lecture_6 (dragged) 2

lecture_6 (dragged) 2 - MA 36600 LECTURE NOTES: MONDAY,...

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Unformatted text preview: MA 36600 LECTURE NOTES: MONDAY, JANUARY 26 3 Note that µ(t0 ) y (t0 ) = 1 · y0 , so we see that the unique solution to the initial value problem must be ￿￿ t ￿ 1 y (t) = µ(τ ) g (τ ) dτ + y0 . µ(t) t0 Existence and Uniqueness Theorems: Nonlinear Case. There is a much stronger result for first order nonlinear equations. Now consider the initial value problem dy = G(t, y ), y (t0 ) = y0 . dt Say that there exists a rectangle ￿ ￿ ￿ ￿ 2￿ R = (t, y ) ∈ R ￿ α < t < β , γ < t < δ such that i. G(t, y ) is continuous on R, ∂G ii. (t, y ) is continuous on R, and ∂y iii. (t0 , y0 ) ∈ R. then there exists a unique solution y = y (t) to the initial value problem. We explain how the linear case above follows from the nonlinear case. Recall that a first order linear equation is in the form dy = G(t, y ) where G(t, y ) = g (t) − p(t) y dt for some functions g (t) and p(t). We compute the partial derivative ∂G G(t, y ) = g (t) − p(t) y =⇒ (t, y ) = −p(t) ∂y ∂G so that both G(t, y ) and (t, y ) are continuous if and only if both p(t) and g (t) are continuous. ∂y ...
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