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ODE Error: "Array indices must be positive integers or logical values"

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Tom Keaton
Tom Keaton on 21 Aug 2019
Commented: Star Strider on 21 Aug 2019
Alright I managed to simplify this down to the most basic I can. I am getting an error in my ode set:
Index in position 1 is invalid. Array indices must be positive integers or logical values: .
Error in bdipuniodefun (line 10)
bdip = [s(4); s(5); s(6); (Ex(s(1),s(2),s(3))); (Ey(s(1),s(2),s(3))); (Ez(s(1),s(2),s(3)))];
Error in odearguments (line 90)
f0 = feval(ode,t0,y0,args{:}); % ODE15I sets args{1} to yp0.
Error in ode15s (line 150)
odearguments(FcnHandlesUsed, solver_name, ode, tspan, y0, options, varargin);
Error in EFieldTest (line 46)
[T,S] = ode15s(@bdipuniodefun, tspan, icv);
What does this mean? My code:
%TSTEP MUST BE ABLE TO GO INTO TFIN AN ODD AMOUNT OF TIMES (Because simulation includes tstep @ 0 so it ends up being even)
tstep = 1; %Defining time step
tfin = 11; %Defining final time
%Create zeros matrices to populate later
intspan = [0:tstep:tfin]; %Total time span
[introw,intcol] = size(intspan);
%Generate matrix that will be populated by positions and velocities (Each matrix within structure "posandvelmat" should be of size 2*introw-1 by 6)
posandvelmat = zeros(2*intcol-1,6);
spart(1) = 0;
spart(2) = 0;
spart(3) = 0;
vpart(1) = 0;
vpart(2) = 0;
vpart(3) = 0;
tic
for t = 0:1:intcol-2 %complete time interval and time step
%Defining relative position and velocity
x = spart(1);
y = spart(2);
z = spart(3);
vx = vpart(1);
vy = vpart(2);
vz = vpart(3);
%Coupled differential equation solver for trajectory within current time step
icv = [x; y; z; vx; vy; vz]; %Initial conditions
tspan = [intspan(t+1) ((intspan(t+2)-intspan(t+1))/2)+intspan(t+1) intspan(t+2)]; %Time span
%Trajectory solver
[T,S] = ode15s(@bdipuniodefun, tspan, icv);
posandvelmat((1+2*t):(3+2*t),(1:6)) = S;
%Redfine the velocity and position components to reference on next if-loop run
spart(1) = posandvelmat((3+2*t),1);
spart(2) = posandvelmat((3+2*t),2);
spart(3) = posandvelmat((3+2*t),3);
vpart(1) = posandvelmat((3+2*t),4);
vpart(2) = posandvelmat((3+2*t),5);
vpart(3) = posandvelmat((3+2*t),6);
end
bdipuniodefun script:
function bdip = bdipuniodefun(t,s)
persistent Ex Ey Ez
if isempty(Ex)
[Ex, Ey, Ez] = E_field_c();
end
bdip = [s(4); s(5); s(6); (Ex(s(1),s(2),s(3))); (Ey(s(1),s(2),s(3))); (Ez(s(1),s(2),s(3)))];
end
E_field_c script:
function [Ex, Ey, Ez] = E_field_c()
syms x y z
E = -2000.*z;
Ex = matlabFunction(E(1));
Ey = matlabFunction(E(2));
Ez = matlabFunction(E(3));
end
  3 Comments
Star Strider
Star Strider on 21 Aug 2019
That still lacks significant clarity.
Note that:
Ex = matlabFunction(E(1));
Ey = matlabFunction(E(2));
Ez = matlabFunction(E(3));
assumes ‘E’ as a vector. It isn’t.
Consider:
syms v x y z
E(v) = -2000.*v;
Ex = matlabFunction(E(1));
Ey = matlabFunction(E(2));
Ez = matlabFunction(E(3));
That at least makes some sense, although ‘Ex’ and the rest are functions without input arguments.
Note that in my version of the code (that has the advantage of at least having the correct syntax):
Ex =
function_handle with value:
@()-2.0e+3
so ‘Ex’ and the rest have no idea what to do with the arguments you give them. Passing them any arguments at all wil throw:
Error using symengine>@()-2.0e+3
Too many input arguments.
Please spend some time with the documentation.

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Answers (1)

Ted Shultz
Ted Shultz on 21 Aug 2019
You are calling bdipuniodefun(t,s) with s having values of 0 (values from icv). Your program then attempts to index to the "0" element of Ex (using s as the index). In MATLAB there is no 0 index, thus the error.
  2 Comments
Ted Shultz
Ted Shultz on 21 Aug 2019
0 is the not the inital condition, but an invalid index.
when you write
Ey(s(1),s(2),s(3))
Matlab things that you are saying that Ey is a three dimentional variable, and that you want the value definded by the location dim1 = s(1), dim2 = s(2) and dim3 = s(3).
I don't belive that is what you are trying to do. I think your notation could be tweaked.

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