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

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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
please help me
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
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回答(2 个)

Jack
Jack 2023-4-3
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');

kavi
kavi 2025-2-5
2.322y''+0.5172y'+0.3124y'^3+0.0914y+0.6889y^3=0;y(0)=0.0872,Y'(0)=0

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