I am using from the following Euler equations :
$$\dfrac{\partial f}{\partial u^{i}}-\dfrac{\text{d}}{\text{d}s}\bigg(\dfrac{\partial f}{\partial u'^{i}}\bigg) =0$$
with function $f$ is equal to :
$$f=g_{ij}\dfrac{\text{d}u^{i}}{\text{d}s}\dfrac{\text{d}u^{j}}{\text{d}s}$$
Firstly, I don't understand the following relation :
$$f=g_{ij} u'^{i}u'^{j}=1$$
(with $u'^{i}=\dfrac{\text{d}u^{i}}{\text{d}s}$)
I know that sum of cosinus directors implies :
$$u'^{1}u'^{1}+u'^{2}u'^{2}+u'^{3}u'^{3}=1$$ without $g_{ij}$ factors.
How to prove that $f=g_{ij} u'^{i}u'^{j}=1$ with $g_{ij}$ factors ?
Secondly, starting from the expression of function $f$ and Euler equations, I would like to get :
$$\dfrac{\text{d}}{\text{d}s}(g_{ij}u'^{j})-\dfrac{1}{2}\partial_{i}g_{jk}u'^{j}u'^{k}=g_{ij}\dfrac{\text{d}u'^{j}}{\text{d}s}+(\partial_{k}g_{ij}-\dfrac{1}{2}\partial_{i}g_{jk})u'^{j}u'^{k}=0$$
but I can't obtain it.
Finally, I should get the general form of geodesics equation, i.e :
$$g_{ij}\dfrac{\text{d}u'^{j}}{\text{d}s}+\Gamma_{ijk}u'^{j}u'^{k}$$
if someone coul explain the different steps to get this final result.
Thanks for your help
UPDATE 1 :
For the first issue, I think that function $f$ can be represented as the curvilinear abscissa $ds^2$ which is equal to :
$$ds^{2}=g_{ij}\text{d}u^{i}\text{d}u^{j}$$
So I got :
$$\text{d}s^{2}/\text{d}s^2=1=f=g_{ij}\dfrac{\text{d}u^{i}}{\text{d}s}\dfrac{\text{d}u^{j}}{\text{d}s}$$
Concerning the second issue, I can write :
$$\dfrac{\partial f}{\partial u^{i}}=\partial_{i}g_{jk}u'^{j}u'^{k}$$
But I can't make appear the factor $\dfrac{1}{2}$ on the right term of above equation.
How to fix this factor $\dfrac{1}{2}$ to get the general form of geodesic equations ??? i.e :
$$g_{ij}\dfrac{\text{d}u'^{j}}{\text{d}s}+\Gamma_{ijk}u'^{j}u'^{k}$$
UPDATE 2 : Concerning the second problem, the issue came from :
$$\dfrac{\partial f}{\partial u'^{i}}=\dfrac{\partial g_{jk}u'^{j}u'^{k}}{\partial u'^{i}}$$ and the following expression is wrong : $$\dfrac{\partial f}{\partial u'^{i}}=g_{ij}u'^{j}$$
because of inversion between $j$ and $k$ index, we count double. So we have :
$$\dfrac{\partial f}{\partial u'^{i}}=\dfrac{\partial g_{jk}u'^{j}u'^{k}}{\partial u'^{i}}$$
$$=g_{jk}\bigg(\dfrac{\partial u'^{j}}{\partial u'^{i}}u'^{k}+u'^{j}\dfrac{\partial u'^{k}}{\partial u'^{i}}\bigg)$$
$$=g_{jk}(\delta^{j}_{i}u'^{k}+\delta_{i}^{k}u'^{j})=2\,g_{ik}\,u'^{k}$$