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Unformatted text preview: Numerical Solution of Ordinary Differential Equations James Keesling 1 Picard Iteration The Picard method is a way of approximating solutions of ordinary differential equations. Originally it was a way of proving the existence of solutions. It is only through the use of advanced symbolic computing that it has become a practical way of approximating solutions. In this chapter we outline some of the numerical methods used to approximate solutions of ordinary differential equations. Here is a reminder of the form of a differential equation. dx dt = f ( t,x 0) x ( t ) = x The first step is to transform the differential equation and its initial condition into an integral equation. x ( t ) = x + Z t f ( ,x ( )) d We think of the right hand side of this equation as an operator G ( x ( t )) = R t f ( ,x ( )) d . The problem then is to find a fixed point, actually a fixed function, for G ( x ( t )). Picard iteration gives a sequence of functions that converges to such a fixed function. We define x ( t ) x and define the sequence inductively by x n +1 = G ( x n ( t )) = x + Z t f ( ,x n ( )) d. In another chapter we describe this approach in more detail and show the convergence to a solution of the initial differential equation. 1 2 Numerical Methods Numerical solutions of differential equations calculate estimates of the solution at a se quence of node points { t ,t 1 ,t 2 ,...,t n } . The initial value is given in the initial conditions x at t . The estimates for the other values { x 1 ,x 2 ,...,x n } are based on estimating the integral Z t n +1 t n f ( ,x ( )) d. We now give a few methods without explaining the derivation. In the following the n th step size is given by h n = t n +1 t n . Euler x n +1 = x n + h n f ( t n ,x n ) Modified Euler x n +1 = x n + h n f t n + h n 2 ,x n + h n 2 f ( t n ,x n ) Heun x n +1 = x n + h n 2 ( f ( t n ,x n ) + f ( t n +1 ,x n + h n f ( x n ,t n ))) RungeKutta x n +1 = x n + h n 6 [ k 1 + 2 k 2 + 2 k 3 + k 4 ] where k 1 = f ( t n ,x n ) k 2 = f t n + h n 2 ,x n + h n 2 k 1 k 3 = f t n + h n 2 ,x n + h n 2 k 2 k 4 = f ( t n + h n ,x n + h n k 3 ) These methods are based on numerical estimates of the integral...
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 Spring '08
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 Calculus, Equations

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