MATLAB: Return Vector of length zero ?

errorfor loopguiguideodeplotvectorwhile loop

I am building a robot simulation by solving the differential equations at different input speed values, the function uses [0 0 0]=[x y theta] as initial conditions then updates the initial conditions every iteration. when i use a for loop to input the number of iterations i want the equation to be solved it works fine!, however when i use a while loop to keep the function solving as i input the speed (through a edit text in gui) i get an error message "KPATH returns a vector of length 0, but the length of initial conditions vector is 3." . How can function return a zero length vector ? or the while loop cannot be used in this case ?
%
**code 
function pushbutton1_Callback(hObject, eventdata, handles)
% hObject    handle to pushbutton1 (see GCBO)
% eventdata  reserved - to be defined in a future version of MATLAB
% handles    structure with handles and user data (see GUIDATA)
global sld1
global sld2 
check = get(handles.checkbox1,'Value');
initials = zeros(1,3); %start at origin
p = par();
tspan = [0 1];
while check == true
    p.WL = sld1 ;p.WR = sld2; 
    sol= ode23(@Kpath, tspan, initials,[],p);
    [t,s] = ode23(@Kpath, tspan, initials,[],p);
    initials = deval(sol,2);
    xlabel('x-position [m]');ylabel('y-position [m]');
    title('robot path');
    grid on 
    plot(s(:,1),s(:,2),'b','linewidth',1.5);
    hold on 
        for j = 1:length(s(:,1))
            q = plot(s(j,1),s(j,2),'ro','MarkerSize',5,'linewidth',1.5);
            axis([-2.5 2.5 -2.5 2.5]);grid on;
            pause(0.01)
            delete(q)
        end
end
function p = par()
         p.L = 0.12;                        %length [m]
         p.r = 0.1;                         %radius of wheel [m]
function dt = Kpath(t,c,p)
          x = c(1);y = c(2);th = c(3);
          dx = (((p.r*p.WL)+(p.r*p.WR))/2) * cos(th);
          dy = (((p.r*p.WL)+(p.r*p.WR))/2) * sin(th);
          dth= ((p.r*p.WL)-(p.r*p.WR))/p.L;
          dt = [dx;dy;dth]

Best Answer

I am guessing here because I cannot run your code.
Defining ‘initials’ as the last row of ‘s’ may be what you want:
initials = s(end,:);
The new initial conditions will be the last row of the previous solution vector.
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