Condsider an incompressible, inviscid, irrotational fluid with constant density $\rho$. Let $\overrightarrow u$ be its velocity field, $p$ its pressure field and $\overrightarrow F$ be an external body force given by some potential $\chi$ so that $ -\nabla \chi = \overrightarrow F$. The momentum equation reads $$ \frac{ \partial \overrightarrow u}{\partial t}= -\nabla(\frac{p}{\rho}+\frac{1}{2}|\overrightarrow u|^2+\chi).$$ I was asked to show that $$\frac{p}{\rho}+\frac{1}{2}|\overrightarrow u|^2+\chi$$ is constant in steady flow, and the implied line of reasoning is to say that it simply follows from $$ 0= -\nabla(\frac{p}{\rho}+\frac{1}{2}|\overrightarrow u|^2+\chi).$$ But I can only see why we may conclude that $$\frac{p}{\rho}+\frac{1}{2}|\overrightarrow u|^2+\chi$$ is a function of time. And we could have this be any function of time, by absorbing it into $\chi$. Am I missing something?

  • $\begingroup$ In steady flow you have $\partial/\partial t=0$ so the right hand side of your equations cannot change as a function of time. $\endgroup$
    – Newbie
    Jan 21, 2022 at 17:02
  • $\begingroup$ @Newbie Here by steady I mean $\partial \overrightarrow u / \partial t = 0$, what does you $\partial / \partial t$ act on? I see that all the gradients in my equations are time independent, but that wouldn't rule out some time dependence in $p/ \rho+u^2/2+\chi$? $\endgroup$ Jan 21, 2022 at 17:12
  • $\begingroup$ Are you worried that your right hand side cannot change as a function of space but can change as a function of time? $\endgroup$
    – Newbie
    Jan 21, 2022 at 17:15
  • $\begingroup$ @Newbie Yes exactly. $\endgroup$ Jan 21, 2022 at 17:16
  • $\begingroup$ Hint: The term involving $|\vec u|^{2}$ cannot change as a function of time since you already have set $\partial\vec u/\partial t=0$ on the left hand side. $\endgroup$
    – Newbie
    Jan 21, 2022 at 17:18

1 Answer 1


Your observation is correct. For steady irrotational flow of an incompressible, inviscid fluid with constant density, the Euler equation of momentum reduces to

$$\nabla \left(\frac{p}{\rho} + \frac{1}{2} |\mathbf{u}|^2 + \chi \right) = 0$$

Integrating the components with respect to the spatial variables, we get the general solution

$$\frac{p}{\rho} + \frac{1}{2} |\mathbf{u}|^2 + \chi = c(t),$$

where the arbitrary function $t \mapsto c(t)$ changes nothing about the flow field and can be absorbed into the pressure. Recall that the pressure field corresponding to a particular velocity field is never uniquely determined since an arbitrary reference pressure can be added. The pressure and body-force potential only affect the velocity through their spatial gradients and adding an arbitrary function of time has no impact.

More generally for irrotational flow where $\nabla \times \mathbf{u} =0$, the velocity field is the gradient of a potential, $\mathbf{u} = \nabla\phi$, and the momentum equation reduces to

$$\frac{\partial\nabla \phi}{\partial t}= -\nabla \left(\frac{p}{\rho} + \frac{1}{2} |\nabla \phi|^2 + \chi \right),$$

which upon integration yields

$$\frac{\partial \phi}{\partial t}+\frac{p}{\rho} + \frac{1}{2} |\nabla \phi|^2 + \chi = c(t)$$

Again, in steady flow where $\frac{\partial \phi}{\partial t} = 0$, the appearance of the time-dependent $c(t)$ is not excluded.

  • $\begingroup$ Thank you very much for your reply! I think this settles it for me, and I also happened to have a chat to the relevant professor who gave me a similar reply. $\endgroup$ Jan 27, 2022 at 23:19
  • $\begingroup$ @LobLikelyhood: You're welcome. $\endgroup$
    – RRL
    Jan 28, 2022 at 3:38

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