I am trying to fill the details from this paper (it's free available) and I get stuck proving the equation (3.16) from the mentioned article. Let me give some context.
Part 1. Setting the notation. Let $M=I \times_f \mathbb{E}_1^2$ the warped product manifold of $I\subseteq \mathbb{R}$ (it is just an open interval) and the Minkowski space of dimension $2,$ $\mathbb{E}_1^2.$ The real-valued function $f$ is always positive. The metric is given by $\tilde{g}=dt^2+f(t)^2(dx^2-dy^2)$ and $\tilde\nabla$ is the Levi-Civita connection of $M.$ Denote by $D$ the Levi-Civita connection of $\mathbb{E}_1^2.$
Part 2. Principal Equations. In order to justify the computations given in the Part 4, I used the following equalities, which a succesfully prove. If $U,V,W$ are lifts of vector fields tangent to $\mathbb{E}_1^2,$ we obtain
$$ \tilde\nabla_U V = D_UV -\tfrac{f'(t)}{f(t)}\tilde{g}(U,V)\partial_t \qquad (2.1a)$$ $$\tilde\nabla_{\partial_t} U=\tilde\nabla_{U} \partial_t=\tfrac{f'(t)}{f(t)}U \qquad (2.1b)$$ $$ \tilde\nabla_{\partial_t} \partial_t=0. \qquad (2.1c)$$
Part 3. Parametrization of a isometric immersed manifold. Now, let $\varphi: M_q \to M$ an isometric immersion, of a surface $M_q$ of index $q=0,1.$ Let $\xi$ be a local unit vector field normal to $M_q,$ and write $\partial_t=T+\Theta \xi,$ where $\Theta=\varepsilon_\xi\tilde{g}(\partial_t,\xi).$ It can be easily shown that $T$ is the tangential projection of $\partial_t$ on the tangent space of $M_q$ at each point. The article only focus on the case where $T$ is a nonnull vector field, so that $\Vert T \Vert \neq 0.$ Now, one can show that there is a chart $(u,v)$ on $M_q$ such that $\tfrac{T}{\Vert T\Vert}=\partial_u$ and set $\varepsilon_1=\tilde{g}(\partial_u, \partial_u).$ On this chart, write $$\varphi(u,v)=(t(u,v),x(u,v),y(u,v)).$$ It is shown that actually $t(u,v)=\varepsilon_1\Vert T \Vert u,$ so that $t_u=\varepsilon_1\Vert T\Vert$ and $t_v=0.$
Then we have $$\partial_u=t_u\partial_t+x_u\partial_x+y_u\partial_y=\varepsilon_1\Vert T \Vert \partial_t+x_u\partial_x+y_u\partial_y.$$
We set $\varphi_u=\varepsilon_1\Vert T \Vert \partial_t+x_u\partial_x+y_u\partial_y=\partial_u,$ so that $\varphi_u-\varepsilon_1\Vert T \Vert \partial_t=x_u\partial_x +y_u\partial_y$ is a tangent vector to $\mathbb{E}_1^2.$ Hence, by the definition of $\tilde{g}$ we have $$\tilde{g}(\varphi_u-\varepsilon_1\Vert T \Vert \partial_t, \varphi_u-\varepsilon_1\Vert T \Vert \partial_t)=\varepsilon_1-\Vert T \Vert^2.$$
Using this observation and (2.1) we have the following computation.
Part 4. Computation where I am stuck.
We assume here that $\Theta$, and hence $\Vert T \Vert,$ is constant. Hence
$$ \tilde\nabla_{\varphi_u} \varphi_u = \tilde\nabla_{\varphi_u-\varepsilon_1\Vert T \Vert \partial_t} (\varphi_u- \varepsilon_1\Vert T \Vert \partial_t)+\tilde\nabla_{\varepsilon_1\Vert T \Vert \partial_t} (\varphi_u- \varepsilon_1\Vert T \Vert \partial_t)+\tilde\nabla_{\varphi_u-\varepsilon_1\Vert T \Vert \partial_t} \varepsilon_1 \Vert T \Vert\partial_t+\tilde\nabla_{\varepsilon_1\Vert T \Vert \partial_t} \varepsilon_1 \Vert T \Vert \partial_t$$
$$ = D_{\varphi_u-\varepsilon_1 \Vert T \Vert \partial_t} (\varphi_u-\varepsilon_1 \Vert T \Vert \partial_t)-\tfrac{f'(t)}{f(t)}\tilde{g}(\varphi_u-\varepsilon_1 \Vert T \Vert \partial_t,\varphi_u-\varepsilon_1 \Vert T \Vert \partial_t)\partial_t+\varepsilon_1\Vert T \Vert\tilde\nabla_{\partial_t} (\varphi_u-\varepsilon_1\Vert T \Vert \partial_t)+\varepsilon_1 \Vert T \Vert \tilde\nabla_{\varphi_u-\varepsilon_1\Vert T \Vert \partial_t}\partial_t+\varepsilon_1^2\Vert T\Vert^2\tilde\nabla_{\partial_t}\partial_t $$ $$ = D_{\varphi_u-\varepsilon_1 \Vert T \Vert \partial_t} (\varphi_u-\varepsilon_1 \Vert T \Vert \partial_t)-\tfrac{f'(t)}{f(t)}\left( \varepsilon_1-\Vert T \Vert^2\right)\partial_t+2\varepsilon_1\Vert T \Vert\tilde\nabla_{\varphi_u-\varepsilon_1\Vert T \Vert \partial_t} \partial_t $$
$$= D_{\varphi_u-\varepsilon_1 \Vert T \Vert \partial_t} (\varphi_u-\varepsilon_1 \Vert T \Vert \partial_t)-\tfrac{f'(t)}{f(t)}\left( \varepsilon_1-\Vert T \Vert^2\right)\partial_t+2\varepsilon_1\Vert T \Vert\tfrac{f'(t)}{f(t)}(\varphi_u-\varepsilon_1 \Vert T \Vert \partial_t)$$
$$=D_{\varphi_u-\varepsilon_1 \Vert T \Vert \partial_t} (\varphi_u-\varepsilon_1 \Vert T \Vert \partial_t)+2\varepsilon_1\Vert T \Vert\tfrac{f'(t)}{f(t)}\varphi_u+\tfrac{f'(t)}{f(t)}\left( -\varepsilon_1+\Vert T \Vert^2-2\varepsilon_1^2\Vert T \Vert^2 \right)\partial_t$$ $$=D_{\varphi_u-\varepsilon_1 \Vert T \Vert \partial_t} (\varphi_u-\varepsilon_1 \Vert T \Vert \partial_t)+2\varepsilon_1\Vert T \Vert\tfrac{f'(t)}{f(t)}\varphi_u-\tfrac{f'(t)}{f(t)}\left( \varepsilon_1+\Vert T \Vert^2 \right)\partial_t$$ $$=D_{\varphi_u-\varepsilon_1 \Vert T \Vert \partial_t} (\varphi_u-\varepsilon_1 \Vert T \Vert \partial_t)+2\varepsilon_1\Vert T \Vert\tfrac{f'(t)}{f(t)}\varphi_u-\varepsilon_1\Vert T \Vert\tfrac{f'(t)}{f(t)}\left( \tfrac{1}{\Vert T \Vert }+\varepsilon_1\Vert T \Vert \right)\partial_t$$ $$=D_{\varphi_u-\varepsilon_1 \Vert T \Vert \partial_t} (\varphi_u-\varepsilon_1 \Vert T \Vert \partial_t)+2\sigma'(u)\varphi_u-\sigma'(u)\left( \tfrac{1}{\Vert T \Vert }+\varepsilon_1\Vert T \Vert \right)\partial_t.$$
I hope that the Parts 1,2,3 are enough for you in order to understand all the steps from my computation above. If not, please let me know.
Part 5. My question In the paper, the equation (3.16) is stated as follows $$\tilde\nabla_{\varphi_u} \varphi_u=\varphi_{uu}+2\sigma'(u)\varphi_u-\sigma'(u)\left( \tfrac{1}{\Vert T \Vert }+\varepsilon_1\Vert T \Vert \right)\partial_t,$$ where I deduce from my computations that $D_{\varphi_u-\varepsilon_1 \Vert T \Vert \partial_t} (\varphi_u-\varepsilon_1 \Vert T \Vert \partial_t)$ must be equal to $\varphi_{uu}$ (I understand that the author means $\varphi_{uu}=t_{uu}\partial_t+x_{uu}\partial_{x}+y_{uu}\partial_{y}=x_{uu}\partial_{x}+y_{uu}\partial_{y}$.) Now, I tried all my ideas in order to show that $D_{\varphi_u-\varepsilon_1 \Vert T \Vert \partial_t} (\varphi_u-\varepsilon_1 \Vert T \Vert \partial_t)=\varphi_{uu},$ but I couldn't get to anything. I tried to use the fact that $\varphi_u-\varepsilon_1 \Vert T \Vert \partial_t=x_u\partial_x+y_u\partial_y$ and the connection $D$ in $\mathbb E_1^2$ is flat, so that $D_{U}V= U(V^1)\partial_x+U(V^2)\partial_y$, being $(V^1,V^2)$ the components of $V$, but it gets me anywhere.
Can you please explain me why such equality holds? Thanks in advance, I would really appreciate it.