Applied Mathematics and Computation 150 (2004) 833–839

www.elsevier.com/locate/amc

On the lateral Cauchy problem

Hakan S�ims�ek

Atat€urk €Universitesi, Fen-Edebiyat Fak€ultesi, Matematik B€ol€um€u, Erzurum T-25240, Turkey

Abstract

Considerable attention is currently begin devoted to Cauchy problems of mathe-

metical physics in view of their increasing importance for the solution of applied

problems [M. Taylor, Partial Differential Equations, Springer, Berlin, 1966]. Uniqueness

of the solution for these problems very important. In this paper we found a bound in the

uniqueness theorem for the Cauchy problem.

� 2003 Elsevier Inc. All rights reserved.

1. Introduction

In this paper deal with the bound in Cauchy problem. We consider the most

challenging hyperbolic equations, where uniqueness in the lateral Cauchy

problem is much less understood.

We consider first the operator

E-m

0096-3

doi:10.

Aðx; t; oÞu ¼ a20o

2j � Duþ

Xbjojuþ cu

where t ¼ xnþ1 and the sum is over j ¼ 1; 2; . . . ; nþ 1, in the cylindrical domain

X ¼ G� ð�T ; T Þ in Rnþ1. We assume that a0 2 C1ðXÞ, a0 > 0, bj, c 2 L1ðXÞ.We will use the following weight function:

/ðx; tÞ ¼ expr2

� �wðx; tÞ ð1:1Þ

with the two choices of w:

w1ðx; tÞ ¼ x21 þ þ x2

n�1 þ ðxn � bÞ2 � h2t2 � s

ail addresses: erzurum.hsimsek@atauni.edu.tr, hsimsek@atauni.edu.tr (H. S�ims�ek).

003/$ - see front matter � 2003 Elsevier Inc. All rights reserved.

1016/S0096-3003(03)00311-4

834 H. S�ims�ek / Appl. Math. Comput. 150 (2004) 833–839

or

w2ðx; tÞ ¼ �x21 � � x2

n�1 � 2ðxn � bÞ2 � h2t2 þ r2 þ 2b2

There we set Xe ¼ X \ fw > eg [2].

Theorem 1.1. Let G be a domain in Rn and c a part of its boundary such that oneof the following three conditions is satisfied:

The origin belongs to G : c ¼ oG ð1:2Þ

G is subset of f�h < xn < 0; jx0j < rg; c ¼ oG \ fxn < 0g ð1:3Þ

G is subset of f�h < xn < 0; jx0j < rg; c ¼ oG \ fxn ¼ 0g ð1:4Þ

Suppose that in the cases (1.2) and (1.3)

h2a20ða0 þ tota0 þ 2a�1

0 jtra0jÞ < a0 þ xra0 � bona0; ha0 6 1 ð1:5Þ

Addition, G � Bð0; hT Þ, b ¼ s ¼ 0 in case (1.2) and hðþ2bÞ < h2T 2, s ¼b2 þ r2 in case (1.3). In case (1.4) we suppose that

3a0 þ jtota0j þ rjra0j þ2r2jra0j

T< 2bona0

r < 2b; r2 < 2hðhþ bÞ; a0r < T on X ð1:6Þ

and we let h ¼ rT .

Let us consider the Cauchy problem:

Au ¼ f on X; ojvu ¼ gj; j6m1 � 1 on C;

oau 2 L2ðXÞ when ja : mj < 1ð1:7Þ

where C part of oX of class Cm1 that is open in oX.

Let w be w1 in cases (1.2) and (1.3) and w2 in case (1.3). Then a solution uto the Cauchy problem (1.8) has the following bound:

koauk2ðXeÞ6CðF þM1�k2 F kÞ when jaj6 1 ð1:8Þ

where C, a 2 ð0; 1Þ depend on e, M2 is kukð1ÞðXÞ and F is kf k2ðXÞþPkgjkðm1�j�ð1=2ÞÞðCÞ the sum is over j6m1 � 1) [1].

The proof of this theorem can be found in [1].

2. Main results

Theorem 2.1. Suppose that either (a) Amðx; nÞ 6¼ 0 for all n 2 R n f0g or (b) thecoefficients of Am are real-valued.

H. S�ims�ek / Appl. Math. Comput. 150 (2004) 833–839 835

If the equalities

Amðx; fÞ ¼ 0; f ¼ n þ isrq;; s 6¼ 0

or

Amðx; fÞ ¼ 0;Xj6 q

ðoAm=onjÞo; ¼ 0

imply that

Xj;k6 q

ðojok/ðoAm=ofjÞðoAm=ofkÞ þ s�1JokAmðoAm=ofkÞÞ > 0

in X, then there are is a constant C such that

sZ

Xjoauj2wdx6C

ZXjAuj2w2 dx; ja : mj < 1

when C < s for all functions C10 ðXÞ.

Theorem 2.2. The left side of Theorem 2.1 is H1 þ H2. Where

H1 ¼ 4r/Xnk¼1

jqkj2

(� h2c4jqkj

2 þ Refq2nþ1gcðrcx� boncÞ

þ 2cðrcnÞnnþ1h2t � c3otcjqnþ1j

2t

)

and

H2 ¼r2

4

Xnj;k¼1

owoAm

oqj

!okw

oAm

oqk

�

Proof of the main results. Let us consider the formulae

H ¼Xnþ1

j;k¼1

ðojok/ÞoAm

oqj

!oAm

oqk

�þ s�1Im

Xnþ1

k¼1

ðokAmÞoAm

oqk

�; ð2:1Þ

where / ¼ expððr=2ÞwÞ, w ¼ x21 þ þ x2

n�1 þ ðxn � bÞ2 � h2t2 � S, Amðx; qÞ ¼�c2q2

nþ1 þ q21 þ þ q2

n, q ¼ n þ isrq/ ¼ n þ is/ðr=2Þð�2x1; . . . ;�2xn�1;2ðxn � bÞ;�2h2tÞ, where �2h2t, in nþ 1 coordinates. rq ¼ ðD1; . . . ;Dq;0; . . . ; 0Þ (Dq in nþ 1th coordinates) and we first try to find

ojok/ ¼ r2

/ojokw þ r4

4/ojwokw

836 H. S�ims�ek / Appl. Math. Comput. 150 (2004) 833–839

by using

okw ¼2xk 16 k6 n� 1

2ðxn � bÞ k ¼ n�2h2t k ¼ nþ 1

8<:

okojw ¼0 j 6¼ k2 i6 j ¼ k6 n�2h2 j ¼ k ¼ nþ 1

8<:

and ok/ ¼ r2/okw.

ojok/ ¼

r/dj;k þ r2/xjxk k6 n� 1; j6 n� 1

r2/xkðxn � bÞ k6 n� 1; j ¼ n

�r2/xkh2t k6 n� 1; j ¼ nþ 1

r2/ðxn � bÞxj k ¼ n; j6 n� 1

r2/ðxn � bÞ2 þ s/ k ¼ n; j ¼ n

�r2/ðxn � bÞh2t k ¼ n; j ¼ nþ 1

�r2/h2txj k ¼ nþ 1; j6 nþ 1

�r2/ðxn � bÞh2t k ¼ nþ 1; j ¼ n

�h2r/ þ s2/h4t2 k ¼ nþ 1; j ¼ nþ 1

8>>>>>>>>>>>>>>>>>>>>><>>>>>>>>>>>>>>>>>>>>>:

oAm

oqj

oAm

oqk¼

4qjqk j; k6 n�4qjqnþ1c

2 j6 n; k ¼ nþ 1

�4qkqnþ1c2 j ¼ nþ 1; k6 n

4c4jqnþ1j2 j; k ¼ nþ 1

8>>><>>>:

ð2:2Þ

oAm

oqj¼ 2qj 16 j6 n

�2c2qnþ1 j ¼ nþ 1

�ð2:3Þ

oAm

oqk¼ 2qk 16 k6 n

�2c2qnþ1 j ¼ nþ 1

�ð2:4Þ

Let us name the first part in H1;

L1 ¼Xnþ1

Jj;k¼1

ðojok/ÞoAm

oqj

!oAm

oqk

�ð2:5Þ

H. S�ims�ek / Appl. Math. Comput. 150 (2004) 833–839 837

Put (2.2)–(2.3) together in (2.5), we get

L1 ¼Xj 6¼k

4r2/xjxkqjqk þXn�1

j¼k;j;k6n�1 or k¼1

ðr/þ r2/x2kÞ4jqkj

2

þXn�1

k¼1;ðj¼nÞr2/xkðxn � bÞ4qjqk þ

Xn�1

k¼1;ðj¼nþ1Þð�r2/xkh

2tÞð�4qkqnþ1c2Þ

þXn�1

j¼1

r2/ðxn � bÞxj4qjqk þ ðr2/ðxn � bÞ2 þ r/Þ4jqnj2

þ ð�r2/ðxn � bÞ/2tÞð�4qnqnþ1c2Þ þ

Xn�1

j¼1

ð�r2/h2txjÞð�4qjqnþ1c2Þ

þ ð�r2/h2tðxn � bÞÞð�4qnqnþ1c2Þ þ ð�h2r/þ r2/h4t2Þð4c4jqnþ1jÞ

L1 ¼ 4r2/Xn�1

k¼1

jqkj2jqnj

2 � h2c4jqnþ1j2

¼Xnk¼1

jqkj2

!

þ 4/r2Xn�1

j 6¼k;j;k6n�1

xjxkqjqk

(þXn�1

k¼1

x2k jqkjj

2 þXn�1

k¼1;ðj¼nÞxkðxn � bÞqnqk

þXn�1

k¼1;ðj¼nÞð � xkh

2tÞð � qkqnþ1Þc2 þXn�1

j¼1

ðxn � bÞxjqjqn þ ðxn � bÞ2jqnj2

þ ðh2tðxn � bÞÞðqnqnþ1c2Þ þ

Xn�1

j¼1

ðth2xjÞðqjqnþ1c2Þ

þ ðh2tðxn � bÞÞðqnqnþ1c2Þ þ h4t2c4jqnþ1j

2

)ð2:6Þ

Now find the imaginary part in H : (call it L2)

okðc2q2nþ1Þ ¼ 2cðokcÞqnþ1

L2 ¼ �Xnk¼1

4cðokcÞq2nþ1qk þ 4c3okcq2

nþ1qnþ1

¼ �4cq2nþ1

Xokcðnn � is/rxkÞ � 4concq2

nþ1ðnn � is/rðxn � bÞÞ

þ 4c3otcq2nþ1ðnnþ1 þ 2is/rth2Þ

¼ �4r/ð�q2nþ1iscr0cx0Þððn� 1Þ coordinateÞ þ 4r/ðisoncq2

nþ1ðxn � bÞÞþ 4r/ðc3otcq2

nþ1is2h2tÞ þ terms in H2

838 H. S�ims�ek / Appl. Math. Comput. 150 (2004) 833–839

combining first two terms we write

L2 ¼ 4r/q2nþ1iscrcx ðn� coordinateÞ þ 4r/isconcq2

nþ1b

¼ 4r/ q2nþ1iscrcx

n� isconcq2

nþ1b� 4cq2nþ1r0cn0 � 4concq2

nþ1nn

þ 4c3ðotcÞjqnþ1j2qnþ1

o

Since

q2nþ1 ¼ ðnnþ1 � is/h2tÞ2 ¼ n2

nþ1 � s2h2rh2t � 2isnnþ1/rh2t

submit by qnþ1 ¼ nnþ1 � is/h2t.Finally, we obtain

L2 ¼ 4r/ficq2nþ1sðrcx� boncÞg þ 2cðrcnÞðsith2Þ � c3otcjqnþ1j

2sih2t

þ real parts

s�1ImL2 ¼ 4r/fRefq2nþ1gcðrcx� boncÞg þ 2cðrcnÞnnþ1h

2t � c3otcjqnþ1j2t:

ð2:7Þ

If we collect all H1 terms in L1 and L2 from (2.6) and (2.7)

H1 ¼ 4r/Xnk¼1

jqkj2

(� h2c4jqkj

2 þ Refq2nþ1gcðrcx� boncÞ

þ 2cðrcnÞnnþ1h2t � c3otcjqnþ1j

2t

)

Then

H2 ¼r2

4

Xnj;k¼1

owoAm

oqj

!okw

oAm

oqk

�

by using

owoAoqj

¼4xjqj 16 j6 n� 1

4ðxn � bÞqn j ¼ n4c2h2tqnþ1 j ¼ nþ 1

8<:

and then we write

H2 ¼ 4r2/jx1q1 þ x2q2 þ þ xnqn � bqn þ h2c2tqnþ1j2

Proof is complete. �

H. S�ims�ek / Appl. Math. Comput. 150 (2004) 833–839 839

References

[1] V. Isakov, in: Inverse Problems for Partial Differential Equations, Appl. Math. Sci., vol. 127,

Springer, Berlin, 1998.

[2] M. Taylor, Partial Differential Equations, Springer, Berlin, 1966.