I have the equation \begin{align} \begin{cases} u_t(\boldsymbol{x},t)-\nabla^2u(\boldsymbol{x},t) = 0, &\boldsymbol{x}\in D, & t>0, \\ u(\boldsymbol{x},t)=0, & \boldsymbol{x}\in\partial D, & t>0, \\ u(\boldsymbol{x},0)=u_0(\boldsymbol{x}), &\boldsymbol{x}\in D, \end{cases} \end{align} where $\boldsymbol{x}=(x_1,x_2,x_3)$.
Does the Laplace operator acts on the vector $\boldsymbol{x}$ AND $t$? I.e. \begin{align} \nabla^2 u(\boldsymbol{x},t)=\frac{\partial^2 u}{\partial x_1^2}+\frac{\partial^2 u}{\partial x_2^2}+ \frac{\partial^2 u}{\partial x_3^2}+\frac{\partial^2 u}{\partial t^2} \end{align} or just \begin{align} \nabla^2 u(\boldsymbol{x},t)=\frac{\partial^2 u}{\partial x_1^2}+\frac{\partial^2 u}{\partial x_2^2}+ \frac{\partial^2 u}{\partial x_3^2} \end{align} If the second one is correct, why is it so?
In this context (heat equation) the Laplacian is taken in the space variables only -- because this is what the heat equation is.
When considering a PDE where the role of Laplacian is not clear, the author will either state the interpretation of Laplacian in text, or use indices to clarify it: $$\nabla_x^2 u(x,t) \quad \text{vs.}\quad \nabla_{x,t}^2 u(x,t)$$ or $$\Delta_x u(x,t) \quad \text{vs.}\quad \Delta_{x,t} u(x,t)$$