How can I plot a second order differential equation with boundary condition using fourth order Runge-Kutta method?
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%%%%%%%%%%%%%%%% Runga-Kutta%%%%%%%%%%%%%%%%
h=0.0001;
xfinal=d;
x(1)=0;
y(1)=0; % initial value of y
y(xfinal)=0; % final value of y
% Let y' = z (f1) and y" = z' (f2);
f1 = @(x, y, z) z;
f2 = @(x, y, z) ky^2*y-(ky*(-2*W*(pi/d)*tan(2*pi*x/d)+2*u0*((pi/d)^2)*cos(2*pi*x/d))*y)/(OP3-ky*u0*(sin(pi*x/d).^2-1/2)+...
B*(OP3-ky*u0*(sin(pi*x/d).^2-1/2)-A*(-2*W*(pi/d)*tan(2*pi*x/d)+2*u0*((pi/d)^2)*cos(2*pi*x/d)-ky*(OP3-ky*u0*(sin(pi*x/d).^2-1/2))))*(1-...
M*(opi^2)/(M*OP3^2-gi*Ti*ky^2)));
for i=1:ceil(xfinal/h)
x(i+1)=x(i)+h;
K1y = f1(x(i), y(i), z(i));
K1z = f2(x(i), y(i), z(i));
K2y = f1(x(i)+0.5*h, y(i)+0.5*K1y*h, z(i)+0.5*K1z*h);
K2z = f2(x(i)+0.5*h, y(i)+0.5*K1y*h, z(i)+0.5*K1z*h);
K3y = f1(x(i)+0.5*h, y(i)+0.5*K2y*h, z(i)+0.5*K2z*h);
K3z = f2(x(i)+0.5*h, y(i)+0.5*K2y*h, z(i)+0.5*K2z*h);
K4y = f1(x(i)+h, y(i)+K3y*h, z(i)+K3z*h);
K4z = f2(x(i)+h, y(i)+K3y*h, z(i)+K3z*h);
y(i+1) = y(i)+(K1y+2*K2y+2*K3y+K4y)*h/6;
z(i+1) = z(i)+(K1z+2*K2z+2*K3z+K4z)*h/6;
end
plot(x,y,'-','linewidth',1)
hold on
1 件のコメント
John D'Errico
2023 年 3 月 17 日
It looks like you already solved the ODE, and plotted it. Where is the problem? (Even so, if this were not homework, as it surely is, you should be using an ODE solver, not writing your own code.)
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