Solve 4th order ODE

I would like to find a solution for the differential equation with boundary conditions (solution for a buckling column on a non-linear elastic foundation):

where E, I, P, k1, and k3 are some constants

I tried to solve thos equation in Wolfram Mathematica. Wolfram Mathematica gives me only a trivial solution.

Could you please kindly help me solve this equation numerically?

function mat4bvp
L=1
xmesh = linspace(0,L,5);
solinit = bvpinit(xmesh, @guess);
sol = bvp4c(@bvpfcn, @bcfcn, solinit);
plot(sol.x, sol.y, '-o')
end
function dydx = bvpfcn(x,y)
dydx = zeros(4,1);
k1=1
k3=1
I=1;
E=1;
P=1
dydx=[y(1)
      y(2)
      y(3)
      -y(2)*P-k1*y(1)+k3*y(1)*y(1)*y(1)]; %4th order equation
end  

  
function res = bcfcn(ya,yb)

res=[ya(1)
     ya(2)
     yb(1)
     yb(2)];
end 


function g = guess(x)
g = [sin(x)
     cos(x)
     -sin(x)
     -cos(x)];
end  

I am trying to solve it using bvp soft, but matlab gives error: unable to solve the collacation equations -- a singular Jacobian encounered.

[Sam Chak]

Hi @user06261999

Can you refer to an example in the bvp4c article, and attempt to write code to solve it?

https://www.mathworks.com/help/matlab/ref/bvp4c.html

Then show us the code if you run into some technical issues.

[user06261999]

Thank you for help! I am attaching my code in Matlab, but I still have a problem with solution.

[John D'Errico]

As far as the question at hand, you need to understand the solution to that equation is TRIVIALLY solved as w(x) == 0. Mathematica told you that, and so will MATLAB. Try it. (Looks like you did.) Does w(x) == 0 satisfy the equation, and all boundary conditions? Of course. So a numerical boundary value solver will tell you the same thing, at least if you get it to work.

[user06261999]

Thank you! But I want to find non-trivial solution for this equation.

[Torsten]

Maybe there are nontrivial solutions as for

w'' + w = 0, w(0) = w(2*pi) = 0

e.g., but it will be difficult to find them.

Numerical methods will most probably give w(x) = 0. Analytical methods are difficult since your equation is nonlinear.

[Sam Chak]

Hi @user06261999

Looks like the trivial solution is the only solution to the specified boundary condition.

You can test, for example

.

The solution is a sinusoidal signal. See what happens if you set the boundary condition

,

,

[Torsten]

@Sam Chak

Looks like the trivial solution is the only solution to the specified boundary condition.

Why do you think so ? You just gave two examples where the trivial solution is not the only solution if tf = 2*pi.

[Sam Chak]

Hi @Torsten

I ran some sampling solutions by fixing & , and a range of non-zero initial conditions for & . But I'm actually unsure whether the trivial solution is the only solution for .

Perhaps the non-trivial solution exists when . That's why I gave the example above. I think @user06261999 can continue searching for the non-trivial solution.

[Torsten]

@Sam Chak

Let's stick to the linear case

y + y'' = 0, y(0) = y(L) = 0

The solution is the trivial solution (and is unique) if L is not a multiple of pi.

Otherwise it has an infinite number of solutions, namely y = a*sin(x).

This doesn't generalize to the nonlinear case, but I think that - if nontrivial solutions exist - they will exist only for special values of L.

The problem is somehow connected with eigenvalue problems for boundary value problems, but I don't have experience in this field.