math572_notes

math572_notes - 1 Thurs 1/6 1 1 IVP for ODEs y = f y y(0 =...

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Unformatted text preview: 1. Thurs 1/6 1 1. IVP for ODEs y = f ( y ) , y (0) = y : find y ( t ) ex y = y , y (0) = y ⇒ y ( t ) = y e t y = sin y , y (0) = y ⇒ y ( t ) = ? def An IVP is well-posed if the following conditions are satisfied. 1. A solution exists. 2. The solution is unique. 3. The solution depends continuously on the data (i.e. y , f ). def f satisfies a Lipschitz condition on a domain D if there exists a constant L st | f ( y )- f ( u ) | ≤ L | y- u | for all y , u ∈ D . note : we typically assume that f ( y ) is smooth, so L = max | f y | thm If f satisfies a Lipschitz condition, then the IVP is well-posed. ex y = ay + b , y (0) = y ⇒ L = | a | , y ( t ) = y e at + b a ( e at- 1) y = y 1 / 2 , y (0) = 0 ⇒ L = ∞ for y ≥ 0 , non-unique solution , y ( t ) = 0 , t 2 4 y = y 2 , y (0) = 1 ⇒ L = ∞ for y ≥ 1 , y ( t ) = 1 1- t , lim t → 1 y ( t ) = ∞ 2 Euler’s method h = Δ t = step size , t n = nh y n = y ( t n ) : exact solution , u n : numerical approximation y = f ( y ) : differential equation u n +1- u n h = f ( u n ) : finite-difference scheme u n +1 = u n + hf ( u n ) , u = y ex y = y , y = 1 u = 1 u 1 = u + hf ( u ) = 1 + h u 2 = u 1 + hf ( u 1 ) = (1 + h ) + h (1 + h ) = (1 + h ) 2 · · · u n = (1 + h ) n t n = 1 ⇒ nh = 1 , y n = y (1) = e = 2 . 7182818 . . . n h u n y n- u n ( y n- u n ) /h 10 0.1 2.5937425 0.1245 1.245 20 0.05 2.6532977 0.0650 1.300 40 0.025 2.6850638 0.0332 1.328 80 0.0125 2.7014849 0.0168 1.344 3 Euler’s method y = y , y (0) = 1 u n +1 = u n + hu n , u = 1 0.5 1 1.5 2 1 2 3 4 5 6 7 8 h=1 0.5 1 1.5 2 1 2 3 4 5 6 7 8 h=1/2 0.5 1 1.5 2 1 2 3 4 5 6 7 8 h=1/4 0.5 1 1.5 2 1 2 3 4 5 6 7 8 h=1/8 note 1. The solid line is the exact solution y ( t ). The circles are the numerical solution u n . The dashed line is the piecewise linear interpolant of u n . 2. For a fixed time t , the error decreases as h → 0. 3. For a fixed stepsize h , the error increases as t → ∞ . 4 convergence proof (special case) y = y , y = 1 ⇒ y ( t ) = e t u n +1 = u n + hu n , u = 1 ⇒ u n = (1 + h ) n consider t = 1 , h = 1 n lim h → u n = lim n →∞ 1 + 1 n ! n = lim n →∞ exp n ln 1 + 1 n !! = exp lim n →∞ n ln 1 + 1 n !! lim n →∞ n ln 1 + 1 n ! = ∞· 0 = lim n →∞ ln (1 + 1 n ) 1 n = lim n →∞ 1 1 + 1 n ·- 1 n 2 · 1- 1 n 2 = 1 ⇒ lim h → u n = e = y (1) ⇒ Euler’s method converges hw : y = ay + b , y (0) = y convergence proof (general case) y = f ( y ) , y (0) = y u n +1 = u n + hf ( u n ) , u : given fix t > , set h = t/n ⇒ t = nh = t n , y n = y ( t n ) goal : lim h → u n = y n y n +1 = y n + hf ( y n ) + τ n , τ n : local truncation error step 1 : estimate τ n y n +1 = y ( t n +1 ) = y ( t n + h ) = y ( t n ) + hy ( t n ) + h 2 2 y 00 ( e t ) y n +1 = y n + hf ( y n ) + h 2 2 y 00 ( e t ) ⇒ τ n = h 2 2 y 00 ( e t ) step 2 : analyze error , e n = y n- u n : error e n +1 = y n +1- u n +1 = y n + hf ( y n ) + τ n- ( u n + hf ( u n )) = y n- u n + h ( f ( y n )- f ( u n )) + τ n = e n...
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math572_notes - 1 Thurs 1/6 1 1 IVP for ODEs y = f y y(0 =...

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