Advanced Calculus fi..

Chapter 2 Differential Calculus of Functions of Several Variables 11 5for .r > .yo and .r sufficiently close to .Yo.ThusIn the same way the same limit is found as .r + .I-,,. ThereforeAgain the argument applies to every point (.r. y) on the graph of f', and we concludethat at each such pointThus the theorem is proved.Rem~rks. The theorem proved here gives inore information about uniqueness thanthe general theorem stated at the end of the previous section: In an appropriateneighborhood of Po: (.r,). yo), (111 points (.I-, J) satisfying the implicit equation are onthe graph of the solution y = f (.r) found. The general theorem can be extended inanalogous fashion.The proof gives some information about the size of the ,I--interval in whicha solution y = f'(.r) can be found. Specifically, one must choose a rectangleE: 1.t - .tnl 5 6. 1y - yol 5 q in which F', > 0 (or F,, < 0) and in which 6 has beenrestricted so that 6 K < q, where K = max g(.r, ?.)Iin E. One sometimes has moreinformation about F that permits one to give a better estimate of the size of the interval.For example. if F, > 0 in E and F is positivc for ?. = yo + 71 and negative for.v = yo - q. then the solulion is defined and unique for 1.1- - .roI < 6. The restrictionIy - 5 11. in general. is needed to ensure the uniqueness of the solution. Forexample. the equation?.' - y sin .r - e-'x + e' sin .r = 0satisfies the hypotheses of the theorem with .r,, = 0. y,, = 0, and a solution is givenby y = sin.\-. However, another solution is y = r'; we exclude this solution byrestricting to a sufficiently small rectangle E about (0, 0).The existence of a unique continuous solution ?; = j'(.r) can also be shown asfollows. With E chosen as above so that F, > O in E. F(.ro, y) is strictly increasingin y, so that F(xo, yo + q) > 0. F(.Y,,. \,,- 11) < 0. By the continuity of F. for 6sufficiently small, F(.r, yo + 11) > 0 and F(.r, yo - q) < 0 for Ix - .rol < 6. By theIntermediate Value Theorem. as above. for each .r with 1.r - .rnl < 6, F(.r, y) = 0for a unique y = f'(.r) between yo - q and y,) + 11. Given c > 0, we can choose r1 sothat rl < t and we have I,f'(.t-) - f'(.ro) < c for I.\- - .w,,l < 6. Thus f is continuousat ro. By applying the same reasoning to each point (.xi, yl) on the graph of f weconclude that f' is continuous for 1.r - ,roi < 6.We can extend the proof to cover the case of G(.r, y, z) = 0 with G = 0 andG, # O at Po: (.xu. yo. Y~)), say G; > O at Po. The proof of the preceding paragraphextends at once to this case to yield a unique continuous solution z = f (.r. y) with