Inverse problem of reconstruction of degenerate diffusion coefficient in a parabolic equation

In e se P oblems
PAPER
In e se p oblem o econs uc ion o degene a e
di usion coe icien in a pa abolic equa ion
To ci e his a icle: Pie ma co Canna sa
e al
2021
In e se P oblems
37 125002
View he a icle online o upda es and enhancemen s.
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ON ASYMPTOTIC “EIGENFUNCTIONS”
OF THE CAUCHY PROBLEM FOR A
NONLINEAR PARABOLIC EQUATION
V A Galak iono , S P Ku dyumo and A A
Sama ski
-
S abiliza ion o solu ions o quasilinea
second o de pa abolic equa ions in
domains wi h non-compac bounda ies
Ruslan Kh Ka imo and La isa M
Kozhe niko a
-
Recons uc ion o degene a e conduc i i y
egion o pa abolic equa ions
Pie ma co Canna sa, Anna Doubo a and
Masahi o Yamamo o
-
This con en was downloaded om IP add ess 150.214.182.233 on 05/12/2024 a 11:47
In e se P oblems
In e se P oblems 37 (2021) 125002 (56pp) h ps://doi.o g/10.1088/1361-6420/ac274b
In e se p oblem o econs uc ion o
degene a e di usion coe icien in a
pa abolic equa ion
Pie ma co Canna sa1, Anna Doubo a2,∗and
Masahi o Yamamo o3,4
1Uni e si y o Rome ‘To Ve ga a’, I aly
2Uni e sidad de Se illa, Dp o. EDAN e IMUS, Spain
3The Uni e si y o Tokyo, Japan
4Hono a y Membe o Academy o Romanian Scien is s, Co espondence membe
o Accademia Pelo i ana dei Pe icolan i, Messina, I aly
E-mail: canna sa@ma .uni oma2.i 9,doubo [email protected] and [email p o ec ed]p
Recei ed 12 June 2021, e ised 25 Augus 2021
Accep ed o publica ion 16 Sep embe 2021
Published 2 No embe 2021
Abs ac
We conside he in e se p oblem o iden i ica ion o degene a e di usion coe -
icien o he o m xαa(x) in a one dimensional pa abolic equa ion by some
ex a da a. We i s p o e by ene gy me hods he uniqueness and Lipschi z s a-
bili y esul s o he iden i ica ion o a cons an coe icien aand he powe
αby knowing in e io da a a some ime. On he o he hand, we ob ain he
uniqueness esul o he iden i ica ion o a gene al di usion coe icien s a(x)
andalso hepowe α o m bounda y da a on one side o he space in e al. The
p oo is based on global Ca leman es ima es o a hype bolic p oblem and an
in e sion o he in eg al ans o m simila o he Laplace ans o m. Finally, he
heo e ical esul s a e sa is ac o y e i ied by nume ically expe imen s.
Keywo ds: in e se p oblems, degene a e pa abolic equa ions, nume ical econ-
s uc ion
(Some igu es may appea in colou only in he online jou nal)
1. In oduc ion
In his pape , we will add ess se e al in e se p oblems o pa abolic equa ions wi h degen-
e acy in he di usion coe icien . Mo e p ecisely, we will conside he weakly degene a e
∗Au ho o whom any co espondence should be add essed.
1361-6420/21/125002+56$33.00 © 2021 IOP Publishing L d P in ed in he UK 1
In e se P oblems 37 (2021) 125002 P Canna sa e al
Cauchy–Di ichle p oblem
⎧
⎪
⎪
⎨
⎪
⎪
⎩
∂ u−∂x(xαa(x)∂xu)= (x, ), (x, )∈(0, )×(0, T),
u(0, )=0, u(, )=0, ∈(0, T),
u(x,0)=u0(x), x∈(0, ),
(1)
whe e 0 <α<1, as well as s ongly degene a e p oblems o he o m
⎧
⎪
⎪
⎨
⎪
⎪
⎩
∂ u−∂x(xαa(x)∂xu)= (x, ), (x, )∈(0, )×(0, T),
(xα∂xu(x, )|x=0=0, u(, )=0, ∈(0, T),
u(x,0)=u0(x), x∈(0, ),
(2)
whe e 1 ⩽α<2. He e, T>0, >0, u0∈L2(0, ), u0=0, ∈L2((0, )×(0, T)) a e gi en,
and a∈C1([0, ]) sa is ies 0 <a⩽a(x)⩽a<+∞ o all x∈(0, ), wi h aand aposi i e
cons an s. No ice ha he bounda y condi ions in (1)and(2) a e di e en . In ac , weakly
degene a e pa abolic p oblems can be s udied ei he wi h Di ichle o Neumann da a whe eas,
when sol ing s ongly degene a e equa ions in L2(0, ), he la e u ns ou o be he na u al
condi ion o impose on he pa o he bounda y whe e degene acy occu s, see ema k 2.1
below.
Ou goal is o de e mine o es ima e he di usion coe icien o he conside ed degene a e
p oblems. Al hough we can discuss mo e gene al cases, we concen a e he e on he one-
dimensional linea equa ion. The ollowing h ee ypes o in e se p oblems o he iden i ica ion
o he unknown di usion coe icien will be conside ed and analyzed.
Linea case: gi en α=1, we will deal wi h he in e se p oblem o iden i ica ion o a cons an
coe icien , a(x)≡a,whe eais an unknown posi i e cons an , om gi en sui able in e nal
da a a some ime (see sec ion 3.1).
Powe -like case: gi en a(x)=1, he in e se p oblem will be he iden i ica ion o he powe
α∈(0, 2) om sui able in e nal obse a ions a some ime (see sec ion 3.2).
Gene al case: he in e se p oblem o he iden i ica ion o a gene al di usion coe icien xαa(x)
om one bounda y obse a ion (see sec ion 4).
In ecen yea s, degene a e pa abolic equa ions ha e ecei ed inc easing a en ion, in iew
o he impo an ela ed heo e ical analysis and p ac ical applica ions such as clima e science
(see Selle s [35], Díaz [12], Ji and Huang [15]), popula ions gene ics (E hie [14]), ision
(Ci i and Man edini [9]), inancial ma hema ics (Black and Scholes [5]), and luid dynamics
(Oleinik and Samokhin [31]). See also Canna sa e al [7] and e e ences he ein.
In e se p oblems a e he kind o p oblems known as no well-posed in he sense o
Hadama d (see [16]). This means ha , ei he a solu ion does no exis , o i is no unique
and/o small e o s in known da a lead o la ge e o s in he compu a ion o he solu ion(s). In
pa icula , he i s heo e ical issues ha we analyze in ou pape a e uniqueness and s abili y.
The main echniques used he e a e usual in con ol and pa ame e iden i ica ion heo y: ools
om eal and complex analysis, ene gy me hods and Ha dy’s inequali y, Laplace and o he
simila in eg al ans o ms, and Ca leman inequali ies.
The li e a u e ela ed o in e se p oblems o degene a e pa abolic PDEs, despi e hei un-
damen al impo ance and p ac ical applica ions, is a he ecen and sca ce. Fo example, he
in e se sou ce p oblem was conside ed in To [38], Canna sa e al [8], Canna sa e al [7],
Deng e al [10] and Hussein e al [18], whe e nume ical econs uc ion was also conside ed.
2
In e se P oblems 37 (2021) 125002 P Canna sa e al
The in e se p oblem o eco e ing he i s -o de coe icien o degene a e pa abolic equa ions
is analyzed in Deng and Yang [11] and Kamynin [24]. An in e se di usion p oblem was con-
side edin[39], whe e a cons an di usion coe icien was eco e ed om measu emen o
second o de de i a i es by means o Ca leman es ima es p o ided in Canna sa e al [8].
As o in e se p oblems o de e mining spa ially a ying coe icien s such as conduc i i ies
and sou ce e ms o non-degene a e pa abolic equa ions, we no e ha he e has been a sub-
s an ial amoun o wo ks. He e we e e o chap e s o he book by Isako [23], and Yamamo o
[40] (see also he e e ences he ein). In pa icula , o he same ype o in e se p oblems o
one-dimensional uni o mly pa abolic equa ions we e e o Mu ayama [30], see also Pie ce
[32], Suzuki and Mu ayama [36].
On he o he hand, mos o he echniques ha ha e been de eloped o ea non-degene a e
pa abolic equa ions cease o be applicable in he degene a e case. Fo ins ance, o s ongly
degene a e ope a o s he ace o he co-no mal de i a e mus anish on he pa o he bounda y
whe e ellip ici y ails, yielding no use ul measu emen 5.
The second main issue o his pape is he econs uc ion me hod o ou in e se p oblems.
The goal is o compu e app oxima ions o he di usion coe icien as solu ions o he in e se
p oblems using he obse a ion da a. To ou bes knowledge, he e a e no wo ks combining he
heo e ical s udies on he uniqueness, s abili y and he nume ical econs uc ion o he di usion
coe icien o a degene a e pa abolic equa ion.
An e icien echnique o achie e his, as shown below, elies on a e o mula ion o he
sea ch o he di usion coe icien as an ex emal p oblem. This is classical nowadays and
has been applied in a lo o si ua ions; see o ins ance La en ie e al [28], Sama skii and
Vabishche ich [34], and Vogel [37].
The pape is o ganized as ollows. In sec ion 2, we conside he well-posedness o he di ec
p oblem (1)and(2). In sec ion 3.1 we p o e, using he ene gy me hods, Lipschi z s abili y and
uniqueness esul s o he linea case co esponding o he cons an coe icien a. Sec ion 3.2
will be de o ed o he powe -like case o he iden i ica ion o he powe αin (1)and(2). Again,
Lipschi z s abili y and uniqueness will be s udied by he ene gy me hods. In sec ion 4a case o
he econs uc ion o he unknown gene al di usion coe icien xαa(x) will be ea ed by means
o he Ca leman es ima es, ob aining he uniqueness esul o bo h αand a(x). Finally, in
sec ion 5, in o de o illus a e heo e ical esul s ob ained in he p e ious sec ions, we pe o m
sa is ac o y nume ical expe imen s co esponding o he conside ed in e se p oblems.
2. P elimina ies
In his sec ion we will conside he well-posedness o he di ec p oblem ela ed o (1)and(2).
We begin wi h ecalling he na u al unc ions spaces whe e he p oblem can be se . Fo any
>0, le us call H=L2(0, ). Fo α∈(0, 2) we will conside he ollowing unc ion spaces:
H1
α(0, )=⎧
⎪
⎪
⎨
⎪
⎪
⎩
u∈H|
0
xα|∂xu|2dx<∞,u(0) =u()=0,0<α<1,
u∈H|
0
xα|∂xu|2dx<∞,u()=0,1⩽α<2
(3)
5Al hough in his pape we conside equa ions ha degene a e on a s ic subse o he bounda y, ou app oach applies
unal e ed when degene acy occu s on he whole bounda y, like in he Budyko–Selle s model [35].
3
In e se P oblems 37 (2021) 125002 P Canna sa e al
and
H2
α(0, )={u∈H1
α(0, )|x→ xα∂xu∈H1(0, )},0<α<2.
Rema k 2.1 (Neumann bounda y condi ion). Fo u∈H2
α(0, ) wi h 1 ⩽α<2, we
ha e xα∂xu|x=0=0. Indeed, i xα∂xu(x)→Lwhen x→0, hen xα|∂xu(x)|2∼L2/xαand,
he e o e L=0o he wiseu/∈H1
α(0, ).
We now adap he classical Poinca ´
e inequali y o he abo e weigh ed spaces.
Lemma 2.2 (Poinca ´
e’s inequali y). Fo all ε>0, he e exis s a cons an Cp=
Cp(ε,)>0such ha o all α∈[0, 2 −ε], he ollowing holds:

0|u|2dx⩽Cp
0
xα|∂xu|2dx∀u∈H1
α(0, ).(4)
P oo o lemma 2.2. Le >0 be gi en. We will dis inguish wo cases.
(a) Assume α=1, o all x∈(0, ), we ha e
|u(x)|⩽
x|∂xu|dy⩽
x
y|∂xu(y)|2dy1/2
x
1
ydy1/2
⩽log 1
−log 1
x1/2
x
y|∂xu(y)|2dy1/2
.
The e o e, we ob ain

0|u(x)|2dx⩽
0
log 1
xdx
0
x|∂xu(x)|2dx
=(1 −log )
0
x|∂xu(x)|2dx.
(b) Assume now ha α=1. We can w i e
|u(x)|⩽
x
yα/2u(y)dy
yα/2
⩽
x
dy
yα/21/2
0
yα|u(y)|2dy1/2
⩽1−α−x1−α
1−α1/2
0
yα|u(y)|2dy1/2
.
The e o e, we ob ain

0|u(x)|2dx=
0
1−α−x1−α
1−αdx
0
xα|∂xu(x)|2dx
=2−α
2−α
0
xα|∂xu(x)|2dx=Cp
0
xα|∂xu(x)|2dx,
he e we ha e used ha α<2. This ends he p oo .

Fo gi en cons an s 0 <a<a, we assume ha
a∈C1([0, ]), 0 <a⩽a(x)⩽a<+∞∀x∈(0, ).(5)
4

In e se P oblems 37 (2021) 125002 P Canna sa e al
P oblems (1)and(2) can be ecas in he abs ac o m
u( )=Au( ) ⩾0
u(0) =u0
(6)
by in oducing he linea ope a o A:D(A)⊂H→Hde ined by
D(A)=H2
α(0, ), Au =∂x(xαa(x)∂xu)∀u∈D(A).(7)
Obse e ha he degene a e coe icien ˜a(x)=xαa(x)iso classC1([0, ]) and sa is ies
lim
x→0x˜a(x)˜a(x)=lim
x→0
xα+1a(x)+αxαa(x)
xαa(x)=α∈(0, 2).(8)
The e o e, we can appeal o [29] o ob ain a well-posedness esul o (6)aswellas o he
nonhomogeneous p oblem
u( )=Au( )+ ( ) ⩾0
u(0) =u0
.(9)
So, one can p o e he ollowing, whe e we ecall ha D((−A)1/2)=H1
α(0, ) o he ope a o
Ade ined in (7).
Theo em 2.3. Assume (5). Then, A is he in ini esimal gene a o o a s ongly con inuous
semig oup o con ac ions, e A. Mo eo e , e A is analy ic. The e o e, o any u0∈H, he solu ion
u(x, )=(e Au0)(x)o p oblem (6)sa is ies:
(a)u∈C([0, ∞); H),
(b) → u(·, )is analy ic as a map (0, +∞)→H,
(c)u( )∈n⩾1D(An)∀ >0.
Fu he mo e, i u0∈D((−A)1/2), hen o any ∈L2(0, T;H) he mild solu ion
u( ):=e Au0+
0
e( −s)A (s)ds( ∈[0, T])
o p oblem (9)belongs o H1(0, T;H)∩C([0, T]; D((−A)1/2)) ∩L2(0, T;D(A)) and sa is ies
he equa ion in (9) o a.e. ∈[0, T].
Consequen ly, he equa ionsin (1)and(2) a e sa is ied in classical sense on (0, )×(0, +∞),
as well as bounda y condi ions, aking in o accoun ema k 2.1. As o he ini ial condi ion, we
ecall ha u0∈L2(0, ) implies ha u∈C0([0, +∞); L2(0, )) and u0∈H1
α(0, ) implies ha
u∈C0([0, +∞); H1
α(0, )). He ea e , we will assume solu ions as smoo h as equi ed.
3. Recons uc ion by ene gy me hods
3.1. The linea case: uniqueness and Lipschi z s abili y
Le us conside he linea case in which a(x)≡a,whe eais a cons an such ha 0 <a⩽a⩽a
and α=1. This co esponds o he p oblem (2) wi h s ong degene acy, ha is w i en as
5
In e se P oblems 37 (2021) 125002 P Canna sa e al
ollows:
⎧
⎪
⎪
⎨
⎪
⎪
⎩
∂ u−a∂x(x∂xu)= (x, ), x∈(0, ), ∈(0, T),
x∂xu(x, )|x=0=0u(, )=0, ∈(0, T),
u(x,0)=u0(x), x∈(0, ).
(10)
In his sec ion we will analyze uniqueness and Lipschi z s abili y o he in e se p oblem o he
iden i ica ion o he cons an coe icien ain (10) om measu emen s ∂ u(x, 0)andx∂xu(x, 0),
o some 0∈(0, T]and o allx∈(0, ), as is indica ed in heo em 3.2. In sec ion 5.1 we will
pe o m nume ical econs uc ions o a ela ed o his in e se p oblem.
Rema k 3.1. On he o he hand, le us no e ha i is also possible o conside o he obse -
a ions, o example, using he bounda y da a ∂xu(, ) o all ∈(0, T). The ela ed in e se
p oblem will be he ollowing: gi en η=∂xu(,·), ind asuch ha he co esponding solu ion
uao (10) sa is ies ∂xu(, )=η( ), o all ∈(0, T). Howe e , uniqueness and s abili y o his
second in e se p oblem a e, o ou knowledge, open ques ions ha we will conside in a o h-
coming pape . None heless, we ha e sa is ac o y nume ical app oaches o his second in e se
p oblem, ha we will p esen in he sec ion 5.2.
Le ui,i=1, 2 be he solu ion o (10) wi h a=ai, ha is osay
⎧
⎪
⎪
⎨
⎪
⎪
⎩
∂ ui−ai∂x(x∂xui)= (x, ), x∈(0, ), ∈(0, T),
x∂xui(x, )|x=0=0ui(, )=0, ∈(0, T),
ui(x,0)=u0(x), x∈(0, ).
(11)
The main esul o his sec ion is he ollowing:
Theo em 3.2. Assume ha a1,a2∈Rsa is y 0<a⩽a1,a2⩽a<+∞(wi h a and a
gi en posi i e cons an s),u
0∈L2(0,),u0=0, and ∈L2((0, )×(0, T)).Le u
1,u2be solu ions
o (11)sa is ying he ollowing hypo hesis: o some 0∈(0, T]and μ>0such ha

0
x|∂xui(x, 0)|2dx⩾μ,i=1, 2.(12)
Then he e exis s a posi i e cons an C =C(a
¯,μ)such ha
|a2−a1|⩽C
0|∂ u1(x, 0)−∂ u2(x, 0)|2+x|∂xu2(x, 0)−∂xu1(x, 0)|2dx1/2
.(13)
Rema k 3.3. No ice ha , as we ha e men ioned be o e, we can ob ain a simila uniqueness
esul o mo e gene al equa ions con aining ze o o de e ms.
Rema k 3.4. Recall ha , since e A is analy ic, ui( ,·)∈H2
1(0, ) o a.e. >0, i=1, 2.
Mo eo e , assuming in addi ion ha ≡0, we ha e he ollowing.
(a) By backwa d uniqueness we know ha , i u0=0, hen he solu ion uo (10) sa is ies

0
x|∂xu(x, )|2dx>0∀ ∈[0, T].(14)
This explains he ole played by assump ion (12) abo e.
6
In e se P oblems 37 (2021) 125002 P Canna sa e al
(b) Since Ais dissipa i e, we ha e ha

0
x|∂xui(x, )|2dx⩾
0
x|∂xui(x,T)|2dx∀ ∈[0, T].
The e o e,
μ=min
i=1,2
0
x|∂xui(x,T)|2dx
is a possible choice o μ o e e y 0∈(0, T].
(b) Ano he way o cons uc μ>0 sa is ying (12)is o akeu0∈D(A), obse ing ha

0
x|∂xu( ,x)|2dx⩾
0
x|∂xu0(x)|2dx−2 
0|(x∂xu0)x|2dx.
The e o e, (12) holds wi h μ=1
2
0x|∂xu0(x)|2dx o all
0< 0<1
4
0x|∂xu0(x)|2dx

0|(x∂xu0)x|2dx.
P oo o heo em 3.2.We assume ha a2⩾a1and se w=u1−u2. Then, we can w i e
∂ w−a1∂x(x∂xw)=(a1−a2)∂x(x∂xu2).
Mul iplying by u2and in eg a ing by pa s his equali y, we ge
(a2−a1)
0
x|∂xu2|2dx=
0
u2∂ wdx+a1
0
x∂xw∂xu2dx
⩽
0|u2|2dx1/21
0|∂ w|2dx1/2
+a1
0
x|∂xw|2dx1/2
0
x|∂xu2|2dx1/2
.
So, by (4), we ha e
|a2−a1|
0
x|∂xu2|2dx
⩽Cp
0|∂ w|2dx1/2
+a1
0
x|∂xw|2dx1/2
×
0
x|∂xu2|2dx1/2
, (15)
which in u n yields he he conclusion in iew o (12) and ends he p oo o his heo em. 
I is clea ha om he p oo o his heo em we can deduce he ollowing uniqueness esul .
Co olla y 3.5. Le us assume ha a1,a2⩾a>0, ≡0, and u0∈L2(0, ), u0=0.I
u1,u2∈C0([0, T]; L2(0, )) a e solu ions o (11)such ha o some 0∈(0, T]we ha e
∂ u1(x, 0)=∂ u2(x, 0)and ∂xu2(x, 0)=∂xu1(x, 0) o all x ∈(0, ), hen a1=a2.
7
In e se P oblems 37 (2021) 125002 P Canna sa e al
Indeed, i is enough o ealize ha we do no need hypo hesis (12) o uniqueness, because
we do no need he cons an in (15) o be uni o m. The e o e, o m (15) and hanks o he
inequali y (14), we ob ain uniqueness.
3.2. The powe -like case: uniqueness and Lipschi z s abili y
In his sec ion we will conside he in e se p oblem o iden i ica ion o he powe αin he
degene a e pa abolic p oblems (1)and(2), assuming o simplici y ha a(x)≡1and ≡0.
Tha is o say, ha we will conside he ollowing p oblems: wi h weak degene acy(0 ⩽α<1)
⎧
⎪
⎪
⎨
⎪
⎪
⎩
∂ u−∂x(xα∂xu)=0, (x, )∈(0, )×(0, T),
u(0, )=0, u(, )=0, ∈(0, T),
u(x,0)=u0(x), x∈(0, ),
(16)
and wi h s ong degene acy (1 ⩽α<2)
⎧
⎪
⎪
⎨
⎪
⎪
⎩
∂ u−∂x(xα∂xu)=0, (x, )∈(0, )×(0, T),
xα∂xu(x, )|x=0=0, u(, )=0, ∈(0, T),
u(x,0)=u0(x), x∈(0, ).
(17)
Simila ly o wha we ha e done in he sec ion 3.1, we will analyze uniqueness and s abili y
o he in e se p oblem o iden i ica ion o he powe αin he abo e p oblems om measu e-
men s ∂ u(x, 0)andx∂xu(x, 0), o some 0∈(0, T]and o allx∈(0, ). In sec ion 5.3 we will
pe o m he nume ical econs uc ion o α o he p oblems (16)and(17).
As we ha e no ed in he p e ious sec ion, i is also possible o conside he in e se p oblem
o iden i ica ion o he powe αwi h only one bounda y obse a ion, o example ∂xu(, ) o
all ∈(0, T). Howe e , he uniqueness and s abili y ela ed o his second in e se p oblem a e,
o ou knowledge, open ques ions. Ne e heless, we ha e sa is ac o y nume ical app oaches
o his second in e se p oblem, ha we will expose in sec ion 5.4.
In his sec ion, we analyze uniqueness and s abili y o he powe αin (17)and(16). Le us
s a wi h a uniqueness esul .
Theo em 3.6. Le assume ha 0<⩽1, u0∈L2(0, ), u0=0and ui,i=1, 2, be he solu-
ion o (17)(o (16)) co esponding o αisuch ha o some 0∈(0, T]we ha e ∂ u1(x, 0)=
∂ u2(x, 0)and ∂xu1(x, 0)=∂xu2(x, 0) o all x ∈(0, ), hen α1=α2.
P oo o heo em—we can assume, wi hou loss o gene ali y, ha α1<α
2. Se ing
w(x, )=u2(x, )−u1(x, ),
we ha e
∂ w−∂x(xα2∂xw)=∂x((xα2−xα1)∂xu1).(18)
Then, mul iplying his equali y by u1and in eg a ing by pa s, we ob ain

0
(xα1−xα2)|∂xu1|2dx=
0
(u1∂ w+xα2∂xw∂
xu1)dx.(19)
8
In e se P oblems 37 (2021) 125002 P Canna sa e al
o a bi a y μ∈(0, 1).
We can e i y ha Aa,αusa is ies (25) in he sense o dis ibu ions. The uniqueness o
solu ion o (25) yields
=Aa,αu∈C1([0, ∞); H1(μ, 1)),
and so
u∈C1([0, ∞); H3(μ,1)).
Thus he p oo o lemma 4.5 is comple e. 
Nex , we in oduce an in eg al ans o m called he Rezni skaya ans o m (see [33], pp
213–215) as ollows:
(Kη)( ):=∞
0
η(τ)G( ,τ)dτ,η∈L∞(0, ∞), (36)
whe e
G( ,τ)=1
√π e−τ2
4 , ,τ>0.
We no e ha
∂k
τG( ,·)∈L∞(0, ∞)∩L1(0, ∞), ∂k
G(·,τ)∈L∞(0, ∞) o all ,τ>0andk∈N.
(37)
Then, he ollowing esul can be p o ed.
Lemma 4.6. Le η∈W2,∞(0, ∞)sa is y η(0) =0.Then
d
d (Kη)( )=Kd2η
dτ2( ), >0.
P oo o lemma 4.6.The p oo is gi en in [33], pp 213–214, and o comple eness we will
he e p esen he p oo . Since ∂ G( ,τ)=∂2
τG( ,τ) o >0andτ>0, and η∈W2,∞(0, ∞),
we ha e
d
d (Kη)( )=∞
0
η(τ)∂ G( ,τ)dτ=∞
0
η(τ)∂2
τG( ,τ)dτ
=η(τ)∂τG( ,τ)−η(τ)G( ,τ)τ=∞
τ=0+∞
0
η(τ)G( ,τ)dτ, >0.
Using ha η(0) =0and∂τG( ,0)=0 o >0, we ob ain
d
d (Kη)( )=∞
0
η(τ)G( ,τ)dτ=Kd2η
d 2( ), >0.
The p oo o lemma 4.6 is comple e. 
15

In e se P oblems 37 (2021) 125002 P Canna sa e al
We will also need he ollowing esul .
Lemma 4.7.
(i)Le ˜ua,αsa is y (32)wi h (a,α,u0)∈A(ε0,ε1,δ0,δ1), and le us se
Va,α(x, ):=(K˜ua,α(x,·))( )=∞
0
˜ua,α(x,τ)G( ,τ)dτ,0<x<1, >0.(38)
Then
∂ Va,α(x, )=∂x(xαa(x)∂xVa,α(x, )) o almos all x ∈(0, 1) and >0
(39)
and
Va,α(0, )=Va,α(1, )=0, >0,
Va,α(x,0)=u0(x), o almos all x ∈(0, 1).
(40)
(ii)Le ua,αbe he solu ion o (25)whe e (a,α,u0)∈A(ε0,ε1,δ0,δ1).Then
ua,α(x, )=∞
0
˜ua,α(x,τ)G( ,τ)dτ,0<x<1, >0.(41)
P oo o lemma 4.7. Pa (ii) ollows eadily om pa (i) by he uniqueness o he solu ion
ua,α(see heo em 2.3 in sec ion 2). The e o e, i su ices o p o e pa (i). Hence o h, we se
(g,h):=1
ε2
g(x)h(x)dx,
wi h a bi a y ε2>0. Fo θ∈C∞
0(ε2,1),wese
˜w( )=(˜ua,α(·, ), θ)=1
ε2
˜ua,α(x, )θ(x)dxand w( )=(Va,α(·, ), θ) o >0.
Then, he ini ial condi ion in (32) and he egula i y (35) yield
˜w∈W2,∞(0, ∞), ∂ ˜w(0) =0
and
w( )=(Va,α(·, ), θ)=1
ε2
Va,α(x, )θ(x)dx
=∞
0
˜w(τ)G( ,τ)dτ=(K˜w)( ), >0.
The e o e, lemma 4.6 implies
dw
d ( )=d
d (K˜w)( )=Kd2˜w
dτ2( )=∞
0
d2
dτ2˜w(τ)G( ,τ)dτ,
16
In e se P oblems 37 (2021) 125002 P Canna sa e al
ha is,
d
d (Va,α(·, ), θ)=∞
0
d2
dτ2(˜ua,α(·,τ), θ)G( ,τ)dτ
=∞
01
ε2
∂2˜ua,α(x,τ)
∂τ2θ(x)dxG( ,τ)dτ
=1
ε2∞
0
∂2˜ua,α
∂τ2(x,τ)G( ,τ)dτθ(x)dx
=K∂2˜ua,α
∂τ2(·, ), θ.
The e o e,
∂Va,α
∂ (·, ), θ=K∂2˜ua,α
∂τ2,θ(42)
o all θ∈C∞
0(ε2,1).
Recall ha
Aa,αw(x, )=∂x(xαa(x)∂xw(x, )), 0 <x<1, >0.
By (33), (35)and(37), we can jus i y he exchange o he ope a ion by Aa,αand ∞
0...dτin
(38), and ob ain
Aa,αVa,α(x, )=∞
0
(Aa,α˜ua,α)(x,τ)G( ,τ)dτ
=K(Aa,α˜ua,α)(x, ), ε2<x<1, >0.
Hence, using (42) and since ˜ua,αis he solu ion o (32), we ha e
∂Va,α
∂ (·, )−Aa,αVa,α(·, ), θ=K∂2˜ua,α
∂τ2−Aa,α˜ua,α(·, ), θ=0
o all θ∈C∞
0(ε2,1)and >0. Consequen ly, since ε2>0 is a bi a y, we each
∂Va,α
∂ (x, )=∂
∂xxαa(x)∂Va,α
∂x(x, )
o almos all x∈(0, 1) and >0. The e o e, (39) holds.
By ˜ua,α(0, )=˜ua,α(1, )=0 o >0, he i s condi ion in (40) is di ec ly seen.
By he change o a iables ξ=τ
2√ ,weha e
Va,α(x, )=∞
0
˜ua,α(x,τ)G( ,τ)dτ=2
√π∞
0
˜ua,α(x,2ξ√ )e−ξ2dξ, >0.
(43)
Since ˜ua,α∈C([0, ∞); C([ε2, 1])), and using ha ∂ ˜ua,α∈L∞(0, ∞;H1(ε2,1))in(35), we
ha e
lim
→0
˜ua,α(x,2ξ√ )=˜ua,α(x,0)
17
In e se P oblems 37 (2021) 125002 P Canna sa e al
o almos all x∈(ε2,1)andξ>0, and by ˜ua,α∈L∞(0, ∞;H1(ε2,1))in(35), we apply he
Lebesgue con e gence heo em o (43) and ob ain
Va,α(x,0)=2
√π∞
0
˜ua,α(x,0)e
−ξ2dξ=˜ua,α(x,0) 2
√π∞
0
e−ξ2dξ=u0(x),
o almos all x∈(ε2, 1), which is he second condi ion in (40). Since ε2>0 is a bi a y, he
p oo o lemma 4.7 is comple e. 
Mo eo e , we p o e
Lemma 4.8. Le us assume ha ∂xua,α(1, )=∂xub,β(1, ) o 0< < 0. Then,
∂x˜ua,α(1, )=∂x˜ub,β(1, ), 0 < <∞.
P oo o lemma 4.8. Since ua,α(x,·) is analy ic in >0 wi h ixed x∈[ε2, 1], by heo em
2.3(b) in sec ion 2, we see ha ∂xua,α(1, )=∂xub,β(1, ) o all >0. The e o e, by (41)we
ha e
∞
0
(∂x˜ua,α−∂x˜ub,β)(1, τ)G( ,τ)dτ=0, >0.(44)
On he o he hand, in he de ini ion (36)o K, we change he a iables in he exp ession o K
by :=1
4 and τ:=√z,andwesee
(Kη)1
4 =
π∞
0
e− z η(√z)
√zdz, >0,
o η∈L∞(0, ∞).
Thus, he uniqueness o he Laplace ans o m implies ha η≡0i (Kη)( )=0 o >0
and η∈L∞(0, ∞). Since (∂x˜ua,α−∂x˜ub,β)(1, ·)∈L∞(0, ∞)by(35), i ollows om (44) ha
∂x˜ua,α(1, )=∂x˜ub,β(1, ) o >0. The e o e, he p oo o lemma 4.8 is comple e. 
4.2.2. Second s ep: uniqueness o he linea ized in e se hype bolic p oblem. He e we es ab-
lish he uniqueness esul o he ollowing hype bolic in e se sou ce p oblem. We ecall ha
a su icien ly small cons an ε1>0 is gi en. Mo eo e le p0>0 be a bi a ily gi en.
Gi en p∈C2([ε1, 1]) wi h p⩾p0>0on[ε1,1],le y=y(x, ) sa is y
⎧
⎪
⎪
⎨
⎪
⎪
⎩
∂2
y=∂x(p(x)∂xy)+∂x( (x)∂xR(x, )), ε1<x<1, >0,
y(1, )=0, >0,
y(x,0)=∂ y(x,0)=0, ε1<x<1,
(45)
whe e R(x, ) is a gi en unc ion.
In wha ollows, we show he uniqueness o an in e se p oblem o de e mining (x) o
ε1<x<1 by bounda y da a ∂xy(1, ) o all >0.
18
In e se P oblems 37 (2021) 125002 P Canna sa e al
P oposi ion 4.9. Le p ∈C2([ε1,1]) wi h p >0on [ε1,1] and le ∈H1(ε1,1)
sa is y (1) =0. We u he assume ha he e exis s a cons an 0>0such ha
⎧
⎪
⎪
⎨
⎪
⎪
⎩
∂ R(x,0)=0, |∂xR(x,0)|⩾ 0>0, ε1⩽x⩽1,
∂ ∂xR,∂ ∂2
xR∈L∞
loc((ε1,1)×(0, ∞)),
∂2
xR(·,0)∈L∞(ε1,1).
(46)
I y ∈H2
loc((ε1,1)×(0, ∞)) sa is ies (45),∂ y∈H2
loc((ε1,1)×(0, ∞)) and
∂xy(1, )=0, >0,
hen (x)=0 o ε1<x<1.
No ice ha in his p oposi ion, as is seen by he p oo , we need no assume ∂xy(1, )=0 o
all >0, bu ∂xy(1, )=0 o 0< <T∗wi h some T∗>0 which is de e mined by p(x), is
su icien . Howe e o ou pu pose we can assume ∂xy(1, )=0 o all >0. The es o his
subsec ion is de o ed o he p oo o p oposi ion 4.9.
P oo o p oposi ion 4.9.The me hodology o he p oo is based on Bukhgeim and
Klibano me hod om [6], and also on he a gumen s by Imanu ilo and Yamamo o om
[21,22]. Recen ly, Huang e al in [17] ha e simpli ied hose a gumen s and omi he cu -o
p ocedu es. He e we apply such a gumen o m [17]. The key idea is ha we ake a sui able
weigh unc ion ha akes smalle alues on he bounda y o a domain in (x, ) whe e da a a e
no gi en, so ha he weigh unc ion can well con ol such unknown bounda y alues and
he e o e he cu -o unc ion is no necessa y.
The p oo will be di ided in o h ee s eps.
Fi s s ep o p oposi ion 4.9: Ca leman es ima es.
Hence o h, wi h a bi a ily chosen cons an s M>0andp0>0, we assume
p∈C2([ε1, 1]), p⩾p0>0on[ε1,1], pC2([ε1,1]) ⩽M.
Fu he mo e, we ix a su icien ly small cons an γ>0 and a su icien ly la ge cons an λ>0.
He e and hence o h, o Mand 0, we choose θ∈(0, 1) su icien ly close o 1 and su icien ly
small δ>0 such ha
ε1<θ+δ<1
is always sa is ied.
We se
ψ(x, ):=(x−θ)2−γ 2,ϕ(x, )=eλψ(x, )
and
Qδ:={(x, ); −∞ < <∞,θ+δ<x<1, ψ(x, )>δ
2},
T0:=(1 −θ)2−δ2
γ.
19
In e se P oblems 37 (2021) 125002 P Canna sa e al
Figu e 1. Se Qδ.
Figu e 2. Se Q−
δ.
Then, we no e ha Qδis bounded by a hype bolic cu e (x−θ)2−γ 2=δ2and x=1, and
(x, )∈Qδimplies θ+δ<x<1and| |<T0,Qδ∩{ =0}=(θ+δ, 1), (47)
see igu e 1.
Hence o h C,C0>0, e c will deno e gene ic cons an s which a e independen o he
pa ame e s>0, bu can be dependen on o he pa ame e s such as λ>0. Since o he
pa ame e s such as λa e ixed, we do no speci y he dependence o C,C0, e c on hem.
We show he ollowing esul :
20

In e se P oblems 37 (2021) 125002 P Canna sa e al
Lemma 4.10 (Hype bolic Ca leman es ima e). The e exis cons an s C >0and le i
be s0>0depending on M and 0,bu independen o s and he choices o p,such ha
Qδ
(s|∂xy|2+s|∂ y|2+s3|y|2)e2sϕdxd
⩽CQδ|F(x, )|2e2sϕdxd +C∂Qδ
(s|∂xy|2+s|∂ y|2+s3|y|2)e2sϕdS
o all s ⩾s0and y ∈H2(Qδ)sa is ying
∂2
y=∂x(p(x)∂xy)+FinQ
δ
wi h some F ∈L2(Qδ).
In pa icula , i y(1, )=∂xy(1, )=0 o −T0< <T0, hen
Qδ
(s|∂xy|2+s|∂ y|2+s3|y|2)e2sϕdxd
⩽CQδ|F(x, )|2e2sϕdxd +Cs3e2seλδ2
y2
H2(Qδ)(48)
o all s ⩾s0and y ∈H2(Qδ).
P oo o lemma 4.10.This is a classical Ca leman es ima e: he p oo can be done simila ly
o heo em 4.2 in Bellassoued and Yamamo o [6] o Imanu ilo [19]. He e we omi he de ails
o he p oo . Es ima e (48) can ollow because ψ(x, )=δ2 o (x, )∈∂Qδ {x=1}and
∂xyL2(∂Qδ)+∂ yL2(∂Qδ)+yL2(∂Qδ)
⩽C(∂xyH1(Qδ)+∂ yH1(Qδ)+yH1(Qδ))⩽CyH2(Qδ)
by he ace heo em. 
Rema k 4.11. Fo Ca leman es ima es o hype bolic equa ions, we usually ha e o assume
a ce ain ex a condi ion o he p incipal pa p(x) e.g., abou he wa e speed, and he ex a
condi ion is es ic i e in highe spa ial dimensions. Howe e , he one spa ial dimension case
is excep ional and does no equi e any ex a condi ion hanks o he smallness o he domain
Qδ. Mo e p ecisely, in he one dimensional case, such an ex a condi ion is desc ibed as

p(x)
2p(x)(x−θ)
<1, θ+δ⩽x⩽1
(see e.g., p oposi ion 2.1 in Imanu ilo and Yamamo o [22]). Since maxθ+δ⩽x⩽1|x−θ|=
1−θis su icien ly small, we see ha

p(x)
2p(x)(x−θ)
⩽M
2 0
(1 −θ)<1.(49)
Hence, he e exis s a cons an μ∗( 0,M)>0such ha i 0<1−θ< μ
∗( 0,M), hen (49)and
so he Ca leman es ima e (48) hold.
21
In e se P oblems 37 (2021) 125002 P Canna sa e al
In iew o (49), he key Ca leman (48) es ima e holds nea x=1 whe e Cauchy da a a e
gi en, ha is, o θ+δ<x<1. Thus, we i s apply he Ca leman es ima e only o θ+
δ<x<1 o p o e ha =0in(θ+δ, 1). This will be done in second s ep o he p oo
o p oposi ion 4.9. Nex , we epea he same a gumen by eplacing x=1byx=θ, and his
will be done in hi d s ep o he p oo o p oposi ion 4.9.
Be o e p oceeding o second s ep, we p o e a Ca leman es ima e o a simple ope a o d
dx.
We se
ϕ0(x)=eλ(x−θ)2
wi h ixed la ge cons an λ>0.
Lemma 4.12. The e exis cons an s s0>0and C >0such ha
s1
θ+δ| (x)|2e2sϕ0(x)dx⩽C1
θ+δ| |2e2sϕ0dx,
o all s ⩾s0and all ∈H1(θ+δ,1)sa is ying (1) =0.
P oo o lemma 4.12.The p oo elies on he es ima ion o h:= esϕ0and he in eg a ion
by pa s, which is a con en ional p oo o he Ca leman es ima e.
We se
h(x):= (x)esϕ0(x),Lh(x):=esϕ0(x)d
dx(e−sϕ0(x)h(x)) =h(x)−sϕ
0(x)h(x).
Then, in eg a ing by pa s and aking in o accoun ha (1) =0, we ha e ha he e exis s
a posi i e cons an Csuch ha
1
θ+δ| |2e2sϕ0dx=1
θ+δ|Lh|2dx=1
θ+δ|h(x)−sϕ
0(x)h(x)|2dx
⩾−2s1
θ+δ
ϕ
0(x)h(x)h(x)dx=−s1
θ+δ
ϕ
0(x)d
dxh2(x)dx
=sϕ
0(x)h2(x)x=θ+δ
x=1+s1
θ+δ
ϕ
0(x)h2(x)dx
=2λsδeλδ2h2(θ+δ)+s1
θ+δ
(2λϕ0(x)
+4λ2(x−θ)2ϕ0(x)h2(x)dx
⩾s1
θ+δ
2λeλ(x−θ)2h2(x)dx⩾Cs1
θ+δ| |2e2sϕ0dx.
Thus, he p oo o lemma 4.12 is comple e. 
Second s ep o he p oo o p oposi ion 4.9.
22
In e se P oblems 37 (2021) 125002 P Canna sa e al
We ake he e en ex ension in o he solu ion yo (45) o(−∞, 0), and we deno e i by he
same le e :
y(x, )=y(x, ), ⩾0,
y(x,− ), <0.
Also, o R,wemake hee enex ensionin <0. Then, using ha y(·,0)=∂ y(·,0)=0and
∂ R(·,0)=0on[ε1,1]by(46), we can e i y ha
⎧
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎩
∂ y,y∈H2
loc((ε1,1)×(−∞,∞)),
∂ R,R∈L2
loc(−∞,∞;H2(ε1, 1)),
∂ ∂xR,∂ ∂2
xR∈L∞
loc((ε1,1)×(−∞,∞))
(50)
and
⎧
⎪
⎪
⎨
⎪
⎪
⎩
∂2
y=∂x(p(x)∂xy)+∂x( (x)∂xR(x, )), ε1<x<1, −∞ < <∞,
y(1, )=∂xy(1, )=0, −∞ < <∞,
y(x,0)=∂ y(x,0)=0, ε1<x<1.
(51)
In (51), we see ha y,∂ y∈H2((ε1,1)×(−T,T)) wi h a bi a ily gi en T>0. Se ing
z:=∂ y∈H2((ε1,1)×(−T,T)),
and using (51), we ha e
⎧
⎪
⎪
⎨
⎪
⎪
⎩
∂2
z=∂x(p(x)∂xz)+∂x( (x)∂ ∂xR(x, )), ε1<x<1, −∞ < <∞,
z(1, )=∂xz(1, )=0, −∞ < <∞,
z(x,0)=0, ∂ z(x,0)=∂x( (x)∂xR(x, 0)), ε1<x<1.
(52)
No ice ha he second and he hi d condi ions in (52) a e ob ained by (51) aking in o accoun
ha ∂2
y∈H1(−T,T;L2(ε1, 1)) and he Sobole embedding such as H1(−T,T;L2(ε1,1))⊂
C0([−T,T]; L2(ε1, 1)). He e, we also no e ha ∂ ∂2
xy,∂ ∂2
xR∈L2(−T,T;L2(ε1, 1)), and so
∂2
xy,∂2
xR∈C0([−T,T]; L2(ε1, 1)) by he Sobole embedding.
Se ing
M0:=zH2((ε1,1)×(−T,T)) =∂ yH2((ε1,1)×(−T,T)),
by he egula i y assump ion (50)ony, we see ha M0<∞.
Now, we apply he Ca leman es ima e (48) o(52) and we ob ain ha he e exis s a cons an
C>0, which is independen o s>0, such ha
23
In e se P oblems 37 (2021) 125002 P Canna sa e al
Qδ
(s|∂xz|2+s|∂ z|2+s3|z|2)e2sϕdxd
⩽CQδ|∂x( (x)∂x∂ R(x, )|2e2sϕdxd +Cs3e2seλδ2
M2
0, (53)
o all la ge s>0.
We se
w:=(∂ y)esϕ=zesϕ.
Then, di ec calcula ions yield
∂2
w=(∂3
y)esϕ+2s(∂2
y)(∂ ϕ)esϕ+(s∂2
ϕ+s2(∂ ϕ)2)) esϕ∂ y.
On he o he hand,
p∂2
xw=p(∂2
x∂ y)esϕ+2ps(∂ ∂xy)(∂xϕ)esϕ+p(s(∂2
xϕ)+s2(∂xϕ)2)) esϕ∂ y,
and also
p∂xw=p(∂x∂ y)esϕ+sp(∂xϕ)(∂ y)esϕ.
The e o e, aking he de i a i e wi h espec o in (45) and mul iplying by esϕ, we can w i e
∂2
w−∂x(p(x)∂xw)=esϕ∂x( (x)∂x∂ R(x, )) +Ps(w), (54)
whe e he e m Ps(w) sa is ies
|Ps(w)|⩽Cs(|∂2
y|+|∂ y|+|∂x∂ y|)esϕ+Cs2|∂ y|esϕ
⩽C(s2|w|+s|∂ w|+s|∂xw|), (x, )∈Qδ, (55)
o all la ge s>0. Fo he las inequali y, we ha e used ha w=(∂ y)esϕand
(∂2
y)esϕ=∂ w−s(∂ ϕ)(∂ y)esϕ=∂ w−s(∂ ϕ)w,
and
(∂x∂ y)esϕ=∂xw−s(∂xϕ)w,
which imply
|∂2
y|esϕ⩽|∂ w|+Cs|∂ y|esϕ
and
|∂x∂ y|esϕ⩽|∂xw|+Cs|∂ y|esϕ,
espec i ely.
On he o he hand, i is no di icul o see ha we can w i e (53)in e mso w=zesϕ=
(∂ y)esϕ
Qδ
(s|∂xw|2+s|∂ w|2+s3|w|2)dxd
⩽CQδ
(∂x( (x)∂x∂ R(x, ))|2e2sϕdxd +Cs3e2seλδ2
M2
0, (56)
24
In e se P oblems 37 (2021) 125002 P Canna sa e al
5. Nume ical esul s
In his sec ion, we will conside nume ical esul s o he p e ious in e se p oblems. We will
ca y ou he econs uc ion o he unknown unc ion aand he powe α h ough he esolu-
ion o some app op ia e ex emal p oblems. This s a egy has been applied in some p e ious
pape s o o he simila p oblems, see [2,3,13,26,27]. The esul s o he nume ical es s ha
ollow p e end o illus a e he heo e ical esul s in he p e ious sec ions. We s a by showing
nume ical expe imen s ela ed o he heo e ical esul o heo em 3.2 om sec ion 3.1, hen
we conside o he ype o obse a ions as indica ed in ema k 3.1 (see sec ions 5.1 and 5.2).
Simila ly, we con inue wi h he heo e ical esul om sec ion 3.2 (see sec ions 5.3 and 5.4).
Finally, we conclude wi h nume ical econs uc ion ela ed o sec ion 4(see sec ion 5.5).
5.1. Nume ical es s o he linea case I
In his sec ion we p esen he nume ical econs uc ion o he cons an coe icien ain (10)by
dis ibu ed measu emen s, ela ed o he heo e ical esul om heo em 3.2. Mo e p ecisely,
we will deal wi h he ollowing in e se p oblem: gi en 0∈(0, T)asin(12), u0=u0(x), γ=
∂ u(·, 0)andβ=u(·, 0), ind a∈Ua
ad such ha he co esponding solu ion uao (10) sa is ies
he addi ional condi ions:
∂ ua(x, 0)=γ(x), x∂xua(x, 0)=β(x), ∀x∈(0, ),
and he admissible se is gi en by Ua
ad ={a:0<a⩽a⩽a}.
We e o mula e he in e se p oblem as he ollowing ex emal p oblem: ind a∈Ua
ad such
ha
J(a)⩽J(a), ∀a∈Ua
ad, (66)
whe e he unc ional J:a∈Ua
ad → Ris gi en by
J(a)=1
2
0|γ(x)−∂ ua(x, 0)|2dx+1
2
0|β(x)−x∂xua(x, 0)|2dx, (67)
wi h ua he solu ion o (10) co esponding o he unknown a.
We ha e pe o med se e al nume ical es s wi h di e en alues o a o eco e : a>1,
a=1anda<1. In hem, we ix T=5, =1andu0(x)=0.5x2(1 −x). In he i s h ee
es s we will ake, o simplici y, (x, )=0. We will deno e by ad he desi ed alue o a ha
we would like o ob ain and by ai he ini ial guess o he op imiza ion algo i hm. Using an
op imiza ion algo i hm om Op imiza ion Toolbox o Ma Lab, he mincon unc ion and
in e io -poin g adien based algo i hm, we ha e compu ed ac. Ou goal is o ob ain ac
as close as possible o ad.
Tes 1. Le us ake ai=0.7as an ini ial guess o he minimiza ion algo i hm and
0=0.2, so ha we ha e compu ed nume ically such ha (12)holds. The desi ed alue o
eco e is ad=1.7. A e nume ical esolu ion o he op imiza ion p oblem (66), we ound
ac=1.7. The nume ical esul s can be seen in igu e 3, whe e he desi ed adand compu ed ac
ha e been ep esen ed. In igu e 4we ha e plo ed he e olu ion o he cos a each i e a ions
o he op imiza ion algo i hm ep esen ed by poin s. In able 1we can see he e olu ion o he
cos when we in oduce andom noises in he a ge .
31

In e se P oblems 37 (2021) 125002 P Canna sa e al
Rema k 5.1 (Nume ical illus a ion o s abili y). In o de o illus a e he s abili y o
he conside ed in e se p oblem in he es 1, we ha e ep esen ed in igu e 5as poin s he
quo ien s
Kj=|a−a∗|
γ(x)−γ∗
j(x)+β(x)−β∗
j(x),j=1, ..., 50, (68)
whe e γ(x)=∂ ua(x, 0), β(x)=x∂xua(x, 0)and o all j,γ∗
j(x)=γ(x)+εj,β∗
j(x)=β(x)+
εjwi h he εj andom pe u ba ions in (0, 0.05). We can see ha all he quo ien s Kja e
bounded.
Tes 2. Le us now ake ai=0.2, 0=0.2and he desi ed alue o eco e ad=1.The
op imiza ion algo i hm gi es ac=1, see igu e 6. In able 2we can see he e olu ion o he
cos when we in oduce andom noises in he a ge .
Tes 3. Le us ake ai=0.7as an ini ial guess o he minimiza ion algo i hm, 0=0.2and
ad=0.2. The nume ical esolu ion o he op imiza ion p oblem gi es ac=0.2(see igu e 7
and able 3).
Tes 4. We conside a non-homogeneous p oblem (10)wi h (x, )=2x . Le us ake ai=
0.7and 0=0.4. The desi ed alue is ad=1.7. The nume ical esolu ion o he op imiza ion
p oblem gi es ac=1.7. Figu e 8shows he e olu ion o i e a ions o he op imiza ion algo i hm
and igu e 9 he e olu ion o he cos . In able 4we p esen he esul s wi h andom noises in
he a ge and igu e 10 ep esen s he solu ion co esponding o he compu ed ac.
5.2. Nume ical es s o he linea case II
In his sec ion we will p esen some nume ical esul s o second in e se p oblem o iden i ica-
ion o a om one bounda y obse a ion ux(1, ), men ioned a he beginning o he sec ion 3.1.
A guing as in he p e ious sec ion, we e o mula e he in e se p oblem as an op imiza ion
p oblem (66), whe e he uncional J:a∈Ua
ad → Ris now gi en by
J(a)=1
2T
0|η( )−∂xua(1, )|2d , (69)
whe e η=∂xu(1, ·) is gi en and uais he solu ion o (10) co esponding o he unknown a.
In wha ollows we p esen some nume ical es s o his in e se p oblem. We will conside
he same da a as in he p e ious sec ion: T=5, =1, u0(x)=0.5x2(1 −x)and (x, )=0. In
o de o simpli y he p esen a ion, we will show jus he e olu ion o he cos and he esul s
wi h andom noises.
Tes 5. We ake ai=0.7as an ini ial guess o he minimiza ion algo i hm. The desi ed
alue is ad=1.7. The esul s o he nume ical compu a ions a e p esen ed in able 5and
igu e 11.
Tes 6. Taking ai=0.2and ad=1, we can see he nume ical esul s in igu e 12 and able 6.
Tes 7. Taking ai=0.7and ad=0.2, igu e 13 and able 7show he nume ical esul s.
32
In e se P oblems 37 (2021) 125002 P Canna sa e al
5.3. Nume ical es s o he powe -like case I
We will p esen in his sec ion some nume ical esul s o he econs uc ion o α o weak
and s ong degene a e cases (16)and(17) om da a ha appea in heo em 3.8.Mo ep e-
cisely, gi en 0∈(0, T)asin(20), u0=u0(x)and<1, in o de o econs uc αwe will
sol e nume ically he ollowing op imiza ion p oblem: ind α∈(0, 2) such ha
J(α)⩽J(α), ∀α∈(0, 2), (70)
whe e he unc ional J: (0, 2) → Ris gi en by
J(α)=1
21
0|γ(x)−∂ uα(x, 0)|2dx+1
21
0|β(x)−x2∂xuα(x, 0)|2dx,
wi h uα he co esponding solu ion o (16)o (17) associa ed o o he unknown powe α.
In he nume ical es s we will ake T=10, u0=0.3x2(1 −x)2and (x, )=0 and we will
use he in e io -poin Ma Lab g adien ee algo i hm om Op imiza ion Toolbox.
5.3.1. Weak degene acy I.
Tes 8. Le us i s assume ha 0<α<1, co esponding o he weak degene acy case (16).
We ake αi=0.8as an ini ial guess o he minimiza ion algo i hm and he desi ed alue o
eco e is αd=0.4. In igu e 14 we can see he compu ed alues o αa each i e a ion o he
op imiza ion algo i hm and igu e 15 shows he e olu ion o he cos . In he able 8,wep esen
he compu ed alues o αcdepending on he alues o close o 1.
5.3.2. S ong degene acy I.
Tes 9. Le us conside he s ong degene acy case wi h 1⩽α<2, i.e. he p oblem (17).
We ake αi=1.6and αd=1.3. The nume ical esul s can be seen in he igu es 16 and 17.In
he able 9, we p esen he compu ed alues o αcdepending on he alues o close o 1.
Rema k 5.2. F om he esul s o he es s 8and 9we can see nume ically ha he hypo hesis
<1 om heo em 3.8 p obably is no eally needed o he econs uc ion o α.Bu he
ques ion whe he we can elimina e his condi ion o m his heo em emains open.
5.4. Nume ical es s o he powe -like case II
We will p esen in his sec ion nume ical esul s o he econs uc ion o α om one obse -
a ion o he o m ∂xu(1, ). As be o e, we will analyze i s weak and hen s ong degene a e
cases. Mo e p ecisely, gi en ,u0=u0(x)andη=∂xu(1, ·), we will sol e nume ically he
ollowing op imiza ion p oblem (70), wi h he unc ional J: (0, 2) → Rgi en by
J(α)=1
2T
0|η( )−∂xuα(1, )|2d ,
whe e uαis he solu ion o (16)o (17) co esponding o α.
In ha ollows, we will ake T=10, =1, u0=0.3x2(1 −x)2and (x, )=0. The compu-
a ions a e pe o med using mincon Ma Lab unc ion o mOp imiza ion Toolbox,combined
wi h in e io -poin g adien algo i hm.
33
In e se P oblems 37 (2021) 125002 P Canna sa e al
5.4.1. Weak degene acy II.
Tes 10. Le us i s conside he weak degene acy case co esponding o he p oblem (16)
wi h 0<α<1.We akeαi=0.2as an ini ial guess o he minimiza ion algo i hm and he
desi ed alue o eco e αd=0.6. The nume ical esul s can be seen in igu es 18 and 19.
Table 10 shows he compu a ions o αcand he cos wi h andom noises in he a ge .
Rema k 5.3 (On he o de o con e gence). We can analyze he o de o con e gence
o αnwhich we ob ain a each i e a ion no he op imiza ion algo i hm o αd. Mo e p ecisely,
we look o posi i e eal numbe s Cand κsuch ha
|en+1|≈C|en|κ,
whe e en=αn−αd. F om igu e 20, we see ha his app oxima e o mula is sa is ied o
κ=1.8064 o some ini e alue o C.
5.4.2. S ong degene acy II. We will conside he e he iden i ica ion o αin he case 1 ⩽
α<2 o s ong degene acy case, ha is o say o he p oblem (17). Fo simplici y o he
p esen a ion, he esul s wi h andom noises in he a ge s a e no p esen ed, because hey a e
simila o hose o he p e ious sec ions.
Tes 11. Le us ake =0.9and αi=1.6as an ini ial guess o he minimiza ion algo i hm.
The desi ed alue o α o eco e is αd=1.3. The nume ical esul s can be seen in he
igu es 21 and 22 and he compu ed alue o αis αc=1.3.
Tes 12. Le us ake =1and αi=1.6and αd=1.3. The nume ical esul s can be seen
in igu es 23 and 24 and he compu ed alue is αc=1.3.
5.5. Nume ical es s o a gene al case
In his sec ion we will p esen nume ical esul s o iden i ica ion o a unc ion a=a(x)in(25)
om ∂xu(1, ) in o de o illus a e he heo e ical pa o m he sec ion 4.
Le us ake =1, T=5, α=0.6andu0(x)=0.3x2(1 −x)2. In o de o econs uc a=
a(x), again we e o mula e he co esponding in e se p oblem as an op imiza ion p oblem:
ind a∈Asuch ha
J(a)⩽J(a), ∀a∈A, (71)
whe e he unc ional J:A → Ris gi en by
J(a)=1
2T
0|η( )−∂xua(1, )|2d ,
whe e uais he solu ion o (25) co esponding o aand Ais gi en by (27). In ac , o he
nume ical econs uc ion we do no need o assume ha ais known nea 1 as in (27).
In o de o p esen some examples, in wha ollows we will analyze wo pa icula cases
o he coe icien a(x): linea and quad a ic unc ions. The nume ical expe imen s will be
pe o med using a g adien based in e io -poin algo i hm o mincon Ma Lab
Op imiza ion Toolbox.
34
In e se P oblems 37 (2021) 125002 P Canna sa e al
Figu e 3. Tes 1,adsolid line, acdashed line.
Le us no ice ha we can also econs uc nume ically bo h αand a om one measu emen ,
bu in o de o simpli y he p esen a ions o he esul s, we will omi he de ails.
Linea unc ion: le us conside ao he o m a(x)=bx +c,whe eband ca e unknown
cons an s. The op imiza ion p oblem (71) is w i en in his case as ollows: ind b,c∈A, wi h
A={(b,c):0<b<b<b,0<c<c<c}such ha
J(b,c)⩽J(b,c), ∀(b,c)∈A.
Tes 13. Le us ake bi=1and ci=1as ini ial guess o he op imiza ion algo i hm and
desi ed alues o eco e bd=5and cd=1.5. A e nume ical app oxima ion o he op imiza-
ion p oblem abo e we ge he compu ed alues bc=5and cc=1.5. The nume ical esul s
we ha e ob ained a e in igu es 25 and 26. Table 11 shows he e olu ion o he cos du ing
he i e a ions o op imiza ion algo i hm and compu ed bcand ccwi h andom noises in he
a ge . Finally, in igu e 27 he solu ion o (25)is ep esen ed o he compu ed alues o he
coe icien s.
Quad a ic unc ion: le us now conside ao he o m a(x)=bx2+cx +h. The op imiza ion
p oblem (71) is w i en in his case as ollows: ind (b,c,h)∈A, wi h A={(b,c,h):0<b<
b<b,0<c<c<c,0<h<h<h}such ha
J(b,c,h)⩽J(b,c,h), ∀(b,c,h)∈A.
35
In e se P oblems 37 (2021) 125002 P Canna sa e al
Figu e 4. Tes 1, he e olu ion o he cos .
Table 1. Cos a he bes s opping and compu ed acwi h andom noises
in he a ge in es 1.
% noise Cos I e a ions ac
5% 1 ×10−517 1.700 831 33
3% 1 ×10−515 1.69 463 291
1% 1 ×10−717 1.704 511 53
0.1% 1 ×10−816 1.700 178 91
0.01% 1 ×10−10 16 1.700 176 65
0.001% 1 ×10−13 15 1.700 003 62
0% 1 ×10−27 20 1.7
Tes 14. Le us ake bi=3.5, ci=2.5and hi=0.5as ini ial guess o he op imiza ion
algo i hm and he desi ed alues o eco e bd=4, cd=3and hd=1. The nume ical esul s
a e shown in he igu es 28 and 29, and he compu ed alues a e bc=4.065 621 97, cc=
2.886 136 22 and hc=1.049 674 27. The nume ical esul s can be seen in igu es 28 and 29.
Table 12 shows he e olu ion o he cos du ing he i e a ions o op imiza ion algo i hm and
compu ed bcand ccwi h andom noises in he a ge .
36

In e se P oblems 37 (2021) 125002 P Canna sa e al
Figu e 5. Tes 1, nume ical illus a ion o s abili y.
37
In e se P oblems 37 (2021) 125002 P Canna sa e al
Figu e 6. Tes 2, he e olu ion o he cos .
Table 2. E olu ion o he cos and compu ed acwi h andom noises in
he a ge in es 2.
% noise Cos I e a ions ac
5% 1 ×10−512 0.975 793 57
3% 1 ×10−613 0.996 128 53
1% 1 ×10−612 0.996 702 60
0.1% 1 ×10−813 1.000 127 71
0.01% 1 ×10−14 10 0.999 951 71
0.001% 1 ×10−13 14 0.999 994 88
0% 1 ×10−24 14 1
38
In e se P oblems 37 (2021) 125002 P Canna sa e al
Figu e 7. Tes 3, he e olu ion o he cos .
Table 3. E olu ion o he cos and compu ed acwi h andom noises in
he a ge in es 3.
% noise Cos I e a ions ac
5% 1 ×10−511 0.198 499 99
3% 1 ×10−611 0.197 690 06
1% 1 ×10−711 0.200 461 40
0.1% 1 ×10−13 9 0.199 976 76
0.01% 1 ×10−11 16 0.199 998 93
0.001% 1 ×10−13 14 0.200 000 28
0% 1 ×10−27 16 0.2
39
In e se P oblems 37 (2021) 125002 P Canna sa e al
Figu e 8. Tes 4, i e a ions o he algo i hm.
Figu e 9. Tes 4, he e olu ion o he cos .
40
In e se P oblems 37 (2021) 125002 P Canna sa e al
Table 9. Compu ed αc o close o1in es 9.
Cos I e a ions αcCos I e a ions αc
0.9 1 ×10−24 18 1.3 1.000 01 1 ×10−22 23 1.300 000 01
0.99 1 ×10−22 24 1.300 000 01 1.0001 1 ×10−23 21 1.3
0.999 1 ×10−21 19 1.300 000 06 1.001 1 ×10−23 25 1.3
0.9999 1 ×10−23 22 1.299 999 99 1.01 1 ×10−24 32 1.3
0.999 99 1 ×10−22 30 1.300 000 01 1.1 1 ×10−22 18 1.299 999 98
11×10−23 27 1.300 000 00 1.2 1 ×10−22 20 1.3
Figu e 18. Tes 10, iden i ica ion o α.
47

In e se P oblems 37 (2021) 125002 P Canna sa e al
Figu e 19. Tes 10, cu en unc ion alues.
Table 10. E olu ion o he cos and compu ed αcwi h andom noises in he a ge in
es 10.
% noise Cos I e a ions αc
5% 1 ×10−714 0.594 706 02
3% 1 ×10−715 0.616 112 17
1% 1 ×10−815 0.596 894 97
0.1% 1 ×10−10 16 0.600 371 82
0.01% 1 ×10−12 16 0.599 976 72
0.001% 1 ×10−14 18 0.599 998 07
0% 1 ×10−21 17 0.6
48
In e se P oblems 37 (2021) 125002 P Canna sa e al
Figu e 20. Tes 10, o de o con e gence.
Figu e 21. Tes 11, iden i ica ion o α.
49
In e se P oblems 37 (2021) 125002 P Canna sa e al
Figu e 22. Tes 11, unc ion alues.
Figu e 23. Tes 12, iden i ica ion o α.
50
In e se P oblems 37 (2021) 125002 P Canna sa e al
Figu e 24. Tes 12, unc ion alues.
Figu e 25. Tes 13,adsolid line, acdashed line.
51
In e se P oblems 37 (2021) 125002 P Canna sa e al
Figu e 26. Tes 13, e olu ion o he cos .
Table 11. E olu ion o he cos and compu ed bcand ccwi h andom noises in he a ge
in es 13.
%Cos I .bccc
51×10−627 4.991 568 07 1.415 068 91
31×10−729 5.070 602 96 1.405 216 65
11×10−829 5.022 230 88 1.474 061 14
0.1 1 ×10−10 29 5.006 369 35 1.495 233 26
0.01 1 ×10−12 29 4.999 897 52 1.500 322 99
0.001 1 ×10−14 29 4.999 940 89 1.500 119 71
01×10−25 39 5 1.5
52

In e se P oblems 37 (2021) 125002 P Canna sa e al
Figu e 27. Tes 13, compu ed solu ion.
Figu e 28. Tes 14,adsolid, acdashed lines.
53
In e se P oblems 37 (2021) 125002 P Canna sa e al
Figu e 29. Tes 14, e olu ion o he cos .
Table 12. E olu ion o he cos and compu ed bcand ccwi h andom noises in he a ge
in es 14.
% Cos I e a ions bccchc
5% 1 ×10−632 4.100 796 95 2.901 687 36 0.991 736 73
3% 1 ×10−734 4.037 814 43 2.866 748 57 1.020 339 48
1% 1 ×10−841 4.059 712 41 2.886 552 92 1.091 051 66
0.1% 1 ×10−10 33 4.068 066 62 2.886 854 41 1.042 748 87
0.01% 1 ×10−12 34 4.065 653 61 2.886 231 99 1.050 383 51
0.001% 1 ×10−12 35 4.065 676 62 2.886 153 26 1.049 531 32
0% 1 ×10−12 33 4.065 621 97 2.886 136 22 1.049 674 27
Acknowledgmen s
The i s au ho was pa ly suppo ed by Is i u o Nazionale di Al a Ma ema ica (GNAMPA 2020
Resea ch P ojec s) and by he MIUR Excellence Depa men P ojec awa ded o he Depa -
men o Ma hema ics, Uni e si y o Rome To Ve ga a, CUP E83C18000100006. The second
au ho was pa ially suppo ed by MICINN, unde G an MTM2016-76690-P.The hi d au ho
was suppo ed by G an -in-Aid o Scien i ic Resea ch (S) 15H05740 and G an -in Aid (A)
20H00117 o Japan Socie y o he P omo ion o Science and by he Na ional Na u al Science
54
In e se P oblems 37 (2021) 125002 P Canna sa e al
Founda ion o China (No. 11771270, 91730303). This pape has been suppo ed by he RUDN
Uni e si y S a egic Academic Leade ship P og am.
Da a a ailabili y s a emen
All da a ha suppo he indings o his s udy a e included wi hin he a icle (and any
supplemen a y iles).
ORCID iDs
Anna Doubo a h ps://o cid.o g/0000-0002-5708-2700
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