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Irrotational flow means $\nabla \times \mathbf{u}=0$; incompressible flow means $\nabla \cdot \mathbf{u}=0$. The given streamfunction automatically satisfies the second condition, since ${\partial }_x{\psi }_y-{\partial }_y{\psi }_x=0$. For an oriented path with unit tangent $(dx/ds,dy/ds)$, choose its right-hand unit normal $\mathbf{n}=(dy/ds,-dx/ds)$. ThenThe opposite normal reverses the sign; the source's unspecified normal orientation must be chosen consistently. The volume flux between two streamlines is thus their streamfunction difference. A closed contour has zero net flux, expressing mass conservation for constant-density flow without a source inside.

For the channel, a rigid boundary has zero normal velocity and hence constant streamfunction. Choose zero on the lower wall. Uniform inflow at infinity must have horizontal velocity $-m$, so there $\psi =-my$. The upper wall then has constant $-m$. The unbroken left wall connects to it and also has constant $-m$. The jump by $m$ between the walls at the corner represents the withdrawn flux. Finally the vorticity is $v_x-u_y=-\Delta \psi$, so irrotationality gives $\Delta \psi =0$.

Write $\psi =-my+w$. Then $w$ vanishes at $y=0,1$ and at infinity, while $w(0,y)=-m(1-y)$ for $0<y<1$. Separation of variables gives the harmonic sine mode in a half-strip $e^{-n\pi x}\sin n\pi y$. The Fourier sine series coefficients of the left boundary areTherefore the corner sink in a semi-infinite channel isFor $x>0$ the series and its differentiated series converge on compact subsets, making the sum harmonic. At $x=0$, ${\sum }_{n\ge 1}\sin (n\pi y)/n=\pi (1-y)/2$ for $0<y<1$, so the left-wall value is exactly $-m$. The other wall values and far-field limit follow directly.

The plus sign printed in the final series is inconsistent with the printed boundary conditions. It would give $m-2my$ on the left wall, not $-m$. The negative sign above is required. As a local flux check, near the corner the wall values give $\psi \sim -2m\theta /\pi$; the radial velocity is $-2m/(\pi r)$ and its outward flux across a quarter-circle is $-m$, precisely withdrawal.