Hello, my name is Nicola Tecli and I am a Field Application Engineer with ST Microelectronics. In this video I'm going to give you an overview of the Digital Power Controller ST-Energy-011. I will start with the high level features of the ST-Energy-011, Digital Combo PFC and LSE Controller. I will then describe more in detail what there is inside the ST-Energy-011 and why it is an excellent choice for high efficiency power suppliers. And finally, I'll briefly show you the evaluation ecosystem available from ST Microelectronics. If you want more information about this device, I'll provide you with references to other videos and documentation at the end of this video. ST-Energy-011 is a Digital Combo IC that integrates both PFC and LSE controllers. It combines key features of typical analog controllers like high voltage start-up, integrated gate drivers, no firmware needed and it can achieve very low standby power. On the other side, it provides the key benefits of digital controllers like high level of configurability, remote monitoring and black box recording with UART and I2C communication interfaces. Let's now take a look at the typical characteristics of target applications for ST-Energy-011. The typical power levels are from 90W to 300W. Usually there is a main CPU that will communicate with ST-Energy-011 via I2C or UART and there are requirements for meeting stringent energy saving regulations. Based on those characteristics, the typical applications are power adapters for computers and for TVs, industrial and medical equipment, LED drivers that require connectivity for remote monitoring and control. Now let's go more in detail on what there is inside this device. In the block diagram, you can see the main building blocks of ST-Energy-011. On the top right, you can see the built-in high voltage start-up circuit that allows this controller to start without the need of an additional housekeeping power supplier. On the top left, you can see the gate drivers for PFC and LSE with 1A peak current for the most demanding of MOSFETs. These two features combined are the key to reduce the bomb of the power supplier. Most digital controllers do not have these blocks and require external components to take care of these functions. On the bottom, you can see the 8-bit microcontroller with 10-bit high-speed ADC to allow for precise control and monitoring of the digital power supply. The device also integrates a non-volatile memory that stores the numerous configuration parameters. Finally, this device has UART and I2C communication interfaces that allow the configuration of the digital power supply as well as remote monitoring or black box logging of the activity. Let's take a look now at the high-level functionality of the two logical macro blocks present in the ST-Energy-011. The first block is the front-end PFC controller. The inputs to this block are the output voltage and the PFC current information coming from the auxiliary winding of the PFC choke. The mode of operation of this PFC is based on a ramp-enhanced constant-on-time, which is a modified version of the standard constant-on-time with the proprietary ST algorithm to compensate for the current cause by the input filter capacitors. This algorithm improves power factor and THD. In order to optimize efficiency, the PFC automatically switches from transition mode at high loads to valet skipping at low loads and burst mode at very low loads. In addition, it also implements an optional skipping area mode that essentially stops the switching activity when the input voltage is around 0 voltage crossing. Skip area mode allows to improve efficiency at the expense of THD. The second macro block in the ST-Energy-011 is the LSE controller. The LSE controller is based on one more ST proprietary control IP called Timeshift control. Timeshift control provides two major advantages compared to a direct frequency control. First, dynamic load performance is greatly improved. Second, it provides anti-capacity mode protection. To evaluate this device, ST provides a fully integrated evaluation platform. In particular, we offer a demo board named EVAL-ST-Energy-011-150 with a GUI that allows customers to easily evaluate the controller on a real and working solution. The demo board delivers 150W at 12V and it accepts wider range AC voltage from 90V to 264V. The demo board comes with an interface board and a dedicated graphical user interface. The interface board provides a later consolation between the PC and the ST-Energy-011 board, as well as the ability to utilize a USB port to communicate with the ST-Energy-011 via the built-in UART. The GUI gives access to all the parameters of the device and it also provides real-time status of the power supply. There are a total of 85 different parameters that allow to tailor the power supply performance. These parameters include PFC tuning, LSE tuning, multiple protections, filtering and burst mode parameters. To close the overview of the ST-Energy-011 evaluation board, I want to quickly mention its performance. As you can see in the main table, the efficiency of the board is good across the full range of output power, but the most impressive performance is at low or no load. The no load consumption of 70mW at 115V AC and 93mW at 230V AC is really low and is very difficult to achieve with any other digital controllers. In order to help anybody who gets this evaluation board, I'm going to show you step-by-step how to access and modify the configuration parameters of ST-Energy-011. First basic step is to connect the interface board to the PC using the supplied USB cable and to the evaluation board using the 6-wire ribbon cable present in the kit. Second, make sure that the evaluation board is not connected to the AC line. Now go to the GUI and click enter ATE mode, present in the ATE mode menu. After this command you will see on the interface board a red LED turning on, which means that the interface board is supplying power to the ST-Energy-011 device. At this point we are ready to access the parameters. Click on NVM operations in the tools menu and then click read from IC on the NVM operations window that just opened. Please note that if you don't click on read from IC, the GUI will show you its own default values for each parameter, but they don't have anything to do with the values in the device. Now you can modify any of the parameters of your interest. When you are done with the changes, you need to write the changes by clicking on write to IC. After you do this, the parameters are written into the numbolatile memory and at next power on they will be used. In order to do a test with this new configuration, you need to exit ATE mode, which is done by clicking on go run mode in the ATE mode menu present on the main GUI window. Now you can connect the IC line and test the board with the new configuration. In this short video I provided you with an overview of the ST-Energy-011 device, but you can find much more information on st.com website and you can also find a longer video on youtube.com. For the video, please search for ST-Energy-011 on youtube.com. While for the documentation, please go on st.com and search for ST-Energy-011 or for the evaluation board EVAL ST-Energy-011-150. Thanks for watching and goodbye.


Hello everyone, welcome to another interesting episode of how to age the hackathon. Do you remember when we said that smart India hackathon is a platform which invites and bring the innovative minds from all across the corners of this vast country India. Today we have one such participant from a beautiful city of Madhavani, Bihar who was not only the winner of one but two editions of smart India hackathon. So let's welcome him and understand how this entire journey was. So without any further ado, Anurag, welcome to the show. Thank you sir. How are you? Sir, I am Mani. I have heard that you come from Madhavani, Bihar. So, anyway Madhavani is very famous for their paintings. You have won two editions of SIH, so this is also a very big thing for your area. Anurag, tell me a basic experience of how your smart India hackathon was. How did you start your smart India hackathon? Sir, actually, the culture of hackathons in our college was very good. My college is in Vitek, in M.I.T.Muchapur, which is a government college in Mujjopur. So, I was in second year at that time, so I saw that my seniors were participating and they were winning there. So, they got a lot of opportunities, like when I got a chance for a start-up, their project became a patent. So, from there, I got motivation that I will also participate next year. So, first of all, I worked on my skills as a team. There were some seniors, some juniors, some of my batchmates. So, the first time I experienced that we got the problems, we worked on them. So, during that work, we learned a lot of skills, we saw a lot of problems. We had the same thing that the project was getting ruined at the last minute. Then, we worked on it. Then, we got an experience of handling pressure. And during our evolution, even then, our mentors, the judges, told us to improve this. So, in a short time, we had to improve, we learned how to do time management there. So, this hackathon has improved my skills a lot. Along with that certificate, it is so valuable that I applied a lot of places, so I got a lot of offers. So, it is a very good platform for any student, for engineering experience. Very good. Okay, Anurag, I have to discuss this with you specially because many people think that in the tier one cities or in the colleges where there is a lot of exposure, if only the students of those colleges participate, then they have the opportunity or the chances of winning. They have this kind of knowledge in the minds of people. So, I would like to, because you, from where you spoke, even there, as you know, there was so much exposure of the hackathon. So, what would you like to say for our first-time participants, which are tier two, tier three cities? It is a very good question, sir. In this, we would like to say that in the year that I participated in our college, in the year 2022, our six teams were selected in the year 2022. So, this does not mean that if the college is tier three, tier two, then its selection will not be there. If you have done a good job, if you have done a good presentation, then your selection will definitely be there. So, this does not depend on the colleges. It depends on your work. It depends on how well you have worked. For example, there are a lot of colleges in Bihar, like Dabangah Engineering College, the government colleges that come in tier three, there is no ranking. From there, a lot of teams have been selected. My previous juniors and seniors, all of them have been selected. The final ones that we have won, we also won in the hardware division in the year 2022. So, I do not think that the college has any effect on this. Your work is the effect. You were right that at the end of the day, your work matters. College does not matter. Here, I would like to come to another question. As you said that your team composition was your seniors as well, some are your juniors and some are your batch mates. So, as a team leader, what suggestions would you like to give? As a team leader, all the points of contact are there. All the jewellery members, all of you would like to discuss the whole problem. So, I had to manage the whole thing. Senior juniors, juniors, juniors, juniors, juniors, juniors, juniors, juniors, juniors, juniors, juniors, juniors. During senior, you have to respect the senior. You cannot force them to do this work. I told them that I would like you to give me this much time. I told them that if you do not do this, I will face these problems. They took this seriously and did their work on time. It is important to have a good team. Sir, it is important to deal with any problem. Anurag, do you have mentors in your team? Yes, there were two mentors in my team, and they guided us very well. They shared their experience like we were making an app, there was a problem. They sorted it out with their experience and gave a lot of new suggestions. Mentors are very good in hackathon. You told me that you had an online experience and once you had an offline experience. How did you notice the major difference because it was definitely online and the second one was offline. How did you notice the major difference? Okay. During online, we got a three-digit service. When the judge gave us some input, we got a good time to solve it. But during offline, you get an evolution in the morning, in the evening, in the night. So you get less time, but the advantage is when you get a presser, you get more output. So the offline solution was better because it visualized you and your competitor was better according to whether it was good or not. The offline solution was better and we were more satisfied with the design.  Okay, tell me what industry experts came because you were in college and the evaluators were industry experts with a lot of experience in the industry. How was your experience and the first time you came and did you understand the benefits of mentoring in the industry? Yes, sir. The industry's mentors have a lot of input. Because you have made something for them. So you will see the presentation that I have done. But what they need is valuable inputs so if you have the results you will get more chances to win. So we did the same we listened to their points and worked on them and gave them the results so we improved in this way. We did the same thing. Finally, before I finish if I ask you directly that 3 or 4 such things you should keep in mind the upcoming teams which will help them towards winning. What will be the result? First of all this presentation should be good which you will submit students will submit that presentation should be good use more flowchart write less than less during the selection if your selection is in top 5 team why?  You will get improvements which you will get in less time. And if you get 5 inputs the jury member said you should do this in 5 minutes so you don't have to do it in 5 minutes if you are doing 3 then do it in 3 minutes so that the competition ends there. If you are able to understand the problems what you have done then this is a very good thing that you will get a lot of learning and you will get a good chance to show your skills. Yes, of course. Anurag, thank you so much you took your time and guided the upcoming teams and we wish you for future projects Thank you so much. Thank you so much. Thank you so much. So people that was Anurag for you so basically what understanding do we get from this talk session was it doesn't matter which city you are coming from it doesn't matter which college you are coming from the basic thing is if you have the talent if you got the idea you can make it through anything and finals has been one of those so for such more interesting and learning videos stay tuned with us and we will be back soon with more enriching and insightful videos for you till then have a good day see you.


Welcome, Sir Eric Kessels, one of the most creative people in the world, I think. I know we were just saying, last saw you in 2016 in Design and Dabble, and you did an absolutely incredible talk. And I think it was about failure. At the time, I think it was when, probably soon after you had launched this book, I imagine. But yeah, how are you doing? Hope all is going well. Yeah, very well. Yeah, thanks. And I heard you just got back from Italy, from doing an incredible talk. And that's sort of, I think this is a fascinating thing. And why I really wanted to chat is you've, unlike a lot of people who, you know, when we interview in the creative world and normally just doing advertising, you do so much more. I know you've recorded music tracks, you've done art exhibitions with photography, written multiple books, and run an incredibly successful advertising agency that's made tons of fantastic brands and great work. So congratulations. How do you do it? Well, do you wake up in the morning and just go, oh, I wonder what I can do today? Is he just naturally a curious person? No, no, I can be also very lazy also. I think to be lazy and a bit bored at sometimes is very good to load the battery. And I don't know, but I mean, I think you need to, I mean, when you really sit down and you have to come up with an idea, it really doesn't work, but it's more that you need to have an antenna for these things. And whenever something passes by, you pick it up and you do something with it. When you are feeling sort of uninspired, are there particular places that you normally go for inspiration? Yeah, I mean, a lot of people say that, for instance, when you take a shower or you're in the toilet or you're in an airplane, that they have the best ideas. But that's also true, I think, because it has more to do with the isolation. So if we literally look at how many minutes a day we are actually very effective in coming up with ideas, it's in a way very little. So you really need to isolate yourself at one point and concentrate and be really on your own with your mind to come up with ideas. And yeah, like for instance, when I'm in an airplane and yeah, to me, it's quite inspiring because there's no phone, there's no internet connection. So you are just down to the basics. And of course, later when you have the spark of an idea, you can use, yeah, you're going online to find materials to find other materials that might help this. But yeah, in basis, you really need to be very isolated. Yeah, I mean, I thought that was interesting looking at a lot of your work. It's often finding these these amazing sort of insights or lateral thinking that perhaps people wouldn't have seen. I remember hearing one of your stories about finding a selection of photos and was it in can of like a lady who like someone who had 400 photographs of one person? Yeah, that was in Barcelona. Yeah, yeah. Yeah, I mean, yeah, my my interest in general is in, I mean, I, to be honest, I'm not really into perfection or perfect ideas or perfect executions. So I always look a little bit for things in the French. And that's also where, you know, in advertising, as an art director, you know, I never used the typical directions or the typical sources to work with, you know, they are when I started, there was a, yeah, really, like, if you worked in advertising, you also worked with advertising photographers or advertising illustrators or and I think since I worked in London for several years that that changed a little bit because there there was already a bit more of an open atmosphere that you could work with anybody. If you would be if you find if that person found you the work interesting. So yeah, from that moment on, I always started to make different combinations and work with people that never did a commercial job before. And yeah, this is very important. And in that case, because of that, I got also interested in, yeah, for instance, in photographs that I found on flea markets, because these were done by amateurs, there were a lot of mistakes in there, even in the family albums when people had their eyes closed or when there was a finger in front of the lens, like that. Yeah, they still stick it stuck it in an album, and that was still there for many years. So yeah, this inspired me always the flaws and the mistakes that you find in people's behavior in people's behavior with the camera or writing things down or recording things. So that that's, I mean, the the imperfect is quite an inspiration. Because, yeah, like when you are completely perfect, and when your idea already has to be from the start perfect, it's not working.  Because nowadays, a lot of people only function and work in their front yard.  It's the finished work or the exhibition. But this is like the end result while, yeah, a lot of people don't really visit their own backyard. And the backyard is a place where there is unfinished projects where you walk around half naked because there's a fence around it. There's a lot of rubbish there because you are kind of embarrassed about your backyard, personal stories, fascinations. There's a dirty shed somewhere with a lot of stuff in there. You know, and these are the places where you really find your ideas and your so the ideas come from deeper passion secrets. You know, ironic things. And yeah, those things are more to be found in the in the backyard. I love the idea of sort of imperfection that you found in photography influencing imperfection that I think I see in your advertising, which makes it sort of stand out a lot. Like you often, I think one of the most famous campaigns you did was for a hotel company where it was sharing all the imperfections. Is it is it still going?  Yeah, idea. I mean, I don't really like advertising also. I mean, to be honest, I really hate it at 95% of the times. Because, you know, like, let's be honest, a lot of stuff is really horrible, you know, like a lot of compromises, you know, it has to be done to please someone, you know, there were a lot of hurdles on the road. That's what you feel, you know, and I think that, yeah, like, it's very stereotypical, the whole industry, and very opportunistic also. But, yeah, like that makes it also quite good to work in to break with these conventions and to really find another way and take side street instead of driving in the traffic jam on the main road, you know, like where everyone is. And that is, yeah, that's what I enjoyed also. You know, you work of course with the biggest talents in a way, because they are all grouping there in that industry because, yeah, there's money to be made. But, yeah, on the other hand, it's also good to be quite stubborn and to be try to do things differently. And that is quite possible if you communicate well about it, you know, I think that, you know, in advertising, a big mistake is that a lot of specialists are very good in communication, but they are actually not so good in communicating themselves about their idea towards a possible client. Or, I mean, there is where they often are afraid because an agency thinks like we have to sell this work because we have to keep this client and we have to do this and that. But, yeah, like, actually, when you are really taking someone by the hand and explaining why you do this and why this is a very good idea and why nobody has done this yet before, many clients will understand this and go for it. Are there certain things that you do to help people when they join your advertising agency or when you meet younger people? Is there a certain thing that you tell them to do to try and make sure that they do more of what you've just been saying? Like, are there sort of bits of advice or wisdom that you give them to make them think more? I mean, one of the most valuable advices is also for people that nowadays, for instance, start their own company or I think it's very important to say no. Like, you have to say many, many times no. You have to say also, of course, yes, a few times, but saying no to things, it's much more important because you are really, I mean, for instance, when you get a work offered and, yeah, it's good money to be made, but it's not really your thing, then it's better to say no because, yeah, like, all the work you do is also, that's your portfolio and that's what you have been working on for a long time. So, when you work for a year on shitty stuff, then, yeah, it's not very likely that the next year you will get suddenly a much better job offered.  And I think that the personality of a single person or of a company is also a bit underrated, of course, because I think that the personality is much more important than your portfolio. You know, like, I had, for instance, a job interview once and I was talking to this designer for first half an hour, then an hour and it was very, you know, yeah, it was really interesting talk and I really liked her, the way she approached things and her character. And in the end, I told her like, okay, when can you start? When can you start working? And suddenly she became very, she was blushing enormously. She became very embarrassed a little bit and she said like, yeah, that's fine, but you haven't seen my work yet. So, I was completely, I forgot completely to ask for her work. And, yeah, in the end, I saw her work. And this was also very, it fitted perfectly. So, it was already what I expected. But the fact that I didn't ask for her work was actually a very good thing because, you know, she had such a illuminating personality that this was, yeah, that was already enough. Yeah. Oh, that's awesome. What was the talking about the power of saying, no, what's the thing that you're most happy that you said no to? Do you think of anything? I mean, we had in Cassus Kramer several times, maybe two or three times, there was a client that was maybe 60% of our income at that time when we were maybe with 15 or 20 people. And, yeah, it started to become quite complicated to work for them. And work was, yeah, starting to go in the wrong direction. And, yeah, at that time, we decided just with a few people in the company to stop with the client, call the client and say, like, let's not do this anymore. And I mean, we informed other people in the company that we would like to do this. And in the end, everyone agreed, even though it could have been very risky for their jobs or for, but in the end, that was always the best decision because you keep a clean slate and you are very, yeah, it's very good that you are very clear about something.  I don't want to make rubbish. And I mean, life is too short for that. I mean, when in the end, you're not able to make good work, maybe you should find another job or, I mean, there's a lot of people complaining in that industry also. And I think like, why? Because you have a lot of opportunities. I mean, they don't see the opportunities, but there is, yeah, when you are clear about the path you want to take, it's good to stick to that.  That's a bit of a utopia.   The other way around it hardly ever works, you know, like, I think we had several times that we knocked on the door of a client that we would like to work for, but that hardly ever worked because you mostly entered the wrong door of that company or the wrong moment. Yeah. Well, I mean, you've obviously dealt with lots of different projects in the past. Was there any sort of particular peculiar briefs that you ever received that, you know, the most interesting or peculiar brief that you've ever received? And how did you kind of approach that? If you don't mind? There is many, yeah, like for instance, I think often like a company or a client comes with a problem and you know, they come to you because they think that you can solve the problem. But in some cases it's even good to think the other way around to make the problem even worse and that that is a solution to really solve the problem. And so those are always quite nice things that were made in the company, like for instance, two people in Cascremum that is, they were working on this job for a school and we did a lot of work for that school. But there was a small job because they had like a lot of problems in the campus with garbage. And there were certain fixed garbage bins there. And yeah, so that's a problem. And you know, you need to solve it. But you know, we decided to make it even worse.  So there were seven permanent insulations that made it very complicated to throw away the garbage. And this worked perfectly because the students, they liked this kind of approach and that we took the piss out of them. And suddenly the whole problem was solved but done by a very kind of upside down solution in a way. That's incredible. I'd love to see that. I'll have to look it up. What's the client name so we can Google it afterwards? Yeah, it was for this school called Koning-Willem-Ain Colleges. It's like a college in Holland.  See where you end up.  Sometimes you have to be comfortable to let go of things that you've created. Have there been anything, any sort of projects that you've worked on or ideas that you've found particularly challenging to let go of? Yeah, I mean sometimes it's, in a way, when you have ideas, you never let them go.  And whenever there's an opportunity, you find a way again. I remember that there was once, I mean, yeah, like we always wanted to make a film where everyone was naked in the film and we tried it once with a fashion company and so that everyone was naked apart from the two people that were wearing the new collection of that company. So then suddenly all the naked people were looking very weirdly at the two people dressed in clothing. But fashion company didn't really go for that. So later we found another, you know, opportunity for that. We worked for a health insurance company and also made a proposed a film where everyone was naked because, you know, with this health insurance you insured your most valuable possession, which is in a way your body. So that's why we showed how vulnerable in a way everyone is without clothes.  But yeah, so we tried this and the client, you know, said like, yeah, sorry, but I'm not going to propose this to my board. And yeah, so they didn't go for it. Two weeks later we proposed another idea with that client and yeah, they liked a new idea. But then he said like, yeah, but I have to admit that in the last two weeks I was still thinking a lot about your first idea. And I want to come back to it. So in the end, we made that first idea and it ran for two years on television and yeah, it was really, really nice. But it's more like to show that sometimes also when you really believe in an idea, you can also show to a client that you're really sick, sick about it, that they don't go for it. And sometimes then you still need to let it go. But yeah, then you put it in the parking and wait for another opportunity. Yeah, have an idea vault is something we often talk about. So this idea of keeping the things that don't work out, maybe they'll work for another client or for another time. Yeah.  And I guess there's, you know, could you live without one or do they both inform each other? Yeah, I mean, since 2001, I have done a lot of things. I have done like more my art work and commercial work. I've done that simultaneous in a way, even though they are completely different things. But since, yeah, maybe three or four years now, I am not so much involved in the or totally not involved anymore in the day to day work in the in the company. And also because like a new generation of people that I have worked with in the past, you know, sometimes for seven or eight years, sometimes for 25 years. Yeah, there was a good opportunity for them to take over and to run this themselves. But they are, you know, they are, it's very nice for me also to see that they are in the DNA of the company, because, you know, I don't know completely everything that they are working on. So sometimes I see something passing. And I think like, wow, that's a nice idea. And then I look on our own website and it's something that they have done. So that that is quite, quite nice. But I mean, yeah, this is very nice that you kind of continue the family and because, yeah, people always stayed and still do for a very long time in the company. And yeah, there's there's many people who have who have been there already now for 25, 26 years. And some people recently a girl left after 20 years working with us. And yeah, so that's a really good sign also. And for, yeah, for the company much better to continue like this than to sell it to, yeah, and people that don't have any connection with and roots with with the style and the ethics of the company. What do you think are some of the reasons why people stay for such a long period of times? It seems to be in the creative world, it seemed like the churn in the creative world tends to be very high in the advertising industry. So, you know, fascinated to learn what what you think helps people to stay longer, the particular things that you do or a particular culture that that you helped grow there. Do you think that work? Yeah, I think yeah, one of the most important things is also honesty, you know, like you have to be very honest and very direct to the people you work with. And yeah, there is no politics in the company.  I mean, at the time when we started this was quite new, not now that's not that new anymore. But anyway, yeah, like in in a group like that, you also see and you feel after only after a few weeks, you feel already if someone fits there or not. And yeah, by now we have kind of an idea, what kind of people that fit there. And yeah, so it's also people that, you know, that have that have certain interesting stories with them, apart from the work they make, also people that are different in what kind of hobbies they have, what kind of fascinations they have. Yeah, I mean, you know, the contrary of what I was just saying of hiring someone that I didn't ask for the portfolio, there was also one case where I looked at the portfolio was very, very good. It was one of the best portfolios I've ever seen. And then I got kind of curious. So I started to discuss with this person more and more why he was doing this. And in the end, it really after an hour, it turned out that he was actually making the portfolio because of that he wanted to work with us or that he wanted to work in some other places where he thought that this work would be really fitting. And but yeah, he admitted also that in a way, yeah, he talking about front yards and backyards, his backyard was was was kind of similar to his front yard.   Like, you should not be normal because it's, I mean, everybody is hyperactive, everybody has a bit ADHD, or it's a bit autistic. So this is normal, you know, like, it's, it's like, it's almost being a controlled madhouse in a way where you have to take care sometimes of people and people take care of themselves and, and of each other. So no, but I think the authenticity of people is very important. And yeah, it's good. So it's like a lot of that work, I guess it comes up front in these interview processes where you just get a good idea of people. Yeah, it's nice. So I guess you probably don't leave that to an HR team in the back office. I'm guessing you do that with with the teams of people who they'll be working with or yeah, it's HR is sorry. Yeah, I think even in a very, very big company that HR is like, completely overrated, I think that the director of the company or should should hire people, you know, like, this makes the culture of a company is very important. Yeah, yeah, makes no sense. And then you've got this amazing business that's also sort of a publishing company now as well, right?  I remember rightly what have you always been fascinated in photography or did it just pop up to him? So my dad was a photographer, I'm a big fan myself. I think, yeah, it has two origins, in a way. One is what I just said in the beginning that I worked with a lot of photography as a art director. And I was asked also in that, you know, working with photographer with for instance, Magnum photographers at one point I was asked to make their edit some of their books. And yeah, so in that case, you, yeah, you are quite deeply into that industry in a way. The other reason is maybe also because yeah, I was in my youth, I was like, when I was 11, my sister was nine. And she yeah, she had an accident and she died because her car drove over her while she was crossing the street and he drove through a red light. And yeah, so this was a very unfortunate thing. And I remember at that time also vividly that my parents, they were looking for her last picture and the last picture that was taken of her. And they found one which was actually a picture from a holiday park where the whole family was. It was like a color image where that was taken by an anonymous photographer who worked in that park and leaving the park, you could buy that print and take it home. So that was the last picture of my sister at that time. And my parents decided to crop that image and make it take another picture of the image in black and white and print it in black and white and enlarge it and put that in the living room. And yeah, that is for me also something very important that, you know, in this case, my parents reappropriated that image and made it like a very, it's a very mundane image. When anybody else would look at that image is very bit out of focus, not really well cropped. It's a terrible image in a way. But for three people in this world, this image is very, very important and almost iconic. And that is also what fascinates me, you know, how you can reappropriate images, take them from its original context, bring it into a new context and then suddenly the meaning changes and suddenly people really start to look at it again instead of consume images. So yeah, you could say that also that we are now living in a period where we consume images in a rapid tempo, but it's quite good to, I think that nowadays, that what interests me that the stories behind a single image or a series of images nowadays may be more important than the image itself. A lot of images are photographs have been made already. There are thousands of memes or ways of people how they photograph things, selfies, their food, their travels, their certain landscapes, certain filters, it's all the same, you know, like, and even in advertising is also the same, like if you could look at online advertising, how a hotel should be photographed from the inside, how a car should be photographed. So these are all repetitions. And for me, it's interesting how I can look for certain outliers and certain things in the fringe and to really put a finger on that and point it out and ask people to look at it. I'm so sorry about your sister and I think there's something sort of lovely in there about the power of imperfection in an image and that almost makes the picture more meaningful, which fits, I guess, with what you were saying about what's your back garden or your backyard look like. It's the imperfections where you often seem to find it. That is my backyard, part of my backyard. Yeah, that's amazing. I mean, what's your most recent project? What's that been focusing on? Yeah, now I'm working for three or four years already on a bigger European project called Europe Archive and it's a website called Europe Archive.eu. And this is a more anthropological project where, yeah, I mean, nowadays people migrate a lot in and out Europe and also within Europe. But the objects that every country lives with and that are sometimes typical for a country and also in the heritage and in the tradition of a company, they also migrate through Europe and they are not really fixed to a certain country anymore. So in this archive, I do this together with Thomas Mylender, a friend of mine, and we collect really objects from that we specifically search online or on flea markets but we buy the objects physically and yeah, we recreate a collective memory of that country. And that is, yeah, we had now in this summer an exhibition with it in Hungary, Hungary. That was the first exhibition and so now this archive is growing and it has like, yeah, we'll have more stops in Europe. I think in, I don't know when this will end, hopefully not soon, but the project is growing a lot and yeah, I mean, it will be also iconic at one point, I hope, because it's really nice to pause these things and collect them and show them what these are sometimes hilarious things. They are not expensive, they are with certain traditions, things that are repaired. I mean, for instance, one of the most brilliant things I found in Romania, it's like a frying pan that the handle was broken, but then they found like a Jesus that was also broken from its cross and they repaired the frying pan with a Jesus handle. So yeah, like these are very strange habits and or yeah, like in, we found in Hungary like a cooking pan on the flea market, but it was filled with mobile phone parts. So it looked almost like a vegetable soup with all these colors in there but yeah, actually it has something to do with the fact how quickly we consume technology at the moment and yeah, then it ends up in a pan, in a red pan in a market and yeah, so there's also on the website, in the archive, in Europe archive, there's all the countries and there you can find also stories with every object that we found. So that is a very, yeah, that's where I'm on and off working on. Yeah, it's incredible. I opened up the site and looked now quickly, it looks fantastic and I love the stories that go with each of the images as well. So yeah, highly recommend it, Europe, archive.eu and very sad that we can't have the UK in there anymore. But yeah, I've got Ireland, so close. We'll have to rectify that at some stage. But then, yeah, the other thing I was going to say was that in so much of your work, there's so much of it is absolutely hilarious as well. Like often the things in the failure book I've got behind me here and the complete amateur guide book that you've got as well as some of those I was cry laughing going through that.  I mean, is it just, you know, just very, very good at observing these things? Because I mean, they're all kind of seem to be out there in the ether. Yeah, but it has also to do with that what we call as so-called like normal life or normal people or ordinary life or ordinary people. Yeah, there you can find the most hilarious things, and life is not normal.  And I mean, that's also what you see, of course, in history of filmmakers or documentary makers or, you know, like there is, yeah, for instance, like in England with Ken Loach, for instance, the filmmaker, yeah, he also focuses on how absurd in a way normal life is. But also in, yeah, you can, when you have an antenna for it, you find things on the street, you sometimes you hear like crazy conversations like a part of a conversation or like a note that someone dropped on the street and, you know, things out of context are always good. And that is also what you point out just now because there also you take it out of its original context, put it in a new one and then suddenly it is something hilarious. So that is often the trick. It's amazing. I know we're running out of time. So I'll try and ask one or two final sort of closing thoughts. But I mean, what sort of advice, I guess, would you give to people who want to be able to look at the world in a slightly different way? I mean, you seem to have a real knack at it. Everything that you do has always been, I mean, personally, I just find it always very inspirational and very interesting. But I don't find many other people who look at the world quite the way that you do. So thank you. This is also, yeah, for me, and that's maybe also the reason why for me, when you say these things, I think like, okay, for me, it's more like the, you know, it's not a planned thing, you know, like with the company, with my art career, you know, when I would have a certain career plan in that, it probably never would happen.  And that is really, in a way, quite a nice thing, not at that moment, because I think like, shit, how many times, how long do I do this and why Kanda have this idea earlier. But at the end, it's very good, because it's good to be, to behave like a beginner. Because, you know, nobody is born with the talent to become a fantastic creative or brilliant artist. I mean, you don't have a talent for that. It's like, really, you need to develop your skills in daring to let go or daring to be very intuitional and not panicking. And of course, these are things that you can learn. But, yeah, like, and these are the important things, why something happens and why you suddenly have an idea, because you leave the space in your head to allow to have that idea. That my brain is not different from yours. But, or from others. But, yeah, sometimes you need to be very, almost naive, stay very naive in that. So, yeah, so those are things, I mean, yeah, you should not be afraid of having not an idea, because but you, if you have the confidence that it will once come, and you have the feeling that you can trust on your intuition. Those are good things. Yeah, it's fascinating. It kind of reminds me of sort of get comfortable with the uncomfortable and be happy or be happy with a beginner's mindset. And I love, yeah, I do love that you often find there's beauty and imperfection. You seem to be an expert at doing that. And I hope that you carry on finding these these beauties and all these wonderful imperfections that we have in the world for many years to come. And thank you for finding them in there. Thank you so much for taking the time. If you anyone listening, please do go check out Eric's website. He's got a brilliant website. You can find lots of his creative work. You can also find links to all of your fantastic books as well. It's just EricKessels.com. Google search will get you there very quickly. But yeah, Eric, thank you so much. Really, really appreciate your time. And yeah, have a good one.


It's time to start. So let me remember our notation. So we have a point x in our n, capital N. Here, we do not distinguish a special variable time with respect to the others for the moment.  Where we know that the coefficients dI and c are sufficiently smooth. And also, f is given and is sufficiently smooth. And we have given also a hypersurface sigma, a smooth hypersurface. And then we have found a way to solve this problem coupled with u equal u bar on sigma. Say an hypersurface contained in omega, omega in open subset of our n. Then we are considering this in omega, in omega. And this on sigma, and at least for, well, depending on the PDE that we are considering, but at least around sigma, we were able to find a solution of class C1. So at the exercise that I gave you the homework was this. So x1 du over dx1 plus 2x2 du over dx2 plus du over dx3 equal to 3 times u. And then this is, say, in R3, say, maybe.  So u bar actually is a function of x1. So could you please tell me the solution that you have? u bar. u bar, yes? Expand into minus x3. u bar, let me see. x1 multiplied by e to the power minus x3. Minus x3? Yeah. Comma. Comma. e to the power minus 2 x3. And now the plus y is going to 3 x3. So let me check. This is correct. Yes, it's correct. So but I would maybe, was everybody able to do this? Maybe it is better that I try. I do it here. Is it OK? So I correct it here. So this is correct, by the way. So we are in the following situation. We have n equal to 3. Sigma is this hyperplane. By the way, this is, again, in a more standard notation, this would have been written as follows. So this is the u over dt plus in a more standard notation, maybe like this, say d of x, rad u of x, equal to 3 u. Where now the notation means that this is the gradient with respect to x1 and x2. v is equal to b1, b2. So this is, of course, a different notation. But you see, usually such kind of PD are written as follows. But anyway, this is the notation. Maybe more similar to the first lecture. So let me come back to this one. So n is equal to 3. So b1 of x is equal to x1, b2 of x is equal to 2x2, b3 of x is equal to 1c of x is equal. With this notation, c of x is equal to minus 3 is a constant. And f is equal to 0. This is the situation of the PD. Of course, when I wrote here, now I have erased, but this b was a different b with this. You see, now this b has three components. But in this way of writing the PD, it has only two components. But I'm using the same symbol, but it is clear what I'm meaning. Anyway, so b has three components. c is constant, f is equal to 0. So let me rewrite down the system of characteristics. So x dot is equal to b of x. y dot is equal to, now I think it is minus c of x, y, OK? Well, but c of x is constant. And then I have x of z. Well, dot means b over ds, x0 and y0. So this is a vector with x1, x2, x3. This is a scalar object. And now x0, let me call this just a point x bar on sigma. And this is just u bar of x bar. This is the system of characteristics. So now this can be written more explicitly. So x1 dot must be equal to x1. x2 dot must be equal to 2x2. And x3 dot is equal to 1. It is clear, by the way, that the important transversality condition is satisfied. Before doing this, maybe it is better to recall that this vector field, as usual, has, so b of x is equal to x1, 2x2, and 1, which is, of course, transverse to the horizontal hyperplane. Because the sigma is just this. It's a plane, sigma horizontal. And b of x has the third component, which is always 1. So b of x is something here, but never parallel to sigma. So we met the important transversality condition between the vector field b and the given hyper surface sigma. So b is transverse to sigma. Then with difference, could you repeat the difference between a plane? Well, first of all, plane is a sort of special case of hyperplane. So when we are in n dimensions, hyperplane means a flat object of a dimension n minus 1. So it's a generalization of a plane in space. So what is a hyperplane? In this case, a plane, because here we are exactly in three dimensions, it is clear. Now, the hyper surface is a sort of plane, but that is an object is a surface, say, which locally, around any point, is like a plane via a map and embedding map. Meaning that locally, so the definition is sort of this. If at point p in sigma, so this is sigma, it means that locally there is a map from an open set u into, say, if this is n minus 1 dimensional, or for instance, two dimensional, then there is a map phi from an open set over 2, say, into p into sigma such that phi of u is equal locally enabled, say, enabled n intersection of sigma. So it's just a piece, a local piece of your object. But n phi, say, in this case, say, is c1. And it has a differential, which is of maximal rank. So it's sort of called an embedding. You take a small disk, and then you deform it into a higher dimensional space. And then you require that if you do this with another phi, then you require that there should be some compatibility condition between the two local parametrizations. You have to think about a sigma like an object which is locally very, very similar to a plane, just a little bit deformed. So this is not a problem here, because our sigma is really a plane in this exercise. So therefore, x1 of s is equal to x1 of 0. And x1 of 0 is a point, let me call it x1, small x1, x1 e to the e to the s. x2 of s is equal x2 of 0, which is, let me call it, a dx1 bar, so x2 bar e to the 2s. And then x3 of s must be equal to s and must be equal to, well, c2. Because at time 0, s is equal to 0. So this is the solution to our first part of the characteristic system. So this is solution of s and this. Next, we have to solve for y. So you see y of s. So y of s is this c was minus 3. So this is equal to e to the 3s times u bar of x bar. So this is the solution to our second, so remember, the system of characteristics consists of three equations. Now this is three equations. And this is just one equation, vector and scalar. So we have now, remember, the delicate point here is now to invert the map. Because we have the solution x, we have found explicitly the solution, which is a solution of depending on what? Depending on s, of course, and on x bar. And then we have also y, which is a solution depending on s and on x bar. Again, you see the third variable can be identified in the previous language with time. So now we have the following map, say, sx bar into xsx bar, which is a local diffimorphism between rectangle here, sort of minus delta, delta around, in this case, around sigma. And then a tubular neighborhood of its image, and is the image of this rectangle through this map, which can be inverted, by the way. So any x can be obtained as s of x and x bar of x, if you want, in a unique way. So we have, and then the solution, remember, u of x, the characteristic, teach us that the solution is y of s of x, x bar of x. So we have to write down this. So we have to explicitly invert this. So we have to solve the following. So x1 is equal to x1 bar into the s. x2 is equal to x2 bar into the 2s, which 3 is equal to s. And so we have simply to find s and x1 bar and x2 bar in dependence of this. So you see, x1 bar is equal to x1 into the minus s. But s is equal to x3. x2 bar is equal to x2 into minus 2s, equal to x2 into minus 2x3. And s is equal to x3. So you see now, given x, we find s of x, x1, and x bar of x. This is x bar of x. So given x, I have s of x, which is equal to x3, and x bar of x, which is this. And so our solution, so u of x, which is equal to y of s of x, x bar of x, is simply equal to now u bar. So first of all, there is e to the 3s. But s is equal to x3. And then there is u bar. x bar 1 is equal to x1 e to the minus x3. And x2 bar is equal to x2 e to the minus 2x3. So this is our initial condition. And this is the solution of our problem, which is correct. It's exactly the same that he said before. So you see the importance of the method of characteristics. So remember that for all the PDs that we have studied up to now, meaning essentially we have studied the following. Let me change a little bit the notation that I prefer. But we understand what this means. u plus c of tx u equal f. For such a kind of linear first-order PD with this new notation, we are able to solve it using the method of characteristics. I hope that this for the moment is OK. I hope that this for the moment is OK. So now let me do some remarks concerning what it would be interesting to do in connection with this PD and that we will not do. Adjust the comment to give you a large view, a more large view on what are now the interesting problems related to this. Even if this is, of course, this is a linear PD, first-order. So maybe the simplest you can imagine. However, still there are difficult problems. One of the problems is that it becomes rather interesting to assume that b, take also f equal to 0 if you wish. And take also c equal to 0 if you want. So just for this very simple first-order homogeneous and without the dependence on the 0 or just this first two objects equal to 0, it is rather interesting to see what happens where b is non-smooth. Also b depending only on x, for instance. So when b is not smooth, say, for instance, also b depending only on x, just even for simplicity, autonomous case, then what we have set up to now doesn't work anymore. It doesn't work because we were based on the system of what is that we know how to solve it, at least when b is continuous. Right? Even more, when b is locally leap sheets, we have uniqueness and so on. Our case was bc1, actually. So bc1 no problems locally. But if now b is not smooth, for instance, say, also discontinuous or less than continuous, then it is really not clear how even to solve this system of what is. And therefore, the method of characteristic is not clear how to implement it. So this is an interesting, further problem related to this first order PDE. OK, this is a remark. Why it could be interesting to take b which is not continuous? For the following, for instance, for the following reason, we can change and generalize that problem to a quasi-linear problem, for instance, of this form, ut plus u, ux equal to 0. OK, this is not linear anymore, right? Because this coefficient does not depend on tx, but depend explicitly on the solution itself. OK? So this is not contained in this. But assume for the moment that you want to look this as a coefficient. Now, it turns out that this equation in general has solutions which are not c1, even possibly also discontinuous. So if u is discontinuous, and if you look to this as a coefficient, now you see that you have ux against the discontinuous function that you could call b, just for simplicity, even if it depends on you. So you have the problem of multiplying this continuous function against something, which is ux in some sense. And so, I mean, this is somehow understanding this. Maybe it is related to also to this nonlinear PDE. That is why one of the reasons for which it is interesting. Another remark related to this, this equation. So I mean, this means that I'm trying to convince you that there are motivations for a further study of that when b is not too smooth, just motivations. We will not do this here, but there are. OK. Then another remark, which will remain just a remark for the moment, is that assume that we have just in one dimension, say this. So the point is that when b is not smooth, what does it mean to be a solution? Why? Well, because if b is not smooth, we expect u to be not smooth. And so if u is not smooth, what is ut? In which sense, I have to understand this equality. Because up to now, there are no problems. There were no problems, because ut, u was c1. And therefore, ut is continuous. And therefore, this is defined at each point. And the same for the gradient. Gradient is continuous. This is defined at each point. And this means that at each point, tx, this must be equal to 0. So there are no ambiguities when u is c1. But when u is not c1, which is maybe more interesting, what is ut? Well, of course, we cannot talk about ut at any point. Maybe at almost any point, if you are lucky. But not at any point, for instance. And the gradient of u also has the same difficulty. So now you see, when this is not smooth, there is even the definition of solution is not clear. Even the very initial definition of solution. You have to declare what is a solution. Otherwise, you cannot continue. So one possibility to declare what is a solution. Now I am anticipating a little bit. But I think it's good for you because this shares some lights in the theory of PDEs. So it's just an anticipation. So one idea could be, assume that I have this. And let me multiply this by a very small function. So let me multiply this by a function phi, say, which is c infinity. So now I don't couple with any initial condition. This is just a qualitative discussion. So let me multiply this by c infinity function with compact support in space and time, whatever. So let me formally do this computation. I multiply this. Now everything is formal. And then I integrate. Maybe I integrate, say, in spacetime. You see, now if I am integrating in spacetime, I can, using the integration by parts in time, I can write at least this first part as minus. Now this is a double integral in spacetime. But first I integrate in time. So I make integration by parts. And then I integrate in space. So this is minus phi t of u, say. And now, by the way, this is well defined. Because phi is very smooth. Compat support. So when this is defined, at least when u is in l1 lock. I mean, this must be an integral in the Lebesgue sense. So compact support, u in l1 on compacts, so l1 lock, this is well defined. So for the moment, this is clear. The point is that I cannot do this trick here. Because you see what happens if I now integrate by part in space. Now assume we have one spacetime dimension. So I want now integrating space. Now what happens? It happens that now I have u, this. This is not so easy to do. Because if b is not smooth, I have the same problem. So I don't know what is bx. So this is another difficulty related to this PDE. And the point is that this PDE is not written in the so-called divergence form. Anyway, this is, you see, this trick of multiplying by phi and integrate by parts is not clear what I have to do. Because now I have bx phi times b phi x. b phi x is OK, but bx phi is not so easy. So what is this? And so you see, maybe for another equation, u, t, let me just modify a little bit the equation. So let me assume now that I consider another first order PDE, where now let me call the unknown rho. And assume that v is a given function, even not smooth. But now I assume that my PDE has this form, which is slightly different from the previous one. This is called the continuity equation, given v. Assume v is given function. And the unknown is rho. This is slightly different from the previous one. For instance, if I multiply by phi as before, and then I integrate, now the situation is slightly better. Because now this is equal to minus phi t rho. And this is equal to minus phi x v rho. Phi has compact support. Assume that we don't have boundary terms and so on. But now these are well-defined, because phi t is smooth with compact support. Phi x is smooth with compact support. To have a meaning of this is just enough to require that rho is in n1 log. Rho in n1 log can be maybe this product v times rho in n1 log, whatever, say, both in n2, rho in n2, v in n2. Anyway, for this continuous function, this has a meaning. And therefore, it seems to be rather natural to define what is the solution of this. It is an object, a function rho, such that this minus this is equal to 0 for any phi in that class. I'm slightly very anticipating the problem of what is a solution and what could be one way to define a solution. These are called generalized solutions. More precisely, they are called a solution in the distributional sense. In the distributional sense. I am not very rigorous, but I'm telling you that a solution in the distributional sense can be defined as follows. A function rho, say in l2, such that this is equal to 0 for any phi smoothed with compact support. You see, in this expression, there are not any more derivatives of rho. All derivatives are on phi. No derivatives on rho. That is the trick. By the way, this trick doesn't work so easily in that case. Because this is so-called equation in divergence form. The derivative with respect to space is outside. And so I can put this x on the test. The phi is called the test function. I can put this x here. But if the derivative is not outside, if I have this term, this trick is not so immediate. So anyway, this is just to tell you that the theory of linear transport equation is rich and continuous. For instance, in the direction of no smooth vector phi is b. Let us continue toward the generalizing for the first order phi is. So let me again insist on the previous problem. So let me write it as, no, in the old form, equal to f. I want to do some geometrical comments on this pd. So we are in the previous notation in our n. So n dimensions, b equal b of x, c equals c of x. So there is no special time in this way of writing things. So we have the following. So now maybe it is better to be slightly more precise. So this is sigma. Now you see sigma is in omega. This is an hyper surface. We have always considered a hyper plane. But now assume that sigma is not an hyper plane. It's just really also curved. It's an hyper surface. So we know I need an open set here, u. And I'm veriting that, at least local phi. So my notation are to call. So now I have this also minus delta. So this is u, by the way. The yellow is u. The image is this. And then I have minus delta delta. And then my notation, I call this a point sigma, point here, sigma. And s is a point here. So the important map is s sigma into x of s. Which is the difference between this and the previous discussion. The previous discussion was that phi was the identity. Because sigma was a plane. So phi was the identity. And it was identifying this yellow object with its image. But this is just the difference. But now that sigma is curved, I need to locally describe it in some way, for instance, using a parametrization phi. So previous phi was the identity. So we have this differmorphism. Let me call this d. So this is the embedding map, phi. But we have a differmorphism taking this sort of big rectangle, or higher dimensional rectangle, into a neighborhood of this d. And this, let me call this n, n. And so d was a differmorphism which was invertible. And the inverse, the differmorphism was given by this. Capital X was a solution of the system of characteristics. Can you remember? So this can be inverted, and so on, and so on. OK. Now let me, so this is the picture at the level of rn. OK, is it clear? Is everything clear? It's very important to have, to write explicitly, the solution of the system of characteristics x dot equal b of x of 0 equal a point here, parametrized, so phi of sigma. So x of 0 is here, and this point is coming from a parameter here. So this was called x bar, but x bar is the image through phi over parameter. So let me call it. So now everything is considered as a function of s and sigma. OK, so x, is it clear? This is very convenient. It is convenient to have a fixed parameter space in a flat parameter space rn minus 1. And then I have an orthogonal component, s, which is the time of parametrization of each orbit. In the physical space, I'm starting from here. This is a point x bar. It's coming from here. And then my vector field was, by the way, transverse. So my vector field was transverse here. And then my capital X are just curves, integral curves of the vector field. This was the situation at the level of rn. Now let me rewrite this PD slightly differently. Of course, the equivalent way, cu minus f equal to 0. Or maybe let me rewrite it in an obvious equivalent way, minus f minus cu equal to 0. What is written here is that the scalar product. Now, again, we have to be a little bit, we have to understand on the role of the symbol. I mean, this dot is the scalar product in r capital N. Now let me write this b against grad u minus 1 equal to 0. Where now b is equal to b f minus cu. Namely, more precisely, b of x and y is a new vector field with one component b of x f of x minus c of x y. So this is rn in rn. This is scalar. And so given b f and c, I can introduce a new vector field in a higher dimensional space. So capital B now is a vector field consisting of a horizontal component, which is this n dimensional, and a vertical component, which is a number. So I'm increasing the dimension because of u, essentially. And then it's somehow natural to define this capital b. This dot here now is the dot in rn plus 1. Sorry for the same notation, dot in rn plus 1, but it is clear. So you see now that this can be written as capital B is now b and this. And now against this new object, grad u minus 1. And you see that this is equivalent clearly to this. Do you agree? But now this is a geometric condition on the graph of u. So maybe it is better that I'm trying to do a sort of picture. So remember, this picture is at the level of rn. Now I have to make a picture in rn plus 1. So I erase this, but you have to understand what I'm doing. So now I need the picture in rn plus 1. I will try to do this in a sort of understandable way. If you don't understand, please let me know that we try to adjust the picture. So I try. Now, so here we have the variable y. Now I am taking, say, n equal to 2, maybe. And this is sigma. Sigma was yellow, so this is sigma. So for simplicity, sigma is just a piece of hyperplane. In this case, we are into dimension. So n minus 1 is 1. So this is a piece of line. It's not curved anymore for simplicity. You should think about a curve here. But for simplicity, for making the picture, I take just a piece of line. Now we have, remember, the vector field b, that here it is horizontal. So this is the vector field in the horizontal plane. It was b, small b. So let me omega is here somewhere here in this horizontal plane. So the previous picture, which was in the blackboard, now must be imagined in this horizontal plane. This is the important thing. The previous picture, above view, now becomes a picture in the horizontal plane. Now this is b in the horizontal plane transverse to sigma. And over this, we have the graph of u bar. u bar was the initial condition. So now let us try to depict the capital b now. What is the vector field capital b? So we are now in one dimension more. So b of x, y is equal to b of x f of x minus c of x, y. So we see the capital b is somewhere here around this. The horizontal part of capital b is the same as small b. But then there is some sort of vertical part also. So maybe capital b has a sort of vertical part. So say, I don't know, something like this. By the way, capital b is non-vertical.  So this is never vertical like this, capital b. And moreover, it is capital b restricted to the graph. This graph is non-tangent to the graph. This capital b, I repeat, if I look at it, just capital b is defined everywhere in this big space. I am interested in it locally around this. But if I restrict it on this graph, the graph of the initial condition, this is non-tangent because small b was transverse to sigma. It's non-tangent to the graph of u bar. OK. What this equation is saying, what is this? So u is a solution now. Assume that we have a solution. We have constructed it, by the way, OK? What is saying this condition? Can you see it? What is this in terms of the graph of the solution u? Is the is a normal vector. Maybe it's not of length 1 because this is not normalized. But it is parallel to the normal to the graph of u, OK? So minus grad u, so remark, minus grad u 1 is parallel to a normal vector to the graph of u. So this equation imposes something. It imposes that b is tangent to the graph of u. This says capital B orthogonal to the normal. Equivalently, capital B tangent to the graph of u, OK? So b tangent to graph u on graph u. So our solution u must be equal to u bar on sigma. And then must have a graph such that capital B is tangent to this graph. So I'm trying to now maybe to, I don't know if this is clear or not, this is a local part of the graph of the solution u. So this is capital B. This is capital B. And this is the graph of u. So this graph is obtained as a union of curves, which are essentially integral curves. So let me rewrite the x dot equal to b of x, y dot equal to f of x minus c of x, y. So the solution to this, there are curves into Rn plus 1. And you see that the tangent to these curves is exactly the vector field capital B. So these are the integral curves of the vector field capital B. Because this is right inside this capital B. So these are the tangent vector to the curve, tangent equal capital B. So this means that the curve has a tangent direction, which is capital B. And therefore, the graph of the solution is just the union of these curves. So the point of the, so there is a geometric interpretation. So you have given this graph. And you want to find another, so this is however co-dimension 2, because this is a curve in R3. And then you look for a surface in R3, which meet this path, because you have to start from here. So you have to touch sigma, this graph of you. Because initially, you have the condition this. And then you have to have this property that the tangent, that the vector field capital B is tangent to the manifold, to the red manifold, which would be the graph of the solution that we have defined. OK, so this is the graph of you. No, this is no, sorry, this is graph of you. No, this is graph of the initial condition. And this is the graph of you. OK, this is the geometric interpretation of our first-order PDE. This is maybe, so is it clear? Now it's not so easy. The important is that the end, the transversality condition of B here, so that we can really somehow, starting from this curve, continue it into a surface with properly suitable properties.  But maybe it is better to have also this in mind to do both at the same time. Is it OK for the moment? OK, this is important because this is the starting point of the next nonlinear PDE. So we pass now to another case, more general, that we write this as follows. So if this is clear, I can continue. I know it is not easy. I know that this system of characteristics is not easy. In particular, this map that you have to invert. But anyway, so now let us now continue by considering the following more general PDE. So let my notation like this. Now I from 1 to n, B i of x. Now B i of x and u of x now times u over dx i of x. OK, and this must be equal to f of x u x. So now this is a non-trivial generalization of the previous case because our coefficients now depend on u and not only on x. And therefore, the equation is nonlinear anymore. Maybe this is called quasi-linear because at the end, you see, the coefficients depend on u, but they do not depend. I mean, the nonlinearity is due of this. It's not the case that you have the higher order derivative square, for instance. This is not the case. So this is called quasi-linear, not really linear, but not nonlinear in the higher derivative. Let me say like this. So if you look, if you freeze this, this is still linear in the higher derivative. However, global is nonlinear in u because the typical case, as I told you, is in the other way, time, space. The typical case that you have to consider is this. Say, OK? And this has also a name. Maybe another way to rewrite this is now. So it is nonlinear. Let us rewrite it like this. However, it has a sort of divergent structure now. Very convenient. This has a name that we will study a little bit. It is called the Berger's Equation. Of course, this equation is a particular case of this. So let us try to say something about this. So we have the previous notation on the diffeomorphism, the embedding map phi, and so on. And so what are now the assumption, transversality assumption? Because now, if you look at the system of characteristics, system of characteristics now is this b of x and y, y dot equal f of x, y. And then there are the initial conditions. So x of 0 is equal to sum phi of sigma. And y of 0 is equal to sum u bar of sum phi of sigma. OK? So this is now the system of characteristics. This gives us, if we are able to solve it, it gives us solution x of sm sigma and y of sm sigma. Now, which is the difference between this system of characteristics and the previous, the linear case? Well, we can see it. These are coupled now. Really, they are coupled. Because in the previous case, first we solve for capital X. And then we found y once we know capital X. Because in the previous case, y was not present here. In the linear case, there were no y. So this was depending just only on x. We solve it.  This was the previous idea. But now you see what happens. In the nonlinear case, unfortunately, we cannot really do this. Because now b depends also on y. So they are really entangled. They are really not coupled. You cannot decouple it so easily, at least. Can you see? So since now here there is both x and y, and here there is both x and y. So this is a system. You have to see each time what you can do. So now we assume on b, f, whatever, that we have a local smooth solution to this system. No, we assume that b is a c1, f is c1, phi is c1. So we have a local solution. We are not interested in the difficult case of this continuous right-hand side. So we have a local solution. But anyway, the point is, how can we state the transversality condition, which in the previous case was just at the horizontal level? Previous case was b transverse to sigma. Now b depends on x and y. And so what do we do? Well, it is natural, in my opinion, to reveal the previous discussion to say that on graph of u bar, capital b is non-tangent. Can you remember the previous discussion? We observed that this condition in the linear case was also saying that this was not tangent to the horizontal part of capital b was not tangent. So we take as a transversality assumption that we take that b of x, y is non-tangent on graph of u bar. On graph of u bar. So meaning that when you put here u bar of x, u bar of phi of sigma, say, then this vector field, just red on this, is non-tangent. Fine. Now, again, our solution, I claim that our solution is, again, in the usual form. So I claim that I can construct one solution, one local C1 solution, again, by this formula, u of x is equal to y of s of x sigma of x. So claim 1, OK? The claim is this. So the usual formula. In this more general case, even on linear, OK? I claim that, and still, we have the change of variable and so on, et cetera, et cetera. So given x, we have s of x and sigma of x. And the claim is that this solves the PDE. Before trying to do this computation, let me write more explicitly what does it mean imposing this transversality condition. So I'm sorry, now I have to erase the picture. So let me erase it. This can be checked as follows, which is a basis for the tangent space to sigma at a point. So you have sigma. Now I'm coming back to the previous picture. The picture number one, not number two. So this is sigma. Or maybe it's not really even necessary to do. So which is a basis for the tangent space to sigma? You have a parametrizing embedding map, phi. Phi is an embedding. So which is the tangent space? Which is the basis at each point for the tangent space? Phi is given. So you have given phi. Remember, phi takes some u into rn and is a local parametrization for sigma. Well, this is in rn minus 1. What do you mean by phi prime? Yeah, exactly. So say d phi over d sigma 1, that we write it like this, d phi over d sigma 1, that we write it like this. Phi over d sigma n minus 1, this is a tangent vector. Phi, remember, phi is phi 1, phi n is OK. Because phi goes from here into a big space, rn. So phi has n components and n minus 1 variables. So I take, say, r2 into r3.  This is phi. And then I have this piece of first surface physically a point here. Then I want to find the tangent space in the physical space. This is generated by two vectors, because it's a two-dimensional now. And each of them is the derivative of the embedding map with respect to one of the, or one of its, variable. The point is that, if I look parametrize, if I, now I have used in the language of parametrization, I can think that I have a parameter here. And I talk about the tangent space here as being the tangent space in the image. This is an embedding. I can't identify this with its image. Anyway. So this is a tangent vector at the image point. The entries of these are points here, of course. But this physically has a meaning of the tangent vector at the image point. This is just first elementary effect in differential geometry. So this is a tangent vector. This is another tangent vector. So these are n minus 1 tangent vectors. Assumptions on phi embeddings, immersion, it says that these are linearly independent. This is an assumption. When I write, when you see what does it mean to be a surface, a hypersurface, this means exactly that these objects are linearly independent in a neighborhood of your point. So this generates, in other words, this generates the tangent space. So now I want that this is non-tangent on graph U bar. So now I take the matrix d phi over d sigma 1, comma, comma. This is a column. I have n minus 1 column. But the column has n components. So I have to take a square matrix n times n. So d phi over d sigma 1, d phi over d sigma n minus 1. And then I have b of sigma U bar phi of sigma. And I have this big matrix. OK. And this big matrix, so now it is suddenly, it is n times n. Because this is another vector with n components. So n minus 1 plus 1, we have n columns. So this is a square matrix. And saying that this is non-tangent means that this must be, the determinant of this must be different from 0. So this is a way to say that for any sigma in U, say, it's our new transversality condition. A little bit more difficult than the previous one. So this is the new form of the transversality condition. So maybe it is better to try to do an exercise because maybe it's not so easy to understand this. So before doing the computation of the, before proving our claim, let us do an exercise together. Because I mean, so the exercise could be the following. So this is after the exercise. I leave this after the exercise. Or maybe if we have no time, because we do the exercise, I leave you as homework that we will do together tomorrow at the beginning if we don't have time. OK. So before doing this, let us do the exercise. Just to understand what we are doing, maybe it is slightly better. So the exercise is the following. Solve u du over dx1 plus du over dx2 is equal to 1. And so this is x1. This is x2. And then on this sigma, u equal u bar, OK? And u bar is equal also to what? Maybe u bar is equal to, yes, u bar of sigma is given also. Initial condition is 1 half of sigma. So which is the situation? Let us try to understand. First of all, this is, if you change the name to the variable, this is du over dt plus u du over dx equal to 1. So this is Berger's equation. du over dt plus u ux equal 1. If you call x2 equal to t, remember that Berger's is ut plus u ui equal to something. So if you call this ut, this is u ux, when x is one-dimensional. So anyway, this is, but usually ut plus ux is coupled with the condition on sigma, which is, and sigma is times 0. So now we have a slightly different situation, because sigma is not here, but it is here. So now, let us try to slowly understand this example so that I think that you can understand better the role of the symbols. So first of all, we are in, of course, n is equal to, this is clear. So sigma is an hyper surface in R2, of course, because it's a segment. So it's one-dimensional in R2. So sigma is one-dimensional. And of course, it's a smooth hyper surface. Obvious is a segment. So more smooth than a segment is impossible. So u bar, now, who is phi? We have to find the embedding map. So this is our manifold, sigma, in the physical space. But we need the embedding map, phi. So we need an open set contained in R, because n minus 1 is R. And we need the map phi, the map phi. Who is phi taking? Who is capital U? And who is phi? Omega is something in R2, is something here around. It's an open set like sort of this. This is omega. Of course, u is equal, say, 0, 1. So this is the parameter space. And now, so phi must go from 0 to 1 into R2. And the image of phi must be your sigma. OK, so the parameter we call this sigma. Now, sigma sigma. Do you agree? Obviously, it takes any point in this interval into a point on sigma. Is it a parameterization, a regular parameterization? Of course, if I differentiate this with respect to sigma, this is 1, 1. Therefore, it's never 0. Therefore, the Jacobian is a maximal rank. In this case, the matrix is just a number. 1 times 1 matrix is non-zero matrix. The parameterization is regular. And the vector 1, 1 generate the tangent space. Obviously. So this is a very simple case, just one dimension, OK? But just to understand, is it clear? Is it clear? No, this one. That's why I'm trying to do first the exercise, so that if you understand the exercise, then you can translate the claim or the theory looking first at the exercise step by step. Because in the theory, sigma is n minus 1 dimensional object into our n. So it's much more difficult. And you have an embedded map, which has several components and so on and so forth. Anyway, let us try to understand at least this example. Now, let us check our transversality condition. So the determinant of what? Now, we have already computed this. We have just 1 here, because n minus 1 is equal to 1. So there is just this. B. I have not written who is B, maybe. So B of x and y, sorry, who is B of x and y? Is it clear? It is clear, because if I remember, I have to write B of x, u of x, B1, u over dx1 plus B2, x, u of x, u over dx2, equal to 1. So if B1 of x and y is equal y, I get exactly this product. And if B2 of x and y is equal 1, I get exactly this 1 here. Is it clear? So B of x and y is this. Now, I can write it also capital B. We don't need it, actually capital B. But capital B is 1, y, 1, 1. We don't need the moment capital B. So let us write down the transversality condition. Transversality condition. So what is this matrix? It's a 2 by 2 matrix, because n is equal to 2. And therefore, we know that this is an n times n matrix. And therefore, we need a 2 by 2 matrix. And what is this 2 by 2 matrix? Tell me what is this 2 by 2 matrix? The first column is 1, 1. That is, this is the tangent, the vector spamming the tangent space to sigma is just one dimensional. And then I have to write down B of sigma u bar of phi of sigma. And u bar is this. U bar of phi of sigma. U bar of phi of sigma. Yes, because u bar is defined, say, at point physical space, x bar, but x bar is coming from sigma here. So let me write u bar of phi of sigma. It has to be extremely precise. Very often, one identifies capital U with its image. But if now, if I want to be really very, very precise, I have to not to identify it, considering the embedding map phi and so on, OK? So which is the second column of this? Exactly, because this is B of x, y. So 1 half sigma and 1. Is it correct? Let me check. Not 1. Ah, yes, 1. No, B of sigma. Yes, u bar of sigma. 1, thank you. So now the determinant of this is 1 minus 1 half sigma. And the determinant of this is non-zero for any sigma between 0 and 1, because our u was 0, 1. So this is never 0 in this u. 1 minus 1 half sigma, OK? Difference from 0 in our interval, 0, 1, OK? Therefore, we can start the theory. We have transversality. And we can at least write down the system of characteristics. And so x dot equal to B of x and y, x of 0 equal to some phi of sigma, y dot equal to 1, and y of 0 is equal to u bar of sigma. And we have to solve this to obtain. So x dot, x1 dot, must be equal to y. x2 dot must be equal to 1. y dot must be equal to 1. And then we have the initial conditions. Phi of sigma was sigma sigma. And this was 1 half of sigma, right? OK, so first of all, x2 of sigma, x2 of s and sigma, by the way, must be equal to s, s equal to 1, x2 of 0, is equal to, so it is s plus sigma. Now, x1 of s and sigma, first of all, y of s and sigma must be equal to, again, s, but now plus 1 half, sorry, plus 1 half of sigma, OK? And x1 of s and sigma now is y. You see now they are coupled. This is an example in which we see exactly the coupling, because for solving for x1, we need to know y. And fortunately, y was independent. So now we can put y here, which is s plus, so x dot 1 is x plus 1 half sigma. This is x dot 1, which gives us also x1 by integration. So x1 of s and sigma equals, sorry, I raised this now, so which gives us this, plus this, plus now, plus sigma, maybe, because for s equal to 0, it was sigma, right? OK. We are in a good shape, because we have, actually, we have the map s sigma, the explicit map x of s sigma.  So the point now is that we have to invert it. And there is no time anymore. So please, homework, we will do this at the beginning of the lecture of tomorrow. So even if you're not able to do this, we do tomorrow. So homework, well, write the solution. So invert this, write ux, and then prove the claim. The claim was that, in general, the solution ux is y of s of x sigma of x, OK? We'll do this tomorrow.


Are we ready to start? Yes, please. Yeah. Honourable Mr Prime Minister, Namaskar. On Ilosaada, Tava Tatteida. It's given me great pleasure to meet you for the first time in this virtual form and together strengthen the partnership between Finland and India. Together with other countries, we have gone through difficult and painful times for over a year. The top item on the agenda of all heads of governments is the COVID-19 pandemic. I want to recognise the efforts of India in extensive vaccination programme and you are among the world leading vaccine manufacturers. In Finland, we're also developing corona vaccines, among them a nasally administered vaccine. This could possibly help in vaccine distribution also in developing countries. The EU is the world's biggest exporter of coronavirus vaccines and provides funding to vaccination programmes in developing countries. In Finland, we have had our corona situation under control, but especially the virus mutations have led to increasing rates of infections and we have had to introduce further restrictive measures. Our aim has been from the beginning of the crisis to prevent the spread of the virus to secure the capacity of healthcare and protect risk groups, especially. Our aims have remained the same, but we have had to adjust our strategy always when we gain new knowledge of the virus and its behaviour. We have also managed with less damage to our economy compared to many other European countries because we have kept the epidemic under control. Our next test is to rebuild our economy strong again and reform its structures to support green, sustainable and digital growth. The pandemic has reminded us that we have to live together with nature. The nature becomes more and more into contact with humankind and I fear that if we don't change our behaviour as humankind to become more sustainable, we will face even worse crises in the future. Simultaneously, when striving for biodiversity, we have to find ambition to prevent climate change. Here we are together too. We have to act together to solve the greatest problem and threat to facing humankind. Finland is part of EU's climate policy and we are preparing for the COP26 climate summit in the autumn. Our national goals are clear. Finland aims to be climate neutral by 2035 and we want to be the first welfare state that is fossil free. Finland has good knowledge in sustainable development and we want to, through green technology, solve climate change problems. I see that we have good commercial cooperation possibilities with India in the field of technology and research. I expect the fresh memorandum of cooperation between the Finnish and Indian ministers of environment that it will be implemented ambitiously. Finland and India are natural partners in the field of new technology. Mobile technology revolutions such as 3G, 4G and 5G were sparked partially with Finnish know-how and digitalisation is a great opportunity for us. Finland and India's co-operate in the development of 6G quantum technologies and developing AI and I wish that our meeting now encourages to deepen cooperation. Finnish know-how in technologies is based on high quality education and equality. India is currently reforming its education system as far as I understand. The development of Finland from a poor rural country to a modern welfare state has been enabled by high quality education. What is even more important to the success of Finland is equality and trust based on that. Finnish women gained universal voting rights with men and the right to stand for election very early on global stage. Finland has partially had a difficult history but it has taught us that success as a society and economy requires including everyone. Women's and girls' rights and position have been the centre of Finland's international activities. Even now in the midst of the pandemic we have to take care of the realisation of equality between genders. The field of cooperation where I believe can do a lot together with India and learn from each other. We can promote international rule based system, sustainable development and combating climate change together. Honourable Prime Minister, I invite you now to take the floor for your introductory remarks. Thank you very much for your remarks. I have received a lot of advice from India for the spread of the virus from COVID-19 to Finland. I congratulate you on the success of the pandemic. During this pandemic India has taken care of the needs of the world along with its domestic struggle. Last year we sent more than 150 countries medicines and other necessary materials. We have reached almost 70 countries with 58 million doses of vaccines in India. I want to assure you that we will continue to support the entire humanity as per our capacity. As I mentioned Finland and India both believe in a rule based partnership between human rights and international basic system. Technology, innovation, clean energy, environment, education. There is a strong connection between us in such areas. In the course of COVID-19, all sectors will be very important for the best economic recovery. Finland is a global leader in the field of clean energy. India is a very important partner. When you worry about the climate, I sometimes tell my friends in joke that we have done justice to the planet. The planet is so angry that we are not allowed to show our face to all human beings. That is why we have to cover our faces by talking to each other on the ground. Because we have done justice to the planet, I sometimes tell my friends in joke. In India, we have committed the most important goal to achieve climate-related goals. In renewable energy, we have achieved 450 gigawatt installed capacity by 2030. We have taken international solar reliance and coalition for disaster and resilient infrastructure initiative. I urge Finland to join ISA, ISA and CDRI. In Finland's capability and expertise, these international organizations will benefit you. Excellency Finland, new and emerging technologies, digital infrastructure, education and skill development are in place. In all these areas, we have the opportunity to cooperate. I am pleased that today, in the field of IT, mobile technology and digital education, we are making a new partnership. Our Ministry of Education is also starting a high-level dialogue. I hope that today's summit will bring more progress in the development of India-F Finland. Excellency, this is our first meeting today. It would be good if we could meet again. But in the last one year, we are all getting used to the technology. But I am pleased that we will soon get the opportunity to meet during India-EU Summit in Portugal, Denmark and India-Nordic Summit. I also give an invitation to travel to India. Whenever you are available, you must come to India. I will finish the initial discussion here. We will talk in the next session. Thank you very much.