查询词典 satisfies

By applying the tools such as modulus of curves and rings, together with Rohde s theorem, it proves that a homeomorphism is a bi Lipschitz map if and only if it satisfies a logarithmically absolute value inequality of rings.

运用曲线族的模和环模等工具，结合 Rohde定理，证明同胚为双 Lipschitz映射的充要条件是：它适合环的对数绝对值不等式；还得到了同胚为拟共形映射的一个充分条件。

We first got the PML equations by making the radius complex, and then proved that the sesquilinear form in the weak formulation of the PML problem satisfies the Grding inequality under the hypothesis of the absorption factor, and further could prove the existence and uniqueness of the solutions of the PML problem, finally, we gave the numerical experiments which show that this method is feasible to a certain extent.

首先通过复化极径得到PML方程，然后在关于吸收参数的假设下，证明了PML问题变分形式中的半双线性形式满足Grding不等式，进而证明了PML问题解的存在惟一性，并给出了数值实验。实验结果表明，该方法有一定的可行性。

We first got the PML equationsby making the radius complex, and then proved that the sesquilinear form in the weak formulation of the PML problem satisfies the Garding inequality under the hypothesis of the absorption factor, and further could prove the existence and uniqueness of the solutions of the PML problem, finally, we gave the numerical experimentswhich show that this method is feasible to a certain extent.

首先通过复化极径得到PML方程，然后在关于吸收参数的假设下，证明了PML问题变分形式中的半双线性形式满足Garding不等式，进而证明了PML问题解的存在惟一性，并给出了数值实验。实验结果表明，该方法有一定的可行性。

If is a sequence constructed by the method of local variations, then any accumulation point of satisfies .

定理3.1。如果是用局部变量法构造的序列，则任何的聚点满足（假设至少是一次连续可导的

Then the relationship which unknown coefficient satisfies from additional condition is derived. We deduce the existence and uniqueness of solution of the pseudoparabolic equation at last.

首先在假定系数已知的条件下，求出方程的形式解，然后再利用附加条件得到未知系数满足的关系式，最终给出了此类反问题解的存在唯一性。

This value is characterized as the unique function that satisfies efficiency, complete symmetry, the quasi-null player property and additivity.

这种价值作为唯一的方程的特点是，满足有效性、完全对称性等特点。

Let G be the domain whose inverse image is the cross section of u definedin (H1) where the mapping is affine,g and ψ are smooth functions and g＞0,thenthere is a unique smooth convex-monotone function which satisfies the followingproblem 〓Furthermore,if 〓 where C is a constant,ψis an affine function,〓 is astrictly interior sub-domain of G in the sense of"parabolic distance",then

若G为按（H1）中定义的u的截影经仿射变换而得的区域，g和〓为光滑函数且〓，则〓存在惟一的凸单调光滑解。若〓常数，〓仿射函数，〓为G在"抛物距离"下的严格内子域，则

The constructed algebraic spline curve not only possesses unwanted components and unexpected singular points but also satisfies C^2 continuity.

该算法得到的通近曲线二次代数样条曲线既不会产生多余的分支和不希望的奇异点，又达到整体C^2连续。

We obtain that if any 〓 is discrete or elementaryand 〓 satisfies Condition A,then the algebraic limit G of group sequence 〓is discrete or elementary.

首先，我们不再仅仅考虑离散非初等群集〓的代数极限G，而是离散群或初等群群集〓的代数极限G，我们对〓上〓变换群中斜驶元及其不动点进行了细致研究，注意到任意一个斜驶元存在一个仅仅含有斜驶元的领域，从而证明了初等群群集〓的代数极限G仍然是初等群，进而我们得到了一个代数收敛定理：如果任一〓是离散群或者初等群并且〓满足条件A，那么，群列〓的代数极限G一定是离散群或者初等群。

To create a caff americano – a coffee that satisfies the American preference for more sips in every cup – Europeans simply add hot water to their espresso.

据说，在欧洲，所谓的咖啡就是指的espresso，不是我们这种america 冲出来的咖啡。

The small blue capacitor has very high chances to go fully shorted compare to electrolytic capacitor which is very rare.

小蓝电容器具有很高的机会去充分地缩短了比较电解电容器，这是非常罕见的。

In the Italian campaigns of 1513 and 1515, when the Swiss won the victories of Novara and Marignan, he acted as army chaplain.

在意大利的运动， 1513年和1515年，当瑞士赢得了胜利，诺瓦拉和marignan ，他担任军队的牧师。