# System of arc-length defined ODEs with ode45

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Michael on 8 Aug 2012
I am working on a code that has a system of ODEs, but I have never worked with systems with ode45. In the part of the code that I have included, S is the arclength (which is basically the time step of this problem), th is the angle (theta) of the graph, R is the x coordinate, and Z is the y coordinate.
When I run the program as shown below, I get simply a matrix full of NaN, even when I change the initial R value to 0.0001 or something.
Any help would be appreciated.
Also, what is the output? I only want to graph R and Z, not theta
function yp=program(S,y)
th=y(1);
R=y(2);
Z=y(3);
dthdS=-sin(th)/R+Z-2*H;
dRdS=cos(th);
dZdS=sin(th);
yp=[dthdS; dRdS; dZdS];
end
[S,Y]=ode45(@program, [0, 1], [0, 0, 0])

Star Strider on 8 Aug 2012
Edited: Star Strider on 9 Aug 2012
As with your 'How can I graph a "system" of ODEs?' post, I suggest:
% dThXYdS(1) = Theta, dThXYdS(2) = R(t) = x, dThXYdS(3) = Z(t)
H = 10;
dThXYdS = @(t,ThXY) [-sin(ThXY(1))/ThXY(2) + ThXY(3) - 2*H; cos(ThXY(1)); sin(ThXY(1))];
x0 = [0.1; 0.1; 0];
Tspan = [0:0.01:2]';
[T ThXY] = ode45(dThXYdS, Tspan, x0);
figure(8)
plot(ThXY(:,2), ThXY(:,3))
xlabel('R(S)')
ylabel('Z(S)')
grid
Except for the axis labels in the plot, I used the variable designation from your previous post (and my previous answer) rather than change it to match your current variable designation.
The reason you are getting a matrix of NaNs is that your initial conditions are [0 0 0]. So R is zero when theta is zero and of course sin(theta) will be zero as well. By convention, (0/0) = NaN. An initial NaN in a recursive calculation such as yours creates a matrix of NaNs.

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Star Strider on 9 Aug 2012
I just checked it to be sure it's the same code I created and ran successfully earlier. I ran it again just now and it worked fine for me. (I'm running it on 2012a, but I doubt that makes any difference.) With axis equal it plots something that looks like a cycloid with about 6 cycles. ThXY is a [201 x 3] double array.
Since I can't reproduce your error, I invite others to run it as well to see if they have problems with it.
Michael on 9 Aug 2012
I guess my error was rounding and scale, because the graph makes a shape, but the tables show otherwise.
This is looking to be similar to the shape I want. Is there a way to limit it so that the solver stops when it starts to go back up (past a half-cycle). Like saying "stop when z(k)<z(k+1)", or do I have to just limit the Tspan?
Star Strider on 9 Aug 2012
You might be able to do that using the Events property described in odeset, the function that creates an options structure for the ODE solvers. I've not yet had occasion to use that option, so I encourage you to experiment.
Also, using format shortEng or format shortE will solve your numerical resolution problem.