Q: Find the geodesic on the circular cylinder of radius 1.
My try: Consider the cylindrical coordinates $(\theta,z)$. Fix two points on the cylinder $x_0=(\theta_0,z_0),x_1(\theta_1,z_1)$. Then the functional can be written as $$ J(\gamma)=\int_0^1\sqrt{\dot{\theta}(t)^2+\dot{z}(t)^2}dt $$ where $\gamma\in U=\{x\in C^1([0,1],\text{cylinder}); x(0)=(\theta_0,z_0),x(1)=(\theta_1,z_1)\}$.
By using the E-L equation, we can find that $$ \gamma^*(t):\begin{cases} \theta=(\theta_1-\theta_0)t+\theta_0\\ z=(z_1-z_0)t+z_0 \end{cases} $$ Now we need to show that the solution of E-L equation is the minimizer of functional. To prove this, we need to check the second derivative (variation) of functional $J$ at $\gamma^*$ is strongly derivative, that is $$ J''(\gamma^*)(h,h)\geq k\Vert h\Vert^2,\forall h\in U $$ where $k$ is postive constant.
Let $h=(h_1,h_2)\in U$, we have $$ J''(\gamma^*)(h,h)=\frac{1}{2}\frac{1}{((\theta_1-\theta_0 )^2+(z_1-z_0 )^2)^{3/2}}\int_0^1[(z_1-z_0)\dot{h}_1- (\theta_1-\theta_0 )\dot{h}_2]^2dt $$ I don't know how to continue