Editing Stream function (section)

=== Condition of existence ===

It's straightforward to show that for two-dimensional plane flow <math>\mathbf{u}</math> satisfies the curl-divergence equation
:<math>
(\nabla \cdot \mathbf{u})\, \hat\mathbf{z} = - \nabla \times (R \, \mathbf{u})
</math>
where <math>R</math> is the <math>3 \times 3</math> [[rotation matrix]] corresponding to a <math>90^\circ</math> anticlockwise rotation about the positive <math>z</math> axis. This equation holds regardless of whether or not the flow is incompressible.

If the flow is incompressible (i.e., <math>\nabla \cdot \mathbf{u} = 0</math>), then the curl-divergence equation gives
:<math>\mathbf{0} = \nabla \times (R\, \mathbf{u})</math>.
Then by [[Stokes' theorem]] the line integral of <math>R\, \mathbf{u}</math> over every closed loop vanishes
:<math>
\oint_{\partial\Sigma} (R\, \mathbf{u}) \cdot \mathrm{d}\mathbf{\Gamma}= 0.
</math>
Hence, the line integral of <math>R\, \mathbf{u}</math> is path-independent. Finally, by the converse of the [[gradient theorem]], a scalar function <math>\psi (x,y,t)</math> exists such that
:<math>R\, \mathbf{u} = \nabla \psi</math>.
Here <math>\psi</math> represents the stream function.

Conversely, if the stream function exists, then <math>R\, \mathbf{u} = \nabla \psi</math>. Substituting this result into the curl-divergence equation yields <math>\nabla \cdot \mathbf{u} = 0</math> (i.e., the flow is incompressible).

In summary, the stream function for two-dimensional plane flow exists if and only if the flow is incompressible.