Opin. Struct. Biol. 5 (1995) 245. W. K.
Olson, Curr. Opin.
Struct. Biol. 6 (1996) 242.
[2] C. W. Wolgemuth, T. R. Powers, R. E. Goldstein, Phys. Rev.
Lett. 84 (2000)
1623.

[3] D. N. McIlroy, D. Zhang, Y. Kranov, M. Grant Norton,
Appl. Phys. Lett. 79
(2001) 1540.
[4] A. Goriely, M. Tabor, Phys. Rev. Lett. 80 (1998) 1564; A.
Goriely, M. Tabor,
Physicalia 20 (1998) 299.
[5] T. McMillen, A. Goriely, J. Nonlin. Sci. 12 (2002) 241.
[6] P. Pieranski, J. Baranska, A. Skjeltorp, Eur. J. Phys. 24
(2004) 613.
[7] M. Nizette, A. Goriely, J. Math. Phys. 40 (1999) 2830.
[8] M. P. do Carmo, Differential Geometry of Curves and
Surfaces, Prentice Hall,
Inc. Englewood Cliffs, New Jersey, 1976.
[9] D. J. Struik, Lectures on Classical Differential Geometry,
2nd Edition, Addison-
Wesley, Cambridge, 1961.
[10] G. Kirchhoff, J. Reine Anglew. Math. 56 (1859) 285.
[11] E. H. Dill, Arch. Hist. Exact. Sci. 44 (1992) 2; B. D.
Coleman, E. H. Dill, M.
Lembo, Z. Lu, I. Tobias, Arch. Rational Mech. Anal. 121 (1993)
339.
28

[12] G. H. M. van der Heijden, J. M. T. Thompson, Nonlinear
Dynamics 21 (2000)
71.
[13] S. Neukirch, M. E. Henderson, J. Elasticity 68 (2002) 95.
[14] I. Tobias, D. Swigon, B.D. Coleman, Phys. Rev. E 61
(2000) 747; B.D. Coleman,
D. Swigon, I. Tobias, Phys. Rev. E 61 (2000) 759.
[15] Y. Sun, J. W. Leonard, Ocean. Eng. 25 (1997) 443; O.
Gottlieb, N. C. Perkins,
ASME J. Appl. Mech. 66 (1999) 352.
[16] A. F. da Fonseca, D. S. Galvão, Phys. Rev. Lett. 92 (2004)
art. no. 175502.
[17] A. Mielke, P. Holmes, Arch. Rational Mech. Anal. 101
(1988) 319.
[18] M. A. Davies, F. C. Moon, Chaos 3 (1993) 93.
[19] G. Domokos, P. Holmes, Int. J. Non-Linear Mech. 28
(1993) 677.
[20] G. Domokos, P. Holmes, B. Royce, J. Nonlinear Sci. 7
(1997) 281.
[21] P. Holmes, G. Domokos, J. Schmitt, I. Szeberényi, Comput.
Methods Appl.

Mech. Engrg. 170 (1999) 175.
[22] A. F. da Fonseca, C. P. Malta, M. A. M. de Aguiar, Physica
A 352 (2005) 547.
[23] A. F. da Fonseca, M. A. M. de Aguiar, Physica D 181
(2003) 53.
[24] A. F. Fonseca, M. A. M. de Aguiar, Phys. Rev. E 63 (2001)
art. no. 016611.
[25] A. Goriely, T. McMillen, Phys. Rev. Lett. 88 (2002) art.
no. 244301.
[26] A whip is a nonhomogenous thread with varying radius of
cross section.
[27] H. Kashimoto, A. Shiraishi, J. Sound and Vibration 178
(1994) 395.
[28] J. M. T. Thompson, A. R. Champneys, Proc. R. Soc.
London, Ser. A 452 (1996)
117.
29
[29] A. Goriely, M. Tabor, Nonlinear Dynamics 21 (2000) 101.
[30] A. Goriely, M. Tabor, Proc. R. Soc. London, Ser. A 453
(1997) 2583.
[31] A. Goriely, P. Shipman, Phys. Rev. E 61 (2000) 4508.

[32] G. H. M. van der Heijden, J. M. T. Thompson, Physica D
112 (1998) 201.
[33] A. Goriely, M. Nizette, M. Tabor, J. Nonlinear Sci. 11
(2001) 3.
[34] N. Chouäieb, PhD Thesis, École Polytechnique Fédérale de
Lausanne, 2003.
[35] W. A. Kronert et al, J. Mol. Biol. 249 (1995) 111.
[36] V. Geetha, Int. J. Biol. Macromol. 19 (1996) 81.
[37] B. de Saint Venant, J. Ec. Polyt. 30 (1845) 26.
[38] The references in [11] can be seen for a complete
derivation of the Kirchhoff
equations.
[39] J. Langer, D. A. Singer, SIAM Rev. 38 (1996) 605.
[40] P. Holmes, G. Domokos, G. Hek, J. Nonlinear Sci. 10
(2000) 477.
30
(a) (b) (c)
Fig. 1. Helical solutions of the Kirchhoff equations using the
Lancret’s Theorem. (a)
circular helix solution for an inhomogeneous rod with constant

bending coefficient
Ba = 1 (47); (b) and (c) Lancret helices for inhomogeneous rod
with bending
coefficient given by Eqs. (48) and (49), respectively. The
parameters, in scaled units,
are M3 = 0.05, Γ = 0.9, and the total length of the rod is L =
130. kF (0) = 0.24 for
the helical solutions displayed in panels (a) and (b), and kF (0)
= 0.22 for panel (c).
(a) (b) (c)
Fig. 2. (a), (b) and (c) are projected curves of the helical
solutions displayed in
Fig. 1a, Fig. 1b and Fig. 1c, respectively. We used Eq. (13) to
obtain the projected
curves.
20 40 60 80 100
s
-0.006
-0.002
0.002
0.006
gλ

Fig. 3. gλ(s) for the free standing helix solution displayed in
Fig. 1b.
31
IntroductionCurves in spaceThe Frenet frame and the Frenet-
Serret equationsThe Fundamental theorem of space
curvesCurves of constant slope: the Lancret's theoremThe static
Kirchhoff equationsHelical solutions of inhomogeneous
rodsRadius and Pitch of the helical solutionStraight and planar
inhomogeneous rodsHelical structure with intrinsic
curvatureConclusionsAppendix: The differential equations for
the curvature and torsionReferences
Math 55H Honors Project Overview - Spring 2018
The project will consist of either two "short" projects or one
"long" project. Possible projects are detailed in Canvas under
the "File" tab. Look for the "Projects" folder. If none of these
projects appeal to you, or if you have your own idea please see
me as soon as possible. Some of the projects are very structured
and require you to complete specific steps or problems. Other
projects are very open ended and just offer you some
mathematical ideas that you need to form into a project.
Roughly, a short project may entail 7 or 8 pages of writing. A
long project would be approximately 15 pages. The project due
date will be Thursday, May 31st. By Thursday, April 5th you
need to submit an "Interim Report" on your progress. This
should consist of your topic and a paragraph telling me what
you plan to do. It should also include a preliminary
bibliography of sources consulted. Of course you may change
your topic after this date; that's okay.
First of all, this is a major, lengthy assignment. To do well you
should start immediately, and work on it every day, if possible.
You will probably need all the days you have been given in

order to complete your project on time.
1. Start today. Let your subconscious work for you. It can do
amazing things. If you immerse yourself in the project,
solutions will come to you at the strangest times.
2. Read the entire project to see what it's all about. Don't worry
too much about details the first time through.
3. Next, read the project very carefully and make a list of any
unfamiliar words or concepts you encounter. If concepts occur
that you're not sure about, you must understand those ideas
before you can do the project. Even if you understand all the
words and terms, don't assume that the project is easy.
4. You may need to do some outside reading. In addition to your
textbook, there are lots of books in the library and, of course,
on the internet that contain information that might be helpful to
you.
5. After you have worked a bit every day on the project, you
will find certain parts easy and you will have completed those
parts. You will have identified the hard parts and begun to zero
in on the obstacles.
6. While I expect you to work independently, I do not expect
that you can work through the project without some assistance. I
encourage you to come and talk to me about your project. This
way I can head you off if you are going in a wrong direction.
7. When you have done the work necessary to complete the
project, you need to prepare it in written form. The paper you
turn in should have a mix of equations, formulas, and prose to
support your conclusions. Use complete sentences. The prose
should be written in order to convey to the reader an
explanation of what you have done.

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