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In hydrodynamics the for a non-viscous flow the velocity (and density) fields are given by the continuity and Euler equations: $$\rho\frac{\partial \vec{v}}{\partial t}+\rho(\vec{v}\cdot\vec{\nabla})\vec{v}=-\nabla p+\rho\vec{F}_\mathrm{body},$$ $$\frac{\partial\rho}{\partial t}+\vec{\nabla}(\rho\vec{v})=0.$$

Now when we then apply this to look at simple 2D flows, we define the streamfunction $\psi(x,y)$ which is found by solving: $$\frac{\partial\psi}{\partial x}=v_y\quad\mathrm{and}\quad\frac{\partial\psi}{\partial y}=-v_x.$$ We can determine this for many kinds of flows (uniform, source, sink, vortex, ...), but when we want to look at a non-lifting flow over a cylinder we add two different kinds of flows (Source + Sink + Uniform).

Can we do this, just add the different velocity fields? I'm not sure if this is correct since the Euler equation is non-linear in the velocities ...

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  • $\begingroup$ @ACuriousMind In principle the tag "superfluidity" should also be present because I look at frictionless flow. $\endgroup$ – Nick Dec 2 '14 at 12:14
  • $\begingroup$ Ah, I should have been more careful, feel free to reintroduce it (and review the pending edit on your post, while you're at it). $\endgroup$ – ACuriousMind Dec 2 '14 at 12:16
  • $\begingroup$ @ACuriousMind No problem at all! I get where the confusion comes from ;). $\endgroup$ – Nick Dec 2 '14 at 12:20
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As you rightly suspected, in general, if $\vec{v}_1$ and $\vec{v}_2$ are two solution of the Euler equation then $\vec{v}_1 + \vec{v}_2$ is not a solution because of the nonlinear term.

However, in many cases, one (or both) of the velocities will make the nonlinear term zero allowing you to add them. For example, in the case of a steady flow in an infinitely long pipe, $\vec{v}$ is independent of $t$, $\phi$ and $z$. (I am using cylindrical coordinates with z axis coincidental with the pipe's axis.) In this case, the left hand side of the equation is zero and one can add the velocities.

Another situation that allows us to add velocities is when they are infinitesimally small and hence terms quadratic in them can be ignored. Typically, one comes across this situation while studying stability of flows against perturbations.

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  • $\begingroup$ Nice one :-). Another general case (which was the one that I was looking for) is for non-viscous, irrotational and incompressible flows. Since the velocity is irrotational it can be written as $\vec{v}=\vec{\nabla}\phi$. Since it's incompressible it has the demand that $\vec{\nabla}\cdot\vec{v}=0$ so that $\nabla^2\phi=0$ which yields a linear equation. $\endgroup$ – Nick Dec 2 '14 at 12:23

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