I had recently ended up with a case where on solving the Laplace equation (for a fluid under certain conditions), the radial dependence turned out to be complex (in general). In such cases, do we work with the real part only? Or should we not neglect the complex part and I am wrong? I can post the exact problem if needed, but this stands as a general question as well. Any hints/tips would be appreciated.

  • $\begingroup$ Was this a numerical solution or an analytical one? If numerical, how complex was it? Is the complex component significant, or very small? $\endgroup$ – tpg2114 May 11 '18 at 18:02
  • $\begingroup$ It was a theoretical derivation and I ended up with a general form of the type $R(r) = C_1 cosh(n log r)+ i C_2 sinh(n log r) $ where n is an integer and r denotes the radial distance . $\endgroup$ – Lelouch May 11 '18 at 18:05
  • $\begingroup$ $R(r)$ denotes the radial component of the flow velocity. $\endgroup$ – Lelouch May 11 '18 at 18:05
  • $\begingroup$ Well, that's the solution for all Laplace equations basically -- often your boundary conditions will eliminate one of the components and you'll be left with either the real or the imaginary part. So without more specific details of your conditions, it's hard to say if your solution will actually have an imaginary part. $\endgroup$ – tpg2114 May 11 '18 at 18:12
  • $\begingroup$ My question is supposing the conditions are such that the complex solution coefficient $C_2 $ not equal to 0, then what happens? $\endgroup$ – Lelouch May 11 '18 at 18:16

In a truly general case, complex solutions are not only possible, but useful. For a 2D, incompressible, irrotational flow, there are two useful functions related to the velocity: the velocity potential and the stream function. Both functions reduce a system of equations for the velocity into a scalar equation of higher order. This results in a Laplace equation.

We can, of course, solve for either the velocity potential or the stream function. But, we can solve for both at the same time, where the velocity potential is the real component of the solution to the Laplace equation, and the stream function is the complex component of the solution to the Laplace equation. See this page for more details

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  • $\begingroup$ Just to be clear, I solve the NSE for the velocity field $u$ and end up with a solution of the form $u = v + if$ , $v$ and $f$ being real valued functions. Then $v$ denotes the actual velocity field while $f$ denotes the stream function? $\endgroup$ – Lelouch May 11 '18 at 18:15
  • $\begingroup$ @Lelouch Are you solving the Navier-Stokes equations, or are you making some assumptions? Because if you have a Laplace equation for the velocity, you're making a lot of assumptions and aren't really solving the Navier-Stokes equations. I'm not trying to be pedantic, but the details really matter when it comes to interpreting the results. If your flow is 2D, and incompressible, and irrotational (which is the most common way one ends up with a Laplace equation), then $u = v + i f$ means $v$ is the velocity potential and $f$ is the stream function. $\endgroup$ – tpg2114 May 11 '18 at 18:19
  • $\begingroup$ And the velocity potential function is orthogonal to the stream function. $\endgroup$ – tpg2114 May 11 '18 at 18:19
  • $\begingroup$ It's also worth noting that you can end up with complex solutions if you are working in a complex domain -- that can happen if you are using conformal mapping to take the solution of a simple flow and apply it to a complex shape via the Joukowsky transformation. $\endgroup$ – tpg2114 May 11 '18 at 18:21

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