# Euler's equation for the motion of incompressible fluid of velocity

Euler's equation for the motion of incompressible fluid of velocity $v(x,y,z,t)$, pressure $P(x,y,z,t)$ and density $\rho(x,y,z,t)$ is $$\frac{\partial v}{\partial t}+(v.\nabla)v=-\frac{1}{\rho}\nabla P+g$$ where $g$ is the acceleration due to gravity,

I want to use the identity $(\nabla \times A)\times A=(A.\nabla)A-\frac{1}{2}\nabla(A.A)$

to show that for a steady flow, Euler's equation reduces to $$(\nabla \times v)\times v=-\frac{1}{\rho}\nabla P-\frac{1}{2}\nabla v^{2}+g$$

so in a steady flow $$\frac{\partial v}{\partial t}=0,(v.\nabla)v=0$$, thus letting $v=A$ we get that $$0=-\frac{1}{\rho}\nabla P+g$$ and $$(\nabla \times v)\times v+\frac{1}{2}\nabla(v^{2})=-\frac{1}{\rho}\nabla P+g$$ rearraging this, we get that $$(\nabla \times v)\times v=-\frac{1}{\rho}\nabla P-\frac{1}{2}\nabla v^{2}+g$$ as required. Is this correct?

$$\frac{\partial v}{\partial t}=0$$
• Ahh okay thanks, what will happen to the $(v.\nabla)v$ term? Commented May 23, 2018 at 18:54
• You just have $$\dfrac{\overrightarrow{\rm grad}(v^2)}{2} - (\nabla\times v)\times v = (v . \nabla)v$$ Commented May 23, 2018 at 19:15