Deducing a $\frac{Dv}{Dt}$=$\frac{∂v}{∂t}$+$\frac{1}{2}$∇ (|v|$^2$) for an irrotational flow.

Question

A flow has been defined in my textbook as irrotational if skew(∇v(x, t)) ≡ 0, i.e. if the velocity gradient is a symmetric tensor.

An exercise left at the end of the chapter tasks one to deduce $\frac{Dv}{Dt}$ = $\frac{∂v}{∂t}$ + $\frac{1}{2}$ $∇(|v|^2$) given a irrotational flow.

I have attached what I have done so far.

I'm confused as to what what I'm doing wrong, I'm pretty sure what I've written at the bottom is correct but it doesn't really tell me much. Is there any way I can use the property of an irrotational flow better to get the desired result? I feel like I ought to use what I know about acceleration to write v. ($∇$)v in terms of skew(∇)v as then we could to simplify the problem further by using the property of an irrotational flow, I'm just unsure how to do so.

Answer 1

The solution given by ryaron is clever but overly complicated. The result can be very easily obtained using the product rule and the symmetry of the deformation tensor.

Let $\mathbf D=\nabla\boldsymbol u$ be the velocity gradient or deformation tensor. We say that $\boldsymbol u$ is irrotational if $\nabla\times \boldsymbol u=0$, which in turn implies that $\mathbf D$ is symmetric.

Now, note that the components of the convective acceleration can be given by $$\big((u\cdot \nabla)u\big)^i=u^j\nabla_j u^i=u_j\nabla^ju^i=u_j(\nabla u)^{ij}=u_jD^{ij}$$

Recall that $(\nabla v)_{ij}=\nabla_jv_i$, NOT $\nabla_iv_j$ !!

On the other hand, $$\left(\frac{1}{2}\nabla(| u|^2)\right)^i=\frac{1}{2}\nabla^i(u^ju_j) \\ =\frac{1}{2}u^j\nabla^iu_j+\frac{1}{2}u_j\nabla^iu^j\\ =u_j\nabla^iu^j \\ =u_jD^{ji}$$

However, when $\mathbf D$ is symmetric, $D^{ji}=D^{ij}$, and hence $(\boldsymbol u\cdot \nabla)\boldsymbol u=\frac{1}{2}\nabla (|\boldsymbol u|^2)$.

Answer 2

An irrotational flow implies:

$\nabla\times v=0$

Take the tensor identity:

$v\times\nabla\times v\implies \\ \epsilon_{ijk}v_j\epsilon_{klm}\frac{\partial v_l}{\partial x_m}=$

$(\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl})v_j\frac{\partial v_l}{\partial x_m}=0$

Since, $\nabla\times v=0$.

Eventually we receive:

$v_j\frac{\partial v_i}{\partial x_j}-v_j\frac{\partial v_j}{\partial x_i}=0$

From which we deduce:

$(v\cdot\nabla)v-\frac{1}{2}\nabla(|v|^2)=0$