为什么会出现“The boundary condition function BCFUN should return a column vector of length 5”的错误信息？

Question

function Y6
solinit = bvpinit(linspace(0,1),[0;3;1;1],2);

sol = bvp4c(@ode, @bc, solinit);
y = sol.y;
time = sol.parameters*sol.x;
ut = -y(4,:);

figure(1);
plot(time,y([1 2],:)','-'); hold on;
plot(time, ut, 'k:');
axis([0 time(1,end) -1.5 3]);
text(1.3,2.5,'x_1(t)');
text(1.3,.9,'x_2(t)');
text(1.3,-.5,'u(t)');
xlabel('time');
ylabel('states');
title('Numerical solution');
hold off;

% -------------------------------------------------------------------------
% ODE's of augmented states
function dydt = ode(t,y,T)
dydt = T*[2*y(2);4*y(4);0;-2*y(3)];

% -------------------------------------------------------------------------
% boundary conditions: x1(0)=11;p2(0)=2; x2(tf)=3; 3*p1(tf)+p2(2)^2=0
function res = bc(ya,yb,T)
res = [ ya(1) - 11; ya(4); yb(2) - 3; 3*yb(3)+yb(4)^2];

I don't know why The boundary condition function BCFUN should return a column vector of length 5 error message is came. Can you explain to me please? Thank you so much

Answer 1

The unknown parameters or constants, here T , also count as components. Thus your state has overall 4+1=5 components, requiring the setting of 5 boundary conditions.

Sometimes that is obvious, for instance in a Sturm-Liouville eigenvalue computation with the eigenvalue as parameter you want to avoid the zero solution, so demand that, eg, the square integral has value 1.

In a ballistic shot with the flight time as parameter where you fix the location of the canon and the target as obvious boundary conditions it is not that obvious what to select as additional BC, but possibilities are to fix the initial speed or the initial angle.

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I might be rusty, but I get from the variation of

L = 3*(x1(T)-9)^2 + integral(0,T, 2*u^2 +p^T*(F(x,u)-x') )

the saddle point conditions

T    : 0 = 12*x2(T)*(x1(T)-9) + 2*u(T)^2    (BC5)
x(t) : 0 = p'(t)+F_x(x,u)^T*p
       p1'(t) = -2*p2(t)                    (DE3)
       p2'(t) = 0                           (DE4)
p(t) : x'(t) = F(x,u)
       x1'(t) = 2*x2(t)                     (DE1)
       x2'(t) = 4*u(t)                      (DE2)
u(t) : 0 = 4*u + p^T*F_u(x,u)
       u(t) = -p2(t)                        (OPT)
x1(T): -p1(T)+6*(x1(T)-9)                   (BC3)
x2(0): 0 = p2(0)                            (BC4)

This can be directly evaluated as p2=-u=0 (BC4+DE4+OPT) are constant, likewise p1 = 6*(x1(T)-9) (BC3+DE3). This in turn forces x1(T)=9 (BC5) As now x2(t)=3 is constant, we get x1(t)=11+6*t . Then the remaining expression of the functional 3*(2+6*T)^2 is minimal for T=-1/3 . For T=0 the minimal value of the functional is L=12 .