18 is unique if w y1 y2 tt0 0 note if w y1 y2 tt0

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Unformatted text preview: ( t0 ) = y 0 . dt y ( t0 ) = y 0 , (6.16) Theorem 3 Given p(t), q (t) continuous on an open interval I . Then there is exactly one solution y = φ(t) of (6.15) with initial conditions (6.16) and t0 ∈ I and the solution exists throughout the interval I . Theorem 4 This result is also true for the inhomogeneous equations (6.1) with initial conditions (6.16) providing g (t) is also continuous on I . For constant coefficient equations, solutions of (6.15) can be found explicitly, are continuous and have continuous first and second derivatives for all t, −∞ < t < ∞. Principle of Superposition and Wronskians We have already proved that if y1 (t) and y2 (t) solve (6.15), then so does y (t) = Ay1 (t) + By2 (t). Question: In what circumstances can we use a linear combination of solutions of (6.15) to find a unique solution to the initial value problem (6.15) and (6.16)? Answer: When a certain functional of y1 (t) and y2 (t), the Wronskian ￿ ￿ W (y1 , y2 ) = y1 (t)y2 (t) − y1 (t)y2 (t) = det ￿ y 1 ( t) ￿ y 1 ( t) ￿ y 2 ( t) ￿ y 2 ( t) (6.17) estimated at t0 is non-zero. Proof: Let y1 (t), y2 (t) solve (6.15). Then so does y (t) = Ay1 (t) + By2 (t). Apply initial conditions y0 = Ay1 (t0 ) + By2 (t0 ) ￿ y0 = ￿ Ay1 (t0 ) + 17 ￿ By2 (t0 ). (6.18) Solve for A and B , ￿ ￿ ￿ ￿ y 0 y 2 ( t0 ) y 1 ( t0 ) y 0 det ￿ det ￿ ￿ ￿ y 0 y 2 ( t0 ) y 1 ( t0 ) y 0 A= ,B = , (W (y1 , y2 ))|t=t0 (W (y1 , y2 ))|t=t0 The solution (A, B ) of (6.18) is unique if W (y1 , y2 )|t=t0 ￿= 0. Note: If W (y1 , y2 )|t=t0 is zero, the solution (A, B ) can still exist but it is not unique. Why? Recall: x−y =1 2x − 2y = b ￿ 1 -1 is zero, either no solutions exist or, if b = 2, then there are infinite 2 -2 solutions (α, α − 1) for all α. Since det ￿ The Wronskian W (y1 , y2 ) satisfies a differential equation. ￿ ￿ Proof: Let W = W (y1 (t), y2 (t)) = y1 (t)y2 (t) − y1 (t)y2 (t) dW dt = = = Thus, ￿￿ ￿￿ y 1 ( t) y 2 ( t) − y 1 ( t) y 2 ( t) ￿ ￿ y1 (t)(−p(t)y2 (t) − q (t)y2 (t)) − y2 (t)(−p(t)y1 (t) − q (t)y1 (t)) − p( t ) W ￿ ￿t ￿ W (t) = W (t0 ) exp − p(t￿ )dt￿ . (6.19) t0 So we have the interesting result that if p(t) is continuous on an interval I , then if W (t0 ) is nonzero, W (t) is nonzero throughout I . This means that no matter what point in I we choose to give initial conditions, the solution of (6.15) and (6.16) is always unique if at any point t0 ∈ I, W (t0 ) ￿= 0. The notion of linear dependence, linear independence and the Wronskian We say two functions f (t) and g (t) are linearly independent (l.d.) on an interval I if there exist two constants k1 and k2 , not both zero, such that k1 f (t) + k2 g (t) = 0 for all t ∈ I. (6.20) We say two functions f (t) and g (t) are linearly independent (l.i.) on an interval I if they are not l.d. on I , i.e. an equation such as (6.20) holds for all t ∈ I and only if k1 = k2 = 0. 18 Example: The solutions e...
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This note was uploaded on 01/22/2014 for the course MATH 254 taught by Professor Indik during the Fall '08 term at University of Arizona- Tucson.

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