Answer to Question #131908 in Differential Equations for VIRENDRA BALKI

Question

State and prove existence and uniqueness theorem.

Answer 1

Theorem 1. (Existence theorem):

Statement :

Suppose that f(x, y) is continuous function in some region R = {(x, y) : |x − x0| ≤ a, |y − y0| ≤ b}, (a, b > 0).

Proof:

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) .

Theorem 2. (Uniqueness theorem):

Statement:

Suppose that f and $\frac{df}{dy}$ are continuous function in R (defined in the existence theorem). Hence, both the f and$\frac{df}{dy}$ are bounded in R.

Proof:-

if |f(x, y)| ≤ K and $\frac{df}{dy}$ ≤ L ∀(x, y) ∈ R Then the IVP (1) has atmost one solution y = y(x) defined in the interval |x−x0| ≤ α where α = min ( a, b K ) . Combining with existence thereom, the IVP (1) has unique solution y = y(x) defined in the interval |x − x0| ≤ α.