On solvability of the Cauchy problem for a loaded system Текст научной статьи по специальности «Математика»

УДК 517.95

On Solvability of the Cauchy Problem for a Loaded System

Yuriy Ya. Belov* Kirill V. Korshun^

Institute of Mathematics and Computer Science, Siberian Federal University, Svobodny, 79, Krasnoyarsk, 660041

Russia

Received 20.01.2014, received in revised form 28.02.2014, accepted 16.03.2014 In this work we investigated the Cauchy problem for a loaded Burgers-type system. Example of mathematical physics inverse problem leading to problem being investigated is given. .Sufficient conditions for existence of solution in continuously differentiable class are obtained.

Keywords: Cauchy problem, inverse problem, Burgers' equation, non-linear system, weak approximation method.

Inverse problems of mathematical physics play important role in science and applications today [1]. Coefficient inverse problems for parabolic equations are problems of finding solutions of differential equation with one (or more) unknown coefficients. These problems often reduce to problems for loaded equations. Loaded differential equations (see [2]) are ones with functionals of solution (e.g. values of solution or its derivatives on lesser-dimensional manifolds) as coefficients or right-hand side.

Existense of solution to special class of loaded two-dimensional parabolic equations has been proved by I. V. Frolenkov and Yu.Ya. Belov (see [3]). Problem being considered in this article arises during generalization of preceding results.

1. Problem formulation

We consider the initial-value problem for loaded system du —

— = ^(t, w(t))Au + v(u • V)u + f(t, x, u, w(t)), (1)

dt

u( 0,x) = <£(x) (2)

in domain n^T] = {(t, x)|0 ^ t ^ T,x £ M"|, where u = (u1(t,x),... ,un(t,x)) are unknown functions. Let w(t) = (u¿(t,xj), Dau¿(t,xj)); i = 1,... ,n; j = 1,... ,r; |a| = 0,... ,p0 be a vector function, with traces of unknown functions and their partial derivatives with respect to spartial variables of order up to p0 at points x1,..., xr £ Rn as its components.

d|a|

Da =

dai x1.....dc

is partial differential operator, where a = (a1,..., an) is multi-index, |a| = a1 + ■■■ + an. Functions ^(t,Q(t)), f = (/i,...,/n), = (^i(x),..., yn(x)) are given ones, v £ R is given coefficient.

* YBelov@sfu-kras.ru tkorshun007@inbox.ru (c Siberian Federal University. All rights reserved

n t

n

We will use following notation:

Cq's(n[ojT]) = \ u = (ui(t,x),.. .,u„(t,x)) dj u.

d j

^,DaUi(t,x) G C(n[o,T]);

^ K, |DaUj(t, x)| ^ K; i = 1,..., n; j ^ q; |a| ^ s; q, s G Z; Kis const

is class of bounded, continuously differentiable functions,

Uj(0) = supx£Rn |Da^(x)|,

Uj(t) = sup5e[Q i] supx£Rn (£,x)|,

n

Uj(t) = max Ui(t), U(t) = 1 + ^ Uj(t)

i= 1

are nondecreasing nonnegative functions. Let p > max(p0, 2), function ^ satisfies

^i(x) G Cp+2(Rn), |Da^(x)| < K1; i = 1,..., n; |a| < p + 2,

(3)

^ h f are continuous in all variables and the following relations are valid for any function

u(t,x) G C(n[ojT]);

Mt,w(t)) > Mo > 0, Vu(t,x) G C 1'p+2(n[ojT]) |Da/i(t,x,U,^)| < K2(1+U(t) + U(t)2), |a| < p + 2. Here and further, K are constants depending only on the initial data. We will prove

(4)

Theorem 1.1. Let the initial data of problem (1), (2) satisfy (3), (4) for some p. Then constant t* exists (t* G (0, T]) for which a solution of problem (1), (2) exists and lies in C 1,p(n[0ji*]) class.

2. An example

We have investigated inverse problem involving finding functions u(t, x), g(t) in Cauchy problem for the Burgers-type equation

Ut(t, x) = M(t)uxx + A(t)uux + B(t)u + g(t)/(t, x),

u(0,x) = u0(x), 0 < t < T, x G R,

which reduces (using overdetermination condition u(t,x0) = ^(t)) to Cauchy problem for loaded parabolic equation

where

Ut(t, x) = ^(t)uxx + A(t)uux + B(t)u + F(t, u), u(0, x) = u0(x), 0 < t < T, x G R,

f(t x)

F(t,u) = ' \ (V>'(t) - B(t)^(t) - M(t)uxx(t,xo) - A(t)^(t)ux(t,xo))

(5)

(6)

f(t, xo)

is functional depending on traces of unknown function and its derivatives at point x0. The problem (5), (6) is the particular case of problem (1), (2) for n =1, U = u(t,x), / = F(t,u),

= U0(x).

Let the initial data of problem the (5), (6) satisfies

uo(x ) G Cp+2(R),

dk uo

dxk

^ Ko — const, k = 0,... ,p + 2,

(7)

A(t),B(t) G C([0,T]), ^(t) G C 1([0,T]), |f(t,xo)| ^,

Ko

|A(t)| + |B(t)| + |V(t)| + |V'(t)| < Ko,

dkf

M(t) > Mo > 0, dxk G C([0,T] x

dxk

< Ko, k = 0, ...,p + 2

(8)

for some p ^ 2. Conditions (3) of Theorem 1.1 are fulfilled by (7). We can check fulfillment of (3) provided (8) are valid;

Vu(t,x) G C 1lP+2([0,T] x R) Vk = 0, ...,p + 2

dk

dXk F (t,u)

<

< K (1 + U2(t) + Ul(t)) < K2 (1 + U(t)).

Thus in is the particular case one can use Theorem 1.1 to prove existence of solution of problem

(5), (6) in C 1'p(n[0ji*]) class.

3. Auxiliary theorem

Theorem 3.1. Let u(t, x) be solution of

ut = ^2 bi(t,x)

du(t, x) dx,-

u(0,x) = U0(x), x G En

in domain G[0 T] = {(t,x)|0 ^ t ^ T, x G En} of C 1,p(G[0 T]) class. Let the conditions

|Da6j(t, x)| < M (p), |a| < p,i = 1,. ..,n; |D aU0(x)| < C(p), |a| < p

are valid. Then u(t, x) satisfies

|Dau(t, x)| < C(p)el(p)M(p)T, |a| < p, where l(p) > 0 depends only on p and does not depend on the initial data.

(9)

4. Proof of Theorem 1.1

We will prove existence of solution of problem (1), (2) using weak approximation method (see [4]). We split the problem into three fractional steps and make time shift by T/3 in traces

of unknown functions and nonlinear terms. This leads to equation system d?/T

= 3M(t,^(t T /s))AuT, t G (tot, (m +1 /3)t], (10)

= 3v(UT(t T /3) •V)uT, t g((m +1 /3)t, (m +2 /3)t] , (11)

di/T

-dtL = 3fi(t — /3, x, U(t — /3, x), w(t — /3)), (12)

2

t G ( (m +2 /3) t, (m + l) t] , uT(t,x)|t<0 = ^i(x); i = 1,..., n; m = 0,..., M - 1; Mt = T. (13)

Let us introduce the following notation

UiT(t) = sup sup |Da<(£,x)|,

££[0,t] x£l„

n

UiT(t) = max UiT(t), UT(t) = 1 +£ UiT(t).

i=1

Zeroth whole step (m = 0) is considered. In first fractional step system (10), (13) is representing n Cauchy problems for parabolic equations, for which the maximum principle can be applied. We differentiate (10), (13) with respect to spartial variables up to (p + 2) times, thus obtaining

UT(t) < Ui(0), UT(t) < U(0), |a| < p + 2, t G (0,T /3]. (14)

In second fractional step (11), (13) is n separate linear first-order partial differential equations

duT duT duT

= 3vu1 (t T /3, x) ^ + • • • + 3v<(t T /3, x) —^, dt dxi dxn

uT|t= 3 = <(T/3, x), i = 1,..., n,

solutions of which satisfy Theorem 3.1, giving us estimate (with K3 equals to l(p + 2) arising in Theorem 3.1)

|DauT(t,x)| < UiT(T/3)eTK3UT/3), |a| < p + 2, t G (t/3,2t /3],

leading to

UT(t) < UT(T/3)eTK3UT/3), t G (T/3,2T /3]. (15)

In third fractional step «T (t, x) are solutions to n separate Cauchy problems for ordinary differential equations with known right-hand sides. Thus «T (t, x) and their derivatives can be expressed explicitly

Da«T(t,x) = «t(-,x)W' 3D«/i (e-T,x,«T(e-T,x),^(e-T)) de,

3 J2t/3 V 3 3 3 /

|a| < p + 2, t G (2t/3, t],

and using (4) can be estimated* by

UT(t) < UT(2t/3)eTK4UT(2T/3), t G (2t/3, t]. (16)

^For detailed derivation of (16), see Appendix 5.

Let t* be nonnegative constant satisfying

e6i*KsU(0) < 2, K5 = max(K3, K4). (17)

We will prove that derivatives {D auT}, |a| < p +2 are bounded uniformly on t in some time interval 0 ^ t ^ t*. Here and further t be arbitrary small (t ^ t*) and for some integer M' = M'(t) equality M't = t* is valid. From (17)

e(2j-i)3TK5u(0) ^ 2, i =1,...,M'. (18)

Using (18) we express from (14)-(16) estimate valid in t G [0, t]

UT(t) < U(0)e3TK5U(0). (19)

We will prove the inequality

UT(iT) < U(0) exp((2i - 1)3tK5U(0)) = K6, i = 1,..., M', (20)

by induction. For i = 1 (20) is valid by (19). Let (20) be valid for some i < M'. Applying our reasoning as in zeroth whole step, we deduce

UT((i + 1)t) < UT(iT)e3TK5UT(iT) < < U(0) exp((2i - 1) • 3tK5U(0))exp(3TKU(0)e(2i-1)3TK5U(0)) <

< U(0) exp((2i + 1) • 3tK5U(0)) = U(0) exp((2(i + 1) - 1) • 3tK5U(0)),

thus validating (20) for i +1. It holds for all i < M' by mathematical induction principle. Since UT(t) is monotonic, from (20) we have

UT(t) < UT(M't) = K6 - const, t G [0,t*].

From the previous inequality it follows that uniform on t

|DauT(t,x)| < K6, (t,x) G n[0,t*], |a| < p + 2, (21)

where n[0,t*] = {(t,x)|0 < t < t*,x G Rn}. Derivatives

d d

— DauT(t,x),-—DauT(t, x), (t, x) G njM", |a| < p, i = 1,. .., n, where n|M0i*] = {(t

, x),t G [0, t*], |xj| ^ M0}, are bounded uniformly on t from (21) and equations (10)-(12), which implies uniform boundedness and uniform equicontinuity (for any M0 > 0) of function sets {DauT}, |a| < p in njM°*].

Applying Arzela-Ascoli theorem about compactness, we show existence of the subsequence uTk (t, x) of sequence uT(t, x), which converges to some vector function u(t, x) with its derivatives Dau(t,x), |a| ^ p. Under the theorem about weak approximation method convergence [4] the vector function u(t, x) is a

solution (of C 1'p(n|0,0i*]) class) to (1), (2) in |xj| ^ M0, and ||DauT - Dauyc(nMo ) ^ 0, |a| < p

for t ^ 0.

Since M0 is arbitrary constant, the vector function u(t, x) is a solution to (1), (2) in whole n[0 t*] domain. Theorem 1.1 proved.

5. Derivation of inequality (15)

We are given with

Da<(t,x) = Da<(— ,x)+ / 3Da/ (e - T, x, UT(e - T,x),w(e - T})

3 j2t / 3 V 3 3 3 /

|a| < p + 2, t e (2t/3,t], Taking absolute value of both sides of the previous equality and using (4) we have

|DauT(t,x)| < DauT(-,x) + f 3K2 (1 + UT(e T /3) + UT(e T /3)2) de-

3 ^/3

Since 2t/3 < e ^ t < t and U(t) is nondecreasing function, it is true that U(e —T /3) < U(2t/3): |DauT(t,x)| < |DauT(2T/3,x)| W' 3K2 (1 + UT(2T/3) + UT(2T/3)2) de.

■'2t/3

Integrand in the previous inequality does not depend on the integration variable. J2t/3 de <T /3.

As UT(t) > 1, it is obvious that UT(2t/3)2 > UT(2t/3) > 1. Thus

|DauT(t,x)| < |DauT(2t/3,x)| + 3tK2UT(2t/3)2. We apply supxeRn first, then sup^^ to both parts of the previous inequality:

UiT(t) < U0T(2T/3) + 3tK2UT(2T/3)2. Taking maxa for |a| < p + 2, and calculating sum for i = 1,..., n, we obtain

UT(t) < UT(2T/3) + 3nTK2UT(2T/3)2. Let K4 be equal 3nK2. We factor out UT(2t/3):

UT(t) < UT(2T/3) • (1 + TK4UT(2T/3)) . Using 1 + x ^ ex we finally get

UT(t) < UT(2T/3) • eTK4Ut(2t/3).

References

[1] V.G.Romanov, Inverse problems of mathematical physics, V.S.P. Intl Science, 1986.

[2] A.I.Kozhanov, Nonlinear loaded equations and inverse problems, Computational Mathematics and Mathematical Physics, 44(2004), no. 4, 657-675.

[3] I.V.Frolenkov, Yu.Ya.Belov, On existence of solution to loaded two-dimensional parabloic equations class with Cauchy data, Nonclassic mathematical physics equations (article collection), 2012, 262-279 (in Russian).

[4] Yu.Ya.Belov, S.A.Cantor, Weak approximation method, Krasnoyarskii gosudarstvennyi universitet, 1999 (in Russian).

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