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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 The jump by $m$ between the walls at the corner represents the withdrawn flux.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.

Let $u={\varphi }_1-{\varphi }_2$. It is a harmonic function in the bounded region and vanishes on its entire boundary. For smooth solutions and boundary, the divergence theorem applied to $u\nabla u$ gives the energy identityThe integrand is nonnegative, so $\nabla u=0$ in $R$. Thus $u$ is constant on each connected component, and its zero Dirichlet boundary condition makes every constant zero. HenceThis proves the Uniqueness of the Dirichlet problem rather than assuming it.

For the separated expression, write $\varphi =F(x)G(y)$. The hyperbolic factor obeys $F^{\prime \prime }={\lambda }^{2}F$ and the trigonometric factor obeys $G^{\prime \prime }=-{\lambda }^{2}G$, sofor all the constants, including $\lambda =0$. In the half-strip take $a>0$ so the region has positive width. For every nonzero integer $n$, the decaying harmonic sine mode in a half-strip isIts second derivatives cancel, it vanishes at $y=0,a$, and it has the prescribed sine trace at $x=0$. The absolute value in the exponential ensures decay for negative as well as positive $n$. For $n=0$, the zero function is the solution. Thus all integer cases are covered, and the decay is uniform across the strip.