Minimizing a functional of two functions with three boundary conditions What are the natural boundary conditions for the following calculus of variations problem:
Minimize:
$$J[y] = \int_0^b (1+(y_1')^2 + (y_2')^2)) \,dx$$
subject to the boundary conditions $$y_1(0) = 0 = y_2(0)$$ and $$b + y_1(b) − y_2(b) = 1.$$
So I used Euler-Lagrange Equations to get a system of two equations. I get 
$y_1'' = 0$ and $y_2'' = 0$ so $y_1 = Ax + B$ and $y_2 = Cx + D$ Using first two boundary conditions I get $y_1 = Ax$ and $y_2 = Cx$ using the third condition I get $b + Ab - Cb = 1$ and I do not know where to go from here.
 A: The boundary conditions allow more  room than usual, since there are only three of them instead of four that one would normally have (a system of 2 ODE of 2nd order). So, the space of solutions of the ODE system is 1-dimensional: 
$$y_1=Ax, \quad y_2=Cx,\quad 1+A-C=0$$
We can still minimize within this 1-dimensional space: 
$$J[y]= b(1+A^2+C^2)$$
Since $C=A+1$, the minimum is at $A=-1/2$, $C=1/2$. 
A: We are given the following 3 essential/Dirichlet boundary conditions (BCs)
$$ y_1(0)~=~0~=~y_2(0), \qquad y_1(b)-y_2(b)~=~1-b.\tag{1} $$
That is 1 BC short of a well-posed variational problem. 
If we vary infinitesimally OP's functional 
$$J[y]~:=~b+ \int_0^b\! dx~ \sum_{i=1}^2 y^{\prime}_i(x)^2,\tag{2}$$ 
while paying attention to boundary contributions. The infinitesimal variations $\delta y_i$ must obey the BCs (1), i.e.
$$ \delta y_1(0)~=~0~=~\delta y_2(0), \qquad \delta y_1(b)~=~\delta y_2(b).\tag{3} $$
We find
$$ \delta J[y]~\stackrel{(2)}{=}~  2\int_0^b\! dx~ \sum_{i=1}^2y^{\prime}_i(x) ~\delta y^{\prime}_i(x)$$
$$~\stackrel{\text{int. by parts}}{=}~  \sum_{i=1}^2\left( y^{\prime}_i(b)~\delta y_i(b)-y^{\prime}_i(0)~\delta y_i(0)  - 2\int_0^b\! dx~ y^{\prime\prime}_i(x) ~\delta y_i(x)\right)$$
$$~\stackrel{(3)}{=}~  \underbrace{(y^{\prime}_1(b)+y^{\prime}_2(b))}_{\text{natural BC}}~ 
\underbrace{\delta y_1(b)}_{\text{essential BC}} 
 - 2\int_0^b\! dx~\sum_{i=1}^2 \underbrace{y^{\prime\prime}_i(x)}_{\text{EL eqs.}} ~\delta y_i(x).\tag{4}$$
The above should vanish at a stationary point. To have a  well-posed variational problem, we must impose one more essential or natural BC. Any other BC would make the variational problem (2) ill-posed.
We read of from eq. (4) that the natural BC is
$$ y^{\prime}_1(b)+y^{\prime}_2(b)~=~0 ,\tag{5}$$
which is OP's main question. 
