Solution of the Dorodnitsin-Ladyzhensky problem

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Abstract

The article is devoted to a rigorous proof of the statement that entropy takes the maximum value on the surface of a body with a blunted nose, streamlined by a supersonic flow, in the presence of a plane of symmetry of the flow. This is obvious for bodies of rotation at zero angle of attack, and it is established by numerical calculations and experimentally at non-zero angles of attack. The proof boils down to the justification that the leading streamline (the current line crossing the shock along the normal) ends on the body. In other words, that the leading streamline and the stagnation line are coincide. Such a proof was obtained by G. B. Sizykh in 2019 for the general spatial case (not only for flows with a plane of symmetry). This rather complicated proof is based on the Zoravsky criterion, which only a narrow circle of specialists has experience using, and is based on the assumption of the continuity of the second derivatives of density and pressure. In this paper, for the practically important case of flows with a plane of symmetry (in particular, the flow around bodies of rotation at a non-zero angle of attack), an original simple proof is proposed, for which the continuity of only the first derivatives of the density and pressure fields is sufficient and it is not necessary to use the Zoravsky criterion.

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About the authors

Grigory B. Sizykh

Moscow Institute of Physics and Technology (National Research University)

Author for correspondence.
Email: o1o2o3@yandex.ru
ORCID iD: 0000-0001-5821-8596
SPIN-code: 5348-6492
Scopus Author ID: 6508163390
ResearcherId: ABI-3162-2020
http://www.mathnet.ru/person112378

Cand. Phys. & Math. Sci; Associate Professor; Dept. of Higher Mathematics

Russian Federation, 141700, Dolgoprudny, Institutskiy per., 9

References

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Supplementary files

Supplementary Files
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1. JATS XML
2. Figure 1. Smooth convex nose in supersonic flow: a — axisymmetric flow; b — non-axisymmetric flow

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3. Figure 2. Closed area $\overline{G}$ and its boundary $\partial\overline{G}$. The stagnation line with the exception of points $A$ and $B$ lies inside $\overline{G}$

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