Accepted Manuscript The supersonic flow past a wedge with large curved boundary
Dian Hu
PII: DOI: Reference:
S0022-247X(18)30107-0 https://doi.org/10.1016/j.jmaa.2018.01.069 YJMAA 22002
To appear in:
Journal of Mathematical Analysis and Applications
Received date:
13 September 2017
Please cite this article in press as: D. Hu, The supersonic flow past a wedge with large curved boundary, J. Math. Anal. Appl. (2018), https://doi.org/10.1016/j.jmaa.2018.01.069
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THE SUPERSONIC FLOW PAST A WEDGE WITH LARGE CURVED BOUNDARY DIAN HU Abstract. In this paper, we construct a piecewise smooth solution for a 2-D supersonic potential flow past a curved wedge, whose boundary is assumed to have large variation. In fact, we originally provide a background solution, which is constructed by assuming the coming flow has limit speed. Its shock coincides with the curved wedge and the flow behind the shock is defined only along the wedge. Then our problem can be treated as a perturbation of this background solution and the solution is obtained by showing certain estimates.
Contents 1. Introduction 2. A Special Case 3. Characteristic Decomposition 4. Proof of Theorem 1.1 Acknowledgments References
1 3 4 7 11 11
1. Introduction In this paper, we employ the following 2-dimensional steady irrotational Euler system to describe the flow fields (ρu)x + (ρv)y = 0, (1.1) vx − uy = 0, where (u, v) denotes the velocity field, ρ the density, p = Aργ the pressure. Since (1.1) is a conservation law, for a piecewise smooth solution, it holds the RankineHugoniot condition [ρu]φ = [ρv], (1.2) [v]φ = −[u], Date: February 5, 2018. 2000 Mathematics Subject Classification. 35L65,35L67,35M10,35B35,76H05,76N10. 1
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DIAN HU
where the bracket means the jump of the corresponding quantities on the discontinuity curve y = φ(x). It is well-known that one can deduce the Bernoulli’s law from (1.1) and (1.2), namely, along each streamline, even across the shock-front, u2 + v 2 c2 q¯2 + = , 2 γ−1 2
(1.3)
dp where c2 = dρ = γp , c is the sonic speed and q¯ is the limit speed. ρ In the 2-dimensional plane, we present an x-axis symmetric wedge, whose upper boundary is defined by W : y = b(x) for x ≥ 0. Its head point locates at the origin (0, b(0)) = (0, 0). A uniform supersonic flow (u, 0) comes from infinity and hits it. Thus a shock S : y = φ(x) arises in front of the wedge. If the wedge has sharp head, then the shock attaches the wedge at the origin. As illustrating in Figure 1, we denote the domain between S and W by Ω.
S
u
→
Ω W
Figure 1. Main Problem Main Problem: For the supersonic coming flow (u, 0), we intend to determine shock S and the flow (u, v) in Ω with the wall condition v ( )(x, b(x)) = b (x). (1.4) u Moreover, the entropy condition need to be held along the shock. Our main theorem in this paper is Theorem 1.1. Assume that the boundary of the wedge y = b(x) is C 2 and there are positive constants a, b, ¯b, B such that 2 ¯ , (1.5) 0 < b < b (x) < b < γ−1 and B |b (x)| ≤ , (1.6) (1 + x)1+a for x ≥ 0. Then there exists a constant 0 > 0, which is depending on the wall W, such that for any |u − q¯| < 0 , (1.7)
THE SUPERSONIC FLOW PAST A WEDGE WITH LARGE CURVED BOUNDARY
3
the Main Problem is solvable. A global supersonic shock exists and the flow in Ω is continuous. Before our discussion, we give some comments on the former works. Main Problem is an old problem and has been studied in many literatures (see [4]). Most of them are devoted to the nonlinear perturbation problem. In fact, assume that the wedge boundary is straight, then one can construct a piecewise constant solution by using the shock polar (see [5]). Based on this background solution, one can obtain a perturbed solution by nonlinear iteration (see [6, 7, 13, 14]). On the other hand, in [1, 2, 3], the author considers this problem for the wedge boundary with some convex properties. For the convex wedge boundary, a global solution is constructed there and its asymptotics are obtained by product (see [1, 2]). For the wedge boundary with some concave properties, there is also a global existence result (see [3]). Here, in this paper, we retreat this problem in another way. Assume the coming flow has limit speed, then we can always construct a special solution for Main Problem with curved wedge boundary. By the conservation of mass, its shock coincides with the wedge boundary. Thus, for the Main Problem, whose coming flow speed is close to the limit speed, we could treat it as a nonlinear perturbation problem again. The characteristic method is applied to attack this problem. We point out that, in our case, the requirement of the wedge boundary is very weak. It needs not to be convex or concave. This paper is organised as follows. In the following section, we introduce the special solution. Then in Sect. 3, we present the characteristic decomposition for (1.1). Finally, we complete the proof of Theorem 1.1 in Sect. 4. 2. A Special Case Since q¯ is the limit speed, for the corresponding density ρ¯, pressure p¯ and sonic c¯, we have ρ¯ = p¯ = c¯ = 0.
(2.1)
In view of above, if the coming flow has limit speed (u, v) = (¯ q , 0), then, the RH condition (1.2) takes the form u(u − q¯) + v 2 = 0,
(2.2)
where (u, v) is the velocity after the shock. It is a circle pasting the origin and centring at ( 2q¯ , 0). For such a case, we have many simplification. First, for the shock y = φ(x), since ρ = ρ¯ = 0, we have φ =
ρv − ρv [ρv] v = = . [ρu] ρu − ρu u
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DIAN HU
It indicates that the flow after the shock is tangent to the shock. It is obviously, since the coming flow satisfies ρu = ρ¯q¯ = 0, there is no mass passing through the shock. Therefore, if the coming flow is (¯ q , 0), then Main Problem can be solved by combining (1.3) and (1.4). The shock S is coincide with the wall W and the domain Ω degenerates to a curve. Particularly, we have flow field q¯ q¯b , ). (2.3) (us , vs )(x, b(x)) := ( 1 + (b )2 1 + (b )2 It can be easily deduced that the flow is subsonic if 2 b > , γ−1 it is supersonic if
0
(2.4)
2 . γ−1
(2.5)
Remark 2.1. The special solution can be defined for more general wedge boundary function b provided that b > 0. Furthermore, we point out that it might be viewed as the background solution for the phenomenon: supersonic flow past a blunt body, (see [4]). In fact, for the pure supersonic shock case, by the discussion in Sect. 4, one can easily prove that the solution of the Main Problem approximates this special solution as u → q¯ (see (4.10)). 3. Characteristic Decomposition Characteristic decomposition: The characteristic decomposition is first introduced in [9] and then utilised to treat the nonlinear 2 × 2 hyperbolic boundary value problem with monotonous properties (see [8, 10, 11, 12]). This approach is powerful and direct. It avoids the discussion of hodograph transformation. In this section, we briefly state the characteristic decomposition used in the proof. The calculation is omitted and can be derived analogously as that in [8]. We rewrite (1.1) in the following non-conservation form ⎧ ⎨(c2 − u2 )u − uv(u + v ) + (c2 − v 2 )v = 0, x y x y (3.1) ⎩uy − vx = 0, it yields the matrix form −2uv u 2 2 + c −u −1 v x
u = 0. 0 v
c2 −v 2 c2 −u2
y
(3.2)
THE SUPERSONIC FLOW PAST A WEDGE WITH LARGE CURVED BOUNDARY
5
If u2 + v 2 > c2 , this is a hyperbolic system with two genuinely nonlinear characteristics √ uv ± c u2 + v 2 − c2 . λ± := u2 − c 2 Thus, we may multiply (3.2) with l± = (1, λ∓ ) on the left and rewrite it in the characteristic form ∂ ± u + λ∓ ∂ ± v = 0
(3.3)
where ∂ ± = ∂x + λ± ∂y . In the following discussion, we normalize the derivatives ∂ ± by defining ∂¯± as follows ∂¯+ := cos α∂x + sin α∂y , tan α := λ+ ,
∂¯− := cos β∂x + sin β∂y ,
tan β := λ− ,
ν := (γ + 1)/[2(γ − 1)],
ω := (α − β)/2,
Ω := m − tan2 ω,
∂¯0 := cos τ ∂x + sin τ ∂y ,
τ := (α + β)/2,
m := (3 − γ)/(γ + 1).
As a result, (3.3) is converted into ∂¯± u + λ∓ ∂¯± v = 0.
(3.4)
In view of ∂¯± u = −λ∓ ∂¯± v, for any function f depending only on u and v, we have ∂¯± f = fu ∂¯± u + fv ∂¯± v = ∂¯± v. (−λ fu + fv ) ∓ depending only on u and v According to the above reduction, if a quantity is a function of u and v, for example c and λ± , then its ∂¯± derivatives can be expressed in terms of (u, v)’s functions multiplying with ∂¯± v. Thus we may express the derivatives ∂¯± of all the quantities via the ∂¯± v, and vice versa. For convenience of the computation, we express these terms by ∂¯± c. Therefore, we have the following expressions ⎧ ⎪ c∂¯+ β = −2ν tan ω ∂¯+ c, c∂¯+ α = −νΩ sin(2ω)∂¯+ c, ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ c∂¯− α = 2ν tan ω ∂¯− c, c∂¯− β = νΩ sin(2ω)∂¯− c, ⎪ ⎪ ⎪ ⎪ ⎨c∂¯± ω = κ−1 sin2 ω + 1 tan ω ∂¯± c, (3.5) ⎪ ⎪ ∂¯+ u = κ−1 sin β ∂¯+ c, ∂¯+ v = −κ−1 cos β ∂¯+ c, ⎪ ⎪ ⎪ ⎪ ⎪ ∂¯− u = −κ−1 sin α∂¯− c, ∂¯− v = κ−1 cos α∂¯− c, ⎪ ⎪ ⎪ ⎪ ⎩c∂¯+ τ = − sin(2ω) ∂¯+ c, c∂¯− τ = sin(2ω) ∂¯− c. 2κ 2κ Secondly, for any function I, we have the Lie derivative relation ∂¯+ ∂¯− I − ∂¯− ∂¯+ I =
1 [(∂¯+ β − cos(2ω)∂¯− α)∂¯+ I − (cos(2ω)∂¯+ β − ∂¯− α)∂¯− I]. sin(2ω) (3.6)
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DIAN HU
Let I = u and I = v. Since ∂¯± u and ∂¯± v can be expressed by ∂¯± c in (3.5), (3.6) yields two equations for ∂¯− ∂¯+ c and ∂¯+ ∂¯− c. Solve the equation of these, we could express ∂¯− ∂¯+ c and ∂¯+ ∂¯− c in terms of ∂¯± c as follows ν(tan2 ω − 1)2 + 2 tan2 ω ¯− ∂ c , c∂¯− ∂¯+ c = ∂¯+ c ν(1 + tan2 ω)∂¯+ c + tan2 ω + 1 ν(tan2 ω − 1)2 + 2 tan2 ω ¯+ ∂ c . c∂¯+ ∂¯− c = ∂¯− c ν(1 + tan2 ω)∂¯− c + tan2 ω + 1
(3.7)
Then, for the solid boundary value condition on the wall (1.4), we have the curvature expression κb :=
¯− ¯+ b ¯0 (τ ) = ∂ τ + ∂ τ . = ∂ (1 + (b )2 )3/2 2 cos ω
Owing to (3.5), it can be reduced to the equation of ∂¯± c as follows ∂¯− c − ∂¯+ c = (γ − 1)qκb .
(3.8)
Finally, for the states along the shock, we have the R-H condition G(u, v) := [ρu][u] + [ρv][v] = 0.
(3.9)
We intend to derive the ∂¯± c relation from the above equation. In fact, let s = arctan φ and define the derivative along the shock by ∂¯s := cos s∂x + sin s∂y = t+ ∂¯+ + t− ∂¯− where t+ =
sin(β−s) sin(β−α)
and t− =
sin(s−α) . sin(β−α)
Using ∂¯s on (3.9), we have
Gu ∂¯s u + Gv ∂¯s v = 0.
(3.10)
Noting that ⎧ ⎨∂¯s u = t ∂¯+ u + t ∂¯− u = κ−1 (t sin β ∂¯+ c − t sin α∂¯− c), + − + − s + − −1 + ⎩∂¯ v = t+ ∂¯ v + t− ∂¯ v = −κ (t+ cos β ∂¯ c − t− cos α∂¯− c), we insert the above equations into (3.10) and convert it into the equation of ∂¯± c as follows t+ (Gu sin β − Gv cos β)∂¯+ c + t− (−Gu sin α + Gv cos α)∂¯− c = 0. u , where k is the angle of the tangent line on the shock polar Let tan k := − G Gv G(u, v) = 0. In the above equation, we have the shock relation for ∂¯± c
sin(β − s) cos(β − k)∂¯+ c + sin(α − s) cos(α − k)∂¯− c = 0.
(3.11)
Characteristic decomposition of the special solution: We calculate the shock relation (3.11) for the special solution introduced in Sect. 2. Noting that
THE SUPERSONIC FLOW PAST A WEDGE WITH LARGE CURVED BOUNDARY
s = τ and 2τ =
π 2
7
+ k, we have t± =
sin ω 1 sin(−ω) = = > 0. sin(−2ω) sin(2ω) 2 cos ω
Moreover, in view of the shock polar (3.10), we have tan( π2 −2s) = tan( π2 −2τ ) = Thus, the shock relation (3.11) has been reduced to − sin α∂¯+ c + sin β ∂¯− c = 0.
Gu . Gv
(3.12)
sin β with 1. For all In order to obtain the estimate of ∂¯± c, we need to compare sin α 2 2 2 the points on the shock polar (2.2), which satisfy u + v > c and v > 0, we have α + β = 2τ ∈ (0, π) and α − β = 2ω ∈ (0, π). These yields −1 <
sin β < 1. sin α
(3.13)
In fact, since α = τ + ω ∈ (0, π), we have sin α > 0. Thus sin α + sin β = 2 sin
α−β α+β cos = 2 sin τ cos ω > 0, 2 2
and sin α − sin β = 2 cos τ sin ω > 0. Above three indicate (3.13). Moreover, if (u, v) is on the shock polar (2.2) and v > 0, then one can calculate that sin β sin β = −1, and lim < 1. (3.14) lim 2 2 2 2 2 2 u +v →c +0 sin α u +v →¯ q −0 sin α
4. Proof of Theorem 1.1 Now, to consider a perturbation of the special solution, we assume that the velocity of coming flow is less than q¯. Indeed, we have u > c and ρu = > 0. Thus the shock polar can be reduced to G (u, v) := (u − u)(u − ) + v 2 = 0, ρ
(4.1)
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DIAN HU
cos(k−α) which is a perturbation of (2.2). For such a case, sin(s−α) is a perturbation of sin(β−s) cos(k−β) sin β . Owing to (3.13), by choosing sufficiently small, we can confirm that, for all sin α the supersonic points (u, v) along (4.1) satisfying 2 v ¯ and v > 0, (4.2) 0 < b − δ0 ≤ ≤ b + δ0 < u γ−1
we have −1 < Km <
sin(s − α) cos(k − α) < KM < 1. sin(β − s) cos(k − β)
sin(s − α) cos(k − α) (4.3) sin(β − s) cos(k − β) ≤ ms < 1. where δ0 << 1, Km , KM , ms are constants. Consider the special solution, the map x → (us , vs ) defines a curve C0 on the (u, v) plane, which is all the supersonic points on the shock polar (2.2) satisfying v = b (x) and v > 0 for some x ≥ 0. (4.4) u Let C0δ be its δ neighbourhood, where δ << 1. We intend to show that there is a constant M > 0, which is determined by the special solution, such that the solution (u, v) can be solved in C0δ , the derivative ∂¯± c satisfy the estimates ± M ∂¯ c < . (4.5) (1 + x)1+a In other words
Proof. First, by the result in [7], we can solve the Main Problem around the origin O. Then, we claim that, C0δ and the estimates (4.5) are invariant for properly chosen constant δ and M . In fact, by choosing small enough, all the points on the shock polar (4.1) satisfying (4.4) must belong to C0δ . Thus we have (u, v) ∈ C0δ and ∂¯± c satisfy (4.5) around the origin. Assume P (x0 , y0 ) is the first point breaking the claim above. Then we have (u, v) ∈ C0δ and ± M ∂¯ c ≤ , (1 + x)1+a for all 0 ≤ x ≤ x0 . Owing to (u, v) ∈ C0δ , we have ρu > Ms > 0 for 0 ≤ x ≤ x0 .
(4.6)
By the conservation of mass, we have φ(x) ρuφ(x) = ρ(x, y)u(x, y)dy.
(4.7)
b(x)
Noting that the derivative φ (x) is determined by C0δ , we have ρuφ(x) ≤ Ms x.
THE SUPERSONIC FLOW PAST A WEDGE WITH LARGE CURVED BOUNDARY
9
So, by (4.7), we combine above with (4.6) to have Ms (φ(x) − b(x)) ≤ Ms x.
(4.8)
φ(x) − b(x) ≤ Ms x.
(4.9)
It yields Then, in view of (3.5) and (u, v) ∈ C0δ , we have |(u, v)(x0 , y0 ) − (u, v)(x0 , φ(x0 ))| + |(u, v)(x0 , y0 ) − (u, v)(x0 , b(x0 ))| ≤
sup
b(x0 )≤y≤φ(x0 )
≤ Ms
sup
|∇(u, v)(x0 , y)| |φ(x0 ) − b(x0 )|
b(x0 )≤y≤φ(x0 )
≤ M x0
+ (∂¯ c(x0 , y), ∂¯− c(x0 , y)) |φ(x0 ) − b(x0 )|
Ms ≤ M Ms (1 + x0 )1+a
On the other hand, we have G (u, v)(x0 , φ(x0 )) = 0 and
v ( )(x0 , b(x0 )) = b (x0 ). u Noting (4.1) is a perturbation of (2.2), we summarise the above three equations to have |(u, v)(x0 , y0 ) − (us , vs )(x0 , b(x0 ))| ≤ M Ms .
(4.10)
where Ms is a constant depending on C0 for δ, << 1. In view of the expression (3.7), we have ± ∓ M Ms ∂¯ ∂¯ c ≤ . (1 + x)2+2a For the point (x0 , y0 ) in Ω, it issues a −characteristic C1− , which finally hits the shock x1 )). Then the point (¯ x1 , φ(¯ x1 )) connects the point (x1 , b(x1 )) on the point (¯ x1 , φ(¯ + xn , φ(¯ xn )) on via a +characteristic C1 . Thus, by this procedure, we obtain a series (¯ ± ± ¯ ¯ the shock S and a series (xn , b(xn )) on the wall W. Denote ∂ c(x0 ) := ∂ c(x0 , y0 ), ∂¯± c(xn ) := ∂¯± c(xn , b(xn )) and ∂¯± c(¯ xn ) := ∂¯± c(¯ xn , φ(¯ xn )). We point out that owing to the narrow estimate (4.9), the length of the characteristic between S and W can be estimated by Ms x. For example, in Figure 4, the length of C1− can be estimated by Ms x1 . Thus integrating along the characteristic and noting the narrow condition (4.9), we have + ∂¯ c(x0 ) − ∂¯+ c(¯ x1 ) <
M Ms , (1 + x1 )1+2a − M Ms ∂¯ c(¯ x1 ) − ∂¯− c(x1 ) < . (1 + x1 )1+2a
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DIAN HU
S
(¯ x1 , φ(¯ x1 ))
C1−
(x0 , y0 )
W
•
C1+ −
(¯ x2 , φ(¯ x2 )) C2
C2+
(x1 , b(x1 ))
(x2 , b(x2 ))
Figure 2. The characteristic between the shock and cone On the other hand, owing to wall relation (3.8) and shock relation (3.11), we have ∂¯− c(x1 ) = ∂¯+ c(x1 ) + (γ − 1)q(x1 )κb (x1 ), and
sin(s − α) cos(k − α) ¯− ∂ c(¯ x1 ) = x1 ). ∂¯+ c(¯ sin(β − s) cos(k − β) Combining the above four, we have + M Ms ∂¯ c(x0 ) ≤ ms ∂¯+ c(x1 ) + ms Ms |κb (x1 )| + ms M Ms + , (1 + x1 )1+2a (1 + x1 )1+2a
which can be rewritten as + ∂¯ c(x0 ) ≤ ms ∂¯+ c(x1 ) + Ms (B + M ) . (1 + x1 )1+a Furthermore, we have + ∂¯ c(xn−1 ) ≤ ms ∂¯+ c(xn ) + Ms (B + M ) . (1 + xn )1+a
(4.11)
Note that the narrow condition (4.9) yields (1 − Ms )xn−1 ≤ xn ≤ xn−1 . Owing to (4.3), by choosing sufficiently small and arguing (4.11) by induction, we deduce + ∂¯ c(x0 ) ≤ Ms (B + M ) . (1 + x0 )1+a Similarly, we have − ∂¯ c(x0 ) ≤ Ms (B + M ) . (1 + x0 )1+a
THE SUPERSONIC FLOW PAST A WEDGE WITH LARGE CURVED BOUNDARY
11
Therefore, let M = Ms (B+2), then by chosen small enough, we have (u, v)(x0 , y0 ) ∈ C0δ and ± M ∂¯ c(x0 , y0 ) < . (4.12) (1 + x0 )1+a This confirms the claim. As a result, we complete our proof of Theorem 1.1.
Acknowledgments Dian Hu was supported by NSF of China (11401209) and Fundamental Research Funds for the Central Universities No. 222201514323.
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DIAN HU
D. Hu: School of Sciences, East China University of Science and Technology, Shanghai, 200237, China. E-mail address:
[email protected]