2015
DOI: 10.1137/130935318
|View full text |Cite
|
Sign up to set email alerts
|

A Gaussian Beam Method for High Frequency Solution of Symmetric Hyperbolic Systems with Polarized Waves

Abstract: Symmetric hyperbolic systems include many physically relevant systems of partial differential equations (PDEs) such as Maxwell's equations, the elastic wave equations and the acoustic equations [26]. In this paper we extend the Gaussian beam method to efficiently compute the high frequency solutions to such systems with constant degeneracy that corresponds to polarized waves, in which the dispersion matrix of the hyperbolic system has eigenvalues with constant degeneracy over the domain of computation. The new… Show more

Help me understand this report

Search citation statements

Order By: Relevance

Paper Sections

Select...
1

Citation Types

0
2
0

Year Published

2015
2015
2019
2019

Publication Types

Select...
4
1

Relationship

0
5

Authors

Journals

citations
Cited by 7 publications
(2 citation statements)
references
References 36 publications
0
2
0
Order By: Relevance
“…Our starting point is [9], and we refer the reader to it for more references to earlier results on superpositions of beams. In addition, some recent effort has also been made to extend the Gaussian beam method to more complex settings such as symmetric hyperbolic systems with polarized waves [3], the Schrödinger equation with discontinuous potentials [4], and wave equations in bounded convex domains [1,2].…”
Section: Introductionmentioning
confidence: 99%
See 1 more Smart Citation
“…Our starting point is [9], and we refer the reader to it for more references to earlier results on superpositions of beams. In addition, some recent effort has also been made to extend the Gaussian beam method to more complex settings such as symmetric hyperbolic systems with polarized waves [3], the Schrödinger equation with discontinuous potentials [4], and wave equations in bounded convex domains [1,2].…”
Section: Introductionmentioning
confidence: 99%
“…We have ∇u GB (x, 0) = −(k(1 + k)x(1) , k(1 + k) −1 x (2) , kx(3) )u GB (x, 0). This givesk −m ||∇u GB (·, 0)|| 2 L 2 = c 1 k 2 (1 + k) 2+r−m (k(1 + k)) −1−r/2 (k/(1 + k)) −(m−r)/2 k −(d−m)/2 + c 2 k 2 (1 + k) −2+r−m (k(1 + k)) −r/2 (k/(1 + k)) −1−(m−r)/2 k −(d−m)/c 3 k 2 (1 + k) r−m (k(1 + k)) −r/2 (k/(1 + k)) −(m−r)/2 k −1−(d−m)/c 1 k 1−d/2 (1 + k) 1−m/2 + c 2 k 1−d/2 (1 + k) −1−m/2 + c 3 k 1−d/2 (1 + k) −m/2 .…”
mentioning
confidence: 99%