MatLab solves BVP with only left-hand side boundaries (How?)

Question

Thanh Hoang 2024-1-10

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2068551-matlab-solves-bvp-with-only-left-hand-side-boundaries-how

编辑： Torsten 2024-1-10

采纳的回答： Torsten

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My code is working but I do not really understand the following.

My ode System consist of three first order ode's:

I dy/dx = z

II dz/dy = 2y + 3z

III dw/dx = 2y + z

I wrote the code and was trying to set right and left hand side boundaries to solve my system, therefore I was setting 6 boundaries in total.

However, it seems like that only 3 boundaries are enough to solve the system. How is this working? I thought each ode requies both left and right hand side boundary to be able to solve.

Best regards,

Thanh

clear all; close all; clc
% Define the initial guess for the solution
solinit = bvpinit(linspace(0, 1, 50), @initial_guess);
% Solve the BVP using bvp4c
sol = bvp4c(@odefun, @bcfun, solinit);
% Extract the solutions from the sol structure
x = sol.x;
y = sol.y(1, :);
z = sol.y(2, :);
w = sol.y(3, :);
% Plot the solutions for y(x), z(x), and w(x)
subplot(3, 1, 1);
plot(x, y);
xlabel('x');
ylabel('y(x)');
title('Solution of ODE for y(x)');
subplot(3, 1, 2);
plot(x, z);
xlabel('x');
ylabel('z(x)');
title('Solution of ODE for z(x)');
subplot(3, 1, 3);
plot(x, w);
xlabel('x');
ylabel('w(x)');
title('Solution of ODE for w(x)');
function dydx = odefun(x, y)
dydx = [ y(2);
    2*y(1) - 3*y(2);
    y(1)*2 + y(2)];
end
function res = bcfun(ya, yb)
res = [ya(1) - 1; ya(2) - 2; ya(3)];% - [yb(1) - 3; yb(2) - 4; yb(3) - 5] right hand side boundaries 
end
function yinit = initial_guess(x)
yinit = [-2;0;0];
end

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Answer 1

Torsten 2024-1-10

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https://ww2.mathworks.cn/matlabcentral/answers/2068551-matlab-solves-bvp-with-only-left-hand-side-boundaries-how#answer_1387261

编辑：Torsten 2024-1-10

A system of odes needs as many boundary conditions as the sum of the degrees of its equations. Since all your equations are first-order odes, it needs 1 + 1 + 1 = 3 boundary conditions in total.

If you had a system of second-order odes, then you are right: it would need 2 + 2 + 2 = 6 boundary conditions.

If you only set left-hand side or right-hand side conditions, it's called an initial value problem (in contrast to a boundary value problem) and should be solved using ode45, e.g., a solver for initial value problems.