Thin Aerofoil theory

Can any one solve the upper equation from the image and obtain lower equation from image?

Image from J D Anderson 5th edition page no. 324

  • 2
    $\begingroup$ Have you tried it yourself? Can you explain where you're getting stuck? $\endgroup$ – fooot May 15 '17 at 16:22
  • $\begingroup$ Of course I have tried! $\endgroup$ – user21819 May 15 '17 at 16:23

Well, there are many solutions to the first equation. We just have to find one which makes physically sense. One of the solutions to this equation has obviously* the form:

$\gamma(\theta) = 2 V_\infty \alpha \cdot\frac{\cos{\theta}-\cos{\theta_0}}{\sin{\theta}}$

Now choose a value for $\theta_0$ that satisfies the Kutta condition. The only solution is $\theta_0=\pi$. Therefore,

$\gamma(\theta) = 2 V_\infty \alpha \cdot\frac{1 + \cos{\theta}}{\sin{\theta}}$

You can check the solution substituting $\gamma(\theta)$ in the original equation. The identities

  • $\int_0^\pi\frac{1}{\cos{\theta}-\cos{\theta_0}}d\theta$ = 0
  • $\int_0^\pi \frac{\cos{\theta}}{cos{\theta}-\cos{\theta_0}}d\theta = \pi$

should be enough to do the check. This integrals are not the easiest, so you would need to find them in a book with trigonometric identities. There, the second one often appears in the general form $\frac{1}{\pi}\cdot\int_0^\pi \frac{\cos{n \theta}}{cos{\theta_0}-\cos{\theta}}d\theta = -\frac{\sin{n \theta_0}}{\sin{\theta_0}}$.

If you want a more formal solution I'd ask at https://math.stackexchange.com/.

*Integrating constants is easy. Therefore I transform the expression inside the integral into a constant. $g(\theta)*f(\theta)=C_1 \Rightarrow g(\theta) = \frac{C_1}{f(\theta)}$. In this case $f(\theta)=\frac{\sin{\theta}}{\cos{\theta}-\cos{\theta_0}}$

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