# solving a second order non linear differential equation using RK 4TH order method

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haridas siddhartha on 3 Apr 2023
Commented: Ramanuja on 3 Mar 2024
Differential equation : h d^2h/dx^2 + (dh/dx)^2 - dh/dx * tan(ax) + c - h * sec^2(ax) * a = 0
Boundary conditions: h(x=0)=h0 and h(x=L)=h0
Dependent variable: h
Independent variable: x
constants: a,c,L,h0
Method to be used : RK 4th order
let y = h and z = dh/dx=dy/dx,dz/dx = a * sec^2(ax) + (1/h) * (z * tan(ax) - z^2 - c) confused how to give boundary conditions
haridas siddhartha on 9 Apr 2023
Torsten on 9 Apr 2023

Jack on 3 Apr 2023
Here's an example code in MATLAB for solving the given differential equation using the RK4 method. Note that you need to define the constants and initial conditions before running the code.
% Define constants and initial conditions
a = 1;
c = 1;
L = 1;
h0 = 1;
N = 1000; % Number of grid points
x = linspace(0, L, N)';
dx = x(2) - x(1);
h = h0*ones(N, 1); % Initial guess for h
dhdx = zeros(N, 1); % Initial guess for dh/dx
% Define the function f(x, y) = [dy/dx, d^2y/dx^2]
f = @(x, y) [y(2); ...
(-y(2)^2 + y(2)*tan(a*x) - c + h0*sec(a*x)^2*a)/h0];
% Solve the differential equation using the RK4 method
for n = 1:10000
k1 = dx*f(x, [h, dhdx]);
k2 = dx*f(x + dx/2, [h + k1(1:N)/2, dhdx + k1(N+1:end)/2]);
k3 = dx*f(x + dx/2, [h + k2(1:N)/2, dhdx + k2(N+1:end)/2]);
k4 = dx*f(x + dx, [h + k3(1:N), dhdx + k3(N+1:end)]);
h = h + (k1(1:N) + 2*k2(1:N) + 2*k3(1:N) + k4(1:N))/6;
dhdx = dhdx + (k1(N+1:end) + 2*k2(N+1:end) + 2*k3(N+1:end) + k4(N+1:end))/6;
end
% Plot the solution
plot(x, h);
xlabel('x');
ylabel('h');
title('Solution of the differential equation');
Ramanuja on 3 Mar 2024
K1 term showing eq error sir