I've been trying to solve this problem but there are no resources that help. I've tried different approaches to solve this problem but every one of them leads to a dead end. I've found one approach that seems promising but I cannot solve the differential equation for $r^\sigma(t)$, which is the path the particle takes.
First of all, I have a vector field$$\Lambda^\alpha(x,y,z)$$ but when $x,y,z$ depends on $t$: $$\Lambda^\alpha(x(t),y(t),z(t))$$
Also I have a path that I am going to vary $$r^\sigma(t)$$ The components are $$r^1(t)=x(t)$$ $$r^2(t)=y(t)$$ $$r^3(t)=z(t)$$
The initial conditions are $$r^\sigma(0)=x^\sigma$$ and $$\frac{dr^\sigma(0)}{dt}=v^\sigma(t)$$
I want to find the path that the position vector and the vector field are pointing in the same direction. So I have to maximize the dot product between $r^\sigma(t)$ and $\Lambda^\alpha(t)$ for each tiny $dr^\sigma(t)$ along $\Lambda^\alpha(t)$, we can write this as: $$A=\int \Lambda_\alpha(t) \ dr^\alpha(t)$$ but we want to integrate over time for the path of stationary action so we get $$A=\int_0^t \Lambda_\alpha(t) \ \frac{ dr^\alpha(t)}{dt}dt$$ Here we can see that: $$\mathcal{L}=\Lambda_\alpha(t) \ \frac{ dr^\alpha(t)}{dt}$$ So using the euler lagrange equations: $$\frac{d}{dt}\frac{\partial \Lambda_\alpha(t) \ \frac{ dr^\alpha(t)}{dt}}{\partial(\frac{dr^\sigma}{dt})}=\frac{\partial \Lambda_\alpha(t) \ \frac{ dr^\alpha(t)}{dt}}{\partial r^\sigma}$$
I don't know how to solve these equations for $r^\sigma$ and I don't know how to implement the initial conditions into this equation. Because the paths should change completely as we implement them in, so it shouldn't be a linear term.
We want to find the stationary points of $$ S[r(t)] = \int \Lambda_\alpha (r^\sigma (t)) \dot{r}^\alpha dt $$ So $$ S[r(t)] = \int L(r^\sigma(t) , \dot{r}^\sigma(t) ) dt $$ where $$ L(r^\sigma(t) , \dot{r}^\sigma(t) = \Lambda_\alpha (r^\sigma (t)) \dot{r}^\alpha $$ The Euler-Lagrange equations tell us that whenever $\delta S = 0$ we have $$ \frac{\partial L}{\partial r^\sigma} = \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{r^\sigma}}\right) $$ For the LHS we have $$ \frac{\partial L}{\partial r^\sigma} = \frac{\partial \Lambda_\alpha}{\partial r^\sigma} \dot{r}^\alpha $$ For the RHS we have \begin{align} \frac{d}{dt} \left(\frac{\partial L}{\partial \dot{r}^\sigma}\right) &= \frac{d}{dt} \left(\Lambda_\alpha (r^\beta (t)) \delta^\alpha_\sigma \right) \\&= \frac{d}{dt} \left( \Lambda_\sigma( r^\beta (t))\right) \\&= \frac{\partial A_\alpha}{\partial r^\sigma} \dot{r}^\sigma \end{align} Putting this together, we have the rather uninspiring equation $$ \frac{\partial A_\alpha}{\partial r^\sigma} \dot{r}^\sigma = \frac{\partial A_\alpha}{\partial r^\sigma} \dot{r}^\sigma $$ What's the moral of the story? Sometimes the Euler-Lagrange equations are no help in solving a variational problem.
How could we have seen this coming? Note that $$ p_\sigma = \frac{\partial L}{\partial \dot{r}^\sigma} = \Lambda_\sigma $$ The canonical momenta are fully constrained. The system has no dynamics.
How else could we work this problem? Sometimes the simplest answer is the easiest. I'm going to assume you want the velocity vector of the particle aligned with the vector field. This is easily translated to a differential equation as $$ \Lambda^\alpha(r^\sigma(t)) = f(r^\sigma(t)) \dot{r}^\alpha $$ where $f(r^\sigma(t))$ is some scale factor that may depend on position.