Discretization and regularization of an inverse problem related to a quasilinear hyperbolic

integrodifferential equation

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Inverse Problems13 (1997) 711–728. Printed in the UK PII: S0266-5611(97)81952-1

Discretization and regularization of an inverse problemrelated to a quasilinear hyperbolic integrodifferentialequation

Jaan JannoInstitute of Cybernetics, Akadeemia tee 21, EE0026 Tallinn, Estonia

Received 18 February 1997

Abstract. An inverse problem for a one-dimensional quasilinear hyperbolic integrodifferentialequation is reduced to a system of hyperbolic and second-kind integral equations. The obtainedsystem is discretized by means of the method of finite differences. The convergence of thedifference scheme in the case of small data is proved and the regularization of the inverseproblem is discussed.

1. Introduction and problem formulation

The determination of relaxation kernels of viscoelastic materials is a problem of boththeoretical and practical interest [1, 13]. Often these kernels are reconstructed fromexperiments of wave propagation. This leads to inverse problems for hyperbolicintegrodifferential equations [1–5, 8, 10].

The problems mentioned are mildly or moderately ill-posed. For the correspondingcorrectness analysis we refer the reader to [1, 2, 4, 5, 8, 10] and references therein. Toregularize these problems one can apply general optimization algorithms [12, 13] basedon the method of Tikhonov (see [11]) and in linear cases modifications of the method ofLavrent’ev with (or without) reducing inverse problems to integral equations [3, 7, 9].

In the preceding paper [6] we proposed a new approach for these problems whichis based on direct discretization. By differentiation we decompose the nonlinear inverseproblem into a linear ill-posed problem of computing derivatives of data and a nonlinearbut well posed system of hyperbolic and integral equations of the second kind. In theformer stage we can use known methods for regularization of linear improper problems andin the latter stage we apply the method of finite differences. The described method is moreeffective than optimization techniques because the difference scheme for the hyperbolicsystem is explicit and we can solve it in one step.

We study the following concrete inverse problem. Determine a pair of functions(R,U)

satisfying the hyperbolic equation

Uxx(x, t)(1+ bUx(x, t))−∫ t

0R(t − s)Uxx(x, s)ds

−a2Utt (x, t) = F(x, t) (x, t) ∈ [0, X] × [0, T ] (1.1)

the initial and boundary conditions

U(x, 0) = A(x) Ut(x, 0) = B(x) x ∈ [0, X] (1.2)

U(0, t) = U(X, t) = 0 (1.3)

0266-5611/97/030711+18$19.50c© 1997 IOP Publishing Ltd 711

712 J Janno

and an additional condition of the following form:

Ux(0, t)+ b0(Ux(0, t))2−

∫ t

0R(s)Ux(0, t − s) ds = G(t) 06 t 6 T . (1.4)

Equation (1.1) describes the motion of the one-dimensional homogeneous rod with nonlinearelastic and weakly linear viscous properties whereas the functionR stands for the relaxationkernel [6]. Condition (1.4) can be interpreted as a stress observation at the left endpoint ofthe rod. The problem (1.1)–(1.4) in the linear case (b = b0 = 0) was studied theoreticallyin [2, 8]. Particularly, the stability of the solution in norms containing derivatives of thedata was shown.

In this paper we study the method of decomposition for the quasilinear problem (1.1)–(1.4) (the corresponding linear case was treated in [6]). In sections 2 and 3 we differentiate(1.1)–(1.4) and discretize it using finite differences. Section 4 gives auxiliary results. Insections 5 and 6 we study the boundedness of the discrete solution and the convergence ofthe difference scheme in the case of small data. Section 7 discusses the regularization ofthe problem (1.1)–(1.4).

2. Differentiation of the inverse problem

Theoretical analysis of inverse problems in viscoelasticity uses preliminary differentiationof these problems in order to apply the contraction principles (see e.g. [1, 8, 10]). In ourpaper we use this technique as a basis for a numerical method. We reduce the problem(1.1)–(1.4) to a system which contains a hyperbolic equation and integral equations of thesecond kind.

Theorem 1. Let the problem (1.1)–(1.4) have a solution(R,U) ∈ C1[0, T ] × C5([0, X] ×[0, T ]). Assume that

1+ bUx(x, t) 6= 0 06 x 6 X 06 t 6 T A′(0) 6= 0

and

F(0, t) = F(X, t) = Fxx(0, t) = Fxx(X, t) = 0 06 t 6 T . (2.1)

Define

V = Uxxx Z = Uxtt (0, ·) P = R′. (2.2)

Then the vector(V , Z,R, P ) is a solution of the following differentiated problem:

Vxx(x, t)

(1+ b

∫ X

00(x, y)V (y, t)dy

)+ 4b

∫ x

0V (y, t)dy Vx(x, t)

+3b(V (x, t))2−∫ t

0R(t − s)Vxx(x, s)ds − a2Vtt (x, t) = f (x, t)

06 x 6 X 06 t 6 T (2.3)

V (x, 0) = α(x) Vt (x, 0) = β(x) 06 x 6 X (2.4)

Vx(0, t) = Vx(X, t) = 0 06 t 6 T (2.5)

Z(t) = 1

a2

[V (0, t)(1+ b

∫ X

00(0, y)V (y, t)dy)−

∫ t

0R(t − s)V (0, s)ds − q(t)

]06 t 6 T (2.6)

R(t) =∫ t

0P(s) ds + ρ 06 t 6 T (2.7)

Discretization and regularization of an inverse problem 713

P(t) = 1

κ0

[− g(t)+ Z(t)

(1+ 2b0

∫ X

00(0, y)V (y, t)dy)+ 2b0

(∫ t

0Z(τ) dτ + κ1

)2

−κ1R(t)−∫ t

0R(s)Z(t − s) ds

]06 t 6 T . (2.8)

Heref = Fxxx α = A′′′ β = B ′′′ q(t) = Fx(0, t)κ0 = A′(0) κ1 = B ′(0) ρ = 1

κ0[κ1+ 2b0κ0κ1−G′(0)] g = G′′

and

0(x, y) =

x − y − (y −X)

2

2Xif y < x

− (y −X)2

2Xif y > x.

(2.9)

Proof. Interpreting (1.1)–(1.4) as formulae forF,A,B,G we see that the smoothnessassumption about(R,U) guarantees the existence of the derivatives of the data included in(2.1), (2.9).

Let us prove (2.5). Boundary conditions (1.3) yieldUtt (0, t) = Utt (X, t) = 0, 06 t 6T . This equality together with (2.1) implies that the right-hand side of (1.1) is equal to zeroif x = 0 or x = X. Thus, equation (1.1) in pointsx = 0 andx = X is a homogeneousVolterra equation of the second kind with respect toUxx(x, ·). Consequently, there holds

Uxx(0, t) = Uxx(X, t) = 0 06 t 6 T (2.10)

which in turn yieldsUxxtt (0, t) = Uxxtt (X, t) = 0. Now in view of (2.1) we see that equation(1.1) differentiated twice byx also has a vanishing right-hand side ifx = 0 or x = X. Weget homogeneous Volterra equations of the second kind for the functionsUxxxx(0, t) andUxxxx(X, t) too. This implies

Uxxxx(0, t) = Uxxxx(X, t) = 0 06 t 6 T .SinceV = Uxxx we have obtained the boundary conditions (2.5).

Integrating the relationUxxx = V and observing the vanishing boundary conditions(1.3), (2.11) we have

Uxx(x, t) =∫ x

0V (y, t)dy Ux(x, t) =

∫ X

00(x, y)V (y, t)dy. (2.11)

In view of (2.12) differentiation of (1.1) and (1.2) immediately implies (2.3), (2.4). Further,differentiating formula (1.1) with respect tox, settingx = 0 and observing (2.11), (2.12)we immediately obtain (2.6). Moreover, computing a derivative from expression (1.4) andsettingt = 0 we get the relationR(0) = ρ for the initial value ofR. Thus, (2.7) holds aswell. Finally, differentiating condition (1.4) twice we derive equation (2.8). Therefore, thetheorem is proved. �

As we see, the differentiated problem (2.3)–(2.8) contains hyperbolic equation (2.3) forV with initial and boundary conditions (2.4), (2.5) and Volterra equations of the secondkind (2.6)–(2.8) forZ,R, P . We will discretize this problem in the next section.

Note that upon solving system (2.3)–(2.8) we have determined the relaxation kernelR

together with its derivative. To evaluate the second component of the solution(R,U) ofproblem (1.1)–(1.4) we can use the simple integration:

U(x, t) =∫ x

0

∫ X

00(y, z)V (z, t)dz dy. (2.12)

714 J Janno

3. Discretization

Let N andM be positive integers and

h = X

Nτ = T

N. (3.1)

We define the following uniform meshes on the intervals [0, X] and [0, T ] :

ωh = {xi = ih : i = 0, . . . , N} ωτ = {tj = jτ : j = 0, . . . ,M}. (3.2)

For values of functionsy ∈ (ωh→ R) we shall use the simplified notation:

yi = y(xi) xi ∈ ωh 06 i 6 N (3.3)

and introduce the following discrete operations being analogues of derivatives:

∂xy : (∂xy)i ≡ ∂xyi = yi+1− yih

i = 0, . . . , N − 1

∂xy : (∂xy)i ≡ ∂xyi = yi − yi−1

hi = 1, . . . , N

3y : (3y)i ≡ 3yi = ∂x∂xyi i = 1, . . . , N − 1.

(3.4)

Further, fory ∈ (ωτ → R) andy ∈ (ωτ → (ωh→ R)) we denote

yj = y(tj ) tj ∈ ωτ 06 j 6 M (3.5)

and define

∂ty : (∂ty)j ≡ ∂tyj = yj+1− yj

τ06 j 6 M − 1

∂t y : (∂t y)j ≡ ∂t yj = yj − yj−1

τ16 j 6 M.

(3.6)

Let us return to the differentiated problem (2.3)–(2.8). We suppose that instead of theexact dataα, β, f , q, ρ, g, κ0, κ1 we know certain approximationsα ≈ α, β ≈ β, f ≈ f ,q ≈ q, ρ ≈ ρ, g ≈ g, κ0 ≈ κ0, κ1 ≈ κ1. According to the notation (3.3), (3.5) we write

fj

i = f (xi, tj ) αi = α(xi) βi = β(xi) qj = q(tj ) gj = g(tj ). (3.7)

We discretize (2.3)–(2.8) making use of the method of finite differences. To this endwe replace the derivatives by formulae (3.4), (3.6) and the integrals by the quadrangle rule.Assuming thatκ0 6= 0 and using the notation

(8v)j

i = 8vji = 2(i − 1)hi∑l=1

vj

l (9v)j

i = 9vji = hN∑l=1

0(xi, xl)vj

l

2(s) = 1 s > 0 2(s) = 0 s < 0

(3.8)

with v = V we obtain the following system:

3Vj

i (1+ b9V ji )+ 2b8V ji (∂x + ∂x)V ji + 3b(V ji )2− τ

j∑l=1

Rj−l3V li − a2∂t ∂t Vj

i = f ji16 i 6 N − 1 16 j 6 M − 1 (3.9)

V 0i = αi ∂t V

0i = βi 06 i 6 N (3.10)

∂xVj

0 = ∂x V jN = 0 16 j 6 M (3.11)

Zj = 1

a2

[Vj

0 (1+ b9V j0 )−2(j − 1)τj∑l=1

Rj−l V l0 − qj]

06 j 6 M (3.12)

Discretization and regularization of an inverse problem 715

Rj = 2(j − 1)τj−1∑l=0

P l + ρ 06 j 6 M (3.13)

P j = 1

κ0

[− gj + Zj (1+ 2b09V

j

0 )+ 2b0

(2(j − 1)τ

j−1∑l=0

Zl + κ1

)2

−κ1Rj −2(j − 1)τ

j∑l=1

Zj−l Rl]

06 j 6 M. (3.14)

In section 6 we will show that the solution(V , Z, R, P ) of system (3.9)–(3.14)approximates the solution of system (2.3)–(2.8) in nodes(xi, tj ), i.e.

Vj

i ≈ V (xi, tj ) Zj ≈ Z(tj ) Rj ≈ R(tj ) P j ≈ P(tj ).Difference scheme (3.9)–(3.14) is explicit and consequently, uniquely solvable. Indeed,from (3.10) we obtain the first two levels forV ji , j = 0, 1 and from (3.12)–(3.14) thevaluesZj , Rj , P j , j = 0, 1. Further, using computedk levels, the hyperbolic scheme (3.9)with (3.11) provides the levelj = k + 1 for V ji and (3.12)–(3.14) values forZi, Rj , P j

with j = k + 1.

4. Auxiliary results

The main aim of this section is to prove a stability theorem concerning difference schemesfor linear hyperbolic integrodifferential equations with given coefficients. This is a necessaryresult for further boundedness and convergence analysis in sections 5 and 6. We begin byintroducing some further notation and proving three lemmas.

For functionsy, z ∈ (ωh→ R) we define the scalar products:

(y, z) := hN−1∑i=1

yizi (y, z] := hN∑i=1

yizi [y, z] := hN∑i=0

yizi (4.1)

and norms:

‖y‖2 :=√(y, y) ‖y]|2 :=

√(y, y] |[y]|2 :=

√[y, y]

‖y‖∞ := max06i6N

|yi |. (4.2)

The functionsy, z ∈ (ωh → R) satisfy the formula for differentiation of product andanalogue of the Green’s first formula:

∂x(yz)i = yi−1∂xzi + ∂xyizi (4.3)

(3y, z) = −(∂xy, ∂xz] + ∂xyNzN − ∂xy0z0. (4.4)

Further,y, z ∈ (ωτ → R) andy, z ∈ (ωτ → (ωh → R)) satisfy the following formula fordifferentiation of the product:

∂t (yj zj ) = yj∂tzj + ∂tyj zj+1 (4.5)

and summation by parts:

τ

l2−1∑j=l1

yj∂tzj = yl2zl2 − yl1zl1 − τ

l2−1∑j=l1

∂tyj zj+1. (4.6)

If yj , zj are vectors, i.e.yj , zj ∈ (ωh → R), 0 6 j 6 M, then formulae (4.5) and (4.6)hold componentwise.

716 J Janno

Lemma 1. For y ∈ (ωτ → (ωh→ R)) the following estimates are fulfilled:

∂xyj

i ∂xyj−1i >

(1− 1

ε

)(∂xy

j

i )2− ετ

2

4(∂t ∂xy

j

i )2

16 i 6 N 16 j 6 M ε > 0 (4.7)

‖yk‖∞ 6 max

{T√X − h,

√X

}[max

16j6k‖∂xyj ]|2+ max

16j6k‖∂t yj‖2

]+ ‖y0‖∞

16 k 6 M. (4.8)

Proof. Inequality (4.7) follows from the relation

∂xyj

i ∂xyj−1i = (∂xyji )2− ∂xyji τ ∂t ∂xyji

if we estimate

∂xyj

i τ ∂t ∂xyj

i 61

ε(∂xy

j

i )2+ ε

4(τ ∂t ∂xy

j

i )2 ε > 0.

Let us prove (4.8). Define

i∗(k) : |yki∗(k)| = min06i6N

|yki |.

Since the relations

|yki∗(k)| 61

(N − 1)hh

N−1∑i=1

|yki | |ykl − yki∗(k)| 6 hN∑i=1

|∂xyki |

hold, we obtain

|ykl | = |yki∗(k)| + |ykl − yki∗(k)| 61

(N − 1)hh

N−1∑i=1

|yki |

+hN∑i=1

|∂xyki | =1

X − hhN−1∑i=1

∣∣∣∣τ k∑j=1

∂t yj

i + y0i

∣∣∣∣+ h N∑i=1

|∂xyki |

6 T

X − h max16j6k

h

N−1∑i=1

|∂t yji | + ‖y0‖∞ + hN∑i=1

|∂xyki |.

Using the Cauchy–Schwarz inequality and observing definitions (4.1), (4.2) we obtain(4.8). �

Lemma 2 (analogue of the Gronwall inequality).Let y, z ∈ (ωτ → R), yj > 0, zj > 0,d > 0 and

yk 6 d2(k − 1)τk−1∑j=0

yj + zk 06 k 6 k0. (4.9)

Then the estimate

max06j6k

yj 6 c2(d) max06j6k

zj 06 k 6 k0 (4.10)

holds, wherec2 is a bounded function ofd.

Discretization and regularization of an inverse problem 717

Proof. In the cased = 0 the assertion is trivial. Letd 6= 0. From (4.9) we immediatelyderive

max06j6k

e−2dtj yj 6 d max06j6k

[2(j − 1)e−2dtj τ

j−1∑l=0

e2dtl max06l6j−1

e−2dtl yl]+ max

06j6kzj

06 k 6 k0.

Since

2(j − 1)e−2dtj τ

j−1∑l=0

e2dtl 6 2(j − 1)e−2dtj

∫ tj

0e2ds ds 6 1

2d

we have

max06j6k

e−2dtj yj 6 12 max

06j6ke−2dtj yj + max

06j6kzj 06 k 6 k0.

Thus

max06j6k

e−2dtj yj 6 2 max06j6k

zj 06 k 6 k0

and due to the estimate

e−2dtk max06j6k

yj 6 max06j6k

e−2dtj yj

we obtain (4.10). �Lemma 3. If

(yk)2+ (zk)2 6 ξk 16 k 6 M (4.11)

whereyk, zk, ξ k > 0 andξk+1 > ξk then(max

16j6kyj + max

16j6kzj)26 4ξk 16 k 6 M. (4.12)

Proof. From (4.11) we have

yk 6√ξk zk 6

√ξk 16 k 6 M. (4.13)

Sinceξk is monotonically increasing, from (4.13) we derive

max16j6k

yk 6√ξk max

16j6kzk 6

√ξk 16 k 6 M.

Summing these inequalities and squaring we get (4.12). �

Now we are ready to formulate and prove a stability theorem concerning differenceschemes for linear hyperbolic integrodifferential equations. Let

nj

0,i nj

1,i nj

2,i Yj

il rj µj

i

16 l i 6 N 16 j 6 M − 1

be certain prescribed quantities and suppose that the vectorvj

i , 06 i 6 N, 06 j 6 M is asolution of the following scheme:

nj

0,i3vj

i + nj1,i (∂x + ∂x)vji + nj2,ivji + hN∑l=1

Yj

il vj

l − τj∑l=1

rj−l3vli − a2∂t ∂t vj

i = µji16 i 6 N − 1 16 j 6 M − 1. (4.14)

718 J Janno

Theorem 2. Let

H := min26j6M

min06i6N

nj

0,i > 0 (4.15)

and the stepsτ andh satisfy the inequality

h > τ σa

√max

26j6M‖nj0‖∞ for someσ > 1. (4.16)

Then for the solution of (4.14) the following estimate is valid

max06j6k

‖vj‖∞ + J k 6 c0(W,H−1)Wk 16 k 6 M. (4.17)

HereJ k is the discrete energy norm:

J k = max16j6k

‖∂t vj‖2+ max16j6k

‖∂xvj ]|2 16 k 6 M (4.18)

W is the bound of the coefficientsn0, n1, n2, Y, r:

W = max16j6M

‖nj0‖∞ + max16j6M−1

‖∂tnj0‖∞ + max16j6M−1

max16i6N

|∂xni0,i | + max16j6M−1

‖nj1‖∞

+ max16j6M−1

‖nj2‖∞ + max16j6M−1

max16l,i6N

|Y jil | + max06j6M

|rj | + τM−1∑j=0

|∂t rj |

(4.19)

Wk is the bound of the initial and boundary values ofv and the right-hand sideµ:

Wk = ‖v0‖∞ + ‖∂xv0]|2+ |[∂tv0]|2+ max16j6M

{|∂xvj0| + |∂xvjN |}

+τM−1∑j=1

{|∂t∂xvj0| + |∂t ∂xvjN |} +2(k − 2)τk−1∑j=1

‖µj‖2 (4.20)

andc0 is a bounded function ofW,H−1 depending also ona, σ,X, T .

Proof. We shall use the method of discrete energy estimates. To this end we multiplyequation (4.14) by the quantity∂t � vji , where�vji is the mean value:

(�v)ji = �vji =vj

i + vj−1i

216 i 6 N − 1 16 j 6 M

thereupon compute the scalar product(·, ·) (formula (4.1)) and sum overj from 1 tok−1.Performing these operations for the first addend in (4.14) on the grounds of formulae (4.3)and (4.4) we obtain

τ

k−1∑j=1

(nj

03vj , ∂t � vj ) = −τ

k−1∑j=1

h

N∑i=1

nj

0,i−1∂xvj

i ∂x∂t � vji − τk−1∑j=1

(∂xvj , ∂xn

j

0∂t � vj ]

+τk−1∑j=1

[nj0,N ∂xvj

N∂t � vjN − nj0,0∂xvj0∂t � vj0] 2 6 k 6 M. (4.21)

Since the relation

∂xvj

i ∂x∂t � vji = 12∂t {∂xvji ∂xvj−1

i }

Discretization and regularization of an inverse problem 719

holds, applying formula (4.6) for the first term on the right-hand side of (4.21) we get

τ

k−1∑j=1

h

N∑i=1

nj

0,i−1∂xvj

i ∂x∂t � vji

= 12h

N∑i=1

nj

0,i−1∂xvj

i ∂xvj−1i |kj=1− 1

2τ

k−1∑j=1

h

N∑i=1

∂tnj

0,i−1∂xvj+1i ∂xv

j

i . (4.22)

Further, for the terma2∂t ∂t vj

i in (4.14) we have the equality

τ

k−1∑j=1

(a2∂t ∂t vj , ∂t � vj ) = a2

2τ

k−1∑j=1

∂t‖∂tvj‖22 =

a2

2(‖∂t vk‖2

2− ‖∂tv0‖22) 26 k 6 M.

(4.23)

Consequently, from (4.14) by virtue of (4.21)–(4.23) we derive the expression

1

2h

N∑i=1

nk0,i−1∂xvki ∂xv

k−1i + a

2

2‖∂t vk‖2

2 =10∑l=1

I kl (4.24)

where

I k1 =1

2h

N∑i=1

n10,i−1∂xv

1i ∂xv

0i +

a2

2‖∂tv0‖2

2

I k2 = τk−1∑j=1

[nj0,N ∂xvj

N∂t � vjN − nj0,0∂xvj0∂t � vj0]

I k3 = −τk−1∑j=1

(∂xvj , ∂xn

j

0∂t � vj ] I k4 = 12τ

k−1∑j=1

h

N∑i=1

∂tnj

0,i−1∂xvj+1i ∂xv

j

i

I k5 = τk−1∑j=1

(nj

1(∂x + ∂x)vj , ∂t � vj ) I k6 = τk−1∑j=1

(nj

2vj , ∂t � vj )

I k7 = τk−1∑j=1

h

N−1∑i=1

h

N∑l=1

Yj

il vj

l ∂t � vji I k8 = −τk−1∑j=1

(µj , ∂t � vj )

I k9 = −τk−1∑j=1

τ

j∑l=1

(rj−l3vl, ∂t � vj ).

Next we are going to estimate the quantitiesI k1 , . . . , Ik9 . By (4.15), (4.16) and the

relation(max26j6M ‖nj0‖∞)−1 6 H−1 there holds

|∂xv1i | =

∣∣∣∣∂xv0i +

τ

h∂tv

0i −

τ

h∂tv

0i−1

∣∣∣∣ 6 c3(H−1)(|∂xv0

i | + |∂tv0i | + |∂tv0

i−1|). (4.25)

Thus, in view of (4.19), (4.20) and the Cauchy–Schwarz inequality we obtain forI k1 theestimate

|I k1 | 6 c4(W,H−1)W 2

k 26 k 6 M. (4.26)

Further, summing by parts inI k2 and observing (4.19), (4.20) we derive

|I k2 | 6 c5W max06j6M

‖vj‖∞Wk

720 J Janno

which by (4.8) and (4.18) implies

|I k2 | 6 c6W(JkWk +W 2

k ) 26 k 6 M. (4.27)

Analogously, in a more straightforward way, using the Cauchy–Schwarz inequality and(4.8) we derive the following estimates forI k3 , . . . , I

k8 :

|I k3 | 6 WJk(τ

k−1∑j=1

J j +Wk

)|I k4 | 6 1

2WJkτ

k−1∑j=1

J j

|I k5 | 6 WJkτk−1∑j=1

J j |I k6 | 6 c7WJk

(τ

k−1∑j=1

J j +Wk

)

|I k7 | 6 c8WJkτ

k−1∑j=1

J j |I k8 | 6 J kWk 26 k 6 M.

(4.28)

For I k9 we first apply formula (4.4), thereupon change the order of summation and useformula (4.6). We obtain

I k9 = −τk−1∑j=1

τ

j∑l=1

rj−l∂t {−(∂xvl, ∂x � vj ] + ∂xvlN � vjN − ∂xvl0 � vj0}

= − τk−1∑l=1

{rk−l [−(∂xvl, ∂x � vk] + ∂xvlN � vkN − ∂xvl0 � vk0]

−r0[−(∂xvl, ∂x � vl ] + ∂xvlN � vlN − ∂xvl0 � vl0]

−τk∑

j=l+1

∂t rj−1−l [−(∂xvl, ∂x � vj ] + ∂xvlN � vjN − ∂xvl0 � vj0]

}.

Estimating this expression by means of the Cauchy–Schwarz inequality we have

|I k9 | 6{

2 max06j6k−1

|rj | + τk−2∑j=0

|∂t rj |}

×{

max16j6k

‖∂x � vj ]|2τk−1∑j=1

‖∂xvj ]|2+ max16j6k

‖ � vj‖∞τk−1∑j=1

(|∂xvjN | + |∂xvj0|)}.

By virtue of (4.8) and (4.19) we get

|I k9 | 6 c9W

[J k(τ

k−1∑j=1

J j +Wk

)+W 2

k

]26 k 6 M. (4.29)

Summing up, relations (4.26)–(4.29) with (4.24) yield

1

2h

N∑i=1

nk0,i−1∂xvki ∂xv

k−1i + a

2

2‖∂t vk‖2

2 6 c10(W,H−1)

[J k(τ

k−1∑j=1

J j +Wk

)+W 2

k

]26 k 6 M. (4.30)

The next step is to estimate the left-hand side of (4.30) from below. Due to (4.7) andassumption (4.15) we have

1

2h

N∑i=1

nk0,i−1∂xvki ∂xv

k−1i > 1

2

(1− 1

ε

)H‖∂xvk]|22−

τ 2ε

8max

26j6M‖nj0‖∞‖∂t ∂xvk]|22 (4.31)

Discretization and regularization of an inverse problem 721

where ε > 0. Further, separating the addend with the subscripti = N from the norm‖∂t ∂xvk]|22 (cf (4.1), (4.2)) and estimating we can write

τ 2‖∂t ∂xvk]|22 64τ 2

h2‖∂t vk‖2

2+ 4hW 2k .

Thus, settingε ∈ (1, σ 2) in (4.31) we obtain

1

2h

N∑i=1

nk0,i−1∂xvki ∂xv

k−1i + a

2

2‖∂t vk‖2

2

> 1

2

(1− 1

ε

)H‖∂xvk]|22+

a2

2

σ 2− εσ 2‖∂t vk‖2

2−hε

2WW 2

k . (4.32)

Combining (4.30) and (4.32) we have

‖∂xvk]|22+ ‖∂t vk‖22 6 c11(W,H

−1)

[J k(τ

k−1∑j=1

J j +Wk

)+W 2

k

](4.33)

where 26 k 6 M. Observing (4.25) and (4.20) we can also derive such an estimate for thecasek = 1

‖∂xv1]|22+ ‖∂t v1‖22 6 c12(H

−1)W 2k .

Thus, (4.33) is valid for valuesk = 1, . . . ,M. In view of lemma 3 and (4.18), estimate(4.33) reads

(J k)2 6 c13(W,H−1)

[J k(2(k − 2)τ

k−1∑j=1

J j +Wk

)+W 2

k

]16 k 6 M.

Solving this quadratic inequality with respect toJ k we derive

J k 6 c14(W,H−1)

(2(k − 2)τ

k−1∑j=1

J j +Wk

)16 k 6 M.

Applying here lemma 2 we obtain the relation

J k 6 c15(W,H−1)Wk 16 k 6 M.

This together with (4.8) implies the assertion of theorem 2. �

5. Boundedness of discrete solution

In this section we show the uniform boundedness of the solution of nonlinear differencescheme (3.9)–(3.14) with respect to stepsh andτ which satisfy a certain inequality condition.This is possible thanks to the assumption that the data of (3.9)–(3.14) are small enough.

Theorem 3. Let

h > τ σa

for someσ > 1 (5.1)

and κ0 6= 0. There exists a positive numberD0 depending ona, b, σ ,X, T , b0, κ0, κ1 suchthat if the following norm of data

D = ‖α‖∞ + ‖∂x α]|2+ |[β]|2+ τM−1∑j=1

‖f ‖2+ max06j6M

|qj | + max06j6M

|gj | + |ρ| + |κ1| (5.2)

722 J Janno

is less thanD0 then the solutionS = (Vj

i , Zj , Rj , P j |0 6 i 6 N, 0 6 j 6 M) of the

difference scheme (3.9)–(3.14) satisfies the estimates

‖S‖∗ 6 c(D) max06j6M

‖9V j‖∞ < 1

|b| min

{1

2, σ − 1

}(5.3)

where the norm‖s‖∗ for vectorss = (vji , zj , rj , pj |0 6 i 6 N, 0 6 j 6 M) is defined bythe formula

‖s‖∗ = max06j6M

‖vj‖∞ + max16j6M

‖∂t vj‖2+ max16j6M

‖∂xvj ]|2+ max

06j6M|zj | + max

06j6M|rj | + max

06j6M|pj | (5.4)

c is an increasing function ofD depending also ona, b, σ , X, T , b0, κ0 and c(0) = 0.

Proof. We approximate the solution of (3.9)–(3.14) by the sequencesm = (vm, zm, rm, pm),m = 0, 1, 2, . . . being solution to the following linearized scheme:

3vj

m,i [1+ b(9vm−1)j

i ] + 2b(8vm−1)j

i (∂x + ∂x)vjm,i + 3bvjm−1,ivj

m,i − τj∑l=1

rj−lm−13v

lm,i

−a2∂t ∂t vj

m,i = f ji 16 i 6 N − 1 16 j 6 M − 1 (5.5)

v0m,i = αi ∂tv

0m,i = βi 06 i 6 N (5.6)

∂xvj

m,0 = ∂xvjm,N = 0 16 j 6 M (5.7)

zjm =1

a2

{vj

m,0[1+ b(9vm−1)j

0] −2(j − 1)τj∑l=1

rj−lm−1v

lm,0− qj

}06 j 6 M (5.8)

rjm = 2(j − 1)τj−1∑l=0

plm + ρ (5.9)

pjm =1

κ0

{− gj + zjm[1+ 2b0(9vm−1)

j

0] + 2b0

(2(j − 1)τ

j−1∑l=0

zlm + κ1

)2

−κ1rjm −2(j − 1)τ

j∑l=1

zj−lm rlm−1

}06 j 6 M (5.10)

for m = 1, 2, . . . and vj0,i = zj

0 = rj

0 = pj

0 = 0. Scheme (5.5)–(5.10) is explicit anddetermines the sequencesm uniquely.

In the first place we derive some auxiliary inequalities for the operators8 and9 (cf(3.8)). Observing the relation∂x0(xi, xl) = 2(xi−1 − xl) we have forw ∈ (ωτ → (ωh →R)) the estimate

max16j6M−1

max16i6N

|∂x(9w)ji | 6 c16 max06j6M

‖wj‖∞. (5.11)

Further, since0(xi, xN) = 0 by the help of the Cauchy–Schwarz inequality we obtain

max16j6M−1

‖∂t (9w)j‖∞ 6 c17 max16j6M

‖∂twj‖2. (5.12)

Moreover, there holds

max06j6k

‖(8w)j‖∞ 6 c18 max06j6k

‖wj‖∞max

06j6k‖(9w)j‖∞ 6 c19 max

06j6k‖wj‖∞ 06 k 6 M.

(5.13)

Discretization and regularization of an inverse problem 723

Let B0 be some positive number less than 1/|b|c19 min{ 12, σ − 1}. Suppose that forsomem > 1 andB ∈ [0, B0) there holds the estimate

‖sm−1‖∗ 6 B. (5.14)

Due to the assumption concerningB0, (5.1), (5.13) and (5.14) the coefficientnj0,i =1 + b(9vm−1)

j

i in (5.5) satisfies assumptions (4.15) and (4.16) withσ replaced by√σ

and there holdsH−1 < B0. Therefore, we can apply theorem 2 for the solutionvm of (5.5).Using inequalities (5.11)–(5.13) withw = vm−1, the relation

max06j6M

|rjm−1| + τM−1∑j=0

|∂t rjm−1| 6 max06j6M

|rjm−1| + T max06j6M

|pjm−1| 6 c20‖sm−1‖∗ (5.15)

(cf (5.4), (5.9)) as well as (5.2), (5.6), (5.7) we can estimate the quantitiesW andWk in(4.19) and (4.20) in termsB andD, respectively. Consequently, theorem 2 implies

max06j6M

‖vjm‖∞ + max16j6M

‖∂t vjm‖2+ max16j6M

‖∂xvjm]|2 6 c0(B, B0)D. (5.16)

Further, by means of (5.16) and estimates of the quantities9vm−1, rm−1 in termsB from(5.8)–(5.10) we derive

max06j6M

|zjm| 6 c21(B)D (5.17)

max06j6k

|rjm| 6 2(k − 1)τk−1∑j=0

|pjm| + D 06 k 6 M (5.18)

|pkm| 6|κ1||κ0|2(k − 1)τ

k−1∑j=0

|pjm| + c22(B)D 06 k 6 M. (5.19)

Applying lemma 2 for (5.19) we have

max06j6M

|pjm| 6 c23(B)D. (5.20)

Inserting (5.20) into (5.18) we get

max06j6M

|rjm| 6 c24(B)D. (5.21)

Summing up, (5.16), (5.17), (5.20), (5.21) in view of (5.4) imply

‖sm‖∗ 6 c25(B)D (5.22)

where c25 is a bounded function ofB. Evidently, we can choose an increasing functionD(B) such thatD(0) = 0 andc25(B)D(B) 6 B holds forB ∈ [0, B0). Thus, assumption(5.14) in view of (5.22) implies‖sm‖∗ 6 B if only D = D(B) andB ∈ [0, B0). Sinces0 = 0 the norm‖sm‖∗ is bounded byB for everym > 0 if B ∈ [0, B0) and D = D(B).Defining D0 = D(B0) and c as the inverse of the functionD we obtain

‖sm‖∗ 6 c(D) for D < D0 m > 0. (5.23)

By the Schauder principle there exists a limit points∗ of the sequencesm which satisfiesestimate (5.23) and the system (3.9)–(3.14). Since (3.9)–(3.14) is solved uniquely, we haves∗ = S and the first inequality in (5.3) holds. To get the second inequality in (5.3) weobserve the esimates

max06j6M

‖V j‖∞ 6 ‖S‖∗ < B0 <1

|b|c19min

{1

2, σ − 1

}and (5.13). This proves theorem 3. �

724 J Janno

6. Error estimate and convergence

In this section we prove a theorem which compares the solutions of the continuous problem(2.3)–(2.8) with the discrete problem (3.9)–(3.14). From this result we conclude theconvergence of the method of finite differences in the case of exact data.

In order to derive the error estimate we first subtract the system (2.3)–(2.8) from (3.9)–(3.14). Denoting

Vj

i = V (xi, tj ) Zj = Z(tj ) Rj = R(tj ) P j = P(tj )fj

i = f (xi, tj ) αi = α(xi) βi = β(xi) qj = q(tj ) gj = g(tj )vj

i = V ji − V ji zj = Zj − Zj rj = Rj − Rj pj = P j − P jand observing (3.8) after some computations we obtain the following system for(V

j

i , zj , rj , pj ):

3vj

i (1+ b9V ji )+ 2b8V ji (∂x + ∂x)vji + 3b(V ji + V ji )vji

+b3V ji 9vji + 2b(∂x + ∂x)V ji 8vji − τj∑l=1

Rj−l3vli − a2∂t ∂t vj

i

= f ji − f ji + εji + τj∑l=1

3V li rj−l 16 i 6 N − 1 16 j 6 M − 1

(6.1)

v0i = αi − αi ∂tv

0i = βi − βi + ζi 06 i 6 N (6.2)

∂xvj

0 = ηj0 ∂xvj

N = ηjN 16 j 6 M (6.3)

zj = 1

a2

[vj

0(1+ b9V j0 )+ V j0 b9vj0 −2(j − 1)τj∑l=1

(Rj−lvl0+ rj−lV l0)+ qj − qj + ϑj1]

06 j 6 M (6.4)

rj = 2(j − 1)τj−1∑l=0

pl + ρ − ρ + ϑj2 06 j 6 M (6.5)

pj = 1

κ0

{gj − gj + zj (1+ 2b09V

j

0 )+ 2b0Zj9vj

0

+2b0

[2(j − 1)τ

j−1∑l=0

(Zl + Zl)+ κ1+ κ1

](2(j − 1)τ

j−1∑l=0

zl + κ1− κ1

)

+κ1rj − (κ1− κ1)R

j −2(j − 1)τj∑l=1

(zj−l Rl − Zj−lr l)− P jκ0+ ϑj3}

06 j 6 M (6.6)

where

εj

i =(

1+ b∫ X

00(xi, y)V (y, tj ) dy

)[Vxx(xi, tj )−3V ji

]+ b3V ji νji

+2b(∂x + ∂x)V ji[2(i − 1)h

i∑l=1

Vj

l −∫ xi

0V (y, tj ) dy

]

+4b∫ xi

0V (y, tj ) dy [Vx(xi, tj )− 1

2(∂x + ∂x)V ji ] + τj∑l=1

Rj−l3V li

Discretization and regularization of an inverse problem 725

−∫ tj

0R(tj − s)Vxx(xi, s)ds + a2∂t ∂tV

j

i − a2Vtt (xi, tj ) (6.7)

ζi = V (xi, 0)− ∂tV 0i (6.8)

ηj

0 = Vx(0, tj )− ∂xV j0 , ηjN = Vx(X, tj )− ∂xV jN (6.9)

ϑj

1 = bV j0 νj0 +∫ tj

0R(tj − s)V (0, s)ds −2(j − 1)τ

j∑l=1

Rj−lV l0 (6.10)

ϑj

2 = 2(j − 1)τj−1∑l=0

P l −∫ tj

0P(s) ds (6.11)

ϑj

3 = 2b0

[Zjν

j

0 +(2(j − 1)τ

j−1∑l=0

Zl + κ1

)2

−(∫ tj

0Z(τ) dτ + κ1

)2]

+∫ tj

0R(s)Z(tj − s) ds −2(j − 1)τ

j∑l=1

RlZj−l (6.12)

and

νj

i = hN∑l=1

0(xi, xl)Vj

l −∫ X

00(xi, y)V (y, tj ) dy.

For the truncation errors (6.7)–(6.12) the following lemma holds which is an immediateconsequence of known error estimates for formulae of numerical differentiation andintegration.

Lemma 4. Let V ∈ C3([0, X] × [0, T ]), Z ∈ C1[0, T ], R ∈ C2[0, T ]. Then

|εji | 6 c26(h+ τ) |ζi | 6 c27τ |ηj0| 6 c28h |ηjN | 6 c29h

|∂tηj0| 6 c30(h+ τ) |∂tηjN | 6 c31(h+ τ) |ϑj1 | 6 c32(h+ τ)|ϑj2 | 6 c33τ |ϑj3 | 6 c34(h+ τ)

wherec26, . . . , c34 are constants depending onX, T , ‖V ‖C3, ‖Z‖C1, ‖R‖C2, b0, κ1.

Now we are ready to prove a theorem concerning the error of the solution of the discreteproblem (3.9)–(3.14).

Theorem 4. Let V ∈ C3([0, X] × [0, T ]), Z ∈ C1[0, T ], R ∈ C2[0, T ], κ0 6= 0 and thestepsτ andh satisfy the inequality

h > τ σa

for someσ > 1. (6.13)

Moreover, as in theorem 3 we assume that the data of the problem (3.9)–(3.14) are smallenough:D < D0. Then the difference of solutionsS = (V , Z, R, P ) andS = (V , Z,R, P )of systems (3.9)–(3.14) and (2.3)–(2.8) can be estimated as follows:

‖S − S‖∗ 6 c(D)E (6.14)

where

E = h+ τ + ‖α − α‖∞ + ‖∂x(α − α)]|2+ |[β − β]|2+ τM−1∑j=1

‖f − f ‖2

+ max06j6M

|qj − qj | + max06j6M

|gj − gj | + |ρ − ρ| + |κ0− κ0| + |κ1− κ1|(6.15)

726 J Janno

the norm‖ · ‖∗ is defined by (5.4) andc is a bounded function ofD depending also ona, b, σ,X, T , b0, κ0, κ0, κ1, κ1, ‖V ‖C3, ‖Z‖C1, ‖R‖C2.

Proof. First we apply theorem 2 for equation (6.1). Due to the second estimate in (5.3)and (6.13) the coefficientnj0,i = 1+ b9V ji in (6.1) satisfies the assumptions (4.15), (4.16)with σ replaced by

√σ andH−1 < (1− min{ 12, σ − 1})−1. Using the first inequality in

(5.3), (5.11)–(5.13), lemma 4 and observing (5.4), (6.2), (6.3), (6.15) we can estimate thequantityW in (4.19) in termsD andWk in (4.20) in termsE, r. Hence, (4.17) gives

max06j6k

‖vj‖∞ + max16j6k

‖∂t vj‖2+ max16j6k

‖∂xvj ]|2 6 c35(D)(E + max06j6k

|rj |) 16 k 6 M.

(6.16)

Estimating (6.4)–(6.6) in a similar way we obtain

max06j6k

|zj | 6 c36(D)(max06j6k

‖vj‖∞ + max06j6k

|rj | + E) 06 k 6 M (6.17)

max06j6k

|rj | 6 2(k − 1)τk−1∑j=0

|pj | + E 06 k 6 M (6.18)

|pk| 6 c37(D)(max06j6k

|zj | + max06j6k

‖vj‖∞ + max06j6k

|rj | + E) 06 k 6 M.

The latter formula, due to (6.16)–(6.18), implies

|pk| 6 c38(D)

(2(k − 1)

k−1∑j=0

|pj | + E)

06 k 6 M.

Applying here lemma 2 we have

max06j6M

|pj | 6 c39(D)E.

Using this inequality in (6.18) and then using (6.18) in (6.16), (6.17) we derive (6.14).Theorem 4 is proved. �

Corollary 1 (convergence of the method of finite differences).Let assumptions of theo-rem 4 be satisfied forV,Z,R. Let κ0 6= 0 and the data of the discrete problem (3.9)–(3.14)be exact and sufficiently small. Then‖S − S‖∗ → 0 ash, τ → 0 provided (6.13) holds inthis process. The difference scheme (3.9)–(3.14) has the first order of accuracy.

7. Regularization

Finally, let us describe the regularization of the preliminary inverse problem (1.1)–(1.4) inthe case of noisy data. Suppose that instead of the exact functionsF,A,B,G we knowcertain approximationsF , A, B, G which have the errorδ, i.e. there holds

‖F − F‖F 6 δ ‖A− A‖A 6 δ ‖B − B‖B 6 δ ‖G− G‖G 6 δ (7.1)

in some norms‖ · ‖F , ‖ · ‖A, ‖ · ‖B and‖ · ‖G.The first stage of solving the inverse problem is linear but ill-posed: in order to provide

scheme (3.9)–(3.14) with necessary data we have to compute the derivatives of the functionsF , A, B, G included in formulae (2.9). At this place we can apply known methods forregularization of linear improper problems. As a result we obtain some approximate values

Discretization and regularization of an inverse problem 727

f , α, β, q, ρ, g, κ0, κ1 for the dataf , α, β, q, ρ, g, κ0, κ1 so that the following estimateshold:

‖α − α‖∞ 6 m(δ) ‖∂x(α − α)]|2 6 m(δ) |[β − β]|2 6 m(δ)

τ

M−1∑j=1

‖f j − f j‖2 6 m(δ) max06j6M

|qj − qj | 6 m(δ) |ρ − ρ| 6 m(δ)

max06j6M

|gj − gj | 6 m(δ) |κ0− κ0| 6 m(δ) |κ1− κ1| 6 m(δ)

(7.2)

where the error functionm(δ) tends to zero ifδ → 0. This means regularization of theprocess of differentiation of the data.

The second stage of solving (1.1)–(1.4) is nonlinear: from explicit difference scheme(3.9)–(3.14) we compute the mesh functionsV ji ≈ Uxxx(xi, tj ), Z

j ≈ Uxtt (0, tj ), Rj ≈R(tj ) and P j ≈ R′(tj ), where 06 i 6 N, 0 6 j 6 M. For these quantities we formulatethe following which is an immediate consequence of theorems 1 and 4.

Theorem 5. Let the problem (1.1)–(1.4) have a solution(R,U) ∈ C2[0, T ] × C6([0, X] ×[0, T ]). Assume that

1+ bUx(x, t) 6= 0 06 x 6 X 06 t 6 T κ0 6= 0, κ0 6= 0

(2.1) hold and the stepsh and τ satisfy inequality (6.13). Moreover, as in theorem 3 weassume the data of the scheme (3.9)–(3.14) are small enough, i.e.D < D0. If inequalities(7.2) are valid then the solution of (3.9)–(3.14) approximates the solution of (1.1)–(1.4) inthe following way:

max06j6M

max06i6N

|V ji − Uxxx(xi, tj )| + max16j6M

{h

N−1∑i=1

|∂t (V ji − Uxxx(xi, tj ))|2} 1

2

+ max16j6M

{h

N∑i=1

|∂x(V ji − Uxxx(xi, tj ))|2} 1

2

+ max06j6M

|Zj − Uxtt (0, tj )|

+ max06j6M

|Rj − R(tj )| + max06j6M

|P j − R′(tj )|

6 c(D)(m(δ)+ h+ τ).Theorem 5 shows that solving the system (3.9)–(3.14) does not increase the order of

error of the differentiated datam(δ). This implies well-posedness of the second stage of thealgorithm of solution.

The system (3.9)–(3.14) provides values for the relaxation kernelR and its derivative.To determine the second componentU of the solution(R,U) from the derivativeUxxx = Vwe can use formula (2.13).

Remark 1. Theorems 3–5 do not have a global nature with respect to the domain of thevariablesx and t. The introduced upper boundD0 for the data depends onX andT andpresumably decreases ifX andT increase. On the other hand, it is physically acceptablefor the restriction on the data to be small because equation (1.1) describes the motion ofthe medium with small deformations (see [6]).

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