# Variational Calculus problem to determine when variation has local minimum

A function $y(x)$ is defined on $[0,1]$ such that $y(0) = y(1) = 0$. Consider the functional $F[y] = \int\limits_0^1 \frac{1}{2}(y')^2+g(y) \ dx$ where $g(y)$ is a function s.t. $g'(0) = 0$. The Euler-Lagrange equation for this is then $g'(y) = y''$. If $y$ satisfies this equation the second variation is then given by $\delta^2F[y,\xi] = \int\limits_0^1 g''(y)\xi^2+(\xi')^2 \ dx$. The question I am doing asks me to calculate the range of values of $g''(0)$ for which $\delta^2F[y,\xi] > 0$.

However as $y$ satisfies the EL equation we have that $g''(y) = \frac{d g'(y)}{dy} = \frac{d y''}{dy} = \frac{d y''}{dx}\frac{dx}{dy} = \frac{y'''}{y'}$. We also know on $[0,1]$ there is a point where $y'(c)= 0$ by the mean value theorem. The Jacobi accessory condition for this problem says that if there is some function $u(x)$ s.t. $u'' = \frac{y'''}{y'}u$ which is not zero anywhere then we must have $\delta^2F[y,\xi] > 0$. The only solution I can find to this is $u = y'$ but we have already shown this has a zero on $[0,1]$. How can I proceed?

It's a bit unclear what OP is asking. It seems OP is asked to consider the second variation $$\delta^2F[y\equiv 0,\xi] ~=~\frac{1}{2} \int_0^1 \!\mathrm{d}x~(\xi^{\prime 2} + g''(0)\xi^2), \qquad \xi(0)~=~0~=~\xi(1), \tag{1}$$ of the constant path $y\equiv 0$, and determine for which $g''(0)$ this is positive.
Well, it's manifestly positive for $g''(0)>0$.
Consider for now on only the case $g''(0)\leq 0$. In physics parlance, this is a harmonic oscillator with time $\Delta t=1$; mass $m=1$; and spring constant $k=-g''(0)$.
To avoid Jacobi point/caustics (which I presume the Jacobi accessory condition is about), the natural frequency $\omega:=\sqrt{\frac{k}{m}}$ should be smaller than the fundamental frequency/lowest harmonic $\frac{\pi}{\Delta t}$.
This leads to the condition $g''(0)>-\pi^2$.