Q2P - Mathematical Methods in Physical Sciences

Question

A fluid flow is called irrotational if ∇×V = 0 where V = velocity of fluid (Chapter 6, Section 11); then V = ∇Φ. Use Problem 10.15 of Chapter 6 to show that if the fluid is incompressible, the Φ satisfies Laplace’s equation. (Caution: In Chapter 6, we used V = vρ, with v = velocity; here V = velocity.)

Step-by-Step Solution

Answer

The function  Φ  satisfy the equation.

Step 1: The velocity of fluid and the equation of continuity.

The velocity of fluid:

 $v=i\frac{dx}{dt}+j\frac{dy}{dt}+k\frac{dz}{dt}$

The equation of the continuity:

  $\nabla ·v+\frac{\partial \rho }{\partial t}=0$

If the fluid is incompressible then, _{$\frac{\partial \rho }{\partial t}=0$} .

Step 2: Determine that if the fluid is incompressible then the function satisfies the Laplace equation.

The fluid is called incompressible,  $\nabla \times v=0$.

In the case of fluid flow, Curl  v at point is a measure of the angular velocity of the fluid in the neighbourhood of the point. When $\nabla \times v=0$  everywhere in the same region, the velocity field  v is called an rotational in that region.

In the case of mathematical conditions, the force is aid to be conservative.

Making  $v=\nabla \varphi$ which is the force of fluid,

Suppose  $\rho$ is the density of a fluid varies from point to point as well as with time such that $\rho =\rho (x,y,z,t)$, along the stream of fluid and x,y,z  are the function of  t and the velocity of fluids is as follows:

 $v=i\frac{\partial x}{\partial t}+j\frac{\partial y}{\partial t}+k\frac{\partial z}{\partial t}$

By the equation of the continuity:

 $\nabla ·v+\frac{\partial \rho }{\partial t}=0$

Fluid flow is called irrational if  $\nabla \times v=0$ where  v is the velocity of the fluid.

Suppose the fluid is incompressible, then $\frac{\partial \rho }{\partial t}=0$.

So, the fluid is rotational, and from equation of continuity as follows:

 $$\begin{gathered} \nabla ·v+0=0 \\ \nabla ·v=0 \end{gathered}$$

From the fluid force:

$v=\nabla \varphi$ 

Putting this equation in above:

$\nabla ·\left(\nabla \varphi \right)=0$ 

So, the equation will be, _{${\nabla }^2\varphi =0$} .

Hence, function $\varphi$  satisfy the equation.