# Velocity and pressure difference in a piston pump

I have a piston pump with incompressible, inviscid fluid. Let me write out the mathematics of the flow field-

$$\text{Continuity Equation: } \frac{\partial u}{\partial x}=0 \\ \text{Euler's Equation:} \frac{\partial u}{\partial t} + u \frac{\partial u}{\partial x} = -\frac{1}{\rho} \frac{\partial p}{\partial x}$$

Simplifying them: $$\frac{du}{dt}=\frac{\Delta P}{\rho L}$$

I put in a harmonic displacement $$x_p=x_0 \sin{\omega t}$$, and correspondingly get $$\Delta P=\rho L x_0 \omega^2 \sin{\omega t}$$ and $$u=x_0 \omega \cos{\omega t}$$.

Note at time $$t=0$$, $$u=x_0 \omega$$ and $$\Delta P=0$$. How is this possible? Velocity is there inspite of pressure difference being 0. Can anyone explain my mistake in the mathematics?

You can imagine this situation using a simpler scenario. The displacement taken by you is simple harmonic. The spring block system for small displacement is also simple harmonic. To start the oscillation in a spring block system you need to provide either kinetic energy to the block at mean position(this is the case described by you, x=0 and v$$\neq$$0 at t=0) or potential energy to the spring by pulling it to some distance from the mean position(In this case v=0 and x$$\neq$$0 at t=0). In an ideal non-dissipative environment, this initial energy provided remains in the system and keeps oscillating between KE and PE. In this case too, the velocity at the mean position is non-zero while the force is equal to zero (F=-kx; x=0).