Adams

A Variable Order Adams-Bashforth-Moulton PECE solver for non-stiff intial value problem (IVP)

Syntax for 1st-Order IVP

adams(dydt, t_start, t_end, nstep, y0)	Input: dydt is a user defined function of the form dydt(t, y) => {some expression of y and t} t_start is a real number for initial time t_end is a real number of end time nstep is an positive integer implies the preferred time steps for the solutions y0 is a real number for the initial value of y when t = t_start Result: A row vector gives the time points of t for the solutions A column vector gives the solutions y for the time points t.
adams(dydt, t_start, t_end, nstep, y0, abserr, relerr)	Input: dydt is a user defined function of the form dydt(y, t) => {some expression of y and t} t_start is a real number for initial time t_end is a real number of end time nstep is an positive integer implies the preferred time steps for the solutions y0 is a real number for the initial value of y when t = t_start abserr is a real number for absolute tolerance relerr is a real number for relative tolerance Result: A row vector gives the time points of t for the solutions A column vector gives the solutions y for the time points t.
adams(dydt, t, y0)	Input: dydt is a user defined function of the form dydt(y, t) => {some expression of y and t} t is a row/column vector which gives the preferred time points for the solutions y0 is a real number for the initial value of y when t = t[0] Result: A row vector gives the time points of t for the solutions A column vector gives the solutions y for the time points t.
adams(dydt, t, y0, abserr, relerr)	Input: dydt is a user defined function of the form dydt(y, t) => {some expression of y and t} t is a row/column vector which gives the preferred time points for the solutions y0 is a real number for the initial value of y when t = t[0] abserr is a real number for absolute tolerance relerr is a real number for relative tolerance Result: A row vector gives the time points of t for the solutions A column vector gives the solutions y for the time points t.

Example I

Solve Problem

y'=0.25*y*(1-y/20)

y(0) = 1

between the time [0,20].

Input

Output

dydt(t, y) => 0.25*y*(1-y/20)

adams(dydt, 0, 20, 5, 1)

dydt(t, y) => 0.25*y*(1-y/20)

adams(dydt, 0, 20, 5, 1, 1.0e-10, 1.0e-10)

dydt(t, y) => 0.25*y*(1-y/20)

t=[0, 4, 8, 12, 16, 20]

adams(dydt, t, 1)

dydt(t, y) => 0.25*y*(1-y/20)

t=[0, 4, 8, 12, 16, 20]

adams(dydt, t, 1, 1.0e-10, 1.0e-10)

Discussion:

Problem has an exact solution of y(t) = 20.0 / ( 1.0 + 19.0 * exp ( - 0.25 * t ) )

Compare the numerical solution above with the exact solution you get

Syntax for High-Order IVP

adams(<dy1dt, ...>, t_start, t_end, nstep, y0)	Input: <dy1dt, dy2dt, ...> is a group of user defined functions of the form dydt(t, y1, y2, ...) => {some expression of t, y1, y2, ...}, which are enclosed with '<>'. These functions defined the all 1st-order IVP differential equations converted from the high-order ODE. t_start is a real number for initial time t_end is a real number of end time nstep is an positive integer implies the preferred time steps for the solutions y0 is a real number for the initial value of y when t = t_start Result: A row vector gives the time points of t for the solutions A matrix gives the solutions y for the time points t whose first column is solution for y1, second row for y2, etc.
adams(<dy1dt, ...>, t_start, t_end, nstep, y0, abserr, relerr)	Input: <dy1dt, dy2dt, ...> is a group of user defined functions of the form dydt(t, y1, y2, ...) => {some expression of t, y1, y2, ...}, which are enclosed with '<>'. These functions defined the all 1st-order IVP differential equations converted from the high-order ODE. t_start is a real number for initial time t_end is a real number of end time nstep is an positive integer implies the preferred time steps for the solutions y0 is a real number for the initial value of y when t = t_start abserr is a real number for absolute tolerance relerr is a real number for relative tolerance Result: A row vector gives the time points of t for the solutions A matrix gives the solutions y for the time points t whose first column is solution for y1, second row for y2, etc.
adams(<dy1dt, ...>, t, y0)	Input: <dy1dt, dy2dt, ...> is a group of user defined functions of the form dydt(t, y1, y2, ...) => {some expression of t, y1, y2, ...}, which are enclosed with '<>'. These functions defined the all 1st-order IVP differential equations converted from the high-order ODE. t is a row/column vector which gives the preferred time points for the solutions y0 is a real number for the initial value of y when t = t[0] Result: A row vector gives the time points of t for the solutions A matrix gives the solutions y for the time points t whose first column is solution for y1, second row for y2, etc.
adams(<dy1dt, ...>, t, y0, abserr, relerr)	Input: <dy1dt, dy2dt, ...> is a group of user defined functions of the form dydt(t, y1, y2, ...) => {some expression of t, y1, y2, ...}, which are enclosed with '<>'. These functions defined the all 1st-order IVP differential equations converted from the high-order ODE. t is a row/column vector which gives the preferred time points for the solutions y0 is a real number for the initial value of y when t = t[0] abserr is a real number for absolute tolerance relerr is a real number for relative tolerance Result: A row vector gives the time points of t for the solutions A matrix gives the solutions y for the time points t whose first column is solution for y1, second row for y2, etc.

Example 2

Solve Problem

y''+ y = 0

y(0) = 1; y'(0) = 0;

between the time [0, 2π].

The above 2nd order ODE can be rewritten to a group of two 1st order ODE

y1' = y2

y2' = -y1

y1(0) = 1; y2(0) = 0;

Input

Output

dy1dt(t, y1, y2) => y2

dy2dt(t, y1, y2) => - y1

adams(<dy1dt, dy2dt>, 0, 2*Pi, 12, [1, 0])

Discussion:

Problem has an exact solution of y(t) = cos(t)

Compare the numerical solution above with the exact solution you get