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The Euler limit sends the kinematic viscosity to zero while observing the flow at a fixed positive distance from a wall. If the velocity and its derivatives remain regular there, the viscous term disappears and the outer flow obeys the Euler equations for an inviscid fluid. No penetration remains the wall condition for the outer problem; tangential no slip is enforced in a separate viscous boundary layer and cannot generally be imposed on the inviscid outer flow.

The incompressible Navier-Stokes equations are $u_t+(u\cdot \nabla )u=-\nabla p/\rho +\nu \Delta u$ and $\nabla \cdot u=0$, with no penetration and no slip at a stationary rigid wall. The Euler limit takes $\nu \to 0$ at fixed distance from the wall, leaving the inviscid Euler equations and a no-penetration condition; imposing tangential no slip on that outer problem generally overdetermines it. The Prandtl limit resolves a shrinking wall layer in which transverse viscous diffusion remains comparable to convection and restores no slip.

For the steady flat-plate boundary layer, $\delta /x\ll 1$, continuity makes $v/U=O(\delta /x)$, and $u_{xx}$ is small compared with $u_{yy}$. Normal momentum makes pressure independent of $y$ to leading order. The constant outer speed makes the outer pressure gradient, and hence $p_x$ throughout the layer, zero. The leading equations are $u{u}_x+v{u}_y=\nu u_{yy}$ and $u_x+v_y=0$. With the stream function convention $u={\psi }_y$, $v=-{\psi }_x$, these giveBalancing $U^2/x$ against $\nu U/{\delta }^2$ gives $\delta =\sqrt{\nu x/U}$. There is no independent streamwise scale for the semi-infinite plate, motivating the similarity solutionIt gives $u=U{f}^{\prime }$ and $v=\frac{1}{2}\sqrt{\nu U/x}(\eta f^{\prime }-f)$. Substitution yields $u{u}_x+v{u}_y=-U^{2}f{f}^{\prime \prime }/(2x)$ and $\nu u_{yy}=U^2{f}^{\prime \prime \prime }/x$, so the Blasius equation and conditions areThe first two conditions impose no penetration and no slip; the last matches the outer stream.

The wall shear on one face is $\tau (x)=\rho \nu u_y(x,0)=\rho U\sqrt{\nu U/x}{f}^{\prime \prime }(0)$. Integration from zero to $L$ gives $2\rho U^{2}L{f}^{\prime \prime }(0)/\sqrt{UL/\nu }$ on one face. With the same flow on both faces, the total drag force per unit width isOur normalization has $f^{\prime }(\infty )=1$, so the displayed denominator is one. The factor four refers to both faces; for a plate wetted on only one face the drag is half this value.