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Detection of glaucoma using retinal fundus images: A comprehensive review

  • Received: 12 December 2020 Accepted: 01 February 2021 Published: 02 March 2021
  • Content-based image analysis and computer vision techniques are used in various health-care systems to detect the diseases. The abnormalities in a human eye are detected through fundus images captured through a fundus camera. Among eye diseases, glaucoma is considered as the second leading case that can result in neurodegeneration illness. The inappropriate intraocular pressure within the human eye is reported as the main cause of this disease. There are no symptoms of glaucoma at earlier stages and if the disease remains unrectified then it can lead to complete blindness. The early diagnosis of glaucoma can prevent permanent loss of vision. Manual examination of human eye is a possible solution however it is dependant on human efforts. The automatic detection of glaucoma by using a combination of image processing, artificial intelligence and computer vision can help to prevent and detect this disease. In this review article, we aim to present a comprehensive review about the various types of glaucoma, causes of glaucoma, the details about the possible treatment, details about the publicly available image benchmarks, performance metrics, and various approaches based on digital image processing, computer vision, and deep learning. The review article presents a detailed study of various published research models that aim to detect glaucoma from low-level feature extraction to recent trends based on deep learning. The pros and cons of each approach are discussed in detail and tabular representations are used to summarize the results of each category. We report our findings and provide possible future research directions to detect glaucoma in conclusion.

    Citation: Amsa Shabbir, Aqsa Rasheed, Huma Shehraz, Aliya Saleem, Bushra Zafar, Muhammad Sajid, Nouman Ali, Saadat Hanif Dar, Tehmina Shehryar. Detection of glaucoma using retinal fundus images: A comprehensive review[J]. Mathematical Biosciences and Engineering, 2021, 18(3): 2033-2076. doi: 10.3934/mbe.2021106

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  • Content-based image analysis and computer vision techniques are used in various health-care systems to detect the diseases. The abnormalities in a human eye are detected through fundus images captured through a fundus camera. Among eye diseases, glaucoma is considered as the second leading case that can result in neurodegeneration illness. The inappropriate intraocular pressure within the human eye is reported as the main cause of this disease. There are no symptoms of glaucoma at earlier stages and if the disease remains unrectified then it can lead to complete blindness. The early diagnosis of glaucoma can prevent permanent loss of vision. Manual examination of human eye is a possible solution however it is dependant on human efforts. The automatic detection of glaucoma by using a combination of image processing, artificial intelligence and computer vision can help to prevent and detect this disease. In this review article, we aim to present a comprehensive review about the various types of glaucoma, causes of glaucoma, the details about the possible treatment, details about the publicly available image benchmarks, performance metrics, and various approaches based on digital image processing, computer vision, and deep learning. The review article presents a detailed study of various published research models that aim to detect glaucoma from low-level feature extraction to recent trends based on deep learning. The pros and cons of each approach are discussed in detail and tabular representations are used to summarize the results of each category. We report our findings and provide possible future research directions to detect glaucoma in conclusion.



    In this paper, we are concerned with the motion of the supersonic potential flow in a two-dimensional expanding duct denoted by Ω (See Fig. 1), which is bounded by the lower wall Γlow={(x,y)|y=f(x),0x<+}, the upper wall Γup={(x,y)|y=f(x),0x<+} and the inlet Γin={(x,y)|x=φ(y),y[f(0),f(0)]}. Here we assume that f(x)C2([0,+)) satisfies

    f(0)>0,f(x)>0,f(x)0,f=limx+f(x)exists, (1)
    Figure 1.  Supersonic flow in 2D convex duct.

    and φ(y)C2([f(0),f(0)]) is an even function which satisfies

    φ(y)=φ(y),φ(±f(0))=0,φ(±f(0))=f(0),φ0. (2)

    At the inlet, the flow velocity is assumed to be along the normal direction of the inlet and its speed is given by q0(y)C1[f(0),f(0)]. Moreover, we require that q0(y) satisfies

    c<c1<q0(y)<c2<ˆq, (3)

    where c1,c2 are positive constants, c is the critical speed of the flow and ˆq is the limit speed of the flow. This means the coming flow is supersonic and does not meet vacuum at the inlet. On the two walls, the flow satisfies the solid wall condition, namely

    vu=tanθ=±f(x),ony=±f(x), (4)

    where θ=arctanvu is the angle of the velocity inclination to the xaxis.

    The supersonic flow in the duct is described by the 2-D steady isentropic compressible Euler equations:

    {x(ρu)+y(ρv)=0,x(ρu2+p)+y(ρuv)=0,x(ρuv)+y(ρv2+p)=0, (5)

    where (u,v),p, and ρ stand for the velocity, pressure, and density of the flow. For the polytropic gas, the state equation is given by p=Aργ, where A is a positive constant and γ>1 is the adiabatic exponent. In addition, the gases are assumed as irrotational. Thus the components (u,v) of the velocity satisfy

    uy=vx, (6)

    then for polytropic gas, the following Bernoulli law holds

    12q2+c2γ1=12ˆq2, (7)

    where q=u2+v2 is the speed of the flow, c=p(ρ) is the sound speed, and ˆq is the limit speed, which is an identical constant over the whole flow. Therefore the density ρ can be expressed by the function of q and the system (5) can be reduced to a 2×2 system with variables (u,v)

    {(c2u2)uxuv(uy+vx)+(c2v2)vy=0,uyvx=0, (8)

    Such problem (8) with initial data (3) and boundary condition (4) has already been studied by Wang and Xin in [25]. In their paper, by introducing the velocity potential φ and stream function ψ, in terms of hodograph transformation, they proved the global solution with vacuum in the phase space (φ,ψ). Inspired by their paper, in order to understand this problem more intuitively, we establish the global existence of such problem in (x,y)-plane, and vacuum will appear in some finite place adjacent to the boundary of duct. In addition, we prove that the vacuum here is not the so-called physical vacuum.

    For better stating our results, we firstly give the description of domains Ωnon and Ωvac, which are the domain before the vacuum appearance and the domain adjoining the vacuum respectively (see Fig. 2 below). Let M and N stand for the first vacuum point on ΓupandΓlow, where xM=xN. Choose a curve whose normal direction coincides with the velocity of the incoming flow at x=xM. Denote it as x=ψ(y) with ψ(±f(xM))=xM. Set Ωnon as the open region bounded by Γin,Γup,Γlowandx=ψ(y). Denote lupandllow as the vacuum boundaries adjacent to ΓupandΓlow respectively. Set Ωvac as the open region bounded by x=ψ(y),lupandllow.

    Figure 2.  A global smooth solution with vacuum in 2D convex duct.

    Our main results in the paper are:

    Theorem 1.1. Assume that fC2([0,+)) satisfies (1), φC2([f(0),f(0)]) satisfies (2) and q0C1[f(0),f(0)] satisfies (3). If q0 satisfies

    |q0(y)|φ(y)1+(φ(y))2q0cq20c2onΓin, (9)

    then two alternative cases will happen in the duct, one contains vacuum, and the other does not. More concretely, the two cases are as follows:

    (i) When vacuum actually appears in the duct, there exists a global solution (u,v)C(¯ΩnonΩvac) C1(¯ΩnonΩvac{lup,llow}) to the problem (8) with (3) and (4). Moreover,

    nc2=0,onlupllow{M,N}, (10)

    here n stands for the normal derivative of vacuum boundary. This means the vacuum here is not the physical vacuum.

    (ii) If vacuum is absent in the duct, the problem (8) with (3) and (4) has a global solution (u,v)C1(ˉΩ).

    Theorem 1.2. Assume that f(x)C2([0,+)) satisfies:

    f(0)>0,f(0)>0,f0,f(x)0. (11)

    If

    |q0(y)|>φ(y)1+(φ(y))2q0cq20c2onΓin, (12)

    then the C1 solution to the problem (8) with (3) and (4) will blow up at some finite location in the duct.

    Remark 1. The condition (9) is equivalent to that the two Riemann invariants are both monotonically increasing along Γin (see Lemma 2.1). It means the gases are spread at the inlet. In fact, this condition is very important to get the global existence of solution. In order to understand this, we have a glimpse of the Cauchy problem for the Burgers equation

    {tu+uxu=0,u(0,x)=u0(x). (13)

    As well known, if the initial data satisfies

    minxRu0(x)<0, (14)

    then the solution u(t,x) must blow up in finite time, and shock will be formed. On the other hand, by introducing the Riemann invariants R±=θ±F(q) with F(q)=q2c2qcdq, the equation (8) is actually equivalent to (20) (see Section 2). Thus, we consider the global solution of following Cauchy problem for (20) with initial data

    {(x+λy)R+=0,(x+λ+y)R=0,R±(0,y)=R0±(y). (15)

    By the results of [16] and [30], there exists a global solution of the Cauchy problem (15) if and only if the two Riemann invariants R0±(y) are both monotonically increasing. Therefore, it seems that posing the condition (9) is reasonable to obtain the global solution. As a contrast, in Theorem 1.2, (12) means that at least one Riemann invariant is strictly monotonic decreasing along some part of the inlet. Thus, the gases have some local squeezed properties at the inlet. Similar to the case in Burgers equation, we prove that the C1 smooth solution to the problem must blow up at some finite location in the straight duct. Actually, for different initial data and structures of boundary, more totally different motions of gases can be found in [7].

    Remark 2. By the effect of expanding duct, as far as we know, it is hard to give a necessary and sufficient condition to ensure that vacuum must form at finite location in the duct. In Proposition 4.1, we will give a sufficient condition such that the vacuum will appear in finite place.

    Remark 3. For the M-D compressible Euler equations, if the gases are assumed irrotational, by introducing the velocity potential φ=u, the Euler system can be changed into a quasi-linear hyperbolic equation (i.e. potential flow equation). It is easy to check that this potential flow equation does not fulfill the "null condition" put forward in [1], [2] and [14]. Thus, in terms of the extensive results of [1], [2] and so on, the classical solution will blow up. On the other hand, if the rotation is involved, in the general case, due to the possible compression of gases, the smooth solutions will blow up and the shock is formed (see [6], [20], [21] and [23]). Meanwhile, if the gases are suitably expanded or expanded into the vacuum, the global solutions can exist(see [4], [5], [10], [11], [22], [24] and [27]-[29]).

    Remark 4. If the initial data contains a vacuum, especially for physical vacuum, the local well-posedness results of the compressible Euler equations have been studied in [8], [9], [12], [13], [18] and so on. But in general, such a local classical solution will blow up in finite time as shown in [3], [26] and the references therein.

    Remark 5. For the case that Riemann invariants are both constants on Γin which is a straight segment vertical to the velocity of the incoming flow, the problem has been solved by Chen and Qu in [5]. Here we extend their work to a more general case.

    Remark 6. The symmetry of φ(y) in (2) and the duct with respect to xaxis is not essential, by the same analysis as in this paper but more tedious computation, the results can be extended to the non-symmetric case. So for the readers' convenience, we only consider the symmetric case.

    The paper is organized as follows. In Section 2, we give some basic structures of the steady plane isentropic flow and discuss the monotonicity of two Riemann invariants R± along the inlet. In Section 3, we use the method of the characteristics to divide the duct into several inner and border regions, and the problem is transformed into some Goursat problems in the corresponding inner regions and some boundary value problems in the corresponding border regions. Then the global C1 solution is obtained before vacuum forms if R± are monotonically increasing along the inlet. Meanwhile, for the case that at least one of R+ or R is strictly monotonic decreasing along some part of the inlet, we show that the C1 solution to the problem will blow up at some finite location in the straight duct. In Section 4, we solve the problem when vacuum appears and obtain that the vacuum here is not the so-called physical vacuum. Combining the results in Section 3 and Section 4, we finally get Theorem 1.1 and Theorem 1.2. At last, We will give a sufficient condition concerning the geometric shape of the duct to ensure the formation of the vacuum.

    In this section, we start with some basic structures of the steady plane isentropic flow, which can be characterized by the Euler system (5).

    Firstly, equation (8) can be written as the matrix form

    (c2u2uv01)x(uv)+(uvc2v210)y(uv)=0. (16)

    The characteristic equation of (16) is

    (c2u2)λ2+2uvλ+(c2v2)=0.

    So for the supersonic flow, (8) is hyperbolic and we can get two eigenvalues

    λ+=uv+cu2+v2c2u2c2,λ=uvcu2+v2c2u2c2. (17)

    Correspondingly, the two families of characteristics in the (x,y) plane are defined by

    dy±dx=λ±. (18)

    By standard computation, the Riemann invariants R± can be defined by

    R±=θ±F(q), (19)

    where F(q)=q2c2qcdq. Since the Jacobian (R+,R)(u,v)=2q2c2q2c>0, then the system (8) without vacuum is equivalent to the following equations

    {(x+λy)R+=0,(x+λ+y)R=0, (20)

    which can also be written as

    ±R=0 (21)

    if we set ± as the differential operators x+λ±y.

    Let A be the Mach angle defined by sinA=c/q, then we have that

    λ±=tan(θ±A). (22)

    Under a proper coordinate rotation, the eigenvalues λ± can always be locally bounded, and the system (21) in the new coordinate is also invariant.

    From now on, we will discuss the monotonicity of the Riemann invariants R± along the inlet, which will play a key role in the following analysis. Note that θ=arctanφ(y) on Γin, then R±=arctanφ(y)±q0cq2c2qcdq on Γin. Some properties can be obtained after direct computation.

    Lemma 2.1. For R± defined on Γin above, we have that R± are both monotonically increasing along Γin if and only if the initial speed q0 satisfies

    |q0(y)|φ(y)1+(φ(y))2q0cq20c2onΓin. (23)

    Proof. Since R±=arctanφ(y)±q0cq2c2qcdq on Γin, then direct computation yields

    R±(y)=φ(y)1+(φ(y))2±q20c2q0cq0(y)onΓin. (24)

    Thus R± are monotonically increasing along Γin if and only if (24) are nonnegative, which is equivalent to (23).

    Due to Lemma 2.1, the following analysis will be focused on these two cases.

    Case I. R± are both monotonically increasing on Γin;

    Case II. At least one of R+, R is strictly monotonic decreasing along some part of Γin.

    The next lemma shows the partial derivatives λ+R and λR+ are positive, which will be often used in the following analysis.

    Lemma 2.2. As the q,c,θ,andA defined above, for the supersonic flow, we have

    λ+R>0andλR+>0. (25)

    Proof. By using (22) and the chain rule, it follows from direct computation that

    λ+R=(γ+1)q24(q2c2)sec2(θ+A),λR+=(γ+1)q24(q2c2)sec2(θA), (26)

    thus (25) holds true for the supersonic flow.

    In this section, the existence of the C1 solution to the problem before vacuum formation will be proved for case Ⅰ. We shall use characteristics to divide the whole region into several inner and border regions, where solutions are constructed successively by the method of characteristics. As illustrated in Fig. 3, we assume that the C characteristic issuing from point A1 intersects the lower wall at point B2 and the C+ characteristic issuing from point B1 intersects the upper wall at point A2. These two characteristics intersect at point C1. Denote the inner region surrounded by A1B1,A1C1,B1C1 as region D1, which is the domain of determination of A1B1.

    Figure 3.  Inner regions and border regions.

    Next we solve the solution in the border region D2(D2, resp.)bounded by A1C1,A1A2,A2C1(B1C1,B1B2,B2C1, resp.). Note that R (R+, resp.) has already been defined in region D2(D2, resp.). By the fixed wall condition R++R=2arctan(±f(x)), we can know the value of R+ (R, resp.) in region D2(D2, resp.).

    Let the C characteristic issuing from A2 and the C+ characteristic issuing from B2 intersect at C2. Define the inner region D3 with boundaries A2C1,B2C1,A2C2 andB2C2. If the C characteristic issuing from A2 approaches the lower wall at B3 and the C+ characteristic issuing from B2 approaches the upper wall at A3, we can get two border regions D4(D4, resp.) with boundaries A2A3,A2C2,A3C2 (B2B3,B2C2,B3C2, resp.). Repeat this process until the vacuum forms. The inner regions D1, D3 and the border regions D2(D2, resp.), D4(D4, resp.) are mainly considered in this section.

    Lemma 3.1. If (9) holds true, then for the C1 solution to the problem we have +R+0,R0 in the duct before vacuum forms.

    Proof. We will prove the conclusion between these inner and border regions separately.

    Step 1. Inner region D1.

    For any fixed two points E,F on Γin and any C characteristic l in region D1, suppose the C+ characteristics issuing from E,F intersect l at E1,F1 and intersect A1A2 at E2,F2 respectively. Since R is constant along the C+ characteristic and R is monotonically increasing along Γin, then the value of R at E1 is no less than the value of R at F1, which implies

    R=ddx(R(x,y(x)))0 (27)

    along l in region D1. Similarly, +R+ is nonnegative in region D1.

    Step 2. Border region D2.

    Next we consider the solution in border region D2 bounded by A1C1,A1A2,A2C1. From step 1, we know that R is monotonically decreasing along A1C1. By the same argument in region D1, we can get that R0 in region D2. Since the value of R at E2 is no less than the value of R at F2, which means that R is monotonically decreasing along A1A2. Due to the solid wall condition

    R++R=2arctanf(x),f(x)0 (28)

    on A1A2, R+ is monotonically increasing along A1A2. Therefore we can get +R+0 in region D2 by the same argument in region D1. By analogous method, we also have +R+0,R0 in region D2.

    Repeating this process in inner region D2i+1(i1) and border region D2i+2 (D2i+2, resp)(i1) until the vacuum forms, thus we prove Lemma 3.1.

    We can ensure that the hyperbolic direction is always in the x-direction by a proper rotation of coordinates if necessary. Without loss of generality, we may assume that u2>c2, which implies that λ+>λ.

    Lemma 3.2. If (9) holds true, then we have yR±0 in the duct before vacuum forms.

    Proof. In terms of the definition of ±, one has

    x=λ+λ+λ+λ,y=+λ+λ. (29)

    Thus we have

    yR+=+R+λ+λ,yR=Rλ+λ. (30)

    Combining this with Lemma 3.1 yields the result.

    Corollary 1. Under the assumption that u2>c2, by (21) and (29), we have that

    |+R+|=λ+λλ2+1|R+|and|R|=λ+λλ2++1|R|. (31)

    Lemma 3.3. Let l=ux+vy stand for the derivative along the streamline, under the assumption that u2>c2, then the speed of the flow is monotonically increasing along the streamline, that is to say lq0.

    Proof. It follows from (29) that l=ux+vy can be expressed as

    l=vuλλ+λ++uλ+vλ+λ. (32)

    which implies that

    lq=vλuλ+λ+q+λ+uvλ+λq=u(vuλ)λ+λ+q+u(λ+vu)λ+λq. (33)

    Since u2>c2, we know that θ±A is away from ±π2. Thus the coefficients of the right hand part of (33) are positive. So it remains to determine the symbol of ±q. Due to (20) and Lemma 3.1, one has that +R+=+(R+R)=2F(q)+q0, which immediately indicates that +q0. Similarly, one can get that q0. Then the proof of Lemma 3.3 is completed.

    Corollary 2. Combining Lemma 3.3 and the Bernoulli law yields that the sound speed is monotonically decreasing along the streamline, that is to say lc0.

    We now focus on the solution in the inner regions D1 and D2i1(i2) bounded by AiCi1,BiCi1,AiCiandBiCi. We only need to discuss the region D2i1(i2), since same method can be used in region D1.

    Figure 4.  Goursat problem in inner region.

    Set R+(x,y)=W+(x,y) on AiCi1 and R(x,y)=W(x,y) on BiCi1. Thus the solution in the region D2i1 should be determined by solving a Goursat problem, which is given by

    {(x+λy)R+=0,(x+λ+y)R=0,inD2i1R+(x,y)=W+(x,y)onAiCi1,R(x,y)=W(x,y)onBiCi1. (34)

    By standard iteration method, the local existence of the problem (34) can be achieved, one can also see [4] and [15] for more details. In order to get the global solution to the problem (34) in the region D2i1(i2), some prior estimates about R±andR± are needed.

    Lemma 3.4. If (9) holds true, then |+R+| is monotonically decreasing along C characteristics and |R| is monotonically decreasing along C+ characteristics before vacuum formation.

    Proof. It comes from the definition of ± that

    +R+=+R++R+=(λ++λ)yR+. (35)

    Due to

    λ++λ=λ+RRλR++R+0 (36)

    and yR+0, then +R+0, which implies +R+ is monotonically decreasing along C characteristics. By the same argument, we can get that R is monotonically increasing along C+ characteristics.

    In terms of Lemma 3.1, one has +R+0 and R0. This means that |+R+| is monotonically decreasing along C characteristics and |R| is monotonically decreasing along C+ characteristics before vacuum formation. Thus we complete the proof.

    Lemma 3.5. Suppose that R±(x,y) are the classical solutions to the problem (34), then R±and±R± are uniformly bounded in the region D2i1(i2) if (9) holds true.

    Proof. Firstly, we give the estimates of |R±|.

    For any point (x,y) in D2i1(i2), by the method of characteristics, we have that

    |R+(x,y)|W+C(AiCi1),|R(x,y)|WC(BiCi1) (37)

    in the region D2i1(i2). To be more specific, we shall point out that W+(x,y)C(AiCi1) and W(x,y)C(BiCi1) listed above only depend on fC1 and R±C(Γin).

    Fix (x0,y0)AiCi1, let y=y(1)(x) be the C characteristic from (x0,y0), which approaches either Γup or Γin at a point (x1,y1). If y1=f(x1), then there exists a C+ characteristic y=y(2)+(x) from (x1,y1), which approaches either Γlow or Γin at a point (x2,y2). If y2=f(x2), then there exists a C characteristic y=y(3)(x) from (x2,y2), which approaches either Γup or Γin at a point (x3,y3). Since the two eigenvalues λ± have the uniform upper or lower bound, then there exists a positive N such that

    x0>x1>...>xN1>xN,xN=φ(yN). (38)

    If y(xN1)=f(xN1) (see Fig. 5), then we get

    W+(x0,y0)=2arctanf(x1)R(x1,y1)=2arctanf(x1)R(x2,y2)=2arctanf(x1)+2arctanf(x2)+R+(x2,y2)=Ni=12arctanf(xi)R(xN,yN). (39)
    Figure 5.  The case that y(xN1)=f(xN1).

    Similarly, if y(xN1)=f(xN1) (see Fig. 6), then we get

    W+(x0,y0)=Ni=12arctanf(xi)+R+(xN,yN). (40)
    Figure 6.  The case that y(xN1)=f(xN1).

    Combining (39) and (40) yields that

    W+C(AiCi1)2Narctanf+max{R+C(Γin),RC(Γin)}. (41)

    Analogously, the estimate of WC(BiCi1) can also be obtained as (41).

    Next, we give the estimates of |±R±|.

    It follows from Lemma 3.4 that

    |+R+(x,y)|+W+C(AiCi1),|R(x,y)|WC(BiCi1). (42)

    As the process shown in the estimates of |R±|, we shall point out that +W+C(AiCi1)andWC(BiCi1) listed above only depend on fC2andR±C1(Γin). To this end, we need to derive the boundary condition of ±R±. It follows from (32) and the solid wall condition that

    vuλλ+λ+R++uλ+vλ+λR=±2uf(x)1+(f(x))2,ony=±f(x). (43)

    By the process of the reflection shown in the estimates of |R±| and Lemma 3.4, if y(xN1)=f(xN1), we can get

    +W+(x0,y0)+R+(x1,y1)=(2uλ+λvuλf1+(f)2uλ+vvuλR)(x1,y1)(2uλ+λvuλf1+(f)2)(x1,y1)(uλ+vvuλ)(x1,y1)×R(x2,y2)=(2uλ+λvuλf1+(f)2)(x1,y1)+(uλ+vvuλ)(x1,y1)×(2uλ+λuλ+vf1+(f)2)(x2,y2)+(uλ+vvuλ)(x1,y1)×(vuλuλ+v)(x2,y2)×+R+(x2,y2) (44)

    Since u2>c2, then the coefficients

    2uf1+(f)2,uλ+vvuλ,λ+λvuλ,λ+λuλ+v

    are uniformly bounded. Denote the maximum one of these bounds as the constant G, thus (44) can be expressed as

    +W+(x0,y0)+R+(x1,y1)G2GR(x1,y1)G2GR(x2,y2)G2+G3+G2+R+(x2,y2)Ni=1Gi+1+GN+R+(xN,yN). (45)

    Similarly, if y(xN1)=f(xN1), then we also have

    +W+(x0,y0)Ni=1Gi+1GNR(xN,yN). (46)

    Combining (45), (46) and (31) yields that

    +W+C(AiCi1)Ni=1Gi+1+GNmax{+R+C(Γin),RC(Γin)}. (47)

    Analogously, the same estimate of WC(BiCi1) holds true.

    Thus, we complete the proof of Lemma 3.5.

    Corollary 3. Based on Lemma 3.5, we have that |R±|M, where M depends on fC2,R±C1(Γin) and the lower bound of the sound speed c in D2i1(i2).

    Proof. Substituting (31) into (47) gives that

    |R±|=λ2+1λ+λ|±R±|1c(Ni=1Gi+1+GNmax{R+C1(Γin),RC1(Γin)}). (48)

    Combining the local existence and the Corollary 3 induces the global C1 solution to the problem (34) in the inner region D2i1(i2) before vacuum forms.

    We now turn to consider the border region D2i(i1), which is bounded by a part of the upper wall AiAi+1, the C characteristic AiCi, and the C+ characteristic Ai+1Ci (see Fig. 7). Then the problem in region D2i(i1) can be generalized as the following problem

    {(x+λy)R+=0,(x+λ+y)R=0,inD2iR(x,y)=Z(x,y)onAiCi,R++R=2arctanf(x)onAiAi+1, (49)
    Figure 7.  Solution in border region.

    where Z(x,y) is a known function. The local existence of the problem (49) can be achieved by [15]. In order to get the global solution to the problem (49) in the region D2i(i1), some prior estimates about |R±|and|R±| are needed.

    Lemma 3.6. Suppose that R±(x,y) are the classical solution to the problem (49), then R±and±R± are uniformly bounded in the region D2i(i1).

    Proof. For any point (x,y) in D2i(i1), by the method of characteristics and the solid wall condition, we have that

    |R(x,y)|ZC(AiCi),|R+(x,y)|2arctanf+ZC(AiCi). (50)

    in the region D2i(i1). It follows from Lemma 3.4 that

    |R(x,y)|ZC(AiCi),|+R+(x,y)|+R+C(AiAi+1). (51)

    Moreover, by (43), we have that

    +R+C(AiAi+1)G2+GRC(AiAi+1)G2+GZC(AiCi). (52)

    By the same process shown in Lemma 3.5, we know that ZC(AiCi) can be controlled by R±C1(Γin). And thus the proof is finished.

    Corollary 4. Based on Lemma 3.6, we have that |R±|M, where M depends on fC2,R±C1(Γin) and the lower bound of the sound speed c in D2i(i1).

    Combining the local existence and the Corollary 4 induces the global C1 solution to the problem (49) in the border region D2i(i1) before vacuum forms. The problem in the border region D2i(i1) can be solved by the same way exactly. After solving the problem in these inner and border regions successively, we have that

    Theorem 3.7. For case I, there exists a global C1 solution to the problem in (¯Ωnon{M,N}) and

    +R+0,R0in(¯Ωnon{M,N}). (53)

    Moreover, there exists a constant C which only depends on the initial data and fC2, such that

    |+R+|+|R|Cin(¯Ωnon{M,N}). (54)

    So far we have established the global C1 solution before vacuum formation for case Ⅰ, where the initial speed q0 satisfies (9). The existence of solution after vacuum formation will be established in Section 4.

    Now we begin to consider the problem in case Ⅱ. We will prove Theorem 1.2.

    Proof of Theorem 1.2. Firstly, we consider the special case f(x)=0, i.e. the duct is straight.

    Suppose that R+ is strictly monotonic decreasing along arc MN on Γin(see Fig. 8). Assume that the C characteristics issuing from M and N intersect Γlow at M1andN1 respectively. Then it follows from the same argument in Lemma 3.1 that +R+<0, which implies that yR+<0 in the region MNM1N1.

    Figure 8.  Blowup in 2D straight duct.

    Let y(0)(x) be the C characteristic from any point (x0,y0) on MN, i.e.

    {dy(0)(x)dx=λ,y(0)(x0)=φ1(x0). (55)

    Then R+ is constant along the curve y=y(0)(x), which means that R+(x,y(0)(x))=R+(x0,φ1(x0)). Differentiating the first equation in (20) with respect to y yields that

    x(R+y)+λy(R+y)+λRRyR+y=λR+(R+y)2.

    Use the second equation in (20), we can get Rx+λRy=(λλ+)Ry, i.e.

    Ry=xR+λyRλλ+.

    Substitute it into the above identity to obtain that

    x(R+y)+λy(R+y)+λRxR+λyRλλ+R+y=λR+(R+y)2. (56)

    Define H0(x):=exp{xx0xR+λyRλλ+λR(s,y(0)(s))ds}, then (56) can be transformed into

    ddx(H0(x)yR+)1=λR+H10, (57)

    where ddx stands for the derivative along the curve y=y(0)(x).

    If y=y(0)(x) intersects Γlow at point (x1,y1), then by integrating (57) from x0 to x1, we get that

    R+y(x1,y1)=yR+(x0,y0)H0(x)(1+yR+(x0,y0)x1x0λR+H10(s,y(0)(s))ds). (58)

    Since R+ is constant along y=y(0)(x), thus the term 1λλ+λR is only the function of R. If we set h0(R)=R01λλ+λRdR, then H0(x) can be expressed by

    H0(x)=exp{h0(R(x,y(0)(x))h0(R(x0,y0)}, (59)

    and H0(x) is bounded.

    Note that yR+(x0,y0)<0 and λR+>0. If we denote the positive lower bound of λR+H10 by K0, then we have that

    R+y(x1,y1)yR+(x0,y0)H0(x)(1+yR+(x0,y0)K0(x1x0)). (60)

    By the boundary condition (43), one has that

    yR(x1,y1)=(vuλuλ+vyR+)(x1,y1)=(cos(θ+A)cos(θA)yR+)(x1,y1). (61)

    The coefficient cos(θ+A)cos(θA)(x1,y1) is greater than 1, which implies that

    yR(x1,y1)<yR+(x1,y1). (62)

    Analogously, let y(1)+(x) be the C+ characteristic from (x1,y1) on N1M1, i.e.

    {dy(1)+(x)dx=λ+,y(1)+(x1)=y1. (63)

    one can obtain the similar equation of yR as (57) by

    ddx(H1(x)yR+)1=λ+RH11, (64)

    where H1(x)=exp{xx1xR++λ+yR+λ+λλ+R+(s,y(1)+(s))ds} and ddx stands for the derivative along the curve y=y(1)+(x). In addition, H1(x) can be written as H1(x)=exp{h1(R+(x,y(1)+(x))h1(R+(x1,y1)} where h1(R+)=R+01λ+λλ+R+dR+. If y=y(1)+(x) intersects Γup at point (x2,y2), integrating (64) from x1 to x2 yields that

    Ry(x2,y2)=yR(x1,y1)H1(x)(1+yR(x1,y1)x2x1λ+RH11(s,y(1)+(s))ds). (65)

    Note that yR(x1,y1)<0 and λ+R>0. If we denote the positive lower bound of λ+RH11 by K1 and let K1=min{K0,K1}, then we have that

    Ry(x2,y2)yR(x1,y1)H1(x)(1+yR(x1,y1)K1(x2x1)). (66)

    By the same process and after series of reflection if necessary, the denominator of (66) will tend to at some finite location in the duct. Thus the C1 solution to the problem will blow up at some finite location in the straight duct.

    Next, we consider the case f(x)0.

    The proof is very similar to the case f(x)=0, except for using following boundary condition to replace (61) by

    ((vuλ)yR+)(x1,y1)((uλ+v)yR)(x1,y1)=2u(x1,y1)f(x1)1+(f(x1))2. (67)

    Thus we have

    yR(x1,y1)=(vuλuλ+vyR+)(x1,y1)+(2uuλ+v)(x1,y1)×f(x1)1+(f(x1))2<yR+(x1,y1). (68)

    This is same as (62). So we omit the details of proof for the case f(x)0.

    Therefore, we complete the proof of Theorem 1.2.

    Due to the conservation of mass, vacuum can not appear in the inner regions of the duct. So if vacuum exists, it must locate on the two walls. On the other hand, since the two walls are the streamlines, Corollary 2 shows that the sound speed is monotonically decreasing along the two walls. When the sound speed decreases to zero, it corresponds to the formation of vacuum.

    First, we list some properties of the vacuum boundary, one can also see [4] and [25] for more details.

    Lemma 4.1. When the vacuum area actually appears, the first vacuum point M and N must form at ΓupandΓlow respectively. Furthermore, the boundary of the vacuum luporllow is a straight stream line which is tangential to ΓuporΓlow at M or N, where f(xM)=f(xN)>0.

    Based on Theorem 3.7, we have obtained the C1 solution to the problem for (x,y)(¯Ωnon{M,N}) and +R+0,R0in(¯Ωnon{M,N}). We now begin to solve the problem when vacuum forms. Set R+(x,y)=˜R+(y) and R(x,y)=˜R(y) on x=ψ(y). Then ˜R±(y)C1(f(xM),f(xM)) and ˜R±(y) are monotonically increasing along x=ψ(y). Since the problem is singular at M and N, we choose Qn+=(xn,yn),Qn=(xn,yn) on x=ψ(y) and ˜Rn±(y)C1[f(xM),f(xM)] such that

    ˜Rn±(y)=˜R±(y),y[yn,yn],f(xM)yn=1n,n=1,2,...|q(y)|ψ(y)1+(ψ(y))2qcq2c2,y[f(xM),yn][yn,f(xM)]. (69)

    Consider the problem in the region Ωvac bounded by x=ψ(y),lupandllow with initial value ˜Rn±(y). The local existence can be obtained by the method in [15]. For any point (x,y)Ωvac, the backC+ characteristic from it must intersect ΓloworΓin and the backC characteristic from it must intersect ΓuporΓin. Then by Theorem 3.7, one has that ±Rn± are uniformly bounded in Ωvac. Thus we can get the global C1 solution Rn± to the problem in Ωvac with initial value ˜Rn±(y).

    Figure 9.  Solution in Ωvac.

    Let y=yn+ stand for the C+ characteristic from (xn,yn) and y=yn stand for the C characteristic from (xn,yn). Then yn+(yn, resp.) tends to lup(llow, resp.) as n and we can set

    R±(x,y)=limnRn±(x,y),(x,y)Ωvac. (70)

    To sum up, we get that

    Theorem 4.2. Assume that ˜R±(y)C1(f(xM),f(xM)) and ˜R±(y) are monotonically increasing along x=ψ(y). Then R±(x,y) defined by (70) is the unique solution to the problem in Ωvac. Furthermore, R±(x,y)C1(¯Ωvac{lupllow}) and

    +R+0,R0in(¯Ωvac{lupllow}). (71)

    Next, we consider the regularity near vacuum boundary. In fact, there are many results about physical vacuum singularity. A vacuum boundary (or singularity) is called physical if

    0<|c2n|<+ (72)

    in a small neighborhood of the boundary, where n is the normal direction of the vacuum boundary. This definition of physical vacuum was motivated by the case of Euler equations with damping studied in [17] and [19]. But the following Theorem shows that the vacuum here is not the physical vacuum.

    Theorem 4.3. Let n stand for the normal derivative on the vacuum boundary, then for any point (x,y) near the vacuum line, we have nc2(x,y) tends to zero as (x,y) approaches to the vacuum line along the normal direction except the vacuum point M and N.

    Proof. For convenience, we only consider the case near the vacuum line llow, which is tangential to Γlow at N. First, we derive the expression of nc2, where n=(n1,n2). Acting n on the Bernoulli law (7) yields that

    nc2=(γ1)qnq. (73)

    By the relation R+R=2F(q), one has that

    xq=qc2q2c2(xR+xR),yq=qc2q2c2(yR+yR). (74)

    Substituting it into (73) gives that

    nc2=(γ1)q2c2q2c2[(n2n1λ)yR++(n1λ+n2)yR]. (75)

    Combining this with (30) yields that

    nc2=(γ1)q2(u2c2)4(q2c2)[(n2n1λ)+R+(n1λ+n2)R]. (76)

    Construct a line NNh vertical to llow at the first vacuum point N, where h represents the distance between N and Nh. Let Ch be the C characteristic through Nh and intersect Γup at Ph. Fix a point S on llow. Make a line from S vertical to llow and it intersects Ch at Sh. Then we have that NhN,ShS as h0. It follows from (35) that

    +R+=λ++λλ+λ+R+. (77)

    Integrating (77) from Ph to Sh yields that

    +R+(Sh)=+R+(Ph)exp{ShPhλ++λλ+λds}. (78)

    This together with (26) yields that

    +R+(Sh)=+R+(Ph)exp{ShPhγ+18c(T1+R++T2R)ds}, (79)

    where

    T1=q2(u2c2)sec2(θA)(q2c2)32,T2=q2(u2c2)sec2(θ+A)(q2c2)32.
    Figure 10.  The regularity near vacuum boundary.

    Since u2>c2, then T1andT2 have the uniform positive lower bound denoted as g0

    We claim that ¯limh0+R+(Sh)=0andlim_h0R(Sh)=0, which immediately implies that limh0nc2(Sh)=0. For convenience, we only prove ¯limh0+R+(Sh)=0, for the case lim_h0R(Sh)=0 can be obtained by the same argument.

    We will use the proof by contradiction to obtain above assertions. Assume that ¯limh0+R+(Sh)=ˉg>0, where ˉg is a positive constant. Then there exists a sequence Sˉgn on ShS such that +R+(Sˉgn)ˉg2>0and|NNˉgn|hn, where Nˉgn is the intersection of the C characteristic Cˉgn through Sˉgn and the segment NNh. Let the C characteristic Cˉgn intersect Γup at Pˉgn. Then it follows from Lemma 3.4 that

    +R+(x,y)+R+(Sˉgn)ˉg2on(x,y)PˉgnSˉgn. (80)

    By the fact that q0 and the Bernoulli law, we can get c0, which implies that the value of c on NˉgnSˉgn is no greater than the value of c at Nˉgn denoted by c(Nˉgn). Thus we have limnc(Nˉgn)=0 and

    SˉgnPˉgnγ+18c(T1+R++T2R)dsSˉgnNˉgnγ+18cT1+R+dsγ+116c(Nˉgn)s0g0ˉg, (81)

    where s0=|NS|. Combining it with (79) yields that

    +R+(Sˉgn)+R+(Pˉgn)exp{γ+116c(Nˉgn)s0g0ˉg}. (82)

    Note that Pˉgn is away from the vacuum as n. If limnPˉgn=P, then limn+R+(Pˉgn)=+R+(P) and +R+(P) is bounded. Then let n in (82), we can get that

    ¯limn+R+(Sˉgn)+R+(P)limnexp{γ+116c(Nˉgn)s0g0ˉg}=0, (83)

    which is a contradiction. Thus, it follows

    ¯limh0+R+(Sh)=0,lim_h0R(Sh)=0. (84)

    This together with (71) gives that

    limh0+R+(Sh)=0,limh0R(Sh)=0. (85)

    Combining it with (76) yields that

    limh0c2n(Sh)=0. (86)

    Note that c2C(ˉΩvac), by the derivative limit theorem, we have that

    c2n=0,onllow{N}. (87)

    Similarly, one has that

    c2n=0,onlup{M}. (88)

    Thus, we prove the Theorem 4.2.

    So far we have established the solution to the problem when vacuum actually appears. When it comes to the case that there is no vacuum in the duct, by the same prior estimates in Lemma 3.5 and Lemma 3.6, one only needs to check whether the duct can be covered by characteristics in finite steps.

    Theorem 4.4. Assume that fC2([0,+)) satisfies (1), φC2([f(0),f(0)]) satisfies (2) and q0C1[f(0),f(0)] satisfies (3), (9). If there is no vacuum in the duct, then there exists a global C1 solution in the duct.

    Figure 11.  The case without vacuum.

    Proof. It remains to check whether the duct can be covered by characteristics in finite steps. Since the duct is convex, then there exists i0 such that we can find an arc EF on Ai0Ai0+1 and f>0 on EF. Let the C characteristics CE,CF intersect Ci0Ai0+1 at E1,F1, intersect Bi0+1Ci0+1 at E2,F2 and intersect Bi0+1Bi0+2 at E3,F3. By Lemma 3.1 and (28), we know that +R+>0 on E1F1 and E2F2. Combining it with +R=0 yields that ~E1~F1,~E2~F2 which are the images of E1F1,E2F2 in (u,v) plane have positive length. That is to say

    |˜Ci0˜Ai0+1|>0,|˜Bi0+1˜Ci0+1|>0. (89)

    Similarly, we can get that

    |˜Ci0+1˜Bi0+2|>0,|˜Ai0+2˜Ci0+2|>0. (90)
    Figure 12.  The image in the (u,v)plane.

    By the induction method, we have that

    |˜Ci0+2(k1)˜Ai0+2k1|>0,|˜Bi0+2k1˜Ci0+2k1|>0,|˜Ci0+2(k1)˜Bi0+2k|>0,|˜Ai0+2k˜Ci0+2k|>0,k=1,2,... (91)

    Suppose that the image of the C+ characteristic through Ci0 in (u,v) plane intersects the critical circle at C0i0 and intersects the limiting circle at C1i0. Then

    C0i0OC1i0=π2μ,θ˜Bi0+1=˜Ci0O˜Bi0+1>0, (92)

    where μ=γ1γ+1. By the same method in [5], we have that

    iπ2μθ˜Bi0+1, (93)

    where . stands for the maximal integer no more than the number inside the bracket.

    Proof of Theorem 1.1. Under the assumptions of Theorem 1.1, Theorem 1.1 comes from Theorem 3.1 and Theorem 4.1-4.3.

    Finally, we give a sufficient condition to ensure that vacuum must form at finite location in the duct.

    Proposition 1. If f>f(0)+2(γ1)MB11+(f(0))2, where MB1 represents the Mach number at B1, then the vacuum will form at the finite locations on Γlow.

    Proof. We will use the proof by contradiction. Assume that there is no vacuum on Γlow. Let dds stand for the derivative along Γlow. By the relation R+R=2F(q), one has that

    dqds=d(R+R)ds12F(q)=qc2q2c2(x(R+R)+tanθy(R+R)). (94)

    This together with (29) and (43) yields that

    dqds=qcq2c2(tanθλλ+λ+R++f(x)1+(f(x))2). (95)

    By (95) and the Bernoulli law (7), we get that

    dcds=(γ1)q22q2c2(tanθλλ+λ+R++f(x)1+(f(x))2). (96)

    Since

    cB1=+0dcds1+(f(x))2dx,

    then

    cB1+0(γ1)q22q2c2f(x)1+(f(x))21+(f(0))2dxqB1(γ1)21+(f(0))2+0f(x)1+(f(x))2dx=qB1(γ1)21+(f(0))2(arctanfarctanf(0))>cB1, (97)

    which is a contradiction. Thus, we complete the proof.

    The authors would like to thank the referee for his (or her) very helpful suggestions and comments that lead to an improvement of the presentation.



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