Answer to Question #135662 in Differential Equations for Deepak sharma

Question

state and prove existence and uniqueness theorem

Answer 1

(Existence theorem): Suppose that f(x, y) is continuous function in some region R = {(x, y) : |x − x0| ≤ a, |y − y0| ≤ b}, (a, b > 0). Since f is continuous in a closed and bounded domain, it is necessarily bounded in R, i.e., there exists K > 0 such that |f(x, y)| ≤ K ∀(x, y) ∈ R. Then the IVP (1) has at least one solution y = y(x) defined in the interval |x − x0| ≤ α where α = min ( a, b K ) (Note that the solution exists possibly in a smaller interval).(Uniquness theorem): Suppose that f and ∂f/∂y are continuous function in R (defined in the existence theorem). Hence, both the f and ∂f/∂y are bounded in R, i.e., (a) |f(x, y)| ≤ K and (b) ∂f ∂y ≤ L ∀(x, y) ∈ R Then the IVP (1) has at most one solution y = y(x) defined in the interval |x−x0| ≤ α where α = min ( a, b K ) . Combining with existence theorem, the IVP (1) has unique solution y = y(x) defined in the interval |x − x0| ≤ α.Condition (b) can be replaced by a weaker condition which is known as Lipschitz condition. Thus, instead of continuity of ∂f/∂y, we require |f(x, y1) − f(x, y2)| ≤ L|y1 − y2| ∀(x, yi) ∈ R .If ∂f/∂y exists and is bounded, then it necessarily satisfies Lipschitz condition. On the other hand, a function f(x, y) may be Lipschitz continuous but ∂f/∂y may not exists. For example f(x, y) = x 2 |y|, |x| ≤ 1, |y| ≤ 1 is Lipschitz continuous in y but ∂f/∂y does not exist at (x, 0)