CC BY
15
MSC 01A70 DOI: 10.14529/mmpl80311
ON THE TELEGRAPH EQUATION WITH A SMALL PARAMETER
J. Kisyiiski, Lublin University of Technology, Lublin, Poland, jan.kisynski@gmail.com
This work is an edited translation, by A. Bobrowski and W. Chojnaeki, of the original Polish text entitled 0 rownaniu telegrafistow z malym parametrem which appeared in Prace Naukowe Politechniki Lubelskiej 212, Matematyka (Research Bulletin of the Lublin University of Technology 212, Mathematics), 1991, vol. 12, pp. 17-40.
Introduction
In this paper we will be concerned with the telegraph equation
d2 u du d 2u
aW + bdt +cu = dX2, (T)
where a > 0, b > ^d c > 0. We will be interested in studying the behaviour of solutions of this equation for t > 0 in the case where the coefficient a is small. In particular, our aim will be to estimate the difference between solutions of (T) and solutions of the diffusion equation
du d2u
bdt + Cu = * (D)
Such estimates have been studied by a number of authors, including M. Zlamal [11-13], J. Kisyiiski [2,3], J. Smoller [7], J. Nedoma [6], and M. Sova [9]. The paper by Zlamal [11] was inspired by a very specific technical problem. The question of estimating the difference between solutions of (T) and (D) has also been considered in the context of neutron transport, see A.M. Weinberg and E.P. Wigner [10, p. 235] and W. Baran [1].
It turned out that it is fruitful to treat (T) as a particular case of a certain type of differential equations in Banach spaces. For, such an approach allows uniform, systematic treatment of a number of boundary-value problems for (T) and obtaining in all these cases the same estimates in terms of the norms of underlying Banach spaces in which the problem is well-posed. We stress here that boundary-value problems that fit into our framework may include the solution u and its spatial derivative but not its time derivative
The difference between the results we present here and those obtained in [3] lies in
c
c > 0
valued function which we denote by V(t,a,b,c,A). This function will critically intervene in what we will call the comparison theorem (see [3, p. 372] and Section 3 in the present paper).
1. Abstract Telegraph Equation
Let £o be a Banach space, let £(t), — to <t< to, be a strongly continuous cosine operator function with values in the space L(E0,E0), and 1 et A be the infinitesimal generator of this cosine operator function, see M. Sova [8] or J. Kisynski [4]. Let
Ei = {u E E0 : the function (—to, to) 9 t M £(t)u E E0 is of class C1 in
Eo
When equipped with the norm
||u||Ei = IMIeq + sup 0<1<t
d£ (t)u
dt
Eo
is a Banach space; see [4, p. 98]. Let us consider the Cartesian pro duct Ei x E0 and let
us agree to denote its elements as column-vectors (") ,u E Ei ,v E E0. Let k be a positive constant. Then the operators
G(t)
£(kt) ¡£(кт )dT
— TO < t < TO,
dC(kt) dt
£(kt)
belong to L(Ei x E0,Ei x E0) and form a one-parameter strongly continuous group with generator
(k0A 0) ■ D(B) = D(A) x Ei'
B
see [4, p. 98]. Now, let a, b and c be fixed scalars with a > 0. By the Dyson-Phillips bounded perturbation theorem, the operator
Bn
(
о 1 \ { о ° ^
a-1(A — c) -a-1b) = l^A °) ^a-1 c a-1b)
generates a strongly continuous one parameter group of operators in Ei x E0. Let us express the operators in this group in the form of operator-valued matrices
exp(tB fSoo(t,a,b,c,A) Soi(t,a,b,c,A)\
exp(tBa,b,c) \Sio(t,a,b,c,A) Sii(t, a, b, c, A) J '
Then the Sij(t, a, b, c, A) are strongly continuous functions of t with values in L(Ei-j, Ei-i).
It is easy to see that the Cauchy problem
du ~dtt
d2u du a^r^ + b——+ cu = Au, — то < t < то,
dt2
u(0) = u0, u1
(T*
£ (0)
has, for any pair of initial conditions u0 E D(A) and ui E E^ a C2(-^, E0)-class solution u(t) with values in D(A), given by the formula
u(t) = S00(t, a, b, c, A)u0 + S0i(t, a, b, c, A)ui.
This solution is unique in the class C2(0,T; E0) on each interval (0,T). To prove this, suppose that v(t) is another solution of (T*). Then, for any arbitrarily fixed t G (0,T),
v(t) - u(t)
Soo(t - t)v(t) + soi(t - t)
dv(r )
dT
t=t
t=0
Using formulae (3) and (4), given in Section 3, for the derivatives of S00 and S11; one may show that the derivative with respect to t of the expression in the brackets is zero, and this implies that v(t) = u(t). Another proof of uniqueness may be obtained by eliminating the first derivative in the differential equation from (T*) with the substitution u(t) = e-2aiv(t) and thereby reducing the problem to the case considered in [4, p. 96].
The operator-valued functions Sj were build from the cosine function «(t) via the group G(t) by applying the Dyson-Phillips bounded perturbation theorem. One can show, and this will be done below, that if b2 > 4ac (a condition which henceforth will be tacitly assumed), then
S01(t, a, b, c,A) = e 2a
/ * ('f<-¡) dT,
(1)
where A = b2 — 4ac and J0(ix) is the Bessel function of type zero with purely imaginary argument, given by the series
*0(ix) =
k=0
(2xL (k!)2 '
An immediate consequence of equality (1) and formula (4) from Section 3 is
A b
dt a
e 2a %
1
1 2a
S00^A +
t
d b
)+e- b i + ^ Û dT
Hence, we see that the solution to (T*) is given by the formula
i-a)
u(t) = e 2atW — J u0 + e 2at
'.VÂ
+ e-2at J0 i-^—Vt2 - t2 M -p ui dT.
I + ^)J0 «( u dT +
) « ( -a )
2a
We now prove (1) by verifying that the right-hand side of the formula is a solution of an appropriate uniquely solvable Cauchy problem. For u1 E E1, we have
[(
c
d2 d
a—- + b——+ c ) S01 (t,a,b,c, A)u1 = AS01 (t, a,b,c, A)u1, dt2 dt
S01(0,a,b,c, A)u1 = 0,
A dt
S01(t, a, b, c, A)u1 = u1,
t=0
t
t
and, as explained above, this system has a unique solution. Let
K(t, т) = e"^ J0 —T^-t^J .
Now, all we need is to check that the integral
Soi(t)ui = K(t, т)M ^ ) ui dT
satisfies (*). The condition S0i(0)u = 0 is obviously met. Moreover,
t
itSoi(t)ui = K (t, t)c{ ja)ui +1 ddr(t,T 71)ui dT =
0
t
dK
e-2^ ( — )ui + dT (t,T— )ui dT,
and so -d |t_0 >S0i(t)ui = ui. We are thus left with checking that the differential equation in system (*) holds. To this end, we calculate
d2 S , ч b t \ t d i t \ idpS"i(t)ui = -2a,e-*«{ts) ui+e-* d«H—Jui +
+ ж (tM{^Ь4ж(tM{ —NdT
"
iva)
and find that
t d2 d \ S a—r^ +b—+c — A) S0i(t)ui
dt2 dt
b b t dK . . E +1й (t,t)
d
M — ) ui +ae a—£[ — )ui +
iva)
a
д2 , д
)
iva)
+ I [a— + b— + cj K (t,T — )ui dt -I K (t,T)A£[ — ) щ dt.
The last integral in this formula can be transformed by means of integration by parts as follows:
t t
J K(t, T)A^ —a) ui dT = a J K(t, т^^ ui dт ""
+ t d ' 4 T=t
=aK (t,T )dT^ ui
aK (t,t)dt^ ( —a) ui- adK (t,T ^
. dK d . / т \ ,
T ^ - "J-* (t'T )dT4 —a)ui =
"
t
T=t
+ a i ^лт(t, T^ ( — ) uidT.
T=0
дт2
t
t
t
Since dK (t, 0) = 0, we thus obtain
( d2 , d A \ / \
I a—^ + o— + c — A ) ¿>oi (t)u1 = a
dt2 dt
t
d2K
b _b_t . dK
2a
dK
2ara + It (t-t)+ S7^
c Ha) +
M ^ I ui dr
+
\ d2K. . BK, . '
(t'T) — alh2 (t'T) + bl)t (t'T) + (t'T)
d2K d2K dK
a^2(t,T) — adT2(t'T) + blt (t'T) +cK(t'T)
iva) iva)
M v )ui dT-
It is now clear that the proof will be complete once we show that
d2K d 2K dK
a—(t't) — a— (t't) + b— (t't) + cK(f t) = 0
O
for
0 < t < t. Let K0(t'T) = J0 (i^Vt2 — t2). Then we have K(t,T) = e-2atK0(t,T),
O
( d b )2 d2 J d b .
a{m — 2a) — aST~2 + b[ßt — 2a + C
(d-A ^
\dt 2a J
Ko(t'T ) = 0'
which is the same as
or
Given that
d2 d2 b2 — 4ac
dt2 — dT2
4a2
Ko(t'T ) = 0'
(
d2 d2 A \
dt2 — dT2 — 402 Ko(t'T) = 0-
O
and
d 2Ko
dt2 VA
(t'T )
+
(
(VA t \
y 2a Vt2 — t2 ) dx2
d2
Jo(ix) +
2a [VW—T2 (Vt—2)
3 ' dx
Jo(ix)
d 2Ko
+
dT2
VA
(t'T )
(VA t \
y 2a Vt2 — t2 ) dx2
d2
c=Vtt—2
Jo(ix) +
-1
2a {Vi—2 (Vt^—T2)
d
3 ' dx
V2—2
Jo(ix)'
O
( d2 Id i , ,
T^ + ~~T — 1 Jo(ix) = 0'
dx2 x dx
0
t
a
2
2
1
t
2
2
and that the latter equation holds can be checked immediately by differentiating the series
(i
J"(IX) = Y, (k!)2 k=" У ''
term by term.
2. Abstract Diffusion Equation
Along with the second order problem (T*) we consider the initial value problem of order one
du
b—:—+ cu = Au, t> 0, dt (D*)
u(0) = u0. The solution of this problem is
u(t) = e-ct exp ^,
where exp(tA), t > 0, is the one-parameter semigroup of operators in the space E0 A
from the previous section, namely,
1 Г т 2
exp(tA) = —=j е- 4T £ (t ) —t, t > 0; (2)
see J. Kisynski [5, p. 9].
3. Reducing the Problem of Estimating the Difference Between Solutions of Problems (T*) and (D*) to a Comparison Theorem
From exp(tBa,b,c)D(Ba,b,c) = D(Ba,b,c) and
-—exp(tBa ,b ,c)( u ) = Ba, b, c exp(tBa, b, c)( ^ ) = exp(tBa, b, c )Ba, b, c( ^) dt ui ui ui
with the latter equality holding for u0 E D(A) and ui E Ei, and taking the form
— fSm(t) S"i(t)\ (щ\ = ( 0 1 \ (S""(t) S"i(t)\ (щ\ =
dt\Si"(t) Sii(t)J \uj \a~i(A - c) -a-ib) \Si"(t) Sn(t)) \uj 'S""(t) S"i(t)\( 0 1 \ fuo Si"(t) Su(t)) V"-i(A - c) -a-i^\ul
Вестник ЮУрГУ. Серия «Математическое моделирование 139
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we deduce the inclusions S00D(A) C D(A), S01E1 C D(A), S10D(A) C E1 and SuEi C E\, and equalities
dS00u0 n n -1
dt
S10u0 = S01a (A — c)u0, (3)
dSoiui bQ
--— = Siiui = S00U1--S01U1, (4)
dt a
dSioUo b
---= a-l(A — c)S00u0--S10u0 = S11a-l(A — c)u0, (5)
dt a
—T1—- = a-l(A — c)S01u1--S11u1 = S10u1--S11u1. (6)
dt a a
If u0 E D(A), then it follows from (3) and (6) applied with u1 = u0 that
S01 a-l(A — c)u0 = a-1 (A — c)S01u0 = S10u0. (7)
S00
an operator in L(E0, E0), which is possible by (4), then, for u0 E D(A), we have
S00Au0 = AS00u0. (8)
By (3) and (5), we see that for u0 E D(A) the function u(t) = S00(t)u0 satisfies the equation
du(t) = b-i( A c)u(t) S (t)b-i, dt
Since u(0) = u0, it follows that
b-1(A — c)u(t) — Sii (t)b (A — c)u0.
S00(t)u0 = exp^b(A — cu0 — J exp ^(A — cSn(r)b 1(A — c)u0 dr. (9)
0
To simplify notation, let
t
U(t,a,b,c, A) = J exp ^ ———— (A — c^ S11(r,a,b,c, A) dr. 0
In view of (9), the difference between the solution
ua(t) = S00(t, a, b, c, A)u0 + S0i(t, a, b, c, A)ui to problem (T*) and the solution
u(t) = exp ^b(A — c^ u0
*
\\ua(t) — u(t) H^ < || U(t, a, b, c, A)b-1(A — c)u0\\Eo + ||S0i(t, a, b, c, A)ui ||Eo (10)
| .|() Bulletin of the South Ural State University. Ser. Mathematical Modelling, Programming
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for u0 E D(A), ui E Ei and t > 0. We will also obtain a similar estimate for the difference — • Since Sii(0)ui = ui, (6) implies that
t
Sii(t)ui = e-atui + j e-a(t-T\Si0(t)ui dT. "
If ui E D(A), then by (3) we may write
Sn(t)ui = e-atui + - I e-a(t-T)S^(t)b-i(A - c)ui —т. (11)
a
If u0 E D(A2) and ui E D(A), then, by (3) and (4),
- e-at(ui - b-i(A - c)u0)
dua(t) du(t) _bt ,
dt dt equals
Si0(t)u0 + Sii(t()b-i(A — c)u0 — exp ^b(A — c^ b-i(A — c)u0 + + (^Sii(t) — e-a^ (uui — b-i(A — c)uo) which in turn, given that, by (5) and (8),
Sio(t)uo + Sii(t)b-i(A — c)uo = b-i(A — c)Soo(t)uo = Soo(t)b-i(A — c)Buo, is equal to
Soo(t)b-i(A — c)uo — exp(^b(A — c)^ b-i(A — c)uo + Sii(t) — e-a^ (ui — b-i(A — c)uoh ,
and this, in view of (9) and (11), finally equals
— U(t, a, b, c, A) [b-i(A — c)f uo +
t
+ bj e-a(t-T)Soi(T,a,b,c,A)b-i(A — c) [ui — b-i(A — c)uo] dT. o
uo E D(A2) ui E D(A)
estimate:
T - ^ - e-at(ui - b-i(A - c)u")
<
U(t, a, b, c, A) [b-i(A - c)] u" + (12)
<
Eo
II Eo
+ sup IS"i(T,a,b,c, A)b-i(A - c) \ui - b-i(A - c)u0] ||E
"< T<t 0
ua(t) -
u(t) U(t, a, b, c, A)
t
Sn(t,a,b,c, A). The case of U will not be treated directly, but will be tackled by means of the auxiliary function V defined by
V(t, a, b, c, A) = U(t, a, b, c, A) + -—— [ U(r,a,b,c, A) dr.
2a I
(13)
U
that
U(t, a, b, c, A) = V(t, a, b, c, A) + f (t-T)V(r, a, b, c, A) dr. (14)
2a J
0
Since < 0, (14) implies immediately that
\\U(t,a,b,c,A)\\ < 2 sup \\V(t,a,b,c, A)\\.
0<t <t
However, this inequality will not be used in the context of the telegraph equation (*) which is of main interest to us, as in this case other, more subtle estimates are available and will
V
The following theorem is the key to estimating the norms of operator-valued functions
V(t, a, b, c, A) and S0\(t, a, b, c, A).
Comparison Theorem. Let a > 0 b > 0 and c > 0 be such that A = b2 — 4ac > 0, and let k > 0. Let Sij(t,a,b,c,k2) be the real functions defined via the equality
(
Soo(t,a,b,c,k2) S0i(t,a,b,c,k2)
2) Soi(t,a,b,c,k)) = / / k20_ c 1
Sl0(t,a,b,c,k ) Sll(t,a,b,c,k2)J 1 1
and let (compare (13))
V(t, a, b, c, k2) := e * Cf' * Sll(t, a, b, c, k2) +
b — VÂ k2
2a
* e * 1 * Sll(t, a, b, c, k ),
where * denotes convolution on the half-line t > 0. If
\\£(t)\\L(Eo , Eo) < M COsh(kt), <t< X,
with M = const > 1, then
exp(y I(A — c)^j
< M expi b (k2 — c) )
L(Eo, Eo) \b J
\\S00(t,a,b,c,A)\\L{Eo, e0) < M S00(t,a,b,c,k'2), \\S0l(t,a,b,c,A)\\L{E0iE0) < M S0l(t,a,b,c,k2),
t > 0,
t > 0, t > 0,
(15)
(16)
(17)
(18)
and
\\V (t,a,b,c, A)\\l(e0 ,e0) < MV (t,a,b,c,k2),
t 0.
(19)
4. Proof of the Comparison Theorem
By (2), we have
oo oo
1 Ft2 1 i
|| exp(tA)\\<—= e- 4t U(t)\\ dT < M—= e- « cosh(kt) dT = M exp(tk2),
and so
exp(^b(A - c))
-ж
< e-cexp(tA)\\ < Me-btebk = Me^t.
Similarly, by (1),
\\S"i(t,a,b,c,A)\\ < / K(t,T)
< —)
—t <
< M J K(t, t) cosh ^ia J dT = MSoi(t, a, b, c, k2), o
because the kernel K(t,T) = e-2atJo (i^Vt2 — t2^ is non-negative for 0 < t < t.
Inequalities (16) and (18) are thus proved. Inequality (17) is established in a similar manner. The proof of (19) is more complicated. According to a well-known result from the theory of semigroups of operators, there is a constant Ao such that for A > Ao we have
кт \
b (b\ + c - A)-1 = (A - b-i(A - c))
i
-xt
exp(^b(A - c)^ dt (20)
and
(A - BaM)-i = I e-xt exp (tBaM) dt.
(21)
Since (15) implies that
A(A - A)-i = e-x £(t) dt for A > к
(22)
it follows that for A so large that aA2 + bA + c> к2 we have
(A - Ba,b,c)
i
(
A01
A a-i(A - c) -a-ib
i
(aA + b)(aA2 + bA + c - A)-i a(aA2 + bA + c - A)-i (A - c)(aA2 + bA + c - A)-i aA(aA2 + bA + c - A)-i
)
Ai
e-xtSii(t, a, b, c, A) = aA(aA2 + bA + c - A)-i foг A > Ai
(23)
»
t
t
Since the Laplace transform of the convolution of two functions (on the half-line 0 < t < to) is the product of the transforms of these functions, equality (13) written as
V(t, a, b, c, A) = exp (^(A — c)) * Sii(t, a, b, c, A) +
b —VA (t \
+--—— * exp I b(A — c) I * Sii(t, a, b, c, A)
and combined with (20) and (23) implies that
e~xtV(t, a, b, c, A) dt = a + b j (A — b"1 (A — c^1 (aA2 + bA + c — A)~l
for A > A2 = max(Ao, A^. Introducing the functions
gi(A) = (a + b)1 g2(A) = (aA2 + bA + c)2,
A +
h(A) = a " ' 2a
we may rewrite the last identity as
(A + $ 2 (aA2 + bA + c)2
oo
-Xtim „ j, „ /i\ i. _ if \\ /W^i M2 A\-1 „ f\ Wr„ M M2 A\-1
o
e-xtV(t, a, b, c, A) dt = h(A)gi(A) ([gi(A)]2 — A)"1 g2(A) (^(A)]2 — A) for A > A2. But now, by (22), we see that
oo oo oo
e-xtV(t, a, b, c, A) dt = h(A) f e-9l(x)t£(t) dt f e-92(x)s£(s) ds
h(A) j f e-9l(x)t-92(x)s (t + s) + £(t — s)] dt ds
(24)
oo
for A > A2. Analogously,
oo oo oo
J e-XtV(t, a, b, c, k2) dt = h(A) j J e-91 (x)t-92(x)s 1[cosh(t + s) + cosh(t — s)] dt ds. (25) o o o
Thus, for each u E Eo such that ||u|| = 1 and each bounded linear functional f on Eo such that II f II = 1, we have
oo
J e-xt [M V(t, a, b, c, k2) — f, V(t, a, b, c, A)u)] dt =
(X (X (26)
= hA U e-9l(x'-92(x)s^it,s)dt ds
oo
for A > A2, while, by (15), the function
s) = 2M[cosh(t + s) + cosh(t - s)] - 2 {f, + s) + C(t - s)]u) is non-negative. Let
F (A)=h(A) II e-"{x,,-"{x"*(t's)d ds
0 0
and let us assume for now that
the function F(-A) is completely monotone in the interval -to < A < -A2, (27) i.e., that d > 0 for all n = 0,1,... and A £ (-to, -A2).
It follows (26) and (27), by virtue of the Post-Widder formula for inversion of the Laplace transform, that
(—1)n /n\n dn-1F(A)
MV(t, a, b, c, k2) - {f, V(t, a, b, c, A)u) = lim (-) A
n^x (n - 1)! Vt/ dAn-i
> 0
T
for each t > 0 Since the only restriction imposed on u and f was ||u|| = \\f || = 1, this results in inequality (19).
The idea of exploiting the Post-Widder formula to obtain estimates as above is taken from the paper by M. Sova [9], who was the first to study the asymptotic behaviour, as a ^ 0, of solutions of initial-value problem of type (T*) in a non-Hilbert space. The idea to use completely monotone functions is due to the author of the present paper; this approach simplifies the analysis and sharpens the estimates. We now present the proof of (27). It suffices to show that
dra[-2i(-A)] > 0 for n = 0,1,... and A £ (-to,-A2), (a)
dAn dn[-92(-X)] dAn
> 0 for n = 0,1,... and A £ {—ж, —A2), (b)
and
d h[ A > 0 for n = 0,1,... and A £ (-to,-A2). (c)
dAn v 2 w
That (a) is true follows immediately from the formula
dn[—gi{—A)] dn (c \ 2 113 2n — 3 (c л-
dAn dAn \b J 2 22
For the proof of (b) we calculate, for A < 0
a—a) 2=---...—a—A\ b 2 2 2 2 b
-92-A)] = — dA{aA2 — bA + c)2 = Q — A (aA2 — bA + c) 2 > 0
and
d2 i / b \ 2 ^^^[-92 — A)] = -a(aX2 - b\ + c)-2 ^^ - aX) {aX2 - bX + c)'
A _3 A 3 _3 _3
= — (aA2 - bX + c) 2 = 4a-2 (¡i - X)-2 (12 - X)-2 > 0, where ¡1 = > 0 and ¡2 = > 0- Since
dk . 3 5 2k + 1
2fe+3
d^ - X)~2 = 2 ^ 2 •••—^ - X)~ 2 for k = 0,1,... ,j = 1, 2 and X < 0, Leibniz's formula implies that
n /»A Ak(,, W —3 An—kt
d2+n[-g2(-X)] = A -3 y- M dk(ii - X)-2 dn-k(¡2 - X)-2 0
dX2+n = 4a ^ UJ dXk dXn-k >
k=0 v 7
for A < 0 and n = 1, 2,....
We are left with showing (c). To this end, we observe that
h(-X) = a --^ 1 1 11 - X 1-1 = a2 ( ¡-Xl) 2 (¡2 - X)-2 .
(- - X) 2 a2 (11 - X)2 (¡2 - X)2 \b - XJ
Since
dk , 13 2k - 1
:(l2 - X) 2 =- •------— (¡2 - X) 2
dAkK^ J 2 2 2 for A < ^d k = 1, 2,..., Leibniz's formula shows that our task reduces to proving that
dk / .. _ A \ 2
JAkl^^) > 0 ior A < 0 and k = 0,1,.... (c*)
Inequality (c*) results from
-= Xi - c = ^ - c >- 0 ™
since b2 - 2ac > b^/A\ the latter condition may easily be checked by taking squares of both sides. By (28), we have
d fii - X\2 = d f I + D - X\2 =1 f I - X \ 2 D b 'W ^ V b « , - Vb ■ D-~X) —)
dX \ c - XJ dX \ b - X J 2\ b + D - X
1 , , 3
=2 a+d—r(b—Ay.
and thus we see that inequality (c*) is true for k = 0,1. To prove it for k = 2, 3,..., one needs to apply Leibniz's formula in a way similar to that used in our proof of (b). Hence, inequality (c) is seen to be true. The proof of the comparison theorem is thereby completed.
5. Consequences of the Comparison Theorem in the Case k = 0
We have
V(t, a, b, c, 0) = e"^ * | ¿^(i, a, 6, c, 0) + f a> c> dr | =
b -y/A
e bt * ^Sn(t,a,b,c, 0) + b ^^Soi(t,a,b,c, = e ^ * u(t),
where
u(t) = S11(t, a, b, c, 0) +--S01 (t, a, b, c, 0)
2a
is the solution of the equation
au'' + bu' + cu = 0
with initial conditions
u(0) = 1,
(0) S' (0) + b - V~A b + b - Va b + Va
u (0) = S11(0) +----=---+
2a a 2a 2a
Therefore we have
ь+УА t
and so
u(t) = e 2a
t t
V(t, a, b, c, 0) = e~^ j eib-^)r dr = e-bt J e-(D+r dr <
0 0
< e-bt< -0=e-bt. D + yfK
Also
J Vir^c, °)dr = J TVit - r,^, 0)*t <
00 t t
fe^-Te-b(t-T) dT = 4=e-bt f e-DT dr.
~ vaj va j
00
k=0
and equality (14), we obtain the estimate
t,- a
\\U(t, a, b, c, A)\\L(Eo,Eo) < e-btM
y/A
1 + ^ Je-Dr dr
0
< e-btM^ (l + -^t),
- VA V VÄ/
<
(29)
for we have b < e u{t) = S01{t,a,b,c, 0) is the solution of the equation
au'' + bu' + cu = 0 satisfying the conditions u{0) = 0 and u'{0) = 1, we have
t
t
Soi(t, a, b, c, 0) = (vi — v2)-i(eVlt — eU2t), where vi = ^^ and v2 = > which
implies that Soi(t, a, b, c, 0) < for t > 0. In view of inequality (18), we thus have
\\Soi(t,a,b,c,A)\\L{Eo,Eo) < Mfort > 0.
(30)
Combining together (10), (12), (29) and (30), we finally obtain the following estimates of
the difference between the solution ua(t) of problem (T*) and the solution u(t) of problem *
uo E D(A) ui E Ei
\\ua(t) - u(t)\\Eo < M^e-c^ 1 + ^ \b-1(A - c)uo\\Eo + \\ui\\Eo
y/K
1
for t > 0;
(ii) and if, furthermore, u0 G D(A2) and u1 G D(A), then also
VÂ
Ie-c ' ('
dua(t) du(t) _b
dt
dt
- e a (u1 - b 1(A - c)u0)
<
e-C M 1 + ^t
VK
Eo
1
(A - c)]2uo\\Eo + \\b-1(A - c)[ui - b-1(A - c)uo]\E
)
for t > 0.
For c =0 these inequalities are identical to those obtained in the paper [3].
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