Reference no: EM13906578

1. Let I = [0, a] and define the three norms in C(I), the maximum norm

||f||₀ = max_x∈I {lf(x)|}

and the weighted norms

||f||₁ = max_x∈I {lf(x)|e^-αx} and ||f||₂ = max_x∈I {|f(x)}|e-x²}_.

Determine the corresponding operator norms ||T||₀, ||T||₁, and ||T||₂ for the operator

(T_f)(x) =  ₀∫^xt, f (t)dt.

2. (a) Use the fixed point theorem with an appropriate choice of T and choice of norm and the results from Question 1 to show that the integral equation

y(x) = 1/2 x² + ∫ty(t)dt, x∈I = [0, a],

has exactly one solution.

(b) Determine the solution

(i) by rewriting the equation as an initial-value problem and solving, and

(ii) by explicitly calculating the successive approximations

y₁ = Ty_0, y₂ = T_y1,.......Y_n+1 = T_yn,.....

beginning with y_o = 0.

3. Prove the existence and uniqueness theorem for the initial-value problem

y' = f(x, y), x ∈ I = |ξ- a, ξ + a|, + y(ξ) = η

using the norm ||y|| = max_x∈I {|y(x)|e-^α|x-ξ|} with α > 0:

3: Theorem. Let f be continuous in the strip S = I x R and satisfy a Lipschitz condition with respect to y. Then the initial value problem (1) has exactly one solution y(x) and the solution exists for ξ - a ≤ x ≤ ξ + a.

4. Find the maximal interval of existence (α, β) for each of the following initial-value problems. If α > -∞ or β< ∞.), discuss the limit of the solution as x → α⁺ or as x → β^- respectively:

(a) y' = y², y(0) = y₀;
(b) y' = sec y, y(0) = 0;
(c) y'= y² - 4, y(0) = 0;
(d) y' = y³. y(0) = y₀ > 0.