Abstract

This article considers a classical solution of the boundary problem for the four-order strictly hyperbolic equation with four different characteristics. Note that the well-posed statement of mixed problems for hyperbolic equations not only depends on the number of characteristics, but also on their location. The operator appearing in the equation involves a composition of first-order differential operators. The equation is defined in the half-strip of two independent variables. There are Cauchy’s conditions at the domain bottom and periodic conditions at other boundaries. Using the method of characteristics, the analytic solution of the considered problem is obtained. The uniqueness of the solution is proved. We have also noted that the solution in the whole given domain is a composition of the solutions obtained in some subdomains. Thus, for the obtained classical solution to possess required smoothness, the values of these piecewise solutions, as well as their derivatives up to the fourth order must coincide at the boundary of these subdomains. A classical solution is understood as a function that is defined everywhere at all closure points of a given domain and has all classical derivatives entering the equation and the conditions of the problem.

Highlights

  • This article considers a classical solution of the boundary problem for the four-order strictly hyperbolic equation with four different characteristics

  • Note that the well-posed statement of mixed problems for hyperbolic equations depends on the number of characteristics, and on their location

  • The equation is defined in the half-strip of two independent variables

Read more

Summary

Introduction

This article considers a classical solution of the boundary problem for the four-order strictly hyperbolic equation with four different characteristics. 136 Proceedings of the National academy of sciences of Belarus, рhysics and mathematics series, 2018, vol 54, no. ( ) где g j j = 1, 4 – произвольные функции с областиями определения D ( g1 ), D ( g3 )= ((−∞,l]), D ( g2 ), D ( g= 4 ) ([0, +∞)), если (t, x)∈Q,и vp – частное решение уравнения (1).

Results
Conclusion

Talk to us

Join us for a 30 min session where you can share your feedback and ask us any queries you have

Schedule a call

Disclaimer: All third-party content on this website/platform is and will remain the property of their respective owners and is provided on "as is" basis without any warranties, express or implied. Use of third-party content does not indicate any affiliation, sponsorship with or endorsement by them. Any references to third-party content is to identify the corresponding services and shall be considered fair use under The CopyrightLaw.