Continuing my previous question Argue that $\gamma_0$ cannot be a maximum of the action we consider once more the one-dimensional harmonic oscillator with the Lagrangian given by: $$L(x,v)=\frac{m}{2}v^2-\frac{k}{2}x^2,k>0$$ and the corresponding action on trajectories $\gamma$ given by the functional: $$S(\gamma)=\int_{t_1}^{t_2}L(\gamma(t),\dot{\gamma(t)})dt.$$
I have found the Euler-Lagrange equation of the system to: $$-kx=m \ddot{x}$$ And the solution to: $$x(t) = c_1 \cos(\sqrt{\frac{k}{m}} t ) + c_2 \sin(\sqrt{\frac{k}{m}} t).$$ We assume that $\gamma_0(t)$ solution of the Euler-Lagrange-equation satisfying $\gamma_0(t_1 = 0) = \xi$ and $\gamma_0(t_2 = T) = \eta$. And variation of the trajectory $\gamma_0$ given as $\gamma(t) = \gamma_0(t) + \nu(t)$ with $\nu$ some curve satisfying $\nu(0) = \nu(T) = 0$. We have to show that the difference between $S(\gamma)$ and $S(\gamma_0)$ is given by: $$\Delta S=S(\gamma)-S(\gamma_0)=\frac{1}{2} \int_0^T m \dot \nu^2(t)-k\nu^2(t)dt$$ But I think we can only get this expression if $S(\gamma_0)=0$. Can anyone help me to show that $S(\gamma_0)=0$? The answer to this question https://physics.stackexchange.com/q/682652/ says that
The classical solution on the SHO is on the form $$x_{\rm cl}(t)~=~A\cos(\omega t) + B\sin(\omega t)$$ An arbitrary virtual path that satisfies the BC is of the form $$ x(t)~=~x_{\rm cl}(t)+ \gamma_0(t) $$ where the fluctuation part is a Fourier series $$ \gamma_0(t)~=~\sum_{n\in\mathbb{N}} a_n\sin\left(n\pi \frac{t-t_i}{\Delta t}\right),\qquad a_n~\in~\mathbb{R} $$
But this can only give us $0$? But I don't understand how I get this expression and if it correct?
We have to use the formula for $\Delta S$ to argue that that $\gamma_0$ cannot be a maximum of the action. Can anyone help me what that argue could be? I'm a bit lost
I think you've miscalculated; you don't need $S(\gamma_0)=0$ because$$\begin{align}S(\gamma)-S(\gamma_0)&=\int_0^T\left[\frac12m\left((\dot{\gamma}_0+\dot{\nu})^2-\dot{\gamma}_0^2\right)-\frac12k\left((\gamma_0+\nu)^2-\gamma_0^2\right)\right]dx\\&=\int_0^T\left[\frac12m\left(2\dot{\gamma}_0\dot{\nu}+\dot{\nu}^2\right)-\frac12k\left(2\gamma_0\nu+\nu^2\right)\right]dx,\end{align}$$which gives the desired result iff$$\int_0^T\left[m\dot{\gamma}_0\dot{\nu}-k\gamma_0\nu\right]dx=0.$$But that integral is$$\int_0^Tm\left[\dot{\gamma}_0\dot{\nu}+\ddot{\gamma}_0\nu\right]dx=[m\dot{\gamma}_0\nu]_{t_1}^{t_2},$$which vanishes as desired because $\nu(t_1)=\nu(t_2)$.