Sentence examples for on the initial function from inspiring English sources

A point of contention in the dinosaur-bird debate centers on the initial function of feathers.

If there exists a function, continuous on with at most the first order discontinuity at the point satisfying, and the inequality (2.7) on, the initial function (3.1).

From the definition of permanence, we need to prove liminf_{trightarrow+infty}x t geq v_{2}, (5.1) where (v_{2}) is some positive constant which does not depend on the initial function φ.

This condition says that, by Theorem 1, every initial problem (1) and (2) with φ satisfying (3) has exactly one solution on an interval [ α 0, σ ] T where σ > t 0. It is also easy to show that this solution depends continuously on the initial function φ.

This condition says that by Theorem 1 every initial value problem (1) and (2) with φ satisfying (3) has exactly one solution on the interval [ τ ( t 0 ), σ ] T, σ > t 0. It is also easy to show that this solution depends continuously on the initial function φ.

First we recall the assumptions imposed on the initial function (u_{0}): u_{0} in C^{2}(G), qquad 0 < alpha_{1} le u_{0} lealpha_{2}, where (alpha_{2} ) is a small number so that (1.1) has global positive solutions by Theorem 2.4.

The advantage of representing a solution of the initial problem (1.1), (1.2) as a solution of (2.6) is that, although (2.6) remains to be a neutral system, its right-hand side does not explicitly depend on the derivative for depending only on the derivative of the initial function on the initial interval.

We next remove condition (3.2) on the initial functions (u^{0}(x)), (v^{0}(x)).

Therefore, we take the expression of (5.3) as the initial function (u_{0}), depicted on the left-hand side of Figure 3.

To emphasize the dependence of the solution on the initial point n0 and the initial functions ϕ, we may use the notation x(n) = x(n; n0, ϕ).

To emphasize the dependence of the solution on the initial point n0 and the initial functions ϕ, we may use the notation x(n) = x(n; n0, ϕ).  . has a unique solution x(n) which is defined for n ∈ ℕ(n0 - d + 1) and satisfies the initial conditions (5).