Finding Stationary Paths of a System

Question

I am working with the below functional

$S[y]=\alpha y(1)^2+ \int_0^1dx \beta y'^2, \:\:y'(0)=0$

with a natural boundary condition at $x=1$ and the constraint

$C[y] =\gamma y(1)^2 + \int_0^1dx w(x)y^2=1$,

where $\alpha$, $\beta$ and $\gamma$ are constants. The task is to show that the stationary paths satisfy an Euler-Lagrange equation

$\beta \frac{d^2y}{dx^2} + \lambda w(x) y=0, \:\:y(0)=0,(\alpha-\gamma \lambda)y(1) + \beta y'(1)=0$,

where $\lambda$ is a Lagrange multiplier. I then have to let $w(x)=1$ and $\alpha=\beta=\gamma=1$ and find the non-trivial stationary paths of this system, along with the eigenfunctions and the values for the associated Lagrange multiplier. Is anyone able to guide me through the process on this one?

Answer 1

The auxiliary functional is: $$\bar{S}[y] =(\alpha -\lambda\gamma)y(1)^2+\int_0^1(\beta y'^2-\lambda\omega(x)y^2 dx$$ Now determine: $\bar{S}[y+\epsilon h]$, with $y(0)=0$ Then the Gateaux differential becomes: $$\triangle \bar{S}[y,h] = 2(\alpha -\lambda\gamma)y(1) h(1) +2 \int_0^1(\beta y'h'-\lambda\omega(x)y)h dx$$ Now integration by parts gives:$$\triangle \bar{S}[y,h] = 2(\alpha -\lambda\gamma)y(1) h(1) + 2\beta y'(1)h(1)-2\int_0^1(\beta y''+\lambda\omega(x)y)h$$ Since $h(0)=0$. On a stationary path $\triangle \bar{S}[y,h]=0$ for all allowed h. Now for $h(1)=0$ as a subset use the fundamental lemma of the calculus of variations to obtain: $$\beta \frac{d^2y}{dx^2}+\lambda\omega(x)y=0$$ Then on a stationary path $2(\alpha -\lambda\gamma)y(1) + \beta y'(1)h(1)=0$ and this must be true for all h, including those for which $h(1) \neq 0$, hence we also need: $$(\alpha -\lambda\gamma)y(1) + \beta y'(1)= 0$$ which is required in your question.