Sentence examples for by solving the initial from inspiring English sources

Exact(2)

Under the assumptions (F1) and (F2) hold, given a solution (,uin W^{2,1}(c,d)) of (4), by solving the initial value problem (28) in ([a,b]]), with ((p,q)= u(c),u'(c))), we can always extend the domain of definition of u to ([a,b]]) as a solution of (4).

end{aligned} The solution (y(1,nh)) and its derivative with respect to t, (partial _{t}y(1,nh)), can be computed directly by solving the initial value problem defined by (3.1) and (3.4) at the nodes ({ nh }_{ninmathbb{Z}_{N}(h^{-1}mu)}).

Similar(57)

We approximate these samples numerically by solving the initial-value problems defined by (1.14) and (2.1) to obtain the approximate values U ˜ ( n h ), n ∈ Z N , i.e., U ˜ ( n h ) = Ω ˜ ( n h ) − K ( n h ).

The desired interpolation curve is just the projection curve, which can be obtained by numerically solving the initial- value problems for a system of first-order ordinary differential equations in the parametric domain for parametric case or in three-dimensional space for implicit case.

For a given interference price ω k, by solving the KKT conditions, the initial optimal solution for (1) is {boldsymbol{p}}_k={left(frac{lambda_k}{omega_k{left|{H}_p{boldsymbol{f}}_kright|}^2}-frac{sigma^2}{{left|{H}_k{boldsymbol{f}}_kright|}^2}right)}^ (21).

The steady-state solution was obtained by solving the transient solution from the initial conditions until a relative convergence of 10-7 wachievedved at each node.

This may be easily assessed by solving the problem from different initial guesses (multistart).

Equations (35), (36) can be obtained by solving the diffusion equation with initial conditions c(l<0, t=0)= c and c(l≥0, t=0)=0.

Each profile g(t), m(t), a i (t), r j (t) is recorded at t= t1,…, t p. The CKE outputs for TDM are g t1),…, g(t p ). Hence each microarray data set provides (p-1) equations linking the n p a r unknown parameters, since the recorded g(t) should be very close to the predicted values ĝ (t ) obtained by solving the ODE (1) with initial value g t1).

To avoid a high payback peak after the control period, the finishing time t f is found by solving the following equation: Figure 13 Initial control period in relation to LT and LU.

We calculated the stationary auxin distribution by solving the Cauchy problem with zero initial data.

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