http___blog.simmakers.com_tag_nonuniform-mesh__lang_bg_1.xml
http___blog.simmakers.com_partitioning-a-segment_.xml
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Предговор: Началната отсечка се разделя със "задължителни" точки (те трябва да бъдат на окото възли), който едновременно да бъде или да не бъде точки на сгъстяване. На всяка част от сегмента на метода на subpartitioning, описани по-долу, може да се прилага. Нека - Е минималната приемлива стъпка, - Да е максимално допустимото стъпка и нека е дължината на сегмента, където . След това минимално възможен брой стъпки равен - , и максимално възможна е - . Ако , тогава проблемът твърдение е невярно. Да приемем, че . Ако двата края на маршрута, не изискват сгъстяване, можем да предприемем единен дял с произволен брой стъпки така че .
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Preface: The initial line segment is partitioned with “obligatory” points (they should be mesh nodes), which can both be or not be points of thickening. On each part of the segment the method of subpartitioning, described below, can be applied. Let – be the minimum acceptable step, – be the maximum acceptable step and let be the length of the segment, where . Then the minimal possible number of steps equals – , and the maximal possible is – . If , then the problem statement is incorrect. Assume that . If both edges of the segment do not require thickening, we can take a uniform partition with any number of steps so that . Let only one end require thickening. Without loss of generality let it be the left one. Suppose ,, where , and are to be determined. The following conditions should be satisfied: Let be specified. We consider the following 2 cases: In the first case let’s find from the equation In second case – from the equation Assuming each equation has a single root on the segment . Accordingly to this, the bisection method can be applied. In both cases one should take from to inclusive. All the partition variants, for which the conditions 1)–3) are fulfilled, are suitable. It can be proved (ex contrario) that if any variant of the described method is found then the problem statement is flawed. Let both ends require thickening. Assume that , , , , where and are to be determined. The following conditions should be satisfied: Consider only 2 cases: In the first case find from the equation , where . In the second case – from the equation , where . Assuming each equation has a single root on the segment . Accordingly to this, the bisection method can be applied. In both cases one should take from to inclusive. All the partition variants, for which the conditions 1) – 3) are fulfilled, are suitable. It can be proved (ex contrario) that if any variant of the described method is found then the problem statement is flawed. There are two open questions left: which variant of the partition to choose if several are possible and what to do if no variant is found. The answer to the first case depends on what is being searched. For example, one is maximizing density at the selected edges or is seeking for the variant with the least possible number of parts of the segment, or a partition with the greatest irregularity. According to this, it is necessary to choose the variant with the greatest , the smallest or with the greatest . If no variant is found, one can find the variant, in which the violation of the conditions 1), 2) is the smallest (as a measure of violation the number , may be considered), then put , and resolve the problem. Remark. When solving the equations in unknown it is necessary to take into account that the error of each ( or in the case of one or two thickenings respectively) should also be small. That is why it is proposed to take raised to the maximal power as the new unknown. Then the error of will be automatically low. In addition, expressions for the right border of the equation root will be significantly simpler: we will have case of one thickening and in case of two thickenings.