# Fluid dynamics: Flow velocity of an incompressible fluid subject to a potential gradient

aclaret
Homework Statement:
Let ##u: R^3 \times R \rightarrow R^3## be flow velocity of incompressible fluid. Let fluid be subject to potential force ##-\nabla \chi##. To prove
$$\frac{d}{dt} \int_{V} \frac{1}{2} \rho \langle u, u \rangle dV + \int_{\partial V} H \langle u, n \rangle dA = 0$$where ##H := \frac{1}{2}\rho \langle u, u \rangle + p + \chi##, and notation ##\langle x, y \rangle## denote standard inner product on ##R^3##.
Relevant Equations:
fluid dynamic, euler's equation of the motion
$$\frac{Du}{Dt} = -\frac{\nabla p}{\rho} - \nabla \chi$$I re-write the Euler equation for incompressible fluid using suffix notation
$$\frac{\partial u_i}{\partial t} + u_j \frac{\partial u_i}{\partial x_j} + \frac{\partial}{\partial x_i} \left(\frac{p}{\rho} + \chi \right) = 0$$what theorems applies to the problem?

Homework Helper
2022 Award
You have integrals involving a volume and a surface. That would suggest the divergence theorem.

aclaret
divergence theorem I can write$$\int_{\partial V} \langle u, n \rangle dA = \int_V \frac{\partial u_j}{\partial x_j} dV$$try multiply equation by ##u_i## and then do implicit summation also over ##i##

$$u_i \frac{\partial u_i}{\partial t} + u_i u_j \frac{\partial u_i}{\partial x_j} + u_i \frac{\partial}{\partial x_i} \left(\frac{p}{\rho} + \chi \right) = 0$$get for second term$$u_i u_j \frac{\partial u_i}{\partial x_j} = \frac{1}{2}u_j \frac{\partial}{\partial x_j} \langle u, u \rangle$$first term also

$$u_i \frac{\partial u_i}{\partial t} = \frac{1}{2} \frac{\partial}{\partial t} \langle u,u \rangle$$do I effect the volume integral to both sides?

$$\int_V \frac{1}{2} \frac{\partial}{\partial t} \langle u,u \rangle dV + \int_V \frac{1}{2}u_j \frac{\partial}{\partial x_j} \langle u, u \rangle dV + \int_V u_i \frac{\partial}{\partial x_i} \left(\frac{p}{\rho} + \chi \right) dV = 0$$can I make use of divergence theorem here?

aclaret
$$\frac{d}{dt} \int_V \frac{1}{2} \rho \langle u,u \rangle dV + \int_V u_i \frac{\partial}{\partial x_i} \left[ \frac{1}{2} \rho \langle u, u \rangle + \left(p + \rho \chi \right) \right] dV = K, \quad K \, \mathrm{= constant}$$I can try integrate second term by parts,$$\int_V u_i \frac{\partial}{\partial x_i} \left[ \frac{1}{2} \rho \langle u, u \rangle + \left(p + \rho \chi \right) \right] dV = \int_{\partial V} \left[ \frac{1}{2} \rho \langle u, u \rangle + \left(p + \rho \chi \right) \right] \langle u, n \rangle dA - \int_{V} \left[ \frac{1}{2} \rho \langle u, u \rangle + \left(p + \rho \chi \right) \right] \frac{\partial u_i}{\partial x_i} dV$$does second term vanish? even if, the first term still incorrect. I wonder if you can tell second hint @pasmith :), I'm in a little confusion ;)

Homework Helper
2022 Award
What is the divergence of an incompressible fluid?

aclaret
aclaret
let me see... clearly from equation of continuity it follow that if ##(\forall x,t) \, \rho(x,t) = \text{constant}##, then ##\nabla \cdot u = 0##. hence, second integral vanish, and get$$\frac{d}{dt} \int_V \frac{1}{2} \rho \langle u,u \rangle dV + \int_{\partial V} \left[ \frac{1}{2} \rho \langle u, u \rangle + \left(p + \rho \chi \right) \right] \langle u, n \rangle dA = K, \quad K \, \mathrm{= constant}$$but this would give me ##H := \frac{1}{2} \rho \langle u, u \rangle + p + \rho \chi##, which tiny bit different to problem statement. would you agree my work is correct, or did I do a mistake somewhere...

Homework Helper
2022 Award
The problem statement says that the fluid is subject ot a potential force of $-\nabla\chi$, not $-\rho\nabla\chi$. Other than that your work is correct.

The constant $K$ is zero; it were not there would be some point where $$\rho \frac{Du}{Dt} + \nabla p + \nabla \chi \neq 0.$$

aclaret
aclaret
parfait! yes I see, thank for point out my mistake with the factor of ##\rho##. I also was not certain how to "convince myself" that ##K## indeed vanish.

i think necessary for me do lot more of these problem to learn the tricks of the trade :), the simple realize that ##\nabla \cdot u = 0## was enough here to reveal how to finish the solution!