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- % 抛体运动——用bvp求解阻力系数
- g=10;k=1;theta=pi/4;n=2;v=10;
- solinit = bvpinit(linspace(0,0.734846711099164,1e4), ...
- @(t)mat4init(t,g,v,theta),k);
- opts_jac = bvpset('FJacobian',@(t,u,k)jac(t,u,k,n), ...
- 'BCJacobian',@(u0,ut,k)bc_ab(u0,ut,k), ...
- 'RelTol',0.1,'AbsTol',0.1,'Stats','on');
- opts = bvpset('RelTol',0.1,'AbsTol',0.1,'Stats','on');
- tic
- sol = bvp4c( @(t,u,k)mat4ode(t,u,k,n,g), ...
- @(u0,ut,k)mat4bc(u0,ut,k,v,theta), solinit,opts_jac);
- toc
- % 输出验证
- fprintf('\n阻力系数k=:%7.3f.\n',sol.parameters)
- t=linspace(0,0.7348,1e3)';
- ut = deval(sol,t)';
- x=ut(:,1);y=ut(:,2);
- plot(x,y)
- % 微分方程
- function du = mat4ode(~,u,k,n,g)
- vx = u(3); vy = u(4);
- v = sqrt(vx^2+vy^2);
- dx = vx;
- dy = vy;
- dvx = -k*v^(n-1)*vx;
- dvy = -g-k*v^(n-1)*vy;
- du = [dx dy dvx dvy]';
- end
- % FJacobian
- function [J,du_k]= jac(~,u,k,n)
- vx = u(3); vy = u(4);
- v = sqrt(vx^2+vy^2);
- dvx_vx = -k*v^(n-1)-k*(n-1)*v^(n-3)*vx*vx;
- dvx_vy = -k*(n-1)*v^(n-3)*vx*vy;
- dvy_vx = -k*(n-1)*v^(n-3)*vx*vy;
- dvy_vy = -k*v^(n-1)-k*(n-1)*v^(n-3)*vy*vy;
- J = [
- 0 0 1 0;
- 0 0 0 1;
- 0 0 dvx_vx dvx_vy;
- 0 0 dvy_vx dvy_vy];
- du_k=[0 0 -v^(n-1)*vx -v^(n-1)*vy]';
- end
- % 边界条件
- function bc = mat4bc(u0,ut,~,v,theta)
- x0=u0(1);y0=u0(2);vx0=u0(3);vy0=u0(4);
- xt=ut(1);yt=ut(2);vxt=ut(3);vyt=ut(4);
- ua=[x0 y0 vx0-v*cos(theta) vy0-v*sin(theta)]';
- bc = [ua(1:4);yt];
- end
- % BCJacobian
- function [a,b,k]=bc_ab(~,~,~)
- a=[ 1 0 0 0 0
- 0 1 0 0 0
- 0 0 1 0 0
- 0 0 0 1 0
- 0 0 0 0 0 ];
- b=[ 0 0 0 0 0
- 0 0 0 0 0
- 0 0 0 0 0
- 0 0 0 0 0
- 0 1 0 0 0 ];
- k=[ 0 0 0 0 0 ]';
- end
- % 初值估计
- function u_t = mat4init(t,g,v,theta)
- vxt=v*cos(theta);vyt=v*sin(theta);
- u_t=[vxt*t vyt*t-0.5*g*t^2 vxt vyt]';
- end
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