Lecture 12 : Well-Posedness of a ODE

[Music] welcome students and in this video we are going to talk about well poess of an OD so what does it mean so first of all we start by considering an OD let's consider a system okay so I'm doing it for a system but uh you do realize that it can be I mean uh you can use the concept for any given OD right so consider the initial value problem let's say initial value problem okay X Prime = to f of x clear huh and x0 = to X not okay so this is given to us right now where and let us understand the start with X not in RN so essentially we are talking about en Crossen system so basically we have a en Crossen system okay now now the point is this uh you see let's let's call this a star now uh the thing is given a system like this what do you want to do with the system so essentially we want to study the system but what do we mean by studying a system so basically what does it mean what does it mean mean to study star okay to study the system start the system above system that is okay and what it means is this essentially you see uh we are basically looking for three main questions so first question so that and that is called the whale posess so basically you want to know whether the system is well posted or not okay and what does it mean so what are the properties which the system should satisfy for it to be well posted okay so properties so first of all a the system system in question star in this question star must have a have a solution okay must have a solution of course in appropriate sense okay appropriate sense you see Solutions can be in many sense right appropriate sense okay must have a solution in appropriate sense okay so first of all it must must have at least one solution that's what I mean to say it means at least one solution one solution okay and whatever you want to call whatever you mean by solution I want talk about that sort of thing so at least one solution should be there b if so first of all Let's uh let's just uh put it this way that once you have found a solution now the thing is if there exist a solution exist exist a solution it should be unique it should should be unique clear so basically there cannot be two solutions to start right so if you have a system or any equation for say okay and uh if you can find out that that system actually admits two solution okay so whatever the meaning of solution is by that meaning if you have let's say two solutions then of course it is is not well poos so basically it you have to have first of all one solution and the solution has to be unique so those are the two properties and the last property is called The Continuous dependence on initial data continuous dependence dependence on the initial data initial data okay so what it means is this see for now I'm not going to make it much so we'll make it much precise just in a minute or so but first for now let's let me just put it this way what it means is let's say if you start with X not okay you get a solution right let's say XT you get now what do you think happens if you start from let's say x not plus Delta X okay so let's say x at the point0 is X right and the system does something so basically you get a XT huh now let's say if you are put starting with x0 which is X Plus Delta X you are also get a you are probably going to get a YT right because you know there's a solution and the solution is unique let's just remember that that is true once those things are true now you see for X not starting from X not you have X of T and if you start from X Plus Delta X you have y of T so what do you think is the relation between XT and YT okay see if x x and Delta so X Plus Delta X is a difference so let's say if Delta X is sufficiently small okay in some sense I'm not putting it like this so basically if it is sufficiently small you expect your XT minus YT should also be you know I mean you can make it sufficiently small right huh uh so we we are actually looking in that direction so essentially what it means is if you change the data a little bit you also expect your solution to change little bit right okay all of this I will make it very precise just in a minute or so but you do understand what the concept of this okay right and this is very very important property and now if the system so basically let me put it this way if start satisfies satisfies a b and c we call it a well poost well poost system of OD yeah system of OD okay right so that's the main idea right now let's say I give you a system right whatever it is x = f of x some system is given you know what that Vector field f is so here f is a vector field from let's say R into RN is a vector field given huh Vector field it is given to you clear okay now the point is this I want to study whether the system is well posted or not so what is the first thing we know need to know whether there's a solution or not if there's a solution whether the solution is going to be unique or not right okay so here we already know we have seen in the last lecture the existence and uniqueness theorem right P's existence and uniqueness theorem which you also called the fundamental theorem basally uniqueness theorem so let me again state it here because anyways we it is very very important and we should always know this okay so what we have is this uh let's say uh if capital F from RN to RN okay the vector field in question is C1 is C1 right then what do you have then then Star right the system you see X Prim = to f of x x0 = to X this this system given f is C1 okay admits a unique admits and unique solution admit and unique solution solution in some NE root of in some NE root of zero okay in some neighbor root let me put it this way nebor root of zero okay and what is this Z this zero this is what the zero okay so essentially this is t equal to Z okay so 0 0 clear okay so in the ne ofus epon plus epon you can actually choose an Epson such that this system has a solution okay that's what we have so essentially let's say we are assuming f is C1 huh which is not a very very strong assumption it's okay so with that we know that there you always have a some I mean solution which is a unique solution in some neighborhood of this thing the origin right now what okay now what about the third property so basically A and B is taken care of f is C1 we know that A and B is satisfied right if f is not C1 then there can be many different cases it may not have a unique solution or it may not have a solution also okay so f is C1 is okay that we know that A and B satisfied okay what about C so before we go on with the c let let us just make it more mathematical what does it mean for a continuous dependence on initial data okay so what it means is this see what I let let me let us write down a uh theorem huh in this thing theorem this theorem will actually uh you know clarify what I mean by continuous dependence on initial data so let Omega subset of RN this is our assumption I am assuming Omega subset of hard here and let and F is a vector field right from Omega to RN this is what we are assuming here Omega to RN is a vector field okay uh with a lip constant K with the lipes constant lip Shish constant key so basically we are looking we have a lip continuous function with the lip constant K let's just put it this way K clear okay so K is in R you do realize that right lip constant okay you remember what when it is lip so f is from Omega to R basically it means that F of xus f of Y has to be the norm has to be bounded by K * Norm of x - y you remember that for all X Y in Omega right for all XY in Omega okay that's what liance means right okay right so we are assuming this things so please remember this thing see the thing is for the well poess we did it for a C1 map huh C1 C1 Vector field now any C1 Vector field is locally LIF right this is locally Li but not may not be Li it is local lipes okay it is local lies now here what I'm saying is you need your vector field to be lip lii continuous right now if that happens let we are assuming y of T and Z of T and Z of T be B Solutions of solutions of f Prime = to f of x right okay which remains in Omega right which should remain you see the thing is you see both the solutions YT and ZT you are getting two solutions right and the thing is those Solutions has to be in Omega because otherwise if is not defined right so we have to have that those Solutions must remain in Omega right because F depend C for every d x of t Okay X of T should be in Omega because you you also have to Define f of XT right for this equation to work clear so this x of T has to be in Omega so basically here I'm starting out with Y and Z those should always be Omega okay and are defined are defined okay on an interval on the interval let's say interval some interval right and let's just put the interval as T to T1 some interval here okay then then for all T in t to T1 T to 2 T1 we have okay now this is what we mean by continuous dependen you see Norm of Y of T see YT and ZT for given T is in RN okay so YT minus ZT should always be bounded by y Y at the point T minus Z at the point t Okay time some exponential map and which depends on K * tus T clear okay so you do understand C if YT minus z t right see T is varying between T and T1 right T is varying in a bounded domain okay so exponential K * T minus t that's a bounded function essentially okay and K is fixed K is fixed that's a li constant let me see if your YT minus ZT at the starting at the point T = to T is sufficiently small of course YT minus Z you can make it as small as you like right you can always find the Delta so given you can always find the Delta so that's what I mean by continuous dependence on initial data so this is what it says right right so uh again let me put it this way in case you forget if the solution it means that if the solutions what are the solutions YT and ZT right YT and ZT clear if the solution YT and ZT start out okay start out sorry start out close together right so if they you know if YT and ZT are close together they're very you know the difference is very small then then they remain remain close close for all time T right sorry uh for all time T closed near T not right remain close for all time for all T near T C near t Okay okay so you will have a small neor of T where it has to be it will be close to each other okay so T not can be zero also doesn't really matter but you do realize what I'm trying to say is this uh if you whenever you are starting out it doesn't really matter which point you starting out with some T not you are starting and then what happens is you always have the solution actually uh is very close to each other right okay so to do this thing this is the proof is extremely easy what we are going to do is we are going to use the groundwell okay inequality and now what I'm going to do is since uh that is done in week one I will just quickly give the statement of Grell okay as I told you this is very very important right grall grall inequality huh let me just write it down and then we are going to use this ground Val the proof I'm not doing because we already you know did it uh okay so let let's say that uh you from 0 a to R be continuous and non- negative continuous and non- negative right non negative non- negative suppose C is greater than equal 0 and K is greater than equal 0 right are such that are such that U is less than equal C + 0 to T K * us DS okay for all T in z a 0 a what does that imply then that will imply that UT can be made bounded by C * exponential KT for all T in Z okay okay so the proof I am of of course I'm not doing the proof you guys we have already talked about the proof so I'm going to skip that part okay now let's understand that what is the proof of this theorem the The Continuous dependence let's just look at the proof of that theorem so proof proof see the thing is it's very simple actually and please remember the trick huh this trick we are going to use it in a lot of places see here I have to show that this quantity particular quantity is less than equal this quantity evaluated at the point T not see times some exponential okay so you do realize that you see I here also I have something like this this so this constant if I can show that this is nothing but YT minus z z t Okay and then I'm more oress done right okay and UT essentially this if I if we can find this we if we can somehow write it as UT then and I can show that this constant will work then we are done right so we have to find out something like this once we have something like this that will imply this right you understand so we are going to try and do that okay so how to do that let's do it okay so we will choose what that Define sorry Define Capital VT yes this I'm defining as the Norm of YT minus ZT here see the thing is T is verying between sorry here T is varying between T and T1 right so basically for that t and T1 VT I will be defining to be the norm of YT minus ZT okay if I do that now you see since what is YT minus ZT you remember you see YT minus ZT are the solutions of of this problem right X Prim = to f of x okay so those also satisfy the integral equations right the corresponding integral equation what is it so YT is given by YT plus T to T to T F of Ys DS and what is z ZT given by ZT minus t 2T capital F of Zs DS you remember when we did the P the we did the exact same thing right okay therefore what do we have we have YT minus Z of T is nothing but y of tus Z of t Okay Plus T 2 t f of y s minus F of Zs DS okay this is what we have therefore you see what is this this is VT right so if I take the norm on both sides okay so taking let me put it this way taking Norms Norm on both side what do we have both side this is nothing but V of T and this will let I can write it at V of T try inquality plus T 2T the norm of f of Y is minus F of Zs DS right this is what I can do here this is triangle inequality plain simple triangle inequality in triangle inequality okay so now you see what do we have VT is less than equal V of T you see this is where we need the lip continuity you see here if I use lip Contin I can say this is bound this can be dominated by a constant lip constant K T 2 T and this y s y- z Norm of y- Z which is nothing but V of s DS right now you see this is in this form okay what is your constant constant is this okay and uh this constants is always there U of S DS this is in this form so that will imply U of T is less thanal C * KT here so this will imply therefore therefore we can say that U sorry V of t v of T equals to uh one second uh I have to match this part so one second let me just do this part uh we have to match it so what I'm going to do is this okay so I'm going to define a new function I just have to shift it a little bit so to make the you know this variable is a problem you see it is from 0 to T right so here it is from T not to T I just have to shift the variable now to make it 0 to T and then I can use a gr wall right so what we have do is this you define uh U of T right U of T Capital U of T to be V V of t + T okay if you do that you see U of 0 is nothing but V of T so this is U of 0 C okay so what do we have then that will imply that U of T which is equals to V of t + t Okay V of t + T and that will be bounded by V of t Okay Plus t to t + t k * vs DS okay U of T you see U of T is V of t + T right so I will just writing V of t + T and that is always dominated by here you see it is I I will just change the variable here so basically from T to t plus T this will change okay so I can write it like this this is what I'm doing here I hope this is clear see this holds for all T this this holds for all T So this holds for tal to t + T also right so I I will just I'm just doing that part here and the variable is from T to t plus T that's what okay so that is there now this is nothing but V of t plus 0 to T okay K U of to D to okay I will just change V this V to U and then the this integral become 0 to T right yeah therefore you see therefore one can say from here one can say that V of t + t t + t Okay is less than equal V of t time exponential KT right that's what we can say that will imply that what is V of T then it will imply that V of T is nothing but it is dominated by V of T not time exponential K tus T clear okay and this is what the conclusion is this is what the conclusion is okay so we have the you know the continuous dependence on the initial data okay so what do we have is essentially you can say that if you have a system which is let's say Li continuous in the whole whatever the domain is okay so if if you have a system such that you know this uh Vector field is continuous in wherever it is defined let's say from Omega def Omega Conant in Omega do you think it is C1 okay let's say if it is also C C1 so basically lip is constant lip is continuous along with that it is C1 right then you can actually say that uh the problem is well poed yeah okay uh so why lip continuous and C1 does lip Contin C1 of course not because FX = to mod x as you know this disc continuous in R but it's not going to be C1 in R okay with this I'm going to end this video [Music] [Music] [Music]