c Allerton Press, Inc., 2015. ISSN 1066-369X, Russian Mathematics (Iz. VUZ), 2015, Vol. 59, No. 6, pp. 23–33. c K.B. Sabitov, 2015, published in Izvestiya Vysshikh Uchebnykh Zavedenii. Matematika, 2015, No. 6, pp. 31–42. Original Russian Text
Initial-Boundary Problem for Parabolic-Hyperbolic Equation with Loaded Summands K. B. Sabitov1* 1
Volga Region Social-Humanitarian Academy, ul. Antonova-Ovseenko 26, Samara, 443090 Russia; Institute of Applied Research of Republic Bashkortostan, ul. Odesskaya 68, Sterlitamak, 453103 Russia Received January 17, 2014
Abstract—In this paper we find necessary and sufficient conditions for uniqueness of solution to the initial-boundary problem for a loaded equation of mixed parabolic-hyperbolic type. The solution is constructed as a sum of a series in eigenfunctions of the corresponding one-dimensional problem on eigenvalues. In the proof of convergence of the series the problem of small denominators arises. Under certain conditions on these problems we obtain an estimate for a small denominator to be separated from zero that allows to prove the existence theorem in the class of regular solutions. DOI: 10.3103/S1066369X15060055 Keywords: mixed-type equation with loaded summands, initial-boundary problem, uniqueness, existence, stability.
1. INTRODUCTION Consider a loaded mixed-type equation ut − uxx + a1 (t)u(x, 0) + a2 (t)u(x, d1 ) = 0, Lu = utt − uxx + b1 (t)u(x, 0) + b2 (t)u(x, −d2 ) = 0,
t > 0; t < 0,
(1)
in a rectangular domain D = {(x, t) | 0 < x < 1, −α < t < β}, where ai (t), bi (t) are given continuous functions, i = 1, 2; α, β, d1 , and d2 are given positive numbers, 0 < d1 ≤ β, −α < −d2 < 0. Following [1, 2], let us set up the following first boundary problem for Eq. (1). Problem. In the domain D find a function u(x, t) satisfying the conditions u(x, t) ∈ C 1 (D) ∩ C 2 (D− ) ∩ Cx2 (D+ ),
(2)
Lu(x, t) = 0, (x, t) ∈ D− ∪ D+ ,
(3)
u(0, t) = u(1, t) = 0, −α ≤ t ≤ β,
(4)
u(x, −α) = ψ(x), 0 ≤ x ≤ 1,
(5)
where ψ(x) is a given smooth enough function, ψ(0) = ψ(1) = 0, D− = D ∩ {t < 0}, D+ = D ∩ {t > 0}. Note that in [3–9] boundary problems for loaded partial differential equations of separate and mixed types are studied. This is due to the fact that loaded equations find applications in various areas of mathematics and science, e.g., fractal and extreme state physics, mathematical biology and mathematical economics [6–8], in the theory of inverse problems for differential equations [8]. For *
E-mail:
[email protected].
23
24
SABITOV
a2 (t) = 0, b2 (t) = 0 problem (2)–(5) is studied in [3]. A similar problem for a loaded mixed elliptichyperbolic type equation is studied in [10]. In this paper we obtain a uniqueness criterion for solution to problem (2)–(5). The solution is constructed as a sum of a series in eigenfunctions of the corresponding one-dimensional problem on eigenvalues. We prove the stability of the solution to problem (2)–(5) with respect to a given function ψ(x). 2. UNIQUENESS OF SOLUTION TO THE PROBLEM Let u(x, t) be a solution to problem (2)–(5). Following [1, 2], consider the functions √ 1 u(x, t) sin λk x dx, λk = πk, k ∈ N. uk (t) = 2
(6)
0
Based on (6), introduce the functions vε (t) =
√ 2
1−ε
(7)
u(x, t) sin λk x dx, ε
where ε > 0 is a small enough number. Differentiating equality (7) by t for t ∈ (0, β) once, and for t ∈ (−α, 0) twice, and taking into account Eq. (1), we get √ 1−ε [uxx (x, t) − a1 (t)u(x, 0) − a2 (t)u(x, d1 )] sin λk x dx, t > 0, (8) vε (t) = 2 ε
vε (t) =
√ 2
1−ε
[uxx (x, t) − b1 (t)u(x, 0) − b2 (t)u(x, −d2 )] sin λk x dx, t < 0.
(9)
ε
We integrate by parts twice in the integrals containing uxx (x, t) on the right-hand sides of equalities (8) and (9). Passing to the limit as ε → 0 taking into account the homogeneous boundary conditions (4), we get uk (t) + λ2k uk (t) + a1 (t)uk (0) + a2 (t)uk (d1 ) = 0, t > 0,
(10)
uk (t) + λ2k uk (t) + b1 (t)uk (0) + b2 (t)uk (−d2 ) = 0, t < 0.
(11)
Differential equations (10) and (11) have common solutions ⎧ 2 ⎨Dk e−λk t − Dk a1k (t) − uk (d1 )a2k (t), uk (t) = A u (−d2 ) ⎩Ak cos λk t + Bk sin λk t + k b1k (t) + k b2k (t), λk λk
t > 0; t < 0,
where Ak , Bk , and Dk are arbitrary constants, uk (d1 ) and uk (−d2 ) are yet unknown constants, t t 2 ai (s)e−λk (t−s) ds, bik (t) = bi (s) sin[λk (s − t)] ds, i = 1, 2. aik (t) = 0
(12)
(13)
0
Substituting t = d1 , and t = −d2 into (12), we get e−λk d1 − a1k (d1 ) , 1 + a2k (d1 )
(14)
λk [Ak cos λk d2 − Bk sin λk d2 ] + Ak b1k (−d2 ) , λk − b2k (−d2 )
(15)
2
uk (d1 ) = Dk
uk (−d2 ) = when for all k ∈ N
1 + a2k (d1 ) = 0, λk − b2k (−d2 ) = 0.
(16)
RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 59 No. 6 2015
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Substituting (14), (15) into (12), we get ⎧ 2 ⎪ e−λk d1 − a1k (d1 ) 2t ⎪ −λ ⎪ k Dk e a2k (t), − Dk a1k (t) − Dk t > 0; ⎪ ⎪ 1 + a2k (d1 ) ⎪ ⎨ A uk (t) = Ak cos λk t + Bk sin λk t + k b1k (t) ⎪ λk ⎪ ⎪ ⎪ λ [A cos λ d − Bk sin λk d2 ] + Ak b1k (−d2 ) ⎪ 2 k k k ⎪ + b2k (t), t < 0. ⎩ λk (λk − b2k (−d2 ))
25
(17)
For function (17) due to (2) the conjugation conditions are fulfilled: uk (0 − 0) = uk (0 + 0), uk (0 − 0) = uk (0 + 0). Conditions (18) are fulfilled only in the case of
a1 (0) a2 (0) + Mk , λk + λk λk
(18)
Dk = Ak , Bk = −Ak
(19)
where e−λk d1 − a1k (d1 ) . Mk = 1 + a2k (d1 ) 2
(20)
Substituting the obtained Bk into (15), we have Here
uk (−d2 ) = Ak Nk .
(21)
λk cos λk d2 + b1k (−d2 ) + λ2k + a1 (0) + a2 (0)Mk sin λk d2 . Nk = λk − b2k (−d2 )
(22)
Substituting (19) and (21) into (17), we get
2 Ak e−λk t − a1k (t) − Mk a2k (t) , t > 0; uk (t) = Ak [cos λk t − λk sin λk t − ωk (t)] , t < 0,
(23)
where ωk (t) = [a1 (0) + a2 (0)Mk ]
sin λk t 1 − [b1k (t) + Nk b2k (t)] . λk λk
In order to find the constants Ak let us use the initial condition (5) and formula (6): √ 1 √ 1 u(x, −α) sin λk x dx = 2 ψ(x) sin λk x dx = ψk . uk (−α) = 2 0
(24)
(25)
0
Then from (23) based on (25) we get Ak =
ψk Δ(k)
(26)
under the condition that for all k ∈ N Δ(k) = cos λk α + λk sin λk α − ωk (−α) = 0. Substituting (26) into (23), we find the final form of the functions ⎧ ψ
2 k ⎪ e−λk t − a1k (t) − Mk a2k (t) , t > 0; ⎨ Δ(k) uk (t) = ψ ⎪ ⎩ k [cos λk t − λk sin λk t − ωk (t)] , t < 0. Δ(k) RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 59 No. 6 2015
(27)
(28)
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SABITOV
Now let us prove the uniqueness of the solution to problem (2)–(5) under conditions (27). Let ψ(x) ≡ 0, then ψk ≡ 0 and from (28) and (6) we get 1 u(x, t) sin λk x dx = 0, k = 1, 2, . . . (29) 0
Equalities (29) due to the completeness of the system of sines {sin λk x} in the space L2 [0, 1] yield u(x, t) = 0 almost everywhere in [0, 1] for any t ∈ [−α, β]. Since by condition (2) the function u(x, t) is continuous in D, it is true that u(x, t) ≡ 0 in D. Let for some α, di , ai (t), bi (t), i = 1, 2, and k = p ∈ N condition (27) be violated, i.e., Δ(p) = cos λp α + λp sin λp α − ωp (−α) = 0.
(30)
In order to find zeros of Δ(p) in view of (24) let us represent it in the following form: Δ(p) = where Kp =
1 Kp sin(λp α + τp ) + [b1p (−α) + Np b2p (−α)], λp λp
2
(31)
λp . Kp
(32)
1 [b1p (−α) + Np b2p (−α)] = Πp (−α). Kp
(33)
λ2p + a1 (0) + a2 (0)Mp
+ λ2p , τp = arcsin
From relation (31) we have sin(λp α + τp ) = −
If the right-hand side of Eq. (33) is greater than one in the absolute value, then it has no solutions, so let us stipulate |Πp (−α)| =
|b1p (−α) + Np b2p (−α)| ≤ 1. Kp
(34)
Hence under condition (34) Eq. (33) is equivalent to the equation α=
b1p (−α) + Np b2p (−α) πn τp (−1)n+1 arcsin + − = f (α), n ∈ N. λp Kp λp λp
(35)
If ωp (−α) = 0, then Eq. (30) has a countable set of solutions 1 n ϕp , n ∈ N, ϕp = arcsin . α= − p λp 1 + λ2 p
If b1p (−α) = b2p (−α) = 0, then from (35) we also get τp n α = − , n ∈ N. p λp If bi (t) = bi = const = 0, i = 1, 2, then bip (−α) = bi (cos λp α − 1)/λp , i = 1, 2, and from (30) we get Δ(p) = 0 if and only if α=
θp b1 + b2 Np n (−1)n arcsin + − , n ∈ N. λp Tp p λp
Here λ2p + b1 + b2 Np , θp = arcsin Tp Tp =
λ2p + b1 + b2 Np
2
2 + λ2p λ2p + a1 (0) + a2 (0)Mp .
In order to consider the general case, let us prove several lemmas. Let ai = max |ai (t)|, bi = max |bi (t)|, i = 1, 2. 0≤t≤β
−α≤t≤0
RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 59 No. 6 2015
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27
Lemma 1. If one of the conditions is fulfilled: a) a2 (t) ≥ 0 for all t ∈ [0, d1 ], b) a2 < π 2 , then there exists a positive constant d0 , depending, generally speaking, on a2 , such that for all k ∈ N the following estimate is true: 1 + a2k (d1 ) ≥ d0 > 0. Proof. Let a2 (t) ≥ 0 in [0, d1 ], then
d1
1 + a2k (d1 ) = 1 +
(36)
a2 (s)e−λk (d1 −s) ds ≥ 1 2
0
for all k ∈ N. In case b) we have d1 1 + a2k (d1 ) ≥ 1 −
1 − e−λk d1 ds ≥ 1 − a2
λ2k 2
−λ2k (d1 −s)
|a2 (s)|e 0
e−λk d1
a2
a2
a2
= 1 + a2
− 2 > 1 − 2 ≥ 1 − 2 > 0. 2 π λk λk λk 2
Lemma 2. If the conditions of Lemma 1 are fulfilled, then there exists a positive constant E1 , depending on d0 , d1 , a1 , such that for all k ∈ N the following estimate is true: |Mk | ≤ E1 /k2 .
(37)
Proof. Based on estimate (36) we have 2 2 e−λk d1 + |a1k (d1 )| 1 −λ2 d1 1 − e−λk d1 1 −λ2 d1 a1
E1 ≤ e k + a1
≤ e k + 2 ≤ 2. |Mk | ≤ 2 d0 d0 d0 k λk λk Lemma 3. If one of the conditions is fulfilled: a) b2k (−d2 ) =
−d 2
b2 (s) sin[λk (s + d2 )] ds = 0, k ∈ N;
0
b) b2 (s) is an increasing non-negative function in [−d2 , 0] and b2 (0) < π 2 /2; c) b2 ≤ π/d2 , then there exists a positive constant d0 , depending on d2 , b2 , b2 (0), such that for all k ∈ N the following estimate is true: λk − b2k (−d2 ) ≥ d0 k.
(38)
Proof. If b2k (−d2 ) ≡ 0, then the constant d0 = π. In case b) based on the second mean value theorem ([11], P. 353) we have 0 b2 (s) sin[λk (s + d2 )] ds λk − b2k (−d2 ) = λk + −d2 0
= λk + b2 (0)
cos[λk (d2 − ξ)] − cos λk d2 sin[λk (s + d2 )] ds = λk + b2 (0) λk −ξ 2 2b2 (0) 2b2 (0) = d0 k, ≥ λk − b2 (0) =k π− ≥k π− λk πk2 π
where −ξ ∈ [−d2 , 0]. If condition c) is fulfilled, then 0
b2 d2 ≥ k(π − b2 d2 ) ≥ d0 k. |b2 (s)| ds ≥ λk − b2 d2 = k π − λk − b2k (−d2 ) ≥ λk − k −d2 Lemma 4. Let the functions a2 (t) and b2 (t) satisfy, respectively, the conditions of Lemmas 1 and 3. Then there exists a positive constant E2 , depending, generally speaking, on d0 , E1 , d2 ,
b1 , |ai (0)|, such that for all k ∈ N the following estimate is true: |Nk | ≤ E2 k. RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 59 No. 6 2015
(39)
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SABITOV
Proof. Based on Lemmas 1–3 and equality (22) we have |Nk | ≤
1
λk + |b1k (−d2 )| + λ2k + |a1 (0)| + |a2 (0)| |Mk | d0 k
b1 d2 |a1 (0)| |a2 (0)|E1 π2k 1 + +1+ + ≤ λ2k λ2k λ2k k2 d0 λk |a1 (0)| |a2 (0)|E1 kπ 2 1 b1 d2 + +1+ + = E2 k. ≤ π2 π2 π2 d0 π
Lemma 5. Let the functions a2 (t) and b2 (t) satisfy, respectively, the conditions of Lemmas 1 and 3, and the function b1 (t) be increasing non-negative in [−d2 , 0]. Then there exist positive constants p0 (p0 ∈ N) and E3 such that for all p > p0 the inequality is fulfilled: |Πp (−α)| ≤ E3 < 1.
(40)
Proof. From the proof of case b) of Lemma 4 it follows that |b1k (−α)| ≤
2b1 (0) 2b2 (0) , |b2k (−α)| ≤ . λk λk
(41)
Then, from (34) due to estimates (39) and (41) we have
2|b1 (0)| 2E2 p|b2 (0)| 1 2|b1 (0)| 2E2 |b2 (0)| + + = . |Πp (−α)| ≤ λ2p λ2p p π2p π2
From this one can see that there exist positive constants p0 (p0 ∈ N) and E3 , such that for all p > p0 inequality (40) is fulfilled. For the solvability of the nonlinear equation (35) it suffices to stipulate the derivative |f (α)| ≤ d = const < 1. Let us calculate the derivative b1p (−α) + Np b2p (−α) (−1)n+1 . f (α) = λp Kp2 − (b1p (−α) + Np b2p (−α))2
Then due to estimates (39)–(41) there exists a natural number p1 , such that for all p > p1 |f (α)| ≤
|b (−α)| + |Np b2p (−α)| 1 1p λp K 2 − (b (−α) + N b (−α))2 p
1p
p 2p
≤
1 2|b1 (0)| + 2pE2 |b2 (0)| 1 2|b1 (0)| + 2pE2 |b2 (0)| < 2 ≤ d < 1. 2 λp K 1 − Π2 (−α) λp λp 1 − E32 p
p
In case Δ(p) = 0 the homogeneous problem (2)–(5), where ψ(x) ≡ 0, has a non-trivial solution up (x, t) = up (t) sin λp x,
(42)
where up (t) is defined by formula (23) for k = p, in which Ap is an arbitrary constant, not equal to zero. Indeed, function (42) by construction satisfies conditions (2)–(4), and due to condition (30) the equality is fulfilled for it: up (x, −α) = up (−α) sin λp x = Ap Δ(p) sin λp x = 0, x ∈ [0, 1]. Thus, we established the following uniqueness criterion. Theorem 1. If there exists a solution to problem (2)–(5) and conditions (16) are fulfilled, then it is unique if and only if for all k ∈ N conditions (27) are fulfilled. RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 59 No. 6 2015
INITIAL-BOUNDARY PROBLEM FOR PARABOLIC-HYPERBOLIC EQUATION
29
3. EXISTENCE OF SOLUTION TO THE PROBLEM Under conditions (16), (27) we will find the solution to problem (2)–(5) as a sum of Fourier series +∞ √ u(x, t) = 2 uk (t) sin λk x,
(43)
k=1
where uk (t) are defined by (28), in which the expression Δ(k) is the denominator. Since under certain conditions on the problem Δ(k) can turn to zero with respect to α, a problem of small denominators [12, 1] arises. Hence, it is necessary to prove the existence of numbers α, di and functions ai (t), bi (t), i = 1, 2, such that for large k the expression Δ(k) is separate from zero. Following [3], let us represent Δ(k) as (44) Δ(k) = 1 + λ2k sin(λk α + γk ) − ωk (−α). Here γk = arcsin
1
,
1 + λ2k
ωk (−α) is defined by formula (24) for t = −α. Lemma 6. a) If α is an arbitrary natural number, then for all k ∈ N |δ(k)| = 1 + λ2k | sin(λk α + γk )| = 1 > 0.
(45)
/ N, then there exist positive b) If α = pq is an arbitrary rational number, where (p, q) = 1, pq ∈ 0 and k0 ( k0 ∈ N) such that for all k > k0 the following estimate is true: constants C 0 > 0. |δ(k)| ≥ C
(46) √ √ c) There exists an infinite set of quadratic irrational numbers M = {α | α = d + p, p ∈ Z, d > −p, d = 2, 3, 5, 7, 8}, for which the estimate (46) is true for all k ∈ N with its own con0 . stant C / N, (p, q) = 1. Let us divide kp Proof. In case a) the validity of estimate (45) is obvious. Let now α = pq ∈ by q with a remainder: kp = sq + r, where s, r ∈ N0 = N ∪ {0}, 0 ≤ r < q. If r = 0, then this case is reduced to the previous one. Let r > 0. Then 1 ≤ r ≤ q − 1, q ≥ 2. The expression δα (k) in this case will take the form πr s 2 + γk . (47) δ p (k) = 1 + λk (−1) sin q q Since γk is an infinitely small series, there exists a finite limit π πr πr + γk = sin ≥ sin > 0. lim sin k q q q k0 from (47) we get Then there exists a natural number k0 ∈ N, such that for all k > |δ p (k)| ≥ q
πr kπ π kπ 0 ≥ C 0 > 0. sin ≥ sin = kC 2 q 2 q
In case c) the fulfillment of estimate (46) is proved similarly to [3]. Lemma 7. Let the conditions of Lemma 5 be fulfilled. Then if for the number α one of conditions a)–c) from Lemma 6 is fulfilled, then there exist positive constants C0 and k0 such that for all k > k0 the following estimate is true: |Δα (k)| ≥ C0 > 0. RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 59 No. 6 2015
(48)
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SABITOV
Proof. Due to Lemmas 1–6, it suffices to estimate from above the expression |ωk (−α)| ≤
1 [|a1 (0)| + |a2 (0)| |Mk | + |b1k (−α)| + |Nk | |b2k (−α)|] λk E1 2|b1 (0)| 2E2 |b2 (0)| 1 + |a1 (0)| + |a2 (0)| 2 + ≤ λk k λk πλk |2E2 b2 (0)| E5 1 |a1 (0)| 2|b1 (0)| |a2 (0)|E2 + . + + = ≤ k π π2 π2 π2 k
This yields that for the stated conditions on the number α there exist positive constants C0 and k0 ∈ N, k0 such that for all k > k0 ≥ 0 − |Δα (k)| ≥ |δα (k)| − |ωk (−α)| ≥ C
E5 ≥ C0 > 0. k
Lemma 8. Let the conditions of Lemma 7 and the corresponding estimate (48) for k > k0 be fulfilled. Then for such k the following estimates are true: H1 |ψk |, t > 0; (49) |uk (t)| ≤ H2 k|ψk |, t < 0, H3 k2 |ψk |, t > 0; (50) |uk (t)| ≤ H4 k2 |ψk |, t < 0, |uk (t)| ≤ H5 k3 |ψk |, t ≤ 0,
(51)
where Hj are positive constants, depending, generally speaking, on ai (t) , bi (t) , |ai (0)|, |bi (0)|, i = 1, 2, α. Proof. Based on formula (28), let us calculate the derivatives uk (t) = −λ2k uk (t) −
ψk (a1 (t) + Mk a2 (t)), Δ(k)
t > 0,
ψk [λk sin λk t + λ2k cos λk t + ωk (t)], t < 0, Δ(k) ψk (b1 (t) + Nk b2 (t)), t < 0. uk (t) = −λ2k uk (t) − Δ(k) uk (t) = −
(52) (53) (54)
The fulfillment of estimate (49) follows immediately from formula (28) and estimates (37), (39), (48). From equalities (52) and (53), based on estimates (37), (39), (48) we get (50). Similarly, from (54), using estimates (49), (48), (39), we get (51). From (43) by formal termwise differentiation we build up the series +∞ √ λk uk (t) cos λk x, −α ≤ t ≤ β, ux (x, t) = 2
(55)
k=1 +∞ +∞ √ ψk (a1 (t) + Mk a2 (t)) sin λk x, t ≥ 0, λ2k uk (t) sin λk x − ut (x, t) = − 2 Δ(k)
(56)
+∞ √ ψk [λk sin λk t + λ2k cos λk t + ωk (t)] sin λk x, t ≤ 0, ut (x, t) = − 2 Δ(k)
(57)
k=1
k=1
k=1
+∞ √ λ2k uk (t) sin λk x, t > 0, uxx (x, t) = − 2
(58)
k=1
RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 59 No. 6 2015
INITIAL-BOUNDARY PROBLEM FOR PARABOLIC-HYPERBOLIC EQUATION +∞ +∞ √ ψk 2 (b1 (t) + Nk b2 (t)) sin λk x, t < 0, utt (x, t) = − 2 λk uk (t) sin λk x − Δ(k) k=1
uxx (x, t) =
k=1 +∞ √ 2 − 2 λk uk (t) sin λk x, k=1
31
(59)
(60)
t < 0.
By Lemma 8, series (43), (55)–(60) are dominated by the numeric series H6
+∞
k3 |ψk |.
(61)
k=k0 +1
If ψ(x) belongs to C 3 [0, 1],
in the segment [0, 1] has a piecewise continuous fourth order Lemma 9. derivative and ψ(0) = ψ(1) = ψ (0) = ψ (1), then 1 (4) (62) ψk = 4 ψk , λk where (4) ψk
√ = 2
1
ψ
(4)
(x) sin λk x dx,
0
+∞
(4)
|ψk |2 ≤ ψ (4) (x) L2 .
(63)
k=1
Proof. Integrating by parts four times in equality (25), we get representation (62). The √ fulfillment of estimate (63) follows from Bessel’s equality with respect to the trigonometric system { 2 sin λk x}. Due to Lemma 9, series (61) is estimated by a numeric series H7
+∞ 1 (4) |ψ |. k k
(64)
k=k0 +1
Due to convergence of series (64) based on Weierstrass criterion series (43), (55)–(57) converge uniformly in D, and series (58)–(60) in the corresponding closed domains D + and D − . Hence, the function u(x, t), defined by series (43), satisfies condition (2). Substituting series (43), (56) and (58) into Eq. (1) for t > 0, and series (43), (59) and (60) into Eq. (1) for t < 0, we verify that function (43) is a solution to Eq. (1) in the set D− ∪ D+ . If for the numbers α mentioned in Lemma 7 the expression Δ(k)=0 for some k=l=k1 , k2 , . . . , kp ≤k0 , where 1 ≤ k1 < k2 < · · · < kp , ki , i = 1, p, p are given natural numbers, then for the solvability of problem (2)–(5) it is necessary and sufficient that the following conditions be fulfilled: √ 1 ψ(x) sin πlx dx = 0, l = k1 , . . . , kp . (65) ψl = 2 0
In this case the solution to problem (2)–(5) is defined as a sum of the series k kp −1 ∞ 1 −1 +··· + + Cl ul (t) sin πlx, uk (t) sin πkx + u(x, t) = k=1
k=kp−1 +1
k=kp +1
(66)
l
where ul (t) is defined by (23) for k = l, Cl are arbitrary constants, in the latter sum the index l takes on the values of k1 , k2 , . . . , kp , finite sums in (66) should be considered zeros, if the upper bound is less than the lower bound. Thus, we have proved Theorem 2. Let the function ψ(x) satisfy the conditions of Lemma 9; the functions ai (t) ∈ C[0, β] and satisfy the conditions of Lemma 1, bi (t) ∈ C[−α, 0] and satisfy the conditions of Lemmas 3 and 5 and the estimate (48) for k > k0 be fulfilled. Then if Δ(k) = 0 for all k = 1, k0 , then problem (2)–(5) has a unique solution, that is defined by series (43); if Δ(k)=0 for some k=k1 , ..., kp ≤k0 , then problem (2)—(5) is solvable if and only if conditions (65) are fulfilled, and the solution in this case is defined by series (66). RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 59 No. 6 2015
32
SABITOV
4. STABILITY OF SOLUTION Consider the following known norms:
u(x, t) L2 [0,1] = u(x, t) L2 =
1
1/2 |u(x, t)|2 dx
, −α ≤ t ≤ β;
0
u(x, t) C(D) = max |u(x, t)|. D
Theorem 3. Let the conditions of Theorem 2 be fulfilled and Δ(k) = 0 for k = 1, k0 . Then for solution (43) to problem (2)—(5) the following estimates are true:
u(x, t) L2 ≤ H8 ψ (x) L2 ,
u(x, t) C(D) ≤ H9 ψ (x) L2 ≤ H9 ψ (x) C[0,1] ,
(67)
where the constants H8 and H9 do not depend on ψ(x). √ Proof. Following [13], since the system { 2 sin πkx} is orthonormal in L2 [0, 1], from (28) and Lemma 8 we have
u(x, t) 2L2 =
+∞
u2k (t) ≤ H22
k=1 (1)
(1)
Since ψk = ψk /λk , ψk =
+∞
k2 |ψk |2 .
(68)
k=1
√ 1 2 ψ (x) cos πkx dx, from (68) we get 0
u(x, t) 2L2 =
H2 π
2 +∞
|ψk |2 ≤ H8 ψ (x) L2 . (1)
k=1
Let (x, t) be an arbitrary point from D. Then using formula (28), based on Lemma 8 we will have |u(x, t)| ≤ H2
+∞
k|ψk |.
(69)
k=1 (2)
(2)
In view of the representation ψk = −ψk /λ2k , where ψk =
√ 1 2 ψ (x) sin λk x dx, from inequality (69) 0
we get +∞ H2 1 (2) H2 ψ ≤ 2 |u(x, t)| ≤ 2 π k k π k=1
+∞ 1/2 +∞ 1/2 1 (2) 2 ψ k k2 k=1
k=1
H2 ≤ √ ψ L2 = H9 ψ (x) L2 . (70) π 6 In the obtaining of estimate (70) we used the Cauchy–Bunyakovsky inequality and the equality
+∞
1/k2 = π 2 /6. From the obtained inequality (70) estimate (67) follows immediately.
k=1
RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 59 No. 6 2015
INITIAL-BOUNDARY PROBLEM FOR PARABOLIC-HYPERBOLIC EQUATION
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REFERENCES 1. Sabitov, K. B. “Tricomi Problem for a Mixed Parabolic-Hyperbolic Equation in a Rectangular Domain,” Mathematical Notes 86, No. 2, 249–254 (2009). 2. Sabitov, K. B. “Nonlocal Problem for a Parabolic-Hyperbolic Equation in a Rectangular Domain,” Mathematical Notes 89, No. 4, 562–567 (2011). 3. Sabitov, K. B. “Initial-Boundary Problem for a Loaded Parabolic-Hyperbolic Equaion,” Dokl. AMAN, Nal’chik 11, No. 1, 66–73 (2009). 4. Nakhushev, A. M. “Loaded Equations and their Applications,” Differ. Equations 19, 74–81 (1983). 5. Nakhushev, A. M. Loaded Equations and their Applications (Nauka, Moscow, 2012) [in Russian]. 6. Dzhenaliev, M. T. To the Theory of Linear Boundary-Value Problems for Loaded Differential Equations (Almaty, 1995). 7. Pulkina, L. S. “A Nonlocal Problem for a Loaded Hyperbolic Equation,” Proc. Steklov Inst. Math. 236, 285– 290 (2002). 8. Kozhanov, A. I. “Nonlinear Loaded Equations and Inverse Problems,” Zh. Vychisl. Mat. Mat. Fiz. 44, No. 4, 694–716 (2004). 9. Khubiev, K. U. “Local and Nonlocal Boundary Problems for Loaded Mixed Type Equations,” Candidate’s Dissertation in Mathematics and Physics (Belgorod, 2009). 10. Sabitov, K. B., Melisheva, E. P. “The Dirichlet Problem for a Loaded Mixed-Type Equation in a Rectangular Domain,” Russian Mathematics (Iz. VUZ) 57, No. 7, 53–65 (2013). 11. Ilyin, V. A. and Poznyak, E. G. Fundamentals of Mathematical Analysis (Fizmatlit, Moscow, 1982), Vol. 1 [in Russian]. 12. Arnold, V. I. “Small Denominators and Problems of Stability of Motion in Classical and Celestial Mechanics,” Russian Mathematical Surveys 18, No. 6, 85–191 (1963). 13. Sabitov, K. B. “Boundary Value Problem for a Parabolic-Hyperbolic Equation with a Nonlocal Integral Condition,” Differential Equations 46, No. 10, 1472–1481 (2010).
Translated by P. A. Novikov
RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 59 No. 6 2015