Non homogeneous Heat Equation: Identification and Regularization for the Inhomogeneous Term

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Non homogeneous Heat Equation: Identification and Regularization for the Inhomogeneous Term

Dang Duc Trong, Nguyen Thanh Long, Alain Pham Ngoc Dinh

To cite this version:

Dang Duc Trong, Nguyen Thanh Long, Alain Pham Ngoc Dinh. Non homogeneous Heat Equation:

Identification and Regularization for the Inhomogeneous Term. Journal of Mathematical Analysis and Applications, Elsevier, 2005, 312, pp.93-104. �hal-00009221�

Non homogeneous Heat Equation: Identication and Regularization for the Inhomogeneous Term

Dang Duc Trong, Nguyen Thanh Long by

Mathematics Department, VietNam National University 227 Nguyen Van Cu, HoChiMinh City, District 5

&

Pham Ngoc Dinh Alain

Mathematics Department, Mapmo UMR 6628, BP 67-59 45067, Orleans cedex

Keywords: Identication, Regularization, error estimations.

AMS Classications: 35K05, 45D05, 65J20

Abstract: We study the nonhogeneous heat equation under the form:

u t

u xx = '(t)f(x) , where the unknown is the pair of functions (u;f) . Under various assumptions about the the function ' and the nal value in t = 1 i.e. g(x), we propose dierent regularizations on this ill-posed problem based on the Fourier transform associated with a Lebesgue measure. For '

0 the solution is unique.

Address for correspondence: Alain Pham Ngoc Dinh (alain.pham@univ- orleans.fr)

I. Introduction

Consider the problem: Find a pair of functions (u;f) satisfying the following equation and boundary and initial values:

?

@u @t + u = '(t)f(x); (t;x)

(0;1)

(0;1) u(1;t) = 0; u x (0;t) = u x (1;t) = 0

u(x;0) = 0; u(x;1) = g(x) (1)

Where ' and g are two given functions.

The previous problem is equivalent to nd a function f satisfying an integral equation of the 1st. kind of Volterra type:

1

g(x) =

0 1

0

N(x;1;;)'()f()dd (2)

where N(x;t;;) [1] is dened by

N(x;t;;) = 1

2

(t

)

exp(

(x

)

4(t

)) + exp(

(x + )

4(t

))

As is well-known the problem (2) (or (1)) is an ill-posed problem and its numerical solution have been discussed by various authors ([2], [3], [4], [5]).

The purpose of this paper is to produce regularized solutions of this problem treated in its form (1) with an error estimates under various hypotheses on the function '(t) and g(x).

For '

0 there is uniqueness of the pair (u;f) solution of (1) (para- graph 2). In paragraph 3 we give two sorts of regularization. In fact it will be shown that if the discrepancy between '(t) (respectively g(x)) and its exact solution '

(t) (respectively g

(x)) is of the order " for the

:

L

;

, then the discrepancy between the regularized solution f " (x) and the exact solution f

(x) is, depending on the degree of smoothness of the exact solu- tion f

(x), of the order

ln 1"

or "

⁼

; 0 < " < 1. The techniques used here is rhe Fourier transform associated with the variational form of (1) and a Lebesgue measure generated by the function '

(t): So the proposed regularization can be applied for an integral Volterra equation of the 1st.

kind of the form (0.2) where the kernel N(x;t;;) is a solution of the heat equation.

II. Uniqueness

Let g;' be known functions in L

(0;1). We consider the problem of identifying a pair (u;f) satisfying

?

dt d < u; >

< u x ; x >= ' < f; >

H

(0;1);

u(1;t) = 0; u(x;0) = 0; u(x;1) = g(x)

(3) where u = u(x;t); f = f(x); (x;t)

[0;1]

[0;1], < :;: > is the inner product in L

(0;1).

2

We rst have

Lemma 1. ^If u

C

([0;1];L

(0;1))

C([0;1];H

(0;1)) ^, f

L

(0;1) satisfy (3) then we have

e

0

g(x)cosxdx =

0

e

^t '(t)dt

0

f(x)cosxdx

C : (4)

Proof

In (3)

, by choosing (x) = cosx, we get

?

dt d

Z

1

0

u(x;t)cosxdx +

0

u x (x;t)sinxdx = '(t)

0

f(x)cosxdx:

In view of the condition u(1;t) = 0, we have (5)

Z

1

0

u x (x;t)sinxdx = u(x;t)sinx

^x _x

0

u(x;t)cosxdx

=

Z

1

0

u(x;t)cosxdx:

Hence, (5) follows that

?

dt d

Z

1

0

u(x;t)cosxdx

0

u(x;t)cosxdx = '(t)

0

f(x)cosxdx:

Integrating this equality from t = 0 to t = 1 and using the conditions u(x;0) = 0; u(x;1) = g(x), we get (4). This completes the proof of Lemma 1.

Now, we consider the uniqueness of the solution of (3). We have

Theorem 1. Let u i

C

([0;1];L

(0;1))

C([0;1];H

(0;1)) , f i

L

(0;1) ( i = 1;2 ) satisfy (3). If '

0 then (u

;f

) = (u

;f

) .

Proof

Put v = u

u

; f = f

f

then v satisfy (3)

subject to conditions v(1;t) = 0; v(x;0) = v(x;1) = 0. Hence, from (4) one has

Z

1

0

e

^t '(t)dt

0

f(x)cosxdx = 0 (6)

3

Put () =

n

ⁿ n!

Z

1

0

'(t)t ⁿ dt; F() =

0

f(x)cosxdx:

We claim that

0. In fact, if

0 then

'(t)t ⁿ dt = 0 for every n = 0;1;2;::: Using Weierstrass theorem, we have '

0, a contradiction.

Hence,

0. It follows that there is a

C and an r > 0 such that

j

()

> 0 for every

< r. From (6) and the latter result, one has F() =

Z

1

0

f(x)cosxdx = 0

;

< r: (7) Since F() is an entire functions, we get in view of (7) that F() = 0 for all

C . Putting

~f=

8

<

: 1

2

f(x) x

(0;1)

1

2

f(

x) x

(

1;0) 0 x

(

1;1) ; we get that F() is the Fourier transform of ~f

F() =

?1

~f(x)e

^ix dx: (8)

From (7), (8), we get ~ f = 0 a.e. on R . It follows that f = 0 a.e. on (0;1).

This completes the proof of Theorem 1.

III. Regularization

We give two regularization results

Theorem 2. Let '

; g

be in L

(0;1) and let (u

;f

) be the exact solution of (3) with '; g replaced by '

; g

. Letting C

; " > 0 , we assume that '; g ^satisfy

k

'

'

< ";

g

g

< "

and '(x) > C

; '

(x) > C

a:e:on (0;1);

4

where

:

is the norm of L

(0;1) . Putting f " (x) =

1

2

Z

"

?

"

e

Z

1

0

g(s)cossds

Z

1

0

e

^t '(t)dt

?1

e ^ix d where (") =

"

; 0 < < 1 , then there exists a positive function (") independent of C

;

g

with lim "

(") = 0 and such that

k

2f "

f

2C

" + 2("):

where C

= _C

0

(1+ C

+

g

) is a positive constant dened in terms of C

and

g

:

If we assume, in addition, that f

H

(0;1) then the function (") can be estimated and can be taken equal to

(") = (1 +

p

2)

p

(")

f

H

;

Proof

From Lemma 1, one has e

0

g

(x)cosxdx =

0

e

^t '

(t)dt

0

f

(x)cosxdx:

It follows that

Z

1

0

f

(x)cosxdx = e

0

g

(x)cosxdx

Z

1

0

e

^s '

(s)ds

?1

: Put

~f

=

8

<

: 1

2

f

(x) x

(0;1)

1

2

f

(

x) x

(

1;0) 0 x

(

1;1) ; We have

F

( ~ f

)() =

?1

~f

(x)e

^ix dx = e

0

g

(x)cosxdx

Z

1

0

e

^s '

(s)ds

?1

:

5

where

(f) is the Fourier transform of f:

F

(f)() =

Z

1

?1

f(x)e

^ix dx:

From Plancherel theorem, we have

k

f "

~f

L

= 1

2

( ~ f

)

(f " )

L

(9) On the other hand, one has,

F

( ~ f

)()

(f " )() =

?1

~f

(x)e

^ix dx

e

0

g(x)cosxdx

Z

1

0

e

^s '(s)ds

?1

= e

0

g

(x)cosxdx

Z

1

0

e

^s '

(s)ds

?1

?

e

0

g(x)cosxdx

Z

1

0

e

^s '(s)ds

?1

:

= e

Z

1

0

e

^s '

(s)ds

?1 Z

1

0

(g

(x)

g(x))cosxdx +e

0

g(x)cosxdx

0

e

^s ('(s)

'

(s))ds

Z

1

0

e

^s '

(s)ds

?1

Z

1

0

e

^s '(s)ds

?1

: We get after arrangements

jF

( ~ f

)()

(f " )()

"

C

(1

e

) + "

g

1

e

C

(1

e

)

2

2

"

C

+ 4"

g

C

4

"

C

(C

+ 1 +

g

) = C

";

1

:(10) for 0 < " < 1. In (8) we have put C

= _C

0

(1 + C

+

g

).

Similarly, for

1, one has

jF

( ~ f

)()

(f " )()

2"

C

+ 4"

g

C

C

";

6

the constant C

having the meaning as before in the case

1.

In either cases, one has

jF

( ~ f

)()

(f " )()

C

"

R (11) Noting that

(f " )() = 0 for

> ("), we get in view of (9), (10), (11) that

k

f "

~f

L

R

= 12

Z

j

"

( ~ f

)()

d + 12

Z

j

<

"

( ~ f

)()

(f " )()

d

2 1

Z

j

"

( ~ f

)()

d + 12

^" "

C

"

(")d

2 1

Z

j

"

( ~ f

)()

d + 1C

(")

"

2 1

Z

j

"

( ~ f

)()

d + C

"

; (12) with (") taken such that (") =

"

as "

0

(0 < < 1).

Putting

( ") = 12

Z

j

"

( ~ f

)()

d we get the rst estimate of Theorem 2.

Now, if f

H

(0;1), one has

F

( ~ f

)() =

0

f

(x)cosxdx

= f

(1)sin

1

Z

1

0

f

(x)sinxdx

= 0:

So, for

R , we have

jF

( ~ f

)()

f

j

+

f

(1)

j

:

On the other hand, since H

(0;1) ,

C[0;1], there exist an x

[0;1] such that f

(x

) =

f

(x)dx. We have

f

(1) = f

(x

) +

Z

1

x

f

(x)dx

7

Hence,

j

f

(1)

Z

1

0

(

f

(x)

+

f

(x)

)dx

s

2

Z

1

0

(

f

(x)

+

f

(x)

)dx

p

2

f

H

;

: Hence,

jF

( ~ f

)()

1 +

2

j

f

H

;

: (13) Combining (12), (13) we get

k

f "

~f

_L

(1 +

2)

f

H

;

Z

j

"

d

+ C

"

(1 +

2)

f

_H

_;

(") + C

"

Since (") =

"

and that

k

2f "

f

2

f "

~f

L

;

we can get the second estimate of Theorem 2. This completes the proof of Theorem 2.

Remark 1 : Choosing = 1=8, we obtain (") =

_" and

k

2f "

f

2 1 +

2

7

p

f

H

;

+ C

!

"

⁼

The last formula gives us the best upper bound for 0 < " < 1 given.

Now we state and prove the last regularization result. We rst put G() =

e

g(x)cosxdx G

() =

e

g

(x)cosxdx () =

e

^t '(t)dt

() =

e

^t '

(t)dt

F() =

f(x)cosxdx F

() =

f

(x)cosxdx

8

We have

Theorem 3 Suppose that '

has the form

'

(t) = (1

t) ^m (a + (1

t)

(t))

where a

= 0;m = 0;1;2;:::;t

(0;1);

L

(0;1): Letting

(0;1=2) , we put

F " () =

G()=() if

()

" ; and

< (");

0 if

()

< " ; or

(") and f " ( x) = 12

^" "

F " ()e ^ix d:

Then, for each

(0;min

;1

2

) there exist a C > 0; > 0 in- dependent of g

;'

and a function (") ^{such that} lim "

(") = 0 ^and that

k

2f "

f

C "

+ (");

where (") =

ln

_" :

Remark 2 In the case '

C ^k [0;1]; k

1, if we put P n '

(t) =

ⁿ

j

'

^j

(1)

j! (t

1) ⁿ ;

(the n-th Taylor polynomial of '

at t = 1) then the condition (14) holds if we have P k '

0. So the class of functions satisfying (14) is very broad.

The proof of Theorem 3 relies on Lemma 2 and Lemma 3 followed.

Lemma 2 If

'

(t) = (1

t) ^m (a + (1

t) (t)) (14) with a

= 0 and

L

(0;1) then

lim

R

1

0

e

^t '

(t)dt

e

=

^m

= m!a:

9

Proof

Put J m () =

0

(1

t) ^m e ^t We prove that

lim J m ()

e

= m! m = 0;1;2;::: (15)

In fact, we shall prove the latter relation by induction. One has _e

J

() =

. So, (15) holds for m = 0. Suppose (15) holds for m = k, we prove (15) for m = k + 1. In fact, one has

J k

= (1

t) ^k

e ^t

1

0

+ k + 1 J ^k

=

1

+ k + 1 J ^k : It follows that

lim J k

()

e

= (k + 1) lim

!+1

J k ()

e

= (k + 1)!:

This completes the proof of (15). Using (15), one has C

such that

Z

1

0

e

^t '

(t)dt

aJ m (

)

Z

1

0

e

^t (1

t) ^m

(t)

dt

k k

L

p

J

m

(2

)

C

L

r

e

^m

as

+

Hence

lim

R

1

0

e

^t '

(t)dt

e

=

^m

= a lim

!+1

J m (

)

e

= m!a:

10

This completes the proof of Lemma 2.

Now we state and prove Lemma 3.

Lemma 3 ^If '

satises (14) then there exist ;

(0;1) ^and C

> 0 such that

m(B )

C

0 < <

: Here B =

R :

()

<

; > 0:

and m(B ) is the Lebesgue measure of B .

Proof

From the lemma 2 and from the analyticity of

, the function

has only nite zeros j ; j = 1;:::;p. We can write

() =

()

^p

j

(

j ) ^m

; where

()

= 0 for every

R : Since

lim J m ()

e

= m!;

we have

lim

() =

: It follows that there exists a C

> 0 such that

j

()

C

R : Hence,

j

()

C

^p

j

j

j

m

R : Without loss of generality, we assume that

<

< ::: < p :

11

Put d = min

s

p

s

s

and j

= _C

1

d

. For s + s

s

s

; s = 1;:::;p; one has

j

()

C

^p

j

j

j

m

C

^m s

s ^m

d ^M

= :

where M s = M

m s

m s

, with M =

^p _j

m j . It follows that B =

R :

()

<

p

s

( s

s ; s + s ) Hence

m(B )

^p

s

2 s = 2d

^p

j

⁼

^m

C

⁼

^m

d ^M=

^m

:

Choosing = min

j

p

m

we complete the proof of Lemma 3.

Now, we turn to the

Proof of Theorem 3

We have

k

f "

~f

L

= 1

2

(f " )

( ~ f

)

L

: On the other hand,

kF

(f " )

( ~ f

)

_L

=

Z

j

"

( ~ f

)()

d +

Z

"

?

"

<"

( ~ f

)()

d +

^"

?

"

"

(f " )

( ~ f

)()

d

I

+ I

+ I

12

We estimate I

; I

. We rst have

j

()

()

Z

1

0

e

^t

'(t)

'

(t)

dt

k

'

'

L

s

Z

1

0

e

^t dt

"

r

e

1 2

: So, if

()

< " then

j

()

< "

r

e

1

2

+ "

("): (16) Now, we have

F

( ~ f

) =

Z

1

0

f

(x)cosxdx:

Hence, one has

jF

( ~ f

)()

f

L

: It follows that

I

f

L

m(B

"

);

where m(A) is the Lebesgue measure of A. Now, we estimate I

. If

j

()

" ; (17)

then by (17), one has

j

()

"

"

r

e

1

2

: (18)

In this case, we have

F

(f " )()

( ~ f

)() = G() ()

G

()

()

= G

()(

()

()) +

()(G()

G

())

()

() :

13

Using (17), (18), one has

jF

(f " )()

( ~ f

)()

g

e

"

^e

+

'

^e

e

"

" ("

"

^e

)

"

e

^e

(

g

+

'

) 1

"

^e

: It follows that

I

"

e

^"

(e

^"

1)(

g

+

'

)

(")

1

"

^e

_"

2

:

So, we have

kF

(f " )

( ~ f

)

_L

Z

j

"

( ~ f

)()

d +

f

L

m(B

"

) +"

e

^"

(e

^"

1)(

g

+

'

)

(")

1

"

^e

_"

2

:

By choosing (") =

ln

_" and using Lemma 3, we shall complete the proof of Theorem 3.

Bibliography

[1] A. Friedman, "Partial dierential equations of parabolic type", En- glewood Clier N.J. Prentice Hall Inc, 1964.

[2] A.N. Tikhonov and V.Y. Arsenin, "Solutions of ill-posed problems"

V.H. Winston and Sons, Washington, D.C., 1977.

[3] L.E. Payne, "Improperly posed problems in PDE" SIAM Publica- tions, Philadelphia, 1971.

[4] D. Colton and R. Kress "Integral equation methods in scattering theory", John Wiley, N.Y., 1983.

[5] P. Linz "Analytical and numerical methods for Volterra equations", SIAM Publications, Philadelphia, 1985.

14