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Suggested derivation of Falkner-Skan equations

Obtaining the Falkner-Skan equations from the boundary-layer equations is somewhat more complicated. We start by eliminating \(v\) using the continuity equation with boundary condition \(v=0\) for \(y=0\).

\[ v = -\frac{\partial}{\partial x} \int_0^y u(x, y) \partial y. \]

Inserting this expression into the momentum equation in the streamwise direction we obtain

\[ u\frac{\partial u}{\partial x} - \frac{\partial u }{\partial y}\frac{\partial }{\partial x}\int_0^y u(x,y)\,dy = U\frac{d U }{d x} + \nu \frac{\partial^2 u}{\partial y^2} \]

This is an integrodifferential equation for the unknown \(u(x,y)\). We seek to simplify this into an ordinary differential equation for the similarity solution \(f(\eta)\). To this end we first make the substitutions \(u(x,y)=U(x)f^{\prime}(\eta)\) and require that

\[\begin{split} \begin{aligned} \eta &= y g(x), \\ \frac{\partial \eta}{\partial x} &= \frac{\eta g\prime}{g}. \end{aligned} \end{split}\]

Using this form for \(u\) we can insert for all terms in the momentum equation. Most derivatives are straightforward, but the term with the integral requires special attention. With \(x\) held constant in the integration we obtain

\[\begin{split} \begin{aligned} -\frac{\partial }{\partial x}\int_0^y u\,dy &= -\frac{\partial }{\partial x}\int_0^y U(x) f^{\prime}(\eta(y))\,dy , \\ &= -\frac{\partial }{\partial x}\left[ U(x)\int_0^y f^{\prime}(\eta(y))\,dy\right] \\ &= -\frac{\partial}{\partial x}\left[ U(x)\int_0^\eta f^{\prime}(\eta)\,\frac{d\eta} {g(x)}\right] \\ &= -\frac{\partial}{\partial x}\left[ \frac{U(x)}{g(x)}\int_0^\eta f^{\prime}(\eta)\,d\eta\right], \\ &= -\frac{\partial}{\partial x}\left[ \frac{U(x)}{g(x)} f(\eta)|^{\eta}_0\right], \\ &=-\frac{\partial}{\partial x}\left[ \frac{U(x)}{g(x)} f(\eta)\right], \\ &=-\left[f(\eta) \frac{\partial}{\partial x}\left(\frac{U}{g} \right) + \frac{U}{g}f^{\prime}(\eta)\frac{\partial\eta}{\partial x} \right] \\ &=-\frac{\eta U f^{\prime} g^{\prime}}{g^2} - \frac{U^{\prime}f}{g} + \frac{Ug^{\prime}f}{g^2} \end{aligned} \end{split}\]

The integration either with respect to \(y\) or \(\eta\) is always such that \(x\) is constant, which permits moving functions \(U, g\) from under the integration.

Next we plug \(\frac{\partial u}{\partial x} = U^{\prime}f^{\prime} + U f^{\prime\prime}\eta\frac{g^{\prime}}{g}\), \(\frac{\partial u}{\partial y}=Uf^{\prime\prime}g\), \(\frac{\partial^2 u}{\partial y^2}=Uf^{\prime\prime\prime}g^2\) and the integral term into the momentum equation, which yields

\[ Uf^{\prime}\left(U^{\prime}f^{\prime} + U f^{\prime\prime}\eta\frac{g^{\prime}}{g}\right) + Uf^{\prime\prime}g\left(-\frac{\eta U f^{\prime} g^{\prime}}{g^2} - \frac{U^{\prime}f}{g} + \frac{Ug^{\prime}f}{g^2}\right) = U\frac{dU}{dx} + \nu Uf^{\prime\prime\prime}g^2 \]

Second and third term cancel out. Dividing by \(\nu U g^2\) we get

\[ f^{\prime\prime\prime} = \left(\frac{U g^{\prime}}{\nu g^3}\right)ff^{\prime\prime} + \left(\frac{dU/dx}{\nu g^2}\right)\left((f^{\prime})^2-ff^{\prime\prime}-1\right) = 0, \]

which is the Falkner-Skan equation on its most general form. Note that we are not, as suggested by White, using Leibniz’ rule to obtain this result. (Thanks to Miroslav Kuchta for performing this detailed derivation!)

A similarity solution is obtained if the coefficients \(\frac{U g^{\prime}}{\nu g^3}\) and \(\frac{dU/dx}{\nu g^2}\) are independent of \(x\). Falkner and Skan obtained such a similarity by choosing

\[ \eta = C y x^a,\, \mathrm{ and }\,\, U = K x^m, \]

where \(m=2a+1\) and \(C\) and \(K\) are constants. The final form of the equation then reads

(70)\[ f^{'''} + f f^{''} + \beta \left( 1 - (f^{'})^2 \right) = 0,\]

where

\[ \beta = \frac{2 m }{1 + m}. \]
Suggested derivation of Blasius equation FEniCS