Download - Some inverse source problems in semilinear fractional PDEs · where (;) is the inner product in X (f;e i) = Z f(x)e i(x) dx: 1This is the Kronecker symbol n;m =1 if n m, else n;m

$: Some inverse source problems in semilinear fractional PDEs · where (;) is the inner product in X (f;e i) = Z f(x)e i(x) dx: 1This is the Kronecker symbol n;m =1 if n m, else n;m$
Some inverse source problems in semilinear fractional PDEs

Marián Slodicka

Ghent University (Belgium)Faculty of Engineering and Architecture

Research Group forNumerical Analysis and Mathematical Modeling

[email protected]://cage.ugent.be/~ms

M. Slodicka (Ghent University) Some inverse source problems in semilinear fractional PDEs 1 / 43

Outline

1 Inverse source problems

2 Fractional calculus

3 ISP for time-fractional parabolic equation

4 Time-fractional hyperbolic ISP


Inverse source problems

Introduction

ν Γ

Ω

ΓD

N

Example: A general linear parabolic PDE with mixed BCs

ut +∇ · (−Adif∇u − aconu) + asouu = f +∇ · f div in Ωu = gDir on ΓD

(−Adif∇u − aconu)Tν − gRobu = gNeu on ΓN

u(x ,0) = u0(x) in Ω

Game of IPsWhat is known/unknown?Additional dataWell-posednessHow to reconstruct missing data?



ISPs in the literature

Consider a PDE of the type Lu(t , x) = f (t , x).Literature research gives many of papers devoted to the recovery of

f (x) based on e.g. on final measurement [Rundell, 1980, Cannon, 1968,Prilepko and Solov’ev, 1988, Solov’ev, 1990, Isakov, 1990, Farcas and Lesnic, 2006,Hasanov, 2007, Johansson and Lesnic, 2007a, Johansson and Lesnic, 2007b]. . .

f (t) based on local or non-local measurement.[Prilepko et al., 2000, Hasanov, 2011, Yang et al., 2011,Hasanov and Slodicka, 2013, Slodicka, 2013, Hazanee et al., 2013, Hasanov, 2011,Hasanov and Pektas, 2013]. . .



Example spectral analysis . . .Let us consider the following homogeneous problem in Ω = (0,1)

−u′′(x) = f (x) x ∈ Ω,u(0) = u(1) = 0. (1)

We denote by A : D(A)→ X the second order differential operator, where

A = − d2

dx2 , D(A) = H2(Ω) ∩ H10 (Ω)

and X = L2(Ω). The spectrum σ(A) of the operator A consists of the eigenvalues

λn = π2n2, n ∈ N.

The corresponding eigenfunctions have the form

en(x) =√

2 sin(nπx), n ∈ N.



. . . Example spectral analysis

The set of all eigenfunctions is an orthonormal complete system in L2(Ω). Thus

(en,em) =

∫Ω

en(x)em(x) dx = δn,m1.

Each function f ∈ L2(Ω) can be written as

f =∞∑i=1

(f ,ei )ei ,

where (·, ·) is the inner product in X

(f ,ei ) =

∫Ω

f (x)ei (x) dx .

1This is the Kronecker symbol δn,m = 1 if n = m, else δn,m = 0.M. Slodicka (Ghent University) Some inverse source problems in semilinear fractional PDEs 6 / 43


Example ISP . . .

IS Problem: Find (u(t , x),h(t)) such that

ut − u′′(x) = h(t)f (x) x ∈ Ω = (0,1),u(0) = u(1) = 0

u(0, x) = 0u(t , x0) = m(t) x0 ∈ Ω

(2)

Question: Is the solution unique?Answer: If yes, then m(t) = 0 =⇒ (u(t , x),h(t)) = (0,0).



. . . Example ISP . . .Take an eigenfunction w of the operator Au = −u′′, i.e. Aw = λw . Set f = w . Seek the solution uin the form u(t , x) = α(t)w(x). From the PDE we get

[α′(t) + λα(t)] w(x) = h(t)w(x)

Thus α solvesα′(t) + λα(t) = h(t); α(0) = 0.

Therefore

α(t) =

∫ t

0e−λ(t−s)h(s) ds.

Non uniqueness: If x0 is a zero point of w(x), i.e. w(x0) = 0, which implies m(t) = 0, but we haveat least 2 solutions

(u,h) = (0,0), (u,h) =

(w(x)

∫ t

0e−λ(t−s)h(s) ds,h(t)

).



. . . Example . . .

Bad choice of a measurement point: Zero point of an eigenfunction.How many eigenfunctions do I have?: ∞Recall that

en(x) =√

2 sin(nπx), n ∈ N.

What are all zero poins of all eigenfunctions?Solving

0 = sin(nπx) =⇒ nπx = mπ m ∈ Z

we getx =

mn.

Zero points of all eigenfunctions are dense in Ω.



. . . Example

Which condition ensures the uniqueness for the ISP? A nonlocal one

(u(t),1)ω :=

∫ω

u(t , x) dx = m(t) ω ⊂ Ω, f ∈ C10 (Ω).

Multiply PDE by 1 and integrate over ω to see

(∂tu(t),1)ω − (u′′(t),1)ω = h(t)(f ,1)ω=⇒h(t) = . . .

Multiply PDE by −u′′ and integrate over Ω to see (m(t) = 0 by uniqueness)

12∂t ‖u′(t)‖

2+ ‖u′′(t)‖2

=(u′′(t),1)ω

(f ,1)ω(f ,u′′) =

−(u′′(t),1)ω(f ,1)ω

(f ′,u′)

Young’s inequality12∂t ‖u′(t)‖

2+ (1− ε) ‖u′′(t)‖2 ≤ Cε ‖u′(t)‖

2.

Grönwall’s lemma implies the uniqueness.M. Slodicka (Ghent University) Some inverse source problems in semilinear fractional PDEs 10 / 43


Rothe’s method as a semigroup

Example 1 ( D(A) = H2(Ω) ∩ H10 (Ω), t ∈ [0,T ],u(0) = u0)

Idea for x ∈ R: ex = limn→∞

(1 +

xn

)n. Set τ = T

n .

continue problem Rothe’s method

∂tu + Au = f (u) δui + Aui =ui−ui−1

τ + Aui = f (ui−1)

S(t) := e−At Sτ (t) := (I + τA)tτ

u(t) = S(t)u0 +

∫ t

0S(t − s)f (u(s)) ds ui = Sτ (ti )u0 +

i−1∑k=0

Sτ (ti − tk )f (uk )τ

Semigroups S(t) := e−At , Sτ (t) := (I + τA)tτ

Error ‖ui − u(ti )‖ ≤ C(∥∥Aβu0

∥∥)τmin1,β


Fractional calculus

Fractional derivatives . . .What do we need?

D0 = I, Aditivity DαDβ = Dα+β , for any α, β ∈ R+

Restriction of the fractional operator to natural numbers coincides with the classical derivative

Dα =dα

dxαfor α ∈ N

Assume that α ≥ 0, x > aRiemann-Liouville

(Dαa y) (x) :=

(DnIn−α

a y)

(x)

=1

Γ(n − α)

(ddx

)n ∫ x

a

y(t)(x − t)α−n+1 dt

Caputo (CDα

a y)

(x) :=(In−αa Dny

)(x)

=1

Γ(n − α)

∫ x

a

y (n)(t)(x − t)α−n+1 dt


Fractional calculus

. . . Fractional derivatives

Theorem 2 (Relationship)

Let α ≥ 0, y ∈ ACn([a,b]). Then

(CDα

a y)

(x) = (Dαa y) (x)−

n−1∑k=0

y (k)(a)

Γ(k − α + 1)(x − a)k−α.

PropertiesAditivity DαDβ = Dα+β , for any α, β ∈ R+

Caputo = Sturm-Liouville if

y (k)(a) = 0 for k = 0, . . . ,n − 1

How to get a priori estimates?


Fractional calculus

Example of a convolution kernelAssume v ∈ C[0,∞) and T > 0. It holds

ddt

[∫ t

0e−(t−s)v(s) ds

]= v(t)−

∫ t

0e−(t−s)v(s) ds.

Soddt

[∫ t

0e−(t−s)v(s) ds

]2

= 2

[∫ t

0e−(t−s)v(s) ds

][v(t)−

∫ t

0e−(t−s)v(s) ds

].

Integration in time over [0,T ] gives

2∫ T

0v(t)

∫ t

0e−(t−s)v(s) ds dt =

[∫ T

0e−(T−s)v(s) ds

]2

+

∫ T

0

[∫ t

0e−(t−s)v(s) ds

]2

dt≥ 0.


Fractional calculus

Positive kernels [Nohel and Shea, 1976]

Let a(t) ∈ L1loc(0,∞) is of positive type if∫ T

0v(t)

∫ t

0v(ξ)a(t − ξ)dξ dt ≥ 0

for any v ∈ C[0,∞) and any T > 0.

A real function a(t) is strongly positive if there exists η > 0 such thatb(t) = a(t)− ηe−t is of a positive type.

Let a(t) ∈ L1loc(0,∞) be not constant, ≥ 0, nonincreasing, convex and

such that da′(t) is not a purely singular measure. Then a(t) is stronglypositive.

In particular, twice-differentiable a(t) satisfying

(−1)k a(k)(t) ≥ 0; 0 < t <∞, k = 0,1,2; a′ 6≡ 0

are strongly positive.M. Slodicka (Ghent University) Some inverse source problems in semilinear fractional PDEs 15 / 43

Fractional calculus

Caputo derivative as a convolutionDefine (Riemann-Liouville kernel)

gβ(t) :=tβ−1

Γ(β), t > 0, β > 0

which is strongly positive definite.

By K ∗ u we denote the usual convolution in time, namely

(K ∗ u(x))(t) =

∫ t

0K (t − s)u(x , s) ds.

The Caputo fractional derivative can be also rewritten as a convolution with gβ

∂αt v(t) =∂αv∂tα

:=

(g1−α ∗ ∂tv) (t), α ∈ (0,1)(g2−α ∗ ∂ttv) (t), α ∈ (1,2)

∂tv(t), α = 1

How to get a priori estimates?M. Slodicka (Ghent University) Some inverse source problems in semilinear fractional PDEs 16 / 43

Fractional calculus

Zacher’s lemma

Let H be a real Hilbert space with a scalar product (·, ·)H with the corresponding norm ‖·‖H .

[Zacher, 2010, Lemma 2.3.2], [Zacher, 2008, Zacher, 2013] proved the following identity

( ddt (k ∗ v)(t), v(t)

)H = 1

2ddt

(k ∗‖v‖2

H

)(t) + 1

2 k(t) ‖v(t)‖2H

+ 12

∫ t

0[−k ′(s)] ‖v(t)− v(t − s)‖2

H ds a.e. t ∈ (0,T ),

which is valid for any k ∈ H1,1([0,T ]) and each v ∈ L2([0,T ],H).

The assumption k ∈ H1,1([0,T ]) is too strong. What now?


Fractional calculus

Crucial lemma . . .

[Slodicka and Šišková, 2016] CAMWAThe following crucial lemma which will play a central role in the proofs.

Lemma 3

Let H be a real Hilbert space with a scalar product (·, ·)H and the corresponding norm ‖·‖H .Assume T > 0, g ∈ L1(0,T ), g′ ∈ L1,loc(0,T ), g′ ≤ 0, g ≥ 0. If v : [0,T ]→ H such that v(0) ∈ H,v ∈ H1((0,T ),H) then∫ ξ

0

(ddt

(g∗ v) (t), v(t))

Hdt ≥ 1

2

(g∗‖v‖2

H

)(ξ) + 1

2

∫ ξ

0g(t) ‖v(t)‖2

H dt

≥ g(T )

2

∫ ξ

0‖v(t)‖2

H dt

for any ξ ∈ [0,T ].


Fractional calculus

. . . Crucial lemma

Proof.

Start from the Zacher’s lemma and replace k by gn(s) := minn,g(s).It holds

g′n(s) ≤ 0, gn(s)→ g(s) a.e. in [0,T ].

Integrate in timePass to the limit for n→∞.


ISP for time-fractional parabolic equation

Inverse source problem . . .Differential operator Ω ⊂ Rd , t ∈ [0,T ]

L(x , t)u = ∇ · (−A(x , t)∇u − b(x , t)u) + c(t)u,A(x , t) = (ai,j (x , t))i,j=1,...,d ,

b(x , t) = (b1(x , t), . . . ,bd (x , t)).

Governing PDE

(g1−β ∗ ∂tu(x)) (t) + L(x , t)u(x , t) = h(t)f (x) +

∫ t

0F (x , s,u(x , s)) ds, (3)

where g1−β denotes the Riemann-Liouville kernel

g1−β(t) =t−β

Γ(1− β), t > 0, 0 < β < 1

IC & BCu(x ,0) = u0(x), x ∈ Ω

(−A(x , t)∇u(x , t)− b(x , t)u(x , t)) · ν = g(x , t) (x , t) ∈ Γ× (0,T ).(4)



. . . Inverse source problem

Find (u(x , t),h(t)) obeying (3), (4).The unknown time-dependent function h(t) will be determined from the following additionalmeasurement

m(t) =

∫Ω

u(x , t) dx = (u(t),1) , t ∈ [0,T ]. (5)



Variational framework . . .

We associate a bilinear form L with the differential operator L as follows

(Lu, ϕ) = L (u, ϕ) + (g, ϕ)Γ , ∀ϕ ∈ H1(Ω),

i.e.L(t) (u(t), ϕ) = (A(t)∇u(t) + b(t)u(t),∇ϕ) + c(t) (u(t), ϕ) .

Throughout the paper we assume that

ai,j ,bi : Ω× [0,T ]→ R, |ai,j |+ |bi | ≤ C, i , j = 1, . . . ,d ,0 ≤ c(t) ≤ C, ∀t ∈ [0,T ],

L(t) (ϕ,ϕ) ≥ C0 ‖∇ϕ‖2, ∀ϕ ∈ H1(Ω), ∀t ∈ [0,T ].

(6)



. . . Variational frameworkIntegrate (3) over Ω, apply Green’s theorem and use (5) to get

(g1−β ∗m′) (t) + c(t)m(t) = h(t) (f ,1)− (g(t),1)Γ +

∫ t

0(F (s,u(s)),1) ds. (MP)

Assuming that (f ,1) 6= 0 we have

h(t) =

(g1−β ∗m′) (t) + c(t)m(t) + (g(t),1)Γ −∫ t

0(F (s,u(s)),1) ds

(f ,1). (7)

The variational formulation of (3) and (4) reads as

((g1−β ∗ ∂tu) (t), ϕ) + L(t) (u(t), ϕ)

= h(t) (f , ϕ) +

(∫ t

0F (s,u(s)) ds, ϕ

)− (g(t), ϕ)Γ

(P)

for any ϕ ∈ H1(Ω), a.a. t ∈ [0,T ] and u(0) = u0.M. Slodicka (Ghent University) Some inverse source problems in semilinear fractional PDEs 23 / 43


Uniqueness

Theorem 4

Let f ,u0 ∈ L2(Ω),∫

Ωf 6= 0, m ∈ C1([0,T ]), F be a global Lipschitz continuous function in all

variables. Assume (6) and g ∈ C([0,T ], Γ).Then there exists at most one solution (u,h) to the (P), (MP) obeyingu ∈ C

([0,T ],L2(Ω)

)∩ L∞

((0,T ),H1(Ω)

)with ∂tu ∈ L2

((0,T ),L2(Ω)

), h ∈ C([0,T ]).

Proof.

Suppose that (ui ,hi ) for i = 1,2 solve (P), (MP).Set u = u1 − u2 and h = h1 − h2.Subtract the corresponding variational formulations from each other.Set ϕ = u(t) in variational formulation and integrate in time over (0, ξ).Use crucial lemma.



Time discretization, Discrete convolutionEquidistant time-partitioning of [0,T ] with a step τ = T/n, for any n ∈ N. Set ti = iτ and denote

zi = z(ti ), δzi =zi − zi−1

τ.

Let us define the discrete convolution in time as follows

(K ∗ v)i :=i∑

k=1

Ki+1−k vkτ.

Please note that this definition allows blow up of K at t = 0. An easy calculation yields

δ (K ∗ v)i =(K ∗ v)i − (K ∗ v)i−1

τ= K1vi +

i−1∑k=1

δKi+1−k vkτ, i ≥ 1 (8)

as (K ∗ v)0 := 0. Similarly we may write

δ (K ∗ v)i = Kiv0 +i∑

k=1

δvk Ki+1−kτ = Kiv0 + (K ∗ δv)i , i ≥ 1. (9)



Discrete crucial lemma

The following technical lemma is a discrete analogy of Lemma 3. It plays a central role byestablishing a priori estimates for ui and hi .

Lemma 5

Let vii∈N and Kii∈N be sequences of real numbers. Assume that K decreases, i.e. Ki ≤ Ki−1for any i. Then

2δ (K ∗ v)i vi ≥ δ(K ∗ v2)

i + Kiv2i , i ∈ N.



Discrete problem

Consider a system with unknowns (ui ,hi ) for i = 1, . . . ,n. At time ti we approximate (P) by

((g1−β ∗ δu)i , ϕ

)+ Li (ui , ϕ) = hi (f , ϕ) +

(i∑

k=1

F (tk ,uk−1)τ, ϕ

)− (gi , ϕ)Γ (DPi)

and (MP) by

(g1−β ∗m′)i + cimi = hi (f ,1) +

(i∑

k=1

F (tk ,uk−1)τ,1

)− (gi ,1)Γ . (DMPi)

Considering uk−1 in the argument of F makes (DPi) linear in ui .The decoupling of ui and hi has been achieved by considering uk−1 in (DMPi).For a given i ∈ 1, . . . ,n solve first (DMPi) and then (DPi). Then increase i to i + 1.



Existence of (ui ,hi)

Lemma 6

Let f ,u0 ∈ L2(Ω),∫

Ωf 6= 0, m ∈ C1([0,T ]), F be a global Lipschitz continuous function in all

variables. Assume (6) and g ∈ C([0,T ], Γ). Then for each i ∈ 1, . . . ,n there exists a uniquecouple (ui ,hi ) ∈ H1(Ω)× R solving (DPi) and (DMPi).

Proof.Resolving (DMPi) for hi we get

hi =(g1−β ∗m′)i + cimi + (gi ,1)Γ −

(∑ik=1 F (tk ,uk−1)τ,1

)(f ,1)

∈ R. (10)

Use Lax-Milgram lemma for (DPi).



Stability analysis . . .

Introduce the following notation

(g1−β ∗‖u‖2

)j

=

j∑k=1

g1−β(tj+1−k ) ‖uk‖2τ.

Lemma 7

Let the assumptions of Lemma 6 be fulfilled. Then there exist positive constants C and τ0 suchthat for any 0 < τ < τ0 we have

(i) max1≤j≤n

(g1−β ∗‖u‖2

)j

+n∑

i=1

g1−β(ti ) ‖ui‖2τ +

n∑i=1

‖ui‖2H1(Ω) τ ≤ C,

(ii) max1≤j≤n

|hj | ≤ C.



. . . Stability analysis . . .

Compatibility condition: Assume that the (3) is fulfilled at t = 0, i.e. (P) holds true for t = 0.Therefore we may also put t = 0 in (MP), which allows us to define h0 as follows

h0 =c0m0 + (g0,1)Γ

(f ,1). (11)

Lemma 8

Let the assumptions of Lemma 6 be fulfilled. Moreover assume (11), u0 ∈ H1(Ω), g ∈ C1([0,T ], Γ),m ∈ C2([0,T ]), ∂tc ∈ L∞(0,T ) and ∂tai,j , ∂tbi ∈ L∞(Ω× (0,T )) for all i , j = 1, . . . ,d. Then thereexist positive constants C and τ0 such that for any 0 < τ < τ0 we have

(i) max1≤j≤n

(g1−β ∗‖δu‖2

)j

+n∑

i=1

g1−β(ti ) ‖δui‖2τ +

n∑i=1

‖δui‖2H1(Ω) τ ≤ C,

(ii) |δhi | ≤ C + Ct−βi for any i = 1, . . . ,n.



Existence of a solution

Theorem 9

Let f ∈ L2(Ω), u0 ∈ H1(Ω),∫

Ωf 6= 0, m ∈ C2([0,T ]), and g ∈ C1([0,T ], Γ). Suppose that F is a

global Lipschitz continuous function in all variables. Assume (6), (11), ∂tc ∈ L∞[0,T ] and∂tai,j , ∂tbi ∈ L∞(Ω× (0,T )) for all i , j = 1, . . . ,d.Then there exists a solution (u,h) to the (P), (MP) obeying u ∈ C

([0,T ],H1(Ω)

)with

∂tu ∈ L2((0,T ),H1(Ω)

), h ∈ C([0,T ]).

Proof.Cauchy, Young inequalities; Lebesgue dominated theoremConvergence (functional analysis)

Noisy data Regularization m ≈ mε ∈ C2



Example . . .

We consider problem (P)-(MP) for Ω = (0.5,3), T = 3 and β = 0.5 with

L (u, ϕ) = (∇u,∇ϕ) ,f (x) = sin x ,

F (x , t ,u) = −4tu exp(

1− u2

sin2 x

),

along with the initial and boundary conditions

u0(x) = 2 sin x ,g(0.5, t) = (t2 + 2) cos 1

2 ,

g(3, t) = (t2 + 2) cos 3,



. . . Example . . .

where the time-dependent measurement is

m(t) =

(cos

12− cos 3

)(t2 + 2

).

One can easily verify that functions

u(x , t) =(t2 + 2

)sin x

andh(t) =

83√π

t32 + t2 − exp

(1− (t2 − 2)2)+ e−3 + 2

solve the given problem.



. . . Example . . .

Discretization parametersΩ is uniformly divided into 50 subintervalsThe solution ui is calculated using a finite element method with Lagrange polynomials of thesecond order used as basis functions.Calculations were made several times for various values of τ .



. . . Example . . .

Figure: Decay of maximal relative error

(a) Logarithm of maximal relative error in time of h fordifferent values of τ . Slope of the line is 0.39529.

(b) Logarithm of maximal relative error in time of u fordifferent values of τ . Slope of the line is 0.99983.M. Slodicka (Ghent University) Some inverse source problems in semilinear fractional PDEs 35 / 43

Time-fractional hyperbolic ISP

Inverse source problem . . .

[Šišková and Slodicka, 2017] APNUMGoverning PDE (x ∈ Ω ⊂ Rd , t ∈ (0,T ))

(g2−β ∗ ∂ttu(x)) (t)−∆u(x , t) = h(t)f (x) + F (x , t ,u(x , t)) (12)

BC & ICsu(x ,0) = u0(x), x ∈ Ω,

∂tu(x ,0) = v0(x), x ∈ Ω,−∇u(x , t) · ν = g(x , t), (x , t) ∈ Γ× (0,T ),

(13)

The unknown time-dependent function h(t) will be determined from the following additionalmeasurement ∫

Ω

u(x , t)ω(x) dx = m(t), t ∈ [0,T ], (14)



Variational setting

Multiplying (12) by the function ω, integrating over Ω, applying the Green theorem and using (14),we obtain

(g2−β ∗m′′) (t) + (∇u(t),∇ω) = h(t) (f , ω)− (g(t), ω)Γ + (F (t ,u(t)), ω) . (MP)

Similarly multiplying (12) by a function ϕ ∈ H1(Ω) and using Green’s theorem, we obtain thevariational formulation of (12) and (13)

((g2−β ∗ ∂ttu) (t), ϕ) + (∇u(t),∇ϕ) = h(t) (f , ϕ) + (F (t ,u(t)), ϕ)− (g(t), ϕ)Γ , (P)

for any ϕ ∈ H1(Ω), a.a. t ∈ [0,T ] and u(0) = u0, ∂tu(0) = v0.The relations (P) and (MP) represent the variational formulation of the ISP (12), (13) and (14).



Time discretization

The discrete convolution is defined by

(K ∗ v)i :=i∑

k=1

Ki+1−k vkτ,

note that by this definition we avoided problems with a blow up if K has a singularity at t = 0. Thenwe can calculate a difference for the discrete convolution as follows

δ (K ∗ v)i =(K ∗ v)i − (K ∗ v)i−1

τ= K1vi +

i−1∑k=1

δKi+1−k vkτ, i ≥ 1, (15)

as(K ∗ v)0 := 0



Schema

On the i−th time-layer we approximate the solution of (P),(MP) by (ui ,hi ), which solves((g2−β ∗ δ2u

)i , ϕ)

+ (∇ui ,∇ϕ) = hi (f , ϕ) + (F (ti ,ui−1), ϕ)− (gi , ϕ)Γ , (DPi)

for ϕ ∈ H1(Ω), with δu0 := v0 and

(g2−β ∗m′′)i + (∇ui−1,∇ω) = hi (f , ω) + (F (ti ,ui−1), ω)− (gi , ω)Γ . (DMPi)



Solvability of the ISP

Theorem 10

Let f ∈ L2(Ω), u0, v0, ω ∈ H1(Ω),∫

Ωfω 6= 0, m ∈ C3([0,T ]), and g ∈ C2([0,T ], Γ). Suppose that F

is a global Lipschitz continuous function in all variables and (11) holds true.Then there exists a solution (u,h) to the (P), (MP) obeying u ∈ C

([0,T ],H1(Ω)

)with

∂tu ∈ C([0,T ],L2(Ω)

)∩ L∞

((0,T ),H1(Ω)

), ∂ttu ∈ L2

((0,T ),L2(Ω)

)and h ∈ C([0,T ]).

Theorem 11

Let f , v0 ∈ L2(Ω),u0, ω ∈ H1(Ω),∫

Ωfω 6= 0, m ∈ C2([0,T ]), F be a global Lipschitz continuous

function in all variables and g ∈ C([0,T ], Γ). Then there exists at most one solution (u,h) to the(P), (MP) obeying u ∈ C

([0,T ],H1(Ω)

), ∂tu ∈ C

([0,T ],L2(Ω)

)∩ L2

((0,T ),H1(Ω)

)with

∂ttu ∈ L2((0,T ),L2(Ω)

)and h ∈ C([0,T ]).



Fractional Dynamical BC

[Šišková and Slodicka, 2018] CAMWA

(g2−β ∗ ∂ttu(x)) (t)−∆u(x , t) = h(t)f (x), x ∈ Ω, t ∈ (0,T ), (16)

The equation (16) is accompanied with the following initial and boundary conditions

u(x ,0) = u0(x), x ∈ Ω,∂tu(x ,0) = v0(x), x ∈ Ω,

u(x , t) = 0, (x , t) ∈ ΓD × (0,T ),− (g2−β ∗ ∂ttu(x)) (t)−∇u(x , t) · ν = σ(x , t), (x , t) ∈ ΓN × (0,T ),

(17)

ISP: Find the couple (u,h) obeying∫Ω

u(x , t)ω(x) dx = m(t), t ∈ [0,T ], (18)



Evolutionary boundary condition

The dynamical BCs model non-perfect contact. (They are not very common in the mathematicalliterature.)

They appear in many mathematical models including heat transfer in a solid in contact with amoving fluid, thermo-elasticity, diffusion phenomena, problems in fluid dynamics, etc. (see[Escher, 1993, Igbida and Kirane, 2002, Chill et al., 2006] and the references therein).ν × E = ν × (∂tB (H)× ν) [Vrábel’ and Slodicka, 2012]∂tβ(u) + (∇u + b(u)) · ν = g [Su, 1993]∂tβ(u)−∆Γu + u +∇u · ν = g [Vrábel’ and Slodicka, 2013]



Boundary measurement

[Šišková and Slodicka, 2019] JCAM(g2−β ∗ ∂ttu(x)) (t)−∆u(x , t) = h(t)f (x) + F (x , t ,u(x , t)),

u(x ,0) = u0(x),∂tu(x ,0) = v0(x),

−∇u(x , t) · ν = γ(x , t)

(19)

The Inverse Source Problem (ISP) we are interested in here consists of identifying a couple(u(x , t),h(t)) obeying (19) and∫

Γ

u(x , t)ω(x)dS = m(t), t ∈ [0,T ], (20)


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