MaplePrimes - answers and comments on Question, Solving system of differential equations with bound on solutions

Events

Preben Alsholm — Fri, 15 Feb 2013 02:55:02 Z

You may want to take a look at the events option:

?dsolve,events

restart;
sys:=diff(x(t),t)= x(t)+y(t) , diff(y(t),t)= x(t)-y(t);
res:=dsolve({sys,x(0)=1/2,y(0)=0},numeric,events=[[x(t)-1,halt],[y(t)-2,halt]]);
plots:-odeplot(res,[x(t),y(t)],0..5);
plots:-odeplot(res,[t,x(t)],0..5);

How to correctly write the event trigger

mapleq2013 — Fri, 15 Feb 2013 07:16:13 Z

Thank you for the suggestion and I think it's very helpful. I read about events and couldn't figure out how to correctly write the event trigger. Naively I think it should be like

events=[[x(t)>0, do_something],[y(t)>0, do_something]]

but the program returns an error message: event1 trigger is of incorrect form: x(t)>0

Also I couldn't understand the way you wrote it either, you wrote

events=[[x(t)-1,halt],[y(t)-2,halt]]

what does x(t)-1 mean for the event trigger?

Rootfinding trigger

Preben Alsholm — Fri, 15 Feb 2013 07:24:01 Z

@mapleq2013 x(t)-1 is a rootfinding trigger, meaning that when during integration x(t) - 1 hits zero the action (in this case 'halt') is activated.
So in the example there are two events both of the type simple rootfinding and both having action 'halt'.

Here is a simple example of a conditional trigger:

sys2:=diff(x(t),t)= y(t) , diff(y(t),t)= -x(t);
res2:=dsolve({sys2,x(0)=1/2,y(0)=0},numeric,events=[[[x(t),y(t)>0],halt]]);
plots:-odeplot(res2,[t,x(t)],0..5);

In this one integration stops when x(t) = 0 provided that also y(t) > 0.

May need "if" statement

mapleq2013 — Fri, 15 Feb 2013 22:54:59 Z

I got how to write events now, thank you! My original problem which was not well formulated couldn't be solved with this though, but this is very still enlightening. The thing is I may need a better conditional like "if" to regulate how the problem behaves. I have re-formulated in the following:

Main Goal: to solve a system of ODE that involves two unknowns dx(t)/dt=f(x,y)+sin(t) and dy(t)/dt=g(x,y) with the condition that x(t)>=0 and y(t)>=0 (initially x(t)=y(t)=0)

Sub-objectives:

a. solve for x(t) and y(t) normally until one or both x(t) and y(t) are <0

b. when x(t) goes below zero solve a new system x(t)=0 and dy(t)/dt=g(x,y), while keeping track of f(x,y)+sin(t), f(x,y)+sin(t) will eventually go positive then we go back solving the original system

c. when y(t) goes below zero solve a new system dx(t)/dt=f(x,y)+sin(t) and y(t)=0, while keeping track of g(x,y), g(x,y) will eventually go positive then we go back solving the original system

d. if both x(t) and y(t) goes below zero we keep track of both f(x,y)+sin(t) and g(x,y), they will eventually go positive, then we go back solving the original system