Can not pass enough input arguments to BVP function using bvp4c

Hi all,

I'm trying to solve the following BVP function, where lambda will is an unknown and a_r is a constant.

y'' + lambda^2 * y = -lambda^2 * a_r defined in the range [0,L].

BC's are: y(0)=0 and y'(L)=0.

By the guidance of tutorial "Solve BVP with Unknown Parameters" I've written the code below.

lambda=2.5899;
solinit=bvpinit(linspace(0,L,10),@mat4init,lambda);
sol=bvp4c(mat4ode,mat4bc,solinit);
%% functions
function dydx=mat4ode(x,y,lambda)
a_r=3*pi/180;
dthdy=[y(2)  -lambda^2*(a_r + y(1))];
end

function res=mat4bc(ya,yb,lambda)
res=[ya(1)
    yb(2)];
end

function thinit=mat4init(x)
global L;
thinit=[sin(x*pi/2/L)
    (pi/2/L)*cos(x*pi/2/L)];
end

As a result, I am getting this error:

Not enough input arguments.

Error in Untitled6>mat4ode (line 7)
dthdy=[y(2)  -lambda^2*(a_r + y(1))];

Error in Untitled6 (line 3)
sol=bvp4c(mat4ode,mat4bc,solinit);

Could you please help me to fix that error?

Thanks

Answer 1

[Jan]

In the line:

sol=bvp4c(mat4ode,mat4bc,solinit);

you call mat4ode without inputs. You want to provide the function handle instead and append the parameter lambda in addition:

sol=bvp4c(@(t,y) mat4ode(t, y, lambda), mat4bc, solinit);

[Abdulkadir Kaya]

Jan,

Thanks for your comment. But after that correction I've received the same for mat4bc. Then, done the same correction for this.

However, I am getting the following error now,

Error using Untitled6>@(t,y)mat4ode(t,y,lambda)
Too many input arguments.

Error in bvparguments (line 105)
    testODE = ode(x1,y1,odeExtras{:});

Error in bvp4c (line 130)
    bvparguments(solver_name,ode,bc,solinit,options,varargin);

Error in Untitled6 (line 3)
sol=bvp4c(@(t,y) mat4ode(t, y, lambda), @(ya,yb)mat4bc(ya,yb,lambda), solinit);

Let me share the new code as well.

lambda=2.5899;
solinit=bvpinit(linspace(0,0.6,10),@mat4init,lambda);
sol=bvp4c(@(t,y) mat4ode(t, y, lambda), @(ya,yb)mat4bc(ya,yb,lambda), solinit);
%% functions
function dydx=mat4ode(x,y,lambda)
a_r=3*pi/180;
dthdy=[y(2)  -lambda^2*(a_r + y(1))];
end

function res=mat4bc(ya,yb,lambda)
res=[ya(1)
    yb(2)];
end

function thinit=mat4init(x)
L=0.6;
thinit=[sin(x*pi/2/L)
    (pi/2/L)*cos(x*pi/2/L)];
end

Thanks.

[Jan]

Two changes:

% solinit = bvpinit(linspace(0,0.6,10), @mat4init, lambda);
  solinit = bvpinit(linspace(0,0.6,10), @mat4init);

% And:
% dthdy=[y(2)  -lambda^2*(a_r + y(1))];
  dydx = [y(2), -lambda^2*(a_r + y(1))];

[Abdulkadir Kaya]

Thank you Jan.

The problem is solved by adding an arbitrary 3rd boundary condition such as

ya(2)-1e6.

The problem is solved.