Boundary Value Problem error bvp4c

I am trying to solve a differential equation using the bvp4c function (using boundary value conditions) but i always get the error "In an assignment A(:) = B, the number of elements in A and B must be the same. " i dont know why this error happens ?

init = bvpinit(linspace(0,2,10),[0 0 0]);
sol = bvp4c(@Kpath1,@bcpath,init);
t = linspace(0,2,100);
BS = deval(sol,t);
plot(t,BS(1,:))

function bv = bcpath(L,R)

bv = [L(1)   L(2)  L(3)-1.2; R(1)-2 R(2)-2 R(3)-1.2];

end

function dt = Kpath1(~,c)
        L = 0.12;                        
        r = 0.1;
        WL = 0.25;WR = 0.25;

          x = c(1);y = c(2);th = c(3);
          dx = (((r*WL)+(r*WR))/2) * cos(th);
          dy = (((r*WL)+(r*WR))/2) * sin(th);
          dth= ((r*WR)-(r*WL))/L;
          dt = [dx;dy;dth];
          pose = [x y th];
  end

Thanks in advance

Answer 1

[Torsten]

You need to return a 3x1 vector instead of a 2x3 matrix in "bcpath".

Best wishes

Torsten.

[Moustafa Aboubakr]

Thanks Torsten, the thing is i have a system of 3 equations [x,y,theta] which are a function of time(t), there are boundary conditions at time (t) = 0, [0 0 1.2] and at time (t) = 2,[2 2 1.2] how can this be written in only a 3*1 vector ?

[Torsten]

Seems you'll have to use "pdepe" instead of "bvp4c".

Best wishes

Torsten.

[Moustafa Aboubakr]

Thanks, but i converted the system of equations from dydt,dxdt into dydx ad dtheta/dx and bvp4c worked fine :) I didnt understand the concept of bvp correctly thnaks again