Runge-Kutta 4 with 2nd Order: Solving Variables (x,y,Q)

[ahmedym_altae]

i would like to programing in MATLAB with runge kutta 4 with 2nd order the variables (x with y and Q) where x distance , y hight of water and Q discharge from the following tow equation

dydx = (s0-(((Q0^2*n^2)/(B^2.*y^(10/3))*(1+2.*y/B)^(4/3))+dQdx))/(1-Q0^2/(g*B^2*y.^3));
dQdx = -2*sqrt(2)/3*ce*(y-w).^1.5*Q0/(B^2*y.^2*sqrt(g));


i traing the following program but some thing error can you correcting to me, with glad

program 1

function kutta4research2(x0,xe, y0,Q0,N)

h = (xe-x0)/N; %the step size
x(1) = x0;
y(1) = y0; %the initial value
Q(1)=Q0;
for i = 1:N
k1y = Q(i);
k1Q = h*f(x(i), y(i), Q(i));
k2y = Q(i)+k1Q*h/2;
k2Q = h*f(x(i)+h/2, y(i)+(k1y)*h/2,Q(i)+k1Q*h/2);
k3y = Q(i)+k2Q*h/2;
k3Q = h*f(x(i)+h/2, y(i)+(k2y)*h/2,Q(i)+k2Q*h/2);
k4y = Q(i)+k3Q*h;
k4Q = h*f(x(i)+h, y(i)+k3y*h,Q(i)+k3Q*h);
x(i+1)=x0+i*h;
y(i+1) = y(i) + (k1y + 2*k2y + 2*k3y + k4y)*h/6;
Q(i+1) = Q(i) + (k1Q + 2*k2Q + 2*k3Q + k4Q)*h/6;
end

[x' y' Q']

%this line of code below is optional if visualization is not needed
plot(x, y, 'r*')

function dydx = f(x, y , Q)
function dQdx = Q(i)

y0=0.17;
s0=0.0005;
Q0=0.143;
n=0.012;
B=0.15;
ce=0.785;
w=0.15;
g=9.81;
dydx = (s0-(((Q0^2*n^2)/(B^2.*y^(10/3))*(1+2.*y/B)^(4/3))+dQdx))/(1-Q0^2/(g*B^2*y.^3));
dQdx = -2*sqrt(2)/3*ce*(y-w).^1.5*Q0/(B^2*y.^2*sqrt(g));

this program error


program 2

function kutta4research3(x0,xe, y0,Q0,N)

h = (xe-x0)/N; %the step size
x(1) = x0;
y(1) = y0; %the initial value
Q(1)=Q0;
for i = 1:N
k1y = Q(i);
k1Q = h*f(x(i), y(i), Q(i));
k2y = Q(i)+k1Q*h/2;
k2Q = h*f(x(i)+h/2, y(i)+(k1y)*h/2,Q(i)+k1Q*h/2);
k3y = Q(i)+k2Q*h/2;
k3Q = h*f(x(i)+h/2, y(i)+(k2y)*h/2,Q(i)+k2Q*h/2);
k4y = Q(i)+k3Q*h;
k4Q = h*f(x(i)+h, y(i)+k3y*h,Q(i)+k3Q*h);
x(i+1)=x0+i*h;
y(i+1) = y(i) + (k1y + 2*k2y + 2*k3y + k4y)*h/6;
Q(i+1) = Q(i) + (k1Q + 2*k2Q + 2*k3Q + k4Q)*h/6;
end

[x' y' Q']

%this line of code below is optional if visualization is not needed
plot(x, y,x,Q ,'r*')

function dy = f(x, y , Q)

y0=0.17;
s0=0.0005;
Q0=0.143;
n=0.012;
B=0.15;
ce=0.785;
w=0.15;
g=9.81;
th=12;
dy = (s0-((((Q0^2*n^2)/B^2.*y^(10/3))*(1+2.*y/B)^(4/3))+2*sqrt(2)/3*ce*(y-w).^1.5*Q0/((B^2*y.^2*sqrt(g)))/(cos(th)-Q0^2/(g*B^2*y.^3))));


this program Q value constant error



program 3


function [t,x,v] = rk4ode2(func, a, b, x0, xd0, h);
% Solution of 2nd order ODE using Runge-Kutta 4th order
% with constant step size. ODE solved is converted to
% two 1st order equations. The RHS of the system is
% dv/dt = func(t, x, v)
% dx/dt = v
% See for example rhs_smd.m for forced spring-mass-damper
%
% USAGE: [t, x, v] = rk4ode2(func,a,b,x0,xd0,h)
%
% input func = name of external function to evaluate the RHS
% of the ODE (eg 'rhs_smd')
% a, b = limits of integration
% x0 = initial condition (position)
% xd0 = initial condition (velocity)
% h = stepsize
%
% output [t, x, v] = solution vectors

t = [a];
x = [x0];
v = [xd0];
i = 1;

while t(i) < b

k1x = v(i);
k1v = feval(func, t(i) , x(i) , v(i) );

k2x = v(i)+k1v*h/2;
k2v = feval(func, t(i)+h/2, x(i)+k1x*h/2 , v(i)+k1v*h/2 );

k3x = v(i)+k2v*h/2;
k3v = feval(func, t(i)+h/2, x(i)+k2x*h/2 , v(i)+k2v*h/2 );

k4x = v(i)+k3v*h;
k4v = feval(func, t(i)+h , x(i)+k3x*h , v(i)+k3v*h );

i = i+1;
t(i) = t(i-1) + h;

x(i) = x(i-1) + (k1x + 2*k2x + 2*k3x + k4x)*h/6;
v(i) = v(i-1) + (k1v + 2*k2v + 2*k3v + k4v)*h/6;

end

this program i can not know how it run


and if possible solution with finite difference and finite element

 

[Valkarie]

?No, Runge-Kutta is the only numerical method that can be used for this system of equations. Finite difference and finite element methods are not applicable here.