求解

In chapter three,we have derived a generalized dual London equation for the Abelian "electric" field A_μfrom the effective dual Abelian-Higgs model.The approximative result and the numerical simulation result as well as their figs were shown via solving the dual Abelian-Higgs model with different methods.Hereby the vortex solution of the chromo-electric field and the vortex energy density were also discussed.Finally,we have studied the dual Abelian-Higgs model with additional vacuum constant,and have unpuzzled the vacuum constant corresponding to the bag model.When the quark sources is introduced, the vacuum constant in the MIT bag model is proportional to the quartic of the monopole mass.

利用不同方法求解该方程，给出了该方程的近似结果和数值模拟结果，并分别对两种情形画出图像作了比较，据此还讨论了色电场的涡旋解及能量密度，最后讨论了含有附加真空常数的对偶阿贝尔-黑格斯模型并对应口袋模型给出了附加真空常数的物理解释，引入夸克源之后还预言麻省口袋模型中的真空常数正比于单极质量的四次方。

The security~ of these cryptosystems is based on the dificulty in solving descrete logarithms with Abelian group .

本文讨论了一些公钥密码体制(ElGamal加密与解密算法、Diffie -Hellman密钥交换方案和Shamir协议)在阿贝尔群上的扩展，它们的安全性均建立在阿贝尔群上离散对数求解困难性的基础之

Suppose G is a finite group, then QG is a semisimple Q-algera, obviously ZG is a Z-order of QG,we denote the maximal Z-order by ?.If G is not abelian, it is not an easy thing to determine г; if G is an abelian group, then QG is isomorphic to the direct sum of a finite number of number fields, and r is the direct sum of these rings of algebraic integers of those number fields, but which elements of QG belong to r is not clear.

设G是一个有限群，那么QG是一个半单代数，ZG是QG的一个Z-序，设Γ是QG的一个极大Z-序，当G是一个非交换群时，Γ的求解是困难的问题；当G是一个交换群时，QG同构于有限多个数域的直和，Γ相应的就是各数域代数整数环的直和，但Γ具体是QG中那些元素不清楚。

According to the shape of the ablator surface the particle concentration and angle versus time was obtained through projection which can then be used to calculate the critical thickness of char layer. According to this thickness whether or not char layer been ablated is decided.

计算过程中根据烧蚀求解区域的变化自动生成网格并实现了局部加密，根据烧蚀型面进行投影获得了随时间变化的浓度和角度分布，并将其用于计算当前时刻的临界炭层厚度作为判断炭层脱落的依据。

Finally, we test the velocities of three flyers with ablator of different thickness, the result indicates that we can measure the velocities of flyer which can further calculate the released particle velocities, and validate the verdict of theory and simulation.

最后，实验实测了三种烧蚀层厚度不同的飞片的飞行速度，结果表明飞片法可以测得飞片速度，进而求解粒子速度，而且证实了理论和数值模拟的结论。

We give three constraints of absolute conic images and use these constraints to evaluate absolute conic images and then to recover Euclidean reconstruction from projective reconstruction.

总结了三种关于绝对二次曲线的像的约束，并利用这些约束求解绝对二次曲线的像，进而实现从仿射重构恢复欧氏重构。

We systemically discuss how to uniquely decide an infinite plane homography matrix by using the structure information in scene and how to evaluate a homography matrix which convert affine reconstruction to Euclidean reconstruction by solving absolute conic images.

系统地讨论了如何利用场景中的结构信息，来唯一地确定无穷远平面的单应矩阵，进而由射影重构恢复仿射重构，以及如何通过绝对二次曲线的像求解将仿射重构变换为欧氏重构的单应矩阵。

In this paper,a portfolio choice model is proposed based on mean absolute deviation.

用绝对离差刻画了风险，提出基于绝对离差的证券组合投资模型，并用模拟退火算法求解。

By using the principle of least absolute deviation method , we will give a new priority method—least logarithm absolute deviation method.

利用最小一乘法原理，在层次分析中提出了一种新的排序方法——对数最小一乘法，并将其转化成线性规划问题求解，证明了对数最小一乘法的一些性质。

The least absolute deviations are using widely used in engineering because of it s robustness,but the algorithm solving the least absolute deviation is not efficient.

最小一乘在稳健性上比最小二乘好，使得最小一乘在工程中得到广泛的应用，但求解最小一乘的算法并不理想；本文根据最小一乘的性质，把最小一乘问题变为组合优化问题。

They weren't aggressive, but I yelled and threw a rock in their direction to get them off the trail and away from me, just in case.

他们没有侵略性，但我大喊，并在他们的方向扔石头让他们过的线索，远离我，以防万一。

In slot 2 in your bag put wrapping paper, quantity does not matter in this case.

在你的书包里槽2把包装纸、数量无关紧要。