Sentence examples for continuation arguments from inspiring English sources

Exact(1)

Hence, it follows from standard unique continuation arguments (cf. [5]) that the gradient of (Psi ) vanishes identically.

Similar(58)

A continuation argument with a moving kernel formula is used to find the instability criteria.

So, from [13, Theorem ] and a unique continuation argument we obtain that in.

The key idea is that in mapping the source calculus to the semantic target calculus, all types are parameterized with an extra continuation argument.

Based on this uniform estimate the global existence is proved by the standard continuation argument together with the local in time existence theory; for example, see [5].

By using a similar argument in [25], one can show the existence and uniqueness of the global solution to problem (1.1 - 1.4 1.1 - 1.4umptions (H1)-(H4) by the technique of Galerkin, the contraction mapping principle, and a continuation argument.

The standard continuation argument then yields that ([delta,varGamma ^)) could not be the maximal interval of existence of (tilde{varvec{u}}), and consequently (varGamma ^*ge T). This concludes the proof.

Our analysis is based on an elementary energy method together with the continuation argument and the key point is how to control the possible growth of the solution caused by the nonlinearity of the equation under consideration.

holds for some positive constant η > 0 which is chosen sufficiently small, and then to use the continuation argument to extend the local solution step by step to a global one.

We derive the Moser Trudinger Onofri inequalities on the 2-sphere and the 4-sphere as the limiting cases of the fractional power Sobolev inequalities on the same spaces, and justify our approach as the dimensional continuation argument initiated by Thomas P. Branson.

Since (theta t,x)), (theta' t,x)) satisfy (3.13), if ϵ and T have been chosen small enough then it follows that (deltatheta (t,x =0) on ([0,T]timesmathbb{R}^{3}). By a standard continuation argument, we can show that (deltatheta t,x =0) in ([0,infty times mathbb{R}^{3}), i.e. (theta t,x =theta' t,x)). This completes the proof of Theorem 1.1.

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