I'm looking more for verification of my answer as I'm not 100% sure how valid my integration is here:
If I have a 2-D incompressible fluid velocity field given by $\vec{U}(x,y,t)=u(x,t)\vec{x}+v(y,t)\vec{y}$, I know that $\nabla \cdot \vec{U} = 0$, call this eqn (1), by the definition of incompressibility.
In order to find out more about the velocity components $u(x,t)$ and $v(y,t)$, i have been told to integrate eqn (1). Is the following integration w.r.t. space valid?:
\begin{align}
&\int \nabla \cdot \vec{U} \mathbb{d} r = \int 0 \\
&\int \left(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right) \mathbb{d}r = C = \text{const.} \\
&\int \frac{\partial u}{\partial x}dx+\int \frac{\partial v}{\partial y}\mathbb{d}y = C \\
&u(x,t) + v(y,t) = C(t)
\end{align}
In other words, the magnitude of the sum of the fluid velocity in the $x$ and $y$ directions are constant everywhere in the fluid at any instant in time?
Is there a better way of obtaining more information?
Best Answer
The usual integral for the divergence of the velocity field is over a volume. Since $u$ does not depend on $y$ and $v$ does not depend on $x$, we have
$$ \begin{align} \int_V \left(\nabla\cdot \vec{U}\right) \mathrm{d}V & = \iint \left(\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y}\right) \mathrm{d} x \mathrm{d} y \\ & = \iint \frac{\partial}{\partial x}u(x,t) \mathrm{d}x \mathrm{d}y + \iint \frac{\partial}{\partial y}v(y,t) \mathrm{d}x \mathrm{d}y \\ & = \int \left[u(x,t) + c_x\right] \mathrm{d}y + \int \left[v(y,t) + c_y\right] \mathrm{d}x \\ & = y\left[u(x,t) + c_x\right] + x\left[v(y,t) + c_y\right] + d\\ & = 0 \end{align} $$
where the constants $c_x$, $c_y$, and $d$ will depend on your boundary conditions. You should also get the same thing if you apply the boundary conditions by taking definite integrals. Without know what the rest of the problem is, it is difficult to say more about this.