= Solution
Write the planar <velocity field> as $\mathbf u=(u,v,0)$, with $\mathbf k=(0,0,1)$. Expanding the <curl> gives
$$
\mathbf u=\nabla\times(\psi\mathbf k)=(\psi_y,-\psi_x,0),
\qquad \nabla\psi=\mathbf k\times\mathbf u=(-v,u,0).
$$
Consequently, for a piecewise smooth path $\gamma$ from the origin to $\mathbf x=(x,y)$,
$$
\boxed{\psi(\mathbf x)=\psi(\mathbf0)+\int_\gamma(\mathbf k\times\mathbf u)\cdot d\mathbf r
=\psi(\mathbf0)+\int_\gamma(u\,dy-v\,dx).}
$$
The arbitrary additive constant $\psi(\mathbf0)$ may be set to zero. The integrand is exactly $\nabla\psi\cdot d\mathbf r$, so the <fundamental theorem for line integrals> proves <path independence> immediately: the difference between two paths is the <line integral> of a <gradient> around a <closed curve>, which is zero.
One can also see the local integrability condition from <Green's theorem>:
$$
\oint(u\,dy-v\,dx)=\iint(u_x+v_y)\,dx\,dy=0,
$$
because the given representation enforces <incompressibility>. This second proof applies whenever the region between the paths lies in the regular fluid domain. The question already assumes a globally defined <streamfunction>; without that assumption, <incompressibility> alone would guarantee a global <streamfunction> on a <simply connected domain>, but a hole carrying nonzero <volume flux> could obstruct it.
Back to article page
