Prandtl's lifting-line equation

Example

Consider a wing of span $2s$ along the axis $-s \le y \le s$, in a uniform flow of speed $V$. Let $c(y)$ be the chord, $\alpha(y)$ the geometric angle of attack and $\alpha_0(y)$ the angle of zero lift of the section at $y$. Assume the sections have the thin airfoil lift slope $2\pi$. Prandtl's lifting-line equation for the circulation $\Gamma(y)$ is $$\Gamma(y) = \pi V c(y) \bigl(\alpha(y) - \alpha_0(y)\bigr) - \frac{c(y)}{4}\, \mathrm{PV}\!\int_{-s}^{s} \frac{\Gamma'(\eta)}{y - \eta}\, d\eta, \quad -s < y < s,$$ with $\Gamma(\pm s) = 0$, where $\mathrm{PV}$ is the Cauchy principal value. The integral term is $\pi V c(y)\, \alpha_i(y)$, where the induced angle is $\alpha_i(y) = \frac{1}{4\pi V}\, \mathrm{PV}\!\int_{-s}^{s} \frac{\Gamma'(\eta)}{y - \eta}\, d\eta$. It is a singular integro-differential equation: the derivative of the unknown appears under a Cauchy kernel, so it is close to a singular integral equation.

Check with the elliptic distribution $\Gamma(y) = \Gamma_0 \sqrt{1 - y^2/s^2}$. Substituting $y = s x$ and $\eta = s t$ gives $$\alpha_i = \frac{\Gamma_0}{4\pi V s}\, \mathrm{PV}\!\int_{-1}^{1} \frac{t}{\sqrt{1 - t^2}\,(t - x)}\, dt = \frac{\Gamma_0}{4 V s},$$ since $\frac{1}{\pi}\, \mathrm{PV}\!\int_{-1}^{1} \frac{t\, dt}{\sqrt{1 - t^2}\,(t - x)} = 1$ for $-1 < x < 1$. So the induced angle is constant along the span. For an untwisted wing with $\alpha$ constant, $\alpha_0 = 0$ and elliptic chord $c(y) = c_0 \sqrt{1 - y^2/s^2}$, the equation reduces to $\Gamma_0 = \pi V c_0 \bigl(\alpha - \Gamma_0/(4 V s)\bigr)$. It gives $\Gamma_0 = \pi V c_0 \alpha / \bigl(1 + \pi c_0/(4 s)\bigr)$.