Sentence examples for optics solution from inspiring English sources

Consider now the geometric optics solution (v_2 y)) for ({tilde{L}}_1^{(m)}u_2=0) with the same initial condition.

Note that the support of geometric optics solution (u_N^{(i)}+u_i^{(N+1)}) is contained in (D(Gamma _2times [T_1,T_2])).

end{aligned} (5.23 Substituting in (5.23) the geometric optics solution (5.1), integrating by parts in s, multiplying by ik and, finally, taking the limit when (krightarrow infty ), we get (5.16) with (v_i^g) replaced by (v_{is}^g).

Substituting the geometric optics solutions in the main formula, we prove in Sects.

The proof of Theorem 1.18 is based on the construction of appropriate complex geometrical optics solutions by Carleman estimates with degenerate weight functions.

Consider the geometric optics solutions (v_{i,k}^g) of the form (5.1), where (g=chi _1 s chi _3(y'),chi _3 y' in C_0^infty (Gamma _2)) is arbitrary.

The paper [87] uses Carleman estimates with weights which are harmonic functions with non-degenerate critical points to construct appropriate complex geometrical optics solutions to prove the result.

This pioneer contribution motivated many developments in inverse problems, in particular in the construction of "complex geometrical optics" solutions of partial differential equations to solve several inverse problems.

The proof is close to arguments in inverse boundary problems, and is based on constructing complex geometrical optics solutions to the Schrödinger equation via a pseudodifferential conjugation argument.

The complex geometrical optics solutions of Theorems 1.5 and 1.6 were also used by A. Nachman [135] and R. Novikov [145] to give a reconstruction procedure of the conductivity from Λ γ.

Assume Λ γ 1 ( f ) = Λ γ 2 ( f ) on V ∀ f ∈ H 1 2 ( ∂ ( Ω ¯ D ) ), supp f ⊂ V Then γ 1 = γ 2. The two dimensional case has special features since one can construct a much larger set of complex geometrical optics solutions than in higher dimensions.