Curl and Cauchy-Riemann Conditions problem

[Saketh]

Problem

The velocity of a two-dimensional flow of liquid is given by

$$\textbf{V} = \textbf{i}u(x, y) - \textbf{j}v(x, y).$$​

If the liquid is incompressible and the flow is irrotational show that

$$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$$​

and

$$\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}$$​

My Work

I evaluated $\nabla \times \textbf{V} = 0$ through a determinant, and ended up with this expression:

$$\textbf{i}\frac{\partial v}{\partial z} + \textbf{i}\frac{\partial u}{\partial z} - \textbf{k}\left ( \frac{\partial u}{\partial y} + \frac{\partial v}{\partial z} \right ) = 0$$​

Through this, I was able to verify:

$$\frac{\partial u}{\partial y} = -\frac{\partial{v}}{\partial x}$$​

I could not verify the other expression. How can I verify the other expression - I've tried everything I can think of. It seems simple, but I am missing something.

Thanks in advance.

 

[nazzard]

Hello Saketh,

to get

$$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$$

you need to use the fact that div V = 0 for incrompressible liquids.

This follows from the continuity equation:

$$\frac{\partial \rho}{\partial t} + \nabla(\rho\,\textbf{V})= 0$$

and using the fact that the liquid is incompressible, hence the mass density $\rho=const.$

Regards,

nazzard

P.S.: There's a small typo in your current solution: one \frac command is missing

 

[Saketh]

Oh, I thought that incompressible just meant that the fluid flow is behaving ideally, and had no mathematical significance.

Now that you tell me $$\nabla \cdot \textbf{V} = 0$$, the answer is obvious. Thanks!