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.)

Short answer

The function Φ satisfy the equation.

Step-by-step solution

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:

role="math" localid="1665141513332" $\nabla ·v+\frac{\partial \rho }{\partial t}=0$

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

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.