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Gradients and Potential Flow Part 2: Streamlines, Harmonics, and Analytic Functions

October 28, 2022 by Oliver Khan Leave a Comment

Last time, we learned that if we have a scalar function φ(x,y) that satisfies Laplace’s equation (∆φ = ∇ • ∇φ = 0), then its gradient can define the velocity field of a well-behaved fluid that is irrotational and has no sources or sinks. Today we will try to find equations for the streamlines that particles have as they move through the fluid flow defined by ∇φ. Here, we can recall that a streamline is defined to be the path a tiny particle takes as it moves through a fluid. To find the equations of the streamlines, we will make use of the complex analysis concepts we developed two blog posts ago. It is also important to note here that we will be restricting our view to two-dimensional fluid flows in this blog post, since the theory behind them is less cumbersome to develop. 

We will begin our discussion with a review of analytic functions, which are complex-valued functions that are differentiable over a certain region of the complex plane. Specifically, if z = x + iy and f(z) = u(x,y) + i v(x,y) is analytic, then u(x,y) and v(x,y) satisfy the Cauchy-Riemann equations: 

∂u/∂x = ∂v/∂y and ∂u/∂y = – ∂v/∂x. 

We showed two blog posts ago that both u(x,y) and v(x,y) are harmonic. In other words, they satisfy Laplace’s equation. Now comes time for a definition. If u(x,y) and v(x,y) are two harmonic functions that can be written as the real and imaginary parts of some complex-valued function f(z) = u(x,y) + i v(x,y), then we will say that u(x,y) and v(x,y) are conjugate harmonics. 

We will now try to find a function ψ(x,y) that is a conjugate harmonic to our velocity potential φ. While this task may appear to lack motivation, it will turn out that this ψ function is exactly what we need to write the equations of the streamlines of the fluid flow defined by ∇φ. Let us suppose that there exists a complex function ϴ(z) such that if z = x + iy, then φ(x,y) is the real part of ϴ(z). We then will want to find a function ψ(x,y) such that ψ(x,y) is the imaginary part of ϴ(z). Furthermore, we will assume that ϴ(z) is analytic. Then we can find ψ(x,y) by solving the following two differential equations, which are in fact precisely the Cauchy-Riemann equations applied to ϴ(z):

∂φ/∂x = ∂ψ/∂y and ∂φ/∂y = – ∂ψ/∂x. 

It luckily turns out that we can continue our discussion without actually solving these equations. Before we can go further it is important to develop a bit more theory on the gradient. If we set ψ(x,y) = c for some constant c, then the equation ψ(x,y) = c defines a curve in two-dimensional space. These curves have a special name: they are called the level sets of ψ. Using some multivariable calculus, we can show that at any point (x0, y0), ∇ψ is perpendicular to the level set ψ(x,y) = ψ(x0,y0) (Note here that ψ(x0,y0) is a constant). 

Therefore, if we can show that ∇ψ is always perpendicular to ∇φ, then (since we are working in two dimensions) we can conclude that the level sets ψ(x,y) = c are always parallel to ∇φ. From there, we can conclude that the level sets of ψ(x,y) always point in the direction of our fluid flow, and hence that the family of curves ψ(x,y) = c precisely characterize the streamlines of the ∇φ vector field. At this point, it might be helpful to draw this idea on a piece of paper. 

To show that ∇ψ and ∇φ are perpendicular, it is enough to show that the dot product between ∇ψ and ∇φ is equal to zero. To do this, we recall the definition of the dot product and make use of the relationship between the partial derivatives of ψ and φ: 

∇ψ • ∇φ = <∂ψ/∂x , ∂ψ/∂y> • <∂φ/∂x, ∂φ/∂y> 

= < – ∂φ/∂y, ∂φ/∂x> • <∂φ/∂x, ∂φ/∂y> 

= – (∂φ/∂y)(∂φ/∂x) + (∂φ/∂y)(∂φ/∂x) = 0. 

We have thus shown that if we have a fluid whose velocity field is defined by ∇φ for some φ satisfying ∆φ = 0, then the level sets of a conjugate harmonic to φ (which we found to be the family of curves ψ(x,y) = c) will precisely describe the streamlines of that flow field. Next week, I will try to deliver some information that is equally rigorous, but hopefully equally rewarding. Until then, please take care.  


Oliver Khan

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