Reference no: EM132160934

Assignment - Prove the given Theorem.

Theorem - Let f and ∂f/∂y be defined and continuous in a domain D ⊂ R x Rⁿ. Let (ξ₀, η₀) ∈ D and y₀(x) := y(x; ξ₀, η₀). Suppose J = [a, b] is a compact interval such that y₀ exists on J, and denote by S_α the set S_α = {(x, y) : x ∈ J, |y - y₀(x)| ≤ α}.

Then there exists on α > 0 with S_α ⊂ D such that the function y(x; ξ, η) is defined in J x S_α (i.e., every solution of the initial value problem with (ξ, η) ∈ S_α exists at least in J), the function y, y_ξ, y_η and their derivation with respect to x, which are denoted by y', y'_ξ, y'_η, are continuous in J x S_α, and

Y_ξ(x; ξ, η) = -f(ξ, η) + _ξ∫^xf_y(t, y(t; ξ, η))y_ξ(t; ξ, η)dt, (1)

y_η(x; ξ, η) = I + _ξ∫^xf_y(t, y(t; (ξ, η)) · y_η(t; ξ, η)dt, (2)

and

y_ξ(x; ξ, η) + y_η(x; ξ, η) · f(ξ, η) = 0. (3)

Remarks -

1. Notation: y, y', y_ξ, f are column vectors; f_y, y_η, y'_η are n x n matrices (cf. the remark in V); IU is the identity matrix; (2) is a linear matrix-integral equation (it is equivalent to n vector-integral equations for the columns y_n,); the product f_y · y_η is a matrix product.

2. The theorem remains valid in the complex case with Cⁿ in place of Rⁿ.

We simplify the proof by extending f continuously and differentiably to the strip J x R and then applying the earlier results (the proof is also valid in the complex case). Let β > 0 be chosen such that S_2β ⊂ D. We determine a real function h(s) ∈ C¹(R) that satisfies

and 0 ≤ h(x) ≤ 1, and we define

f*(x, y) = f(x, y₀(x) + (y - y₀(x))h(|y - y₀(x)|)).

Note - Prove the above Theorem and go in detailed. Please try to do it.