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Let the inverse problem (1.1)–(1.4) possess a solution, say ({u, alpha, f}).
(a) Let the inverse problem (1.1)–(1.4) possess a solution, say ({u, alpha, f}).
Then (3.2) would possess a solution which is bounded on the entire real line, which contradicts the fact that the polynomial P ( ξ ′, t, q ) has only roots with nonzero imaginary part.
If the equation (Tx = x) does not possess a solution, then we attempt to resolve the problem of finding an element x such that x is in proximity to Tx.
If the fixed point equation T x = x does not possess a solution, then d ( x, T x ) > 0 for all x ∈ A. In such a situation, it is our aim to find an element x ∈ A such that d ( x, T x ) is minimum in some sense.
For the Neumann problem to possess a solution in R E 1, p ( D, w ) Open image in new window, it is necessary and sufficient that 1 π ı ∫ T u 1 ( c | c ′ | d ζ ζ = 0. Open image in new window.
Similar(43)
We prove that there existsε0so that for all 0<ε<ε0, Eq. possesses a solution having exactly one maximum pointxε∈Δ, such thatV(xε)→cand ∇V(xε)→0 asε→0.
Assume that (36) possesses a solution.
That is, the functional equation (1.3) possesses a solution.
then the equation Ax = μx possesses a solution in.
It follows from Corollary 2.5 that, and so the equation Ax = μx possesses a solution in.
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