1 | 2 | 3 | 4 | 5 | 6 | 7 | ||
1 | 1 | 1 | ||||||
2 | 0 | 1 | 1 | |||||
3 | 0 | 1 | 1 | 2 | ||||
4 | 0 | 1 | 2 | 2 | 5 | |||
5 | 0 | 2 | 4 | 5 | 5 | 16 | ||
6 | 0 | 5 | 10 | 14 | 16 | 16 | 61 | |
7 | 0 | 16 | 32 | 46 | 56 | 61 | 61 | 271 |
This paper focuses on two-dimensional continuous subsonic-sonic potential flows in a semi-infinitely long nozzle with a straight lower wall and an upper wall which is convergent at the outlet while straight at the far fields. It is proved that if the variation rate of the cross section of the nozzle is suitably small, there exists a unique continuous subsonic-sonic flows in the nozzle such that the sonic curve intersects the upper wall at a fixed point and the velocity of the flow is along the normal direction at the sonic curve. Furthermore, the sonic curve is free, where the flow is singular in the sense that the flow speed is only Hölder continuous and the flow acceleration blows up. Additionally, the asymptotic behaviors of the flow speed at the far fields is shown.
Citation: Mingjun Zhou, Jingxue Yin. Continuous subsonic-sonic flows in a two-dimensional semi-infinitely long nozzle[J]. Electronic Research Archive, 2021, 29(3): 2417-2444. doi: 10.3934/era.2020122
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This paper focuses on two-dimensional continuous subsonic-sonic potential flows in a semi-infinitely long nozzle with a straight lower wall and an upper wall which is convergent at the outlet while straight at the far fields. It is proved that if the variation rate of the cross section of the nozzle is suitably small, there exists a unique continuous subsonic-sonic flows in the nozzle such that the sonic curve intersects the upper wall at a fixed point and the velocity of the flow is along the normal direction at the sonic curve. Furthermore, the sonic curve is free, where the flow is singular in the sense that the flow speed is only Hölder continuous and the flow acceleration blows up. Additionally, the asymptotic behaviors of the flow speed at the far fields is shown.
The Euler numbers
1+∑n≥1Enxnn!=tanx+secx. |
This is the sequence A000111 in [20]. In 1877 Seidel [19] defined the triangular array
En,k=En,k−1+En−1,n+1−k(n≥k≥2) | (1) |
with
E1,1E2,1→E2,2E3,3←E3,2←E3,1E4,1→E4,2→E4,3→E4,4⋯=10→11←1←00→1→2→2⋯ | (2) |
The first few values of
1 | 2 | 3 | 4 | 5 | 6 | 7 | ||
1 | 1 | 1 | ||||||
2 | 0 | 1 | 1 | |||||
3 | 0 | 1 | 1 | 2 | ||||
4 | 0 | 1 | 2 | 2 | 5 | |||
5 | 0 | 2 | 4 | 5 | 5 | 16 | ||
6 | 0 | 5 | 10 | 14 | 16 | 16 | 61 | |
7 | 0 | 16 | 32 | 46 | 56 | 61 | 61 | 271 |
André [1] showed in 1879 that the Euler number
DU4={2143,3142,3241,4132,4231}. |
In 1933 Kempener [14] used the boustrophedon algorithm (2) to enumerate alternating permutations without refering to Euler numbers. Since Entringer [7] first found the combinatorial interpretation of Kempener's table
Theorem 1 (Entringer). The number of the (down-up) alternating permutations of
DUn,k:={σ∈DUn:σ1=k}. |
According to Foata-Schützenberger [9] a sequence of sets
The Springer numbers
1+∑n≥1Snxnn!=1cosx−sinx. |
Arnold [2,p.11] showed in 1992 that
1ˉ23,1ˉ32,1ˉ3ˉ2,213,2ˉ13,2ˉ31,2ˉ3ˉ1,312,3ˉ12,3ˉ21,3ˉ2ˉ1, |
where we write
S1,−1S2,2←S2,1S3,−3→S3,−2→S3,−1S4,4←S4,3←S4,2←S4,1⋯⇕12←10→2→316←16←14←11⋯S1,1S2,−1←S2,−2S3,1→S3,2→S3,3S4,−1←S4,−2←S4,−3←S4,−4⋯⇕11←03→4→411←8←4←0⋯ |
where
Sn,k={Sn,k−1+Sn−1,−k+1if n≥k>1,Sn,−1if n>k=1,Sn,k−1+Sn−1,−kif −1≥k>−n. | (3) |
Theorem 2 (Arnold). For all integers
Sn,k:={σ∈Sn:σ1=k}. |
Moreover, for all integers
Sn,k=#{σ∈DUn(B):σ1=k}. |
Similarly, the numbers
-6 | -5 | -4 | -3 | -2 | -1 | 1 | 2 | 3 | 4 | 5 | 6 | ||
1 | 1 | 1 | |||||||||||
2 | 0 | 1 | 1 | 2 | |||||||||
3 | 0 | 2 | 3 | 3 | 4 | 4 | |||||||
4 | 0 | 4 | 8 | 11 | 11 | 14 | 16 | 16 | |||||
5 | 0 | 16 | 32 | 46 | 57 | 57 | 68 | 76 | 80 | 80 | |||
6 | 0 | 80 | 160 | 236 | 304 | 361 | 361 | 418 | 464 | 496 | 512 | 512 |
This paper is organized as follows. In Section 2, we shall give the necessary definitions and present our main results. The proof of our theorems will be given in Sections 3-4. In Section 5, we shall give more insightful description of two important bijections. More precisely, Chuang et al.'s constructed a
Let
For each vertex
Definition 3. Given an increasing 1-2 tree
Let
Tn,k={T∈Tn:Leaf(T)=k}. |
Donaghey [5] (see also [3]) proved bijectively that the Euler number
Theorem 4 (Gelineau-Shin-Zeng). There is an explicit bijection
Leaf(ψ(σ))=First(σ) |
for all
Let
Hetyei [12,Definition 4] defined recursively André permutation of second kind if it is empty or satisfies the following:
(ⅰ)
(ⅱ)
(ⅲ) For all
It is known that the above definition for André permutation of second kind is simply equivalent to the following definition. Let
Definition 5. A permutation
For example, the permutation
τ[1]=1,τ[2]=12,τ[3]=312,τ[4]=3124,τ[5]=31245. |
Foata and Schützenberger [10] proved that the Euler number
A4={1234,1423,3124,3412,4123}. |
Remark. Foata and Schützenberger in [10] introduced augmented André permutation is a permutation
σj−1=max{σj−1,σj,σk−1,σk}andσk=min{σj−1,σj,σk−1,σk}, |
there exists
Definition 6. A permutation
By definition, an André permutations is always a Simsun permutation, but the reverse is not true. For example, the permutation
τ[1]=1,τ[2]=21,τ[3]=213,τ[4]=2134,τ[5]=25134. |
Let
RS3={123,132,213,231,312}. |
As for
An,k:={σ∈An:σn=k},RSn,k:={σ∈RSn:σn=k}. |
Some examples are shown in Table 3.
Foata and Han [8,Theorem 1 (ⅲ)] proved that
Theorem 7. For positive integer
Leaf(T)=Last(ω(T)) | (4) |
for all
Whereas one can easily show that the cardinality
Stanley [22,Conjecture 3.1] conjectured a refinement of Purtill's result [18,Theorem 6.1] about the
Theorem 8 (Hetyei).
For all
#An,k=#RSn−1,k−1. | (5) |
In the next theorem, we give a bijective proof of the conjecture of Stanley by constructing an explicit bijection.
Theorem 9. For positive integer
Last(σ)−1=Last(φ(σ)) | (6) |
for all
Given a permutation
σ[1]=ˉ4,σ[2]=ˉ4ˉ1,σ[3]=2ˉ4ˉ1,σ[4]=2ˉ4ˉ13,σ[5]=2ˉ4ˉ135. |
Some examples of
Definition 10. A type
For example, all type
Our second aim is to show that these two refinements are new Arnold families. Recall that the sequence
Sn,k:={σ∈DUn(B):σ1=k}. |
Theorem 11. For all
ψB:Sn,k→T(B)n,k, | (7) |
ωB:T(B)n,k→A(B)n,k. | (8) |
Thus, for all
Sn,k=#A(B)n,k=#T(B)n,k. | (9) |
In particular, the two sequences
Hetyei[12,Definition 8] defined another class of signed André permutations.
Definition 12 (Hetyei). A signed André permutation is a pair
We write
Conjecture 13. For all
Sn,k=#A(H)n+1,n+2−k. |
Since the last entry of any permutation in the family
Definition 14. A permutation
Let
Theorem 15. For positive integer
Last(σ)−1=Last(φ(B)(σ)) | (10) |
for all
Remark. Ehrenborg and Readdy [6,Section 7] gave a different definition of signed Simsun permutation as follows: A signed permutation
12,21,ˉ12,2ˉ1,1ˉ2,ˉ21,ˉ1ˉ2,ˉ2ˉ1 |
are Simsun permutations, we note that it is not an Arnold family.
First of all, we prove Theorem 7, in order to show that
Given
For example, if the tree
12,21,ˉ12,2ˉ1,1ˉ2,ˉ21,ˉ1ˉ2,ˉ2ˉ1 |
then
Given
πi={σi−1ifi∉{i1,…,iℓ},σik−1ifi=ik−1fork=2,…,ℓ. | (11) |
We show that
σa=πa+1,σa+1=πa+1+1,…,σc−1=πc−1+1, and σc≤πc+1. |
Hence a triple
Consider the running example
Remark. Considering the bijection
ψ(τ)=T∈T9,7,ω(T)=σ=684512937∈A9,7,φ(σ)=π=57341286∈RS8,6, |
where
One can extend the above mapping
Remark. This bijection preserves the
ababaaba=cddcd. |
For the cd-index of a Simsun permutation
aababaab=cddcd. |
Given a
ψB(σ)=π−1(ψ(πσ)) |
through the unique order-preserving map
For example, in the case of
π=(ˉ8ˉ4ˉ3ˉ125679123456789). |
So we have
π=(ˉ8ˉ4ˉ3ˉ125679123456789). |
In Subsection 3.1, we define the bijection
ωB(T)=π−1(ω(π(T))) |
through the unique order-preserving map
For example, in the case of
ωB(T)=π−1(ω(π(T))) |
we obtain
We summarize four interpretations for Entringer numbers
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In 2012, Chuang et al. [4] construct a bijection
Algorithm A.
(A1) If
(A2) Otherwise, the word
ρ(T)=ω⋅ρ(T′), |
where the subword
(a) If the root of
(b) If the root of
As deleted only
Remark. Originally, in [4], the increasing 1-2 trees on
Theorem 16. The bijection
Proof. Suppose that we let
The root of
To record the left-child
It is clear that all vertices in the minimal path in a tree become the right-to-left minimums in a permutation under
The bijection
Given an increasing 1-2 tree
Algorithm B. Gelineau et al. described the bijection
|DUn|=|Tn|=1, |
we can define trivially
(B1) If
π′j={πj+2,if πj+2<k−1,πj+2−2,if πj+2>k. |
We get
π′j={πj+2,if πj+2<k−1,πj+2−2,if πj+2>k. |
We get the tree
(B2) If
(a) If
π′j={πj+2,if πj+2<k−1,πj+2−2,if πj+2>k. |
(b) If
π′j={πj+2,if πj+2<k−1,πj+2−2,if πj+2>k. |
Algorithm C. We define another bijection
If
For
(C1)
v1<u1<v2<u2<⋯<vj−1<uj−1<vj |
Decomposing by the maximal path from
● Graft
● Flip the tree at vertex
● Transplant the trees
● Graft
We can illustrate the above transformation by
v1<u1<v2<u2<⋯<vj−1<uj−1<vj |
(C2) If
● Graft
● Transplant the trees
● Graft
We can illustrate this transformation by the following
v1<u1<v2<u2<⋯<vj−1<uj−1<vj |
We note that the vertex
Example. We run the new algorithm to the examples
d5(σ)=(3),d4(σ)=(6,2),d3(σ)=(9,1),d2(σ)=(8,5),d1(σ)=(7,4). |
By Algorithm C, we get five trees sequentially
d5(σ)=(3),d4(σ)=(6,2),d3(σ)=(9,1),d2(σ)=(8,5),d1(σ)=(7,4). |
with
a(4)=3,a(3)=2,a(2)=9,a(1)=5,b(4)=3,b(3)=2,b(2) does not exist,b(1)=5. |
Thus, the increasing 1-2 tree
Theorem 17. The two bijections
Proof. It is clear that (C2) is equivalent to (B1). Since the rule (B2a) just exchange two labels, but does not change the tree-structure, it is enough to show that (C1) is produced recursively from (B1) and (B2b).
Assume that
a(4)=3,a(3)=2,a(2)=9,a(1)=5,b(4)=3,b(3)=2,b(2) does not exist,b(1)=5. |
Due to
a(4)=3,a(3)=2,a(2)=9,a(1)=5,b(4)=3,b(3)=2,b(2) does not exist,b(1)=5. |
Since
a(4)=3,a(3)=2,a(2)=9,a(1)=5,b(4)=3,b(3)=2,b(2) does not exist,b(1)=5. |
Since (C2a) is produced from the rule (B1) and (B2b), then Algorithm C follows Algorithm B.
The first author's work was supported by the National Research Foundation of Korea (NRF) grant funded by the Korea government (MSIP) (No. 2017R1C1B2008269).
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1 | 2 | 3 | 4 | 5 | 6 | 7 | ||
1 | 1 | 1 | ||||||
2 | 0 | 1 | 1 | |||||
3 | 0 | 1 | 1 | 2 | ||||
4 | 0 | 1 | 2 | 2 | 5 | |||
5 | 0 | 2 | 4 | 5 | 5 | 16 | ||
6 | 0 | 5 | 10 | 14 | 16 | 16 | 61 | |
7 | 0 | 16 | 32 | 46 | 56 | 61 | 61 | 271 |
-6 | -5 | -4 | -3 | -2 | -1 | 1 | 2 | 3 | 4 | 5 | 6 | ||
1 | 1 | 1 | |||||||||||
2 | 0 | 1 | 1 | 2 | |||||||||
3 | 0 | 2 | 3 | 3 | 4 | 4 | |||||||
4 | 0 | 4 | 8 | 11 | 11 | 14 | 16 | 16 | |||||
5 | 0 | 16 | 32 | 46 | 57 | 57 | 68 | 76 | 80 | 80 | |||
6 | 0 | 80 | 160 | 236 | 304 | 361 | 361 | 418 | 464 | 496 | 512 | 512 |
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1 | 2 | 3 | 4 | 5 | 6 | 7 | ||
1 | 1 | 1 | ||||||
2 | 0 | 1 | 1 | |||||
3 | 0 | 1 | 1 | 2 | ||||
4 | 0 | 1 | 2 | 2 | 5 | |||
5 | 0 | 2 | 4 | 5 | 5 | 16 | ||
6 | 0 | 5 | 10 | 14 | 16 | 16 | 61 | |
7 | 0 | 16 | 32 | 46 | 56 | 61 | 61 | 271 |
-6 | -5 | -4 | -3 | -2 | -1 | 1 | 2 | 3 | 4 | 5 | 6 | ||
1 | 1 | 1 | |||||||||||
2 | 0 | 1 | 1 | 2 | |||||||||
3 | 0 | 2 | 3 | 3 | 4 | 4 | |||||||
4 | 0 | 4 | 8 | 11 | 11 | 14 | 16 | 16 | |||||
5 | 0 | 16 | 32 | 46 | 57 | 57 | 68 | 76 | 80 | 80 | |||
6 | 0 | 80 | 160 | 236 | 304 | 361 | 361 | 418 | 464 | 496 | 512 | 512 |
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