M427K Handout: Examples and the Existence and UniquenessTheoremSalman ButtFebruary 8, 2006Example problemsSolve the differential equationsdydx=sin xy, y(π/2) = 1sin 2tdx + (2x cos 2t − 2t)dt = 0The Existence and Uniqueness TheoremThis theorem (see p. 110 of the text) is not a theorem to stress over. Professor Cecil wentthrough the majority of its proof during class yesterday. You do not need to know the details ofthat proof – I believe his intention was to give a complete story of differential equations and toalso provide the proof for those of you who may be more interested in mathem atics. If you foundthe proof interesting or have any questions about it, come see me or Professor Cecil in office hours.But you all should know that you will not be required to know the proof of the theorem. Whatyou will be required to know is the state ment of the theorem and some basic arguments about themethod of successive approximation, which we will review today.Bear in mind that not every differential equation is solvable. But that does not mean a solutiondoes not exist. This is a mathematical subtlety that may not be obvious at first. There is a largedivide in math between knowing something exists and actually constructing it. You must cometo grips with this idea if you are to understand the motivation for the existence and uniquenesstheorem. This is perhaps where some modesty is helpful: you cannot solve every differentialequation, but that does not mean a solution does not exist. Perhaps one day these equations willbe solvable after new mathematics have been developed, but many of these problems will remainunsolvable. But that does not mean their solutions do not exist – we just cannot find them. Thinkabout these statements and let this distinction between existence and construction sink in.Now concerning the actual theorem, it provides sufficient conditions for a differential equationto e xist and be unique. More precisely, it statesTheorem 1. For the first order initial value problemy0= f(x, y), y(0) = 0, (1)1suppose f and ∂f/∂y are both continuous in a rectangle R in ty -plane, R : |t| ≤ a, |y| ≤ b. Thenthere is some interval (c, d) in the interval (−a, a) in which there exists a unique solution y = φ(t)of the initial value problem (1).So what does this mean?Now the problems you will be asked to do require not an intimate knowledge of this theoremand its proof. Instead you will need to know how to work with one of the tools used in its proof:the method of successive approximations, also known as Picard’s iteration method. This methodworks as follows: We have our initial value problem (1). If we let φ(t) = y(t) be our solution, thenwe haveNow in order to find this solution, we create a sequence of functions {φn} that will hopefullyconverge to some limit function˜φ. First a few notes:1.2.We create our sequence by defining our φn’s recursively:φ0(t) =φ1(t) =φ2(t) =.........φn+1(t) =Some