I am trying to solve a 4th order linear ODE with boundary conditions. using bvp4c. The code is the following:

 function mat(solver)
if nargin<1 
    solver = 'bvp4c' ;
end
bvpsolver = fcnchk(solver);
pr=4;A=0.5;B=0.5;k1=0.4;Ec=2;del=1;N=1;
w=(3*N)/(3*N+4);
solinit = bvpinit(linspace(0,1,5),[0 1 0 ]);
sol = bvpsolver(@nanoode,@nanobc,solinit); 
for Bnew = infinity+1:maxinfinity
    solinit = bvpxtend(sol,Bnew); 
    sol = bvp4c(@nanoode,@nanobc,solinit);
function dfdeta = nanoode(eta,y)
         dfdeta = [
                    y(2)
                    y(3)
                    y(4)
                    (1/k1)*(y(1)^-1)*((y(2)^2)-(y(1)*y(3))-(k1*(y(3)^2))+(2*k1*y(2)*y(4)))
                   ];

end function res1 = nanobc(y0,yinf) res1 = [y0(1)-0 y0(2)-1 yinf(2)-0 ];

end

but i get the following error

Error in ==> Walter>nanoode at 27 dfdeta = [

Error in ==> bvparguments at 116 testODE = ode(x1,y1,odeExtras{:});

Error in ==> bvp4c at 120 [n,npar,nregions,atol,rtol,Nmax,xyVectorized,printstats] = ...

Error in ==> Walter at 11 sol = bvpsolver(@nanoode,@nanobc,solinit);

[Jan]

Please edit your question an post a complete copy of the error message. Some important details are missing.

[Torsten]

You have 4 ODEs, but only 3 initial and boundary conditions. This is not possible.

Best wishes

Torsten.

Answer 1

[Jan]

A line ending with "[" is no valid Matlab syntax. Append at least the continuations characters '...'. Although it is allowd to omit the continuation marks and let assume Matlab a vertical concatenation when a line break appears, better write this explicitly to be as clear as possible:

function dfdeta = nanoode(eta,y)
dfdeta = [ y(2); ...
           y(3); ...
           y(4); ...
           (1/k1)*(y(1)^-1)*((y(2)^2)-(y(1)*y(3))-(k1*(y(3)^2))+(2*k1*y(2)*y(4)))];