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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.

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2 Answers2

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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 $A=-1/2$, $C=1/2$.

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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.

Qmechanic
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