text:
ANVIL's STEVE 'LIPS' KUDLOW: 'I Have No Regrets'
In a recent interview with France's United Rock Nations, ANVIL frontman Steve "Lips" Kudlow was asked if he had any regrets about the band's career path thus far. He responded (hear audio below): "You can always go 'shoulda, woulda, coulda,' but at the end of the day, if anything had changed… To be honest with you, I think it worked out the way that it should and the best way that it could. If I'd made it to begin with, right now we wouldn't be sitting here. I wouldn't be sitting here at 64 years old with something to do with my life, and I'm making money and I'm not fucking broke. Okay? A lot of musicians who had made it when they were 20 years old, by the time they're 64, they've got none of the fucking money in the bank. They're fucking broke. They were once rock stars. Anything they had, they had and lost. They can never get it back. No fucking thanks.
"I went and I worked a regular job, and I raised a family," he continued. "And I've got real estate. I am in fucking the best place I can possibly be. Because now I've got a house that's my own. I've got a family that's my own, that didn't fall apart. I got all that done. And now I'm starting to make it in music. So none of it is under threat of being wrecked because I'm on drugs and fucking it up on the road. I'm old enough and mature enough not to be drinking and fucking getting wasted and wasting my life away, like you do when you're young, when you think that there's endless tomorrows. Every day and every fucking minute counts. And that's the beautiful thing about where I am and what it is, because I really respect where I am right now and I respect the fans for loving ANVIL and I don't think of them as being replaceable: 'I'll replace you. I'll write another album and I'll replace you. I'll do better later.' There is no later. It's all a moment. It's all now. Live it now. Make it the best it can be now. And those kind of things come with age and wisdom.
"So what I'm saying is everything ended up where it had to end up and where it should end up, and I'm grateful for it. So I have no regrets, and that's where it all is."
ANVIL's latest album, "Legal At Last", was released in February via AFM Records.
A Toronto, Ontario, Canada-based group, ANVIL was formed in 1978 by childhood friends, Kudlow and drummer Robb Reiner. Both came from Jewish families and had been playing music together since their teens. ANVIL is considered to have been an influence on many metal bands, including METALLICA, MEGADETH, SLAYER and ANTHRAX.
Though initially a four-piece band, ANVIL's current lineup includes Kudlow, Reiner and bassist Chris Robertson.
ANVIL gained popularity and new fans since the 2008 theatrical documentary, "Anvil: The Story Of Anvil", which is currently available on Netflix.
Tags: anvil
PRODUCT OF HATE Releases Epic Cinematic Video For 'Redemption'
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PETER IWERS Has 'No Regrets' About Leaving IN FLAMES
MÖTLEY CRÜE Is Officially 40 Years Old Today
DAVE LOMBARDO Looks Back On Filling In For METALLICA's LARS ULRICH At 2004 DOWNLOAD Festival: 'It Was A Moment In History'


Globe Garden is an ergonomic chair for waiting rooms and office use made of beech wood and fabric covering. The original Garden chair, designed by Peter Opsvik 29 years ago, is now re-launched in a new version by Moment AB. In the updated version the branches of the tree continues down and become supporting footrests. Garden is an icon, an art object and eye catcher. But it also offers numerous comfortable sitting positions.
With its spectacular height the new Garden is adapted to function in public spaces, such as forums and meeting points, since the person waiting can both see and be seen in the crowd.


Migraine is a type of headache that lasts for hours or days. There are migraines that are accompanied by auras such as seeing streaks of light or temporary blindness. These are considered as a warning that a migraine may occur. When you are suffering from migraine, you will tend to become overly sensitive to light. You will also opt to stay in a quiet place since migraine can cause you to be annoyed by both soft and loud sound.
You have to be aware about the migraine headache symptoms. Aside from light and sound sensitivity, you will feel pain on either one side or both sides of your head. This is typically described as pulsating pain. The pain that is associated with migraine can be aggravated when you move. With this condition, your activities of daily living may be affected.
Migraine headache symptoms usually last for about four hours. This may extend up to 72 hours. The symptoms may happen for a couple of times in a month. With some people, migraine may be less frequent. You can take note the number of times that you experienced migraine in a month.
You should take note that migraine headache symptoms can be experienced with or without a warning. This is the basis of the two types of migraines. A classic migraine presents its symptoms after an aura. On the other hand, a common migraine is experienced without a warning.
Aside from an aura, there are symptoms that you can feel hours or a day before you experience a migraine. These are called as a feeling of prodrome. You may feel that you have an intense energy before you have your migraine. You can also feel either extremely elated or irritable. There are people that feel unexplainable thirst or unusual cravings. You have to remember that you may or may not feel these symptoms.
Some migraine episodes go away without being treated. However, you still have to keep track of the symptoms that you felt. You should also take note of the things that you did that may have helped in relieving you from the symptoms. The length of your migraine attack should also be listed. These are important points that you have to discuss with your doctor during your consultation.
There are also other symptoms that you may misconceive as those of a migraine attack. If you feel a sudden headache that is really painful, you should rush to the nearest hospital. This may be related to other serious medical conditions. A headache that is accompanied by fever or mental confusion is another medical emergency. A migraine-related headache does not present with seizures, stiff neck or rashes. If you experience headache with the mentioned symptoms, you should consult a doctor at once. Aside from these, if you had a head injury and suddenly experienced headache, you should not mistake this as a migraine. Migraine is aggravated by movements but if your headache is worsened by coughing, this is a different case.
Health and Fitness Goals - Are You as Fit as You Want to Be?
If you don't like your answrs to these questions, you must change your lifestyle! There is a direct relationship between well-being and health. If your body is strong, your attitude will be, too. If your nutrition is high in value, your alertness will increase. And if your stamina is increased, so will your ability to accomplish more of your goals. Nice little circle.
The primary benefit of white tea is that it is less processed than the other teas from the Camellia sinensis plant. White tea is harvested early in the Spring while the tender young buds and leaves still have a fine silver fuzz on them. The white appearance gives this specialty crop its name.
Because it is not fermented like black tea or oolong tea, white tea retains more of the natural polyphenols (potent antioxidants). White Tea vs Green Tea The Benefit of White Tea - If you are one of those people who want the health benefits of green tea, but you just don't care for the taste, you should check out this benefit: white tea doesn't have that "green" or "grassy" flavor. It is mellow, but crisp, with a slightly sweet tone.
Turmeric Enlivens Food And Helps Ease Inflammation - If your idea of a great evening includes digging into a rich curry at your local Indian restaurant, you've got lots of company. Indian food has gone from exotic fare to firm favorite in American cuisine.
Indian Curry An added bonus to the flavorful experience is turmeric, the spice that gives Indian delicacies that distinctive reddish-orange hue. In its native land, turmeric is also prized by practitioners of Ayurveda (India's traditional health system) as a whole-body cleanser. Now scientists are deciphering this plant's healing powers, especially that of its primary chemical constituent, curcumin.
What they've found: You can use a spice that fires the tongue to cool the fire of inflammation.


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Katherine Haynes, Head
Katherine Haynes, BA (Warwick) MEd (Birmingham) NPQH, Head
Katherine read Maths at Warwick, followed by MEd and then taught in the Midlands, becoming head of Maths at Edgbaston High School, followed by Warwick School, where she first started out as a school inspector and took the professional training scheme for headship. Her appointment at John Lyon made her the first woman ever to head an HMC boys' day school. Katherine continues to work as a school Reporting Inspector, and, in her limited free time, enjoys gardening and travelling.
Michael Gibson
Michael Gibson, BA, MSc (Trinity, Dublin), Bursar
In the early days of his distinguished career at Diageo, Michael juggled the challenge of a full-time job with playing rugby at the highest level, gaining 10 international caps for Ireland. Michael joined John Lyon in March 2009 as Commercial Director, taking up the role of Bursar in September of the same year. He sits on the School's Finance and General Purposes Committee, Estates Committee and Strategy Committee and has overall responsibility for the support staff and efficient running of the School.
Andy Sims
Andy Sims, MA, MEng (Cantab), Deputy Head
Andy read Engineering at Churchill College, Cambridge, where he also completed his PGCE. Before teaching, Andy worked in the design and applied research of renewable energy for AECOM, London. He started his teaching career at Whitgift School in 2009 where he was also Assistant Head of Sixth Form and led the Senior STEM programme. In 2012, he was appointed Head of Physics at Blundell's School where he coordinated Oxbridge Science Preparation. An avid sportsman, Andy was Head of Cross Country in both schools, coached various teams and organised outdoor pursuit events and trips. He joined John Lyon in 2015. He is a Governor of Alpha Preparatory School. As Deputy Head, Andy has joint responsibility for the day-to-day running of the School, specialising in academic and ICT strategy. He oversees the School Calendar, staffing, Sport and Art.
Jonathan Pepperman
Jonathan Pepperman, MA (Cantab), Deputy Head
Jonathan always knew he wanted to teach. After securing his degree in History from Selwyn College, Cambridge he started his career at Ryde School teaching History and Politics before moving on to be Head of History and Politics and then Head of Upper College at Cheltenham College, where he worked with Housemasters on pastoral matters and oversaw the Oxbridge provision. Jonathan is a believer in an all-round education: on top of his academic commitments, he has coached cricket, hockey and football, been a member of the CCF and sung in school choirs. In 2013 Jonathan joined John Lyon as Head of Sixth Form becoming Deputy Head in 2015. Alongside his joint responsibility for running the School, Jonathan's primary focus is on the School's pastoral provision, with oversight of the Sixth Form, Higher Education and Careers, Admissions, Music and Drama.
Tim Lewis, MA, MEng (Cantab), Director of Studies
Tim read Engineering at Trinity Hall, Cambridge, and his first teaching role was at Reading Blue Coat School. He joined John Lyon as Head of Mathematics in 2014 before becoming Director of Studies in 2017. In this role, Tim has overall responsibility for the day-to-day running of academic life at John Lyon, including the curriculum and timetable. Tim has led expeditions to Ecuador, Mongolia and Morocco as well as coordinating the School's annual ski trip.
Darren Boylan
Darren Boylan, BEd (Queen's, Belfast), Senior teacher (Admissions) and Designated Safeguarding Lead
After finishing his BEd in Business Education at Queen's University Belfast, Darren began his teaching career in the UK at St George's School, Harpenden as a teacher of Business, before moving to Duff Miller sixth form college as Head of Business Education. He joined John Lyon as Head of Business in 2013. Darren's roles here have included Head of Year and Head of Sixth Form. He is most proud of the excellent community at John Lyon throughout the School. He sees John Lyon as the perfect environment to give students an excellent all round education – not only achieving excellent academic results but developing boys into young men who make their parents proud. Darren believes John Lyon to be a happy and busy place that allows each individual to flourish in a safe environment.
Jonathan Rowe
Jonathan Rowe, MA (UEA), Head of Sixth Form
Jonathan read History at Reading University before completing a Master's degree in History at the University of East Anglia. He began his teaching career in 2004 as a History Teacher at Plume School in Essex, before moving to Farlingaye High School in Suffolk. From here Jonathan moved to Colchester Sixth Form College where he led the Ancient History Department and taught Modern History and Classical Civilisation. He then moved to Battle Abbey School in East Sussex, undertaking numerous roles, including Head of History, Head of Lower School, Deputy Head of Boarding and Head of Sixth Form. Jonathan has a strong interest in assessment. He is a Principal Examiner for A-Level History and a fellow member of the Chartered Institute of Educational Assessors. Jonathan joined John Lyon in 2018. He has overall responsibility for the day-to-day running of the Sixth Form.
Jonathan Peel
Jonathan Peel, BA (King's College, London), Senior Teacher (Staff)
Jonathan graduated from King's College London with a degree in Classics. This was followed by a Postgraduate Diploma in Opera Studies from the RSAMD in Glasgow. His career as an opera singer saw him working as a soloist throughout Europe and in the USA, notably in Cologne, Naples and Seattle, as well as at the Royal Opera House in London. He qualified as a teacher of English in 2005 and has worked at John Lyon since 2014, previously as Head of English. He is an edu-blogger and his blog of English teaching resources has received well over a million views and is regularly listed among the most influential blogs of its kind.

redpajama_set_name: RedPajamaCommonCrawl


Showing results for tags 'kamas'.
A few days ago my Wife (IGN: DecapitatedAngel) shown me an image of someone who is scamming and using her name. Please be aware that "Gg-easy" is NOT Decapitatedangel or one of her "New accounts" Here is a SS. You will see the C&P of the message in the screenshot (see below). Also as a rule of thumb ; Do some research. This person is claiming to be Decapitatedangel on a NEW account that was JUST created because her account was hacked (which is wasn't). If you look into the character info and info page you will notice ALOT of holes in that story. http://www.dofus.com/en/mmorpg/community/directories/character-pages/1220134500002-gg-easy 1. Decapitatedangel Main is a 200 Eni ; If I was hacked and had to restart I would choose a class I love ; why make a osa? 2. If the account was JUST created ; how is it a lvl 149 in a few mins? Sure it can be done quick ; but not within minutes of making it. 3. Decapitatedangel is the leader of a guild. Why join a random guild in INT? And not put the alt in the guild you own? 4. If the leader of the guild was hacked ; how is she online and still leader of the guild? 5. Why bother gearing an account that you created just to get help? And why use that account for Votes? 6. The account wasn't recently created. Either way. There is always a "Vote for me" scam going on. Please avoid them ; and never blame the person that person is claiming to be. Do some research first. Doesn't really take more then common sense. Finally. When clicking links NEVER enter your log in information and use Ankama Shied. Always check the Domain AND subdomain AND folder. Domains are the "Main" part to look at. example. Someone may send you a link like login.dofus.com The Domain DOFUS.Com is correct. The subdomain could be correct but if it isn't you would be redirected to Dofus.com so the site would be likely safe to use. If the site however is Dofus.Gladiator.com the Dofus SEEMS legit but thats a subdomain and isn't monitored where as the actual domain is Gladiator.com and is being used for phishing. It can be misleading but anyone can have dofus.something.something. But NOONE else can have Something.DOFUS.com so please be safe.
Best places to perceptor hunt!?
As the title says name what you consider to be the most profitable place to hunt perceptors in.
Hello everyone . I had very long break from game , not sure but ~1 year. Before that I used to get kamas from making a lot of DP souls then seeling them , or just by making kwakwa dungeons . Now I have some free time and finally I am able to back. I have got few accounts : 200 enu , 200 eni , 200 sac , 200 iop, 200 sadi ,199 panda etc. I want to learn how to do new dungeons and enjoy the game. But there is a one big problem , i literally have no idea how to get kamas nowadays! Please guys , try to be helpfull maybe I will be able to help some of you. Which profession is best? or maybe there is a dungeon hunt like few years ago was Crackler for diamond stones and 1 year ago like kwakwa.. Please guys help me !
Kamas? Kamas! What are some good ways to earn 'em?
(Edit: for those popping in from 'Activity today' or something similar, this thread is for Wakfu, not Dofus!) Obligatory "How do I Kama" thread. So far I've been sailing for 150 levels on kamas from drops and treasure hunting and doing just fine, but soon I'll be getting to the point where things need to be runed and where equipment prices increase exponentially, and I'm wondering - how do you all go about earning kamas? Of course drops are a huge source of income, especially if you run dungeons repeatedly for fragments and tokens, but does that really manage to cover everything? Are professions a steady source of income as well? I'm particularly curious about mining/kama minting - are the 1150/1250 kama recipes worth the metal they require? Newb questions here, but soon I'll have to buckle down and work on making kamas. Better late than never!
Discover the Makeover Mystery Box!
ALL MY OGRINES :((((( The Mystery Box concept is simple: when you open it, you discover a random item inside! All you have left to do is hope to have the same luck as Ecaflip. Here's what you can find in a Makeover Mystery Box: Common items: Slait Set (value: 700 OG) Noke Set (value: 700 OG) Slugly Set (value: 700 OG) Uncommon items: Tetra Set (value: 700 OG) Black Quaquack Set (value: 700 OG) Khamelerost Dragoturkey (value: 1000 OG) Rare items: Ougicle Set (value: 2600 OG) Platypus Set complete with pet (value: 3500 OG) Offer a Cake emote (value: 5000 OG) And if luck is really on your side, you might go home with a four-emote pack (Offer a Cake + Reckless + Write + Greed for a cumulated value of 18500 OG) or even a cosmetic pack containing all the Parasymbic items (all four), all the Livinitems (all four items), 10 mimisymbics and color change and face change potion, for a cumulated value of 42500 Ogrines. Try your luck! The Mystery Box is only available until Sunday, June 1, at 11:59 PM (Paris Time). To open your Mystery Box after you buy one, go to your pending gifts page!
Hey guys, what is your best Dofus trash can finds. My best on solar was 157 kamas and some blop resources. On oto, i found a gobbal cape, which is pretty good for that server.
Hello,we have introduced our guide before and we are here to introduce a new part of it! The Contests page,its a brand new part of our dofus guide its here to help all the low levels and med levels who make little amounts of kamas,easy challenges you can do to gain a particular amount of kamas We have made 2 contests so far,1 is done and the winner won a prize of 250kk, Our second is one that you can participate in right now here http://dofficiency.wordpress.com/contests to win 300kk! -These contests are only available in the server Rushu for now -We have done those to support the low leveled who are in need of such contests,if your a high level and you make millions and you think this is useless,its not for you.


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Home»News»Grade 2 Share New Track, "Only Ones I Trust"
Grade 2 Share New Track, "Only Ones I Trust"
UK punk band Grade 2 have just released "Only Ones I Trust," an outtake from the band's 2019 album Graveyard Island. Co-written and produced by Rancid's Tim Armstrong, the band hopes this feel good track with stomping choruses will "unify those who have seen struggles this year, because together in this game we are damned and we are all the same. We aim to have you singing your heart out, for now just in your living room but hopefully at a show soon!"
Grade 2 is Sid Ryan (vocals/bass), Jack Chatfield (vocals/guitar), and Jacob Hull (drums). Formed in 2013, the band met at school where they would spend their lunch break playing covers of classic punk tunes together in the music room. Since their formation, the band has released several EPs and three studio albums; Mainstream View (2016) and Break The Routine (2017), and Graveyard Island (2019). The bands' Hellcat debut Graveyard Island was recorded and produced by Rancid's Tim Armstrong at Armstrong's Shiprec Studios and mixed by The Interrupters' Kevin Bivona.


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Global Gab
GFA Podcast
Global From Asia News Column Section
Latest New Labour Rules Make Indonesian Trade Unions Rally
New Labour Rules Make Indonesian Trade Unions Rally
The Indonesia Government is prioritizing the revamp of several labour rules. The trade unions have caught wind of these planned changes and are taking to the streets to rally against it.
Relaxing Rules
The Indonesian government plans to relax several established labour rules. These rules have been in place for more than 15 years. The Indonesian government is trying to change the 2003 Labour Law.
The 2003 Labour Law is very popular among workers. For one, it ensures generous severance pay. In fact, it is said to be one of the most generous severance packages in the whole world.
Investors have stated that the 2003 Labour Law is one that negatively affects the economy. Indonesia is one of Southeast Asia's biggest economies. This law affects investors' decisions to invest, and the decision for companies to hire workers formally.
President Joko Widodo, Indonesia's President, is dead set on having this bill passed. It is one of his biggest priorities in his second term, and one that will make foreign investors happy.
Despite that though, the popularity of the bill will make it a hard fought battle during the upcoming parliamentary debate.
President Widodo has more than 70% control of the parliamentary seats. However, this does not mean that the parliamentary debate will go smoothly for this important bill.
Rallying Together
Currently, President Widodo is set to hand over the bill to parliament for debate and discussion. If it is approved, then it will be legally drafted.
Indonesia's Chief Economic Affairs Minister referred to the incoming bill as an "omnibus bill." The "omnibus bill" is a term used by people in power to put points across various industries together for discussion. This is convenient, as a single vote can approve all the points.
People also say that various points are placed within an omnibus bill in order to hide certain points.
Members from various trade unions have banded together in order to battle this upcoming bill. They currently plan to meet up and take to the streets. They will be protesting in front of parliament on Monday, right after the lawmakers make their way back from recess.
The Deputy Chairman of the Indonesian Confederation of Workers sheds light on why they are against the bill. He says that this bill will drive the workers into poverty. It will not only affect them, but also their children and grandchildren.
The coalition of unions predicts that the upcoming bill will decrease the severance pay. This will also move Indonesia's hiring towards more outsourced avenues. They also say that it will do away with the protection of basic human rights.
The Indonesia government is quick to reply to these sentiments. They counter that this bill will improve job creation for new people entering the workforce. They also promise to provide approaches that will make all concerned unions happy.
President Widodo has urged lawmakers to take up to three months to finish all their deliberations.
Tags:Indonesia, Indonesia Labour Rules, Indonesia Trade Unions, President Widodo
India Business Owners Protest Massive Amazon Investment
China Set to Unveil New Digital Currency DCEP
Taiwan President Tsai Ing-Wen Re-elected; Beats Beijing Candidate
Foreign Teaching Materials Not Allowed in China
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Ravi / Chandra / Kuja / Budha / Guru / Sukra / Sani / Rahu Kethu - Navagrahaalaku Okkokka Pusthakam... Chakradatta of Chakrapanidatta (Sanskrit Text with English Translation) by Prof. P.V.Sharma. This book by Chakrapanidatta describes many diseases and their treatment procedures.
If searched for the ebook Bhaisajya Ratnavali of Govinda Dasji Bhisagratna (3 volume set) by Dr. Kanjiv Lochan in pdf form, in that case you come on to the faithful website.


Whether you are receiving care in your own home, a nursing home, or a retirement community, the Centura Health at Home family is committed to making your surroundings a healing sanctuary.
Centura Health at Home will work with you and your physician to come up with a plan of care that meets your needs and helps you heal.
In addition to providing excellent clinical care, Centura Health at Home knows that faith and inner strength can help with healing. That's why we give special attention to your spiritual well-being. Centura Health at Home will spend time with you and your family or caregiver to explain your condition and how we'll treat it. Before we leave, we'll make sure all your questions are answered. Centura Health at Home is dedicated to providing the best care possible, so our staff is fully trained and screened before they ever reach your door. And we're accredited by The Joint Commission, which means we've demonstrated our ability to provide quality care.
What criteria do I need to meet to qualify for home care services?
In addition to having a physician's referral, you need to have a safe home environment to provide care in, and be a willing patient. Medicare patients must also be homebound, meaning that it requires considerable and taxing effort to leave home.
How often will a nurse come to visit me?
Centura Health at Home will work closely with your physician to develop a treatment plan that meets your needs. And it may not always be a nurse who comes to help you.
Who other than a nurse might come to my home?
Physical, occupational and speech therapy for rehabilitation programs.
Certified nursing aides for help with personal care like bathing and meal preparation.
Social workers to help you find community resources for additional help.
Nondenominational chaplains to visit with you about spiritual issues.
In most cases yes. Centura Health at Home accepts Medicare and many private insurances. Our health benefit advisors will help you determine if your insurance will pay for the cost of care.


text:
\section{Introduction}
In this paper, we continue the investigation of
the \it constant astigmatism equation (CAE) \rm
$$\numbered\label{CAE}
z_{yy} + (\frac{1}{z})_{xx} + 2 = 0,
$$
the Gauss equation of \it surfaces of constant astigmatism \rm immersed in Euclidean space. These surfaces are defined as having constant
(but nonzero) difference between principal radii of curvature.

Historical roots of the constant astigmatism surfaces can be traced back in works of the 19th century mathematicians
\cite{Bia a, Bia b, Bia I, Bia II, Ri, Lip, vLil}, though the surfaces were nameless at that time. For more detailed history we refer to,
e.g., \cite{P-S}.

In 2009, after about a century in oblivion, the subject of constant astigmatism surfaces
has been resurrected by Baran and Marvan in the work \cite{B-M I} concerning the systematic search for integrable classes of Weingarten surfaces. In the paper, the surfaces gained their name and Equation \eqref{CAE} was obtained as well.
Recently, the Equation \eqref{CAE} has been examined by several authors \cite{H-M I, P-Z, H-M II, M-P, H-M III}.

The main result of this paper is Proposition~\ref{prop1}, where
an algebraic formula \eqref{prop1eq} provides arbitrary many solutions of Equation \eqref{CAE} from a given seed.
The proof is based on the observation that superposition formulas \cite[Eq.~21]{H-M I} (see also equations \eqref{supCAE} below)
can be conveniently written in a matrix form and, therefore, their iteration coincides with the composition of linear transformations,
i.e. matrix multiplication. Consequently, the same conclusion holds in the case of multisoliton solutions,
i.e. solutions of the CAE having their counterparts in the well known multisoliton solutions of the sine-Gordon equation.
Since the algebraic formula for the $n$-soliton sine-Gordon solution is well known, the corresponding solution of the CAE
can be routinely computed using Proposition~\ref{propSoliton}.

The results of Proposition~\ref{prop1} in combination with \cite[Prop.~3]{H-M I} also enable us
to construct arbitrarily many constant astigmatism surfaces by purely algebraic manipulations and differentiation
once an initial step (including an integration) is successfully performed.

The most important results from the earliest history of the subject of constant astigmatism surfaces were,
undoubtedly, obtained by Bianchi \cite{Bia a, Bia b, Bia I, Bia II}.
He showed (see also \cite{Ri}) that evolutes (focal surfaces) of constant astigmatism surfaces are pseudospherical, i.e. with constant negative Gaussian curvature. Conversely, if one equips a pseudospherical surface with parabolic geodesic coordinates and takes the corresponding involutes, then they are of constant astigmatism. Bianchi also succeeded in finding some of the constant astigmatism surfaces explicitly \cite[Eq.~(30)]{Bia a}.

A remarkable class of constant astigmatism surfaces was studied by Lipschitz \cite{Lip}
and its subclass was later investigated by von~Lilienthal \cite{vLil}.
Lipschitz parameterised his surfaces by spherical coordinates of the Gaussian image, see Figure~\ref{Lipsurf}.
Recently, \cite{H-M II},
we showed that the solutions of the constant astigmatism equation that correspond
to the Lipschitz class of surfaces, are the Lie
symmetry invariant solutions and constitute a four-dimensional manifold.
The counterpart sine-Gordon solutions are
shown to be Lie symmetry invariant as well.
\begin{figure}[ht]
\begin{center}
\includegraphics[scale=0.17]{Lip.jpg}
\caption{Lipschitz surfaces of constant astigmatism.}\label{Lipsurf}
\end{center}
\end{figure}

The aforementioned geometric link to pseudospherical surfaces served as a tool for
deriving a nonlocal transformation between Equation \eqref{CAE} and the well-known sine-Gordon equation
$$\numbered\label{SG}
\omega_{\xi\eta} = \frac12 \sin 2\omega,
$$
which is the Gauss equation of pseudospherical surfaces parameterised by asymptotic coordinates. For detailed description of the transformations see \cite{B-M I}.

By an \it orthogonal equiareal pattern \rm (Sadowsky~\cite{Sad I,Sad II})
we mean a parameterisation $x,y$ such that the corresponding metric is of the form
$$
z \mathrm{d} x^2 + \frac{1}{z} \mathrm{d} y^2,
$$
$z$ being an arbitrary function of $x,y$.
An associated {\it slip line field} is a parameterization $\xi,\eta$ such that the angle between
$\partial_x$ and $\partial_\xi$ as well as the angle between
$\partial_y$ and $\partial_\eta$ is equal to $\pi/4$.
In \cite{H-M I} we observed that to every surface of constant astigmatism
there corresponds an orthogonal equiareal pattern on the Gaussian sphere. Note that
the same result was obtained by Bianchi~\cite[\S375, eq.~(20)]{Bia II} in the context of pseudospherical congruences.
We also showed that solutions of the sine-Gordon equation \eqref{SG}
correspond to slip line fields on the same Gaussian sphere.

The method of \cite{H-M I} for generating solutions of the constant astigmatism equations and corresponding surfaces of constant astigmatism  has its origin in
B\"acklund transformation for the sine-Gordon equation.
Let $\omega$ be a solution of the sine-Gordon equation \eqref{SG}. Its B\"acklund transformation,
$\omega^{(\lambda)}$,
is given by the system
$$\numbered\label{BT}
\omega^{(\lambda)}_\xi  = \omega_\xi + \lambda\sin(\omega^{(\lambda)} + \omega), \qquad
\omega^{(\lambda)}_\eta = -\omega_\eta + \frac{1}{\lambda}\sin(\omega^{(\lambda)} - \omega),
$$
$\lambda$ being called a \it B\"acklund parameter. \rm
The famous Bianchi permutability theorem~\cite{Bia b}, see also \cite{Bia II, R-S},
and the \it superposition formula \rm
$$\numbered\label{supSG}
\tan\frac{\omega^{(\lambda_1\lambda_2)} - \omega}{2} = \frac{\lambda_1 + \lambda_2}{\lambda_1 - \lambda_2} \tan\frac{\omega^{(\lambda_1)} - \omega^{(\lambda_2)}}{2}
$$
enable us to compute solution $\omega^{(\lambda_1\lambda_2)}$, the B\"acklund transformation of $\omega^{(\lambda_1)}$
with B\"acklund parameter $\lambda_2$, by algebraic manipulations.
Therefore, when a seed solution $\omega$ is given, the integration of the system \eqref{BT} needs to be done only once and the further iteration of the B\"acklund transformation can be performed purely algebraically.

Pairs of sine-Gordon solutions related by B\"acklund transformation induce solutions of the constant astigmatism equation.
In \cite[Def.~4]{H-M I} we defined the \it associated potentials \rm $f^{(\lambda)}, x^{(\lambda)}, y^{(\lambda)} $
corresponding to a pair of sine-Gordon solutions $\omega, \omega^{(\lambda)}$. In terms of
$
g^{(\lambda)} = `e^{f^{(\lambda)}}
$
the associated potentials are given by
$$\numbered\label{aspot}
g^{(\lambda)}_\xi = g^{(\lambda)}\lambda \cos(\omega^{(\lambda)} + \omega),
\qquad
g^{(\lambda)}_\eta = g^{(\lambda)}\frac1\lambda \cos(\omega^{(\lambda)} - \omega),
\\
x^{(\lambda)}_\xi = \lambda g^{(\lambda)} \sin(\omega^{(\lambda)} + \omega),
\qquad
x^{(\lambda)}_\eta = \frac1\lambda g^{(\lambda)} \sin(\omega^{(\lambda)} - \omega),
\\
y^{(\lambda)}_\xi = \frac{\lambda \sin(\omega^{(\lambda)} + \omega)}{g^{(\lambda)}}  ,
\qquad
y^{(\lambda)}_\eta = - \frac{\sin(\omega^{(\lambda)} - \omega)}{\lambda g^{(\lambda)}} .
$$
Expressing $z^{(\lambda)} = 1/{g^{(\lambda)}}^2 $ in terms of $x^{(\lambda)}$ and $y^{(\lambda)}$ one obtains a solution of the CAE.

In the same paper (Prop. 5) we succeeded in extending the superposition principle~\eqref{supSG}
to solutions of the CAE. Let $\omega, \omega^{(\lambda_1)}, \omega^{(\lambda_2)},
 \omega^{(\lambda_1\lambda_2)}$ be four sine-Gordon solutions
related by the Bianchi superposition principle~\eqref{supSG}.
Then the associated potentials
$g^{(\lambda_1\lambda_2)}$, $x^{(\lambda_1\lambda_2)}$, $y^{(\lambda_1\lambda_2)}$
corresponding to the pair $\omega^{(\lambda_1)},\omega^{(\lambda_1\lambda_2)}$
are related to the associated potentials
$g^{(\lambda_2)}$, $x^{(\lambda_2)}$, $y^{(\lambda_2)}$ corresponding to the pair
$\omega,\omega^{(\lambda_2)}$ by formulas
$$\numbered\label{supCAE}
g^{(\lambda_1\lambda_2)} =
\frac{-\lambda_1\lambda_2}{\lambda_1^2+\lambda_2^2-2\lambda_1\lambda_2\cos(\omega^{(\lambda_1)} - \omega^{(\lambda_2)})}g^{(\lambda_2)},
\\
x^{(\lambda_1\lambda_2)}
 = \frac{\lambda_1 \lambda_2}{\lambda_1^2 - \lambda_2^2}
   (x^{(\lambda_2)}
     -
      \frac{2 \lambda_1 \lambda_2 \sin(\omega^{(\lambda_1)} - \omega^{(\lambda_2)})}
         {\lambda_1^2 + \lambda_2^2
          - 2 \lambda_1 \lambda_2 \cos(\omega^{(\lambda_1)} - \omega^{(\lambda_2)})
          } g^{(\lambda_2)}),
\\
y^{(\lambda_1\lambda_2)}
 = \frac{\lambda_1^2 - \lambda_2^2}{\lambda_1\lambda_2} y^{(\lambda_2)}
   - \frac{2 \sin(\omega^{(\lambda_1)} - \omega^{(\lambda_2)})}{g^{(\lambda_2)}}.
$$
The above formulas can be regarded as a starting point of our approach in this paper.

The contents of this paper are as follows.
In Section~2 we observe that equations~\eqref{supCAE} can be conveniently
exposed in matrix form and the iteration of the superposition principle for the CAE reduces to mere matrix multiplication.
In Section~3 we handle the multisoliton case, i.e. we deal with solutions of the CAE coming from the well known $n$-soliton solutions of the sine-Gordon equation.
Section~4 deals with constant astigmatism surfaces and slip line fields.
In Section~5 we study in detail the $n$-soliton case for $n = 1,2,3$ producing exact solutions of the CAE, constant astigmatism surfaces and slip line fields, whilst
the last section is devoted to the subcase when all B\"acklund parameters are equal to $1$.

\section{Solutions of the CAE}
Let $\omega^{[0]} = \bar{\omega}^{[0]} $ be some seed solution of the sine-Gordon equation.
Fix B\"acklund parameters $\lambda_1, \dots, \lambda_{k + 1}$ and let us denote
$$\numbered\label{defk}
\omega^{[k]} = \omega^{(\lambda_1\lambda_2\ldots\lambda_k)}, \qquad
\bar{\omega}^{[k]} = \omega^{(\lambda_2\lambda_3\ldots\lambda_{k+1})},
$$
see the diagram (a part of the well known Bianchi lattice)
\begin{displaymath} \numbered \label{lattice}
    \xymatrix{
\omega^{[0]} \ar[r]^{\lambda_2} \ar[d]_{\lambda_1} 	& \bar{\omega}^{[1]} \ar[r]^{\lambda_3} \ar[d]_{\lambda_1} & \bar{\omega}^{[2]} \ar[d]_{\lambda_1} \ar[r]^{\lambda_4}  & \bar{\omega}^{[3]} \ar[d]_{\lambda_1} \ar[r]^{\lambda_5}  & \bar{\omega}^{[4]} \ar[d]_{\lambda_1} \ar[r]^{\lambda_6} & \dots
\\
\omega^{[1]} \ar[r]^{\lambda_2}  & 	\omega^{[2]}  \ar[r]^{\lambda_3} & 	\omega^{[3]} \ar[r]^{\lambda_4} & 	\omega^{[4]} \ar[r]^{\lambda_5} & 	\omega^{[5]} \ar[r]^{\lambda_6} & \dots
}
\end{displaymath}
Let $g^{[j]}, x^{[j]}, y^{[j]}$ denote the associated potentials \eqref{aspot} corresponding to the pair $\bar{\omega}^{[j-1]}, \omega^{[j]}$.
In this notation, the superposition formulas \eqref{supCAE}
turn out to be
$$\numbered\label{difeq}
x^{[j+1]}
 = \frac{\lambda_{j+1} \lambda_1}{\lambda_{j+1}^2 - \lambda_1^2}
   (x^{[j]}
     -
      \frac{2 \lambda_{j+1} \lambda_1  \sin(\bar{\omega}^{[j]} - \omega^{[j]})}
         {\lambda_{j+1}^2 + \lambda_1^2
          - 2 \lambda_{j+1} \lambda_1 \cos(\bar{\omega}^{[j]} - \omega^{[j]})
          } g^{[j]} ),
\\
y^{[j+1]}
 = \frac{\lambda_{j+1}^2 - \lambda_1^2}{\lambda_{j+1}\lambda_1} y^{[j]}
   - \frac{2 \sin(\bar{\omega}^{[j]} - \omega^{[j]})}{g^{[j]}},
\\
g^{[j+1]} = \frac{-\lambda_{j+1}\lambda_1}{\lambda_{j+1}^2+\lambda_1^2-2\lambda_{j+1}\lambda_1\cos(\bar{\omega}^{[j]} - \omega^{[j]})}g^{[j]}.
$$
The above formulas constitute recurrence relations for the quantities $x^{[n]}, y^{[n]}, g^{[n]}$ with the initial conditions
$$ \numbered\label{ic}
x^{[1]} = x_1, \quad y^{[1]} = y_1, \quad g^{[1]} = g_1.
$$
\begin{proposition}\label{prop1}
Let $x_1, y_1, g_1$ be the associated potentials corresponding to the pair $\omega^{[0]}, \omega^{[1]}$ of sine-Gordon solutions.
Let $S^{[j]}$ be $4\times 4$ matrices with entries defined by formulas
$$\numbered\label{Smatrix}
S^{[j]}_{11} = \frac{\lambda_{j+1} \lambda_1}{\lambda_{j+1}^2 - \lambda_1^2}, \quad
S^{[j]}_{13} = -\frac{\lambda_{j+1}^2 \lambda_1^2}{\lambda_{j + 1}^2 - \lambda_1^2} \cdot \frac{2  \sin(\bar{\omega}^{[j]} - \omega^{[j]})}
{\lambda_{j+1}^2 + \lambda_1^2 - 2 \lambda_{j+1} \lambda_1 \cos(\bar{\omega}^{[j]} - \omega^{[j]})}, \\
S^{[j]}_{22} = \frac{\lambda_{j+1}^2 - \lambda_1^2}{\lambda_{j+1}\lambda_1}, \quad
S^{[j]}_{24} = - 2 \sin(\bar{\omega}^{[j]} - \omega^{[j]}), \\
S^{[j]}_{33} = \frac{1}{S^{[j]}_{44}} = \frac{-\lambda_{j+1}\lambda_1}{\lambda_{j+1}^2+\lambda_1^2-2\lambda_{j+1}\lambda_1\cos(\bar{\omega}^{[j]} - \omega^{[j]})}
$$
all the other entries being zero. Let
$$\numbered\label{prop1eq}
(
\begin{array}{c}
x^{[n]} \\
y^{[n]} \\
g^{[n]} \\
1/g^{[n]}
\end{array}
)
=
\prod_{i = 1}^{n - 1} S^{[i]}
(
\begin{array}{c}
x_1 \\
y_1 \\
g_1 \\
1/g_1
\end{array}
).
$$
Then $x^{[n]}, y^{[n]}, g^{[n]}$ are the associated potentials corresponding to the pair $\bar{\omega}^{[n-1]}, \omega^{[n]}$. Moreover,
if $z^{[n]} = 1/{g^{[n]}}^2 $, then $z^{[n]}(x^{[n]},y^{[n]})$ is a solution of the constant astigmatism equation \eqref{CAE}.
\end{proposition}
\begin{proof}
One easily observes that relations \eqref{difeq}
can be obtained by acting of $4\times 4$ matrix $S^{[j]}$ with the only nonzero entries \eqref{Smatrix}
on the column vector $X^{[j]} = (x^{[j]}, y^{[j]},g^{[j]}, 1/g^{[j]})$. The recurrent formula \eqref{difeq} with initial conditions
\eqref{ic} then can be conveniently written as
$$
X^{[j + 1]} = S^{[j]}X^{[j]}, \qquad X^{[1]} = X_1 = (x_1, y_1,g_1, 1/g_1)^T.
$$
Hence
$$
X^{[n]} =  \prod_{i = 1}^{n - 1} S^{[i]} X_1.
$$
\end{proof}

\begin{remark} \rm
Expanding the matrix formula \eqref{prop1eq} we have
$$
x^{[n]} = x_1 \lambda_1^{n-1}
\prod_{i=1}^{n-1}\frac {\lambda_{i+1}}{
\lambda_{i+1}^{2}-\lambda_{1}^{2}} \\
-
2\sum _{i=1}^{n-1}
 \left( \frac{
\lambda_{i+1}\lambda_1^{n-i+1} g^{[i]} \sin ( \bar{\omega}^{[i]}-\omega^{[i]} ) }
 { \lambda_{i+1}^{2} + \lambda_1^{2}-2\lambda_{i+1}\lambda_1\cos ( \bar{\omega}^{[i]}-\omega^{[i]} )
}
\prod _{p=1}^{n-i}{\frac{\lambda_{n-
p+1}}{\lambda_{{n-p+1}}^{2}-\lambda_1^2}}
 \right),
\\
y^{[n]} =
\frac{y_1}{\lambda_1^{n-1}}
\prod _{i=1}^{n-1}{\frac {\lambda_{{i+1}}^{2}-\lambda_{{1}}^{2}}{
\lambda_{{i+1}}}}
-
2\sum _{i=1}^{n-1}
(
\frac{\sin ( \bar{\omega}^{[i]}-\omega^{[i]} )}
{\lambda_{{1}}^{n-i-1} g^{[i]} }
\prod _{p=1}^{n-i-1}{\frac {\lambda_{{n-p
+1}}^{2}-\lambda_{{1}}^{2}}{\lambda_{{n-p+1}}}}
),
\\
g^{[n]} = \lambda_1^{n-1} g_1 \prod _{i=1}^{n-1} \frac{-\lambda_{i+1}}{\lambda_{i+1}^2+\lambda_1^2-2\lambda_{i+1}\lambda_1\cos(\bar{\omega}^{[i]} - \omega^{[i]})}.
$$
\end{remark}

\section{Multisoliton solutions}
Let $\omega^{[0]} = 0$. Let us define $\lambda_{kl}^+ := \lambda_k + \lambda_l,\;\; \lambda_{kl}^- := \lambda_k - \lambda_l$ and
$$
a_i := \exp(\lambda_i \xi + \frac{\eta}{\lambda_i} + c_i).
$$
Solving System \eqref{BT}, we get one-soliton solutions
$$
\omega^{[1]} = 2\arctan a_1, \qquad
\bar{\omega}^{[1]} = 2\arctan a_2
$$
and, applying the superposition principle \eqref{supSG} to the triple $\omega^{[0]}, \omega^{[1]}, \bar{\omega}^{[1]}$,
we easily obtain the two-soliton solutions
$$\numbered\label{twosol}
\omega^{[2]} = 2\arctan\frac{\lambda_{12}^+ (a_1-a_2)}{\lambda_{12}^- (1+a_1 a_2)}, \qquad
\bar{\omega}^{[2]} = 2\arctan\frac{\lambda_{23}^+(a_2-a_3)}{\lambda_{23}^-(1+a_2a_3)}.
$$
An exact analytic $n$-soliton solution, in our notation $\omega^{[n]}$, of the sine-Gordon equation has been obtained by
several authors \cite{Abl, Bry, Cau I, Cau II, Cau III, Hir}, see also \cite{A-I, S-B}.
The formula best suited for this paper can be found e.g. in \cite{Cau III} and is of the form
$$\numbered\label{SGmultisol}
 \omega^{[n]} =
\frac12 \arccos  \varphi^{[n]},
$$
where
$$\numbered\label{phi}
\varphi^{[n]} = 1 - 2\frac{\partial^2}{\partial \xi \,\partial\eta} \ln \det M
$$
$M$ being the $n\times n$ matrix with entries
$$
M_{ij} = \frac{1}{\lambda_i + \lambda_j}
(\sqrt{a_i a_j} + \frac{1}{\sqrt{a_i a_j}}).
$$
Note also that $\bar{\omega}^{[n]}$ arises from $\omega^{[n]}$ by increasing all lambdas' indices by one, namely
$$\numbered\label{SGmultisolbar}
\bar{\omega}^{[n]} =
 \frac12 \arccos \bar{\varphi}^{[n]},
$$
where
$$\numbered\label{phibar}
\bar{\varphi}^{[n]} = 1 - 2\frac{\partial^2}{\partial \xi \, \partial\eta} \ln \det \bar{M}
$$
$\bar{M}$ being given by
$$
\bar{M}_{ij} = \frac{1}{\lambda_{i + 1} + \lambda_{j + 1}}
(\sqrt{a_{i+1}a_{j+1}} + \frac{1}{\sqrt{a_{i+1}a_{j+1}}}).
$$

\begin{definition} \rm
By a \it $j$-soliton solution of the constant astigmatism equation \rm
we shall mean a triple $(x^{[j]}, y^{[j]}, g^{[j]})$ formed by associated potentials corresponding to the
$j$-soliton solution $\omega^{[j]}$ and the $(j-1)$-soliton solution $\bar{\omega}^{[j-1]}$ (see Diagram \eqref{lattice}) of the sine-Gordon equation.
\end{definition}
\begin{remark} \rm
To obtain a solution of the CAE explicitly, one would have to express $z^{[j]} = 1/{g^{[j]}}^2$ in terms of $x^{[j]}$ and $y^{[j]}$.
However, this is almost never possible in terms of elementary functions, see examples in Section~5 and~6.
\end{remark}

A one-soliton solution of the CAE is easy to construct.
Following \cite{Bia a, Bia II}, see also \cite[Prop. 4]{H-M I}, $x^{[1]} = x_1$ and $g^{[1]} = g_1$ can be obtained by differentiation, namely
$$\numbered\label{CA1solxg}
g_1 = \frac{\mathop{}\!\mathrm d \omega^{[1]}}{\mathop{}\!\mathrm d c_1} = \frac{2 a_1}{1+a_1^2}, \qquad
x_1 = -\frac{\mathop{}\!\mathrm d \ln g_1}{\mathop{}\!\mathrm d c_1} = \frac{a_1^2-1}{a_1^2 + 1}.
$$
For $y^{[1]} = y_1$ we have the system
$$
y^{[1]}_\xi = \frac{\lambda_1\sin(\omega^{[1]} + \bar{\omega}^{[0]})}{g_1} = \lambda_1,  \\
y^{[1]}_\eta = -\frac{\sin(\omega^{[1]} - \bar{\omega}^{[0]})}{\lambda_1 g_1} = -\frac{1}{\lambda_1}
$$
with the general solution
$$\numbered\label{CA1soly}
y_1 = \lambda_1 \xi - \frac{\eta}{\lambda_1} + k_1,
$$
$k_1$ being an arbitrary constant.
Setting $z_1 = 1/g_1^2$, eliminating $\xi,\eta$ and dropping the lower indices, one reveals the \it von~Lilienthal solution \rm
$$\numbered\label{vLsol}
z = \frac{1}{1 - x^2},
$$
see Figure \ref{onesol} on p.~11.

\begin{proposition} \label{propSoliton}
Let us denote
$A^{[j]} = 2 \bar{\varphi}^{[j]} \varphi^{[j]}$ and
$B^{[j]} = 2 \sqrt{({\bar{\varphi}^{[j]}}^2 - 1) ({\varphi^{[j]}}^2 - 1)}$, where
$\varphi^{[j]}$ and $\bar{\varphi}^{[j]}$ are defined by \eqref{phi} and \eqref{phibar} respectively.
Then the $n$-soliton solution of the CAE is given by the formula
$$\numbered\label{nsolCAE}
(
\begin{array}{c}
x^{[n]} \\
y^{[n]} \\
g^{[n]} \\
1/g^{[n]}
\end{array}
)
=
\prod_{i = 1}^{n - 1} S^{[i]}
(
\begin{array}{c}
x_1 \\
y_1 \\
g_1 \\
1/g_1
\end{array}
),
$$
where the only nonzero entries of matrices $S^{[j]}$ are given by
$$\numbered\label{Smatrixsol}
S^{[j]}_{11} = \frac{\lambda_{j+1} \lambda_1}{\lambda_{j+1}^2 - \lambda_1^2}, \qquad
S^{[j]}_{13} =  \frac{\lambda_{j+1}^2 \lambda_1^2}{\lambda_1^2 - \lambda_{j+1}^2} \cdot \frac
{\sqrt{2-A^{[j]}-B^{[j]}}}
{\lambda_1^2+\lambda_{j+1}^2 - \lambda_{j+1}\lambda_1 \sqrt{2+ A^{[j]} + B^{[j]}} } \\
S^{[j]}_{22} = \frac{\lambda_{j+1}^2 - \lambda_1^2}{\lambda_{j+1}\lambda_1}, \qquad
S^{[j]}_{24} = -\sqrt{2-A^{[j]}-B^{[j]}}, \\
S^{[j]}_{33} = \frac{1}{S^{[j]}_{44}} =
\frac
{-\lambda_{j+1} \lambda_1}
{\lambda_1^2 + \lambda_{j+1}^2 -  \lambda_{j+1} \lambda_1 \sqrt{2+A^{[j]}+B^{[j]}} }.
$$
\end{proposition}

\begin{proof}
Formulas \eqref{Smatrixsol} follow from plugging \eqref{SGmultisol} and \eqref{SGmultisolbar} into \eqref{Smatrix}
and employing trigonometric identities.
\end{proof}

\section{Surfaces of constant astigmatism}
Let $\mathbf r $ be a pseudospherical surface corresponding to a sine-Gordon solution $\omega$. Its B\"acklund transformation $\mathbf r^{(\lambda)}$ is given by the formula
$$\numbered\label{BTsurf}
\mathbf r^{(\lambda)} = \mathbf r
 + \frac{2\lambda}{1 + \lambda^2}(\frac{\sin(\omega - \omega^{(\lambda)})}{\sin(2\omega)} {\mathbf  r_{\xi}}
 + \frac{\sin(\omega + \omega^{(\lambda)})}{\sin(2\omega)} {\mathbf r_{\eta}}).
$$
Let us define (cf. \eqref{defk})
$$
\mathbf r^{[k]} = \mathbf r^{(\lambda_1\lambda_2\ldots\lambda_k)}, \qquad
\bar{\mathbf r}^{[k]} = \mathbf r^{(\lambda_2\lambda_3\ldots\lambda_{k+1})}.
$$
Then we have the recurrence relation
$$\numbered\label{BTsurfseq}
\mathbf r^{[j+1]} =
\mathbf r^{[j]}
+ \frac{2\lambda_{j+1}}{1 + \lambda_{j+1}^2}(\frac{\sin(\omega^{[j]} - \omega^{[j+1]})}{\sin(2\omega^{[j]})} {\mathbf  r^{[j]}_{\xi}}
+ \frac{\sin(\omega^{[j]} + \omega^{[j+1]})}{\sin(2\omega^{[j]})} {\mathbf r^{[j]}_{\eta}})
$$
with the initial condition $\mathbf r^{[0]} = \bar{\mathbf r}^{[0]} = \mathbf r_0$. Surfaces $\bar{\mathbf r}^{[i]} $ are obtained from $\mathbf r^{[i]} $ simply by increasing all lambdas' indices by one and replacing
$\omega^{[i]}$ with $\bar{\omega}^{[i]}$. The iteration process is shown in the diagram below, cf. \eqref{lattice}.
\begin{displaymath}
    \xymatrix{
{\bf r}^{[0]} \ar[r]^{\lambda_2} \ar[d]_{\lambda_1} 	& \bar{{\bf r}}^{[1]} \ar[r]^{\lambda_3} \ar[d]_{\lambda_1} & \bar{{\bf r}}^{[2]} \ar[d]_{\lambda_1} \ar[r]^{\lambda_4}  & \bar{{\bf r}}^{[3]} \ar[d]_{\lambda_1} \ar[r]^{\lambda_5}  & \bar{{\bf r}}^{[4]} \ar[d]_{\lambda_1} \ar[r]^{\lambda_6} & \dots
\\
{\bf r}^{[1]} \ar[r]^{\lambda_2}  & 	{\bf r}^{[2]}  \ar[r]^{\lambda_3} & 	{\bf r}^{[3]} \ar[r]^{\lambda_4} & 	{\bf r}^{[4]} \ar[r]^{\lambda_5} & 	{\bf r}^{[5]} \ar[r]^{\lambda_6} & \dots
}
\end{displaymath}
Substituting $\lambda = 1$ into \eqref{BTsurf} one gets what is called a \it complementary \rm pseudospherical surface
$$
\mathbf r^{(1)} = \mathbf r
 + \frac{\sin(\omega - \omega^{(1)})}{\sin(2\omega)} {\mathbf  r_{\xi}}
 + \frac{\sin(\omega + \omega^{(1)})}{\sin(2\omega)} {\mathbf r_{\eta}}.
$$
Obviously, the surfaces $\mathbf r^{[j]}$ and $\bar{\mathbf r}^{[j-1]}$ become complementary when substituting
$\lambda_1 = 1$ into~$\mathbf r^{[j]}$.

According to \cite[Prop. 3]{H-M I}, the common involute, $\tilde{\bf r}^{[j]}$,
of a pair of complementary pseudospherical surfaces, $ \mathbf r^{[j]}|_{\lambda_1 = 1} $ and $ \bar{\mathbf r}^{[j-1]} $,
is of constant astigmatism and is given by the equation
$$
\tilde{\bf r}^{[j]} =  \bar{\mathbf r}^{[j-1]} - \tilde{\mathbf n}^{[j]} \ln g^{[j]} |_{\lambda_1 = 1},
$$
where $g^{[j]} $ is determined by \eqref{prop1eq} and  $\tilde{\mathbf n}^{[j]}$, a unit normal of the constant astigmatism surface, is simply
$$
\tilde{\mathbf n}^{[j]} = \left. \mathbf r^{[j]} \right|_{\lambda_1 = 1} - \bar{\mathbf r}^{[j-1]} .
$$
If the surfaces $\mathbf r^{[j]}|_{\lambda_1 = 1} $ and $ \bar{\mathbf r}^{[j-1]} $ are $j$-soliton and $(j-1)$-soliton pseudospherical surfaces respectively, then the corresponding common involute, $\tilde{\bf r}^{[j]}$, will be called a \it $j$-soliton surface of constant astigmatism. \rm

Let us also remark that $\tilde{\mathbf n}^{[j]}(\xi,\eta)$ parameterises a unit sphere by slip lines (see the Introduction,
for details see \cite{H-M I}).

\section{Examples of multisoliton solutions}
In this section we provide explicit formulas for some multisoliton solutions of the constant astigmatism equation
as well as corresponding constant astigmatism surfaces.

Firstly, let us introduce a notation. Let us define
$$
\alpha := \xi - \eta, \qquad \beta := \xi + \eta
$$
which is nothing but the space and time coordinates in which sine-Gordon equation is of the form
$\omega_{\beta\beta} - \omega_{\alpha\alpha}  = \sin \omega$.
Also recall that
$$
a_i := \exp(\lambda_i \xi + \frac{\eta}{\lambda_i} + c_i), \qquad
\lambda_{kl}^+ := \lambda_k + \lambda_l,\qquad \lambda_{kl}^- := \lambda_k - \lambda_l
$$
and, in order to have short formulas, let us define
$$
a := \exp( \xi + \eta + c), \qquad
\lambda_{kl}^\oplus := \lambda_k^2 + \lambda_l^2,\qquad
\lambda_{kl}^\ominus := \lambda_k^2 - \lambda_l^2, \\
\lambda_{k}^+ := \lambda_k + 1,\qquad \lambda_{k}^- := \lambda_k -1, \qquad
\lambda_{k}^\oplus := \lambda_k^2 + 1,\qquad \lambda_{k}^\ominus := \lambda_k^2 -1.
$$

\subsection{One-soliton solutions}
In Section 3 we constructed the one soliton solution of the CAE corresponding to the pair $\omega^{[0]} = 0$ and
$\omega^{[1]} = 2\arctan a_1$. It belongs to the von~Lilienthal class.
Let us proceed to the corresponding surfaces of constant astigmatism.
The family of well known one-soliton Dini's surfaces is
$$
\bar{\mathbf r}^{[1]} =
(
\frac{4 \lambda_2 a_2 \cos\alpha }{\lambda_2^+ (a_2^2 + 1)},
\frac{4 \lambda_2 a_2 \sin\alpha }{\lambda_2^+ (a_2^2 + 1)},
\beta-\frac{2 \lambda_2 (a_2^2-1)}{\lambda_2^+ (a_2^2 + 1)}
).
$$
Substituting $\lambda_1 = 1$ into another Dini's surface
$$
\mathbf r^{[1]} = (
\frac{4 \lambda_1 a_1 \cos\alpha }{\lambda_1^+ (a_1^2 + 1)},
\frac{4 \lambda_1 a_1 \sin\alpha }{\lambda_1^+ (a_1^2 + 1)},
\beta-\frac{2 \lambda_1 (a_1^2-1)}{\lambda_1^+ (a_1^2 + 1)}
)
$$
we obtain the pseudosphere
$$\numbered\label{pseudosphere}
\mathbf r^{[1]}|_{\lambda_1 = 1} =
(
\frac{2 a \cos\alpha }{a^2 + 1},
\frac{2 a \sin\alpha }{a^2 + 1},
\beta+ \frac{1 - a^2}{1 + a^2}
).
$$
Note that in this case the `seed surface' $\mathbf r^{[0]} = \mathbf r_0$ is degenerated and coincides with the $z$-axis $(0,0,\beta)$.

The Gaussian map of corresponding constant astigmatism surface is
$$
\tilde{\mathbf{n}}^{[1]} = \mathbf r^{[1]}|_{\lambda_1 = 1} - \mathbf{r}_0 =
( \frac{2a \cos\alpha}{a^2 + 1},
\frac{2a \sin\alpha}{a^2 + 1},
\frac{1 - a^2}{1 + a^2}
),
$$
forming the net of $45^{\circ}$ loxodromes on the unit sphere. The family of one-soliton constant astigmatism surfaces is then
$$\numbered\label{vLsurf}
\tilde{\mathbf{r}}^{[1]} = \mathbf{r}_0 -  \tilde{\mathbf{n}}^{[1]} \ln (k g^{[1]})|_{\lambda_1 = 1} = (
\begin{array}{c}
\frac{-2a}{1+a^2} \ln(\frac{2k a}{1+a^2})\cos\alpha \\
\frac{-2a}{1+a^2} \ln(\frac{2k a}{1+a^2})\sin\alpha \\
\beta + \frac{a^2-1}{a^2 + 1}\ln(\frac{2k a}{1+a^2})
\end{array}
),
$$
$k$ being a real constant.
The surfaces coincide with the von~Lilienthal class \cite{vLil}, see Figure \ref{onesol}; for detailed description and pictures see \cite{B-M I}.
Evolutes of the surface $\tilde{\mathbf{r}}^{[1]}$
are the pseudosphere $\mathbf{r}^{[1]}|_{\lambda_1 = 1}$ and the $z$-axis $ \mathbf{r}_0$.

\begin{figure}[ht]
\begin{center}
\includegraphics[scale=0.25]{01.jpg}
\caption{Pseudosphere $\mathbf r^{[1]}(\xi,\eta)|_{\lambda_1 = 1}$ (upper left);
von~Lilienthal surface $\tilde{\mathbf{r}}^{[1]}(\xi,\eta)$ for $\ln k = 0.3$ (upper right);
Gaussian map $\tilde{\mathbf{n}}^{[1]}(\xi,\eta)$ (lower left);
solution $z = 1/(1-x^2)$ of the CAE (lower right).}\label{onesol}
\end{center}
\end{figure}

\subsection{Two-soliton solutions}\label{TwosolEx}
The two-soliton solutions of the sine-Gordon equation are given by \eqref{twosol}
and the two soliton solution of the CAE corresponding to the pair  $\bar{\omega}^{[1]}$ and $\omega^{[2]}$
is
$$\numbered\label{CA2sol}
x^{[2]} = \frac{-\lambda_2  \lambda_1}{{\lambda_{12}^\ominus}} \cdot
\frac{
  {\lambda_{12}^+}^2 (a_1^2-a_2^2) + {\lambda_{12}^-}^2 (a_1^2 a_2^2 - 1)}
{
{\lambda_{12}^+}^2 (a_1^2+a_2^2)
+  {\lambda_{12}^-}^2 (a_1^2 a_2^2+1)
-8 {\lambda_1} {\lambda_2} a_1 a_2 }, \\
y^{[2]} =
-\frac{\lambda_{12}^\ominus}{\lambda_1^2\lambda_2} (\lambda_1^2 \xi - \eta)
+
\frac{2 (1+a_1 a_2)(a_1-a_2)}{a_1 (1+a_2^2)} , \\
{g^{[2]}} =
\frac
{-2 \lambda_1 \lambda_2  a_1 (1+a_2^2)}
{{\lambda_{12}^+}^2 (a_1^2+a_2^2)
+  {\lambda_{12}^-}^2 (a_1^2 a_2^2+1)
-8 {\lambda_1} {\lambda_2} a_1 a_2}, \\
z^{[2]} = \frac{1}{{g^{[2]}}^2} = (
\frac{{\lambda_{12}^+}^2 (a_1^2+a_2^2)
+  {\lambda_{12}^-}^2 (a_1^2 a_2^2+1)
-8 {\lambda_1} {\lambda_2} a_1 a_2 }
{2 \lambda_1 \lambda_2  a_1 (1+a_2^2)})^2.
$$
The triple $z^{[2]}$, $x^{[2]}$, $y^{[2]}$ provides a solution of the CAE in parametric form; the plot
can be seen in Figure \ref{CAsol}.
Eliminating $\xi, \eta$ one obtains an implicit formula for the function $z(x,y) = z^{[2]}(x^{[2]},y^{[2]})$, namely
$$
y =
2 \ln a_2  -
\frac{\lambda_{12}^\oplus \ln a_1}{\lambda_1  \lambda_2 }
+
\frac{2 (1+a_1 a_2)(a_1-a_2)}{a_1 (1+a_2^2)} ,
$$
where
$$
a_1 = \frac{
-(x^2{\lambda_{12}^{\ominus}}^2 - {\lambda_{1}^2} {\lambda_{2}^2})^2 z^2
-2 {\lambda_{12}^+}^4
(x^2{\lambda_{12}^-}^4 - {\lambda_{1}^2} {\lambda_{2}^2})
 z
+2 K {\lambda_{1}} {\lambda_{2}} {\lambda_{12}^+}^2 \sqrt{z}-{\lambda_{12}^\ominus}^4}
{(x\lambda_{12}^{\ominus} +{\lambda_{1}} {\lambda_{2}})^2 (4  \lambda_{1}^2 \lambda_{2}^2 z^{\frac32} +
 K z )+4 \lambda_{1}^2 \lambda_{2}^2 {\lambda_{12}^\ominus}^2 \sqrt{z}+
K{\lambda_{12}^\ominus}^2 },
$$
$$
a_2 =
\frac{4\lambda_2^2\lambda_1^2\sqrt{z} +K}
{{\lambda_{12}^\ominus}^2 + (x^2{\lambda_{12}^\ominus}^2-\lambda_1^2\lambda_2^2)z},
\qquad
K = 16\lambda_2^4\lambda_1^4{z} - [{\lambda_{12}^\ominus}^2+ (x^2{\lambda_{12}^\ominus}^2 - \lambda_1^2\lambda_2^2)z]^2 .
$$

\begin{figure}[ht]
\begin{center}
\includegraphics[scale=0.32]{CA23sol.jpg}
\caption{Two soliton solution (top) and three soliton solution (bottom) of the CAE, $\lambda_1 = 1.2$, $\lambda_2 = 1.5$, $\lambda_3 = 1.8$, $c_i = 0$.  Right parts of the figures show the behavior around the origin.}\label{CAsol}
\end{center}
\end{figure}

Using \eqref{BTsurfseq} one obtains the two-soliton pseudospherical surface
$$
\mathbf{r}^{[2]} =
\frac{4\lambda_{12}^\ominus}{\lambda_{1}^\oplus \lambda_{2}^\oplus R_1 }
[
(
\begin{array}{c}
R_2 \\
R_3 \\
0
\end{array}
) \sin\alpha
+
(
\begin{array}{c}
R_3 \\
-R_2 \\
0
\end{array}
) \cos\alpha
]
+
(
\begin{array}{c}
0 \\
0 \\
\beta +  \frac{R_4}{\lambda_{1}^\oplus \lambda_{2}^\oplus R_1 }
\end{array}
) ,
$$
where
$$\numbered \label{P}
R_1  =   {\lambda_{12}^+}^2 (a_1^2+a_2^2)
+  {\lambda_{12}^-}^2 (a_1^2 a_2^2+1)
-8 {\lambda_1} {\lambda_2} a_2 a_1, \\
R_2 = 2 {\lambda_1} {\lambda_2} (1+a_1 a_2) (a_1-a_2), \\
R_3  =  {\lambda_2}  {\lambda_1^\ominus} a_2 (1+a_1^2) - {\lambda_1} {\lambda_2^\ominus} a_1 (1+a_2^2), \\
R_4 =  2  {\lambda_{12}^\ominus} {\lambda_{12}^+} ({\lambda_1} {\lambda_2}-1) (a_1^2-a_2^2)
- 2  {\lambda_{12}^\ominus} {\lambda_{12}^-} ({\lambda_1} {\lambda_2}+1) (a_1^2 a_2^2 - 1),
$$
for picture see Figure \ref{PSsurf}.
\begin{figure}[ht]
\begin{center}
\includegraphics[scale=0.17]{PS.jpg}
\caption{From the left: One-, two- and three-soliton pseudospherical surfaces, $\bar{\mathbf r}^{[1]}, \mathbf{r}^{[2]}, \mathbf{r}^{[3]}$ respectively,
$\lambda_1 = 1$, $\lambda_2 = 1.5$, $\lambda_3 = 1.8$, $c_i = 0$.}\label{PSsurf}
\end{center}
\end{figure}
The Gaussian map of corresponding constant astigmatism surface $\tilde{\mathbf{r}}^{[2]}$ is
$$
\tilde{\mathbf{n}}^{[2]} = \mathbf{r}^{[2]}|_{\lambda_1 = 1} - \bar{\mathbf r}^{[1]} =
(
\begin{array}{c}
N_1\sin\alpha + N_2\cos\alpha \\
N_2\sin\alpha - N_1\cos\alpha \\
N_3
\end{array}
),
$$
where
$$
N_1 = -2\frac{{\lambda_2^\ominus} R_2'}{\lambda_2^\oplus R_1'}, \quad
N_2 = -2\frac{{\lambda_2^\ominus}R_3'}{\lambda_2^\oplus R_1'}
-\frac{4 a_2 \lambda_2}{{\lambda_{2}^\oplus} (a_2^2 + 1)}, \quad
N_3 = \frac{R_4'}{2 {\lambda_{2}^\oplus} R_1' } + \frac{2 \lambda_2 (a_2^2-1) }{{\lambda_{2}^\oplus} (a_2^2 + 1)}.
$$
Obviously, $R'_i$ arises from $R_i$ by substituting $\lambda_1 = 1$, namely
$$
\numbered
R'_1  =   {\lambda_{2}^+}^2 (a^2+a_2^2)
+  {\lambda_{2}^-}^2 (a^2 a_2^2+1)
-8 {\lambda_2} a a_2, \\
R'_2 = 2 {\lambda_2} (1+a a_2) (a-a_2), \\
R'_3 = - {\lambda_2^\ominus} a (1+a_2^2), \\
R'_4 =  -2  {\lambda_{2}^\ominus} {\lambda_{2}^+} ( {\lambda_2}-1) (a^2-a_2^2)
- 2  {\lambda_{2}^\ominus} {\lambda_{2}^-} ( {\lambda_2}+1) (a^2 a_2^2 - 1).
$$
Recall that $\tilde{\mathbf{n}}^{[2]}(\xi,\eta)$ parameterises the unit sphere by slip lines,
for a picture see the left side of Figure~\ref{n2}.
One can easily check that in the case when $c_i = 0$, the slip lines net is symmetric with respect to the point $\tilde{\mathbf{n}}^{[2]}(0,0) = (1,0,0)$, i.e. if
$\tilde{\mathbf{n}}^{[2]}(\xi,\eta) = (n_x, n_y, n_z)$, then $\tilde{\mathbf{n}}^{[2]}(-\xi,-\eta) = (n_x, -n_y, -n_z )$. In the picture,
we observe the sphere from the positive part of the $x$-axis and the symmetry of the pattern can be clearly recognized.
Some parts of the sphere are multiply covered and one can see singularities (envelopes) of slip lines.
We leave open the question (suggested by a few numerical calculations)
whether coordinate lines $\xi = \rm const$ and $\eta = \rm const$ converge to
the poles, i.e. points $(0,0,\pm 1)$.
\begin{figure}[ht]
\begin{center}
\includegraphics[scale=0.4]{N2finalt.jpg}
\caption{Slip line field $\tilde{\mathbf{n}}^{[2]}$, $\lambda_2 = 1.5$, $c_i = 0$ (left) and its limit $\tilde{\mathbf{n}}^{\{2\}}$, $c = 0$ (right) with coordinate lines $\xi = 0$ and $\eta = 0$ highlighted (thick black curves).}\label{n2}
\end{center}
\end{figure}

To obtain an associated orthogonal equiareal pattern one needs to invert the transformation $(x,y) \leftrightarrow (\xi,\eta)$ given by first two equations in \eqref{CA2sol}, which is not possible in terms of elementary functions. The parameterisation
$\tilde{\mathbf{n}}^{[2]}(x,y)$ then would provide orthogonal equiareal net sought.

The corresponding family of two-soliton constant astigmatism surfaces having evolutes $\mathbf{r}^{[2]}|_{\lambda_1 = 1}$ and a Dini surface $\bar{\mathbf r}^{[1]} $ is then
$$
\tilde{\mathbf{r}}^{[2]} = \bar{\mathbf r}^{[1]} - f \tilde{\mathbf{n}}^{[2]}  =
(
\begin{array}{c}
\tilde{R}_1\sin\alpha + \tilde{R}_2\cos\alpha \\
\tilde{R}_2\sin\alpha - \tilde{R}_1\cos\alpha \\
\beta - \tilde{R}_3
\end{array}
),
$$
where
$$
f = \ln (k g^{[2]})|_{\lambda_1 = 1}, \quad
\tilde{R}_1 = 2 f \frac{{\lambda_2^\ominus} R_2'}{{\lambda_2^\oplus} R_1'}, \quad
\tilde{R}_2 = 2 f
\frac{ {\lambda_2^\ominus} R_3'}{{\lambda_2^\oplus} R_1'} +
\frac{4 \lambda_2 a_2}{{\lambda_2^\oplus} (a_2^2 + 1)}( f + 1   ), \\
\tilde{R}_3 =  \frac12 f \frac{R_4'}{{\lambda_2^\oplus}R_1'}
+ \frac{2 \lambda_2 (a_2^2-1)}{{\lambda_2^\oplus} (a_2^2 + 1)}(f + 1).
$$
A part of the surface can be seen in the left side of Figure~\ref{CA23}. One can observe cuspidal edges
obviously related to singularities of corresponding slip line field (Figure~\ref{n2}).
\begin{figure}[ht]
\begin{center}
\includegraphics[scale=0.5]{RR.jpg}
\caption{Two soliton surface $\tilde{\mathbf{r}}^{[2]}$ (left) and three soliton surface $\tilde{\mathbf{r}}^{[3]}$ (right)
of constant astigmatism, $\lambda_2 = 1.5$, $\lambda_3 = 1.8$, $c_i = 0$, $k = 1$.}\label{CA23}
\end{center}
\end{figure}

\subsection{Three-soliton solutions}
The 3-soliton solution, $(x^{[3]},y^{[3]},g^{[3]})$,
of the CAE corresponding to the pair
$$
\omega^{[3]}
= 2\arctan(
\frac{
  {\lambda_{12}^+} {\lambda_{13}^+} {\lambda_{23}^-} a_1
- {\lambda_{12}^+} {\lambda_{13}^-} {\lambda_{23}^+} a_2
+ {\lambda_{12}^-} {\lambda_{13}^+} {\lambda_{23}^+}  a_3
+ {\lambda_{12}^-} {\lambda_{13}^-} {\lambda_{23}^-}  a_1 a_2 a_3}
{
 {\lambda_{12}^-} {\lambda_{13}^+} {\lambda_{23}^+}  a_1 a_2
-{\lambda_{12}^+} {\lambda_{13}^-} {\lambda_{23}^+}  a_1 a_3
+{\lambda_{12}^+} {\lambda_{13}^+} {\lambda_{23}^-} a_2 a_3
+{\lambda_{12}^-} {\lambda_{13}^-} {\lambda_{23}^-} }
) \\
$$
and $\bar{\omega}^{[2]}$ can be easily obtained using Proposition \ref{propSoliton},
explicit formulas being too lengthy to be written here. The graph of
the solution $z(x,y) = z^{[3]}(x^{[3]},y^{[3]})$ can be seen in Figure \ref{CAsol},
a multivaluedness of the function $z$ being clearly identified.
In the right side of the figure (values of $z$ near the point $(0,0,0)$) one can observe that at least eight values of $z$
may correspond to one particular choice of $x$ and $y$.

Proceeding to the constant astigmatism surface we firstly compute the three-soliton pseudospherical surface (see Figure \ref{PSsurf})
$$
\mathbf{r}^{[3]} = \mathbf{r}^{[2]}
+ \frac{2\lambda_3}{\lambda_3^2 + 1}
(
\frac{\sin (\omega^{[2]} - \omega^{[3]})} {\sin 2 \omega^{[2]} }
\mathbf{r}^{[2]}_\xi
+
\frac{\sin (\omega^{[2]} + \omega^{[3]})} {\sin 2 \omega^{[2]} }
\mathbf{r}^{[2]}_\eta
).
$$
The Gaussian map of corresponding constant astigmatism surface is then
$$
\tilde{\mathbf{n}}^{[3]} = \mathbf{r}^{[3]}|_{\lambda_1 = 1} - \bar{\mathbf r}^{[2]}
$$
forming a net of slip lines, for picture see the left side of Figure~\ref{n3}.
\begin{figure}[ht]
\begin{center}
\includegraphics[scale=0.4]{N3finalt.jpg}
\caption{Slip line field $\tilde{\mathbf{n}}^{[3]}$, $\lambda_2 = 1.5$, $\lambda_3 = 1.8$,  $c_i = 0$ (left)
and its limit $\tilde{\mathbf{n}}^{\{3\}}$, $c = 0$ (right) with coordinate lines $\xi = 0$ and $\eta = 0$ highlighted (thick black curves).}\label{n3}
\end{center}
\end{figure}

Finally, we construct the family of constant astigmatism surfaces having evolutes $\mathbf{r}^{[3]}|_{\lambda_1 = 1}$ and $\bar{\mathbf r}^{[2]} $.
They are
$$
\tilde{\mathbf{r}}^{[3]} = \bar{\mathbf r}^{[2]} -  \tilde{\mathbf{n}}^{[3]} \ln (k g^{[3]})|_{\lambda_1 = 1},
$$
$k$ being a constant. A picture of rather complicated surface can be seen in the right side of Figure~\ref{CA23}.

\section{Examples of multisoliton solutions with $\lambda = 1$}
Computing the limit for $\lambda_i \rightarrow 1$ allows us to iterate the B\"acklund transformation with $\lambda = 1$ and construct
solutions of the sine-Gordon equation and corresponding pseudospherical surfaces. Slip-line fields are also available.
However, taking the limit for $(\lambda_1,\lambda_{2}) \rightarrow (1,1)$ and $(c_1,c_2) \rightarrow (c,c)$ in \eqref{CA2sol} yields
$$
\lim x^{[2]} = \infty, \qquad \lim y^{[2]} = 0, \qquad \lim g^{[2]} = \infty.
$$
Fortunately, finite limits result when applying the scaling symmetry
$
\mathcal S_s (x,y,g) \rightarrow (sx, y/s, sg)
$$
of the CAE to the triple $(x^{[2]}, y^{[2]}, g^{[2]})$, the constant $s$ being suitably chosen.
In what follows we iterate the B\"acklund transformation with $\lambda = 1$ (the case examined e.g. in \cite{S-B}) and we
find the corresponding solutions of the CAE as the limits of the solutions obtained in previous section.

Let us denote the $k$-th iteration of the B\"acklund transformation with
 $\lambda = 1$ by $\omega^{\{k\}}$, according to the diagram
\begin{displaymath}
    \xymatrix{
\omega^{\{0\}} \ar[r]^{1} 	& \omega^{\{1\}} \ar[r]^{1} & \omega^{\{2\}} \ar[r]^{1}  & \omega^{\{3\}} \ar[r]^{1} & \omega^{\{4\}} \ar[r]^{1} & \dots .
}
\end{displaymath}
The corresponding pseudospherical surface is, obviously, denoted by $\mathbf r^{\{k\}}$. We also extend the notation
to the solutions of the constant astigmatism equation, i.e. $x^{\{k\}}, y^{\{k\}}, g^{\{k\}}$.

\subsection{One-soliton solutions}
Starting with the zero solution $\omega^{\{0\}} = 0$, the one-soliton solution of the sine-Gordon equation is
$$
\omega^{\{1\}} = 2\arctan a =
2\arctan \mathrm e^{\xi + \eta + c}
$$
and the one-soliton pseudospherical surface, $\mathbf r^{\{1\}}$, is the pseudosphere \eqref{pseudosphere}.

The one-soliton solution of the CAE can be easily found by substituting $\lambda_1 = 1$
into Equations \eqref{CA1solxg} and \eqref{CA1soly}.
Actually, since the solution does not depend on $\lambda_1$ (it disappears when eliminating parameters $\xi, \eta$ from
\eqref{CA1solxg} and \eqref{CA1soly}),
we obtain precisely the von~Lilienthal solution \eqref{vLsol}.

Corresponding constant astigmatism surfaces are exactly those given by \eqref{vLsurf}.

\subsection{Two-soliton solutions}\label{2sollim}
Taking the limit of $\omega^{[2]}$ for $\lambda_i \rightarrow 1$ and $c_i \rightarrow c$ yields
$$
\omega^{\{2\}} = 2\arctan\frac{2a\alpha}{1+a^2} =
2\arctan\frac{2\mathrm e^{\xi + \eta + c}(\xi - \eta)}{1+\mathrm e^{2(\xi + \eta + c)}}.
$$
The corresponding two-soliton pseudospherical surface is
$$
\mathbf{r}^{\{2\}} =
\frac{1}{a^4+2(2 \alpha^2+1) a^2 + 1}
(
\begin{array}{c}
4a \alpha(1+a^2) \sin\alpha + 4 a (1+a^2) \cos\alpha
 \\
4 a (1+a^2) \sin\alpha  - 4 a \alpha (1+a^2) \cos\alpha
 \\
\beta (a^4+2(2 \alpha^2+1) a^2 + 1) - 2(a^4-1)
\end{array}
)
$$
and the slip-line field on the constant astigmatism surface's Gaussian sphere is
$$
\tilde{\mathbf{n}}^{\{2\}} = \mathbf{r}^{\{2\}} - \mathbf{r}^{\{1\}} =
\frac{1}{a^6+(4 \alpha^2+3) (a^4+ a^2) + 1}
\\
\times
(
\begin{array}{c}
4a(a^2+1)^2 \alpha\sin\alpha + 2a[a^4+ 2(1-2\alpha^2)a^2 + 1]\cos\alpha \\
2a[a^4+ 2(1-2\alpha^2)a^2 + 1] \sin\alpha - 4a(a^2+1)^2 \alpha\cos\alpha \\
(1-a^2)[a^4+2(1-2 \alpha^2) a^2 + 1]
\end{array}
),
$$
see the right side of the Figure \ref{n2}. The picture indicates some symmetries of the pattern. Indeed, if
$c = 0$ and $\tilde{\mathbf{n}}^{\{2\}}(\xi,\eta) = (n_x, n_y, n_z)$, then
\begin{itemize}
\item $\tilde{\mathbf{n}}^{\{2\}}(-\xi,-\eta) = (n_x, -n_y, -n_z)$
(symmetry with respect to the point $\tilde{\mathbf{n}}^{\{2\}}(0,0) = (1, 0, 0)$) ,
\item $\tilde{\mathbf{n}}^{\{2\}}(\eta,\xi) = (n_x, -n_y, n_z)$
(symmetry with respect to the zero meridian -- the image of the line $\xi = \eta$ ),
\item $\tilde{\mathbf{n}}^{\{2\}}(-\eta,-\xi) = (n_x, n_y, -n_z)$
(symmetry with respect to the equator -- the image of the line $\xi = -\eta$ ).
\end{itemize}
Coordinate lines of the slip line field $\tilde{\mathbf{n}}^{\{2\}}$ converge to the poles, namely
$$
\lim_{\xi\rightarrow \pm \infty} \tilde{\mathbf{n}}^{\{2\}} = \lim_{\eta\rightarrow \pm \infty} \tilde{\mathbf{n}}^{\{2\}} = (0,0,\mp 1).
$$
According to the definition of slip lines, the corresponding orthogonal equiareal pattern has the same symmetries and the same
limit behavior. Unfortunately, no picture is provided; to do that one has to reparameterise the sphere which requires expressing
$\xi, \eta$ in terms of $x^{\{2\}}, y^{\{2\}}$, i.e. inverting the transformation described by first two equations in \eqref{CA2sollim}.
Plugging the functions $\xi(x,y), \eta(x,y)$ into $\tilde{\mathbf{n}}^{\{2\}}(\xi, \eta)$
then would provide the orthogonal equiareal net $\tilde{\mathbf{n}}^{\{2\}}(x,y)$.

The two soliton solution $(x^{\{2\}}, y^{\{2\}}, g^{\{2\}}) $ of the CAE can be obtained by taking a limit for
$\lambda_i \rightarrow 1$ and $c_i \rightarrow c$ of the
triple $(x^{[2]}, y^{[2]}, g^{[2]}) $ scaled by the factor
$s = -(\lambda_2 - \lambda_1)^2$, i.e.
$$
(x^{\{2\}}, y^{\{2\}}, g^{\{2\}}) =
\mathop{\lim_{\lambda_i \rightarrow 1}}_{c_i \rightarrow c} \mathcal{S}_{-(\lambda_2 - \lambda_1)^2} (x^{[2]}, y^{[2]}, g^{[2]}).
$$
We have
$$\numbered\label{CA2sollim}
x^{\{2\}}
= \frac{4 \mathrm e^{2(\xi + \eta + c)}(\xi - \eta)}{\mathrm e^{4(\xi + \eta + c)}+2[2(\xi-\eta)^2+1] \mathrm e^{2(\xi + \eta + c)} +1}, \\
y^{\{2\}}
= \frac{\mathrm e^{2(\xi + \eta + c)} [\xi + \eta -(\xi - \eta)^2 + c]  + \xi + \eta + (\xi - \eta)^2 + c}{\mathrm e^{2(\xi + \eta + c)}+1}, \\
g^{\{2\}} =  \frac{2 \mathrm e^{\xi + \eta + c} (\mathrm e^{2(\xi + \eta + c)}+1) }{\mathrm e^{4(\xi + \eta + c)}+ 2[2 (\xi - \eta)^2+1] \mathrm e^{2(\xi + \eta + c)} +1}.
$$
Then $z^{\{2\}} = 1/{g^{\{2\}}}^2$ as a function of $x^{\{2\}},y^{\{2\}}$ is a solution of the CAE.
Eliminating parameters $\xi,\eta$
provides the implicit formula for the function $z = z(x,y) = z^{\{2\}}(x^{\{2\}},y^{\{2\}})$, namely
$$
y =
\frac{L^2-1}{L^2+1}
(\frac{ x z}{ z x^2+1 })^2
- \ln{L} ,
\qquad
L = \frac{z x^2+1}{
\sqrt{ z  - (zx^2 + 1)^2  }
- \sqrt{z}}.
$$
Finally, we write down the constant astigmatism surface $\tilde{\mathbf{r}}^{\{2\}}$ having evolutes $\mathbf{r}^{\{1\}} $
and $\mathbf{r}^{\{2\}}$.
For brevity, let us denote $\ln (k g^{\{2\}})$ by $f$. Then
$$
\tilde{\mathbf{r}}^{\{2\}} = \mathbf{r}^{\{1\}} - f\tilde{\mathbf{n}}^{\{2\}}
= (
\begin{array}{c}
\tilde{R}_1\sin\alpha + \tilde{R}_2\cos\alpha \\
\tilde{R}_2\sin\alpha - \tilde{R}_1\cos\alpha \\
\beta + \tilde{R}_3
\end{array}
),
$$
where
$$
\tilde{R}_1 = -\frac{4 a(a^2+1) f \alpha }{a^4+2(2 \alpha^2+1) a^2 + 1}, \\
\tilde{R}_2 = -2a\frac{ (f-1) a^4-2[2 \alpha^2+(2 \alpha^2-1) f+1] a^2+f-1}
{a^6+(4 \alpha^2+3) (a^4+ a^2)+1}, \\
\tilde{R}_3 = \frac{(f-1) a^6 - [(4 \alpha^2-1) f+4 \alpha^2+1](a^4 - a^2)-f+1}
{a^6+(4 \alpha^2+3) (a^4+ a^2)+1}.
$$
The results from this example coincide with ones obtainable using a
\it reciprocal transformation \rm for the constant astigmatism equation introduced in \cite{H-M III},
albeit the parameterisation of the results is different.
For instance, one can observe apparent similarity between the surface $\tilde{\mathbf{r}}^{\{2\}}$
in Figure \ref{CA2lim} and the surface \cite[Sect.~8, Ex.~5, Fig.~3]{H-M III}.
\begin{figure}[ht]
\begin{center}
\includegraphics[scale=0.32]{CAlim.jpg}
\caption{Two soliton surface, $\tilde{\mathbf{r}}^{\{2\}}$, of constant astigmatism, $k = 1$, $c = 0$.
The central part of the surface is zoomed in the right. }\label{CA2lim}
\end{center}
\end{figure}

\subsection{Three-soliton solutions}
The limit of $\omega^{[3]}$ for $\lambda_i \rightarrow 1$ and $c_i \rightarrow c$ yields
$$
\omega^{\{3\}} = 2\arctan\frac{a(2\alpha^2 + 2\beta+1) + a^3}{a^2(2\alpha^2 -2\beta + 1 ) + 1}
$$
or, alternatively \cite{S-B},
$$
\omega^{\{3\}} = - 2\arctan(\frac{1}{a}\cdot\frac
{\beta+\frac12\sinh2(\beta+c)+\cosh^2(\beta+c)+\alpha^2}
{\beta+\frac12\sinh2(\beta+c)-\cosh^2(\beta+c)-\alpha^2}).
$$
The corresponding three-soliton pseudospherical surface is
$$
\mathbf{r}^{\{3\}} = \mathbf{r}^{\{2\}}
+
(
\frac{\sin (\omega^{\{2\}} - \omega^{\{3\}})} {\sin 2 \omega^{\{2\}} }
\mathbf{r}^{\{2\}}_\xi
+
\frac{\sin (\omega^{\{2\}} + \omega^{\{3\}})} {\sin 2 \omega^{\{2\}} }
\mathbf{r}^{\{2\}}_\eta
)
$$
and the slip-line field on the constant astigmatism surface's Gaussian sphere is given by
$$
\tilde{\mathbf{n}}^{\{3\}} = \mathbf{r}^{\{3\}} - \mathbf{r}^{\{2\}},
$$
formulas being too lengthy to be written here, hence omitted. However, the picture can be seen in the right side of Figure~\ref{n3} and
the properties of the pattern (symmetries when $c = 0$ and limits of the coordinate lines for $\xi,\eta \rightarrow \pm \infty$)
are exactly same as in the two soliton case from the previous subsection \ref{2sollim}.

In order to obtain the three soliton solution $(x^{\{3\}}, y^{\{3\}}, g^{\{3\}})$
we rescale the triple $(x^{[3]}, y^{[3]}, g^{[3]})$ by the scaling factor $s = (\lambda_2 - \lambda_1)^2(\lambda_3 - \lambda_1)^2$
and then compute the limit for $\lambda_i \rightarrow 1$ and $c_i \rightarrow c$, i.e.
$$
(x^{\{3\}}, y^{\{3\}}, g^{\{3\}}) =
\mathop{\lim_{\lambda_i \rightarrow 1}}_{c_i \rightarrow c} \mathcal{S}_{(\lambda_2 - \lambda_1)^2(\lambda_3 - \lambda_1)^2}
(x^{[3]}, y^{[3]}, g^{[3]}).
$$
The result is
$$
x^{\{3\}} =
  \frac{-4 a^2 [(\alpha^2-\beta) a^2-\alpha^2-\beta]}
{a^6 + [4 (\alpha^2-\beta)^2 + 8 \alpha^2 + 3] a^4 + [4 (\alpha^2+\beta)^2 + 8 \alpha^2 + 3] a^2 + 1}, \\
y^{\{3\}} = \frac{2 \alpha}{3} \cdot
\frac{ (\alpha^2-3 \beta+\frac32) a^4 - 2( \alpha^4-\alpha^2+3\beta^2 -\frac32) a^2+\alpha^2+3 \beta+\frac32}
{ a^4+2(2 \alpha^2+1) a^2 + 1}, \\
g^{\{3\}} =
\frac{2a[a^4+2(2 \alpha^2+1) a^2 + 1]}{a^6 + [4 (\alpha^2-\beta)^2 + 8 \alpha^2 + 3] a^4 + [4 (\alpha^2+\beta)^2 + 8 \alpha^2 + 3] a^2 + 1}.
$$

Finally, the constant astigmatism surface $\tilde{\mathbf{r}}^{\{3\}}$, part of which can be seen in Figure~\ref{CA3lim},
is
$$
\tilde{\mathbf{r}}^{\{3\}} = \mathbf{r}^{\{2\}} - \ln(k g^{\{3\}})\tilde{\mathbf{n}}^{\{3\}},
$$
$k$ being a real constant.
\begin{figure}[ht]
\begin{center}
\includegraphics[scale=0.38]{CAlim3.jpg}
\caption{Three soliton surface, $\tilde{\mathbf{r}}^{\{3\}}$, of constant astigmatism, $k = 1$, $c = 0$.}\label{CA3lim}
\end{center}
\end{figure}

\ack

The author was supported by Specific Research grant SGS/1/2013 of the Silesian University in Opava and
wishes to thank Michal Marvan for his guidance and valuable advice.
The author is indebted to Iosif S.~Krasil'shchik for reading the manuscript and
would also like to thank Petr Blaschke for useful discussions.

\section {References}

redpajama_set_name: RedPajamaArXiv


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About Collaborative Law
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Cynthia L. Mackenzie
Cynthia L. Mackenzie is a lawyer of 30 years' experience. After articling with a large London law firm, she was then employed by a boutique personal injury firm. Shortly thereafter, she joined forces with her formidable father, Charles L. Mackenzie, Q.C. to form the law office of Mackenzie & Mackenzie in general practice. For the past 20 years, she has practiced primarily in the area of family law. In the year 2001, she became a panel lawyer with the Office of the Children's Lawyer and has had the privilege of representing children in custody and access and child protection proceedings for the past 18 years. Cynthia has gained invaluable insight into the effect that separation and divorce have on children, which experience has allowed her to offer real life experience and guidance to separating parents.
Cynthia obtained collaborative family law training over 15 years ago in order to expand the scope of her practice to offer clients the ability to resolve family law disputes in a different way, and found that it has the ability to reduce conflict and damage to families caused by protracted litigation. Collaborative law is an excellent option for couples who wish to negotiate terms of domestic contracts, such as marriage contracts or separation agreements, with two experienced lawyers dedicated to crafting custom made solutions which fit the needs of that particular family. Cynthia will advocate for a fair and reasonable solution in a safe and private process at a pace controlled by the parties. Cynthia is married and has three grown children, two dogs and a cat. She was born in London Ontario and obtained her high school diploma at London Central Secondary School. She obtained her B.A. (Hons) in English Literature from McGill University. She obtained her law degree from the University of Western Ontario in 1987 and is now a sole practitioner at the law office of Mackenzie Law Professional Corporation.
Mackenzie Law Professional Corporation
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Don't be caught out by extra charges when you're trying to get a good deal on your broadband.
It can seem like we're all constantly being bombarded with adverts for the latest broadband deals, some for just a couple of quid a month and others even claiming to be free. But are they really all they seem?
While these deals can seem like great value at first glance, another look reveals that they often have several charges that might not be immediately obvious. With this in mind, we're taking a look at a selection of the best current deals* from some the UK's biggest broadband providers and seeing if they're really all they seem, or if you could end up paying more than you'd bargained for.
On top of the £5 a month charge for broadband, you'll also have to pay £17.99 a month line rental. If you're not currently a BT customer, you'll have to get a BT Home Hub wireless router. This is technically free but you pay £6.95 to have it delivered. As the contract is for 12 months, you'll pay a total of £282.83 over the year for the broadband deal.
It sounds too good to be true and of course you can't get broadband without paying anything. It's only free for the first 12 months of the 18-month contract, and then it's £9.99 a month for the remaining six months. You'll also have to pay £16.99 a month line rental but if you're willing to shell out for a year's line rental in one go, this will be £185.88 (the equivalent of £15.99 a month). Once again, your wireless router is free but it will cost £6.99 in delivery charges. That means that you'll pay a total of £372.75 for the 18 months – or £354.75 if you buy the year's line rental at once.
Like the Plusnet deal, this isn't completely free either, just for the first 12 months. After this time, the broadband will cost £7.50 a month for the final six months of the 18-month contract. Line rental is £17.70 a month, and new customers will have to pay £6.75 for the delivery of their router. The total cost for 18 months will be £370.35.
This is another 18-month contract where you'll only get the headline deal of £4.99 a month for the first nine months, and then £17.50 a month for the remaining nine. Line rental is £16.99 a month and you'll have to pay £49.95 for the router to be installed. In total, you'll pay £558.18 for 18 months of fibre broadband.
It probably won't come as a surprise to hear that this deal isn't totally free, just like the TalkTalk and Plusnet offers. You do only have to sign up for 12 months though, so you won't have to pay any extra for any more months of broadband. Line rental is £17.40 a month and the Sky Hub router delivery is £6.95. For the year, you'll pay £215.75.
Broadband deals are constantly changing, so it's likely that these offers won't be around for long. In general, when you're checking broadband offers, make sure to check how much the line rental is. You should also see how long the contract is, and if that means you'll have to pay any extra after the deal runs out.
Finally, it's also worth looking at the costs for getting the broadband company's wireless router. It might be worth going for a deal with your current supplier rather than a slightly better one with another, as this means you won't have to pay for a new router to be delivered.


text:
This Feb. 20, 2019, photo shows the Oroville Dam spillway in Oroville, Calif. California officials say the flood-control spillway at the nation's tallest dam is expected to be deployed as early as Tuesday, April 2, 2019, for the first time since it was rebuilt after it crumbled during heavy rains two years earlier, forcing nearly 2000,000 people to evacuate. (AP Photo/Rich Pedroncelli)
Water flows on California dam's spillway after 2017 crisis
OROVILLE, Calif. (AP) — Officials at the nation's tallest dam unleashed water down a rebuilt spillway Tuesday for the first time since it crumbled two years ago and drove hundreds of thousands of California residents from their homes over fears of catastrophic flooding.
Water flowed down the spillway and into the Feather River as storms this week and melting snowpack are expected to swell the lake behind Oroville Dam in the foothills of the Sierra Nevada, said Molly White, principal engineer with the California Department of Water Resources.
The spring storms follow a very wet winter that coated the mountains with thick snowpack, which state experts will coincidentally measure Tuesday to determine the outlook for California's water supplies. Heavy winter rain and snow has left the state drought-free for the first time since December 2011, experts say.
The dam's main spillway "was designed and constructed using 21st century engineering practices and under the oversight and guidance from state and federal regulators and independent experts," Joel Ledesma, deputy director of the department's State Water Project, said in a statement.
"We spent the last two years restoring full functionality of the spillway. We expect it to run as designed," Ledesma said during a news conference.
The original spillway on the 770-foot-high (235-meter) dam, which is 150 miles (241 kilometers) northeast of San Francisco, was built in the 1960s.
In early 2017, storms drenched the state and the massive spillway broke apart as it carried heavy flows.
Dam operators reduced the flow and allowed water to run down an emergency spillway — essentially a low area on the reservoir's rim — but the flow began eroding the earthen embankment that had never been used. Authorities suddenly had to order an evacuation of nearly 200,000 people living in communities downstream.
The threat of a dam collapse that would unleash a torrent of water did not happen, however, and people were allowed to go home days later.
In January 2018, an independent panel of dam safety experts released a nearly 600-page report that blamed the crisis on "long-term and systemic failures" by California dam managers and regulators to recognize inherent construction and design flaws in the dam.
Repairs have cost $1.1 billion. California requested about $639 million from the Federal Emergency Management Agency for the fixes, but the federal government has rejected $306 million of those reimbursements. U.S. officials say the dam's upper gated spillway was damaged prior to the heavy rain two years ago.
Local water agencies are already paying some of the repair costs, and they would cover anything not paid by the federal government.
Rodriguez reported from San Francisco.
Accidents and disasters
Dam breaches