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Choose $U$ and $Q$ as the speed and source strength in the same kinematic velocity convention. With $r=(x^2+y^2{)}^{1/2}$ and $\vartheta =\arg (x+iy)$, superposition of uniform potential flow and a two-dimensional point source giveswhere $(u,v)=\nabla \Phi =({\psi }_y,-{\psi }_x)$. The angular branch cut records the volume flux from the source; the logarithm's arbitrary reference length only changes an additive constant in the velocity potential.

The stagnation point lies on the negative axis, where $u=U+Q/(2\pi x)=0$. Thus the limiting upstream reach of source fluid isThe stagnation point is approached asymptotically, rather than reached in finite time by a parcel on the upstream axis. The dividing streamlines through it have $\psi =\pm Q/2$ when the angle is taken in $(-\pi ,\pi )$. Far downstream $\vartheta \to 0$, so these streamlines approach $y=\pm Q/(2U)$. Consequently the source fluid occupies the Rankine half-body between them, with downstream width $2b=Q/U$. In the upper half-plane its boundary can also be parametrized by $y=b(1-\vartheta /\pi )$, $x=y\cot \vartheta$, $0<\vartheta <\pi$.

For arrival at the downstream axis, the direct streamline $y=0$, $x>0$ has speed $U+Q/(2\pi x)$. Its transit time from the ideal point source isThe first source fluid at a downstream cross-section lies on this symmetry axis: for positive $x$, the horizontal source contribution $Qx/[2\pi (x^2+y^2)]$ is maximal at $y=0$, and off-axis paths cannot advance through positive $x$ more quickly. A far-away parcel passing the inlet line at the same release time has transit time $x/U$. The time advance is thereforeThis is an advance, not the source parcel's absolute transit time. If a source parcel is released at time $t_0$ while the distant comparison parcel crosses $x=0$ at time zero, the advance relative to that parcel is $\tau -t_0$; the printed formula uses simultaneous releases. If $U,Q$ instead denote Darcy velocity and bulk-area source flux in a medium of uniform porosity $\varphi$, their ratio still gives the same $b$, but both parcel times acquire a factor $\varphi$. No porosity is specified here, so the displayed time formula fixes the kinematic convention.