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The Prandtl–Glauert transformation is a mathematical technique which allows solving certain compressible flow problems by incompressible-flow calculation methods. It also allows applying incompressible-flow data to compressible-flow cases.
Einstein's equations can also be solved on a computer using sophisticated numerical methods. [1] [2] [3] Given sufficient computer power, such solutions can be more accurate than post-Newtonian solutions. However, such calculations are demanding because the equations must generally be solved in a four-dimensional space.
Given the difficulty of constructing explicit small families of solutions, much less presenting something like a "general" solution to the Einstein field equation, or even a "general" solution to the vacuum field equation, a very reasonable approach is to try to find qualitative properties which hold for all solutions, or at least for all ...
The elements of an arithmetico-geometric sequence () are the products of the elements of an arithmetic progression (in blue) with initial value and common difference , = + (), with the corresponding elements of a geometric progression (in green) with initial value and common ratio , =, so that [4]
Solving the geodesic equations is a procedure used in mathematics, particularly Riemannian geometry, and in physics, particularly in general relativity, that results in obtaining geodesics. Physically, these represent the paths of (usually ideal) particles with no proper acceleration , their motion satisfying the geodesic equations.
We obtain the distribution of the property i.e. a given two dimensional situation by writing discretized equations of the form of equation (3) at each grid node of the subdivided domain. At the boundaries where the temperature or fluxes are known the discretized equation are modified to incorporate the boundary conditions.
But the solution at z = 0 is identical to the solution we obtained for the point x = 0, if we replace each γ by α + β − γ + 1. Hence, to get the solutions, we just make this substitution in the previous results. For x = 0, c 1 = 0 and c 2 = 1 − γ. Hence, in our case, c 1 = 0 while c 2 = γ − α − β. Let us now write the solutions ...
(,) and (,) are given, 2. (,) is given and () is real on the real axis, 3. only (,) is given, 4. only (,) is given. He is really interested in problems 3 and 4, but the answers to the easier problems 1 and 2 are needed for proving the answers to problems 3 and 4.