M4.5 Application on Analytic Functions

Let vector field \(V\)\(=v_x\hat{i}+v_y\hat{j}\) represents velocity vector field. If the fluid is irrotational then \(\nabla\times V=0\). Also the velocity of an irrotational flow can be represented by gradient of a scalar function \(\phi\). i.e., $$ V=\nabla \phi=\frac{\partial \phi}{\partial x}\hat{i}+\frac{\partial \phi}{\partial y}\hat{j}\,\,\,\,-(1).$$ \(\phi\) is called as velocity potential and the family of curves \(\phi(x,y)=c\) are called as equipotenttial lines

If the fluid is incompressible then \(div V=0\), using (1), we will get \(\nabla^2 \phi=0\), hence \(\phi\) is harmonic. In other words an incompressible, irrotational flow is governed by Laplace's equation.

If \(\phi\) is harmonic then there exists conjugate harmonic function \(\psi(x,y)\) called as stream function such that \(w(z)=\phi(x,y)+i\psi(x,y)\) is analytic. Where the family of curves \(\psi(x,y)=c'\) are called as stream lines
From orthogonal property we can say that stream lines and equipotential lines are orthogonal as shown in the following figure.

Equipotential Lines Vs Stream Lines

Airflow streamlines behind tall building(PC:https://sesamenet.eu)

Problems
01. Find the complex potential of a fluid flow if the stream function is given by \(\psi=\frac{-y}{x^2+y^2}\). Also find the velocity potential \(\phi\).
Ans:
Let complex potential of a fluid is given by \(f(z)=\phi(x,y)+i\psi(x,y)\Longrightarrow f'(z)=\phi_x+i\psi_x\)
Since \(f(z)\) is analytic, \(\phi,\psi\) satisfies CR-equations\(\Longrightarrow f'(z)=\psi_y+i\psi_x\)
Given \(\psi=\frac{-y}{x^2+y^2}\Longrightarrow \psi_x=\frac{2xy}{(x^2+y^2)^2}\,\,\,\, \&\,\,\, \psi_y=\frac{y^2-x^2}{(x^2+y^2)^2}\)
\(\Longrightarrow f'(z)=\frac{y^2-x^2}{(x^2+y^2)^2}+i\frac{2xy}{(x^2+y^2)^2}\)
Now put \(x=z,y=0 \Longrightarrow f'(z)=\frac{-z^2}{(z^2)^2}+i\cdot 0 \Longrightarrow f'(z)=\frac{-1}{z^2}\)\(\Longrightarrow f(z)=\frac{1}{z}\)
Now put \(z=x+i y\Longrightarrow f(z)=\frac{1}{x+i y}=\frac{x-iy}{x^2+y^2}=\frac{x}{x^2+y^2}-i\frac{y}{x^2+y^2}\)
Hence \(\phi = \frac{x}{x^2+y^2}\)
Mathematica Notebook Sourcecode which shows the above problem Graphically.
02. If \(\phi+i\psi\) represents the complex potential of an electric field if \(\psi=(x^2-y^2)+\frac{x}{x^2+y^2}\) find the complex potential as a function of \(z\) and also find \(\phi\).
Ans:
Let complex potential of a fluid is given by \(f(z)=\phi(x,y)+i\psi(x,y)\Longrightarrow f'(z)=\phi_x+i\psi_x\)
Since \(f(z)\) is analytic, \(\phi,\psi\) satisfies CR-equations\(\Longrightarrow f'(z)=\psi_y+i\psi_x\)
Given \(\psi=(x^2-y^2)+\frac{x}{x^2+y^2}\Longrightarrow \psi_x=2x+\frac{y^2-x^2}{(x^2+y^2)^2}\,\,\,\, \&\,\,\, \psi_y=-2y-\frac{2xy}{(x^2+y^2)^2}\)
\(\Longrightarrow f'(z)=-2y-\frac{2xy}{(x^2+y^2)^2}+i\left(2x+\frac{y^2-x^2}{(x^2+y^2)^2}\right)\)
Now put \(x=z,y=0 \Longrightarrow f'(z)=i\left(2z-\frac{1}{z^2}\right) \Longrightarrow\)\( f(z)=i\left(z^2+\frac{1}{z}\right)\)
Now put \(z=x+i y\Longrightarrow f(z)=i\left((x+iy)^2+\frac{1}{x+i y}\right)=i\left((x^2-y^2+i2xy)+\frac{x-iy}{(x+iy)(x-iy)}\right)=\left(-2xy+\frac{y}{x^2+y^2}\right)+i\left((x^2-y^2)+\frac{x}{x^2+y^2}\right)\)
Hence \(\phi = -2xy+\frac{y}{x^2+y^2}\)