3Uchinchi renessansyosh olimlari: zamonaviy vazifalar,
innovatsiya va istiqbol Young Scientists of the Third Renaissance: Current
Cha'Les.hnnoeon.and^en
INVERSE PROBLEM FOR SUBDIFFUSION EQUATION
M. D. Shakarova
PhD student Institute of Mathematics named after V.I. Romanovskiy. Uzbekistan.
shakarova2104 @gmail .com
Let p e (0,1]. We study the inverse problem of finding functions {u( x, t), f (x)} that satisfy the following problem
Dfu (x, t ) -Au (x, t ) = f (x) g (t ), x eQ, t e (0,T ],
u( xt ) |SQ= 0, u( x,0) = ç( x),
(1)
i
Ju(x,t)dt = ^(x), x eQ.
Here f(x), g(t) and cp(x) are continuous functions in the domain Q<=l ;V and Dp stands for the Caputo fractional derivative.
In order to prove the existence of solutions of forward and inverse problems, it is necessary to study the convergence of the following series:
I hk |2, r> N,
k=i 2
(2)
where hk are the Fourier coefficients of function h(x).
The theorem of V.A. Il'in states that, if function h (x ) satisfies the conditions
I ?k
[ N
h(x)eW^2J\Q) and h(x),A/?(j),....,AMJ/?(x)e^(Q),
(3)
then the number series (2) converges. Here [ a ] denotes the integer part of the number a .
Similarly, if in (2) we replace r by r+2, then the convergence conditions will have the form:
H(x)gW22 (n) and /7(x),A/7(x),....,AL4J h{x)^Wl2{Q). (4) Lemma 1. Let p e (0,1], g(t) e C[0, t] and g(t) ^ 0, t e [0, t]. Then there is constant C0 > 0, depending on T, such that for all k:
C
I Pk ,p(T )!> C°,
(5)
May_15,_2024_
0
262
Uchinchi renessans yosh olimlari: zamonaviy vazifalar,
innovatsiya va istiqbol Young Scientists of the Third Renaissance: Current
where
Pk,
i
,(T ) = j (T - ^)pEpp+l (-Ak (T - nY) g (n)dn.
Lemma 2. Let p e (0,1], g(t) e C1[0, T] and g(0) * 0. Then there exist numbers m0 > 0 and k0 such that, for all T < m0 and k > k0, the estimate (5) hold. Here constant C0 depend on m0 and k0.
Theorem 1. Let pe (0,1], g(t) e C[0,t] and g(t) * 0, t e [0, t] . Moreover let function p(x) satisfy condition (3) and x) satisfy condition (4). Then there exists a unique solution of the inverse problem (1):
1
f ( x) = I
k=i Pk,P(T)
¥k -VkTEP2(-AkTP) vk(x)
/( x t ) = I
k=i
<PEp(-AktP) +
bk pp(t )
Pk pp(T )
¥k -PkTEp,i(-AkTPp
vk( x),
where
t
bkpp (t) = j (t - n)P-1Eppp (-A (t -n)P )g(n)dn.
Theorem 2. Let p g (0,1], g (t ) g C1[0, T ], g (0) * 0 and T is sufficiently small. Moreover, function p(x) satisfy condition (3) and y/(x) satisfy condition (4).
If set B0 p is empty, i.e. pKp(T) * 0, for all k, then there exists a unique
solution of the inverse problem (1).
If set B0p is not empty, i.e. pkp(T) = 0, then for the existence of a solution
to the inverse problem, it is necessary and sufficient that the following conditions
¥k = TEp,2(-\Tppk, k g b0,p be satisfied. In this case, the solution to the problem (1) exists, but is not unique:
1
f ( x) = I
k^B0pk ,p(T )
¥k -PkTEp,2(-AkTp) Vk(X) + I fkVk(x)
keB,
0,p
œ
u^t) = I[PkEp,i(-Akt p) + bkp )fk K (x)
k=1
where fk, k g B0 are arbitrary real numbers.
REFERENCES
0
0
263
May 15, 2024
Uchinchi renessansyosh olimlari: zamonaviy vazifalar,
innovatsiya va istiqbol Young Scientists of the Third Renaissance: Current _Challenges, Innovations and Prospects
1. V.A. Il'in. On the solvability of mixed problems for hyperbolic and parabolic equations // Russian Math. Surveys. {\bf 15}:2, 97-154 (1960).
264
May 15, 2024
