Sentence examples for as a second problem from inspiring English sources

As a second problem for illustrating the proposed loss function approach, machine group selection problem from Wang et al. (2000) is considered.

As a second problem, we consider the interaction of two separated solitary waves having different amplitudes and traveling in the same direction.

As a second problem, we will discuss the existence of solutions for the following fractional differential equation with the boundary conditions (1.2): D q c x ( t ) = f ( t, x ( t ), c D r x ( t ), c D r + 1 x ( t ) ), t ∈ [ 0, T ], T > 0, 2 < q ≤ 3, 0 < r ≤ 1. (1.3).

As a second problem we discuss in Section 4 an initial value problem for hybrid fractional sequential integro-differential equations, left { textstylebegin{array}{l} D^{alpha} [frac{D^{omega}x t -sum_{i=1}^{m}I^{beta_{i}}h_{i}(t,x(t))}{f(t, x(t))} ]=g(t,x(t),I^{gamma }x(t)), quaD^{omega}x t -sum_{i=1}^{m}ega}x(0)=0, end{array}dI^{beta_{i} right.

As a second problem in describing EMR use, distinguishing EMR from other software or media appeared as a problem in the interviews (theme 4).

As a second problem in cyclic graphs, it may then happen that a formerly removable edge in R becomes non-removable because certain paths may have been interrupted by preceding deletions of other removable edges.

In fact, he has a second problem as well, but in Pépinesque fashion makes little of it.

A second problem arises as a result of strong signals of association among cohorts utilized to expand the control group.

As a first problem, we consider a Caputo type fractional differential equation D q c x ( t ) = f ( t, x ( t ) ), n − 1 < q ≤ n, t ∈ [ 0, 1 ], (1.1).

As a first problem, the nonhomogeneous terms in the coupled fractional differential system depend on the fractional derivatives of lower orders only.

As a first problem, we consider begin{aligned}& D^{alpha}x t) = fbigl t,x^{t}bigr),quad 1leq tleq e, 1< alpha< 2, end{aligned} (1) begin{aligned}& x t) = chi(t), quad 1-rleq tleq1, end{aligned} (2) begin{aligned}& x t) = psi(t),quad eleq tleq e+h, end{aligned} (3) where (D^{alpha}) is the Hadamard fractional derivative.