 Welcome to the tutorial on general non-linear equations. In this tutorial we are going to solve 4 Cauchy problems for general non-linear equations and we will highlight some of the important and unique features of each of these examples. So let us move to the first problem. So let us solve the Cauchy problem for this equation. So first thing you have to identify what is the capital F which defines this equation. So that is p square minus 3 q square minus z. And next second step is to parameterize the given Cauchy data. Cauchy data given here is ux0 equal to x square for x in R therefore x equal to fs equal to s. This is in the notation that we have used in the proof of the existence and uniqueness theorem. It is helpful to stick to the notation done. Now we need to solve the step 1 is to solve characteristic strip ODE's. F is given like this. So what is fp? It is 2p. fq is minus 6q and pfp plus qfq is 2p square minus 6q square and using the equation, equation recall is p square minus 3q square minus z equal to 0. pq and z should satisfy that equation. So this equal to 2z. And the equation does not depend on x. Therefore the derivative with respect to fx is 0. Similarly derivative with respect to fy is 0. f does not depend on x and y. So what we have to compute is fz. fz is minus 1, minus 1 into minus p is minus p is plus p. Similarly minus 1 into minus q is q. So this is the system of characteristic strips, equation for characteristic strips. Now we have to solve this characteristic strip ODE with some initial data. But as you know the Cauchy data gives us only initial conditions for x, y, z we need to find for p and q and that is what is called finding an initial strip. So our completion of gamma to an initial strip. Gamma is given by here and then we need to find psq as solutions to this system. Often it is a case that the second equation is gives us information more quickly because first equation generally is a non-linear function. And second one most of the times because the nature of the f and g that we choose one of them is 0. For example f prime g prime in this example g prime is a let us see what that means in this example it is p f prime s is 1 plus q into g prime s is 0. So I do not write that equal to h prime s that is 2s which means we know now ps. So that is p of s equal to 2s. Now we need to find qs. Now I will use this equation. The equation is p square minus 3 q square minus z in the place of z it is hs equal to 0. But now I have found what is p square that is 4s square minus 3q square hs is s square equal to 0. So that will give us 3q square equal to 3s square and that will give us q is equal to plus or minus s. Therefore we have got two pairs ps qs ps of course is always 2s other one is qs is s or minus s. So thus we have two initial strips what are those s 0 s square this is fgh part and then pq that is 2s comma s second strip is second initial strip is 0 s square 2s minus s. So what we do is we will solve the question problem using this initial strip using the second one goes similarly and we will give the finally the answer for the second strip but we will take this as the initial strip. Step 3 is finding a candidate solution which means solving characteristic strip ODE with the initial strip. I have recalled here what is a characteristic ODE and this is the initial strip initial conditions given by the first initial strip I have chosen qx equal to s here. So we have to identify which equation is easy to solve for example the equation where p and q involve only and p and q therefore dp by dt equal to p this I will demonstrate here clearly the solutions later on I will straight away write the answers d by dt equal to p therefore the solution looks like it will be a constant times e power t right dy by dx equal to y means solution is e power x into constant but now when t equal to 0 I want the initial condition to be satisfied. So a must be 2s sorry so this equal to 2s e power t similarly we have dq by dt equal to q that will give us that q of ts equal to s into e power t so we know p and q so we can substitute these values here for x and y in the equation for x and y so therefore dx by dt equal to 2p that is 4s e power t and here dy by dt is minus 6q that is equal to minus 6s e power t now we can integrate and we get x of ts is equal to 4s e power t integral of this is same plus that constant it should be such that x of 0s is equal to s when I put t equal to 0 I get 4s from here so I need to subtract 3s from here. So now this satisfies the initial condition similarly for y this is minus 6s e power t integral will be again e power t so minus 6s e power t plus constant but at t equal to 0 I want initial condition to be 0 therefore minus 6s e power t I add 6s now this satisfies so this is x ts this is y ts. Now we need to find s and t in terms of x and y before that let us also solve dz by dt to solve dz by dt we do not need any information because it involves only z that gives us that z ts is equal to e power 2t into constant and that constant has to be s square. So therefore what I need to know to find the solution is t and s right in terms of x and y that is what the theorem says solve for t and s from these equations so how do we solve that so let me recall what is x ts, y ts we have from here you can observe that 3x plus 2y equal to 3s so that implies s equal to s of xy equal to 3x plus 2y by 3 this is done. Now what about t, t is equal to t of xy that is equal to logarithm of 4x plus 2y divided by 4 by 3 into 3x plus 2y of course only when this makes sense this is the how the s and t turn out to be now if I want to find my solution u of xy what I need to do our definition of u of xy is z of t of xy comma s of xy what is our z of ts recall that is s square e power 2t. So e power 2t is what we want we have found t here so why do not we compute e power 2t from here or sometimes is much easy to compute s e power t because that is what is there in our equations for x and y if you notice this combination comes s c power t s c power t. So if you multiply x and y suitably and add and do something then you can get it of this so you get an expression for s e power t directly then you can substitute here and get the answer. So let us do that so what we get is s e power t is actually x plus y by 2 x plus y by 2. So y by 2 will make it minus 3 s e power t plus 3 s when you add 3 s 3 s gets cancelled minus 3 s e power t 4 s e power t so you get s e power t so this is what the expression you got therefore u of xy is equal to x plus y by 2 whole square this is the solution. So if we chose that is what it is done now suppose you chose the other initial strip so this I leave it for you to worked out what is that initial strip s 0 s square p s equal to 2 s q s is minus s we get the solution as u xy is equal to x minus y by 2 whole square. Let us have remark first point is that z equal to u of xy okay so what is that one is x plus y by 2 whole square other one is x minus y by 2 whole square so z equal to u of xy in both cases contains the datum curve gamma okay that is what solution is global with respect to datum curve is true. So if you actually compute the Jacobian at 0, 0 that is actually 0 so this happens despite that okay that means if you have one point of singularity it may happen that you have a global solution and what is the domain of this solution? Domain of this solution is r 2 what better we can expect so this is also global with respect to domain okay equation is non-linear but still nice things happen for non-linear equations also it can happen okay. So with this observation what I want to say is that tools used may not be applicable but it does not prevent equations to have or from having global solutions okay tools of course when j 0 is equal to 0 you cannot apply your theorem existence uniqueness theorem you cannot apply because transfer of the condition fails but despite that we had global solution means tools may fail us but something else actually happens and because tools are only sufficient conditions isn't it application of inverse function theorem or implicit function theorem they guarantee you certain thing can be done they do not say it cannot be done if you do not satisfy these conditions. So therefore nice things can happen in this example base distinct base characteristics distinct base characteristics do not intersect they do not intersect so 3x plus 2y equal to 3s this is the equation of family of base characteristics they are all parallel lines this is in the case of the first initial strip and in the case of second initial strip it was 3x minus 2y equal to 3s this is a second initial strip so they do not intersect okay let us move on to the next problem this looks likely more complicated okay so what is f here p square plus q square plus 2 into p minus x into q minus y minus 2z and the datum curve ux 0 equal to 0 therefore this is s g is 0 h is 0 s in 0 1 so what is fp from here it is 2p the first term from here it is 2p into q minus y which is 2 into p plus q minus y what is fq it is 2q from here and from here it is 2q into 2 into p minus x now there is no p here 2p that is coming from p square plus 2 into q minus y q minus now here is 2q plus 2 into p minus x that is 2 into p plus q minus x now dz by dt after some simplifications will become I am not doing a computation please do it 2x into q minus y plus 2y into p minus x and dp by dt will turn out to be 2 into p plus q minus y 2 into p plus q minus x okay so this is the characteristic strip of ODE is for characteristic strips now completion of gamma 2 initial strip so as before this equation gives us p into f prime s is 1 plus q into g prime is 0 equal to h prime is 0 so that gives us straight away p of s is 0 so once p of s is 0 we have to substitute in the equation and get the equation for qx so next thing is f of f g h p q equal to 0 we need to solve but what is that it is f of small fs is s g is 0 h is 0 p we just found is 0 and q 0 so we have to substitute this quantities in the equation what we get is q square because p is 0 q square plus 2 into minus s into q equal to 0 so that implies if you take q common q minus 2s equal to 0 that implies q of s it has 2 solutions s and 2s therefore what are the initial strips 2 initial strips are possible say as we discussed earlier initial strips have to be smooth functions otherwise our procedure does not go through and of course initial strips are going to be restriction of the strips coming from the equation right therefore they must be smooth if the solutions are smooth so we will not take for example we could also take qs is equal to s if s is rational and 2s if it is irrational sorry not this is 0 yeah we can do this but we will not do that because that those are bad functions they are not smooth functions okay so f g h that is s 0 0 here one more 0 should be there this is p is also 0 so comma p for q also 0 is one strip another thing is 0 0 f g h p is 0 and it is 2s these are the 2 strips which are possible so let us work with this strip working with this is similar so we will not do this we will do with this so we have to solve the system of equations visibly the system is very complicated right everything is coupled like anything right p q y appear p q x appear and so on but you observe dx by dt is same as dp by dt okay that simplifies slightly right because dx by dt equal to probability that means d by dt of x minus p is 0 that means x minus p is always constant that means x and p differ by a constant so similarly for y and q okay so we have to make some simplifications so first observation is as I did x minus p d by dt is 0 so that gives us p ts minus x ts I do not know both of them but this relation is there is constant and what that constant should be at t equal to 0 what is it at t equal to 0 p 0 and x is yes so this is minus s so therefore that gives us p ts equal to x ts minus s similarly we need to find about q so y minus q d by dt is 0 so that will give us y minus q is constant and what that constant should be here I am taking q equal to 0 so it must be 0 so therefore I get q ts equal to y ts therefore the equation for x we may write it as 2x minus 2s so that is nothing but dx by dt minus 2x equal to 2s and integrating this what we get is x ts equal to s is minus 2s x ts equal to s so please do this computations there is integration of this ODA similarly the equation for y y dash okay let me do on the next page okay equation for y is y dash equal to 2y plus 2s and integrating this we get y ts is equal to s into e power 2t minus 1 so from these 2 expressions what is that expression for x x ts equal to s this will give us an expression for s function of x y is actually x and what about y y I do not know but let me write e power 2t using this s equal to x so y by x plus 1 e power 2t is x plus y is x so now equation for z z prime equal to 4z minus 2sy on integration what we get is z ts is equal to s square by 2 into e power 2t minus 1 whole square therefore u of xy equal to x square by 2 into y by x whole square that will give us y square by 2 that is a solution and solution using the other initial strip what was that s 0 0 p was 0 q was 2s gives u of xy is equal to y by 2 into 4x minus 3y so here both solutions are global with respect to domain and of course once it is global with respect to domain it is also global with respect to datum curve normally the equation global solutions so message is that normally the equations does not mean things are always going to be bad let us look at this Cauchy problem u x cube minus ui equal to 0 so f is p cube minus q this is s 0 2s root s that is the initial curve Cauchy data now this is very simple because f depends only on p and q f p is 3 p square f q is minus 1 p of p plus q of q is 3 p cube minus q that is equal to we can keep it as it is for now and f x is 0 f z is 0 so these are 0 this is 0 now we need to do this extension so always concentrate on this p f prime is 1 plus q into g prime is 0 so do not even write equal to 2s root s derivative with respect to s and that you can see is actually 3 root s 3 root s now what is q by our equation p cube minus q is 0 that implies that q is equal to p cube that is 3 root s whole cube that is 27s root s so what is the initial strip we have got only one initial strip here s 0 2s root s p s is 3 root s q s is 27s root s so we need to solve this system of equations now here p and q are the simplest equations so that gives us p ts is a constant and that has to be 3 root s similarly q ts has to be constant that has to be 27s root s and y ts is also easy y of ts is minus t plus constant when t equals 0 it should be 0 therefore this is minus t and we have to now compute x and z so dx by dt is 3p square that is 3 into 3 root s whole square that is 27s so therefore that implies x ts is 27s t plus constant so when t equal to 0 I want it to be s therefore that must be s 27s t plus s now let us go to dz by dt that is 3p cube minus q actually that is 2p cube because p cube minus q we have chosen it to be 0 so that is actually 2p cube so which is nothing but 2 into 3 root s cube 27 54 s root s therefore z of ts 54 s root s into t because this is constant does not depend on t when t equal to 0 I want it to be 2s root s because it is 0 at t equal to 0 so that if you take 2s root s common 1 plus 27t is what we have so next step is to find out t as a function of x y and s as a function of x y okay so x ts is equal to s into 1 plus 27t and y ts is equal to minus t so that will give us t is equal to t x y that is very simple equal to minus y and s equal to s of x y equal to x by 1 minus 27y okay obviously y should not be equal to 1 by 27 so something is going to happen when y is 1 by 27 therefore solution to the Cauchy problem is given by u of x y equal to capital Z of t x y comma s x y and that will be equal to 2x root x divided by 1 minus 27 y square root of course x should be positive right and y should be less than 1 by 27 because the initial data is given on the x axis and y equal to 1 by 27 is here solution is not meaningful at y equal to 1 by 27 so you have to choose either this domain or this domain we choose this domain because this is where it is in touch with the Cauchy data we want to solve solution nearby this gamma 2 now let us give a comment on base characteristic curves base characteristic curves they are given by this equation okay for each s in 0 to 1 we have considered s in 0 to 1 so for each fixed s it is a straight line this straight line is like this for reference I have written x axis and y axis and this is a line y equal to 1 by 27 this point something is happening the base characteristic curves are straight line they always pass through this point 0 comma 1 by 27 lies on them that means all of them are meeting up at this point slopes are 1 by 27 yes minus 1 by 27 yes okay as you are coming closer they are becoming steeper and steeper like that so all of them are meeting at this point and u is not defined at y is equal to 1 by 27 so this example tells us that intersecting base characteristic curves pose a problem stop or prevent a solution from being global that is the problem global with respect to domain and solution of course is global with respect to datum curve in this example though existence and uniqueness theorem asserts local existence near with respect to datum curve that kind of solution it asserts we observe in this example that solution is actually global with respect to the datum curve solution is actually global with respect to datum curve but it is not global with respect to the domain because these base characteristics are meeting up we will also see in examples using burgers equation in the next lecture that this is indeed one of the main problems even there and the last Cauchy problem for today is this equation u x plus half v y square equal to 1 and this is the datum curve and f as usual p plus q square by 2 minus 1 and once again it does not depend on x and z so these are 0 f p is 1 f q is q p of p plus q of q is p plus q square now this is p into f prime is 0 so it is not there q into g prime is 1 and that is equal to h s so q is determined now use the equation and get the p p is equal to 1 minus q square by 2 by the equation that is equal to 1 minus 2 s whole square 4 s square by 2 that is 2 s square so this is p so we have got p s and q s only one initial strip f is 0 g is s h is s square p is 1 minus 2 s square and q is 2 s now as before p t s is 1 minus 2 s square it is a constant q t s is 2 s what else is easy x t s is easy it is t plus some constant and that constant has to be 0 because it should satisfy 0 condition so x t s is t now what is y t s y t s is going to be q is a constant right 2 s so 2 s into t but at t equal to 0 it should be s so it is this and d z by d t we need to write down slightly d z by d t is p plus q square p is 1 minus 2 s square q square is 4 s square so that is equal to 1 plus 3 s square therefore that gives z t s equal to 1 plus 3 s square into t when t equal to 0 it should be s square sorry this is not 3 it is 2 1 plus 2 s square so this is not 3 it is 2 1 plus 2 s square okay so t is here s is here where is s t is here from here we can get for s this is s into 1 plus 2 t so what we get is t is equal to t of x y is equal to x and s is equal to s of x y is equal to y by 1 plus 2 x so when you substitute in the formula for z the solution we obtain will be u x y after simplification it is x plus y square by 1 plus 2 x so once again the solution is global with respect to datum curve but not global with respect to domain as we see formula has a denominator 1 by 1 plus 2 x somewhere so there is a problem at the point when the denominator is 0 so it turns out that at that point base characteristics meet up so equations for base characteristics is y by s equal to 1 plus 2 x that implies y is equal to 2 s x plus s all of them pass through minus half 0 when s is minus half y is 0 so whatever maybe the s ls goes through this point so that is this point minus half comma 0 passing through this maybe like that this is for s positive these are family correspond to positive s's and for the other one for negative s's is going to be like that and so on okay so all of them pass through so base characteristics meet up at this point therefore we see a trouble in the formula for the solution 1 by 1 plus 2 x so the solution surface will be singular at this point and that is cleared by the formula so base characteristics meet we expect troubles yeah so this completes the tutorial on Cauchy problems for non-linear equations please solve as many problems as you can and in each of the problem analyze if you can find out some expression for the base characteristic curves and see whether they meet or not meet etc consider all these considerations thank you