ON A DARBOUX PROBLEM FOR A THIRD ORDER HYPERBOLIC EQUATION WITH MULTIPLE

CHARACTERISTICS

O. JOKHADZE

Abstract. A Darboux type problem for a model hyperbolic equation
of the third order with multiple characteristics is considered in the
case of two independent variables. The Banach space*C*^{◦}^{2,1}*α* ,*α**≥*0,
is introduced where the problem under consideration is investigated.

The real number*α*0 is found such that for *α > α*0 the problem is
solved uniquely and for*α < α*0it is normally solvable in Hausdorff’s
sense. In the class of uniqueness an estimate of the solution of the
problem is obtained which ensures stability of the solution.

*§* 1. Formulation of the Problem

In the plane of independent variables *x, y* let us consider a third order
hyperbolic equation

*u**xxy*=*f,* (1.1)

where*f* is the given and*u*the unknown real function.

Straight lines*y* =*const*form a double family of characteristics of (1.1),
while*x*=*const* a single family.

Let*γ**i* : *ϕ*=*ϕ**i*, 0*≤r <* +*∞*, *i* = 1,2, 0*≤ϕ*1 *< ϕ*2 *≤* * ^{π}*2, be two rays
coming out of the origin of the coordinates

*O(0,*0) written in terms of the polar coordinates

*r, ϕ. The angle formed by these rays will be denoted by*

*D*:

*ϕ*1

*< ϕ < ϕ*2, 0

*< r <*+

*∞*. Let

*P*

_{1}

^{0}and

*P*

_{2}

^{0}be the points at which

*γ*1 and

*γ*2 intersect respectively the characteristics

*L*1(P

^{0}) :

*x*=

*x*0 and

*L*2(P

^{0}) :

*y*=

*y*0 coming out of an arbitrarily taken point

*P*

^{0}(x0

*, y*0)

*∈D.*

Equation (1.1) will be considered in the rectangular domain *D*0 : 0*< x <*

1991*Mathematics Subject Classification. 35L35.*

*Key words and phrases.* Darboux problem, hyperbolic equation, characteristic line,
Banach space, normal solvability, stability of solution, integral representation, regular
solution.

469

1072-947X/95/900-0469$7.50/0 c**1995 Plenum Publishing Corporation

*x*0, 0 *< y < y*0, bounded by the characteristics *x*= 0, *x*=*x*0 and *y* = 0,
*y*=*y*0.

Equations of segments *OP*_{1}^{0}, *OP*_{2}^{0} of the rays *γ*1*, γ*2 will be written in
terms of rectangular coordinates *x, y* as *OP*_{1}^{0} : *y* = *ρ*1*x, 0* *≤* *x* *≤* *x*0;
*OP*_{2}^{0}:*x*=*ρ*2*y, 0≤y* *≤y*0, where*ρ*1= tg*ϕ*1,*ρ*2 = ctg*ϕ*2 and 0*≤ρ*1*ρ*2=

tg*ϕ*1

tg*ϕ*2 *<*1.

For equation (1.1) we shall consider a Darboux type problem formulated
as follows: Find in *D*0 a regular solution*u*:*D*0 *→R,* *R≡*(*−∞,*+*∞*), of
equation (1.1), satisfying on the segments*OP*_{1}^{0} and*OP*_{2}^{0} the conditions

(M1*u**xx*+*N*1*u**xy*+*P*1*u**x*+*Q*1*u**y*+*S*1*u)|**OP*_{1}^{0} =*f*1*,* (1.2)
(M2*u**xx*+*N*2*u**xy*+*P*2*u**x*+*Q*2*u**y*+*S*2*u)|**OP*_{2}^{0} =*f*2*,* (1.3)
(M3*u**xx*+*N*3*u**xy*+*P*3*u**x*+*Q*3*u**y*+*S*3*u)|**OP*_{2}^{0} =*f*3*,* (1.4)
where*M**i**, N**i*,*P**i**, Q**i*,*S**i**, f**i*,*i*= 1,2,3, are given continuous real functions.

A function*u*:*D*0*→R*continuous in*D*0together with its partial deriva-
tives *D*_{x}^{i}*D*_{y}^{j}*u,* *i* = 0,1,2, *j* = 0,1, *i*+*j >* 0, *D**x* *≡* _{∂x}* ^{∂}* ,

*D*

*y*

*≡*

_{∂y}*and satisfying equation (1.1) in*

^{∂}*D*0and conditions (1.2)–(1.4) is called a regular solution of problem (1.1)– (1.4).

It should be noted that the boundary value problem (1.1)–(1.4) is the natural development of the well-known classical formulations of Goursat and Darboux problems (see, for example, [1]–[4]) for second-order linear hyperbolic equations. Variants of Goursat and Darboux problems for one hyperbolic equation and systems of second order, and also for systems of first order, are investigated in some papers (see, for example, [5]–[13]). Note that the results obtained in [7]–[9] are new even for one equation and in a certain sense bear a complete character.

Initial-boundary and characteristic problems for a wide class of hyper- bolic equations of third and higher order with dominating derivatives are treated in [14]–[19] and other papers.

*Remark* 1.1. Conditions (1.2)–(1.4) take into account the hyperbolic na-
ture of problem (1.1)–(1.4) as they contain only the derivatives dominated
by the derivative*D*_{x}^{2}*D**y**u.*

*Remark* 1.2. Since the family of characteristics *y* =*const* is the double
one for the hyperbolic equation (1.1), two conditions (1.3), (1.4) are given
on the segment*OP*_{2}^{0}.

*Remark* 1.3. The above problem could be formulated also for an angular
domain bounded by the rays*γ*1,*γ*2and the characteristics*L*1(P^{0}),*L*2(P^{0})
of (1.1) under the same boundary conditions (1.2)–(1.4), but, as is well
known, the solution *u(x, y) of the thus formulated problem continues into*
the domain *D*0 as the solution of the original problem (1.1)–(1.4). The
problem formulated in the form of (1.1)–(1.4) is convenient for further in-
vestigations provided that equation (1.1) contains dominated lowest terms
and the Riemann function is effectively used.

Let us introduce the functional spaces
*C**◦**α*(D0)*≡ {u*:*u∈C(D*0), u(0) = 0, sup

*z**6*=0,z*∈**D*0

*|z|*^{−}^{α}*|u(z)|<∞},*
*α≥*0, z=*x*+*iy,*

*C**◦**α*[0, d]*≡ {ϕ*:*ϕ∈C[0, d], ϕ(0) = 0,* sup

0<t*≤**d*

*t*^{−}^{α}*|ϕ(t)|<∞},*
*α≥*0, d >0.

For*α*= 0 the above classes will be denoted by*C** ^{◦}* (D0) and

*C*

*[0, d], respec- tively.*

^{◦}Obviously,*C*^{◦}*α*(D0) and*C*^{◦}*α*[0, d] will be the Banach spaces with respect
to the norms

*kuk*^{◦}

*C**α*(D0)= sup

*z**6*=0,z*∈**D*0

*|z|*^{−}^{α}*|u(z)|,* *kϕk*^{◦}

*C**α*[0,d]= sup

0<t*≤**d*

*t*^{−}^{α}*|ϕ(t)|,*
respectively.

It is easy to see that the belonging of the functions *u* *∈C** ^{◦}* (D0) and

*ϕ∈C*

*[0, d] to the spaces*

^{◦}*C*

^{◦}*α*(D0) and

*C*

^{◦}*α*[0, d], respectively, is equivalent to the fulfillment of the following inequalities:

*|u(z)| ≤c*1*|z|*^{α}*, z∈D*0*,* (1.5)

*|ϕ(t)| ≤c*2*t*^{α}*, t∈*[0, d], c*i**≡const >*0, i= 1,2. (1.6)
The boundary value problem (1.1)–(1.4) will be investigated in the space

*C**◦* ^{2,1}*α* (D0)*≡ {u*:*D*^{i}_{x}*D*^{j}_{y}*u∈C*^{◦}*α*(D0), i= 0,1,2, j= 0,1*},*
which is Banach with respect to the norm

*kuk*^{◦}

*C*^{2,1}*α* (D0)=
X2
*i=0*

X1
*j=0*

*kD*^{i}_{x}*D*^{j}_{y}*uk*^{◦}

*C**α*(D0)*.*

To consider (1.1)–(1.4) in the class *C*^{◦}^{2,1}*α* (D0) it is required that
*f* *∈C*^{◦}*α* (D0), *M*1*, N*1*, P*1*, Q*1*, S*1 *∈* *C[0, x*0], *M**i**, N**i**, P**i**, Q**i**, S**i* *∈* *C[0, y*0],
*i*= 2,3,*f*1*∈C*^{◦}*α*[0, x0],*f**i* *∈C*^{◦}*α*[0, y0], *i*= 2,3.

*§*2. Integral Representation of a Regular Solution of the
Class *C*^{◦}^{2,1}*α* (D0)of Equation (1.1)

The integration of equation (1.1) along the characteristics enables us to prove that the following lemma is valid.

Lemma 2.1. *The formula*

*u(x, y) =*
Z*x*
0

(x*−ξ)ϕ(ξ)dξ*+
Z*y*
0

*ψ(η)dη*+*x*
Z*y*
0

*ν*(η)dη+

+
Z*x*
0

Z*y*
0

(x*−ξ)f*(ξ, η)dξdη, (x, y)*∈D*0*,* (2.1)

*establishes the one-to-one correspondence between regular solutions* *u(x, y)*
*of the class* *C*^{◦}^{2,1}*α* (D0) *of equation* (1.1) *and values* *ϕ* *∈C*^{◦}*α* [0, x0], *ψ, ν* *∈*
*C**◦**α* [0, y0]. Note that *ϕ(x) =* *u**xx*(x,0), 0 *≤* *x* *≤* *x*0*,* *ψ(y) =* *u**y*(0, y),
*ν(y) =u**xy*(0, y),0*≤y≤y*0*.*

*Proof.* Introduce the notation*u(x,*0)*≡ϕ*1(x), 0*≤x≤x*0,*u(0, y)≡ψ*1(y),
*u**x*(0, y)*≡ν*1(y), 0*≤y* *≤y*0. It is obvious that*ϕ*1(0) =*ϕ*^{0}_{1}(0) =*ψ*1(0) =
*ν*1(0) = 0. Further, by integrating (1.1) twice and one time along the
characteristics*y*=*const*and*x*=*const, respectively, we obtain*

*u(x, y) =ϕ*1(x) +*ψ*1(y) +*xν*1(y) +
+

Z*x*
0

Z*y*
0

(x*−ξ)f*(ξ, η)dξdη, (x, y)*∈D*0*.* (2.2)

Let*ϕ*^{00}_{1}(x)*≡ϕ(x), 0≤x≤x*0, *ψ*_{1}* ^{0}*(y)

*≡ψ(y),*

*ν*

_{1}

*(y)*

^{0}*≡ν*(y), 0

*≤y*

*≤y*0. Then (2.2) takes the form of (2.1).

From *u* *∈C*^{◦}^{2,1}*α* (D0) it follows that *ϕ∈C*^{◦}*α* [0, x0], *ψ, ν* *∈C*^{◦}*α* [0, y0]. To
prove the converse statement note that by (1.5), (1.6) we have the estimates

*|ϕ(x)| ≤c*3*x*^{α}*, x∈*[0, x0], *|ψ(y)| ≤c*4*y*^{α}*,* *|ν*(y)*| ≤c*5*y*^{α}*, y∈*[0, y0],

*|f*(x, y)*| ≤c*6*|z|*^{α}*, z≡*(x, y)*∈D*0*, c*2+i*≡const >*0, i= 1,2,3.

Setting*ζ*= (ξ, η), from formula (2.1) we have

*|u(x, y)| ≤c*3*x*0

Z*x*
0

*ξ*^{α}*dξ*+*c*4

Z*y*
0

*η*^{α}*dη*+*c*5*x*0

Z*y*
0

*η*^{α}*dη*+

+c6*x*0

Z*x*
0

Z*y*
0

*|ζ|*^{α}*dξdη≤c*7*|z|*^{α}*,*

where*c*7*≡**α+1*^{1} (c3*x*^{2}_{0}+*c*4*y*0+*c*5*x*0*y*0) +*c*6*x*^{2}_{0}*y*0*>*0,*z∈D*0.

Hence*u∈C*^{◦}*α*(D0). In a similar but relatively simpler manner it is proved
that *D*^{i}_{x}*D*^{j}_{y}*u∈C*^{◦}*α* (D0), *i* = 0,1,2, *j* = 0,1,*i*+*j >* 0, and therefore *u∈*
*C**◦* ^{2,1}*α* (D0).

*§* 3. Reducing Problem (1.1)–(1.4) to a System of
Integro-Functional Equations

Substituting (2.1) into conditions (1.2)–(1.4) gives us

*M*1(x)ϕ(x) +R*x*

0[P1(x) +*S*1(x)(x*−ξ)]ϕ(ξ)dξ*+*Q*1(x)ψ(ρ1*x)+*

+S1(x)R*ρ*1*x*

0 *ψ(η)dη*+ [N1(x) +*xQ*1(x)]ν(ρ1*x) + [P*1(x)+

+xS1(x)]R*ρ*1*x*

0 *ν(η)dη*=*f*e1(x), 0*≤x≤x*0*,*
*Q**i*(y)ψ(y) +*S**i*(y)R*y*

0 *ψ(η)dη*+ [N*i*(y) +*ρ*2*yQ**i*(y)]ν(y)+

+[P*i*(y) +*ρ*2*yS**i*(y)]R*y*

0 *ν(η)dη*+*M**i*(y)ϕ(ρ2*y) +*R*ρ*2*y*
0 [P*i*(y)+

+S*i*(y)(ρ2*y−ξ)]ϕ(ξ)dξ*=*f*e*i*(y), i= 2,3, 0*≤y≤y*0*,*

(3.1)

where

*f*e1(x)*≡f*1(x)*−M*1(x)R*ρ*1*x*

0 *f*(x, η)dη*−N*1(x)R*x*

0 *f*(ξ, ρ1*x)dξ−*

*−P*1(x)R*x*
0

R*ρ*1*x*

0 *f*(ξ, η)dξ dη*−Q*1(x)R*x*

0(x*−ξ)f*(ξ, ρ1*x)dξ−*

*−S*1(x)R*x*
0

R*ρ*1*x*

0 (x*−ξ)f*(ξ, η)dξ dη, 0*≤x≤x*0*,*
*f*e*i*(y)*≡f**i*(y)*−M**i*(y)R*y*

0 *f*(ρ2*y, η)dη−N**i*(y)R*ρ*2*y*

0 *f*(ξ, y)dξ*−*

*−P**i*(y)R*ρ*2*y*
0

R*y*

0 *f*(ξ, η)dξ dη*−Q**i*(y)R*ρ*2*y*

0 (ρ2*y−ξ)f*(ξ, y)dξ*−*

*−S**i*(y)R*ρ*2*y*
0

R*y*

0(ρ2*y−ξ)f*(ξ, η)dξ dη, i= 2,3, 0*≤y≤y*0*.*
(3.2)

Transferring all the integral terms contained in system (3.1) to the right- hand side, we have

*M*1(x)ϕ(x) +*Q*1(x)ψ(ρ1*x) +R*1(x)ν(ρ1*x) =F*1(x), 0*≤x≤x*0*,* (3.3)
*Q*2(y)ψ(y) +*R*2(y)ν(y) +*M*2(y)ϕ(ρ2*y) =F*2(y), 0*≤y≤y*0*,* (3.4)
*Q*3(y)ψ(y) +*R*3(y)ν(y) +*M*3(y)ϕ(ρ2*y) =F*3(y), 0*≤y≤y*0*,* (3.5)
where

*F*1(x)*≡f*e1(x)*−*
Z*x*

0

[P1(x) +*S*1(x)(x*−ξ)]ϕ(ξ)dξ−S*1(x)

*ρ*1*x*

Z

0

*ψ(η)dη−*

*−*[P1(x) +*xS*1(x)]

*ρ*1*x*

Z

0

*ν(η)dη,* 0*≤x≤x*0*,*

*F**i*(y)*≡f*e*i*(y)*−S**i*(y)
Z*y*
0

*ψ(η)dη−*[P*i*(y) +*ρ*2*yS**i*(y)]

Z*y*
0

*ν(η)dη−*

*−*

*ρ*2*y*

Z

0

[P*i*(y) +*S**i*(y)(ρ2*y−ξ)]ϕ(ξ)dξ, i*= 2,3, 0*≤y≤y*0*,*
*R*1(x)*≡N*1(x) +*xQ*1(x), 0*≤x≤x*0*,*

*R**i*(y)*≡N**i*(y) +*ρ*2*yQ**i*(y), i= 2,3, 0*≤y≤y*0*.*
Rewrite equations (3.4) and (3.5) as follows:

*Q**i*(y)ψ(y) +*R**i*(y)ν(y) =*F**i*(y)*−M**i*(y)ϕ(ρ2*y),*

*i*= 2,3, 0*≤y* *≤y*0*.* (3.6)
Assuming that

∆(y)*≡*

*Q*2(y) *N*2(y)
*Q*3(y) *N*3(y)

*6*= 0, 0*≤y≤y*0*,* (3.7)
we find from system (3.6) that

*ψ(y) =a*1(y)*−b*1(y)ϕ(ρ2*y),*

*ν*(y) =*a*2(y)*−b*2(y)ϕ(ρ2*y),* 0*≤y≤y*0*,* (3.8)
where

*a*1(y)*≡*∆^{−}^{1}(y)[F2(y)R3(y)*−F*3(y)R2(y)],
*b*1(y)*≡*∆^{−}^{1}(y)[M2(y)R3(y)*−M*3(y)R2(y)],
*a*2(y)*≡*∆^{−}^{1}(y)[F3(y)Q2(y)*−F*2(y)Q3(y)],

*b*2(y)*≡*∆^{−}^{1}(y)[M3(y)Q2(y)*−M*2(y)Q3(y)], 0*≤y≤y*0*.*

Note that here and in what follows the upper index*−*1 means the inverse
value.

Let

*M*1(x)*6*= 0, 0*≤x≤x*0*.* (3.9)
If the obtained expressions for the functions *ψ(y),ν(y), 0≤y* *≤y*0, are
substituted from (3.8) into equality (3.3), then we shall have

*ϕ(x)−a(x)ϕ(τ*0*x) =F*(x), 0*≤x≤x*0*,* (3.10)
where*a(x)≡M*_{1}^{−}^{1}(x)[Q1(x)b1(ρ1*x)+R*1(x)b2(ρ1*x)],F*(x)*≡M*_{1}^{−}^{1}(x)[F1(x)

*−Q*1(x)a1(ρ1*x)−R*1(x)a2(ρ1*x)], 0≤x≤x*0,*τ*0*≡ρ*1*ρ*2.
Simple calculations lead to

*F*(x) =R*x*

0 *K*1(x, ξ)ϕ(ξ)dξ+R*τ*0*x*

0 *K*2(x, ξ)ϕ(ξ)dξ+

+K3(x)R*ρ*1*x*

0 *ψ(η)dη*+*K*4(x)R*ρ*1*x*

0 *ν(η)dη*+*F*4(x), 0*≤x≤x*0*,*
*a*1(y) =R*ρ*2*y*

0 *K*5(ξ, y)ϕ(ξ)dξ+*K*6(y)R*y*

0 *ψ(η)dη+*

+K7(y)R*y*

0 *ν(η)dη*+*F*5(y), 0*≤y≤y*0*,*
*a*2(y) =R*ρ*2*y*

0 *K*8(ξ, y)ϕ(ξ)dξ+*K*9(y)R*y*

0 *ψ(η)dη+*

+K10(y)R*y*

0 *ν(η)dη*+*F*6(y), 0*≤y≤y*0*,*

(3.11)

where*K*1(x, ξ), 0*≤x≤x*0, 0*≤ξ≤x,K*2(x, ξ), 0*≤x≤x*0, 0*≤ξ≤τ*0*x,*
*K**i*(x), 0*≤x≤* *x*0, *i* = 3,4, *K**i*(ξ, y), 0 *≤ξ* *≤ρ*2*y, 0* *≤y* *≤* *y*0, *i* = 5,8,
*K**i*(y), 0 *≤* *y* *≤y*0, *i* = 6,7,9,10, expressed in terms of the *M*1*, N**i**, P**i*,
*Q**i**, S**i**, ρ*1*, ρ*2, *i* = 1,2,3, are continuous kernels of the integral terms
contained on the right-hand sides of system (3.11), while the functions*F**i*,
*i*= 4,5,6, denote the following values:

*F*4(x)*≡M*_{1}^{−}^{1}(x)*f*e1(x) + [M_{1}^{−}^{1}(x)R1(x)∆^{−}^{1}(ρ1*x)Q*3(ρ1*x)−*

*−M*_{1}^{−}^{1}(x)Q1(x)∆^{−}^{1}(ρ1*x)R*3(ρ1*x)]f*e2(ρ1*x)+*

+[M_{1}^{−}^{1}(x)Q1(x)∆^{−}^{1}(ρ1*x)R*2(ρ1*x)−*

*−M*_{1}^{−}^{1}(x)R1(x)∆^{−}^{1}(ρ1*x)Q*2(ρ1*x)]f*e3(ρ1*x),* 0*≤x≤x*0*,*
*F*5(y)*≡*∆^{−}^{1}(y)R3(y)*f*e2(y)*−*∆^{−}^{1}(y)R2(y)*f*e3(y), 0*≤y≤y*0*,*
*F*6(y)*≡*∆^{−}^{1}(y)Q2(y)*f*e3(y)*−*∆^{−}^{1}(y)Q3(y)*f*e2(y), 0*≤y≤y*0*.*

(3.12)

Introducing the notation

(Kϕ)(x)*≡ϕ(x)−a(x)ϕ(τ*0*x),* 0*≤x≤x*0*,* (3.13)

equalities (3.8), (3.10) due to (3.11) we can write as

(Kϕ)(x) =R*x*

0 *K*1(x, ξ)ϕ(ξ)dξ+R*τ*0*x*

0 *K*2(x, ξ)ϕ(ξ)dξ+

+K3(x)R*ρ*1*x*

0 *ψ(η)dη*+*K*4(x)R*ρ*1*x*

0 *ν*(η)dη+*F*4(x), 0*≤x≤x*0*,*
*ψ(y) =*R*ρ*2*y*

0 *K*5(ξ, y)ϕ(ξ)dξ+*K*6(y)R*y*

0 *ψ(η)dη+*

+K7(y)R*y*

0 *ν*(η)dη*−b*1(y)ϕ(ρ2*y) +F*5(y), 0*≤y≤y*0*,*
*ν*(y) =R*ρ*2*y*

0 *K*8(ξ, y)ϕ(ξ)dξ+*K*9(y)R*y*

0 *ψ(η)dη+*

+K10(y)R*y*

0 *ν*(η)dη*−b*2(y)ϕ(ρ2*y) +F*6(y), 0*≤y≤y*0*.*

(3.14)

*Remark* 3.1. It is obvious that if conditions (3.7), (3.9) are fulfilled, then
in the class *C*^{◦}^{2,1}*α* (D0) problem (1.1)–(1.4) is equivalent to the system
of equations (3.14) with respect to the unknowns *ϕ* *∈C*^{◦}*α* [0, x0], *ψ, ν* *∈*
*C**◦**α*[0, y0].

*§* 4. Invertibility of the Functional Operator *K* Defined by
Equality (3.13)

Assume that conditions (3.7), (3.9) are fulfilled. Set*σ≡a(0) and* *α*0*≡*

*−*log*|σ|/*log*τ*0(σ*6*= 0).

Lemma 4.1. *Let either* *γ*1 *or* *γ*2 *be the characteristic of equation* (1.1)
(i.e.,*τ*0= 0).Then the equation

(Kϕ)(x) =*g(x),* 0*≤x≤x*0*,* (4.1)
*has a unique solution in the spaceC*^{◦}*α*[0, x0] *for allα≥*0.

The proof follows from the fact that under the assumption of the lemma
*K* is the identity operator in the space*C*^{◦}*α*[0, x0].

Lemma 4.2. *Let the straight lines* *γ*1*,* *γ*2 *be not the characteristics of*
*equation*(1.1) (i.e.,0*< τ*0*<*1)*andσ6*= 0. Then for*α > α*0 *equation*(4.1)
*has a unique solution in the space* *C*^{◦}*α* [0, x0] *and for the inverse operator*
*K*^{−}^{1} *we have the estimate*

*|*(K^{−}^{1}*g)(x)| ≤cx*^{α}*kgk*^{◦}

*C**α*[0,x]*,* (4.2)

*where the positive constantc* *does not depend on the function* *g.*

*Proof.* We introduce into consideration the operators
(Γϕ)(x) =*a(x)ϕ(τ*0*x),* 0*≤x≤x*0*, K*^{−}^{1}=*I*+

X*∞*
*j=1*

Γ^{j}*,* (4.3)

where*I* is the identity operator. It is easy to see that the operator*K*^{−}^{1} is
formally inverse to the operator*K. Thus it enough for us to prove that the*
Neuman series*I*+P_{∞}

*j=1*Γ* ^{j}* converges in the space

*C*

^{◦}*α*[0, x0].

By the definition of the operator Γ from (4.3) we have (Γ^{j}*ϕ)(x) =*
*a(x)a(τ*0*x). . . a(τ*_{0}^{j}^{−}^{1}*x)ϕ(τ*_{0}^{j}*x), 0* *≤* *x* *≤* *x*0. The condition *α > α*0 is
equivalent to the inequality*τ*_{0}^{α}*|σ|<*1. Therefore by virtue of the continuity
of the function *a*and the equality *a(0) =* *σ* there are positive numbers *ε*
(ε < x0),*δ*and*q*such that the inequalities

*|a(x)| ≤ |σ|*+*δ, τ*_{0}* ^{α}*(

*|σ|*+

*δ)≡q <*1 (4.4) will hold for 0

*≤x≤ε.*

It is obvious that the sequence*{τ*_{0}^{j}*x}*^{∞}*j=0* uniformly converges to zero as
*j→ ∞*on the segment [0, x0]. Therefore there is a natural number*j*0such
that

*τ*_{0}^{j}*x≤ε,* for 0*≤x≤x*0*, j≥j*0*.* (4.5)
We can take as*j*0, say,*j*0=h_{log} _{εx}* _{−}*1

0

log *τ*0

i+ 1, where [p] denotes the integral
part of the number*p.*

Let max

0*≤**x**≤**x*0*|a(x)| ≡* *β. By virtue of (4.4), (4.5) the following estimates*
hold for*j > j*0,*g∈C*^{◦}*α*[0, x0]:

*|*(Γ^{j}*g)(x)|*=*|a(x)a(τ*0*x). . . a(τ*_{0}^{j}^{0}^{−}^{1}*x)| · |a(τ*_{0}^{j}^{0}*x). . . a(τ*_{0}^{j}^{−}^{1}*x)| · |g(τ*_{0}^{j}*x)| ≤*

*≤β*^{j}^{0}(*|σ|*+*δ)*^{j}^{−}^{j}^{0}(τ_{0}^{j}*x)*^{α}*kgk*^{◦}

*C**α*[0,x]*≤*

*≤β*^{j}^{0}(*|σ|*+*δ)*^{−}^{j}^{0}

*τ*_{0}* ^{α}*(

*|σ|*+

*δ)*

*j*

*x*^{α}*kgk*^{◦}

*C**α*[0,x]=*c*0*q*^{j}*x*^{α}*kgk*^{◦}

*C**α*[0,x]*,*(4.6)
where*c*0*≡β*^{j}^{0}(*|σ|*+*δ)*^{−}^{j}^{0}.

For 1*≤j≤j*0 we have

*|*(Γ^{j}*g)(x)| ≤β** ^{j}*(τ

_{0}

^{j}*x)*

^{α}*kgk*

^{◦}*C**α*[0,x]*≤β*^{j}*x*^{α}*kgk*^{◦}

*C**α*[0,x]*.* (4.7)

Now by (4.6) and (4.7) we eventually have

*|ϕ(x)|*=*|*(K^{−}^{1}*g)(x)| ≤ |g(x)|*+

*j*0

X

*j=1*

(Γ^{j}*g)(x)*

+

X*∞*
*j=j*0+1

(Γ^{j}*g)(x)*

*≤*

*≤*
1 +

*j*0

X

*j=1*

*β** ^{j}*+

*c*0

X*∞*
*j=j*0+1

*q** ^{j}*

*x*

^{α}*kgk*

^{◦}*C**α*[0,x] =

= 1 +

*j*0

X

*j=1*

*β** ^{j}*+

*c*0

*q*^{j}^{0}^{+1}
1*−q*

*x*^{α}*kgk*^{◦}

*C**α*[0,x]*,*

from which we obtain the continuity of the operator *K*^{−}^{1} in the space
*C**◦**α*[0, x0] and the validity of estimate (4.2).

*Remark* 4.1. If *σ* = 0, then the inequality*τ*_{0}^{α}*|σ|* *<*1 is fulfilled for any
*α* *≥* 0 and, as seen from the proof, in that case Lemma 4.2 holds for all
*α≥*0.

Lemma 4.3. *Let the straight lines* *γ*1*,* *γ*2 *be not the characteristics of*
*equation* (1.1) (i.e.,0*< τ*0*<*1)*and* *σ6*= 0. Then equation(4.1)*is solvable*
*in the spaceC*^{◦}*α*[0, x0]*forα < α*0*and the homogeneous equation correspon-*
*ding to*(4.1)*has in the said space an infinite number of linearly independent*
*solutions, i.e.,* dim Ker*K*=*∞.*

*Proof.* The condition *α < α*0 is equivalent to the inequality *τ*_{0}^{α}*|σ|* *>* 1.

Therefore, as in proving Lemma 4.2, there are positive numbers*ε*1(ε1*< x*0),
*δ*1and*q*1 such that the inequalities

*|a*^{−}^{1}(x)*| ≤*(*|σ| −δ*1)^{−}^{1}*,* *|σ| −δ*1*>*0, τ_{0}* ^{α}*(

*|σ| −δ*1)

*≡q*

^{−}_{1}

^{1}

*>*1 (4.8) will hold for 0

*≤x≤ε*1.

It is easy to see that the operator Γ from (4.3) is invertible and
(Γ^{−}^{1}*ϕ)(x) =a*^{−}^{1}(τ_{0}^{−}^{1}*x)ϕ(τ*_{0}^{−}^{1}*x),* 0*≤x≤τ*0*ε*1*.*
Rewrite (4.3) in the equivalent form

*ϕ(x)−*(Γ^{−}^{1}*ϕ)(x) =−*(Γ^{−}^{1}*g)(x),* 0*≤x≤τ*0*ε*1*.* (4.9)
Obviously, for any*x*from the interval 0*< x < τ*0*ε*1there exists a unique
natural number*n*1=*n*1(x) satisfying the inequalities

*τ*0*ε*1*< τ*_{0}^{−}^{n}^{1}*x≤ε*1*.*

It is easy to verify that

*n*1(x) =hlog *ε*^{−}_{1}^{1}*x*
log *τ*0

i*≥*log *ε*^{−}_{1}^{1}*x*

log *τ*0 *−*1. (4.10)

Similarly, for *ε*1 *< x* *≤* *x*0 there exists a unique natural number *n*2 =
*n*2(x) satisfying the inequalities

*τ*0*ε*1*≤τ*_{0}^{n}^{2}*x < ε*1*.*
Clearly,*n*2(x) =h

1*−*^{log} ^{ε}^{−}

1
1 *x*
log *τ*0

i .

One can easily verify that any continuous solution on the half-interval
0*< x≤x*0 of equation (4.1) or (4.9) is given by the formula

*ϕ(x) =*

*ϕ*^{0}(x), *τ*0*ε*1*≤x≤ε*1*,*
(Γ^{−}^{n}^{1}^{(x)}*ϕ*^{0})(x)*−*P*n*1(x)

*j=1* (Γ^{−}^{j}*g)(x),* 0*< x < τ*0*ε*1*,*
(Γ^{n}^{2}^{(x)}*ϕ*^{0})(x)*−*P*n*2(x)*−*1

*j=0* (Γ^{j}*g)(x), x > ε*1*,*

(4.11)

where *ϕ*^{0} is an arbitrary function from the class *C[τ*0*ε*1*, ε*1], satisfying the
condition*ϕ*^{0}(ε1)*−a(ε*1)ϕ^{0}(τ0*ε*1) =*g(ε*1).

Let us show that the function *ϕ* given by (4.11) belongs to the class
*C**◦**α*[0, x0] for*g∈C*^{◦}*α*[0, x0]. The arbitrariness of*ϕ*^{0}implies that Lemma 4.3
holds for equation (4.1).

By (4.8), (4.10) the estimates

*|*(Γ^{−}^{n}^{1}^{(x)}*ϕ(x)|*=*|a*^{−}^{1}(τ_{0}^{−}^{1}*x)a*^{−}^{1}(τ_{0}^{−}^{2}*x). . . a*^{−}^{1}(τ_{0}^{−}^{n}^{1}^{(x)}*x)ϕ(τ*_{0}^{−}^{n}^{1}^{(x)}*x)| ≤*

*≤*(*|σ| −δ*1)^{−}^{n}^{1}^{(x)}*kϕ*^{0}*k**C[τ*0*ε*1*,ε*1] *≤τ*_{0}^{αn}^{1}^{(x)}*kϕ*^{0}*k**C[τ*0*ε*1*,ε*1]*≤*

*≤τ*

log*ε**−*1
1 *x*
log*τ*0 *−*1

*α*

0 *kϕ*^{0}*k**C[τ*0*ε*1*,ε*1] =*τ*_{0}^{−}^{α}*ε*^{−}_{1}^{α}*x*^{α}*kϕ*^{0}*k**C[τ*0*ε*1*,ε*1] (4.12)
hold for 0*< x < τ*0*ε*1.

In a similar manner for 0*< x < τ*0*ε*1 and 1*≤j≤n*1(x) we have

*|*(Γ^{−}^{j}*g)(x)| ≤*(*|σ| −δ*1)^{−}* ^{j}*(τ

_{0}

^{−}

^{j}*x)*

^{α}*kgk*

^{◦}*C**α*[0,x0]=

= [τ_{0}* ^{α}*(

*|σ| −δ*1)]

^{−}

^{j}*x*

^{α}*kgk*

^{◦}*C**α*[0,x0]=*q*^{j}_{1}*x*^{α}*kgk*^{◦}

*C**α*[0,x0]*.*
Hence it follows that

*n*X1(x)
*j=1*

(Γ^{−}^{j}*g)(x)**≤** ^{n}*X

^{1}

^{(x)}

*j=1*

*q*^{j}_{1}
*x*^{α}*kgk*^{◦}

*C**α*[0,x0] *≤*

*≤* *q*1

1*−q*1

*x*^{α}*kgk*^{◦}

*C**α*[0,x0]*.* (4.13)

By (4.12) and (4.13) we conclude that the function*ϕ*given by (4.11) and
being a solution of equation (4.1) belongs to the class*C*^{◦}*α*[0, x0].

*§* 5. Proof of the Main Results

Theorem 5.1. *If at least eitherγ*1*orγ*2*is the characteristic of equation*
(1.1) (i.e., *τ*0 = 0) *and conditions* (3.7), (3.9) *are fulfilled, then problem*
(1.1)–(1.4)*has a unique solution in the class* *C*^{◦}^{2,1}*α* (D0)*for all* *α≥*0.

Theorem 5.2. *Let conditions* (3.7), (3.9) *be fulfilled and the straight*
*linesγ*1*,γ*2*be not the characteristics of equation* (1.1) (i.e., 0*< τ*0*<*1). If
*the equalityσ*= 0*holds, then problem* (1.1)–(1.4)*is uniquely solvable in the*
*class* *C*^{◦}^{2,1}*α* (D0) *for all* *α≥*0. If however *σ6*= 0, then problem (1.1)–(1.4)
*is uniquely solvable in the class* *C*^{◦}^{2,1}*α* (D0) *for* *α > α*0*, while for* *α < α*0

*problem* (1.1)–(1.4) *is normally solvable in Hausdorff ’s sense in the class*
*C**◦* ^{2,1}*α* (D0)*and its index* *κ*= +*∞. In particular, the homogeneous problem*
*corresponding to* (1.1)–(1.4) *has an infinite number of linearly independent*
*solutions.*

*Proof.* Rewrite system (3.14) in terms of the new unknown functions
*ψ(y) +b*1(y)ϕ(ρ2*y)≡ω(y), ν(y) +b*2(y)ϕ(ρ2*y)≡λ(y),* 0*≤y≤y*0*,*
as

(Kϕ)(x) =R*x*

0 *K*11(x, ξ)ϕ(ξ)dξ+*K*3(x)R*ρ*1*x*

0 *ω(η)dη+*

+K4(x)R*ρ*1*x*

0 *λ(η)dη*+*F*4(x), 0*≤x≤x*0*,*
*ω(y) =*R*ρ*2*y*

0 *K*12(ξ, y)ϕ(ξ)dξ+*K*6(y)R*y*

0 *ω(η)dη+*

+K7(y)R*y*

0 *λ(η)dη*+*F*5(y), 0*≤y≤y*0*,*
*λ(y) =*R*ρ*2*y*

0 *K*13(ξ, y)ϕ(ξ)dξ+*K*9(y)R*y*

0 *ω(η)dη+*

+K10(y)R*y*

0 *λ(η)dη*+*F*6(y), 0*≤y≤y*0*,*

(5.1)

where

*K*11(x, ξ)*≡K*1(x, ξ) +*K*_{1}* ^{∗}*(x, ξ),

*K*

_{1}

*(x, ξ)*

^{∗}*≡*

(*K*2(x, ξ)+ρ^{−}_{2}^{1}[K3(x)b1(ρ^{−}_{2}^{1}*ξ)+K*4(x)b2(ρ^{−}_{2}^{1}*ξ)],* 0*≤ξ≤τ*0*x,*
0, τ0*x < ξ≤x,*

*K*12(ξ, y)*≡K*5(ξ, y) +*ρ*^{−}_{2}^{1}[K6(y)b1(ρ^{−}_{2}^{1}*ξ) +K*7(y)b2(ρ^{−}_{2}^{1}*ξ)],*
*K*13(ξ, y)*≡K*8(ξ, y) +*ρ*^{−}_{2}^{1}[K9(y)b1(ρ^{−}_{2}^{1}*ξ) +K*10(y)b2(ρ^{−}_{2}^{1}*ξ)].*

*Remark* 5.1. If *ρ*2 = 0, then*ψ(y)≡ω(y),* *ν(y)≡λ(y), 0≤y* *≤y*0, and
the introduction of the new unknown functions*ω* and*λ*is superfluous.

We can rewrite system (5.1) in terms of the new independent variables
*x*=*x*0*t,y*=*y*0*t,* *ξ*=*x*0*τ,η*=*y*0*τ*, 0*≤t, τ≤*1, as

(K*ϕ)(t) =*e R*t*

0*K*e11(t, τ)*ϕ(τ)dτ*e +*K*e3(t)R*τ*1*t*

0 *ω(τ)dτ*e +
+*K*e4(t)R*τ*1*t*

0 *λ(τ)dτ*e +*F*e4(t), 0*≤t≤*1, 0*<*^{ρ}^{1}_{y}^{x}_{0}^{0} *≡τ*1*<*1,
e

*ω(t) =*R*τ*2*t*

0 *K*e12(τ, t)*ϕ(τ*e )dτ +*K*e6(t)R*t*

0*ω(τ)dτ*e +
+*K*e7(t)R*t*

0e*λ(τ)dτ*+*F*e5(t), 0*≤t≤*1, 0*<*^{ρ}^{2}_{x}^{y}_{0}^{0} *≡τ*2*<*1,
*λ(t) =*e R*τ*2*t*

0 *K*e13(τ, t)*ϕ(τ)dτ*e +*K*e9(t)R*t*

0*ω(τ)dτ*e +
+*K*e10(t)R*t*

0e*λ(τ*)dτ +*F*e6(t), 0*≤t≤*1,

(5.2)

where the functions with waves are expressions of the corresponding func-
tions in terms of the variables *t* and *τ, for example,* *ϕ(x) =* *ϕ(x*0*t)* *≡*

e

*ϕ(t),* *ω(y) =* *ω(y*0*t)* *≡* *ω(t),*e *K*e11(t, τ) *≡* *x*0*K*11(x0*t, x*0*τ),* *K*e12(τ, t) *≡*
*x*0*K*12(x0*τ, y*0*t),K*e3(t)*≡y*0*K*3(x0*t),K*e6(t)*≡y*0*K*6(y0*t), 0≤t,τ≤*1.

Let*T**i*(*ϕ,*e *ω,*e *λ),*e *i*= 1,2,3, be the linear integral operators acting by the
formulas

*T*1(*ϕ,*e *ω,*e e*λ)(t)≡*R*t*

0*K*e11(t, τ)*ϕ(τ)dτ*e +*K*e3(t)R*τ*1*t*

0 *ω(τ)dτ*e +
+*K*e4(t)R*τ*1*t*

0 e*λ(τ)dτ,* 0*≤t≤*1, 0*< τ*1*<*1,
*T*2(*ϕ,*e *ω,*e e*λ)(t)≡*R*τ*2*t*

0 *K*e12(τ, t)*ϕ(τ*e )dτ +*K*e6(t)R*t*

0*ω(τ)dτ*e +
+*K*e7(t)R*t*

0e*λ(τ)dτ,* 0*≤t≤*1, 0*< τ*2*<*1,
*T*3(*ϕ,*e *ω,*e e*λ)(t)≡*R*τ*2*t*

0 *K*e13(τ, t)*ϕ(τ*e )dτ +*K*e9(t)R*t*

0*ω(τ)dτ*e +
+*K*e10(t)R*t*

0e*λ(τ)dτ,* 0*≤t≤*1.

(5.3)

*Remark* 5.2. The integral operators *T**i*, *i* = 1,2,3, acting by formulas
(5.3) are Volterra type operators.

To prove Theorems 5.1 and 5.2 we shall solve system (5.2) for the un-
known functions *ϕ*e *∈C*^{◦}*α*, *ω*e *∈C*^{◦}*α*, e*λ* *∈C*^{◦}*α*, using the method of successive
approximations.

Set*ϕ*e0(t)*≡*0,*ω*e0(t)*≡*0,*λ*e0(t)*≡*0, 0*≤t≤*1, and for*n≥*1,

(K*ϕ*e*n*)(t) =
Z*t*
0

*K*e11(t, τ)*ϕ*e*n**−*1(τ)dτ+*K*e3(t)

*τ*1*t*

Z

0

e

*ω**n**−*1(τ)dτ+

+*K*e4(t)

*τ*1*t*

Z

0

e*λ**n**−*1(τ)dτ+*F*e4(t), 0*≤t≤*1, (5.4)

e
*ω**n*(t) =

*τ*2*t*

Z

0

*K*e12(τ, t)*ϕ*e*n**−*1(τ)dτ+*K*e6(t)
Z*t*
0

e

*ω**n**−*1(τ)dτ+

+*K*e7(t)
Z*t*

0

e*λ**n**−*1(τ)dτ+*F*e5(t), 0*≤t≤*1, (5.5)

e*λ**n*(t) =

*τ*2*t*

Z

0

*K*e13(τ, t)*ϕ*e*n**−*1(τ)dτ +*K*e9(t)
Z*t*
0

e

*ω**n**−*1(τ)dτ+

+*K*e10(t)
Z*t*
0

e*λ**n**−*1(τ)dτ+*F*e6(t), 0*≤t≤*1, (5.6)

where the operator*K* acts by (3.13).

Using estimate (4.2) and taking into account Remark 5.2, we shall prove below that under the assumptions of Lemma 4.1 or Lemma 4.2 we have the estimates

*|ϕ*e*n+1*(t)*−ϕ*e*n*(t)*| ≤ML*^{n}

*n!t*^{n+α}*,* (5.7)

*|eω**n+1*(t)*−*e*ω**n*(t)*| ≤ML*^{n}

*n!t*^{n+α}*,* (5.8)

*|*e*λ**n+1*(t)*−*e*λ**n*(t)*| ≤ML*^{n}

*n!t*^{n+α}*,* (5.9)

where *M* =*M*(M*i**, N**i**, P**i**, Q**i**, S**i**, f**i**, i*= 1,2,3, f, c, ρ1*, ρ*2) *>*0,*L*=*L(M**i**,*
*N**i**, P**i**, Q**i**, S**i**, i*= 1,2,3, c, ρ1*, ρ*2)*>*0 are sufficiently large positive numbers
which do not depend on *n* and which are to be defined, while *c* is the
constant from (4.2).

*Proof of Estimates* (5.7)–(5.9). Since a restriction is imposed on*f, f**i*, *i*=
1,2,3, we have *F*e3+i *∈C*^{◦}*α*, *i*= 1,2,3. Indeed, by (1.6) we have *|f*1(x)*| ≤*
*k*1*x** ^{α}*,

*k*1

*>*0,

*α≥*0,

*x∈*[0, x0]. Further, the first of equalities (3.2) gives

*|f*e1(x)*| ≤k*1*x** ^{α}*+

*k*4

*c*6

*ρ*1

*x*

Z

0

(x^{2}+*η*^{2})^{α/2}*dη*+*k*4*c*6

Z*x*
0

(ξ^{2}+*ρ*^{2}_{1}*x*^{2})^{α/2}*dξ*+

+k4*c*6

Z*x*
0

*ρ*1*x*

Z

0

(ξ^{2}+*η*^{2})^{α/2}*dξdη≤k*5*x*^{α}*, x∈*[0, x0], α*≥*0,

where*k*4*≡k*4(M1*, N*1*, P*1*, Q*1*, S*1),*k*5*≡k*5(k1*, k*4*, c*6*, x*0*, α, ρ*1) are the com-
pletely defined positive numbers.

Hence we conclude that*f*e1*∈C*^{◦}*α*. The case*f*e2*,f*e3*∈C*^{◦}*α*is proved similarly.

Taking into account the expressions of the functions*F*3+i,*i*= 1,2,3, from
(3.12), it is now easy to establish that *F*e3+i *∈C*^{◦}*α*, *i*= 1,2,3. Therefore by
(1.6) the following estimates hold: *|F*e3+i(t)*| ≤θ*3+i*t** ^{α}*or

*t*

^{−}

^{α}*|F*e3+i(t)

*| ≤θ*3+i,

*i*= 1,2,3,

*α≥*0,

*t*

*∈*[0,1]. If in this inequality

*t*is replaced by

*s∈*[0, t], then by the definition of a norm in the space

*C*

^{◦}*α*[0, t] we shall have

*kF*e3+i*k*^{◦}

*C**α*[0,t] *≤θ*3+i *i*= 1,2,3, *∀t∈*[0,1]. (5.10)
Since*ϕ*e0(t)*≡ω*e0(t)*≡*e*λ*0(t)*≡*0, 0*≤t≤*1, and under the assumptions
of Lemma 4.2 estimate (4.2) holds, from (5.4), (5.10) we shall have

*|ϕ*e1(t)*−ϕ*e0(t)*|*=*|ϕ*e1(t)*|*=*|*(K^{−}^{1}*F*e4)(t)*| ≤*

*≤ct*^{α}*kF*e4*k*^{◦}

*C**α*[0,t]*≤cθ*4*t*^{α}*.* (5.11)
(5.5), (5.10) in turn imply

*|eω*1(t)*−ω*e0(t)*|*=*|eω*1(t)*|*=*|F*e5(t)*| ≤θ*5*t*^{α}*.* (5.12)
Similarly, (5.6), (5.10) give

*|*e*λ*1(t)*−*e*λ*0(t)*|*=*|λ*e1(t)*|*=*|F*e6(t)*| ≤θ*6*t*^{α}*.* (5.13)
Assuming that estimates (5.7)–(5.9) hold for *n,n >*0, let us prove that
they are valid for*n*+ 1 for sufficiently large*M* and*L.*

Denote by *K*e the largest of the numbers sup

(t,τ)*∈*[0,1]*×*[0,1]*|K*e1i(t, τ)*|*,
*i*= 1,2,3, sup

*t**∈*[0,1]*|K*e*i*(t)*|*,*i*= 3,4,6,7,9,10.

From (5.4) we have

*K(ϕ*e*n+2**−ϕ*e*n+1*)(t) =*T*(*ϕ*e*n+1**−ϕ*e*n**,*e*ω**n+1**−ω*e*n**,λ*e*n+1**−*e*λ**n*)(t),(5.14)
where

*T*(*ϕ*e*n+1**−ϕ*e*n**,*e*ω**n+1**−ω*e*n**,λ*e*n+1**−*e*λ**n*)(t)*≡*
Z*t*
0

*K*e11(t, τ)(*ϕ*e*n+1**−ϕ*e*n*)(τ)dτ+

+*K*e3(t)

*τ*1*t*

Z

0

(*ω*e*n+1**−ω*e*n*)(τ)dτ+*K*e4(t)

*τ*1*t*

Z

0

(e*λ**n+1**−*e*λ**n*)(τ)dτ.