# Relationship between weak and covariance stationary

I have read that the definition of weak stationary is :

$$Mean(t) = mean(t + \tau)\\ Cov(t_1,t_2) = cov(t_1-t_2,0)\\ E[|x(t)|^2] < \infty$$

In this definition of weakly stationary, can the variance change as a function of time as long as the covariance is a constant function of $$\Delta t$$?

I have also read the definition of covariance stationary is: $$\exists \mu : E(x_n)=\mu \,\forall \, n>0\\ \forall \, j \geq 0 \,\exists \,\gamma_j : cov(x_n,x_{n-j})=\gamma_j \, \forall \, n > j$$

Are these definitions identical?

I have also read that variance must be a constant function of time if data is stationary or weakly stationary, but these definitions mention covariance and not variance. Does variance have to be constant for data to be stationary, or not? Is covariance always stationary if variance is constant?

The two definitions are nearly the same, except that the second concerns the process after $$t=0$$. And, in the first one, the notation is abused. It should be \begin{align} & E[x(t)]=E[x(t+\tau)]\\ &\operatorname{cov}(x(t_1),x(t_2))=\operatorname{cov}(x(t_1-t_2),x(0))\end{align}
The variance is a special case of the covariance where $$t_1=t_2$$, i.e. $$\operatorname{cov}(x(t_1),x(t_1))=\operatorname{var}(x(t_1))=\operatorname{var}(x(0))$$ which means variance is also constant. When variance is constant, it doesn't mean that the process is covariance stationary; but the other way is true as you can see.