 Solution

Put the stagnation point at the origin and the local tangent plane at $z=0$. A regular Taylor expansion has no linear term because $\mathbf{u}(0)=0$. The impermeable tangent plane imposes ${\partial }_z\varphi =0$ at $z=0$ to leading order, excluding mixed quadratic terms $xz,yz$. For the illustrated axisymmetric approach, rotational symmetry in the tangent plane makes the $x^2$ and $y^2$ coefficients equal and removes $xy$. Finally the Laplace equation fixes the normal coefficient to minus twice the tangential one. Up to an irrelevant additive constant,For incoming flow from $z>0$, $W>0$. This is a local leading form under the stated symmetry, not the exact universal potential near every possible wall stagnation point; a generic tangential strain can have unequal principal rates.

A streamline solves $\dot{x}=2Wx$, $\dot{y}=2Wy$, $\dot{z}=-4Wz$, hence $x=A{e}^{2Wt}$, $y=B{e}^{2Wt}$, $z=C{e}^{-4Wt}$. Eliminating the parameter givesEach streamline lies in a meridional plane. With cylindrical radius $s=\sqrt{x^2+y^2}$, its shape is $z=C_1/s^2$. The symmetry axis and wall are limiting invariant streamlines. A particle exactly on the incoming axis approaches the stagnation point asymptotically.