t 6 3 t 2 n n By math induction assume true for n k φ k 1 t Z t 2 s 1 φ k s ds

# T 6 3 t 2 n n by math induction assume true for n k

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+ t 6 3! + ... + t 2 n n ! , By math induction, assume true for n = k φ k +1 ( t ) = Z t 0 2 s (1 + φ k ( s )) ds = Z t 0 2 s (1 + s 2 + ... + s 2 k k ! ) ds = t 2 + t 4 2! + t 6 3! + ... + t 2 k +2 ( k + 1)! which is what we needed to show The limit exists if the series converges or lim n →∞ φ n ( t ) exists Joseph M. Mahaffy, h [email protected] i Lecture Notes – Existence and Uniqueness — (11/23) Introduction Linear Differential Equation Nonlinear Differential Equation Existence and Uniqueness Picard Iteration Uniqueness Examples Picard Iteration - Example 3 Apply the Ratio test lim k →∞ t 2 k +2 ( k + 1)! k ! t 2 k = t 2 k + 1 0 which shows this series converges for all t Since this is a Taylor’s series, it can be integrated and differentiated in its interval of convergence. Thus, it is a solution of the integral equation Note that this is the Taylor’s series for φ ( t ) = e t 2 - 1, which can be shown to satisfy the IVP Joseph M. Mahaffy, h [email protected] i Lecture Notes – Existence and Uniqueness — (12/23) Subscribe to view the full document.

Introduction Linear Differential Equation Nonlinear Differential Equation Existence and Uniqueness Picard Iteration Uniqueness Examples Picard Iteration - Example 4 First 4 Picard Iterates -1.5 -1 -0.5 0 0.5 1 1.5 0 0.5 1 1.5 2 2.5 3 t φ ( t ) Picard Iterates φ 1 φ 2 φ 3 φ 4 Joseph M. Mahaffy, h [email protected] i Lecture Notes – Existence and Uniqueness — (13/23) Introduction Linear Differential Equation Nonlinear Differential Equation Existence and Uniqueness Picard Iteration Uniqueness Examples Example - Uniqueness 1 Example - Uniqueness - Suppose there are two solutions, φ ( t ) and ψ ( t ) satisfying the integral equation φ ( t ) - ψ ( t ) = Z t 0 2 s ( φ ( s ) - ψ ( s )) ds Take absolute values and restrict 0 t A/ 2 ( A arbitrary). then | φ ( t ) - ψ ( t ) | = Z t 0 2 s ( φ ( s ) - ψ ( s )) ds Z t 0 2 s | φ ( s ) - ψ ( s ) | ds A Z t 0 | φ ( s ) - ψ ( s ) | ds for 0 t A/ 2 Joseph M. Mahaffy, h [email protected] i Lecture Notes – Existence and Uniqueness — (14/23) Introduction Linear Differential Equation Nonlinear Differential Equation Existence and Uniqueness Picard Iteration Uniqueness Examples Example - Uniqueness 2 Let U ( t ) = R t 0 | φ ( s ) - ψ ( s ) | ds , then U (0) = 0 and U ( t ) 0 for t 0 U ( t ) is differentiable with U 0 ( t ) = | φ ( t ) - ψ ( t ) | We have the differential inequality U 0 ( t ) - AU ( t ) 0 , 0 t A/ 2 Multiplying by positive function e - At , then integrating gives d dt ( e - At U ( t ) ) 0 , 0 t A/ 2 , e - At U ( t ) 0 , 0 t A/ 2 Hence, U ( t ) 0 with A arbitrary. It follows that U ( t ) 0 or φ ( t ) = ψ ( t ) for each t , so the functions are the same, giving uniqueness Joseph M. Mahaffy, h [email protected] i Lecture Notes – Existence and Uniqueness — (15/23) Introduction Linear Differential Equation Nonlinear Differential Equation Existence and Uniqueness Picard Iteration Uniqueness Examples Existence and Uniqueness Theorem 1 We leave the details of the proof of the Existence and Uniqueness Theorem to the interested reader, but give a sketch of the key steps 1 Restrict the time interval | t | ≤ h a Since f is continuous in the the rectangle R : | t | ≤ a, | y | ≤ b , the function f is bounded on R , so there exists M such that | f ( t, y ) | ≤ M ( t, y ) R Let h = min ( a, b M )  • Fall '08
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