Questions about elements in function ode45(@(t,x) x.*(4-x), [0,5], -2:4:6)

In ode45(@(t,x) x.*(4-x), [0,5], -2:4:6),

why do we have to add t in our @function even though t is not used. For the initial y0 conditions, what does -2:4:6 mean? Is it the initial conditions: -2, 4, 6, or is it the initial conditions -2, 2, 6 (adding by 4)? How come when I graph this in matlab it doesn't show the initial conditions? Instead it gives me the error: Failure at t=2.746521e-01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (8.881784e-16) at time t. Explain the error and how to fix it.

Answer 1

[Sam Chak]

Hi @Anna

The kind of error message usually implies that the singularity has occured somewhere around s.

The analytical solution for , with is given by

.

The solution has a denominator. And if you try solving , you'll get .

So, even though you set the simulation to 5 sec, it cannot get pass 0.274653 sec.

Further analysis shows that has two equilibrium points: one at and the other at . You will get a clear picture of the region of attraction with the vector field diagram. If your initial condition , you will get a stable convergence to .

clear all; clc
[X, Y] = meshgrid(-1:10/14:7, -1:10/14:7);
S = Y.*(4 - Y);
L = sqrt(1 + S.^2);
U = 1./L;
V = S./L;
quiver(X, Y, U, V, 0.3)
axis square

hold on

x = linspace(0, 7, 701);
y = 20*exp(4*x)./(5*exp(4*x) - 1);
plot(x, y, 'r')

hold off

Also see @Walter Roberson's rich explanation on the ode45 solver. 👍

Answer 2

[Walter Roberson]

why do we have to add t in our @function even though t is not used

Imagine the opposite: imagine a system in which ode45 somehow knew to omit passing in t when t is not used in the function. What would that look like?

Well, for one thing in such a case you would have @() with a single parameter, like @(x) x.*(4-x) . So when ode45 received the function, it would hypothetically look and see that the function only accepted a single parameter. But how would ode45 know whether it was the time variable or the boundary conditions that were being omitted? Would you expect ode45() to look at the name of the single parameter to figure out which of the two parameters it was? Perhaps you would expect that... but if you look at the ode45() documentation, the representative variable names it uses in the documentation are t and y not t and x . So it might look at the @(x) and see a single parameter, and might compare the x it sees to t and decide they are not the same and form the hypothesis that t was omitted. But then it would have to look at the x and compare it to the expected y and determine that they are not the same, so it would have to form the hypothesis that neither t nor y were passed, and that instead what was being passed was an extra parameter (extra parameters can happen in some cases.)

That suggests that relying on parameter names to determine whether a parameter was omitted is not a good idea.

Suppose further that instead of @(x)x.*(4-x) what you had passed in was @MyFunction -- a handle to a real function instead of a handle to an anonymous function. Would you expect that ode45 would use some way to look at the dummy parameter names built into the real function? If so then what would happen in the case where the function was compiled into a .mex* file and so there are no names for the parameters?

ode45*() and all other MATLAB functions skill all that guessing about the intended meaning, by working with a strict positional notation. ode45() is not going to fish through the headers to figure out which parameters it needs to pass based on name: ode45() is just always going to pass the given function two parameters based just on position. The receiving function is not required to use what is passed in.

For the initial y0 conditions, what does -2:4:6 mean?

Same as if you saw that in some other context: it forms a numeric row vector starting with -2 and adding 4 each time until the addition would exceed 6. So -2, then (-2+4) --> +2, then (-2+4+4) --> +6 then it looks ahead to (-2+4+4+4) ---> 10 and sees that exceeds the upper bound 6 so it stops before putting 10 on the list. The final list is [-2, +2, +6] so a vector of three boundary conditions will be passed to the ode function.

[Walter Roberson]

@Sam Chak's reply about the analytic solution is a good one.

[Anna]

And when I plot the differential equation how come the initial values -2, 2, and 6 were not plotted?

[Walter Roberson]

you did not show us your plot code.

[Anna]

I just entered ode45(@(t,x) x.*(4-x), [0,5], -2:4:6) as my code and pressed enter to plot it.

[Walter Roberson]

Your plot ends around -10^14 . On that scale, -2, +2, and +6 are not visible. If you

ylim([-20 20])

you can see them clearly.