Solve the higher order of boundary condition.

Question

Nur 2024-1-17

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2070851-solve-the-higher-order-of-boundary-condition

回答： SAI SRUJAN 2024-1-24

How to incorporate higher-order boundary conditions into the matrix of the Keller box method?

1 个评论
显示 -1更早的评论隐藏 -1更早的评论

Torsten 2024-1-17

Since you always have to reduce higher-order equations/systems to a first-order system before starting to solve it, "higher-order boundary conditions" don't exist. Or what exactly do you mean ?

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

SAI SRUJAN 2024-1-24

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2070851-solve-the-higher-order-of-boundary-condition#answer_1396126

在 MATLAB Online 中打开

Hi Nur,

I understand that you are trying to handle higher-order boundary conditions when using the Keller box method to solve differential equations.

Convert higher-order ordinary differential equations (ODEs) into a system of first-order ODEs, once we have the system of first-order ODEs and the corresponding boundary conditions, we can use the following MATLAB functions to solve the systems of ODEs. Here are some MATLAB functions that might be used:

'bvp4c' or 'bvp5c': These functions are designed for solving boundary value problems (BVPs) for systems of ODEs. They are well-suited for problems with two-point boundary conditions.
'fsolve': This function can be used to solve systems of nonlinear algebraic equations, which is what you get after discretizing a system of ODEs with the Keller box method.

Please follow the below example to understand the working of 'bvp4c',

function example
    function dydx = diff_eqn(x, y)
        dydx = [y(2); -y(1)]; % Example: y'' + y = 0
    end
    function res = bc(ya, yb)
        res = [ya(1) - alpha; yb(1) - beta]; % Example: y(0) = alpha, y(1) = beta
    end
    solinit = bvpinit(linspace(0, 1, 10), [0, 0]);
    sol = bvp4c(@diff_eqn, @bc, solinit);
end