Sentence examples for is continuous up from inspiring English sources

Suppose u s is continuous up to at least one continuity point z 1 ∈ S Open image in new window of u0, then u0 z1) = u s (z1) + c, implying c = u0 z1) - u s (z1).

To complete the proof of the theorem, it needs only to examine that u is continuous up to the boundary of Ω.

The behavior of such gapless surface states that exist on the topologically non-trivial side is continuous, up to and at the gap closing.

If (x_{0}inpartialOmega ) is a regular point (see [4, p.25] for the definition), then u is continuous up to (x_{0}).

Since ∂Ω satisfies the Wiener criterion, we use the sub-super solution method similarly to [4] and get that the positive solution (u_{0}) of (1.2) is continuous up to ∂Ω.

Since (u_{n}) is bounded and g is continuous, up to subsequence, there is (t_{0}geq0) such that gbigl(|u_{n}|^{p}bigr to gbigl(t_{0}^{p} bigr geqalpha_{0}quad text{as } ntoinfty, and so lim_{ntoinfty} int_{Omega}|nabla u_{n}|^{p-2}nabla u_{n}nabla u_{n}-u),dx=0.

In Theorem 1.3, it is not necessary to assume that and are continuous up to.

The solution is continuous irrespective of the time segment (step) and the derivatives are continuous up to orderN-1 whereNis the order of the series expansion.

It follows from (3.6) and (3.14) that the function (N^) is analytic in (mathbb{C}_), and all of its derivatives are continuous up to the real axis.

By finding a barrier function related to the sub-Laplacian L, we prove that the Perron solutions for linear Dirichlet problems are continuous up to the boundary.

The top-order coefficients of the operator A ( x, D ) are assumed to be continuous up to the boundary in each subdomain Ω k but may have jumps across the interface Γ.