How should I show the following stationary path with a natural boundary condition at $x=0$?

Question

Let $a>0, b>0, c>0$ be constants. Show that the stationary path of the functional $$S[y]=\int_{0}^{a}(y'^2+2by)dx, \quad y(a)=0,$$ with a natural boundary condition at $x=0$ and subject to the constraint $$C[y]=\int_{0}^{a}ydx=C,$$ is given by $$y(x)=\frac{3C}{2a^3}(a^2-x^2).$$

Here's my work:

Consider the functional $S[y]=\int_{0}^{a}(y'^2+2by)dx, y(a)=0,$ with a natural boundary condition at $x=0$ where $a>0, b>0, c>0$ are constants.

Then the auxiliary functional is $\bar{S}[y]=\int_{0}^{a}(y'^2+2by+\lambda_{1}y)dx, y(a)=y(c)=0$ where $\lambda_{1}$ is the Lagrange multiplier.

By definition, the Euler-Lagrange equation is $\frac{d}{dx}(\frac{\partial F}{\partial y'})-\frac{\partial F}{\partial y}=0, y(a)=A, y(b)=B$ for the functional $S[y]=\int_{a}^{b}F(x, y, y')dx, y(a)=A, y(b)=B.$

Let $F(x, y, y')=y'^2+2by+\lambda_{1}y.$

Note that $\frac{\partial F}{\partial y'}=2y'$ and $\frac{\partial F}{\partial y}=2b+\lambda_{1}.$

This gives $\frac{d}{dx}(\frac{\partial F}{\partial y'})=2y''.$

Thus, the Euler-Lagrange equation is $\frac{d}{dx}(\frac{\partial F}{\partial y'})-\frac{\partial F}{\partial y}=0\implies 2y''-2b-\lambda_{1}=0.$

Observe that $2y''-2b-\lambda_{1}=0\implies 2y''=2b+\lambda_{1}\implies y''=b+\frac{\lambda_{1}}{2}.$

I know I need to find the general solution first by solving this differential equation but how should I solve this?

Answer 1

You are totally on the right track, and all that is left is to use the boundary conditions and the constraint to pin down some coefficients.

Note that $b$ doesn't appear in the answer, and this is naturally so since $$ S[y]=\int_0^a (y'^2+2by)dx = \int_0^a y'^2\,dx + 2b\int_0^a y\,dx= \int_0^a y'^2\,dx + 2bc, $$ since the constraint is $C[y]=\int_0^a y\,dx=c$. (I think your $C$ should be $c$ from your first sentence.)

Therefore you can just use $\int_0^a y'^2\,dx$ as your functional, and $b$ never comes into play. (You can keep it if you want to, but the final answer won't change.)

Therefore, your calculation of the Euler-Lagrange equation with Lagrange multiplier gives $$ y'' = \frac{\lambda}{2}. $$ (I dropped your subscript 1, and dropped $b$ as explained above.) Integrating with respect to $x$ twice, the solution would be $$ y = \frac{\lambda}{4}x^2 + \alpha x + \beta, $$ where $\lambda, \alpha, \beta$ are constants to be determined. We do this by the boundary conditions and the constraint.

The natural boundary at $x=0$ means that for $p=\frac{\partial L}{\partial y'}=2y'$, we should have $p(0)=0$, that is, $2y'(0)=0$. Here $L=y'^2$ is the Lagrangian (the integrand in the functional). See Difference between "essential boundary conditions" and "natural boundary conditions"?

Therefore the three conditions for $y= \frac{\lambda}{4}x^2 + \alpha x + \beta$ are $$ \begin{cases} 2y'(0)=0 &\Longrightarrow \alpha=0, \text{ so}\\ y(a)=0&\Longrightarrow \frac{\lambda}{4}a^2 + \beta = 0,\\ \int_0^a y\,dx=c&\Longrightarrow \frac{\lambda}4\frac{a^3}3 + \beta a=c. \end{cases} $$ Solving the last two equstions gives $$ \lambda=-\frac{6c}{a^3},\quad \beta=\frac{3c}{2a}. $$ Therefore $$ y=\frac{\lambda}4 x^2 + \beta = \frac{3c}{2a^3}(a^2-x^2). $$