 In this lecture and the next lecture, we are going to discuss how to solve first order partial differential equations in the sense that how to obtain general solutions. So, in today's lecture, we will discuss about general solutions to linear and semi-linear equations because it is easier and we see how to extend these ideas to the case of quasi-linear equation in the next lecture. So, the outline of today's lecture is first we discuss a useful idea in finding general solutions and then we move on to linear equations with constant coefficients because that is a simplest case and from there we get an idea and try to solve linear equations with variable coefficients. Semi-linear equations follows very similarly to linear equations because the part where the first order partial derivatives appear in both linear and semi-linear equation they look alike. So, useful idea in finding general solutions. So, general solutions to linear and semi-linear equations can be found using very simple ideas. We will see in the next lecture how to generalize or extend these ideas to the case of quasi-linear equations. The method resulting out of these ideas is a famous Lagrange's method for finding general solutions of quasi-linear equations. So, let us discuss a linear and semi-linear equations now and what is the key idea behind this method. So, first of all observe that the partial differential equation is linear in first order derivatives. The terms involving first order derivatives look like a directional derivative. We will see more on this in the next slides. Now, what happens is change of independent variables will then convert our PDE into an ODE and we are experts in solving ODE's. That is why ODE's are easier to solve. So, that is why this strategy we will follow. Recall linear equation of first order is of this form AXYUX plus BXYUY equal to CXYU plus DXY. This we denote by L and the coefficients A, B, C, D are C1 functions of omega 2. Omega 2 is a subset of R2. Of course, there is another way to think of this as a special case of quasi-linear equations in which case A of XYZ will appear but A of XYZ is independent of Z. So, asking it that A of XYZ is C1 of omega 3 is same as asking A of XYZ is C1 of omega 2 because it does not depend on Z. And we do not want both these coefficients of the partial derivatives to vanish at the same time in the sense at the same point in omega 2. So, at every point in omega 2 at least one of them should be non-zero. And then semi-linear equation of first order looks like this AXYUX plus BXYUY the left-hand side is exactly same as that of L whereas the right-hand side C of XYU that is a semi-linear equation. And assumptions natural are A, B are C1 of omega 2 and C is C1 of omega 3. Of course, omega D are subsets of Rd equal to 1 3 omega 2 is a projection of omega 3 to the XY plane and A square plus B square is not equal to 0 on the domain omega 2. Let us see an observation that first order partial derivatives appear in the same way for both L and S we have already observed this namely this is the part AXYUX plus BXYUY this is the left-hand side in both L and SL. If either A or B is a 0 function what does that mean imagine A is 0 then this above expression is simply BXYUY and the equation would look like BXYUY equal to CXYU plus D in the case of L and BXYUY equal to C of XYU in the case of SL. That means there is no derivative with respect to X will appear if A is 0 in other words it is a ODE it reduces to ODE. Therefore, solving L and SL in such a case is same as solving ODE it is no different from solving ODE. Now we ask a question do not we like if it is true that either A or B is 0 function in every same linear equation? Answer is yes of course we like it but we know that we cannot expect that but we can make it happen. We can make it happen after a change of variables we will we are going to do that now. First let us understand the simplest case it is a linear equation with a constant coefficients. So, let us look at this equation AUX plus BUI equal to 0 coefficients are constants so A and B are real numbers at least one of them is non-zero real number. If one of them is 0 it is a ODE so we do not want to discuss so you can as well assume both A, B are non-zero of course it is not needed for the purposes of our lecture but you can imagine that both are non-zero numbers that is when it is not clear what you want to what you can do immediately. If any one of them is 0 it was ODE you have already solved. So, we place this assumption that one of these A, B must be non-zero real number. Let us look at an example of course in this example I made it B equal to 0 and A equal to 1 this is just an example this will guide us in further proceedings. So, let us look at this example UX equal to 0 its general solution is given by UXY equal to FY where F is any differentiable function because you need to differentiate this and check that derivative is 0 so derivative should exist that is reasonable to assume F is a differentiable function though it does not matter here even if F is just a function not even continuous does not matter because when you write the partial derivatives expression with respect to UX all difference quotients are 0. So, how did we arrive at the above answer it looked like ODE in the variable X. So, AUX plus BUY equal to 0 means the directional derivative of U in the direction A, B is 0 in terms of the notations it is like this DAB U equal to 0 this looks similar to UX equal to 0 UX equal to 0 is nothing but D 1 0 the vector AB is 1 0. So, it is a directional derivative of U in the direction of 1 0 that is the partial derivative with respect to X. So, of course DAB U equal to 0 is more general than UX equal to 0. But important thing is that it is a directional derivative we know the directional derivative is like one variable derivative. It looks like one of the other variables is not there in the equation exactly like UX equal to 0 Y is absent. So, in this case we have to find out what are the variables one of them is there in this equation but another one is not there we need to identify in this case it is very simple X is there because partial derivative of U with respect to X is there Y is not in the equation that is very obvious here. Now, let us understand UX equal to 0 more closely UX equal to 0 it means that U is constant on every line which is parallel to X axis parallel to X axis means Y equal to constant at the equation. So, solution we saw U of XY equal to F of Y. So, if Y is constant UX is constant. So, in general we expect by analogy in this case U is constant on every line which is parallel to X axis now in this case I expect U is constant on any line having the direction AB. In the case of UX the direction is 1 0 right parallel to X axis has the direction 1 0. So, therefore I expect from this experience that solutions of DAB U equal to 0 are constant on any line having the direction AB. We are going to prove this. So, in the next slide we compare these two equations side by side. First let us look at UX equal to 0 and AB we need to make some special numbers. So, AB we take 1 1. So, now UX equal to 0 solution is FY. So, whenever Y equal to 1 that is a line parallel to X axis it has a direction 1 0 and U is F of 1 U is constant. Now if you take any other line parallel to X axis that means any line having the direction 1 0 will look like Y equal to K for some K and U on that is F of K. So, it is constant on each line parallel to X axis. Now in this case UX plus UI equal to 0 this is having the direction 1 1. 1 1 is this direction the line Y equal to X has the direction 1 1. So, U should be constant along each of these lines parallel to the line Y equal to X. Any line parallel to Y equal to X looks like Y minus X equal to K for some K and U is equal to F of Y minus X. You can check by substituting in this equation that this U is indeed a solution to this equation. So, we have obtained a general solution in the case of UX plus UI equal to 0. So, we need to find what that is we got from AB equal to 1 1 if it is general AB what will this be we will see that formula. Now how do I extend these ideas to linear equations but with variable coefficients how will we do that? So, what to do that is the first question. So, notice that for every point in omega 2 this AXY UX plus BXY UI is also a directional derivative of U at the point XY. But the catch is that the direction depends on the point XY it is not constant A comma B as in the constant coefficient case here the direction is AXY comma BXY it varies from point to point this is the difference otherwise it is also directional derivative. Now can we generalize from constant coefficient case and say that U is constant along each line having this direction we would like to say that of course what it means that a line having this direction is questionable because this changes from point to point right in the constant coefficient case it was just A comma B it never depended on XY but now it depends that is why I have put in quotes can we expect such a result? Having this direction means this is a slope Y coordinate by X coordinate B by A that is a slope of the line what line XY is varying B by A will vary from point to point. So, as the direction depends on the point answer to the above question is as the direction depends on the point XY in omega 2 we cannot talk of lines but an infinitesimal version of lines works what is that we are going to present that soon it means that we need to consider curves with slopes B by A we cannot say lines all the time consider curves now lines are also after all curves having a constant slope that is why they are lines but here we have to admit now more general things than lines so we admit curves with slopes B by A that is of course one condition is that we have A in the denominator so we must assume that A is not 0 in the domain let us assume that we will assume that and we will also show that does not make much difference if you have not assumed that A is not equal to 0 throughout it is enough at a point that A is non-zero we know that every point X naught Y naught in omega 2 one of them is non-zero A or B right so we assume that A is non-zero and we can do the things if A is 0 we will work with B so that can be done. Okay so this means what curves with these slopes means you have to look at solutions of this ODE this is the slope of the curve at a point X Y and that I am saying should be B by A. So if B and A were constants the solutions are straight lines so now these are curves so d Y by dx equal to B X Y by A X Y okay so we have got hold of analogous things to straight lines which are curves with slopes B by A okay so we assume that A is not equal to 0 on omega 2 we will come back to this discussion after we finished doing this analysis okay where we say that does not matter you do not have to assume that it is non-zero throughout omega 2 because of certain things which will come into the analysis soon okay there is nothing that you gain by assuming that A is not 0 and throughout omega 2 we will come back to that. So let zeta be a solution to this okay defined on some interval J in X okay whenever this is a solution one observation is that U is constant along this curve X, zeta of X X, zeta of X as X varies describes a curve and U of X comma zeta of X is constant whenever zeta is a solution to this ODE we can verify that so d by dx of U of X zeta X we want to compute this now X dependence is there in both the coordinates here as well as here both the components so first you have to differentiate U with respect to X and derivative of X with respect to X is 1 then differentiate U with respect to Y and differentiate zeta of X with respect to X that is what the chain rule says okay so that will give us a zeta dash of X d zeta by dx through this dependence we get this. So as I told you chain rule will keep coming definitely throughout this lecture on the next one but chain rule if you remove then I think you cannot do PDE so we can we may also say that this is one of the most fundamental results in differential calculus it does not look like a big result to us but it plays a role everywhere okay. So question good but tell me how to get a formula for solutions okay you have observed that if you have a solution that will be constant along curves which are solutions of this ODE fine but from there how do you get formula for the solution that is a question the answer is that do not worry using this family of solutions to one because it is a ODE right so its solutions will be a one parameter family so using that family we convert our PDE into ODE okay we convert our PDE into ODE and ODE is may be solved to get solutions. So now you are as good as your ability to solve ODCs we will do this shortly of course this assumes that you are able to solve this equation one get that family and then converting the PDE into ODE is very simple and your ability to solve the ODE it hinges on that but definitely there is an algorithm okay. So we follow a common strategy to solve L and SL because as we observed the LHS in both the equations and L and SL are is the same both are the same. So now we have to do what is called change of coordinates sometimes called change of variables. So suppose that we have a change of coordinates from x, y to xi eta and vice versa change of variables always like you are going from x, y description to xi eta and you should be able to come back only then it is useful otherwise things are lost okay. So change of variable always both sides and vice versa given by this. So xi equal to small phi of x, y eta equal to small xi of x, y and x equal to capital phi of xi eta y equal to capital xi of xi eta it is very important to write this kind of setup whenever you do change of variables so that you will not have confusion. This is another part change a chain rule will be used chain rule as such is easy okay very easy you can apply but here when you do change of coordinates that is where your real test of understanding will come and to be careful better you always use this kind of notations okay. Normally people write xi equal to xi of x, y eta equal to eta of x, y avoid that that is the first message okay. So this is the change of coordinates obviously it means that a certain domain has this change of coordinates in R2. A function now under this change of coordinates u of x, y gets transformed to a new function call it new function new notation w of xi eta do not call u of xi eta that will cause confusion w of xi eta and vice versa of course by this formula u of x, y right but x equal to phi xi eta y equal to xi xi eta so you substitute you get a function of xi eta similarly here u x, y equal to w of phi x, y xi x, y. Now we illustrate this in a picture see here we have the rectangular coordinates x, y and we have gone from here to xi eta coordinate system through these functions small phi, small psi xi equal to constant of this blue curves eta equal to constant of this black curves. From here you can come back via this transformation and if you have a function defined from here x, y thing to R you can define a function from xi eta description to R okay. Basically it is the address that you are changing with respect to the new coordinates earlier to describe a point in the plane you are giving a x address and y address x coordinate and y coordinate. Now for example you are using a different coordinate system if you want to describe this may be eta equal to some number c1 this may be xi equal to some number c2 so c1, c2 will uniquely fix you for example look at our globe you can easily give coordinates of any location using the longitude and latitude numbers right. It is exactly like that so it is a different change of here it is a change of coordinates so you need to give address fully full address okay. So how PDE changes under change of variables very important we need to do that so AUX plus BUY BXY UI is our LHS in both L and SL equations. Now we will change it to xi eta coordinates very easy we have to look at the relations that we have between U and W and differentiate and find out what is UX what is UY substitute for UX and UY similarly for X and Y here you substitute then you will get the new expression. So UXY, UX is derivative of W look X appears both in this location and this location so differentiate W with respect to the first variable which is xi at this point phi X, Y, psi X, Y, phi X, Y, psi X, Y and then differentiate phi with respect to X at the point X, Y then differentiate W with respect to the second coordinate that is W eta at this point phi X, Y, psi X, Y this is chain rule and then differentiate psi with respect to X at the point X, Y. Similarly you do for UY now substitute UX, UY in this expression AUX plus BUY so you get this expression what have you gained nothing because earlier it was AUX plus BUY now you got something into W xi plus something else in W eta now the coefficients are all functions of xi eta you have to convert them into xi eta because phi X is a function of X, Y, A is also function of X, Y but you know the expression of X, Y in terms of xi eta substitute that you get this. So these are the functions evaluated at this points X equal to this, Y equal to this. So we have got one expression in X, Y coordinates equal to another expression you may call this capital A of xi eta W xi plus capital B of capital B of xi eta into W eta. So we have not gained anything but this is where we can play because xi eta is something that I want to make a choice so that it will be useful for me convenient to me. In fact we wanted to convert to ODE so we would like to remove one of these derivatives how do I remove one of the derivatives just ask that xi eta should be such that this guy is 0 or maybe this guy is 0 then it becomes a ODE in one of the variables that is the idea. So choose xi so that this features only derivative with respect to xi, I decided to let only derivative with respect to xi be there. So that means I do not want this term therefore I demand this guy is 0 choose xi such that this is 0 this forces xi to satisfy this. Of course if you have said there is I do not want derivative with respect to eta you would have said this is equal to 0 but if you see equation is the same right A phi X plus B phi Y equal to 0 so equation is the same equation okay. So let us find a xi with this property now this is also first order PDE right we are trying to solve one first order PDE we ended up with another first order PDE. So what have we gained thankfully this equation is a homogeneous equation so that helps us right hand side is 0 linear homogeneous how does that help? Earlier we observed that every solution of this particular equation is constant along solutions of some ODE which is this we observed this. So from theory of ODE's we know that solutions of the above ODE this ODE form a one parameter family. So assume that solutions are given implicitly by eta of X Y equal to K where K is a parameter. Why one parameter family because this is the first order ODE one initial condition you give and you get a solution that is a parameter okay. Same thing we can express differently but that is why I am assuming that they are given in this form. Once you have this form you set psi equal to eta psi of X Y equal to eta of X Y so you know psi now now you need to still find phi. So you have set psi like this it means your PDE the transformed equation is good because you do not have this term okay but phi is still there. So we need to find phi only then we will have a change of coordinates we will do that. Now having chosen psi choose phi to be a C 1 function such that this Jacobian is never 0 the Jacobian is never 0. The Jacobian condition above will guarantee that this function X Y going to phi X Y psi X Y has a local inverse and phi I psi being C 1 functions the inverse will also be C 1 function that means we have a diffeomorphism. So we have to apply what is called inverse function theorem therefore this is the second most important theorem in differential calculus of multivariables inverse function theorem which is equivalent to another theorem which is called implicit function theorem. You can prove one from each other so that is why both are equivalent and you must be thorough with application of these theorems how to apply these theorems. So the inverse function may be expressed now that X as a function of xi eta and Y as a function of xi eta you can do that. But if you look at the inverse function theorem the assumption will be at some point X 0 Y 0 Jacobian is non-zero then a neighborhood of X 0 Y 0 you have this as a diffeomorphism. So inverse function theorem conclusions are what are called local even if you have the Jacobian everywhere non-zero you cannot say that throughout the domain X Y on which this non-zero this defines a inverse function diffeomorphism you cannot that is why local inverse is always a conclusion from the standard inverse function theorem. What is local inverse it means each point there is an open set on which the above conclusions hold and the open set need not be omega 2. Thus the LHS of both L and SL which is here AUX plus BUY becomes this okay where we are just writing where these coefficients are evaluated at they are evaluated at this W xi is at xi eta. So the right hand side is function of xi eta. So we can give a name to that let us call it A of xi eta as this quantity. Therefore this LHS of L and SL become simply A xi eta W xi of xi eta. So no eta derivative here so it is going to become ODE. The linear equation now becomes this this is LHS right hand side C of X Y you express as xi eta by this change of variables use W and D also you express using the change of variables D of X Y capital D of xi eta that is a relation. Now you may write it like this it is a ODE right it is a linear ODE therefore it can be solved. Now let us see what happens to semi-linear equations. The PDE will transform to this because right hand side there is nothing to do much it is just C of small C of X Y U earlier now it becomes capital C of xi eta W. Now you write like this and the equation is non-linear ODE. The ease of finding a solution depends heavily on the non-linearity the capital C and capital A. But we know that it has solution as the right hand side is a local ellipses function solutions are there. Once the function W is determined function U is known now we know right the correspondence between U and W. Now small remark on our assumption that we assumed this right A is not equal to 0 on omega 2 and one more thing we assumed on the way after choosing psi we said choose another phi C 1 function such that the Jacobian is never 0 on omega 2. However we met inverse function on the way that theorem even if you with go with the global conditions like this this one it will give you only local conclusions there is a problem of the inverse function theorem. So we did not gain much despite our global assumptions that tells that there is no need to assume globally. Assume that A is non-zero at some point x naught y naught and the conclusion that you get will be in an hybrid around x naught y naught. In any case even if you assumed A is not equal to 0 in omega 2 when you went to inverse function theorem it only gave you conclusions around the point x naught y naught. So as a consequence the change of coordinates you will find in a neighborhood of that point whichever point you have fixed here I have fixed x y. So but there are also global inverse function theorems this is just for you to be aware of global conclusions are possible from global inverse function theorem. For this I refer you to this book of multivariable real analysis volume 1 it is called differentiation it is authored by Wister, Matt and Kolk they also have a second volume which is about integration these are very slightly fat books but they write very nicely. So let us now specialize to constant coefficients. So solutions to d by d x equal b by a because we have not yet solved a u x plus b u y equal to 0 so far now we are going to solve. So equal to b by a are given by b x minus a y equal to k where k is a parameter choose psi as b x minus a y okay solutions okay ad y equal to b dx so b after integrating you get b x minus a y equal to constant that is how we get psi. Now you have to choose phi the only condition on phi it should be a c 1 function and the Jacobian non-zero. So this is the Jacobian psi I know the derivatives I have put in here this is psi x this is psi y you want this to be non-zero in this case there is a easy choice for phi it is a x plus b y what is the relation between like a x plus b y equal to constant and b x minus a y equal to constant they are perpendicular they are like x axis and y axis right okay so note that this change of coordinates x y going to a x plus b y b x minus a y you may call this as psi equal to phi of x y this is eta equal to psi of x y these are global it is an invertible it is a linear transformation not to it is a invertible it is global this is linear no linear if things are invertible you will get global. But in the general situation these are non-linear equations okay special case of constant coefficients let us continue the equation a u x plus b u y now becomes a square plus b square and w xi equal to 0. Now I need not even bother to change this into some new coordinates because this is non-zero right you can cancel and conclude w xi equal to 0 and solution it means it is a function of eta alone right w xi equal to 0 it is a o d e you see from a u x plus b u y equal to 0 we have got a o d e w xi equal to 0 of course after change of variable. So this implies that w is simply a function of eta now what is eta we can substitute back get the u u of x y equal to w of phi x y psi x y but w I know is only arbitrary function of second coordinate that is f of psi of x y and what is your psi of x y b x minus a y. So you just connect it back to our original equation u x equal to 0 that example u x equal to 0 equation means a is 1 b 0. So therefore it is arbitrary function of minus y arbitrary function of minus y is same as arbitrary function of y so we got back. So observe that solutions do not depend on the choice of phi at all. Now you see b x minus a y equal to constant the function is constant on that line. So this again means that u is constant on any line which is having the direction of a b. Now let us look at an example x u x plus y u y equal to 2 u here I have said using change of coordinates reduce this equation and find the general solution. So this is a recall of the change of variables we have to do this and then go on substitute for u x u y expressions inside this get the expression x phi x plus y phi y w psi plus x psi x plus y psi y e w eta now set this equal to 0 find solutions of this. So that means divide by dx into y by x and solutions are given by x by y equal to constant that is a family therefore psi of x y is psi x by y now we have to find phi choose phi equal to y and Jacobian condition will be satisfied. So we have got now psi and eta this is a function of phi of x y eta is a function of x y psi of x y is equal to x by y and then you get the relation between x and y and psi and eta then the LHS becomes y w psi because we made sure that w eta coefficient is 0 so y w psi but what is y y is psi so it is psi w psi is the LHS what is RHS it is 2 u so it is 2 w so psi w psi equal to 2 w. So w psi equal to 2 by psi into w it suggests that we should avoid psi equal to 0 so solution will be this w psi eta equal to f of eta into xi square after solving this. So this is a way to solve using change of variables and then going back to x and y we get u x y equal to y square f of x by y now it is clear that it is defined for all x and every y such that x by y belongs to domain of f. So you choose any arbitrary function that you like of one variable and x by y should make sense that means y should be non-zero and then it should be in the domain of f so you get so many solutions. Now this is another example where the right hand side is u square non-linear equation right but as a PDE it is a semi-linear equation LHS is same so we can skip the computations finally you end up with w psi equal to 1 by psi w square actually what you get is psi w psi equal to w square so I take it this psi. If you solve you get this answer w psi equal to minus 1 by f eta plus log mod psi f is an arbitrary function going back to x y coordinates we get this expression so this clearly tells y equal to 0 has to be avoided because logarithm is there and once y is non-zero it is fine. So and x by y should belong to domain of f so choose any f of one variable function of one variable with some domain and then this solution is valid for all those x y such that y is non-zero x by y belongs to domain of f and log mod y plus f of x by y is non-zero. So domain of u may be written down that is not important one can always write down. So summarize what we did is that linear and semi-linear equations were transformed to ODE's by a change of variables and general solutions were obtained. Two examples were presented to concretely illustrate the general algorithm that we have presented for finding general solutions to L and SL. So in the next lecture we are going to see how these ideas can be generalized or extended to the case of quasi-linear equations. Quasi-linear equations will pose a new troubles we will see how to overcome that. Thank you.