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I want to know a concrete derivation of 3D Stokes stream function.

The statement is, for example in 3D spherical coordinates (with symmetry in rotation about the $z$-axis), if $$\nabla \cdot u=0\tag{1}$$ which is $${1\over r^2}{\partial \over \partial r}(r^2 u_r)+{1\over r \sin\theta}{\partial \over \partial \theta}(\sin\theta u_\theta)=0,\tag{2} $$ then there is a scalar $\psi$ which satisfies $$u_r={1\over r^2\sin\theta}{\partial \psi\over \partial \theta}, \qquad u_\theta =-{1\over r \sin\theta}{\partial \psi\over \partial r}.\tag{3} $$

It is easy to show that those form satisfies the continuity equation, but i've found it hard to prove that $\psi$ exists and that it is the only solution.

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  • $\begingroup$ Since the creeping flow equations are linear, it is the only solution. $\endgroup$ Commented Feb 28, 2024 at 12:24

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  1. Define functions$^1$ $$ f(r,\theta)~:=~r^2\sin\theta~u_r(r,\theta) \quad\text{and}\quad g(r,\theta)~:=~r\sin\theta~u_{\theta}(r,\theta). \tag{A}$$

  2. Next consider the one-form $$ \eta ~:=~ f{\rm d}\theta -g{\rm d}r. \tag{B}$$

  3. The incompressible flow/zero divergence condition $$\vec{\nabla}\cdot\vec{u}~=~0\tag{C}$$ is equivalent to $$ \frac{\partial f}{\partial r}+\frac{\partial g}{\partial\theta}~=~0, \tag{D}$$ which in turn is equivalent to the closedness condition $${\rm d}\eta~=~0.\tag{E}$$

  4. Poincare lemma shows that the one-form $$ \eta~=~{\rm d}\psi \tag{F}$$ is locally an exact differential, or equivalently, $$ f~=~\frac{\partial\psi}{\partial\theta} \quad\text{and}\quad g~=~-\frac{\partial\psi}{\partial r}. \tag{G}$$ Here $\psi$ is Stokes stream function. This proves OP's eq. (3).

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$^1$ A similar trick works for 3D cylindrical coordinates.

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  • $\begingroup$ Thank you for the outline. I think I got everything else but I don't know how to prove 3. I am not familiar with differential forms and I have no idea how to manipulate divergence with differential forms. I have read this part of the wikipedia, but still have no clue. $\endgroup$ Commented Feb 28, 2024 at 11:35
  • $\begingroup$ I updated the answer. $\endgroup$ Commented Feb 28, 2024 at 12:01
  • $\begingroup$ Now I understand! Thank you for your help! $\endgroup$ Commented Feb 28, 2024 at 12:09
  • $\begingroup$ So, is it right to understand (3)->(4) with this? $\endgroup$ Commented Feb 28, 2024 at 12:10

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