 Welcome back to the series of lectures on ODE and its applications. Today, we will discuss about the existence and the uniqueness of solution of initial value problems. You have already seen that in the theory of first order and second order differential equations, the need of existence and uniqueness of solution and this session, we deal with an initial value problem that is a differential equation of the form d y by d x is equal to f of x y. So, call this as a given differential equation, where f is a function is defined on r cross r to r is a function not necessarily linear. So, this model can handle both linear and non-linear differential equations. Along with the differential equation, we also give an additional initial condition y at x 0 is equal to y 0. So, call this as equation number 2. This equation 1 along with the initial condition 2 is known as an initial value problem and equation 1 plus 2, this form an initial value problem is an initial value problem and commonly written as IVP initial value problem. Now, this model, this first order differential equation could be a model derived from a physical system and a very important question. Once a model of this form is framed is, does there exist a solution to this equation? If the solution exist, will that solution be unique and also will the solution varies continuously with respect to the initial condition y 0. So, these are the major three problems, which we will discuss in details in this session. See, first let us see what you mean by a solution of this differential equation. So, equation of IVP, so equation 1 and 2, this form an IVP, the solution of IVP. The solution of IVP is a function say y x, which is differentiable and satisfies the equation 1 and the initial condition 2. So, a solution, since the equation is dy by dx is equal to f of x y, so we require that the solution y has to be a differentiable function, it has to satisfy the differential equation. So, therefore, it is a differentiable function and also satisfies the given initial condition y at x 0 is equal to y 0. In other words, we are looking for a function, so we look for a function y x, which is differentiable, we look for a function y x, which is differentiable and starts from the initial point x 0, y 0, starts from the initial point x 0, y 0, that is if this point is x 0 and this point is y 0, say this is a point x 0, y 0. We are looking for a function, which starts from x 0, y 0 and whose slope of y x at any point x y in the x y plane is dy by dx at x y, which is given by the differential equation f of x y. So, therefore, the solution of a differential equation 1 with the initial condition 2 is a function that starts from the initial point x 0, y 0, of course, that is differentiable and the slope at any point is given by dy by dx is equal to f of x y, which is coming from the differential equation f of x y is given to us. So, slope at any point x y is known to us and the initial point is known to us, we are looking for the function, which starts from x 0, y 0, which starts from x 0, y 0 and whose slope is at any point is f of x y. Now, we deal with the well poseness of this problem, well poseness of mathematical model. So, the mathematical model is given by the initial value problem. So, let the mathematical model is given by the differential equation dy by dx is equal to f of x y with initial condition y at x 0 is y 0. So, this model is well posed if the following three properties are satisfied, if the following three properties are satisfied. So, this well poseness was raised by the famous French mathematician J Hadamard. So, the well poseness problem was introduced by Hadamard and Hadamard raised the issue of well poseness, which says that given a mathematical model, if the model satisfies the following three properties, then the model is well posed. The first property is there exist a solution to the initial value problem, there exist a solution. So, that is the existence problem, which is known as the existence problem. So, given a mathematical model, the first question is to decide whether there exist a solution or not, that is existence problem. Second problem is a solution unique, the solution is unique. So, that is uniqueness problem. If the solution exist, will there be more than one solution or the solution is unique and the third problem or third property of a well posed problem is the solution. So, behavior, so solutions behavior changes continuously with respect to the initial condition, the initial condition. So, these three problems, the first one is there exist a solution to the mathematical model. The second problem is whether the solution is unique and third problem is whether the solution changes continuously with respect to the initial condition, initial condition here it is y 0. So, the third problem is known as the stability problem. So, the existence problem first, then the uniqueness problem and the third one is a stability problem. So, if a model is not satisfying these three properties, so any model, which is not satisfying these three properties is known as a ill posed problem. Ill posed a problem does not satisfy the hard amount, does not satisfy the hard amount well posed conditions. Now, we will discuss the problem of existence, uniqueness and stability of some symbol differential equations, symbol initial value problems. So, let us take an example, so example one. So, consider the first order, consider the first order homogeneous linear differential equation, homogeneous linear differential equation given by d y by d x is equal to 2 y along with an initial condition say y at 0 is 3. You have already seen how to solve this linear differential equation in the previous lectures. If you separate separating out the variables and integrating, so separating out the variables and integrating, we obtain d y by y is equal to 2 d x. If you separate out the variables and if you integrate adding a constant c, we get ln of y is equal to 2 x plus c. Taking exponential on both sides, you get y of x is equal to e to the power 2 x plus c, which is equal to say the first constant is c is 0, you get c into e to the power 2 x, where c is equal to e to the power c 0. Therefore, the solution of this differential equation, I have not applied the initial condition, the solution of this differential equation can be obtained easily by separating out the variables or by the method of separation of variables. So, y x is equal to c e to the power 2 x, which is a family of solutions or curves. So, one parameter c, this is a solution for every value of c. So, c is an arbitrary constant. So, the given linear differential equation has got infinitely many solutions and for every c, the given expression y, this y is equal to c into e to the power 2 x is a solution. So, if you plot the solution, this is one solution or one value of c. So, this is c 1 e to the power 2 x and this could be another solution for c 2 e to the power 2 x and you have a family of, one parameter family of solutions. So, these are all solutions, a family of solutions to the given linear differential equation and this solution is known as a general solution. So, y x is equal to c e to the power 2 x is known as a general solution to the differential equation d y by d x is equal to 2 y. So, I have not applied the given initial condition. So, once I give an initial condition, so applying the initial condition y at 0 is 3, then this becomes y at 0 is equal to 3, which is equal to, we can plug in the value of, value to x as 0. So, c into e to the power 2 into 0, so which is c or in other words c is equal to 3. So, therefore, we get the solution y x is equal to 3 e to the power 2 x. So, this is a solution of the given differential equation satisfying the given initial condition. Say if this point is 3, say then 3 e to the power 2 x. So, this is a solution if this point is 3 and for a particular value of c, the general solution becomes a particular solution and if you give a specific value to the arbitrary constant c, then the solution we obtain is known as a particular solution. A particular solution, so y of x is equal to 3 e to the power 2 x is a particular solution and similarly y of x is equal to 5 e to the power 2 x, they are all particular solutions by putting various values. So, initially it is c is equal to 3 and second case c is equal to 5. So, a particular solution is obtained by giving a particular value to the general solution. Now, for a given differential equation, there can be another type of solution which is known as a singular solution. So, three types of solutions we have discussed, one is a general solution and a particular solution and a third type of solution is singular solution. So, general solution we obtained and in the general solution, there is a parameter a parameter that is a parameter family of solution. So, in our case, it was y is equal to c e to the power 2 x and particular solution say c is equal to you particularize the value of c, c is equal to 10, then y is equal to 10 e to the power 2 x is a particular solution and a singular solution is a solution which cannot be obtained from the general solution. But in our in the previous example, we have only these two types of solution, general solution and a particular solution and singular solution does not exist for the previous example. Now, we consider an example. So, example is called a 2. So, consider the non-linear differential equation say d y by d x is equal to y minus 3 the whole square. So, we have seen that this equation is obviously a non-linear differential equation. So, how to solve it? By the method of separation of variables, by the method of separation of variables, separating out the variables and integrating you get this is d y y minus 3 square d y is equal to d x and integrating plus adding a constant c. So, integral of 1 by y minus 3 the whole square is minus 1 by y minus 3 which is equal to x plus c. So, this simplifying we get y is equal to you take y minus 3 over there and minus 1 by x plus 3 there. So, y is equal to 3 minus 1 by x plus c. So, this is a general solution for every value of c, every value of c this is a solution. Therefore, this is a general solution. So, this is a general solution and if you give particular values to c say for example, y is equal to 3 minus 1 by x is a particular solution. So, where c is equal to 0 when c is equal to 0 you get a particular solution. So, this is a particular solution. Now, you can verify easily that y is equal to 3, once you substitute y is equal to 3 is a constant function into this equation and this equation is satisfied is also a solution, but that is not obtained from the general solution and therefore, this is a singular solution. So, y is equal to 3 is a singular solution, is a singular solution. So, this example we see three types of solution general solution and a particular solution and also a singular solution and we can give see further examples say another example say example call it 3. So, consider a differential equation d y by d x is equal to y square minus 4, y square minus 4 again by the method of separation of variables and we can integrate and we can easily see that y x is equal to 2 plus 2 some constant c times e to the power 4 x divided by 1 minus c e to the power 4 x is a general solution. I will leave this as an exercise to show that y is a general solution and once you specialize say c when c is equal to 0, then we obtain y is equal to 2 is a particular solution. Now, this can also be shown that observe that y x is equal to minus 2 if you look at the function constant function y x is equal to minus 2 also satisfies given differential equation. So, y is equal to minus 2 is also a solution which cannot be obtained from the general solution. So, therefore, this y is equal to minus 2 this solution that cannot be obtained from the general solution. So, hence so therefore, this is a singular solution is a singular solution. Now, we look into initial value problem where an initial value problem need not have a solution or another initial value problem have more than one solution or an initial value problem has a unique solution these three situations. So, initial value problem cases no solution infinitely many solution and unique solution. So, these three situations. So, we will look into some examples of initial value problems that has a unique solution and infinitely many solution and no solution all these situations will. So, let us first consider a linear case. So, later on we will give sufficient condition to ensure under what condition an initial value problem has a unique solution and under what condition a solution has a solution and solution is unique and all such things we will deal with later. So, example 4 just to get a feel of existence and non-existence of solutions we take a few more examples. So, consider the initial value problem say d y by d x is equal to say 2 by x y with an initial condition y at 0 is 0. So, obviously this is a linear equation. So, linear differential equation and it is a homogeneous linear homogeneous and variable coefficient. So, this coefficient is variable 2 by x. So, it is a and there is no term which does not depend upon y. So, therefore, it is a homogeneous and obviously it is linear with respect to the unknown function y. So, therefore, it is linear differential equation. Now, let us look into the solution again by the method of separation of variables method of separation of variables. We have d y by y is equal to 2 by x into d x separating out the variables and integrating and adding a constant call it c 0. So, you can write this is a ln of y this is 2 ln of x plus a constant c 0. So, which is ln of y plus a constant y which is ln of x square 2 ln of x is ln of x square plus c 0 and bring this term over here ln of y minus ln of x square is equal to c 0 that is ln of y by x square is equal to c 0 or taking exponential on both sides get y by x square is equal to e to the power c 0. So, call this as c. So, therefore, y is equal to c into x square is the solution is a general solution where c is an arbitrary parameter. For every value of c y is equal to c x square is a solution and in order to satisfy the initial condition y at c 0 is obviously satisfied y at c 0 is c into c 0 square which is c 0 is obviously satisfied for all values of c. So, therefore, y is equal to c x square is a solution to the initial value problem. It is satisfying the differential equation at the same time it is also satisfying the initial condition for all values of c. So, y is equal to c x square is a solution to the initial value problem for every value of c. So, what does it say? It says that the initial value problem has infinitely many solutions. So, if you look at the solution, so if you this is a parabola y is equal to c x square for one value of c and for another value of c this is another parabola y is equal to c 1 x square y is equal to c 2 x square and for any value of c. So, this is also another solution. So, this is a family of solutions of parabolas. You get infinitely many parabolas happens to be solutions of the initial value problem. So, therefore, for an initial value problem there is a possibility that the solution is not unique it may have infinitely many solutions. Now, if you look at the same problem the same equation with a different initial condition. So, look at the same problem. So, if the example is 5 then d y by d x is equal to 2 by x y. Now the initial condition is changed y at 0 is 1 and as we have already noticed that all solution of this initial value this differential equation should pass through the origin. So, it cannot pass through the point 0 1. So, 0 1 is a point you are looking for you want the solution to pass through the point 0 1 when x is equal to 0 and this is not possible. So, there is no solution this is 1 which is passing through the point 0 1 and is also a solution to this differential equation. So, no solution satisfies the initial condition y at 0 is equal to 1. Why because y of x is c into x square and you want y at 0 is equal to c into x is 1 0 square and you want this to be 1 which is not possible. So, therefore, the moral of the story is no solution to the initial value problem d y by d x is equal to 2 by x y the initial condition y at 0 is equal to 1. And one more thing we observe that if we change the initial condition for the same problem say it is a example 6 the same differential equation d y by d x is equal to 2 by x y and y at 1 is 2 say for example I am giving a condition at 1 which is a non-zero condition the value is non-zero. So, it can be shown easily that y of x we have already seen the solution is of the form c x square and you are looking for y at 1 is equal to c into 1 square which you want this to be 2. So, this implies that c is equal to 2. So, therefore, y is equal to 2 x square y is equal to 2 x square y is equal to 2 x square is a unique solution. So, this example the same problem with a different initial condition d y by d x is equal to 2 by x into y with a different initial condition y at 1 is equal to 2 has a unique solution. So, same differential equation which is giving rise to three different situations where no solution and where there is unique solution and also there is infinitely many solution. Now, we look into the solution of a differential equation where it is defined. So, our equation is d y by d x is equal to f of x y, initial conditions y at x 0 is y 0, the domain of in which the solution is defined. So, for a initial value problem of this form there may be a solution starting from the point x 0 x 0 y 0 and that solution may exist only for some interval on the x axis. So, this is x 0 and this is y 0 and this is a point from where we start x 0 y 0 looking for a solution and the solution is y x and it can in some many different many of the initial value problem this interval call it x 1. The solution exist in a small interval x 0 x 0 x 1 this is called the domain of solution domain of solution y and it can happen that the solution is defined on the entire x axis and some problems it is defined only on some small interval or only on some particular bounded set in the x axis. So, for that also we look into some examples. So, example 7. So, consider an initial value problem. So, consider an initial value problem given by d y by d x is equal to y square and initial condition is y at 0 is equal to b a positive number. This is a non-linear differential equation the initial condition is y at 0 is some number b we are starting at this point 0 b and we are looking for a solution y. So, from the initial condition the point x is equal to 0. Let us look at the solution again by the method of separation of variables. So, by the method of separation of variables. So, we get separating out the variables this is 1 by y square d y is equal to d x and integrating adding a constant c and integral of 1 by y square is minus 1 by y is equal to x plus c. So, this just by simplifying you get y is equal to minus 1 by x plus c. So, the solution is y is equal to minus 1 by x plus c. Now, the initial condition y at 0 is b. So, this implies that y at 0 is equal to b which is equal to minus 1 by c. So, this implies that c is equal to minus 1 by c. So, plugging this value to the solution y of x is equal to minus 1 by x minus 1 by b which is equal to b by 1 minus b x. So, this is the solution to the initial value problem. So, this is the solution to the initial value problem. So, the solution is y of x is equal to b by 1 minus b x. So, look at the solution. So, starting from 0 b as you increase x if x is less than 1 by b, see if x is equal to 1 by b then the solution blows up. So, therefore, the solution exists. So, this exists for x strictly less than 1 by b. So, therefore, if b is 1 then 1 by b is 1. So, this is 0 to 1. So, for example, b is equal to 1 then solution exists the domain of solution is 0 to 1. But, if b is a very large number, so if b is large, so 1 by b is small and hence the domain of existences, domain of existences from 0 is very small in double. Note that although the equation looks to be a very nice equation, the initial value problem looks nice on the entire real line. Starting from 0 b the solution y x exists in a small interval 0 1 by b. So, 0 1 by b if b is large, it is a small interval. So, solution exists if b is large then the solution exists on a small domain to the right of x is equal to 0. Now, looking at another program example, call it example 8. Here we have a small interval here non-uniqueness. So, non-uniqueness of non-linear initial value problem. So, consider the initial value problem is given by d y by d x is equal to 3 y to the power 2 by 3 and the initial condition is y at 0 is 0. So, this is also a non-linear differential equation with a given initial point 0 0, initial condition is 0 0. So, you are looking for a solution that starts from 0 0 and again by the method of separation of variables. So, this is d y 1 by 2 to the power 3 by 2. So, 3 is equal to d x and integrating and adding a constant c. So, which is integral 1 by 3 y to the power minus 2 by 3 d y which is equal to integral d x is x plus c. So, integral of this one y to the power minus 2 by 3 plus 1 or divided by minus 2 by 3 plus 1, which is y to the power 1 by 3 is equal to x plus c. So, this implies that the solution y satisfies. The solution is given by y of x is equal to x plus c the whole cube. Now, applying the initial condition y at 0 is 0 which implies that 0 plus 3 whole cube. So, implying that c itself is 0. So, that means we get a solution y x is equal to x cube for the initial value problem. So, the solution looks like a function y is equal to x cube. So, this is a solution and we can also see that. So, consider some other functions. So, consider the function for k is a positive constant y x defined y, y x is equal to 0 for x less than or equal to k and greater than or equal to 0 and x minus k the whole cube for x strictly greater than k. So, if you consider these functions for every value of k positive this gives a family of functions y k. So, this for every k you have this function and you can show that or you call it is a y k and obviously the derivative if you differentiate it the derivative of this one. So, this looks like up to k say this is k up to k the value of the function is 0 from k the function is a smooth. So, x minus k cube and if you is a smooth function if you differentiate it y prime x is 0 for x between 0 and x minus k. So, x minus k and k and its value is 3 times x minus k square for x greater than k. For a different k for a different k. So, this is another k call it k 1 k 1 and another k. So, the value of the function up to k 2 is 0 then it is x minus k 2 cube and you can see that the derivative y prime x satisfies because of this form this satisfies a given differential equation this is 3 times y to the power 2 y 3. So, for all x greater than 0 and it satisfies y at 0 is 0. So, you can say this is you can show that this is a family of curves. So, therefore, and also you can see that y x is equal to 0 is also a solution. So, y x is equal to 0 0 function is also a solution which is a singular solution otherwise you get a family of solution. The conclusion is the given initial value problem has initial value problem has infinitely a many solutions. In this lecture what we have discussed is we have taken a few initial value problems and we post the well postness of Hadamard and have seen that there are problems in which the initial value problem has no solution and there are problems in which the initial value problem has a unique solution and there are initial value problems having unique solutions. And now how to characterize these solutions these properties that we will see in the next lecture. Bye.