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So let's say we have a circular pipe with a variable cross section with an inviscid incompressible fluid flowing through it. The areas of the two ends are $A_1$ and and $A_2$. Let's say we also control the pressure at the two ends $p_1$ and $p_2$ with $p1>p2$. The unsteady Bernoulli's equation written down for the two ends states

$(p_2 - p_1) + \frac{\rho}{2} (u_2^2 - u_1^2) + \rho \frac{d}{dt} (\phi_2 - \phi_1) =0$

Use conservation of mass we can write $u_1 = \frac{A_2}{A_1} u_2$. We can put this into the above equation to get

$(p_2 - p_1) + \frac{\rho}{2} u_2^2 (1 - (\frac{A_2}{A_1})^2) + \rho \frac{d}{dt} (\phi_2 - \phi_1) =0$

To find the fixed points we set the time derivatives equal to 0 and solve for $u_2^*$.

$ u_2^* = \pm \sqrt{ \frac{2}{\rho} \frac{(p_1 - p_2)}{(1 - (\frac{A_2}{A_1})^2)}}$

Now of $A_1>A_2$ everything is fine and we get a stable fixed point. However if $A_2>A_1$ the expression under the square root is negative and the fixed point is destroyed. The flow diverges to infinity. This is obviously nonphysical. So what went wrong?

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If $A_2>A_1$, due to continuity of flow, fluid will decelerate and thus pressures will obey $p_2>p_1$. It is not possible to set up stationary flow for every pair of pressure values.

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