Two derivations are presented below. The first is a
heuristic argument, based on physical insight. The second is a formal and technical one, requiring basic
vector analysis and
complex analysis.
Heuristic argument For a heuristic argument, consider a thin airfoil of
chord c and infinite span, moving through air of density \rho. Let the airfoil be inclined to the oncoming flow to produce an air speed V on one side of the airfoil, and an air speed V + v on the other side. The circulation is then :\Gamma = Vc - (V + v)c = -v c.\, The difference in pressure \Delta P between the two sides of the airfoil can be found by applying
Bernoulli's equation: :\begin{align} \frac {\rho}{2}(V)^2 + (P + \Delta P) &= \frac {\rho}{2}(V + v)^2 + P,\, \\ \frac {\rho}{2}(V)^2 + \Delta P &= \frac {\rho}{2}(V^2 + 2 V v + v^2),\, \\ \Delta P &= \rho V v \qquad \text{(ignoring } \frac{\rho}{2}v^2),\, \end{align} so the downward force on the air, per unit span, is :L' = c \Delta P = \rho V v c = -\rho V\Gamma\, and the upward force (lift) on the airfoil is \rho V\Gamma.\, A
differential version of this theorem applies on each element of the plate and is the basis of
thin-airfoil theory.
Formal derivation {{ math proof First of all, the force exerted on each unit length of a cylinder of arbitrary cross section is calculated. Let this force per unit length (from now on referred to simply as force) be \mathbf{F}. So then the total force is: : \mathbf{F} = -\oint_C p \mathbf{n}\, ds, where
C denotes the borderline of the cylinder, p is the
static pressure of the fluid, \mathbf{n}\, is the
unit vector normal to the cylinder, and
ds is the arc element of the borderline of the cross section. Now let \phi be the angle between the normal vector and the vertical. Then the components of the above force are: : F_x = -\oint_C p \sin\phi\, ds\,, \qquad F_y = \oint_C p \cos\phi\, ds. Now comes a crucial step: consider the used two-dimensional space as a
complex plane. So every vector can be represented as a
complex number, with its first component equal to the real part and its second component equal to the imaginary part of the complex number. Then, the force can be represented as: : F = F_x + iF_y = -\oint_Cp(\sin\phi - i\cos\phi)\,ds . The next step is to take the
complex conjugate of the force F and do some manipulation: : \bar{F} = -\oint_C p(\sin\phi + i\cos\phi)\,ds = -i\oint_C p(\cos\phi - i\sin\phi)\, ds = -i\oint_C p e^{-i\phi}\,ds. Surface segments
ds are related to changes
dz along them by: : \begin{align} dz &= dx + idy = ds(\cos\phi + i\sin\phi) = ds\,e^{i\phi} \\ {} \Rightarrow d\bar{z} &= e^{-i\phi}ds. \end{align} Plugging this back into the integral, the result is: : \bar{F} = -i\oint_C p \, d\bar{z}. Now the
Bernoulli equation is used, in order to remove the pressure from the integral. Throughout the analysis it is assumed that there is no outer force field present. The mass density of the flow is \rho. Then pressure p is related to velocity v = v_x + iv_y by: : p = p_0 - \frac{\rho |v|^2}{2}. With this the force F becomes: : \bar{F} = -ip_0\oint_C d\bar{z} + i \frac{\rho}{2} \oint_C |v|^2\, d\bar{z} = \frac{i\rho}{2}\oint_C |v|^2\,d\bar{z}. Only one step is left to do: introduce w = f(z), the
complex potential of the flow. This is related to the velocity components as w' = v_x - iv_y = \bar{v}, where the apostrophe denotes differentiation with respect to the complex variable
z. The velocity is tangent to the borderline
C, so this means that v = \pm |v| e^{i\phi}. Therefore, v^2 d\bar{z} = |v|^2 dz, and the desired expression for the force is obtained: : \bar{F}=\frac{i\rho}{2}\oint_C w'^2\,dz, which is called the
Blasius theorem. To arrive at the Joukowski formula, this integral has to be evaluated. From complex analysis it is known that a
holomorphic function can be presented as a
Laurent series. From the physics of the problem it is deduced that the derivative of the complex potential w will look thus: : w'(z) = a_0 + \frac{a_1}{z} + \frac{a_2}{z^2} + \cdots . The function does not contain higher order terms, since the velocity stays finite at infinity. So a_0\, represents the derivative the complex potential at infinity: a_0 = v_{x\infty} - iv_{y\infty}\,. The next task is to find out the meaning of a_1\,. Using the
residue theorem on the above series: : a_1 = \frac{1}{2\pi i} \oint_C w'\, dz. Now perform the above integration: : \begin{align} \oint_C w'(z)\,dz &= \oint_C (v_x - iv_y)(dx + idy) \\ &= \oint_C (v_x\,dx + v_y\,dy) + i\oint_C(v_x\,dy - v_y\,dx) \\ &= \oint_C \mathbf{v}\,{ds} + i\oint_C(v_x\,dy - v_y\,dx). \end{align} The first integral is recognized as the
circulation denoted by \Gamma. The second integral can be evaluated after some manipulation: : \oint_C(v_x\,dy - v_y\,dx) = \oint_C\left(\frac{\partial\psi}{\partial y}dy + \frac{\partial\psi}{\partial x}dx\right) = \oint_C d\psi = 0. Here \psi\, is the
stream function. Since the
C border of the cylinder is a streamline itself, the stream function does not change on it, and d\psi = 0 \,. Hence the above integral is zero. As a result: :a_1 = \frac{\Gamma}{2\pi i}. Take the square of the series: : w'^2(z) = a_0^2 + \frac{a_0\Gamma}{\pi i z} + \cdots. Plugging this back into the Blasius–Chaplygin formula, and performing the integration using the residue theorem: : \bar{F} = \frac{i\rho}{2}\left[2\pi i \frac{a_0\Gamma}{\pi i}\right] = i\rho a_0 \Gamma = i\rho \Gamma(v_{x\infty} - iv_{y\infty}) = \rho\Gamma v_{y\infty} + i\rho\Gamma v_{x\infty} = F_x - iF_y. And so the Kutta–Joukowski formula is: : \begin{align} F_x &= \rho \Gamma v_{y\infty}\,, & F_y &= -\rho \Gamma v_{x\infty}. \end{align} }} ==Lift forces for more complex situations==