Numerically solving ODE’s

Lecture 5

Differential Equations •  Ordinary Differential Equations (ODE)

–  Function(s) of one independent variable (e.g. f(x), g(t)) •  Partial Differential Equations

–  Function(s) of two or more independent variables: f(x,y)

•  Can have coupled sets of ODE –  E.g. f(x), g(x) depend on each other

•  ODE’s (relatively) easy to solve –  PDE’s can be harder

The Pendulum

θ l

mg

F=ma in direction of θ:

If θ is small, then sinθ ~ θ

Solution: θ=θ0 cos((g/l)1/2 t)

Initial conditions θ=θ0 at t=0

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d2θdt 2

+glsinθ = 0

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d2θdt 2

+glθ = 0

The linear Pendulum

Solution: θ=θ0 cos((g/l)1/2 t)

Initial conditions θ=θ0 at t=0

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d2θdt 2

+glθ = 0

The (non-linear) Pendulum

For large θ, the ODE is non-linear:

How do we solve this?

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d2θdt 2

+glsinθ = 0

Euler’s method Take a general first-order ODE:

0 t1 t2 t3

x0 x1

x2

x3

Approximate x(t1=h) as:

Approximate x(t2=2h) as:

and so on…

In general:

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dxdt

= f (x, t)

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x1 = x0 + hf (x0,0)

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x2 = x1 + hf (x1,h)

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xn+1 = xn + hf (xn,tn )

Apply Euler to Pendulum 2nd order ODE:

First, re-write in dimensionless form:

Define new time variable:

To get:

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d2θdt 2

+glsinθ = 0

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τ = t(l /g)−1/ 2

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d2θdτ 2

+ sinθ = 0

Apply Euler to Pendulum 2nd order ODE:

Split into two coupled 1st order ODE’s:

Apply Euler’s method to both equations

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d2θdτ 2

+ sinθ = 0

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dθdτ

= y

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dydτ

= −sin(θ)

from math import *

def euler(dt, y, theta): dtheta = dt * y dy = dt * (-sin(theta)) y = y + dy theta = theta + dtheta return (y,theta)

theta = 0.5 y = 0 t = 0 tstop = 40.0 dt = 0.05

while (t < tstop): (y,theta) = euler(dt, y, theta) t = t + dt print t, theta, y

Python program to solve Pendulum using Euler’s method

Set initial conditions

Use euler function to go from n to n+1

Function that does one Euler step

Apply Euler to Pendulum 800 steps

Apply Euler to Pendulum 800 steps

4000 steps

Phase space diagram

θ

dθ/dt Pendulum at one moment in time

Phase space diagram 800 steps

Phase space diagram 800 steps

4000 steps

Improved Euler’s method

0 t1 t2 t3

x1 x2

x3

Instead of using slope just at xn:

Use an average slope:

But xn+1 appears on RHS?!

Solution: use Euler method to get an approximate xn+1

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xn+1 = xn +12h( f (xn,tn ) + f (xn+1,tn+1))

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xn+1 = xn + hf (xn,tn )

Improved Euler’s method

c = hf(x,t)

xnew = x + c

c = hf(x,t) d = hf(x+c, t+h) xnew = x + (c+d)/2

Original Euler method

Improved Euler method

Apply Improved Euler to Pendulum 800 steps

Euler

4000 steps Euler (red)

800 steps Improved Euler (blue)

Improved Euler: Phase space diagram 800 steps

Euler

4000 steps Euler (red)

800 steps Improved Euler

High amplitude pendulum

High amplitude pendulum

High amplitude pendulum

High amplitude pendulum

High amplitude pendulum

Summary •  Euler method

–  slope at xn Error ~ h

•  Improved Euler –  slope averaged over 2 points – Error ~ h2

•  Runge-Kutta – Slope averaged over 4 points – Error ~ h4

•  Phase diagram (velocity vs. position)