DSpace at VNU: On a backward parabolic problem with local Lipschitz source

Clair, Inverse Heat Conduction, Ill-Posed Problems, Wiley–Interscience, New York, 1985.. Carasso, Logarithmic convexity and “slow evolution” constraint in ill-posed initial value problem

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On a backward parabolic problem with local Lipschitz source

Nguyen Huy Tuana,b, ∗, Dang Duc Tronga

a

Faculty of Mathematics and Computer Science, University of Science, Vietnam National University,

227 Nguyen Van Cu, Dist 5, HoChiMinh City, Viet Nam

b Institute for Computational Science and Technology at Ho Chi Minh City (ICST),

Quang Trung Software City, Ho Chi Minh City, Viet Nam

a r t i c l e i n f o a b s t r a c t

Article history:

Received 2 October 2012

Available online 20 January 2014

Submitted by Goong Chen

Keywords:

Nonlinear parabolic problem

Quasi-reversibility method

Backward problem

Ill-posed problem

Contraction principle

We consider the regularization of the backward in time problem for a nonlinear

parabolic equation in the form u t +Au(t) = f (u(t), t), u(1) = ϕ, where A is a positive self-adjoint unbounded operator and f is a local Lipschitz function As known, it

is ill-posed and occurs in applied mathematics, e.g in neurophysiological modeling

of large nerve cell systems with action potential f in mathematical biology A new

version of quasi-reversibility method is described We show that the regularized

problem (with a regularization parameter β > 0) is well-posed and that its solution

U β (t) converges on [0, 1] to the exact solution u(t) as β → 0+ These results extend some earlier works on the nonlinear backward problem.

© 2014 Elsevier Inc All rights reserved.

1 Introduction



u(t), t

where the function f is defined later and the operator A is self-adjoint on a dense space D(A) of H

f (u(t), t) = u u2

L2(0,l) then a concrete version of problem(1)is given as

⎧

⎪

⎪

u t − Δu = uu2

L2(0,l) , (x, t) ∈ (0, l) × (0, 1), u(0, t) = u(l, t) = 0, t ∈ (0, 1),

(2)

* Corresponding author.

E-mail address:tuanhuy_bs@yahoo.com (N.H Tuan).

0022-247X/$ – see front matter © 2014 Elsevier Inc All rights reserved.

The first equality in problem(2) is a semilinear heat equation with cubic-type nonlinearity and has many applications in computational neurosciences It occurs in neurophysiological modeling of large nerve cell

u(0) In practice, u(1) is known only approximately by ϕ ∈ H with u(1) − ϕ  β, where the constant β is

f : H × R → H is a global Lipschitz function with respect to the first variable u, i.e there exists a positive

with the backward parabolic equations included the local Lipschitz source f

Lipschitz function f The techniques and methods in previous papers on global Lipschitz function cannot

establish the following approximation problem

v β (t), t

{λ k } of A; i.e Aφ k = λ k φ k Without loss of generality, we shall assume that

0 < λ1< λ2< λ3< · · · , lim

k→∞ λ k =∞.

k=1 v, φ k φ k, we define

∞

k=1

βλ k + e −λ k



v, φ k φ k ,

∞

βλ k + e −λ kt −s

, 1

v, φ k φ k

Then, problem(5)can be rewritten as the following integral equation

1



t

v β (s), s

This paper is organized as follows In the next section we outline our main results Its proofs will be given

2 The main results

f (v1, t) − f(v2, t) K p v1− v2,



f (v1, t) − f(v2, t), v1− v2



f (v1, t) − f(v2, t), v1− v2  f(v1, t) − f(v2, t)  v1− v2

Sof(v1, t) − f(v2, t), v1− v2  −Kv1− v22 This means that (H2) is true

have

f (u) − f(v) = u u2− vv2

=  u 2(u − v) + vu2− v2

u2+vu + v2

u − v.

ϕ < p, we can choose K p = 3p2 It follows that condition (H1) holds We verify (H2)

g(u, v) =

f (u) − f(v), u − v=

u u2− vv2, u − v

(u − v)u2+ v

u2− v2

, u − v

=u − v2u2+

v

u2− v2

, u − v

and

g(u, v) =

f (u) − f(v), u − v=

u u2− vv2, u − v

u

u2− v2

+v2(u − v), u − v

=u − v2v2+

u2− v2

, u − v.

Adding two equalities, we get

u2+v2

u2− v2

, u − v

=u − v2

u2+v2

u2− v2

u + v, u − v

=u − v2

u2+v2

u2− v22

 0.

Consequently, we have

g(u, v)u2+v2

Now we state main results of our paper Its proofs will be given in the next section

that (H1), (H2), (H3) hold Then the problem

v β (t), t

has uniquely a solution U β ∈ C1([0, 1]; H).

u(t) = ∞

k=1 u(t), φ k φ k such that

E2= 1

 0

∞

k=1

λ2k e 2λ ku(s), φ k2

ds < ∞.

Then

U β (t) − u(t) M β t ln e

β

where M = 2e (2L+1)(1−t) E + e L(1 −t) .

Remark 1.

between the exact solution and the approximation solution has the form

β

First, we shall prove some inequalities which will be used in the main part of our proof

also have

βx + e −x



βx + e −x



βx + e −x



= 0, x > x β Moreover, for x > 0, we have

βx + e −x



β

,

max

 1

βx + e −x , 1



β

.

β , + ∞) Since m(0) = 0, m(ln(1

βx + e −x



Consider the function

βx + e −x



, ∀x ∈ (0, x β ),

for 0 < β < 1 Computing the derivative of g(x), we get

βx + e −x .

g(x)  g ln1

β



β

.

And, we have

max

 1

βx + e −x , 1



= eln+(βx+e−x1 )

β

Lemma 4 For any 0 < β < 1, we have

A β   ln β −1 lne

β

.

A β v 2=

∞

k=1



βλ k + e −λ k



2 v, φ k 2

β −1 lne

β

k=1

 v, φ k 2

β −1 lne

β

v2.

β

So, if 0  s − t  h  1 then

G β (t, s)(v) 2

=

∞

k=1

βλ k + e −λ k2t −2s

, 1 v,φ k 2

β

v2,

Lemma 6 Suppose v β ∈ C([0, 1]; H) is the unique solution of the integral equation (6) Then v β ∈

∞

k=1

βλ k + e −λ kt−1

, 1

ϕ, φ k φ k

−∞ k=1

t

βλ k + e −λ kt−s

, 1

f

v β (s), s

, φ k



ds



φ k

We get by direct computation

d

∞

k=1

ln+

βλ k + e −λ k

× maxβλ k + e −λ kt−1

, 1

ϕ, φ k φ k

−

∞

k=1

t

ln+

βλ k + e −λ k

βλ k + e −λ kt−s

, 1

f

v β (s), s

, φ k



ds



φ k

+

∞

k=1



f

v β (t), t

, φ k



φ k

=

∞

k=1

ln+

βλ k + e −λ k

v β (t), φ k



φ k + f

v β (t), t

v β (t), t

.

k=1 ϕ, φ k φ k = ϕ Hence, v β is the solution of problem(5) 2

KM P , we put

N =



M 2 − KM1 − P





where [x] is the integer part of the real number x and K M is the Lipschitz constant in (H1) with respect

to M For ϕ i ∈ H, ϕ i   P , we define

T i= 1− ih, i = 0, 1, , N,

L i=

v ∈ C[T i+1 , T i ]; H

, v(T i ) = ϕ i , sup

T i+1 tT i

v(t)  M,

J i v = G β (t, T i )ϕ i −

T i



t

v(s), s

ds.

Then the operator J i has a unique fixed point on L i

J i v(t)  G β (t, T i)  ϕ i  +

T i



G β (t, s) f

v(s), s ds.

FromLemma 5, one has for Ti − h  t  T i

J i v(t) β t−T i



ϕ i  +

T i



t

f

v(s), s ds.

f

ϕ i  + MK M h

From (13), we have N  2β −1 K

1

N  M 2 − 1 KM −P

M K M

− 1

KM − P



2KM1

β −β2  e1

J i v1(t) − J i v2(t) T i

t

G β (t, s) f

v1(s), s



T i



t

β −h K M v1(s) − v2(s) ds

T i+1 tT i

v1(t) − v2(t)

−h K

T i+1 tT i

v1(t) − v2(t)

2 T i+1suptT i v1(t) − v2(t)

2T i+1suptT i

v1(t) − v2(t) .

Lemma 8 Assume that f satisfies (H2), (H3) Let 0  τ  1 and let u β ∈ C1([τ, 1]; H) satisfy

u β (t), t

, τ < t < 1,

Then for τ  t  1

Proof One has

1 2

d

dt u β 2+A β u β , u β  =f

u β (t), t

, u β

It follows that

1

2 u β(1) 2

+ 1



t



ds −1

2 u β (t) 2

 −L

1



t

u β (s) 2

ds.

So we have

1

2 u β (t) 2

1



t



+ L u β (s) 2

ds.

It follows that

u β (t) 2

A β  + L

1



t

u β (s) 2

ds.

Gronwall’s inequality gives

u β (t) 2

Proof of Theorem 1 We shall prove by induction that the equation

v(t), t

(19)

u 

0(t) = A β u0(t) + f

for T k  t  1 Put ϕ k = u(t k) From Lemma 7, we can find uk ∈ C([T k+1 , T k ]; H) such that J k u k = u k.

then

β

Here we recall that the constant L is in (H2).

Proof For a > 0, we put

u β (t) − v β (t)

.

By direct computation, we get

d

f

u β (t), t

.

It follows that



e a(t−1)

f

u β (t), t

− fv β (t), t

, w β (t)

.



e a(t−1)

f

u β (t), t

− fv β (t), t

, w β (t)

f

u β (t), t

− fv β (t), t

, u β (t) − v β (t)

FromLemma 4, we have

β

β (t) 2

,

which gives



β

β (t) 2

.

It follows that

1 2

d

dt w β (t) 2

 a w β (t) 2

− ln β −1 lne

β

β (t) 2

.

Then, we get

w β(1) 2

− w β (t) 2

 2

1



a − L − ln β −1 lne

β

β (s) 2

ds.

We choose a = L + ln(β −1(ln e

w β (t) 2

 w β(1) 2

=ϕ − ω2.

Hence, we get

u β (t) − v β (t) e a(1 −t) ϕ − ω = e L(1 −t) β t −1 lne

β

u(t) − v β (t) e (2L+1)2 (1−t) β t ln e

β

where we recall

E =

0

∞

k=1

λ2

k e 2λ ku(s), φ k2

ds.

⎧

⎨

⎩

d



v β (t), t

,

and

⎧

⎨

⎩

d

dt u(t) + A β u(t) = (A β − A)u(t) + fu(t), t

, u(1) = ϕ.

For any b > 0, let

v β (t) − u(t).

Then by direct calculation

d

b(t −1)

v β (t) − u(t)+ e b(t −1)

v  β (t) − u  (t)

v β (t), t

+ A β u(t) − fu(t), t

− e b(t −1) (A

β − A)u(t)

f

v β (t), t

− fu(t), t



z  β (t) + A β z β (t) − bz β (t), z β (t)

e b(t −1)

f

v β (t), t

− fu(t), t

, z β (t)

− e b(t−1)

(A β − A)u(t), z β (t)

This means that

d

dt z β (t) 2

+ 2b z β (t) 2

e b(t −1)

f

v β (t), t

− fu(t), t

, z β (t)

− 2e b(t−1)

(A β − A)u(t), z β (t)

.



e b(t −1)

f

v β (t), t

− fu(t), t

, z β (t)

and



β

β (t) 2

.

d

dt z β (s) 2

+ 2e b(s−1)

(A β − A)u(s), z β (s)

 2b z β (s) 2

− 2L z β (s) 2

− 2 ln β −1 lne

β

.

z β(1) 2

− z β (t) 2

+ 1



t

2e b(s−1)

(A β − A)u(s), z β (s)

1



t

z β (s) 2

ds.

This can be rewritten as

z β (t) 2

 2L

1



t

z β (s) 2

ds +

1



t

2e b(s−1)

(A β − A)u(s), z β (s)

ds.

1



t

2e b(s−1)

(A β − A)u(s), z β (s)

ds

1



t

e 2b(s −1) (A β − A)u(s) 2

ds +

1



t

z β (s) 2

ds



1



t

(A β − A)u(s) 2

ds +

1



t

z β (s) 2

k=1 u(s), φ k φ k, it follows that 1



t

(A β − A)u(s) 2

ds =

1



t

∞

k=1

βλ k + e −λ k

u(s), φ k2

ds

= 1

λ <x

ln2

1 + βλ k e λ ku(s), φ k2

ds +

1

λ x

λ2ku(s), φ k2

Here the positive number x β is defined inLemma 3 We have

1



t

λ k <x β

ln2

1 + βλ k e λ ku(s), φ k2

ds

1



t

∞

k=1

β2λ2k e 2λ ku(s), φ k2

ds

1

 0

∞

k=1

λ2k e 2λ ku(s), φ k2

ds

1



t

λ k x β

λ2ku(s), φ k2

ds =

1



t

λ k x β

e −2λ k λ2k e 2λ ku(s), φ k2

ds

1



t

λ k x β

λ2k e 2λ ku(s), φ k2

ds

1

 0

∞

k=1

λ2k e 2λ ku(s), φ k2

From (24),(25), (26)and(27), we obtain

e 2b(t −1) v β (t) − u(t) 2

 (2L + 1)

1



t

e 2b(s −1) v β (s) − u(s) 2

ds + 2β2E2.

It implies that

e 2bt v β (t) − u(t) 2

 (2L + 1)

1



t

e 2bs v β (s) − u(s) 2

ds + β2e 2b E2

= (2L + 1)

1



t

e 2bs v β (s) − u(s) 2

ds + E2 lne

β

e 2bt v β (t) − u(t) 2

 2e (2L+1)(1 −t) lne

β

E2.

It can be rewritten as

v β (t) − u(t) 2

 2e (2L+1)(1 −t) β 2t lne

β

E2.

v β (t) − u(t) 2e (2L+1)2 (1−t) β t lne

β

E.

U β (t) − u(t)  U β (t) − v β (t) + v β (t) − u(t)

 e L(1−t) β t−1 lne

β

ϕ − ϕ β  + 2e (2L+1)

2 (1−t) β t lne

β

E

β

2e (2L+1)2 (1−t) E + e L(1−t)

,

Acknowledgments

This work is supported by Vietnam National University HoChiMinh City (VNU-HCM) under Grant

No B2014-18-01

The authors would like to thank the anonymous referees for their valuable suggestions and comments leading to the improvement of our manuscript

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