POLONICI MATHEMATICI LXV.2 (1997)
On boundary-value problems for partial differential equations of order higher than two
by Jan Popio lek (Bia lystok)
Abstract. We prove the existence of solutions of some boundary-value problems for partial differential equations of order higher than two. The general idea is similar to that in [1]. We make an essential use of the results of our paper [12].
1. The problem. Let x = χ
p(t), 0 < t ≤ T , p = 1, 2, be equations of non-intersecting curves on the (x, t) plane.
In this paper we prove the existence of a solution of the problem (1) L u(x, t) ≡
n+2
X
i=0
X
m j=0a
ij(x, t)D
xiD
tju(x, t) − D
nxD
m+1tu(x, t) = f (x, t), where (x, t) ∈ S
T= {(x, t) : χ
1(t) < x < χ
2(t), 0 < t ≤ T }, T = const < ∞, n, m ∈ N
0≡ N ∪ {0}, n + m > 0 (for n = m = 0 equation (1) is a parabolic equation of second order, the theory of which is well known), satisfying the initial conditions
(2) D
tlu(x, 0) = 0, χ
1(0) ≤ x ≤ χ
2(0), l = 0, 1, . . . , m, and the boundary conditions
(3) B
plu(χ
p(t), t) ≡
rpl
X
k=0
b
pkl(t)D
xku(χ
p(t), t) = g
pl(t),
where 0 < t ≤ T , p = 1, 2, l = 1, . . . , l
0= [(n + 3)/2] (denotes the greatest integer function), 0 ≤ r
p1< r
2p< . . . < r
lp0≤ n + 1, r
lp∈ N
0, b
prpl,l
(t) ≥ b
0= const > 0.
We distinguish the following four cases:
1) r
pl0
< n + 1, p = 1 or p = 2, n is odd,
1991 Mathematics Subject Classification: Primary 35G15; Secondary 45D05.
Key words and phrases: partial differential equation, boundary-value problem, Vol- terra integral equation.
[139]
2) r
pl0< n + 1, p = 1 or p = 2, n is even, 3) r
pl0
= n + 1, p = 1 or p = 2, n is odd, 4) r
pl0= n + 1, p = 1 or p = 2, n is even.
We shall exactly analyse cases 1) and 3). The argument in the remaining cases is similar. Note that in cases 1) and 3) we have to put [(n − 1)/2]
boundary conditions on one of the curves χ
pand [(n − 1)/2] + 1 on the other.
Boundary-value problems in rectangular domains and for particular cases of the operator L and of the boundary operators B
plhave been consid- ered in many papers (see [2], [3], [4], [10] and [15]). In [14] the boundary-value problem for the equation
D
n+2xu − D
xnD
tu = f (x, t, u, . . . , D
xn+1u) was examined. Paper [13] was devoted to the equation
L(D
x+ D
t)
nu(x, t) = f (x, t),
where L ≡ D
t− a(x, t)D
2x+ b(x, t)D
x+ c(x, t). In [5] some boundary-value problems for the equation
(D
2x− D
t)(aD
x+ bD
t+ c)u(x, t) = 0
were investigated, where a, b, c are constants and a · b 6= 0. Moreover, in [11]
Cauchy’s problem for equation (1) was examined.
Note that particular cases of equation (1) describe the propagation of waves in a compressible viscous medium (see [3], [6], [17]) and some problems of magneto-hydrodynamics (see [8], [9]).
2. Assumptions. We make the following assumptions:
(A.1) There are constants a
0and a
1such that
0 < a
0≤ a
n+2,m(x, t) ≤ a
1for (x, t) ∈ S
T(S
Tdenotes the closure of S
T).
(A.2) The coefficients a
ij(i = 0, 1, . . . , n + 2, j = 0, 1, . . . , m) are contin- uous in S
Tand satisfy the H¨older condition with respect to x with exponent α (0 < α ≤ 1); moreover, a
n+2,msatisfies the H¨older condition with respect to t with exponent
12α.
(A.3) The functions χ
p(p = 1, 2) have continuous derivatives up to order n
∗= [(n + 1)/2] in [0, T ] and the highest derivatives satisfy the H¨older condition
|∆
t[χ
(np ∗)(t)]| ≤ const
(∆t)
α/2if n + 1 is even,
(∆t)
(α+1)/2if n + 1 is odd,
where ∆
t[χ(t)] ≡ χ(t + ∆t) − χ(t), t, t + ∆t ∈ [0, T ], α ∈ (0, 1].
(A.4) The function f (x, t) is defined and continuous for (x, t) ∈ S
T, and satisfies the inequalities
|f(x, t)| ≤ M
f, |∆
xf (x, t)| ≤ m
f|∆x|
α,
where ∆
xf (x, t) ≡ f(x + ∆x, t) − f(x, t), (x, t), (x + ∆x, t) ∈ S
T, M
f, m
f= const > 0, α ∈ (0, 1].
(A.5) The functions g
pl, p = 1, 2, l = 1, . . . , l
0, are defined and have continuous derivatives D
tνg
pl(ν = 0, 1, . . . , M = [d
r/2], d
r= n−r
lp+2m+1) in [0, T ] and satisfy the conditions
|∆
t[D
tMg
pl(t)]| ≤ M
g(∆t)
α/2if d
ris even, (∆t)
(α+1)/2if d
ris odd, and D
tνg
pl(0) = 0, where M
g= const > 0, 0 < α ≤ 1.
(A.6) The functions b
pkl, p = 1, 2, l = 1, . . . , l
0, k = 0, 1, . . . , r
lp, are defined in [0, T ] and have continuous derivatives up to order M.
R e m a r k. Without restricting generality, we can assume b
prpl,l
(t) ≥ b
0≡ 1.
3. Solution of the problem. In all cases 1)–4) we shall seek a solution of the problem (1)–(3) in the form
(4) u(x, t) = X
2 σ=1l0
X
q=1 t
\
0
Λ
rσq(x, t; χ
σ(τ ), τ )ϕ
σq(τ ) dτ + ZS
T(x, t),
where ϕ
σqare unknown functions, Λ
rσqare the fundamental solutions of (1) constructed in [12] and
(5) Z
ST(x, t) =
\\
St
Λ
0(x, t; y, τ )f (y, τ ) dy dτ.
3.1. C a s e 1). Observe that the function u given by (4) satisfies equation (1) and initial conditions (2). Boundary conditions (3) lead to the system of equations
(6) g
pl(t) = X
2 σ=1l0
X
q=1 t
\
0
B
plΛ
rσq(χ
p(t), t; χ
σ(τ ), τ )ϕ
σq(τ ) dτ + z
pl(t), where z
pl(t) = B
plZ
ST(χ
p(t), t), 0 < t ≤ T , p = 1, 2, l = 1, . . . , l
0.
By Lemma 3 of [12] we obtain (7) D
rp
xl
w
rpq(χ
p(τ ), t; χ
p(τ ), τ )
=
0, 1 ≤ l < q,
(−1)
n−rpl√
π[a(τ )]
(n−rpl)/2Γ
−1(d
r/2)(t − τ)
dr/2−1, q ≤ l ≤ l
0, (p = 1, 2, l, q = 1, . . . , l
0), where d
r= n − r
pl+ 2m + 1 and the functions w
rpl
are defined by formula (6) of [12], and a(τ ) = a
n+2,m(χ
p(τ ), τ ).
Using the definition of the operator I
κ([12], (25)) and (7) we can write (8)
t
\
0
D
rp
xl
w
rpq(χ
p(τ ), t; χ
p(τ ), τ )ϕ
pq(τ ) dτ = c
plqI
dr/2([a(t)]
(n−rpl)/2ϕ
pq(t)) (p = 1, 2, l, q = 1, . . . , l
0, 0 < t ≤ T ), where
(9) c
plq=
0, 1 ≤ l < q, (−1)
n−rpl√
π, q ≤ l ≤ l
0. By (8) and (9) we can rewrite system (6) in the form (10)
l0
X
q=1
c
plqI
dr/2([a(t)]
(n−rpl)/2ϕ
pq(t))
+ X
2 σ=1l0
X
q=1 t
\
0
K
pσlq(t, τ )ϕ
σp(τ ) dτ + z
pl(t) = g
pl(t), where
(11) K
pσlq(t, τ ) = B
plΛ
rqσ(χ
p(t), tχ
p(τ ), τ )
− 0 if σ 6= p or σ = p and 1 ≤ l < q,
D
rp
xl
w
rpq(χ
p(τ ), t; χ
p(τ ), τ ) if σ = p and q ≤ l ≤ l
0, (p, σ = 1, 2, l, q = 1, . . . , l
0, 0 < t ≤ T ).
(10) is a system of first-kind Volterra equations. Using the method given by Baderko [1] and the properties of the operator R
1/2defined by formula (14) of [12], we reduce this system to a system of second-kind Volterra equations. Applying to both sides of (10) the operator R
d1/2r, where d
r= n − r
lp+ 2m + 1, by Lemma 4 of [12], we obtain
(12)
l0
X
q=1
c
plq[a(t)]
(n−rpl)/2ϕ
pq(t) + X
2 σ=1l0
X
q=1
R
d1/2rh
t\0
K
pσlq(t, τ )ϕ
σq(τ ) dτ i + R
d1/2r[z
pl(t)] = R
d1/2r[g
pl(t)], p = 1, 2, l = 1, . . . , l
0, 0 < t ≤ T.
By Theorem 1 of [12],
(13) |D
tνK
pσlq(t, τ )| ≤ const (t − τ)
(dr−2ν+α)/2−1(ν = 0, 1, . . . , M = [d
r/2], d
r= n − r
pl+ 2m + 1, p, σ = 1, 2, l, q = 1, . . . , l
0, 0 ≤ τ < t ≤ T , 0 < α ≤ 1).
We consider two cases: (i) d
ris even, (ii) d
ris odd.
In case (i) the function K
pσlqsatisfies condition (18) of Lemma 4 of [12]
with N = d
r/2 and ̺ = α/2; hence, in view of formula (19) of [12] we have (14) R
d1/2rh
t\0
K
pσlq(t, τ )ϕ
σq(τ ) dτ i
=
t
\
0
D
dtr/2K
pσlq(t, τ )ϕ
σq(τ ) dτ.
In case (ii), K
pσlqsatisfies the same condition with N = (d
r− 1)/2 and
̺ = (α + 1)/2; hence, by formula (20) of [12] we get (15) R
d1/2rh
t\0
K
pσlq(t, τ )ϕ
σq(τ ) dτ i
=
t
\
0
R
1/2[D
tdr/2K
pσlq(t, τ )]ϕ
σq(τ ) dτ.
By (14) and (15) system (12) can be written in the form (16)
l0
X
q=1
c
plq[a(t)]
(n−rlp)/2ϕ
pq(t)
+ X
2 σ=1l0
X
q=1 t
\
0
K
pσlq(t, τ )ϕ
σq(τ ) dτ + z
pl(t) = g
pl(t) (p = 1, 2, l = 1, . . . , l
0, 0 < t ≤ T ), where
(17) K
pσlq(t, τ ) =
( D
tdr/2K
pσlq(t, τ ) if d
ris even, R
1/2[D
(dt r−1)/2K
pσlq(t, τ )] if d
ris odd, (18) z
pl(t) = R
d1/2r[z
pl(t)],
(19) g
pl(t) = R
d1/2r[g
pl(t)].
Now, we estimate the functions K
pσlq, z
pland g
pl. In case (i), by Theorem 1 [12], we have
(20) |D
dtr/2K
pσlq(t, τ )| ≤ const (t − τ)
α/2−1, 0 ≤ τ < t ≤ T, (21) |∆
tD
tdr/2K
pσlq(t, τ )| ≤ const (∆t)
β/2(t − τ)
µ−1, 0 ≤ τ < t ≤ t + ∆t ≤ T , 0 < β ≤ α ≤ 1, µ = min{α/2, 1 − α/2}.
Analogously, in case (ii), we get
(22) |D
(dt r−1)/2K
pσlq(t, τ )| ≤ const (t − τ)
(1+α)/2−1, 0 ≤ τ < t ≤ T, (23) |∆
tD
t(dr−1)/2K
pσlq(t, τ )| ≤ const (∆t)
(1+α)/2(t − τ)
µ−1, 0 ≤ τ < t ≤ t + ∆t ≤ T , µ = min{α/2, 1 − α/2}.
From (22) and (23) it follows that the functions D
t(dr−1)/2K
pσlqsatisfy the assumptions of Lemma 6 of [12], and therefore
(24) |R
1/2[D
t(dr−1)/2K
pσlq(t, τ )]| ≤ const (t − τ)
α/2−1, 0 ≤ τ < t ≤ T, (25) |∆
tR
1/2[D
(dt r−1)/2K
pσlq(t, τ )]| ≤ const (∆t)
β/2(t − τ)
µ−1, 0 ≤ τ < t ≤ t + ∆t ≤ T , 0 < β ≤ α ≤ 1, µ = min{α/2, 1 − α/2}.
Combining (20), (21), (24) and (25), we arrive at
(26) |K
pσlq(t, τ )| ≤ const (t − τ)
α/2−1, 0 ≤ τ < t ≤ T,
(27) |∆
tK
pσlq(t, τ )| ≤ const (∆t)
β/2(t − τ)
µ−1, 0 ≤ τ < t ≤ t + ∆t ≤ T, p, σ = 1, 2, l, q = 1, . . . , l
0, 0 < β ≤ α ≤ 1, µ = min{α/2, 1 − α/2}.
Now, we examine the function g
plgiven by (19). If d
ris even, by (A.5) the function g
plsatisfies the assumptions of Lemma 5 of [12] with N = d
r/2, and so
g
pl(t) = D
dtr/2g
pl(t), 0 ≤ τ < t ≤ T.
If d
ris odd, by (A.5), g
plsatisfies the assumptions of that lemma with N = (d
r− 1)/2, and thus
g
pl(t) = R
1/2[D
t(dr−1)/2g
pl(t)], 0 ≤ τ < t ≤ T.
Hence
(28) g
pl(t) =
( D
dtr/2g
pl(t) if d
ris even, R
1/2[D
t(dr−1)/2g
pl(t)] if d
ris odd, (d
r= n − r
lp+ 2m + 1, p = 1, 2, l = 1, . . . , l
0, 0 < t ≤ T ).
From (28) and (A.5) in case (i) we obtain
(29) |∆
tg
pl(t)| ≤ const (∆t)
α/2, 0 ≤ t < t + ∆t ≤ T, g
pl(0) = 0.
In case (ii) we have
|∆
tD
(dr−1)/2g
pl(t)| ≤ const (∆t)
(1+α)/2, 0 ≤ t < t + ∆t ≤ T, D
(dt r−1)/2g
pl(0) = 0,
hence, by Lemma 2 of [16], we also get (29).
It remains to investigate the function z
plgiven by (18). Using (5) and Lemma 5 of [12], we obtain
z
pl(t) =
( D
dtr/2z
pl(t) if d
ris even, R
1/2[D
t(dr−1)/2z
pl(t)] if d
ris odd,
(d
r= n − r
pl+ 2m + 1, p = 1, 2, l = 1, . . . , l
0, 0 < t ≤ T ); hence, by Lemma 8 of [12], we find
(30) |∆
tz
pl(t)| ≤ const (∆t)
α/2, 0 ≤ t < t + ∆t ≤ T, z
pl(0) = 0.
Now, we return to system (16). Multiplying both sides by [a(t)]
−(n−rpl)/2we obtain
(31)
l0
X
q=1
c
plqϕ
pq(t) + X
2 σ=1l0
X
q=1 t
\
0
K
pσlq(t, τ )ϕ
σq(τ ) dτ + z
pl(t) = g
pl(t) (p = 1, 2, l = 1, . . . , l
0, 0 < t ≤ T ), where
K
pσlq(t, τ ) = [a(t)]
−(n−rpl)/2K
pσlq(t, τ ), z
pl(t) = [a(t)]
−(n−rlp)/2z
pl(t),
g
pl(t) = [a(t)]
−(n−rpl)/2g
pl(t), a(t) = a
n+2,m(χ
p(t), t).
Using assumptions (A.1), (A.2) it can be proved that the functions K
pσlq, z
pland g
plsatisfy the estimates (26), (27), (29) and (30) respectively.
Now, we treat system (31) as an algebraic system with respect to the functions ϕ
pq, p = 1, 2, q = 1, . . . , l
0. Its determinant is of the form
W =
c
p110 0 . . . 0 c
p21c
p220 . . . 0 c
p31c
p32c
p33. . . 0 .. . .. . .. . . .. .. . c
pl0,1c
pl0,2c
pl0,3. . . c
pl0,l0.
Hence, in view of (9), we have W = c
p11c
p22. . . c
pl0,l0
= (−1)
nl0−(r1p+rp2+...+rpl0)( √
π)
l06= 0 on one of the curves χ
p(see §1) and
W = c
p11c
p22. . . c
pl∗−1,l∗−1
c
pl∗+1,l∗+1
. . . c
pl0,l06= 0 on the other. Cramer’s formulae yield
(32) ϕ
pq(t) + X
2 σ=1l0
X
q=1 t
\
0
K e
pσlq(t, τ )ϕ
σq(τ ) dτ + ez
pl(t) = g e
pl(t), where
K e
pσlq(t, τ ) =
l0
X
v=1
A
plvK
pσvq(t, τ ), ez
pl(t) =
l0
X
v=1
A
plvz
pv(t),
e g
pl(t) =
l0
X
v=1
A
plvg
pv(t), A
plv= C
lvp/W,
p = 1, 2, l = 1, . . . , l
0, 0 < t ≤ T (C
lvpdenotes the algebraic complement of c
plvin W).
It is easy to see that e K
pσlq, ez
pland g e
plsatisfy the same estimates as K
pσlq, z
pland g
plrespectively. Thus, (32) is a system of second-type Volterra integral equations with weak singularities and hence it has a solution of the form (33) ϕ
pl(t) = g e
pl(t) − ez
pl(t) +
X
2 σ=1l0
X
q=1 t
\
0
K
pσlq
(t, τ )[ g e
σq(τ ) − ez
σq(τ )] dτ, where K
pσlqdenote the resolvent kernels of the e K
pσlq, p, σ = 1, 2, l, q = 1, . . . , l
0. Moreover, the estimates (26), (27), (29) and (30) imply
(34) |∆
tϕ
pl(t)| ≤ const (∆t)
β/2, ϕ
pl(0) = 0
(p = 1, 2, l = 1, . . . , l
0, 0 ≤ t < t + ∆t ≤ T , 0 < β ≤ α ≤ 1).
3.2. C a s e 3). Without losing generality we may assume that on both the curves χ
p, l
0−1 conditions are posed given by the operators B
pl, p = 1, 2, l = 1, . . . , l
0− 1, with 0 ≤ r
p1< r
2p< . . . < r
lp0−1
< n + 1, and moreover, one more condition given by B
pl0with r
1l0= n + 1 is posed on χ
1.
Now, we rewrite formula (4) in a form more suitable for further consid- erations:
u(x, t) =
t
\
0
Λ
n+1(x, t; χ
1(τ ), τ )ϕ
1l0(τ ) dτ (35)
+ X
2 σ=1l
X
0−1 q=1t
\
0
Λ
rqσ(x, t; χ
σ(τ ), τ )ϕ
σqdτ + Z
ST(x, t),
where the functions Λ
rσqfor σ = 1, 2, q = 1, . . . , l
0−1 are defined by formula (7) of [12] and
(36) Λ
n+1(x, t; y, τ ) = Λ
r1∗(x, t; y, τ )
((x, t), (y, τ ) ∈ S
T), where r
1∗is a positive integer with 0 ≤ r
∗1≤ n, r
1∗6= r
1lfor l = 0, 1, . . . , l
0− 1.
Applying to both sides of (35) the operator B
1l0given by (3), we get (37) B
1l0u(x, t) =
t
\
0
B
1l0Λ
r1∗
(x, t; χ
1(τ ), τ )ϕ
1l0(τ ) dτ +
X
2 σ=1l
X
0−1 q=1t
\
0
B
1l0Λ
rσq(x, t; χ
1(τ ), τ )ϕ
σq(τ ) dτ + B
1l0Z
ST(x, t).
By (5) and Lemma 2 of [12] we can write B
1l0Λ
r1∗
(x, t; χ
1(τ ), τ ) = P
m[D
xω
χ1(τ ),τ(x, t; χ
1(τ ), τ )]+B
1l0w
r1∗
(x, t; χ
1(τ ), τ ) ((x, t) ∈ S
T). Consider the integral
J
m(x, t) =
t
\
0
P
m[D
xω
χ1(τ ),τ(x, t; χ
1(τ ), τ )]ϕ
1l0(τ ) dτ (m ∈ N
0).
We investigate its behaviour as x → χ
1(t), (x, t) ∈ S
T. For m = 0 we have J
0(x, t) =
t
\
0
D
xω
χ1(τ ),τ(x, t; χ
1(τ ), τ )ϕ
1l0(τ ) dτ.
This is a heat potential of second kind which has the following property ([7], p. 1085):
(38) lim
x→χ1(t)
J
0(x, t) = −
r π
a(t) ϕ
1l0(t) + J
0(χ
1(t), t), (x, t) ∈ S
T,
where a(t) = a
n+2,0(χ
1(t), t).
For m > 0 the integral J
mcan be written in the form J
m(x, t) =
t
\
0
\tτ
(t − ζ
m)
m−1(m − 1)! D
xω
χ1(τ ),τ(x, ζ
m; χ
1(τ ), τ )dζ
mϕ
1l0(τ ) dτ.
It follows that
J
m(x, t) =
t
\
0
(t − ζ
m)
m−1(m − 1)! J
0(x, ζ
m) dζ
m, and hence, by (38), we obtain
(39) lim
x→χ1(t)
J
m(x, t) = −
t
\
0
(t − ζ
m)
m−1(m − 1)!
r π
a(t) ϕ
1l0(ζ
m) dζ
m+ J
m(χ
1(t), t) ((x, t) ∈ S
T, m ∈ N).
Making use of the definition of the operator I
κ(see (25) in [12]), formulae (38) and (39) can be written in the form
(40) lim
x→χ1(t)
J
m(x, t) = −I
mr π
a(t) ϕ
1l0(t)
+ J
m(χ
1(t), t) ((x, t) ∈ S
T, m ∈ N
0), where a(t) = a
n+2,m(χ
1(t), t).
Passing to the limit x → χ
1(t) in (37), we have g
1l0(t) = − I
mr π
a(t) ϕ
1l0(t)
+
t
\
0
K
11l0l0(t, τ )ϕ
1l0(τ ) dτ (41)
+ X
2 σ=1l
X
0−1 q=1t
\
0
K
1σl0q
(t, τ )ϕ
σq(τ ) dτ + z
1l0(t), where K
11l0l0(t, τ ) = B
1l0Λ
r1∗
(χ
1(t), t; χ
1(τ ), τ ), K
1σl0q(t, τ ) = B
1l0Λ
rσq(χ
1(t), t;
χ
σ(τ ), τ ), σ = 1, 2, q = 1, . . . , l
0− 1, 0 < t ≤ T , the operators B
1l0are defined by formula (34) of [12] and the functions z
1l0are given by relation (42) of [12].
Applying R
2m1/2to both sides of (41), by Lemmas 4 and 5 of [12], we obtain (42) −
r π
a(t) ϕ
1l0(t) +
t
\
0
K
11l0l0(t, τ )ϕ
1l0(τ ) dτ
+ X
2 σ=1l
X
0−1 q=1t
\
0
K
1σl0q(t, τ )ϕ
σq(τ ) dτ + z
1l0(t) = g
1l0(t), 0 < t ≤ T,
where K
11l0l0(t, τ ) = D
mtK
11l0l0(t, τ ), K
1σl0q(t, τ ) = D
tmK
1σl0q(t, τ ), z
1l0(t) = D
tmz
1l0
(t), g
1l0(t) = D
tmg
1l0
(t), σ = 1, 2, q = 1, . . . , l
0− 1.
Using Theorem 2 of [12] we find the estimates
|K
11l0l0(t, τ )| ≤ const (t − τ)
α/2−1, 0 ≤ τ < t ≤ T, (43)
|K
1σl0q(t, τ )| ≤ const (t − τ)
α/2−1, 0 ≤ τ < t ≤ T, (44)
|∆
tK
11l0l0(t, τ )| ≤ const (∆t)
β/2(t − τ)
µ−1, 0 ≤ τ < t ≤ t + ∆t ≤ T, (45)
|∆
tK
1σl0q(t, τ )| ≤ const (∆t)
β/2(t − τ)
µ−1, 0 ≤ τ < t ≤ t + ∆t ≤ T, (46)
where σ = 1, 2, q = 1, . . . , l
0− 1, 0 < β ≤ α ≤ 1, µ = min{α/2, 1 − α/2}.
Similarly, using Lemma 9 of [12], we have
(47) |∆
tz
1l0(t)| ≤ const (∆t)
α/2, 0 ≤ t < t + ∆t ≤ T, z
1l0(0) = 0, moreover, in view of assumption (A.5), we get
(48) |∆
tg
1l0(t)| ≤ const (∆t)
α/2, 0 ≤ t < t + ∆t ≤ T, g
1l0(0) = 0.
Observe that equation (42) can be written in the form (49) ϕ
1l0(t) +
t
\
0
K e
11l0l0(t, τ )ϕ
1l0(τ ) dτ
+ X
2 σ=1l
X
0−1 q=1t
\
0
K e
1σl0q(t, τ )ϕ
σq(τ ) dτ + ez
1l0(t) = g e
1l0(t),
where e K
11l0l0(t, τ ) = − p
a (t)/π ·K
11l0l0(t, τ ), e K
1σl0q(t, τ ) = − p
a (t)/π ·K
1σl0q(t, τ ), ez
1l0(t) = − p
a (t)/π · z
1l0(t), g e
1l0(t) = − p
a (t)/π · g
1l0(t), σ = 1, 2, q = 1, . . . , l
0− 1, 0 < t ≤ T .
From assumptions (A.1) and (A.2) it follows that K
11l0l0, K
1σl0q, z
1l0and g
1l0
satisfy inequalities (43)–(48) respectively. This means that if we treat the functions ϕ
σq, σ = 1, 2, q = 1, . . . , l
0− 1, as parameters, then (49) is a second-kind Volterra equation with respect to ϕ
1l0. Because the singularity of the kernel of this equation is weak one can solve it.
Imposing on the function u, given by formula (35), the remaining bound- ary conditions (3) given by the operators B
p1, B
p2, . . ., B
pl0−1with 0 ≤ r
p1<
r
2p< . . . < r
lp0−1
< n + 1 (p = 1, 2, l
0= [(n + 3)/2]), we obtain the following system of integral equations:
(50) X
2 σ=1l
X
0−1 q=1t
\
0
B
plΛ
rqσ(χ
p(t), t; χ
p(τ ), τ )ϕ
σq(τ ) dτ
+
t
\
0
B
plΛ
r∗1(χ
p(t), t; χ
1(τ ), τ )ϕ
1l0(τ ) dτ + z
pl(t) = g
pl(t),
p = 1, 2, l = 1, . . . , l
0− 1, 0 < t ≤ T .
System (50) is a system of first-kind Volterra integral equations with 2(l
0− 1) equations and 2(l
0− 1) unknown functions ϕ
σq, σ = 1, 2, q = 1, . . . , l
0− 1. Now, we apply to system (50) the method presented in subsec- tion 3.1 to obtain
(51) ϕ
pl(t) + X
2 σ=1l
X
0−1 q=1t
\
0
K e
pσlq(t, τ )ϕ
σq(τ ) dτ
=
t
\
0
K e
11l0l0(t, τ )ϕ
1l0(τ ) dτ − e g
pl(t) − ez
pl(t), p = 1, 2, l = 1, . . . , l
0− 1, 0 < t ≤ T .
The functions e K
pσlq, g e
pland ez
plsatisfy inequalities (26), (27), (29) and (30), respectively, thus (51) is a system of second-kind Volterra integral equations with weak singularities.
Finally, we are able to find a solution of system (49), (51) in the form ϕ
pl(t) = g
pl(t) − z
pl(t)
(52)
+ X
2 σ=1l
X
0−1 q=1 t\
0