the equation is of second order, the problem is well defined. In spite of the lack of a firm theoretical basis for the ITU method, a trial-error by dichotomy (or binary search) can also be used. The next figure shows the results of numerical simulations. The curve labeled as bvp is the trajectory found by correcting the elevation angle. The other two are from a fixed step and a variable step (chosen in accordance with the ITU recommendations) solutions without the elevation angle correction. The nominal elevation angle for this case is −0.5-degree. The numerical results obtained at 22.5 GHz were: Note the way the solution bvp bends over the straight line. A consequence of this property is that the ray can reach locations situated below the horizon of S. This is consistent with observations. The trajectory is a Concave function is a consequence of the fact that the gradient of the refraction index is negative, so the Eikonal equation implies that the second derivative of the trajectory is negative. From the point where the ray is parallel to ground, relative to the chosen coordinates, the ray goes down but relative to ground level, the ray goes up. Often engineers are interested in finding the limits of a system. In this case, a simple idea is to try some low elevation angle and let the ray reach the desired altitude. This point of view has a problem: if suffice to take the angle for which the ray has a tangent point of lowest altitude. For instance with the case of a source at 5 km altitude, of nominal elevation angle −0.5-degree and the target is at 30 km altitude; the attenuation found by the boundary value method is 11.33 dB. The previous point of view of worst case leads to an elevation angle of −1.87-degree and an attenuation of 170.77 dB. With this kind of attenuation, every system would be unusable! It was found also for this case that with the nominal elevation angle, the distance of the tangent point to ground is 5.84 km; that of the worst case is 2.69 km. The nominal distance from source to target is 6383.84 km; for the worst case, it is 990.36 km. There are many numerical methods to solve boundary value problems. For the Eikonal equation, due to the good behavior of the refraction index just a simple Shooting method can