How do we show this variety is normal? I just saw my question is closed. I think I need to provide some details.
Suppose $X$ is a variety over a field $k$ and $f: X\to S=\mathbb{A}_{k}^{1}=\operatorname{Spec}(k[t])$ is a flat morphism. Assume the restriction of the morphism $f$ over $\operatorname{Spec}k[t,1/t]$ is smooth and the fiber over $t=0$ is an integral scheme. I want to show that $X$ is normal.
After thinking for a long time, I think we may use Serre’s $R_{1}+S_{2}$ to prove it. For $R_{1}$, it is clear, since the fiber over $0$ is integral, the singular point in this fiber cloud just be an proper closed subset of the fiber. Since other fibers are smooth, all singular locus could just be a codimension 2 subset. So it is regular codimension.
Notice we do not use flat to prove $R_{1}$, so I think we need to use flat to prove $S_{2}$. But I do not how to do it.
 A: Note that you did actually use flatness for $R_1$: this gives you that $X\to \Bbb A^1$ is open, so the fiber over $0$ is of positive codimension. We will be using flatness for $S_2$, though.
For $S_2$, note first that this condition is satisfied on the preimage of $\Bbb A^1_k\setminus \{0\}$ in $X$: this is smooth. Therefore we just need to check this condition on the fiber $X_0$. If $X_0$ is empty, we're done, else let $x\in X_0$ be arbitrary. By flatness, the image of $t$ along the map $k[t]_{(t)}=\mathcal{O}_{\Bbb A^1_k,0}\to \mathcal{O}_{X,x}$ is not a zero-divisor. But $\mathcal{O}_{X,x}/t \cong \mathcal{O}_{X_0,x}$ - if this is a field, then the height of $\mathfrak{m}_x$ is one, and if not, we can pick a non-zero-divisor in the maximal ideal of $\mathcal{O}_{X_0,x}$ by the assumption $X_0$ is integral. So in the first case, we've shown that the depth of $\mathcal{O}_{X,x}$ is at least the height of $\mathfrak{m}_x$ by finding a regular sequence of length one, and in the second, we've shown it's at least 2 by finding a regular sequence of length two - thus we've shown that $X$ satisfies $S_2$ and is normal.
