"Too many input arguments" error while using bvp4c.

I am trying to find the solution to the following differential equations (which comes up while finding the similarity solution for boundary layer) using bvp4c. Here, .

Here m is a known parameter. I convert this third-order ode to three first order ode's as follows.

, , .

This gives,

The boundary conditions are: .

The code is as given below, with initial guess :

m = -0.05; % n = -0.5; Pr = 0.72;
init = bvpinit(linspace(0,50,200),[0 0 0.5], m); % 0 0.5]);
sol = bvp4c(@(Z) bvp_rhs(Z, m), @(Zi, Zf) bvp_bc(Zi, Zf, m), init);

eta = linspace(0,50,200); BS = deval(sol, eta);
plot(eta, BS(1,:))

function rhs = bvp_rhs(Z, m) %, n, Pr)
    rhs = [Z(2); Z(3); -((m + 1)/2)*Z(1)*Z(3) - m*(1 - Z(2)^2)]; %Z(5); -((m + 1)/2)*Pr*Z(1)*Z(5) - n*Pr*(1 - Z(4))*Z(2)];
end

function bc = bvp_bc(Zi, Zf)
    bc = [Zi(1) - 0; Zi(2) - 0; Zf(2) - 1]; %Zi(4) - 0; Zi(5) - 0;  Zf(4) - 1];
end

Here since is achieved at , I have considered .

I keep on getting the following error message and I am not sure why.

Error using untitled2>@(Z)bvp_rhs(Z,m)

Too many input arguments.

Error in bvparguments (line 96)

testODE = ode(x1,y1,odeExtras{:});

Error in bvp4c (line 119)

bvparguments(solver_name,ode,bc,solinit,options,varargin);

Error in untitled2 (line 3)

sol = bvp4c(@(Z) bvp_rhs(Z, m), @(Zi, Zf) bvp_bc(Zi, Zf, m), init);

untitled

Error using untitled>@(Z)bvp_rhs(Z,m)

Too many input arguments.

Error in bvparguments (line 96)

testODE = ode(x1,y1,odeExtras{:});

Error in bvp4c (line 119)

bvparguments(solver_name,ode,bc,solinit,options,varargin);

Error in untitled (line 3)

sol = bvp4c(@(Z) bvp_rhs(Z, m), @(Zi, Zf) bvp_bc(Zi, Zf, m), init);

Can anyone please help me with this?

If I have to further extend this code to couple the following ode with the first one, withn and as known parameters,

such that and and additional boundary conditions . Can the commented portion in the code work when appropriately added in the required places?

Answer 1

[Walter Roberson]

bvp4c expects the first parameter to be an ode function. That ode function is expect to take two parameters -- current time and current boundary conditions/] But you are passing @(Z) bvp_rhs(Z, m) which is an anonymous function that expects one parameter, not two.

In MATLAB, parameters are passed positionally, not by name of the variable, so there is no possibility that MATLAB could look at the @(Z) and guess from the name of the variable that you were wanting to skip the time parameter. If you do not want to pass the time parameter into later code then use

sol = bvp4c(@(t, Z) bvp_rhs(Z, m), @(Zi, Zf) bvp_bc(Zi, Zf, m), init);

which would accept a first parameter but then just not do anything with it.

[Mukul]

I do need the time parameter (eta in this case). The parameter m should not change. So I don't want to pass it.

If I modify the code as following I am getting a new error:

m = -0.05; % n = -0.5; Pr = 0.72;
init = bvpinit(linspace(0,50,200),[0 0 0.5]); % 0 0.5]);
sol = bvp4c(@(eta, Z) bvp_rhs(eta, Z, m), @(Zi, Zf) bvp_bc(Zi, Zf, m), init);

Warning: Unable to meet the tolerance without using more than 3333 mesh points.
The last mesh of 1774 points and the solution are available in the output argument.
The maximum residual is 165.673, while requested accuracy is 0.001.

Warning: Results may be inaccurate because of an ill-conditioned Jacobian whose reciprocal condition number is 3.99159e-39.

eta = linspace(0,50,200); BS = deval(sol, eta);
plot(eta, BS(1,:))

function rhs = bvp_rhs(eta, Z, m) %, n, Pr)
    rhs = [Z(2); Z(3); -((m + 1)/2)*Z(1)*Z(3) - m*(1 - Z(2)^2)]; %Z(5); -((m + 1)/2)*Pr*Z(1)*Z(5) - n*Pr*(1 - Z(4))*Z(2)];
end

function bc = bvp_bc(Zi, Zf, m)
    bc = [Zi(1) - 0; Zi(2) - 0; Zf(2) - 1]; %Zi(4) - 0; Zi(5) - 0;  Zf(4) - 1];
end

I am not at all sure what I am doing wrong.

[Torsten]

How should MATLAB know that you additionally want to pass "m" to your ODE function ? You must indicate this in your list of inputs to the function.

So use

sol = bvp4c(@(eta,Z)bvp_rhs(eta,Z,m), @bvp_bc, init);

instead of

sol = bvp4c(@bvp_rhs, @bvp_bc, init);

And I think you made a mistake in the transformation to a first-order system.

Shouldn't it read

-((m + 1)/2)*Z(1)*Z(2) - m*(1 - Z(2)^2)

instead of

-((m + 1)/2)*Z(1)*Z(3) - m*(1 - Z(2)^2)

?

And why do you set

Zf(2) - 1

as a boundary condition if you want to prescribe Z_2(Inf) = 0 ?

[Mukul]

The transformation looks okay to me.

I had incorrectly put the BC instead of in my original question. I have editted it there.

Sorry for the mistakes. I am new to MATLAB and am still learning.

I have attached the expected result for different values of m.