Transcript for:
Exploring Partial Differential Equations

so good morning my name is Giovanni Bella teeny I'm this I'm giving a course on partial differential equations and functional analysis so partial differential equations and functional analysis so this is a very huge very huge topic so it's it's we have to choose to choose various subjects and we cannot talk about everything of course so the idea of the course is essentially to divide it into two parts the first part we will study some special kind of partial differential equations in particular we will start with the first part of the course we will start with first-order PD and linear and non-linear and then we will pass to study the second order second order P DS and this means that we will focus attention on the class equation where this symbol means the laplacian so this is the basic so-called elliptic equation then maybe in not exactly in this order but this will be the other equation that we studied is the heat equation heat equation which is the standard example of parabolic PD and then we will pass to say something about the wave equation so this is the standard example of hyperbolic second-order PV so this so after this this kind of studies then we will pass the second part of the course so the second part of the course is functional analysis this is more abstract therefore the work that will be values results but essentially we will study just some some of the properties of Hilbert spaces where here the difficulty will be that this is a vector space of infinite dimension so this is exactly the difficult point then the Banach space is and then we will see what how much time it remains to continue the part of functional analysis so I'm going to say anything more because I still don't know how much time we'll have for instance to say something about Fourier transform maybe okay books that are suggested so concerning this part there is a book by Evans titled partial differential equations equations this is a standard reference at the moment for the first part of the course this book contains a lot of the a lot of things a lot of informations we will not at all cover the content of this book but it is a very good reference in my opinion for for this and then another very good reference concerning this more apt part of course in this book there are also there is also there are also several informations on function analysis so you can use this maybe also to cover some of the topics in the second part of the of the course but on the other hand very standard and important reference on this is the book of bridge this may be the title is functional analysis and partial differential equations functional analysis and PD and of course also in this book you can find a lot of informations about p d--'s so ok these are more or less standard standard references there are a lot there are a lot of other references so when I will use some other reference that we let you know and I will write on the blackboard the book that I'm using at the moment at that moment ok so please feel free to ask everything you don't understand okay that's important because also probably for all for all of us okay if somebody does not understand something it is better to come to to repeat it or okay so so let us start with P DS first order please so first order P DS now I will start from probably from the simplest one so a so let me given a vector in RN so this is a constant vector and consider to begin this partial differential equation okay so what is this U is a function of T and X so T say is bigger or equal than zero and X is in RN u is a scalar because we don't want in this course to talk about systems so U is so this is just one PD and not two or three couple PDS this is another problem so it's just one PD and this is the gradient here is the gradient with respect to space so X is equal to x1 xn and this is equal to B over the sine so this is the scalar product between the vector B and this vector here now this dot means the scalar product so and UT is the UT is the derivative with respect to u 2 T ok so in the sequel I I will also change the notation but for the moment let us use this as a notation for for the variables so in this way in some sense I am giving the first variable some sort of special role we will see in the examples what this does mean so concretely this this is the I do sigh to zero okay B of course is b1 B and in components okay so this is called a linear transport equation we will see why in a minute we will see why this is called transport equation it is linear because you see the derivatives here appear linearly this is clear it is of course first order because we have just only first derivatives of the unknown which is a function U it is homogeneous it is emoji Gnaeus because there is not the right hand side is identically zero okay so as it happens in ordinary differential equation so we don't have anything on the right hand side so this is called the whole genes and the coefficients are constant okay because the coefficients are one here in front of you t1 and then the vector B VI in front of the you idea this is not a function is just a constant okay so constant coefficients the coefficients are 1 B if you want so one be in r1 plus n and we are taking the scalar product between this time space value gradient of U against u 1 1 B so you see scalar product of this against you one against one be ok and this now this dot means the scalar product in air 1 plus n ok is this clear is everything ok so this is maybe one of maybe the simplest PD that we can imagine but still it is this worthwhile to to studied so to understand its its structure okay so what what oh and we look for look for the point is to look for a c1 solution c1 in time space social yc1 but well it is clear because we write the equation in this form so we take partial derivatives and if the function is c1 these partial derivatives are continuous and so this is well well defined at each point everywhere at each point so the regularity of the solution we look for is exactly this so what one one has to study so this is you see this is an equation without any kind of say boundary condition this is a three equation but soon we will couple this with a condition on some of some sort of boundary condition see this very soon we will do this and this is also this is also an issue because it's not clear in general which kind of boundary condition you have to impose on that specific equation this is not completely obvious why one type of boundary condition and not another one for instance hmm so so now we will discuss also that issue so which are the main problems that you that one face when studying a new PD so that the main problems are the following so first of all one the standard way to to proceed is one look for special solution special solutions explicit look for special explicit maybe solutions means that you look for solutions with some kind of symmetry for instance I mean you look for maybe radial solutions or solutions which are independent of time or something like that okay special special not not general solution but this is this is an important point but maybe sometimes you're not able or maybe you'll find one extra solution but then you have a problem on domain open domain with some strange boundary condition again you are not able to find a special solution to that so what do you do now of course we have the problem of existence of a solution but existence of a solution means that you have to exactly define what do you mean by a solution so this means that existence of a solution in some class and of course the larger the class is the easier is to find a solution the larger the class the better these somehow of course when you large the class if your solution is not very smooth for instance not see one but say lip sheets or less then you have to declare what do you mean by being a solution because maybe you cannot differentiate point twice hmm so if you large your class because you want to find existence you look for a very large class but then you have to define what do you mean by a solution because if your solution is not differentiable then what is what does it mean solving the PDS is not clear ok but this is strictly related by with another problem which is uniqueness of the solution so sometimes it is very useful to know that the solution is unique but then if you want the solution is unique your class should be not very large because it is more difficult to have uniqueness in a large class so if your class here is too large maybe you lose uniqueness to many solutions of the same PD and so here this smaller say the class the better it is so you have to find a compromise between 0.2 and point 3 to have it at the same time existence and uniqueness is it clear and then there is maybe the more difficult the most difficult point usually so these two these two points are connected so you cannot really solve somehow separately very often you have to look at two and three almost at the same time and point four is is the regularity of solutions mem this means the following assume that you are able to prove existence of a solution and uniqueness for instance at the same time then maybe your solution is much more smooth then you expect only from point two for some mysterious reasons it happens very often that the solution is much more smooth than you expect and this is probably the more deep part of all the story and again point four is somehow is strictly related to these two and you cannot think of point four separately from point two and three somehow you have to have in mind all these three points at the same time so when you look for a class here for finding for a solution then you have to imagine in advance which could be the regularity that you would like to have for your solution so I mean these these are all together and this is the difficult point of all the story okay so for the moment we have a simple say reasonable PD and we start for looking for special for explicit solutions just to understand okay so now we concentrate on point one by the way it is in point to where you you look for this this large class it is a point where functional analysis enter because in the in that large class you need non smooth functions in general may be Sobel of spaces for instance so in this part in point two here there is functional analysis entering the story in particular may be Sobel spaces and distribution distributions theory of distribution which are two typical arguments understand the structure of the sobel space is a typical problem in functional analysis I usually need bitten Banach spaces okay so now let us start from this PD here so what do we see from here this says a question 1 the question 1 says that U has sort of directional derivative which is constant do you see it can you see it I mean here it is written that there is a sort of directional derivative of U which is constant in time space Y you see if I take the time space gradient of U along the direction one comma B I see that this derivative is is zero this is exactly the scalar product between between the time space gradient and the direct and and the vector 1 comma B and this is here it is written that this scalar product is zero so this means that the derivative of U along the direction the vector along the direction of the vector 1 B is 0 so it is natural to introduce a function let me call this Z of s maybe with my notation yes which is this U of T plus s X plus s V so this is U of T X plus s1 B where s at the moment is an exterior parameter and this is this is looking at a possible solution of your problem along the line passing through the point TX and parallel to the vector 1 B ok so in my pictures now what I do is usually the following I put X here and T here so notice this kind of choice T here is considered as the first variable but when I draw it in in a graph I put it vertically we just a matter of convention just the convention ok so now so let me let me so now I have the vector 1b say and then and then let me compute the derivative of Z with respect to s which is of course UT at the point TX plus s 1 B plus the gradient of U at the point TX plus s1 be scalar product will be and this is 0 okay so this means that Z is constant so this means that if I have say if this this this is the direction of the vector 1b then this means if I take a line here parallel to this vector and all lines say and I read the function along this line passing to the point TX what I see is that here the U is constant on this on these lines okay okay so now so what we can do is to couple this with with a boundary condition so with my notation now let me call it s so let me introduce some notation here so Sigma is equal to t equal to 0 into this is so this is Sigma okay and then let me take a function U bar into c1 of Sigma even so this is given maybe you can you read it here okay so now I am giving an hypersurface for instance this hyperplane okay and then I assign the following new condition here I look for a solution to the following problem so you t PQ T plus B dot radio equal to zero in in zero plus infinity x RN you in c1 say zero plus infinity x RN u equal u bar on sigma before say c1 maybe let me close it so now i assume that i want to find a solution of the three equation in the half space positive time solution of class c-1 maybe it would be enough to assume you continuous up to the boundary I've written c1 up to the boundary but just for so continuous up to the boundary and then I want you u bar X given Y bar is given exactly on this boundary so now this is this is a problem with the boundary condition more precisely in this case it is an initial value problem it should very problem meaning that I assume what happens at time zero you bar and then I start the flow so this in this form with this Sigma this is called an initial value problem so now I want to find this solution so how can I argue so I assume that I have a point here so let me write it as the point X and T so I have a generic point here in the half space and this is di and X and then I want to find you so a solution U at the point t index what do I do so what I know is that the solution here through this point is constant where along the line parallel to the vector one B and passing through this point so so now I fix this point here in the positive half space I take the unique line parallel to one be passing through this point and then I know that U is constant on this line and therefore in particular its value is exactly the same as its value at time 0 so the idea is you it is equal to its value the value the value of u bar because I want you to be u equal to u bar at time zero but at a different point of course which point not not the point XE but the point in this this point you we're at the point so that I've indicated maybe by PI of T X so this is so this is the X and this is a PI of T X this notation stands for say projection point in some sense even if it is well it's a sort of notation so at the point PI of T X so that the line passing through PI of T X bar in the direction 1 be in the direction one we contains the X okay so I have to find so given T X now I have to find the PI of T X so given T X in 0 plus infinity say maybe this x RN find PI of T X into our n and s of T X into R such that T X is equal to PI of T X 0 maybe plus s of T X 1 is it clear the point a few STX is the parameter s I somehow I have introduced an exterior parameter which say is the parameter parameterizing your line okay is s of T X is just a number is this written here is the parameter along your line parameterizing your line and so of course this point here to describe it into into a new system of coordinates what do you need you need this point here and then the parameter s such that if I move from here of the quantity s along the direction then I reach exactly this point which is which is what is written here in this equation say hmm is it clear is it okay up up to now okay so now I can solve maybe I can you read it maybe I did okay so then I can solve this you see what happens here well let me write this so T X must be equal to 0 plus STX time 1 so STX comma and then PI of T X plus s of T X times me and remember that V is a vector s is a number pi is a vector okay and this is the first component these are n components 1 n 1 M C is it ok well s must be equal to t s is equal to T and then X is equal to PI of T X plus T B because now s is equal to B s is equal to T so if I impose second component the second block of components X equal to this Plus this I know already know that s is equal to T from the first equation so let me let me substitute here T and so then I have this equal to this and then finally I can recover PI of T X which is my unknown essentially in this procedure I am inverting a map I'm inverting a map we this will be more clear in the in the next in the sequel of the lecture but one point is exactly to invert the change of variables okay in any case PI of T X you see given T and X PI of T X is X minus TB explicit so the inverse map say s of T X is equal to T and n is explicit and and this is the expression and so we have found that necessary solution of two interesting because we have found an explicit solution solution of two is necessarily U of T X at any point in the half space is equal to u bar of X minus TB so look at this expression u bar is given from the problem B is given and this is an explicit solution the idea is simply the following what is the value of U here is the value of u bar here that's simple idea because U is constant along exactly this line so it's very same however it is interesting okay so this is some remarks in the order I think they are quite important so if everything is clear now I erase just keep the equation and the solution okay so at the end the exterior parameter s parametrizing the line is identified with time s is equal to T at the end okay okay remarks let me let me write this in in the order okay remark one may be remarked one is that you is c1y you ec1 because you bar was assumed to be c1 on the map on the subspace on the half plane on the hyper plane so on the hyper plane by assumption you bar was c1 and therefore you is a composition of C 1 and C infinity in particularly C 1 okay so this condition is satisfied it is clear that at time 0 if I put here T equal to 0 so you see 1 up to the boundary up to Sigma if I put t equal to 0 here it is clear that u is equal to u bar so u 0 X equal u bar X now and this means the third condition and then by construction you solves this so U is clearly a solution of our problem it is it is very smooth because and as exactly the same smoothness so same smoothness rottenness as u bar what does it mean is this this is obviously in this case but it is an indication of the following phenomenon that such a kind of equation does not regularize initial conditions if your initial condition is c1 your solution is c1 and not more than C 1 naught C 2 or city is just C 1 so there is not a regular izing effect of the equation the question has not a regular izing effect the reason for this is that this is first-order actually linear PD the question is how can I prove it that this equation is not a regular rising effect this this sentence means if your initial condition as I saw as some kind of degree of smoothness is say C 3 then the solution is C 3 and not more than C 3 this means the sentence ah yes you mean uniqueness well this is a proof because if you have a solution the question is what about uniqueness of the solution okay so well if I have a solution you of this problem an essentially U is equal to that this is what I have proven so any other solution you have of this problem C 1 and that etc etc must be equal to that because my customer I have proven this this SNU this I mean you have if you have a solution then your solution is explicit and necessities that one this is also uniqueness so okay same smoothness as u bar next remark why I insist on this because this kind of this kind of PDS are in some sense worse than other PDS exactly because they do not regularize your initial condition on the other hand other equations such as for instance typically the heat equation second order PDE has an immediate regularizing effect if you start with something which is not smooth at any positive time it immediately becomes extremely smooth and that is an irregular izing equation but this is not hmm this is not so this is a somehow a point that is makes makes interesting the subsequent theories are also but more it is not so so so trivial as it seems this remarks not so trivial because of all of its consequences okay then remark - we have used in the proof the following argument the following assumptions that the vector 1 D is transverse to Sigma what does it mean transverse means that non tangent 1 be non tangent to cigarette but well this is obvious why it is obvious because Sigma was this hyperplane right this was time first coordinate put vertically and this is space okay and Sigma was this red yellow when was this red hyper surface and the vector one B is of course non tangent exactly because there is one here so this one component independent of of what B is this one component making me makes it non tangent is transverse is not like this is this okay and this makes the procedure work because we have used where we have if if this were tangent then our procedure clearly does not work so this is another remark transverse to Sigma third remark maybe the third remark is that this is a transport equation linear transport equation in the sense that we are transporting the initial condition so we have you bar and at subsequent times we are transporting the value of u bar somehow in some sense because the value here is u bar here so in some sense this is then yes then there is a more delicate point this is more much more delicate if you bar is non smooth for instance like this this continues with constant values 1 0 assume that this is you bar the graph of you bar 1 0 now if you put your bar here then you is not anymore c1 so the point is is it reasonable or not if you bar has this kind of profile discontinuous is it reasonable or not to accept this right-hand side as a solution of our PD in some sense not c1 and it is not so easy to understand is it reasonable that you bar of X minus TB solves to in some weak sense of course not in point yc1 sense because if you is discontinuous i cannot differentiate so easily at each point I cannot so somehow that the answer is yes still it is reasonable but this requires a notion of solution of saying what does it mean that it is continuous object solves this what does it mean exactly what I have to put here in place of UT and grade you in order that this makes sense so this so this is long story and so I don't want to touch now this question just a hint that even in this very special case you can ask a non trivial questions like this okay finally the last remark okay a notice remark five the lines lines the lines parallel to one be our solution our solve the following the following point capital X dot equal one B and say X dot of 0 say equal to a given X if you want notation is X bar sorry so if I solve this summer capital X is a vector in M plus 1 so X bar now is a vector in RN maybe I could call it PI but let me be let me be free of using different notation as as far as this is understandable okay so now I have a point X bar here time and then I have to solve this ordinary differential system of ordinary differential equations now it is a very easy system so so X capital X let me write it as X 0 X 1 X n naught so this is the first say time component with the index 0 and this is the these are the usual space components and so X 0 of s from here is equal to s and X 1 X n of s are equal to X bar p SB and so it is clear that this is a parameterization X of s capital X of s is a parameterization of the line passing through this point in the direction 1 B so these lines in particular we have obtained them looking at the PD but in particular it one can observe that these lines are obtained as solutions of a system of Audis which is this ok so this is called the system of characteristics characteristics it is a system of all these now it is not really an OD because just an integration I mean there is not capital X on the right-hand side so it's an immediate OD but anyway just a remark to keep in mind for the future development okay now so let me pass so we have understood everything about I think the the simplest at PD let me now modify it and make it a little bit just a little bit more difficult so so under the previous assumptions now let me consider this folder the following problem so let be given now let F be a right hand side so say a source term in c1 say zero plus infinity x RN and assume that I want to study the following problem so UT plus B grad u equal to F so now this is non-homogeneous case so linear first order PDE with constant coefficients coefficients but non-mo genius non-homogeneous and so assume that I have again at the pond so this is in the in the half space and this is well this is a little bit too much but anyway doesn't matter the smoothness enough and then you equal you bar on Sigma with the same assumptions on Sigma and you bar as in the previous case okay Sigma is the hyper plane horizontal hyper plane and you body c1 okay under its own Tolliver plane so now this is of course Lena as before non-homogeneous because there is this source this is given is on the right hand side does not depend on U is given function of T and X mm-hmm so now the point is assume that okay now f is c1 maybe is too much because you see you is c1 and so UT and grad u are continuous so maybe it would be more natural to assume that F is continuous doesn't matter let let me let me I don't maybe to be more precise one could write here c0 anyway doesn't matter so now I want to to solve this so I mean it's natural as before to consider so how can I solve is it now it is not so clear that we have still the transport interpretation as before now maybe it's not so clear indeed and however I can proceed as before so I can introduce as before the function Z of s okay so which was U of T plus s for a given T X and then it was X plus s be for a given TX and let me compute the derivative with respect to s so Z dot of s is equal to UT evaluated at T plus s X plus s B plus the gradient of U plus B scalar product the gradient of U T plus s X plus s B and now this is not anymore if u is a solution it is not anymore equal to 0 it is not but it is equal to it is equal to F at those points F T plus s X plus s B ok so now so now we can we can look at this and we can integrate so now tia is an ex are given and we can integrate this this between minus T and 0 pattern this in zero plus infinity in the open half space usually this is a way to write down a PD with the cost with with the boundary condition so you want the PD to be to be satisfied in the interior of your domain hmm let me draw a rough picture you want the PD to be satisfied in the interior in this case the interior is the half space open half space T positive and then you ask for a boundary condition on the boundary and this in this case the boundary is just the horizontal bottom of the half space and then of course you need you to be sufficiently smooth inside and you need that it makes sense that you has a trace here so say c1 inside and c0 up to the boundary for instance I have written for simplicity c1 up to the boundary you see here there is a square parenthesis but it would be enough to write c1 in the interior so round parenthesis and then c0 with the square parenthesis to be more precise okay hmmm this is because I want the PD to have a sense in the classical meaning inside all the rivet is on the boundaries does it there is a continuous trace and so everything is well-defined this is just for this okay now so we are here and so we can integrate now I integrate between minus T so that s T plus minus T becomes 0 and then 0 so that T plus s becomes T so I integrate this between minus T and zero and I end up of course Z is equal to Z of 0 minus Z minus P ok and which is in turn equal to my Z of 0 is U of T X Z of minus T is equal to U of 0 X minus TB is it ok after now agree ok so from this I deduce that my solution U of T X so let me continue this here so U of T X is equal to this right-hand side plus this okay so I did use I did use the following U of T X is equal to by the way the trace of you on the hyperplane is equal to u bar so I can substitute this with u bar at X minus DV plus okay so now let me change variable cell let me call this as say Sigma T is fixed so I consider this as a change of variable for s and so this becomes equal to u bar of X minus TB now when s is equal to minus T Sigma 0 so when s is equal to 0 Sigma is T t plus s is just Sigma and then s is equal s is equal to Sigma minus T now Sigma minus T be in this Sigma so let me just for maybe elegance let me call Sigma equal to S okay so this is our solution again so we have found another explicit solution much more complicated than the previous one of course the previous one was this when F is equal to zero so you see when F is equal to zero this is the previous solution so we are clearly generalizing the case when F is nonzero and we have this strange expression integral so we don't have a clear interpretation as before using just only the initial condition U bar so now you have the point X here and the point P here and so your solution here at this in the future at this time T and at this point X is equal to your initial condition in the past say X minus TB but then there is another contribution which uses all the values of f in between 0 and D so s is in between 0 and T so I need to know the values of M of F for all x in between 0 and T and then here X plus s minus T when s is equal to 0 this is just X minus T and when s is equal to T this is just X so I am using the values of f in an intermediate region here somehow on the left of X and for all times T okay so this is my this is somehow another interesting result so when we put the right-hand side here we we have we lose some somehow this transport interpretation and we use the values of F in a big region here okay so yeah I have maybe another remark before leaving this non-homogeneous problem so let me consider again the the system of Audis sweet and as before so X dot of s equal to 1 B X of 0 equal X bar equal 0 X bar maybe ok and then let me consider also the following mu it's zero equal u bar of X bar so now I have the previous system of characteristics that I called characteristics so these systems give you the line now the line parameterize with s equal to D so this is this is the previous system with s equal to T passing through I get the line passing through X bar and then I have another now this is one or D these are n plus 1 or D this is just one because this is a scholar and then I and so and then I look at this together and what I see well they are not really coupled why because I can solve X explicitly here then I can put the solution here and so I can find y just by integration so this actually they are not really coupled this one equation one problem one Cauchy problem is not coupled with this system but anyway let let me consider this ok so therefore Y of S is explicit because it is the integral so an end moreover may be look at this part here so define so solve this solve this for X giving you a solid solution the obvious solution which is next depending on what on s and also on your initial condition X bar okay so this is again the line passing through X bar through 0 X bar in the direction 1 let me call this more implicitly as X depending on s the exterior parameter which are here this time and X bar and then consider the following object Y of s equal to u of X of s X bar so I claim that Y of s solve this so assume that you have given X assume that you you're given your solution you define Y and then I claim that twice of this and then I want to go back assume that there were a FX I have y and I want to find you of course I know what is you okay I am saying this because for more complicated equations I cannot explicitly find the lines they are not lines anymore they are curves and so on and in that case I have to write more implicitly the solution so in view of having more complicated problems where the solution will be more implicit now let me make this kind of comments that will help in more difficult cases so so let us start from now from this system find the solution define this and look assume that U is the solution of the PD and look at what kind of OD Y salts okay so Y of S is nothing that is nothing else U of S X plus X bar plus s B because we know that X of s X bar is exactly as X bar plus X is B okay this is equal to X capital X as X bar Q's so Y dot is equal to UT evaluated at s X bar plus B plus B cut color product value at s X bar plus s B since you by assumption for the moment is a solution of our PD this is equal to F s X bar plus s B okay which is f of X do you agree so we have discovered that again given capital X given you solution of the PD this function y satisfies this equation so given you solution of the D of took maybe chose written to solve for X define y as here then why salsa three okay it is clear that Y of zero is equal to u 0 X bar so that Y salts s Y is also 3 so now however that we have we are more interested in the opposite direction so it is somehow so let me stress the following fact this is extremely important for the next lectures a dissolution of this very easy system of Audis for the moment is of course a function of the parameter s but it is very important to consider it as a function also on of the initial condition X bar is the initial condition so I consider the solution X as a function of its parameter and the initial condition okay this is very very important when you write when you look in the books chapters on the sister Nakano concerning the system of characteristics you will see that the solution of this system of all these are always indicated denoted also the independence also of the initial point you start the trajectory not only that very often more generally it happens that but this is not the case here but just to let you know it happens that you can write the characteristics capital X also as a function of the initial condition 0 I don't write it here because I start from 0 before but if I want to start the problem at some some TZ t naught then the solution depend on s x-bar and T naught okay so and very often you see in the books that capital X is is capital X of s T naught and X by okay but now T naught is zero so I don't I don't use this interpretation but it's important on the other hand to have this dependence on the initial condition okay so now so now here given you solve X define y and y substring now of course this is not our problem our problem is to find you I don't want you to be given I want to find you so what do I do so the idea is maybe also the other way around works namely so and this is finally the beginning of the method of what water I'm slowly coming to the method of characteristics for first-order P DS very slowly so and this is maybe it works also the following I mean assume find X solving this find y solving this because once you know X then you can solve for y and then try to find you try to define U at the point TX ok so now u of t X what is the reasonable definition for U of T X in your opinion after this discussion so you know X because b is given X bar is given then you know you know why because f is given so x and y are given the trajectories the lines are given the y is given now you want to find you what could be a reasonable definition for you in order that you solves your PD y yes exactly now the point is this what I would do I put inside the why why remember so sorry so sorry once once and so let me call the solution of this again let me stress that Y is a function of s but is a function also of X bar so now once I've said this what could be the definition now I have to put something here here there should be T and X because the end X is here so if I want to write an equality I have to put the same variables on the left and on the right okay so what do I put here yeah and so so you know why depend on s right but s is a function of T and X and also X bar as you said is a function of T and X right so this is say a possible definition again we have some how to invert a so a change of variable okay so let me check that this is a solution of so check that this solves our problem okay so what is y of s DX so you see at the end that s in this case will be equal to T we know this so we have a change of variable essentially the idea is the fall we have a change of variable for which any point here can be expressed as a projection point here and the parameter s and conversely if I give you s and the projection point then I can recover T and X so there is this change of variable now you don't see very well this change of variable exactly because the situation is too simple and because this is flat but as soon as we will take an I per surface Sigma which is curved this change of variable will be more easy to understand now because they hear there are some identifications because of this they make the same the change of variable a little bit hidden but okay so let me write this so this is X sub R T and X so this is equal to u bar of X bar plus the integral u Bar of X bar of TX plus the integral in between 0 and s of T X of f of X of Sigma V Sigma so is equal to u bar of X bar of the X plus the integral 0 s of T X now X of Sigma is given Sigma X bar of T X plus Sigma B and what do we know we know that s of T X is equal to T and X bar of T X is equal to X minus TB okay this we go remember so what do we have what I've done so I have defined you in this form okay now what is y why is the integral between 0 and Sigma or F in the Sigma plus its initial value so Y is equal to its initial value plus the integral in between 0 and s of F Sigma and so on ok now I substitute my expression of s and X so that finally I find that that is equal to u bar now and in place of X is I write X minus DB which is this okay Plus now we know that s of DX is equal to T so in place of the X 3 MoMA s I put 3 here then I have F what Sigma Sigma comma now X bar is X minus TB plus Sigma be X minus TB plus Sigma be ok and this is our previous solution that compatible what with what we have already found ok so the message for today is for this ok we have solved the PD explicitly but maybe there is a more general way to solve the PD the general way is you want to solve the PD ok construct a system of OD X dot equal write down another another OD and then define you as follows and that is your solution here everything is explicit but this so the idea is to solve our PD what we do is to solve Oddie's and then with the solution of these we construct the solution of the PD and then this is the so-called method of characteristics which works more generally than this and we will study more in detail this matter than the next tomorrow in the next lectures ok you you