Torquil Macdonald Sørensen wrote:
Hi, I'm getting very different results when solving the following
initial value ODE problem in Matlab and Octave:
dy/dt=1/sqrt(y^2 + 1)+y-y^2 on t \in [0,10] with y(0) = 0
From looking at the equation, I believe that the Matlab solution is the
correct one, so I'm wondering if I have not converted the Matlab-file
correctly to Octave? Or does the Octave algorithm not work in this
situation? It does not change when I lower the values for the absolute
and relative tolerances with lsode_options, so maybe I have done
something wrong and I'm solving a different ODE in Octave.
Here are the Octave and Matlab programs that I thought were solving the
same equation:
*** Octave version ***
function nonlinear_de
% Time span
t = linspace(0,10,100);
% Initial condition
y0=[0];
% Solve the DE
lsode_options('absolute tolerance', 0.0001);
[y, istate, msg] = lsode("ode",y0,t);
istate
msg
% Plot the solution
plot(t,y(:,1),'-')
% Differential equation
function dydt = ode(t,y)
dydt = [ 1/sqrt(y(1)^2 + 1)+y(1)-y(1)^2];
*** Matlab version ***
function nonlinear_de
% Time span
tspan=[0 10];
% Initial condition
y0=[0];
% Solve the DE
[t,y] = ode45(@ode,tspan,y0);
% Plot the solution
plot(t,y(:,1),'-')
% Differential equation
function dydt = ode(t,y)
dydt = [ 1/sqrt(y(1)^2 + 1)+y(1)-y(1)^2];
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The arguments of your differential function should be swapped for the
code using lsode. From "help lsode":
The first argument, FCN, is a string, or cell array of strings,
inline or function handles, that names the function to call to
compute the vector of right hand sides for the set of equations.
The function must have the form
XDOT = f (X, T)
Changing your Octave code to read
function dydt = ode(y,t)
instead of
function dydt = ode(t,y)
gives results in agreement with ode45 using the other program.
You might also want to look at odepkg in Octave-Forge, it includes ode45
and other handy solvers.