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Surface temperature determination from borehole measurements: regularization by cardinal series
Nonlinear Analysis 50 (2002) 1055 – 1063
www.elsevier.com/locate/na
Surface temperature determination from borehole measurements: regularization by cardinal series Dinh Ngoc Thanha , Nguyen Van Nhanb , Pham Ngoc Dinhc; ∗ , Tran Thi Lea a Department
of Mathematics, HoChiMinh City National University, College of Natural Sciences, 227 Nguyen Van Cu Str., Dist.5, HoChiMinh City, Viet Nam b University of Economics, HoChiMinh City National University, 59C Nguyen Dinh Chieu Str., Dist.3, HoChiMinh City, Viet Nam c MAPMO, UMR 6628, Department of Mathematics, Orl) eans University, BP 6759 45067 Orl)eans, Cedex 2, France Received 3 June 2000; accepted 14 May 2001
Keywords: Heat equation; Borehole measurements; Ill-posed; Regularization; Truncated integration; Error estimates
1. Introduction Consider the heat equation @u 8u − (1) = 0; (x; y) ∈ R × R + ; t ¿ 0; @t where we have taken the heat conductivity to be 1 for notational convenience. The problem is to determine the temperature v(x; t) = u(x; 0; t) on the surface y = 0 of the half plane (x; y) ∈ R × R + from temperature measurements u(x; 1; t) on the line y = 1; given the initial temperature [1] u(x; y; 0) = f(x; y):
(2)
As is known, the problem is ill-posed. Here, we shall regularize it by truncated integration which is a continuous analogue of the truncated expansion series method used by Ang and Hai in [2]. ∗
Corresponding author. E-mail address: [email protected] (P.N. Dinh).
c 2002 Elsevier Science Ltd. All rights reserved. 0362-546X/02/$ - see front matter PII: S 0 3 6 2 - 5 4 6 X ( 0 1 ) 0 0 8 0 2 - 1
1056
D.N. Thanh et al. / Nonlinear Analysis 50 (2002) 1055 – 1063
Assume the exact solution v0 of the problem corresponding to the exact data g0 to be in Sobolev space H (R2 ) for some ¿ 0, and that g − g0 L2 6 : We construct a regularized solution v in the form of a Cardinal series with the following error estimate (Theorem 1): E= − 2 for small; v − v0 L 6 C ln ln E= and Theorem 3 shows how error estimates can be further improved by strengthening regularity on v0 . 2. Convolution equation First, we shall formulate the problem as a convolution equation. Let (x − )2 + (y − )2 1 exp − G(x; y; t; ; ; ) = 2(t − ) 4(t − ) (x − )2 + (y + )2 −exp − : 4(t − )
(3)
Note that the function G satisEes the following identities: @G G + G + = 0; (4) @ and @ (5) ∇ · (u∇G − G∇u) + (Gu) = 0: @ Integrate Eq. (5) over the domain −R ¡ ¡ R; 0 ¡ ¡ R; 0 ¡ ¡ t − and apply Stokes’ formula, then letting R → ∞; → 0+ ; y = 1; we obtain the following integral equation in v(; ): t ∞ 1 v(; ) (x − )2 + 1 d d = g(x; t); (6) exp − 4(t − ) 2 0 −∞ 4(t − ) where
g(x; t) = −
0
∞
∞
−∞
f(; )G(x; 1; t; ; ; 0) d d + u(x; 1; t):
(7)
Now, we put
2 1 exp − x + 1 ; 4t w(x; t) = t 2 0;
t ¿ 0;
(8)
t 6 0;
and convert Eq. (6) into an equation of convolution type (u ∗ w)(x; t) = 2g(x; t):
(9)
D.N. Thanh et al. / Nonlinear Analysis 50 (2002) 1055 – 1063
1057
Taking the Fourier transform of both sides of (9), we formally have v(; ˆ ) w(; ˆ ) = g(; ˆ ); where ’(; ˆ ) =
1 2
Now, we have w(; ˆ ) =
1 2
R2
(10)
’(x; t)e−i(x+t) d x dt:
∞
2 1 −x + 1 exp + ix + it d x dt; 2 4t −∞ t 2 ∞ 1 1 x exp − + it dt exp − + ix d x: t2 4t 4t −∞
0
∞
∞ 1 2 0 √ Putting X = x= 2t, we obtain 2 ∞ √ ∞ −X 2 =2 iX (√2t) x exp − + ix d x = 2t e e dX; 4t −∞ −∞ √ 2 = 2 te−t
=
2
2
since the Fourier transform of e−X =2 is e−u =2 . Substituting (12) into (11), we get ∞ 1 1 −3=2 2 w(; ˆ ) = √ t exp − + it − t dt: 4t 0 Applying the following formula ([3], p. 340) ∞ 2 i (1) x+ d x = i e−i =2 H− x −1 exp ( ) 2 x 0 (Im ¿ 0; Im( 2 ) ¿ 0) we obtain
i i 2 dt; t exp (2 + 2i ) t + 2 4t( + i2 ) 0 −1=4 1 i (1) = √ i ei=4 H1=2 ((i − 2 )1=2 ): 4( + i2 )
1 w(; ˆ ) = √
∞
−3=2
Now, by formula ([3], p. 967)
2 eiz (1) H1=2 (z) = ; z i we have √
w(; ˆ ) = i5=4
√
2
( + i2 )1=4
√
2
(i − 2 )1=2
where we have used the identity ei=4 = i1=2 .
ei
√
i−2
i
√ 2 = 2e− −i
(11)
(12)
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D.N. Thanh et al. / Nonlinear Analysis 50 (2002) 1055 – 1063
Hence
√ 2 2v(; ˆ ) e− −it = g(; ˆ ):
(13)
It is seen that if a solution v exists, then v is given by v(; ˆ ) = 12 e
2
−i
g(; ˆ ):
(14)
The foregoing relation shows that solutions of the problem do not always exist for arbitrary g in L2 ≡ L2 (R2 ), and that the problem admits at most one solution. If a solution exists, it is given by √ 1 2 g(; ˆ ) e −i ei(x+t) d d (15) v(x; t) = 4 R2 We shall construct a regularized solution that is stable with respect to variations in the measured data. 3. Error estimates, cardinal series We have the following Theorem 1. Let g0 be such that Eq. (9) has the solution v0 ∈ L2 : Let g − g0 L2 6 : Assume that
v0 H ≡ Put gˆ = gˆ · !D ;
(16) 1=2
R2
(1 + 2 + 2 ) |vˆ0 |2 d d
6
E : 2
R E= 1=4 D = [ − a; a] ; with a = √ ; R = 2 ln ln (E=) 2 2
in which !D is the characteristic function of D . Put √ 1 2 gˆ (; ) e −i ei(x+t) d d v(x; t) = 4 R2 Then; we have the following error estimate: E E= for 6 : v − v0 L2 6 C ln− ln (E=) 10 √ 2 Proof. First, we have |e −i | = e& , where 1=2 2 + (4 + 2 )1=2 &(; ) = : 2
(17)
(18)
(19)
(20)
D.N. Thanh et al. / Nonlinear Analysis 50 (2002) 1055 – 1063
In view of the inequality (
ab)2 6
a2
1059
b2 , we deduce that
(2 + (4 + 2 )1=2 )2 6 2(24 + 2 ); which gives &(; ) 6
24 + 2 2
1=4 :
Hence, we get for (; ) in D 1=4 4 1=4 4 2a + a2 R + R2 R &(; ) 6 = 6 1=4 : 2 4 2 On the other hand, for || ¿ a or || ¿ a, we have R2 1 + 2 + 2 ¿ a2 = : 2 Now, in view of Plancherel’s theorem, we have √ √ 1 1 2 2 |e −i |2 |gˆ0 |2 d d; |e −i |2 |g− ˆ gˆ0 |2 d d+ ˆ vˆ0 2L2 = v−v0 2L2 =v− 4 Dc 4 D 1 1 2& 2 e |gˆ − gˆ0 | d d + e2& |gˆ0 |2 d d: 6 4 D 4 Dc From (14) we have |vˆ0 |2 = 14 e2& |gˆ0 |2 : Hence
1 2 2R (1 + 2 + 2 ) |vˆ0 |2 d d; exp 1=4 gˆ − gˆ0 2L2 + 2 4 2 R 2 R 2
2 E 2 E 1 + 2 : 6 2 4 ln (E=) R
v − v0 2L2 6
For 6 E=10, we have R = 21=4 ln(E==ln2 E=) 6 21=4 ln E=: Hence for 6 E=10; 1=ln2 (E=) 6 2 =2 =R2 which implies that E2 v − v0 2L2 6 (2 + 2 =2 ) 2 ; 4R E= E2 ; 6 (1 + 2 =2 ) ln−2 4 ln (E=) or equivalently v − v0 L2 6 C ln where
−
E=
ln (E=)
;
E 1 + 2 =2 : 2 which completes the proof of Theorem 1. C=
1060
D.N. Thanh et al. / Nonlinear Analysis 50 (2002) 1055 – 1063
The following theorem gives a representation of the regularized solution v in Theorem 1 as a Cardinal Series. Theorem 2. If v is de>ned as in (19); there v can be represented by a double Cardinal series as follows: ∞
v(x; t) =
2v(mh; nk) S(m; h) S(n; k)(t);
(21)
n=−∞|m|6|n|
where h = =a; R is de>ned as in (18); and S(p; d)(z) is the Sinc function given by S(p; d)(z) =
sin[(z − pd)=d] ; (z − pd)=d
p = 0; ± 1; ± 2; : : : ; d ¿ 0:
(22)
The series in (21) converges in L2 . Proof. First we have 2 − ; : supp vˆ ⊂ D = h h Now, the Fourier series for vˆ is
v(; ˆ ) ≈ Cmn e−i(mh+nh) ;
(23)
m; n
in which h2 Cmn = 2
D
v(; ˆ )ei(mh+nh) d d;
= h2 v(mh; nh): Let |m| 6 |n| and substitute Cmn into (23) v(; ˆ ) =
∞
Cmn e−i(mh+nh) ;
n=−∞|m|6|n|
= h2
∞
v(mh; nh)e−i(mh+nh) ;
n=−∞|m|6|n|
for (; t) ∈ D . Here, the double series converges in L2 . Now, equality (24) implies that 1 v(x; t) = v(; ˆ )ei(x+t) d d; 2 R2 ∞
h2 = v(mh; nh)e−i(mh−x+nh−t) d d; 2 D n=−∞ |m|6|n|
(24)
D.N. Thanh et al. / Nonlinear Analysis 50 (2002) 1055 – 1063
=
1061
=h =h ∞ 1 v(mh; nh) he−i(mh−x) d he−i(nh−t) d; 2 n=−∞ −=h −=h |m|6|n|
=
∞ 1 v(mh; nh) S(m; h)(x) S(n; h)(t); 2 n=−∞ |m|6|n|
where we have used the following formula ([4], p. 92) =d 1 d ei(pd−z) d = S(p; d)(z); p = 0; ± 1; ± 2; : : : ; d ¿ 0: 2 −=d This completes the proof of Theorem 2. Now, strenghthening regularity conditions on the exact solution of the problem, we obtain the following theorem. Theorem 3. Let (16) hold. Suppose that E2 exp(25=4 (1 + 2 + 2 )1=2 ) |vˆ0 |2 d d 6 ; 4 R2
(25)
for some ¿ 0: √ Let gˆ be de>ned as in (18) where a = R= 2; and E 21=4 ln :
+1 Let v(x; t) be de>ned as in (19): Then; we get the following error estimate: R=
(26)
v − v0 L2 6 √12 E 1=( +1) =( +1) :
(27)
Proof. As in the proof of Theorem 1, we have 1 1 2 2& 2 v − v0 L2 = e |gˆ − gˆ0 | d d + e2& |gˆ0 |2 d d; 4 D 4 R2 \D 6
1 exp(23=4 R)2 + exp(−23=4 R ) 4 exp(25=4 (1 + 2 + 2 )1=2 ) |vˆ0 |2 d d; × R2
6
E 2=( +1) 2 2 =( +1) 2 + E ; 42=( +1) 4E 2 =( +1)
=
1 2 =( +1) 2=( +1) E : 2
Hence v − v0 L2 6 √12 2 =( +1) E 2=( +1) : which completes the proof of Theorem 3.
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D.N. Thanh et al. / Nonlinear Analysis 50 (2002) 1055 – 1063
Remarks. (1) The regularized solution v in Theorem 2 can also be represented by the Cardinal series as in Theorem 3. (2) It is seen that v(x; t) is completely recovered from its values at the lattice points (mh; nk); m; n = 0; ± 1; ± 2; : : : :
4. Numerical results We shall take a numerical approximation of (7) in the form ∞ ∞ g(xl ; ti ) = − f(; )G(xl ; 1; ti ; ; ; 0) d d + u(xl ; 1; ti ); 0
−∞
(28)
and, where ti = ihT ; xl = − 10 + lhX ; i = 1; : : : ; N; l = 1; : : : ; N : To calculate the double integral in (28) we have used the rectangle rule which gives good accuracy if one integrates on the interval [−10; 10], the integrand G(x; 1; t; ; ; 0) and its partial derivatives are quite small for |x| ¿ 10. In the main program the Fourier transform of ’(; ) 1 ’(x; ˆ t) = ’(; )e−i(x+t) d d; 2 R2 has been calculated for each point (x; t) ∈ [−0:5; 0:5]×[0; 0:6] with a step hX = hT = 0:02, the function ’(; ) being stored in a (1000; 500) array, the double integral being calculated as previously.
Fig. 1.
D.N. Thanh et al. / Nonlinear Analysis 50 (2002) 1055 – 1063
1063
The Fourier transform of the regularized solution v (x; t), for given ¿ 0 is calculated from formula vˆ (; ) = exp( 2 + i)(fˆ !D )(; ); D = [ − a; a] × [ − a; a]; f chosen such that fˆ − fˆ 0 L2 (R2 ) 6 , f0 designating the function g in (7). If fˆ 0 is the r.h.s of (10) corresponding to the exact solution v0 , then we choose f in the following form: (i − it) ˆ t) = fˆ (x; t) + : f(x; 0 (x2 + 1)(t 2 + 1) Thus we have
fˆ − fˆ 0 = √ : 2 To solve numerically (28) we have considered the functions f(x; t) = |cos x| cos 2t;
u(x; 1; t) = 0:
For = 10−9 which corresponds approximately to a = 3:5, for |x| 6 0:5 and 0 6 t 6 0:6 with a step hX = hT = 0:02, we End that max |vˆ (xi ; tj ) − vˆ0 (xi ; tj )| 6 10−5 :
i; j=1;50
The Fig. 1 gives the surface corresponding to this parameter. References [1] R.S. Anderssen, V.A. Saull, Surface temperature history determination from borehole measurements, Math. Geol. 5 (1973) 269–283. [2] D.D. Ang, D.D. Hai, On the backward heat equation, Ann. Polonici Math. 52 (1991) 39–42. [3] I.S. Gradshteyn, I.M. Ryznik, Table of Integrals, Series and Products, Academic Press, New York, 1996. [4] F. Stenger, Numerical methods based on Sinc and analytic functions, Springer, New York, 1993.
www.elsevier.com/locate/na
Surface temperature determination from borehole measurements: regularization by cardinal series Dinh Ngoc Thanha , Nguyen Van Nhanb , Pham Ngoc Dinhc; ∗ , Tran Thi Lea a Department
of Mathematics, HoChiMinh City National University, College of Natural Sciences, 227 Nguyen Van Cu Str., Dist.5, HoChiMinh City, Viet Nam b University of Economics, HoChiMinh City National University, 59C Nguyen Dinh Chieu Str., Dist.3, HoChiMinh City, Viet Nam c MAPMO, UMR 6628, Department of Mathematics, Orl) eans University, BP 6759 45067 Orl)eans, Cedex 2, France Received 3 June 2000; accepted 14 May 2001
Keywords: Heat equation; Borehole measurements; Ill-posed; Regularization; Truncated integration; Error estimates
1. Introduction Consider the heat equation @u 8u − (1) = 0; (x; y) ∈ R × R + ; t ¿ 0; @t where we have taken the heat conductivity to be 1 for notational convenience. The problem is to determine the temperature v(x; t) = u(x; 0; t) on the surface y = 0 of the half plane (x; y) ∈ R × R + from temperature measurements u(x; 1; t) on the line y = 1; given the initial temperature [1] u(x; y; 0) = f(x; y):
(2)
As is known, the problem is ill-posed. Here, we shall regularize it by truncated integration which is a continuous analogue of the truncated expansion series method used by Ang and Hai in [2]. ∗
Corresponding author. E-mail address: [email protected] (P.N. Dinh).
c 2002 Elsevier Science Ltd. All rights reserved. 0362-546X/02/$ - see front matter PII: S 0 3 6 2 - 5 4 6 X ( 0 1 ) 0 0 8 0 2 - 1
1056
D.N. Thanh et al. / Nonlinear Analysis 50 (2002) 1055 – 1063
Assume the exact solution v0 of the problem corresponding to the exact data g0 to be in Sobolev space H (R2 ) for some ¿ 0, and that g − g0 L2 6 : We construct a regularized solution v in the form of a Cardinal series with the following error estimate (Theorem 1): E= − 2 for small; v − v0 L 6 C ln ln E= and Theorem 3 shows how error estimates can be further improved by strengthening regularity on v0 . 2. Convolution equation First, we shall formulate the problem as a convolution equation. Let (x − )2 + (y − )2 1 exp − G(x; y; t; ; ; ) = 2(t − ) 4(t − ) (x − )2 + (y + )2 −exp − : 4(t − )
(3)
Note that the function G satisEes the following identities: @G G + G + = 0; (4) @ and @ (5) ∇ · (u∇G − G∇u) + (Gu) = 0: @ Integrate Eq. (5) over the domain −R ¡ ¡ R; 0 ¡ ¡ R; 0 ¡ ¡ t − and apply Stokes’ formula, then letting R → ∞; → 0+ ; y = 1; we obtain the following integral equation in v(; ): t ∞ 1 v(; ) (x − )2 + 1 d d = g(x; t); (6) exp − 4(t − ) 2 0 −∞ 4(t − ) where
g(x; t) = −
0
∞
∞
−∞
f(; )G(x; 1; t; ; ; 0) d d + u(x; 1; t):
(7)
Now, we put
2 1 exp − x + 1 ; 4t w(x; t) = t 2 0;
t ¿ 0;
(8)
t 6 0;
and convert Eq. (6) into an equation of convolution type (u ∗ w)(x; t) = 2g(x; t):
(9)
D.N. Thanh et al. / Nonlinear Analysis 50 (2002) 1055 – 1063
1057
Taking the Fourier transform of both sides of (9), we formally have v(; ˆ ) w(; ˆ ) = g(; ˆ ); where ’(; ˆ ) =
1 2
Now, we have w(; ˆ ) =
1 2
R2
(10)
’(x; t)e−i(x+t) d x dt:
∞
2 1 −x + 1 exp + ix + it d x dt; 2 4t −∞ t 2 ∞ 1 1 x exp − + it dt exp − + ix d x: t2 4t 4t −∞
0
∞
∞ 1 2 0 √ Putting X = x= 2t, we obtain 2 ∞ √ ∞ −X 2 =2 iX (√2t) x exp − + ix d x = 2t e e dX; 4t −∞ −∞ √ 2 = 2 te−t
=
2
2
since the Fourier transform of e−X =2 is e−u =2 . Substituting (12) into (11), we get ∞ 1 1 −3=2 2 w(; ˆ ) = √ t exp − + it − t dt: 4t 0 Applying the following formula ([3], p. 340) ∞ 2 i (1) x+ d x = i e−i =2 H− x −1 exp ( ) 2 x 0 (Im ¿ 0; Im( 2 ) ¿ 0) we obtain
i i 2 dt; t exp (2 + 2i ) t + 2 4t( + i2 ) 0 −1=4 1 i (1) = √ i ei=4 H1=2 ((i − 2 )1=2 ): 4( + i2 )
1 w(; ˆ ) = √
∞
−3=2
Now, by formula ([3], p. 967)
2 eiz (1) H1=2 (z) = ; z i we have √
w(; ˆ ) = i5=4
√
2
( + i2 )1=4
√
2
(i − 2 )1=2
where we have used the identity ei=4 = i1=2 .
ei
√
i−2
i
√ 2 = 2e− −i
(11)
(12)
1058
D.N. Thanh et al. / Nonlinear Analysis 50 (2002) 1055 – 1063
Hence
√ 2 2v(; ˆ ) e− −it = g(; ˆ ):
(13)
It is seen that if a solution v exists, then v is given by v(; ˆ ) = 12 e
2
−i
g(; ˆ ):
(14)
The foregoing relation shows that solutions of the problem do not always exist for arbitrary g in L2 ≡ L2 (R2 ), and that the problem admits at most one solution. If a solution exists, it is given by √ 1 2 g(; ˆ ) e −i ei(x+t) d d (15) v(x; t) = 4 R2 We shall construct a regularized solution that is stable with respect to variations in the measured data. 3. Error estimates, cardinal series We have the following Theorem 1. Let g0 be such that Eq. (9) has the solution v0 ∈ L2 : Let g − g0 L2 6 : Assume that
v0 H ≡ Put gˆ = gˆ · !D ;
(16) 1=2
R2
(1 + 2 + 2 ) |vˆ0 |2 d d
6
E : 2
R E= 1=4 D = [ − a; a] ; with a = √ ; R = 2 ln ln (E=) 2 2
in which !D is the characteristic function of D . Put √ 1 2 gˆ (; ) e −i ei(x+t) d d v(x; t) = 4 R2 Then; we have the following error estimate: E E= for 6 : v − v0 L2 6 C ln− ln (E=) 10 √ 2 Proof. First, we have |e −i | = e& , where 1=2 2 + (4 + 2 )1=2 &(; ) = : 2
(17)
(18)
(19)
(20)
D.N. Thanh et al. / Nonlinear Analysis 50 (2002) 1055 – 1063
In view of the inequality (
ab)2 6
a2
1059
b2 , we deduce that
(2 + (4 + 2 )1=2 )2 6 2(24 + 2 ); which gives &(; ) 6
24 + 2 2
1=4 :
Hence, we get for (; ) in D 1=4 4 1=4 4 2a + a2 R + R2 R &(; ) 6 = 6 1=4 : 2 4 2 On the other hand, for || ¿ a or || ¿ a, we have R2 1 + 2 + 2 ¿ a2 = : 2 Now, in view of Plancherel’s theorem, we have √ √ 1 1 2 2 |e −i |2 |gˆ0 |2 d d; |e −i |2 |g− ˆ gˆ0 |2 d d+ ˆ vˆ0 2L2 = v−v0 2L2 =v− 4 Dc 4 D 1 1 2& 2 e |gˆ − gˆ0 | d d + e2& |gˆ0 |2 d d: 6 4 D 4 Dc From (14) we have |vˆ0 |2 = 14 e2& |gˆ0 |2 : Hence
1 2 2R (1 + 2 + 2 ) |vˆ0 |2 d d; exp 1=4 gˆ − gˆ0 2L2 + 2 4 2 R 2 R 2
2 E 2 E 1 + 2 : 6 2 4 ln (E=) R
v − v0 2L2 6
For 6 E=10, we have R = 21=4 ln(E==ln2 E=) 6 21=4 ln E=: Hence for 6 E=10; 1=ln2 (E=) 6 2 =2 =R2 which implies that E2 v − v0 2L2 6 (2 + 2 =2 ) 2 ; 4R E= E2 ; 6 (1 + 2 =2 ) ln−2 4 ln (E=) or equivalently v − v0 L2 6 C ln where
−
E=
ln (E=)
;
E 1 + 2 =2 : 2 which completes the proof of Theorem 1. C=
1060
D.N. Thanh et al. / Nonlinear Analysis 50 (2002) 1055 – 1063
The following theorem gives a representation of the regularized solution v in Theorem 1 as a Cardinal Series. Theorem 2. If v is de>ned as in (19); there v can be represented by a double Cardinal series as follows: ∞
v(x; t) =
2v(mh; nk) S(m; h) S(n; k)(t);
(21)
n=−∞|m|6|n|
where h = =a; R is de>ned as in (18); and S(p; d)(z) is the Sinc function given by S(p; d)(z) =
sin[(z − pd)=d] ; (z − pd)=d
p = 0; ± 1; ± 2; : : : ; d ¿ 0:
(22)
The series in (21) converges in L2 . Proof. First we have 2 − ; : supp vˆ ⊂ D = h h Now, the Fourier series for vˆ is
v(; ˆ ) ≈ Cmn e−i(mh+nh) ;
(23)
m; n
in which h2 Cmn = 2
D
v(; ˆ )ei(mh+nh) d d;
= h2 v(mh; nh): Let |m| 6 |n| and substitute Cmn into (23) v(; ˆ ) =
∞
Cmn e−i(mh+nh) ;
n=−∞|m|6|n|
= h2
∞
v(mh; nh)e−i(mh+nh) ;
n=−∞|m|6|n|
for (; t) ∈ D . Here, the double series converges in L2 . Now, equality (24) implies that 1 v(x; t) = v(; ˆ )ei(x+t) d d; 2 R2 ∞
h2 = v(mh; nh)e−i(mh−x+nh−t) d d; 2 D n=−∞ |m|6|n|
(24)
D.N. Thanh et al. / Nonlinear Analysis 50 (2002) 1055 – 1063
=
1061
=h =h ∞ 1 v(mh; nh) he−i(mh−x) d he−i(nh−t) d; 2 n=−∞ −=h −=h |m|6|n|
=
∞ 1 v(mh; nh) S(m; h)(x) S(n; h)(t); 2 n=−∞ |m|6|n|
where we have used the following formula ([4], p. 92) =d 1 d ei(pd−z) d = S(p; d)(z); p = 0; ± 1; ± 2; : : : ; d ¿ 0: 2 −=d This completes the proof of Theorem 2. Now, strenghthening regularity conditions on the exact solution of the problem, we obtain the following theorem. Theorem 3. Let (16) hold. Suppose that E2 exp(25=4 (1 + 2 + 2 )1=2 ) |vˆ0 |2 d d 6 ; 4 R2
(25)
for some ¿ 0: √ Let gˆ be de>ned as in (18) where a = R= 2; and E 21=4 ln :
+1 Let v(x; t) be de>ned as in (19): Then; we get the following error estimate: R=
(26)
v − v0 L2 6 √12 E 1=( +1) =( +1) :
(27)
Proof. As in the proof of Theorem 1, we have 1 1 2 2& 2 v − v0 L2 = e |gˆ − gˆ0 | d d + e2& |gˆ0 |2 d d; 4 D 4 R2 \D 6
1 exp(23=4 R)2 + exp(−23=4 R ) 4 exp(25=4 (1 + 2 + 2 )1=2 ) |vˆ0 |2 d d; × R2
6
E 2=( +1) 2 2 =( +1) 2 + E ; 42=( +1) 4E 2 =( +1)
=
1 2 =( +1) 2=( +1) E : 2
Hence v − v0 L2 6 √12 2 =( +1) E 2=( +1) : which completes the proof of Theorem 3.
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D.N. Thanh et al. / Nonlinear Analysis 50 (2002) 1055 – 1063
Remarks. (1) The regularized solution v in Theorem 2 can also be represented by the Cardinal series as in Theorem 3. (2) It is seen that v(x; t) is completely recovered from its values at the lattice points (mh; nk); m; n = 0; ± 1; ± 2; : : : :
4. Numerical results We shall take a numerical approximation of (7) in the form ∞ ∞ g(xl ; ti ) = − f(; )G(xl ; 1; ti ; ; ; 0) d d + u(xl ; 1; ti ); 0
−∞
(28)
and, where ti = ihT ; xl = − 10 + lhX ; i = 1; : : : ; N; l = 1; : : : ; N : To calculate the double integral in (28) we have used the rectangle rule which gives good accuracy if one integrates on the interval [−10; 10], the integrand G(x; 1; t; ; ; 0) and its partial derivatives are quite small for |x| ¿ 10. In the main program the Fourier transform of ’(; ) 1 ’(x; ˆ t) = ’(; )e−i(x+t) d d; 2 R2 has been calculated for each point (x; t) ∈ [−0:5; 0:5]×[0; 0:6] with a step hX = hT = 0:02, the function ’(; ) being stored in a (1000; 500) array, the double integral being calculated as previously.
Fig. 1.
D.N. Thanh et al. / Nonlinear Analysis 50 (2002) 1055 – 1063
1063
The Fourier transform of the regularized solution v (x; t), for given ¿ 0 is calculated from formula vˆ (; ) = exp( 2 + i)(fˆ !D )(; ); D = [ − a; a] × [ − a; a]; f chosen such that fˆ − fˆ 0 L2 (R2 ) 6 , f0 designating the function g in (7). If fˆ 0 is the r.h.s of (10) corresponding to the exact solution v0 , then we choose f in the following form: (i − it) ˆ t) = fˆ (x; t) + : f(x; 0 (x2 + 1)(t 2 + 1) Thus we have
fˆ − fˆ 0 = √ : 2 To solve numerically (28) we have considered the functions f(x; t) = |cos x| cos 2t;
u(x; 1; t) = 0:
For = 10−9 which corresponds approximately to a = 3:5, for |x| 6 0:5 and 0 6 t 6 0:6 with a step hX = hT = 0:02, we End that max |vˆ (xi ; tj ) − vˆ0 (xi ; tj )| 6 10−5 :
i; j=1;50
The Fig. 1 gives the surface corresponding to this parameter. References [1] R.S. Anderssen, V.A. Saull, Surface temperature history determination from borehole measurements, Math. Geol. 5 (1973) 269–283. [2] D.D. Ang, D.D. Hai, On the backward heat equation, Ann. Polonici Math. 52 (1991) 39–42. [3] I.S. Gradshteyn, I.M. Ryznik, Table of Integrals, Series and Products, Academic Press, New York, 1996. [4] F. Stenger, Numerical methods based on Sinc and analytic functions, Springer, New York, 1993.