I am currently learning tensor calculus and came up with an interesting exercise for myself. Try to use calculus of variations to derive the equation of the shortest path between two points for any possible metric tensor. However, I am new to tensor calculus and it has been a while since I used calculus of variations and I ended up with a result I was unable to verify even using something as simple as a 2d cartesian space. So I feel I might have made a mistake but am not sure. So could someone more familiar with the topics please look over what I did to see if I made a mistake.
First to find the length of a path I used the arc length integral: $$L=\int^{t_2}_{t_1}\left|\frac{d\vec{x}}{dt}\right|dt$$ Where $\vec{x}(t_1)=\vec{a}$ and $\vec{x}(t_2)=\vec{b}$. So basicaily it the above integral gives me the arc length of a path paramtized by $t$ from some arbitary position vector $\vec{a}$ to some arbitrary position vector $\vec{b}$. Then let: $$M\left(\dot x^1,\dot x^2,...,\dot x^n\right)=\left|\frac{d\vec{x}}{dt}\right|=\sqrt{\dot x^i\dot x^jg_{ij}}$$ $$L=\int^{t_2}_{t_1}M\left(\dot x^1,\dot x^2,...,\dot x^n\right)dt$$ Now the standard procedure for calculus of variations: $$X^i=x^i+\epsilon f^i\;\;\;\;\;f^i(t_1)=f^i(t_2)=0\;\;\;\;\;\dot X^i=\dot x^i+\epsilon \dot f^i$$ $$L=\int^{t_2}_{t_1}M\left(\dot X^1,\dot X^2,...,\dot X^n\right)dt$$ $$\frac{dL}{d\epsilon}=\int^{t_2}_{t_1}\frac{\partial M}{\partial \dot X^i}\frac{\partial \dot X^i}{\partial \epsilon}dt=\int^{t_2}_{t_1}\frac{\partial M}{\partial \dot X^i}\dot f^i dt=\frac{\partial M}{\partial \dot X^i}f^i\Biggr|_{t_1}^{t_2}-\int^{t_2}_{t_1}\frac{d}{dt}\frac{\partial M}{\partial \dot X^i} f^i dt=-\int^{t_2}_{t_1}\frac{d}{dt}\frac{\partial M}{\partial \dot X^i} f^i dt$$ Setting $\frac{dL}{d\epsilon}=0$ to find a stationary point. $$\int^{t_2}_{t_1}\frac{d}{dt}\frac{\partial M}{\partial \dot X^i} f^i dt=0$$
Then by the fundamental lemma of calculus of variations and setting $\epsilon = 0$: $$\frac{d}{dt}\frac{\partial M}{\partial \dot x^i}=0$$ $$\frac{\partial M}{\partial \dot x^i}=c_i$$ Where $c_i$ is a set of arbitary constants. Now to find $\frac{\partial M}{\partial \dot x^i}$: $$\frac{\partial }{\partial \dot x^k}\sqrt{\dot x^i\dot x^jg_{ij}}$$ Applying chain rule and then product rule: $$\frac{1}{2\sqrt{\dot x^i\dot x^jg_{ij}}}\frac{\partial }{\partial \dot x^k}\dot x^i\dot x^jg_{ij}=\frac{1}{2\sqrt{\dot x^i\dot x^jg_{ij}}}\left(\frac{\partial \dot x^i}{\partial \dot x^k}\dot x^jg_{ij}+\dot x^i\frac{\partial \dot x^j}{\partial \dot x^k}g_{ij}+\dot x^i\dot x^j \frac {\partial g_{ij}}{\partial \dot x^k}\right)=\frac{1}{2\sqrt{\dot x^i\dot x^jg_{ij}}}\left(\dot x^jg_{kj}+\dot x^i g_{ik}+\dot x^i\dot x^j \frac{\partial g_{ij}}{\partial \dot x^k}\right)=\frac{2\dot x^ig_{ki}+\dot x^i\dot x^j \frac{\partial g_{ij}}{\partial \dot x^k}}{2\sqrt{\dot x^i\dot x^jg_{ij}}}$$ I suspected $\frac{\partial g_{ij}}{\partial \dot x^k}$ goes to zero but wasen't sure how to verify it (but then again didn't really try). So just incase there is some is some obscure metric tensor where it doesn't go to zero I left it in. Hence: $$\frac{2\dot x^ig_{ki}+\dot x^i\dot x^j \frac{\partial g_{ij}}{\partial \dot x^k}}{2\sqrt{\dot x^i\dot x^jg_{ij}}}=c_k$$ $$2 \dot x^ig_{ki}+\dot x^i\dot x^j \frac{\partial g_{ij}}{\partial \dot x^k}=2c_k\sqrt{\dot x^i\dot x^jg_{ij}}$$ Then I wanted to verify this via cartesian coordinates where this equation should give me a straight line. Since in cartesian coordinates the metric tensor is the kronecker delta, hence: $$2 \dot x^i \delta_{ki}=2c_k\sqrt{\dot x^i\dot x^j\delta_{ij}}$$ $$\dot x^k=c_k\left|\dot{\vec{x}}\right|$$ $$\dot{\vec{x}}=\vec{c}\left|\dot{\vec{x}}\right|$$ I am didn't manage to fand any analytic solution to that equation and I don't know any numerical method for such a equation (though if anyone knows one that could work please let me know). So did I do any mistakes?
Edit: I just realized after posting this that I can verify that it works in cartesian coordinates since I know the expected result. That being a straight line. Hence let $\vec{x}=\vec{a}t+\vec{b}$ Where $\vec{a}$ and $\vec{b}$ are constant vectors. Hence: $$\dot{\vec{x}}=\vec{a}$$ $$\vec{a}=\vec{c}\left|\vec{a}\right|$$ Since $\vec{a}$ is constant the equation holds. Still, I would like to know if I made any mistakes just to be sure.
Edit 2: Actually I did just manage to decuple the coupled differential equations in two dimensions. $$\dot x_1=c_1\sqrt{\dot x^2_1 + \dot x^2_2}\;\;\;\;\dot x_2=c_2\sqrt{\dot x^2_1 + \dot x^2_2}$$ $$\dot x^2_1=c^2_1\dot x^2_1 + c^2_1\dot x^2_2\;\;\;\;\dot x^2_2=c^2_2\dot x^2_1 + c^2_2\dot x^2_2$$ $$\dot x^2_1\left(1-c^2_1\right)=c^2_1\dot x^2_2$$ $$\dot x^2_1=\frac{c^2_1}{\left(1-c^2_1\right)}\dot x^2_2$$ $$\dot x^2_2=c^2_2\dot x^2_1 + c^2_2\dot x^2_2=c^2_2\dot x^2_1 + \frac{c^2_1c^2_2}{\left(1-c^2_1\right)}\dot x^2_2$$ $$\dot x^2_2\left(1-c^2_2-\frac{c^2_1c^2_2}{\left(1-c^2_1\right)}\right)=0$$ $$\dot x_1=0\;\;\;\;\dot x_2=0$$ So this would imply that $\vec{x}$ is a constant, which is a valid solution as in euclidian space a vector is a straight line. Now i would just have to define it as pointing from the intial piont to final one and it would be a the shortest path between. Not really the solution I expected as it isn't paramatized. But it works. And a paramatized solution does work too as shown in the previous edit. SO I guess I did everything right. Again the reason I asked is that I am still new to the topic and am still learning, hence I wanted to double check.