Hi Hugo,
When dealing with differential equations that include complex boundary conditions, such as a Robin condition, symbolic solutions can be difficult to obtain explicitly. In these situations, a numerical approach is often more practical. The process begins by converting the second-order differential equation into a system of first-order equations. Then, you define the boundary conditions.
To solve the problem numerically, use a boundary value problem solver like MATLAB's bvp4c or bvp5c. These solvers are designed to handle complex boundary conditions effectively. Once the solver runs, it iterates to find a solution that satisfies both the differential equation and the boundary conditions.
Please review the following code for your reference:
% Define the differential equation as a system of first-order ODEs
function dydr = odefun(r, y, K, gamma)
dydr = [y(2); (y(1) - gamma - K*y(2)/r)/K];
end
% Define the boundary conditions
function res = bcfun(ya, yb, K, alpha, beta, r1, r2)
res = [K*ya(2) - alpha*ya(1) - beta; yb(1) - 4.8];
end
% Initial guess for the solution
init_guess = @(r) [1; 0]; % Adjust as necessary
% Setup the boundary value problem
solinit = bvpinit(linspace(r1, r2, 10), init_guess);
sol = bvp4c(@(r, y) odefun(r, y, K, 10), @(ya, yb) bcfun(ya, yb, K, 1, 0, r1, r2), solinit);
For reference:
- ya(1) corresponds to the value of the function (u) at the boundary (r1).
- ya(2) corresponds to the derivative of the function (u) with respect to (r) at the boundary (r1).
I hope it resolves your query!
