# On Tricomi Problem of Chaplygin’s Hodograph Equation

Authors: Meng Xu; Li Liu; Hairong Yuan
Journal: Abstract and Applied Analysis
 (2015)

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## Abstract

The existence and uniqueness results for the Tricomi problem of Chaplygin’s hodograph equation are shown, in the case that the domain considered is close to the parabolic degenerate line, by adopting the energy integral methods and choosing judiciously suitable multipliers.


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## Body

## 1. Introduction

In this paper we consider the Tricomi problem of the following second-order linear partial differential equation. Consider(1)Lϕ≜Ktϕθθ+ϕtt-Ptϕt=0,where(2)Kt=t2-t1-t21-μ21-t2,Pt=t2-t1-t1-μ21-t2,and μ∈(0,1) is a constant. Equation (1) is of elliptic type for 0<t<1 and 0≤θ<2π, hyperbolic type for 1-1/μ<t<0 and 0≤θ<2π, and parabolic degenerate on the line {t=0}. We are interested in this equation because it is actually an equivalent form of Chaplygin’s hodograph equation (with Φ=Φ(u,v) as the unknown). (In this paper we will use the subscripts like ϕt and ϕtt to denote the partial derivatives ∂ϕ/∂t and ∂2ϕ/∂t2)(3)c2-v2Φuu+2uvΦuv+c2-u2Φvv=0,where the function c=c(u,v) (called sonic speed in gas dynamics) is given by the Bernoulli law [1, page 23] (4)μ2u2+v2+1-μ2c2=c∗2,with c∗ being a positive constant, μ=(γ-1)/(γ+1), and γ>1 the adiabatic exponent for polytropic gas. One can easily show that, by taking t=1-u2+v2/c∗ and θ=arctan⁡(v/u), (3) is transformed to (1), with ϕt,θ=Φ(c∗(1-t),θ) (cf. [2, page 72]).The significance of Chaplygin’s hodograph equation (3) lies in the fact that it is the hodograph transform of the following compressible Euler equations of isentropic irrotational flows:(5)vx-uy=0,ρux+ρvy=0,where ρ is the density of mass of the flow and (u,v) is the velocity of the flow along the (x,y) coordinates of the Euclidean plane. Since in this case the sonic speed c=ρ(γ-1)/2, then ρ is a function of (u,v) given by the Bernoulli law. Some fundamental problems in gas dynamics, such as detached shocks in supersonic flow past blunt bodies and subsonic jets (cf. [1, 3]), could be considered more favorably by using hodograph equation (3) (or (1)) rather than Euler equations (5), because the latter are generally a quasi-linear mixed elliptic-hyperbolic system, which is still far beyond the ability of present-day analytical tools to study.In a previous work [2], the authors have studied a mixed boundary value problem of (3) in the sonic circle {u2+v2<c∗2}, with an artificial Dirichlet boundary condition on part of the sonic line {u2+v2=c∗2}, to understand the regularity and behavior of solutions of (3) in the elliptic region and near the degenerate line. Now, we continue our project in this paper to investigate theTricomi problem of (1), that is, to find a function ϕ=ϕ(θ,t) satisfying (in certain sense to be specified later) (1) in a planar domain D which is simply connected, containing a segment of the θ-axis, and bounded by the characteristic curves (by definition, a characteristic curve of (1) satisfies equation --Kt2dt2+dθ2=0 for t<0)Γ2 and Γ3 lying in the lower half plane {t<0} and a Jordan curve Γ1 lying in the upper half plane {t>0}, with Dirichlet conditions on Γ1 and Γ3 (see Figure 1). Here (6)Γ3:θ=-∫0t-Krdr,t<0emanates from the origin O and intersects the horizontal line {t=t1} at a point P(θ1,t1), where t1<0 is sufficiently small. The characteristic curve (7)Γ2:θ=∫0t-Krdr+θ0,t<0emanates from the point P and intersects the θ-axis at a point A(θ0,0). The arc Γ1 has two endpoints O=(0,0) and A=(θ0,0). The Dirichlet conditions on Γ1 and Γ3 are, respectively,(8)ϕ=fsonΓ1,ϕ=gsonΓ3,where s is the arc-length parameter of the boundary curve ∂D=Γ1∪Γ2∪Γ3 so that the point (θ(s),t(s)) moves counterclockwise on ∂D as s increases. Then the outward unit normal along ∂D is given by n=(n1,n2)=(dt/ds,-dθ/ds). Note that one can require Γ1∪Γ3 to be piecewise smooth except at the point O, where at best the curve is C1,1/2. (We thank a referee for pointing out this fact. Here as usual we use Ck,α(Ω) to denote the Hölder space of k-times continuously differential real-valued functions on Ω whose kth order derivatives are all Hölder continuous with the exponent α∈(0,1).) Let ϕ~ be a given function in the standard Sobolev space H2(D). The functions f and g are the traces of ϕ~ on Γ1 and Γ3, respectively. Then it is obvious that their union (9)f∪gs=fsθs,ts∈Γ1,gsθs,ts∈Γ3belongs to H1(Γ1∪Γ3).Figure 1
The domainD and its boundary in the formulations of the Tricomi problem.It is well known that the Tricomi problem was firstly proposed and studied by Tricomi in [4] for the now so-called Tricomi equation tϕθθ+ϕtt=0, by using singular integral equations and the matching technique. Tricomi’s study of this problem was mainly motivated to understand second-order mixed elliptic-hyperbolic type equations from a purely mathematical point of view. Later it was discovered that the Tricomi equation may be considered in certain sense as a simple approximation near the sonic line of Chaplygin’s hodograph equation in transonic aerodynamics (cf. [5]) and the Tricomi problem is physically relevant to determining some flow field in transonic flows, such as the detached bow shock and the mixed subsonic-supersonic flow ahead of a blunt body [6]. More general linear mixed elliptic-hyperbolic equations and more general formulations of boundary conditions (such as generalized Tricomi problem, Frankl problem, and generalized Frankl problem) were also considered. For example, Morawetz [7] proved the uniqueness for smooth solutions using Noether’s theorem on conservation laws A=∬(K(σ)ϕθ2+ϕσ2)dσdθ for the equation K(σ)ϕθθ+ϕσσ=0(σK(σ)≥0). Rassias [8] studied weak solutions for the equation (10)Kσϕθθ+ϕσσ+rσ,θϕ=fσ,θ,σKσ≥0.Osher [9] showed the existence for Lavrentiev-Bitsadze equation sgn⁡(σ)ϕθθ+ϕσσ=0. Aziz and Schneider [10] investigated the existence of weak L2 solutions of the Gellerstedt problem and the Gellerstedt-Neumann problems for the equation (11)Kσuθθ+uσσ+λu=fσ,θ,λ=constant<0.Lupo et al. [11] proved existence of weak solutions for Tricomi problem with closed Dirichlet boundary conditions. Lupo et al. [12] considered the existence, uniqueness, and qualitative properties of weak solutions to the degenerate hyperbolic Goursat problem on characteristic triangles for linear and semilinear equations of Tricomi type. See also, for example, [13–16] for works on the nonlinear Tricomi problems. We recommend the introduction in the monograph [17] for a review of the status of mixed-type equations around the 1970s. Morawetz [18] also reviewed the existence and uniqueness theorems for mixed-type equations and their applications to transonic flows, and Chen [19] introduced more recent progress.However, to the best of our knowledge, there is not any result on the Tricomi problem of Chaplygin’s hodograph equation (1), which is relevant to many physical problems in transonic aerodynamics. So we will devote this work to establishing some basic properties of such problems. The main result is the following theorem.Theorem 1.
There is a positive constantɛ0 determined only by μ such that if the domain D is contained in the strip t<ɛ0, then the Tricomi problem (1) and (8) has a quasi-regular distributional solution. Furthermore, the solution is unique in C2(D)∩C1(D¯).For the definition of quasi-regular distributional solution, see Definition2. The constant ɛ0 is given in (77).Our proof depends on the classical energy methods, or thea-b-c method of multipliers (see [8, 20, 21]). Besides the method of singular integral equations, these seem to be the only general way to study well-posedness of mixed-type equations. (However, see also [22] for regularity of solutions of Tricomi equation by using the methods from harmonic analysis.) Although the idea of energy method is rather simple, it is usually very technical to choose appropriate multipliers to a physically interesting equation, like (1), as shown in this paper.We remark that there is another type of mixed elliptic-hyperbolic equations, firstly studied by Maria Cinquini-Cibrario, now called Keldysh type, whose canonical form ist2m+1ϕtt+ϕθθ=0(m=0,1,…) (see [23, page 11]). An up-to-date review of studies of Keldysh-type mixed equations was presented in [24]. It is possible now to study directly many boundary value problems of quasi-linear Keldysh-type equations; for example, see [5, 25, 26] for studies of steady continuous subsonic-supersonic flows in (approximate) de Laval nozzles.The rest of the paper is organized as follows. In Section2 we will define a quasi-regular distributional solution to our Tricomi problem and show that it satisfies the equation and boundary conditions in the ordinary sense if it is a classical solution. We will establish the uniqueness of classical solutions in Section 3. Finally, in Section 4, the existence of a quasi-regular distributional solution is proved by the dual method in functional analysis.


## 2. Definition of Solutions

Denote the linear operatorL∗ by (12)L∗ψ≜Ktψθθ+ψtt+Ptψt,ψ∈Dom⁡L∗,with (13)Dom⁡L∗≜w∈C2D¯:wΓ1∪Γ2=0.It is a formal dual operator of L.Definition 2.
LetF∈L2(D) and f=ϕ~Γ1 and g=ϕ~Γ3 for some ϕ~∈H2(D). A function ϕ∈L2(D) is aquasi-regular distributional solution of the equation(14)Lϕ=FinDsubjected to boundary conditions (8), if(15)F,ψL2D=ϕ,L∗ψ-∫Γ1Kψθn1+ψtn2fds+∫Γ3Kgθn1+gtn2ψ-Kψθn1+ψtn2g-Pgψn2dsfor all ψ∈Dom⁡(L∗)⊂L2(D).Now we show that a quasi-regular distributional solutionϕ satisfies (1) and boundary conditions (8) in the classical pointwise sense if it belongs to C2(D)∩C1(D¯). In fact, using integration by parts and (15), we get, for all ψ∈Dom⁡(L∗), that(16)F,ψL2D+∫Γ1Kψθn1+ψtn2fds-∫Γ3Kgθn1+gtn2ψ-Kψθn1+ψtn2g-Pgψn2ds=ϕ,L∗ψ=Lϕ,ψ-∮∂DKϕθn1+ϕtn2ψ-Kψθn1+ψtn2ϕ-Pϕψn2ds.Choosing particularly that ψ∈C0∞(D), all the three boundary integrals vanish, and (16) is reduced to (17)F,ψL2D=Lϕ,ψfor  anyψ∈C0∞D.Since C0∞(D) is dense in L2(D), we get(18)Lϕ=Fin  theL2-sense,and hence Lϕ=F almost everywhere in D.Next, by employing (16) and (18), we have, for all ψ∈Dom⁡(L∗), (19)∮∂DKϕθn1+ϕtn2ψ-Kψθn1+ψtn2ϕ-Pϕψn2ds=-∫Γ1Kψθn1+ψtn2fds+∫Γ3Kgθn1+gtn2ψ-Kψθn1+ψtn2g-Pgψn2ds.Since ψΓ1∪Γ2=0, it follows that (20)∫Γ3Kϕθn1+ϕtn2ψ-Kψθn1+ψtn2ϕ-Pϕψn2ds-∫Γ1∪Γ2Kψθn1+ψtn2ϕds=-∫Γ1Kψθn1+ψtn2fds+∫Γ3Kgθn1+gtn2ψ-Kψθn1+ψtn2g-Pgψn2ds. Therefore, (21)∫Γ3Kϕ-gθn1+ϕ-gtn2ψ-Kψθn1+ψtn2+Pψn2ϕ-gds-∫Γ1Kψθn1+ψtn2ϕ-fds-∫Γ2Kψθn1+ψtn2ϕds=0for all ψ∈Dom⁡(L∗). Taking that ψ vanishes in a neighborhood of Γ1 and Γ2, we infer that ϕΓ3=g, and furthermore, for all ψ∈Dom⁡(L∗), there holds(22)∫Γ1Kψθn1+ψtn2ϕ-fds+∫Γ2Kψθn1+ψtn2ϕds=0.Recall that ψΓ1∪Γ2=0, and we have (23)0=dψ=ψtdt+ψθdθ=ψtn1-ψθn2dsonΓ1∪Γ2.Hence we get ψtn1=ψθn2 on Γ1∪Γ2, or(24)ψt=Nn2,ψθ=Nn1,onΓ1∪Γ2,where N is a normalizing factor. Thus, we have (25)Kψθn1+ψtn2Γ1∪Γ2=Kn12+n22Nby using (24). Since Kt>0 on Γ1 and Γ2 is a characteristic curve, then Kn12+n22Γ1>0 and [Kn12+n22]Γ2=0. Thus, we have(26)∫Γ1Kn12+n22Nϕ-fds=0by using (22). Since N is an arbitrary function, we see ϕΓ1=f.


## 3. Uniqueness of Classical Solutions

Assume thatϕ1,ϕ2∈C2(D)∩C1(D¯) are two solutions of Tricomi problem (1) and (8), and take(27)ϕ=ϕ1-ϕ2.Then ϕ solves(28)Lϕ=Ktϕθθ+ϕtt-Ptϕt=0inD,ϕΓ1∪Γ3=0.We will show that ϕ≡0 in D.Set(29)I≜2∬DLϕ·at,θϕ+bt,θϕt+ct,θϕθdtdθ=∬D2aKtϕϕθθ+2aϕϕtt-2aPtϕϕt+2bKtϕtϕθθ+2bϕtϕtt-2bPtϕt2+2cKtϕθϕθθ+2cϕθϕtt-2cPtϕθϕtdtdθ,where a(t,θ), b(t,θ), and c(t,θ) are sufficiently smooth functions to be determined (cf. Remark 4). Since (30)2aKtϕϕθθ=2aKϕϕθθ-2aKϕθ2-aθKϕ2θ+aθθKϕ2,2aϕϕtt=2aϕϕtt-2aϕt2-atϕ2t+attϕ2,2aPtϕϕt=aPtϕ2t-aPttϕ2,2bKtϕtϕθθ=2bKϕtϕθθ-bKϕθ2t+bKtϕθ2-2bθKϕtϕθ,2bϕtϕtt=bϕt2t-btϕt2,2cKtϕθϕθθ=cKϕθ2θ-cθKϕθ2,2cϕθϕtt=2cϕθϕtt-cϕt2θ+cθϕt2-2ctϕtϕθ,we obtain that (31)0=I=∬DaθθK+att+aPttϕ2dtdθ-∬D2aK-bKt+cθKϕθ2+2bθK+ct+cPtϕθϕt+2a+bt+2bPt-cθϕt2dtdθ+∮∂D2aϕKϕθn1+ϕtn2-Kaθn1+atn2ϕ2-aPtϕ2n2ds+∮∂Dcn1-bn2Kϕθ2+2bKn1+cn2ϕθϕt+bn2-cn1ϕt2ds≜I1+I2+J1+J2.The goal is to show that all integrals I1, I2, J1, and J2 are nonnegative by choosing suitable functions a, b, and c.One observes that the integralI1≥0 if(32)aθθK+att+aPtt≥0inD,and the integral I2≥0 if the following conditions hold in D:(33)2aK-bKt+cθK≤0,bθK+ct+cPt2-2aK-bKt+cθK2a+bt+2bPt-cθ≤0.By using (28), we have(34)J1=2∫Γ2aϕKϕθn1+ϕtn2ds-∫Γ2Kaθn1+atn2ϕ2ds-∫Γ2aPtϕ2n2ds≜J11+J12+J13.Since n2=-n1-K on Γ2, it follows that (35)dϕΓ2=ϕtdt+ϕθdθ=ϕtn1-ϕθn2ds=ϕt+ϕθ-Kn1ds=-Kϕt-Kϕθn1ds-K=-Kϕθn1+ϕtn2ds-K. Then (36)J11=2∫Γ2aϕKϕθn1+ϕtn2ds=-2∫Γ2a-Kϕdϕ=-∫Γ2a-Kdϕ2=-a-Kϕ2AP+∫Γ2ϕ2da-K.Recall that ϕ(A)=0 and ϕ(P)=0, we have(37)J11=∫Γ2ϕ2da-K.We also note that (38)daΓ2=atdt+aθdθ=atn1-aθn2ds=at+aθ-Kn1ds=-Kat-Kaθn1ds-K=-Kaθn1+atn2ds-K,and hence(39)J12=-∫Γ2Kaθn1+atn2ϕ2ds=∫Γ2ϕ2-Kda.Therefore, we get(40)J11+J12=∫Γ2ϕ2da-K+-Kdaby using (37) and (39). It follows that(41)J1=∫Γ2ϕ2da-K+-Kda-aPtn2dsby using (34). Since n2<0 and dθ=-Kdt=-n2ds>0 on Γ2, we have (42)da-KΓ2=a-Ktdt+a-Kθdθ=at+aK′2K-Kdt+-Kaθdθ=at+aK′2K+aθ-Kdθ,-KdaΓ2=-Katdt+aθdθ=at+aθ-Kdθ.Thus (43)da-K+-Kda-aPtn2ds=2at+2aθ-K+aK′2K+aPtdθonΓ2.So J1≥0 by using (43) provided that(44)at+aθ-K+aK′+2KPt4K≥0onΓ2.Next, observe that the integralJ2≥0 if(45)J2=∮∂DQds=∫Γ1∪Γ3Qds+∫Γ2Qds≜J21+J22≥0, where(46)Q=cn1-bn2Kϕθ2+2bKn1+cn2ϕθϕt+bn2-cn1ϕt2is a quadratic form of ϕθ and ϕt. Since ϕΓ1∪Γ3=0, we have (47)0=dϕ=ϕtdt+ϕθdθ=ϕtn1-ϕθn2dsonΓ1∪Γ3,which implies that, by similar analysis as in Section 2, we can set(48)ϕt=N~n2,ϕθ=N~n1,onΓ1∪Γ3,with N~ being a normalizing factor. Thus, we obtain that (49)QΓ1∪Γ3=cn1-bn2Kϕθ2+2bKϕθ2n2+cn1ϕt2+bn2-cn1ϕt2=cn1+bn2Kϕθ2+ϕt2=cn1+bn2Kn12+n22N~2.Since K(t)>0 on Γ1 and Γ3 is characteristic, then [Kn12+n22]Γ1>0 and [Kn12+n22]Γ3=0. Thus, we have (50)QΓ1∪Γ3=QΓ1=cn1+bn2Kn12+n22N~2≥0⟹J21≥0provided that(51)cn1+bn2≥0onΓ1.Also, sinceΓ2 is characteristic, we infer that [Kn12+n22]Γ2=0. Moreover, we have (52)bKn1+cn22-cn1-bn2K·bn2-cn1=b2K+c2Kn12+n22=0onΓ2.Since KΓ2<0, we have QΓ2≥0 by using (46), and then J22≥0, provided that(53)bn2-cn1≥0onΓ2.Recall that n2=-n1-K on Γ2 and n1>0 on Γ2; then (53) is equivalent to(54)c+b-K≤0onΓ2.Therefore, by (32), (33), (44), (54), and (51), we summarize the requirements on the multipliers a, b, and c as follows:(55a)aθθK+att+aPtt≥0inD,(55b)2aK-bKt+cθK≤0inD,(55c)bθK+ct+cPt2-2aK-bKt+cθK2a+bt+2bPt-cθ≤0inD,(55d)at+aθ-K+aK′+2KP4K≥0onΓ2,(55e)c+b-K≤0onΓ2,(55f)cn1+bn2≥0onΓ1.Our task below is to find sufficient conditions such that inequalities (55a)–(55f) hold. Set D+=D∩{t>0} to be the elliptic region and set D-=D∩{t<0} to be the hyperbolic region. We will actually choose (56)a=eλθt2-σinD,b=c=0inD+¯,b=4aKK′-2KPtinD-,c=-b-KinD-,where λ can be taken as -1 and σ is to be (1-μ2)/4. What is left is to choose ɛ0 announced in Theorem 1 sufficiently small (depending only on μ that appeared in the coefficients of (1)), as computations shown below.Elliptic Region
D
+. First of all, we specify (57)c=b=0inD+to meet the requirement of (55f). Thus, remember that K≥0 in D+, and inequalities (55a)–(55c) are reduced to (58)a≤0,aθθK+att+aPtt≥0inD+.Thus, if a=eλθφ(t), (55a) is transformed to(59)λ2φtK+φ′′t+φtP′+φ′tP≥0.Next, we choose φ(t)=t2-σ<0. Then φ′t=2t, φ′′t=2, and (59) is simplified as (60)2+2tP+P′+λ2Kt2-σ≥0.It is easy to see that this holds for sufficiently small t, provided that σP′<3/2, which is exactly(61)σ1-μ21+1-t2+μ21-1-t22<321-t21-μ21-t22.By fixing σ=(1-μ2)/4, a sufficient condition for this inequality is to take a small ɛ1 (depending only on μ) and then require that(62)t<min⁡ɛ1,σ.Hyperbolic Region
D
-. Now we choose (63)c=-b-KinD- to satisfy (55e). Then (64)bθK+ct+cPt=12-KbtK+bK′-cθK+btK+2bKPt-cθK, and (55c) becomes (65)0≥bθK+ct+cPt2-2aK-bKt+cθK2a+bt+2bPt-cθ=1-4KbtK+bK′-cθK-btK+2bKPt-cθK2-8aKbtK+bK′-cθK-btK+2bKPt-cθK+16a2K2=1-4KbK′-2bKPt2-8aKbK′-2bKPt+16a2K2=1-4KbK′-2bKPt-4aK2. Therefore, we must choose(66)b=4aKK′-2KPtinD-. Note that, by using (2), direct computation yields (67)K′t=21-μ2+μ2t22-t21-t31-μ21-t22>0,(68)K′t-2KtPt=21-μ21-t22-t21-t31-μ21-t22>0for t<ɛ2, where ɛ2 is a small positive constant determined by μ. Hence (66) is well defined.Next, we still choosea=eλθ(t2-σ) in D-. Since condition (62) is valid as required, then, as shown above, (55a) holds automatically. Hence, (55d) becomes (69)eλθ2t+λeλθt2-σ-K+eλθt2-σK′+2KP4K≥0onΓ2.That is,(70)8tK+4λt2-σ-KK+t2-σK′+2KP≤0onΓ2.By fixing λ=-1, this is valid for t<ɛ3 with ɛ3 being a small positive constant. Here we still used continuity and the facts that K′>0 and KP≥0 on Γ2.In order to get (55b), we need to have (71)2aK-btK-bK′+bθ-K3/2≤0.Since (66) implies that (72)2aK-bK′=-2aK-2bKPt=-2Ka+bPt=-2aKK′+2KPK′-2KP≤0,we only need to guarantee that (73)bθ≤0bt≤0,inD-.In fact, by using (68), we have (74)b=2eλθ1-μ2·t2-σt2-t1-μ21-t21-t1+2t-t2.It is obvious that (75)bθ=2λeλθ1-μ2·t2-σt2-t1-μ21-t21-t1+2t-t2≤0forλ=-1.Direct computation yields that(76)bt=2eλθt2-σ1-μ2·2t22-t/t2-σ+2-2t1-μ21-t2+2μ2t2-t1-t1-t1+2t-t2-t2-t1-μ21-t21-6t+3t21-t21+2t-t22<0for -ɛ4<t<0 as desired. Here ɛ4 is a small positive constant determined by μ.Finally we see that the positive constantɛ0 should be chosen so that(77)ɛ0≤min⁡ɛ1,ɛ2,ɛ3,ɛ4,σ.This finishes the choice of multipliers and we proved that I1=0, I2=0, J1=0, and J2=0. (To guarantee existence claimed in Section 4,  ɛ0 might need to be chosen further smaller, according to the construction of multipliers in Section 4.1, but anyway it is in essence determined by the parameter μ that appeared in (1)).Finally, observe that (62) actually guarantees the stronger property that (78)aθθK+att+aPtt>0inD.Hence, I1=0 implies that ϕ≡0 as desired.Remark 3.
Note that in the above we have chosenb and c to be only continuous across the degenerate line {t=0}. This is harmless for our earlier computations since, by applying integration by parts separately in D+ and D- and then summing up, the resultant line integrals on D∩{t=0} are cancelled.Remark 4.
There are some other ways to choose the multipliers. For example, we may set(79)a=b=c=0inD+¯,a=eλθtσinD-,b=4aKK′-2KPtinD-,c=-b-KinD-,where λ≤0 and σ=1/(2k+1) (k∈N).
The other way is to choose(80)a=b=c=0inD+¯,a=eλθarctan⁡σtinD-,b=4aKK′-2KPtinD-,c=-b-KinD-,where λ≤0 and σ>2/1-μ2 is sufficiently large.
However, in both cases, as before, we needD to be quite close to the line {(t,θ):t=0}. So the restriction on smallness of ɛ0 required in Theorem 1 is still not removed.


## 4. Existence of Quasi-Regular Distributional Solutions

In this section, we firstly indicate how to obtain a priori estimate for our Tricomi problem and then use this estimate to show the existence of a quasi-regular distributional solution by a dual method in functional analysis.


### 4.1. A Priori Estimate

We now prove that(81)ψW1,2D≤CL∗ψL2D,∀ψ∈Dom⁡L∗.Similar to the analysis in the previous section, we have(82)
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s
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.Since ψΓ1∪Γ2=0, it follows that(83)J1∗=∫Γ32a∗ψKψθn1+ψtn2-Kaθ∗n1+at∗n2ψ2+a∗Pn2+b∗P′n2+c∗P′n1ψ2ds≜J11∗+J12∗+J13∗.Observing that dθ=--Kdt, or -n2ds=--Kn1ds, and hence n2=-Kn1 on Γ3, we have (84)dψΓ3=ψtn1-ψθn2ds=Kψθn1+ψtn2ds-K.Then (85)J11∗=2∫Γ3a∗-Kψdψ=a∗-Kψ2OP-∫Γ3ψ2da∗-K.Remember ψO=0 and ψP=0, and we get(86)J11∗=-∫Γ3ψ2da∗-K.Next, using(87)da∗Γ3=at∗n1-aθ∗n2ds=Kaθ∗n1+at∗n2ds-K,we may write(88)J12∗=-∫Γ3Kaθ∗n1+at∗n2ψ2ds=-∫Γ3ψ2-Kda∗.Henceforth, using (83)–(88), (89)J1∗=∫Γ3-da∗-K--Kda∗+a∗Pn2+b∗P′n2+c∗P′n1dsψ2≥0,provided that(90)-da∗-K--Kda∗+a∗Pn2+b∗P′n2+c∗P′n1ds≥0onΓ3.Since n2<0 and dθ=--Kdt=-n2ds>0 on Γ3, we have (91)da∗-KΓ3=-at∗+a∗K′2K-aθ∗-Kdθ,-Kda∗Γ3=-at∗+aθ∗-Kdθ.Thus, using (90) and dθΓ3≥0, J1∗≥0 provided that(92)2at∗+a∗K′2K-2aθ∗-K-a∗P-b∗P′-c∗P′-K≥0onΓ3.The integralJ2 is nonnegative if (93)J2∗=∮∂DQ∗ds≡∫Γ1∪Γ2Q∗ds+∫Γ3Q∗ds≜J21∗+J22∗≥0, where (94)Q∗=c∗n1-b∗n2Kψθ2+2b∗Kn1+c∗n2ψθψt+b∗n2-c∗n1ψt2 is a quadratic form with respect to ψθ and ψt. Similarly to get (24), we have (95)Q∗Γ1∪Γ2=c∗n1-b∗n2Kψθ2+2b∗Kψθ2n2+c∗n1ψt2+b∗n2-c∗n1ψt2=c∗n1+b∗n2Kψθ2+ψt2=c∗n1+b∗n2Kn12+n22N2.Since K(t)>0 on Γ1 and Γ2 is a characteristic curve, then [Kn12+n22]Γ1>0 and [Kn12+n22]Γ2=0. Thus, we have (96)Q∗Γ1∪Γ2=Q∗Γ1=c∗n1+b∗n2Kn12+n22N2≥0⟹J21∗≥0provided that(97)c∗n1+b∗n2≥0onΓ1.SinceΓ3 is characteristic, then [Kn12+n22]Γ3=0. Moreover, we have (98)b∗Kn1+c∗n22-c∗n1-b∗n2K·b∗n2-c∗n1=b∗2K+c∗2Kn12+n22=0 on Γ3. Because KΓ3<0, we see Q∗Γ3≥0 (then J22∗≥0) provided that(99)b∗n2-c∗n1≥0onΓ3.Since n2=n1-K on Γ3, then (99) is equivalent to(100)b∗-K-c∗n1≥0onΓ3.Note that n1<0 on Γ3, so (99) or (100) is equivalent to(101)c∗-b∗-K≥0onΓ3.Therefore we conclude, with the further help of Young’s inequality, that(102)I1∗+I2∗≤I∗≤∬D2L∗ψ·a∗ψ+b∗ψt+c∗ψθdtdθ≤∬Dλ1a∗ψ2+λ2b∗ψt+c∗ψθ2+1λ1+1λ2L∗ψ2dtdθprovided that (92), (97), and (101) hold. (Here λ1 and λ2 are two positive constants that can be chosen arbitrarily small.) Therefore, we have the a priori estimate (81), if (92), (97), (101), and (103a)a∗P′+aθθ∗K+att∗-at∗P-b∗P′t-cθ∗P′-λ1a∗2≥ϵ1>0inD,(103b)N1≜-2a∗K+b∗Kt-cθ∗K-λ2c∗2≥0inD,(103c)N2≜2b∗P-2a∗-bt∗+cθ∗-λ2b∗2≥ϵ2>0inD,(103d)N1N2-c∗P-bθ∗K-ct∗-λ2b∗c∗2≥0inDare valid for some multipliers a∗, b∗, and c∗. Here ϵ1 and ϵ2 are two positive constants.As a matter of fact, we can choose the functionsa∗, b∗, and c∗ such that (104)a∗=eλθt2-σinD,b∗=c∗=0inD+¯,b∗=4a∗KK′+2KPtinD-,c∗=b∗-KinD-,to guarantee that conditions (92), (97), (101), and (103a)–(103d) hold, if λ<0 and 0<σ<(1-μ2)/2 were chosen similarly as in Section 3, and ɛ0 are taken appropriately small. The verification is very similar to that in the previous section and therefore we omit the details.


### 4.2. The Proof of the Existence

By our assumption, there exists a functionϕ~∈H2(D) such that ϕ~Γ1=f and ϕ~Γ3=g. Next, take (105)ϕ¯=ϕ-ϕ~,and then ϕ¯ satisfies(106)Lϕ¯=Lϕ-ϕ~=-Lϕ~∈L2DinD,ϕ¯Γ1∪Γ3=0.Next, we show that there is a quasi-regular distributional solutionϕ¯∈L2(D) of Tricomi problem (106).In fact, letran⁡(L∗) be the range (image) of the operator L∗ defined on Dom⁡(L∗). For F≜-Lϕ~∈L2(D), we define a linear functional on ran⁡(L∗) by(107)T:ran⁡L∗⟶R,L∗ψ⟼F,ψfor  anyψ∈Dom⁡L∗.Here we consider ran⁡(L∗) as a linear subspace of L2(D). Using estimate (81), we have(108)TL∗ψ=F,ψ≤FL2DψL2D≤FL2DψW1,2D≤CFL2DL∗ψL2D.Thus, the functional T is bounded.Sinceran(L∗) is a linear subspace of L2(D), by Hahn-Banach theorem, there exists a linear functional T¯:L2(D)→R as an extension of T that preserves the operator norm.Thus, by Riesz representation theorem, there isF¯∈L2(D) such that (109)T¯w=∬DwF¯dtdθ=F¯,w,T¯=F¯L2D. Therefore, (110)F¯,w=T¯w=Tw∀w∈ran⁡L∗.Take w=L∗ψ. Then, for all ψ∈Dom⁡(L∗), we have (111)F¯,L∗ψ=T¯L∗ψ=TL∗ψ=F,ψ.Therefore, F¯ is a quasi-regular distributional solution of Tricomi problem (106) by Definition 2.Finally, by using (105), it is obvious that ϕ=ϕ¯+ϕ~∈L2(D) is a quasi-regular distributional solution of (1) with boundary conditions (8). This finishes the proof of Theorem 1.


## 4.1. A Priori Estimate

We now prove that(81)ψW1,2D≤CL∗ψL2D,∀ψ∈Dom⁡L∗.Similar to the analysis in the previous section, we have(82)
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.Since ψΓ1∪Γ2=0, it follows that(83)J1∗=∫Γ32a∗ψKψθn1+ψtn2-Kaθ∗n1+at∗n2ψ2+a∗Pn2+b∗P′n2+c∗P′n1ψ2ds≜J11∗+J12∗+J13∗.Observing that dθ=--Kdt, or -n2ds=--Kn1ds, and hence n2=-Kn1 on Γ3, we have (84)dψΓ3=ψtn1-ψθn2ds=Kψθn1+ψtn2ds-K.Then (85)J11∗=2∫Γ3a∗-Kψdψ=a∗-Kψ2OP-∫Γ3ψ2da∗-K.Remember ψO=0 and ψP=0, and we get(86)J11∗=-∫Γ3ψ2da∗-K.Next, using(87)da∗Γ3=at∗n1-aθ∗n2ds=Kaθ∗n1+at∗n2ds-K,we may write(88)J12∗=-∫Γ3Kaθ∗n1+at∗n2ψ2ds=-∫Γ3ψ2-Kda∗.Henceforth, using (83)–(88), (89)J1∗=∫Γ3-da∗-K--Kda∗+a∗Pn2+b∗P′n2+c∗P′n1dsψ2≥0,provided that(90)-da∗-K--Kda∗+a∗Pn2+b∗P′n2+c∗P′n1ds≥0onΓ3.Since n2<0 and dθ=--Kdt=-n2ds>0 on Γ3, we have (91)da∗-KΓ3=-at∗+a∗K′2K-aθ∗-Kdθ,-Kda∗Γ3=-at∗+aθ∗-Kdθ.Thus, using (90) and dθΓ3≥0, J1∗≥0 provided that(92)2at∗+a∗K′2K-2aθ∗-K-a∗P-b∗P′-c∗P′-K≥0onΓ3.The integralJ2 is nonnegative if (93)J2∗=∮∂DQ∗ds≡∫Γ1∪Γ2Q∗ds+∫Γ3Q∗ds≜J21∗+J22∗≥0, where (94)Q∗=c∗n1-b∗n2Kψθ2+2b∗Kn1+c∗n2ψθψt+b∗n2-c∗n1ψt2 is a quadratic form with respect to ψθ and ψt. Similarly to get (24), we have (95)Q∗Γ1∪Γ2=c∗n1-b∗n2Kψθ2+2b∗Kψθ2n2+c∗n1ψt2+b∗n2-c∗n1ψt2=c∗n1+b∗n2Kψθ2+ψt2=c∗n1+b∗n2Kn12+n22N2.Since K(t)>0 on Γ1 and Γ2 is a characteristic curve, then [Kn12+n22]Γ1>0 and [Kn12+n22]Γ2=0. Thus, we have (96)Q∗Γ1∪Γ2=Q∗Γ1=c∗n1+b∗n2Kn12+n22N2≥0⟹J21∗≥0provided that(97)c∗n1+b∗n2≥0onΓ1.SinceΓ3 is characteristic, then [Kn12+n22]Γ3=0. Moreover, we have (98)b∗Kn1+c∗n22-c∗n1-b∗n2K·b∗n2-c∗n1=b∗2K+c∗2Kn12+n22=0 on Γ3. Because KΓ3<0, we see Q∗Γ3≥0 (then J22∗≥0) provided that(99)b∗n2-c∗n1≥0onΓ3.Since n2=n1-K on Γ3, then (99) is equivalent to(100)b∗-K-c∗n1≥0onΓ3.Note that n1<0 on Γ3, so (99) or (100) is equivalent to(101)c∗-b∗-K≥0onΓ3.Therefore we conclude, with the further help of Young’s inequality, that(102)I1∗+I2∗≤I∗≤∬D2L∗ψ·a∗ψ+b∗ψt+c∗ψθdtdθ≤∬Dλ1a∗ψ2+λ2b∗ψt+c∗ψθ2+1λ1+1λ2L∗ψ2dtdθprovided that (92), (97), and (101) hold. (Here λ1 and λ2 are two positive constants that can be chosen arbitrarily small.) Therefore, we have the a priori estimate (81), if (92), (97), (101), and (103a)a∗P′+aθθ∗K+att∗-at∗P-b∗P′t-cθ∗P′-λ1a∗2≥ϵ1>0inD,(103b)N1≜-2a∗K+b∗Kt-cθ∗K-λ2c∗2≥0inD,(103c)N2≜2b∗P-2a∗-bt∗+cθ∗-λ2b∗2≥ϵ2>0inD,(103d)N1N2-c∗P-bθ∗K-ct∗-λ2b∗c∗2≥0inDare valid for some multipliers a∗, b∗, and c∗. Here ϵ1 and ϵ2 are two positive constants.As a matter of fact, we can choose the functionsa∗, b∗, and c∗ such that (104)a∗=eλθt2-σinD,b∗=c∗=0inD+¯,b∗=4a∗KK′+2KPtinD-,c∗=b∗-KinD-,to guarantee that conditions (92), (97), (101), and (103a)–(103d) hold, if λ<0 and 0<σ<(1-μ2)/2 were chosen similarly as in Section 3, and ɛ0 are taken appropriately small. The verification is very similar to that in the previous section and therefore we omit the details.


## 4.2. The Proof of the Existence

By our assumption, there exists a functionϕ~∈H2(D) such that ϕ~Γ1=f and ϕ~Γ3=g. Next, take (105)ϕ¯=ϕ-ϕ~,and then ϕ¯ satisfies(106)Lϕ¯=Lϕ-ϕ~=-Lϕ~∈L2DinD,ϕ¯Γ1∪Γ3=0.Next, we show that there is a quasi-regular distributional solutionϕ¯∈L2(D) of Tricomi problem (106).In fact, letran⁡(L∗) be the range (image) of the operator L∗ defined on Dom⁡(L∗). For F≜-Lϕ~∈L2(D), we define a linear functional on ran⁡(L∗) by(107)T:ran⁡L∗⟶R,L∗ψ⟼F,ψfor  anyψ∈Dom⁡L∗.Here we consider ran⁡(L∗) as a linear subspace of L2(D). Using estimate (81), we have(108)TL∗ψ=F,ψ≤FL2DψL2D≤FL2DψW1,2D≤CFL2DL∗ψL2D.Thus, the functional T is bounded.Sinceran(L∗) is a linear subspace of L2(D), by Hahn-Banach theorem, there exists a linear functional T¯:L2(D)→R as an extension of T that preserves the operator norm.Thus, by Riesz representation theorem, there isF¯∈L2(D) such that (109)T¯w=∬DwF¯dtdθ=F¯,w,T¯=F¯L2D. Therefore, (110)F¯,w=T¯w=Tw∀w∈ran⁡L∗.Take w=L∗ψ. Then, for all ψ∈Dom⁡(L∗), we have (111)F¯,L∗ψ=T¯L∗ψ=TL∗ψ=F,ψ.Therefore, F¯ is a quasi-regular distributional solution of Tricomi problem (106) by Definition 2.Finally, by using (105), it is obvious that ϕ=ϕ¯+ϕ~∈L2(D) is a quasi-regular distributional solution of (1) with boundary conditions (8). This finishes the proof of Theorem 1.



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*Source: 754781-2015-11-04.xml*