NumericalODE

# NumericalODE - Numerical Solution of Ordinary Dierential...

This preview shows pages 1–3. Sign up to view the full content.

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 0 ) = x 0 The first step is to transform the differential equation and its initial condition into an integral equation. x ( t ) = x 0 + Z t 0 f ( τ, x ( τ )) We think of the right hand side of this equation as an operator G ( x ( t )) = R t 0 f ( τ, x ( τ )) . 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 0 ( t ) x 0 and define the sequence inductively by x n +1 = G ( x n ( t )) = x 0 + Z t 0 f ( τ, x n ( τ )) dτ. In another chapter we describe this approach in more detail and show the convergence to a solution of the initial differential equation. 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
2 Numerical Methods Numerical solutions of differential equations calculate estimates of the solution at a se- quence of node points { t 0 , t 1 , t 2 , . . . , t n } . The initial value is given in the initial conditions x 0 at t 0 . 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 ( τ, x ( τ )) dτ. 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 ))) Runge-Kutta 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 R t n + h n t n f (
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern