# 10.2.3.1: Existences of Stream Functions

- Page ID
- 1289

The potential function in order to exist has to have demised vorticity. For two dimensional flow the vorticity, mathematically, is demised when

\[

\label{if:eq:zeroVortisity}

\dfrac{\partial U_x}{\partial y} -

\dfrac{\partial U_x}{\partial x} = 0 \tag{80}

\]

The stream function can satisfy this condition when

Stream Function Requirements

\[

\label{if:eq:streamRequirement}

\dfrac{\partial}{\partial y} \left( \dfrac{\partial \psi}{\partial y} \right) +

\dfrac{\partial}{\partial x} \left( \dfrac{\partial \psi}{\partial x} \right) = 0

\Longrightarrow

\dfrac{\partial^2\psi}{\partial y^2} +

\dfrac{\partial^2\psi}{\partial x^2} = 0 \tag{81}

\]

Example 10.4

Is there a potential based on the following stream function

\[

\label{canItBePotential:streamFun}

\psi = 3\,x^5 - 2\,y \tag{82}

\]

Solution 10.4

Equation (81) dictates what are the requirements on the stream function. According to this equation the following must be zero

\[

\label{canItBePotential:check}

\dfrac{\partial^2\psi}{\partial y^2} +

\dfrac{\partial^2\psi}{\partial x^2} \overset{?}{=} 0 \tag{83}

\]

In this case it is

\[

\label{canItBePotential:theCheck}

0 \overset{?}{=} 0 + 60\,x^3 \tag{84}

\]

Since \(x^3\) is only zero at \(x=0\) the requirement is fulfilled and therefore this function cannot be appropriate stream function.

### Contributors

Dr. Genick Bar-Meir. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or later or Potto license.