Q:

Bernoulli equation constant for water waves

A fluid fills $\{(x,y,z) \in\mathbb{R}^3:y \leq0\}$. The flow $\textbf{u}=(u,v,w)$ is given by $u=k\sin(\omega t)\cos(kx)e^{-|k|y}$, $v=-|k|\sin(\omega t)\sin(kx)e^{-|k|y}$, $w=0$, for $y\leq0$. We assume no gravity/body forces, and the flow is irrotational, unsteady, and incompressible, so we may use Bernoulli's equation in the form:
$$\nabla\left(\frac{1}{2}\rho|\textbf{u}|^2 + p +\rho\frac{\partial \phi}{\partial t}\right)=0$$ 
where $\phi$ is the vector potential, $\rho$ is the constant density.  
This tells us that $\frac{1}{2}\rho|\textbf{u}|^2 + p +\rho\frac{\partial \phi}{\partial t}$ does not vary in space, i.e. is constant.  In the example in the textbook, this is simply rearranged to give 
$$p = -\frac{1}{2}\rho|\textbf{u}|^2-\rho\frac{\partial \phi}{\partial t}$$
I do not understand why we can assume the constant is 0, and rearrange for this, instead of having that 
$$p = C-\frac{1}{2}|\textbf{u}|^2-\rho\frac{\partial \phi}{\partial t}$$
where $C$ is a constant.  Why do we have the freedom to just say that the constant is $0$?  

A:

Since the derivative of a constant is zero, once the gradient operator is removed there will be a constant on the rhs of the equation.  That constant when subtracted from the other constant which appears from the integration of the lhs are assumed to cancel.   
I am not an expert but it intuitively made sense to me that way.