A NNALI DELLA
S CUOLA N ORMALE S UPERIORE DI P ISA Classe di Scienze
C ARLO D OMENICO P AGANI
On an initial-boundary value problem for the equation w
t= w
xx− xw
yAnnali della Scuola Normale Superiore di Pisa, Classe di Scienze 4
esérie, tome 2, n
o2 (1975), p. 219-263
<http://www.numdam.org/item?id=ASNSP_1975_4_2_2_219_0>
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for the equation Wt = Wxx2014 xWy (*).
CARLO DOMENICO PAGANI
(**)
Summary. -
Thefollowing problem
is considered : tofind
afunction
’tv (x, y, t) in theregion
Q ----{O
t T, y > 0, - oo x +oo}
which is a solutionof
the equa-tion nit = wxx - xw.,, with the
boundary
conditions : w(x, y, 0) =f (x,
y), wheref
isa given
function defined for
- oo x + 00, y >- 0 and w(x, 0, t) = 0for
x > 0, 0 t T. We prove the existence anduniqueness of
a solution w, which is a con-tinuously differentiable function of
t, with values in a certain Banach space. The2cniqueness
assertionfollows from
a suitable apriori
estimateof
the solution; theexistence theorenz is
proved by
meansof
thesemigroup theory,
as anapplication of
the Hille-Yosida theorem. Then the main partof
the paper contains thestudy of
thestationary equation :
u~~ - xu, - ku =f (k
> 0). This is af orward-baek-
ward
parabolic equation,
and it is studied via aWiener-Hopf technique.
Thisprocedure requires
thestudy of
the associatedparabolic equation of
evolution:uxx-Bxluy-ku = f (k >
0) andof
a certainintegral equation of
aWiener-Hop f
type.1. - Introduction.
In this paper a
boundary problem
for thesimple-looking equation
is considered in the
region Q - {O
Ct T, y
>0,
- oo x+
the solu-tion w satisfies the initial condition
w(x,
y,0)
=y)
and theboundary
condition
w(x, 0, t)
=h(x, t) only
for x > 0(and
every 0 tT).
(*)
This research waspartly developed
at theUniversity
of California atBerkeley
(supportedby
ascholarschip
of the italianC.N.R.)
andpartly
at the Politecnico di Milano(Italy) (supported by
the « contratto di ricerca sulleEquazioni
Funzio-nali >> from the C.N.R.
(**)
Istituto di Matematica del Politecnico - Via Bonardi, 9 - Milano(Italy).
Pervenuto alla Redazione il 4 Dicembre 1974.
Such a
problem
arises whenstudying
arandomly
acceleratedparticle moving
on thehalf-line y >
0(see [1]
for theanalogous stationary problem);
then eq.
(1.1)
is the backwardequation
for the vector Markov process(r(t), v(t)),
wherer(t)
is theposition
of theparticle
at timet, v(t)
is itsvelocity.
We can notice also that eq.
(1.1)
is a linearized model of the(onedimen- sional)
kineticequation,
when the collision term issimply approximated by
the differentialoperator 02/0X2.
In this case the function w has themeaning
of the(deviation
from theequilibrium
ofthe)
distribution func- tion of the molecules at thetime t, position y,
andvelocity
x; the gas fills thehalf-space y
>0,
has the initial distributionf,
and has a distribution hon the
boundary y = 0,
but this distribution isassigned only
for theemerging
molecules(x > 0 ) .
From a mathematical
point
ofwiew, ( 1. ~ )
is anequation
of the kindwhere rank Such
equations (sometime
called in the literatureultraparabolic)
have been consideredespecially
in connection with theCauchy problem:
see, e.g.,Kolmogorov [4],
Weber[13],
Il’in[5],
Oleinik[6]
and others.
Conversely,
y there are few papersdealing
with mixed(initial
and
boundary) problems:
see[2], [11 ] ;
see also[7]
recalled below. On the otherhand,
we note thatboundary problems
for the so-calledelliptic-parabolic equations (i.e. , (second order) equations
withnonnegative
characteristicform)
which areobviously
a wider class then(1.2),
have beenlargely
con-sidered
by
various authors(see
the book[7]
for a detailedbibliography):
in
particular
Kohn andNirenberg [3] give
an existence anduniqueness
theorem for the first
boundary
valueproblem
in Sobolev spaces; but some restrictionsimposed
to the boundaries make their resultsinapplicable,
e.g., to the
problem
outlined above for eq.(1.1),
as well as for the mostinteresting problems arising
in thetheory
of brownian motion or in kinetictheory.
When
discussing
eq.(1.1),
our aim is to find solutionsw(t)
which arecontinuously
differentiable functionsof t,
with values in a certain Banach space; w, as a vector in this space, hasgeneralized
derivatives wx, and w,,, which are squareintegrable
with suitableweights.
In such aclass,
auniqueness
theorem will beeasily proved:
it will follow from an apriori
estimate of the solution. An existence theorem will be
proved by
means ofthe
semigroup theory,
as anapplication
of the Hille-Yosida theorem.Thus,
we have to discuss first the
stationary equation,
This is a forward-backward
parabolic equation,
and we will treat itby
the same
technique already
used in[10] (see
also[8]),
which involves the discussion of anintegral equation
of theWiener-Hopf type.
In Sect. 2 we list some functional spaces and state some trace theorems.
In Sect. 3 we discuss an evolution
equation
associated to(1.3),
i.e.The results obtained in this section are
preliminary
to the discussion of eq.(1.3)
in Sect. 4.Finally,
in Sect.5,
the resultspreviously
obtained areapplied
to thestudy
of eq.(1.1).
2. - Function spaces and trace theorems.
Let us define first some functional spaces used in the
following.
B2 willdenote the whole
plane
and(when
thisplane
is referred to anorthogonal
cartesian
system) ~+
is thehalf-plane Y+
thehalf-plane
.R denotes the real
line, R+(_~
thepositive (negative)
half-line.denotes,
asusual,
the space ofinfinitely
differentiablefunctions,
defined on
R2,
withcompact support.
where G is an open subset ofR2,
is the space of the restrictions to G(the
closure ofG)
of functionsbelonging
toC-(R2).
Z(G)
is the set ofcomplex-valued functions u,
defined onG,
such thatuxxg XUy
(the
derivatives of u are taken in the sense ofdistributions) belong
toL2(G) .
We can make two
simple
remarks :i) Z(G), equipped
with the norm,is a
(complete)
Banach space:ii)
is dense inZ(G), provided
G is suf-ficiently smooth;
the cases of main interest for us will be G =X+
or G =Y+ . W( Y+)
is a subset ofZ( Y+) ;
it is definedby
the functions u with theproperties
W+(Y+)
isquite analogous
to theprevious
one; it is definedby
func-tions such that
For
w’( Y+)
andW+( Y+)
we can make remarksanalogous
to those made forZ,
i.e.:i) W(Y+)[~W"+(Y+)]
is a(complete)
Banach space(equipped
withthe norm
[and
ananalogous
definition for||.||W+(Y+)])
andii) C- -
is dense inW(Y+){W+(Y+)]
We will use also the notation
Y+)
for the set of functions defined in the firstquadrant
ofbelonging
toY+),
with the sameproperty ii)
in the definition ofW+( Y+) .
We will consider the usual Sobolev spaces
(I
is an open subset ofR ; s real)
and the spacethe set of functions cp, defined on
I,
such that thefollowing
norm
1
is finite.
REMARK. Let us
point
out thatc L2(R+);
the assertion may be deduced from thefollowing inequality
which may be
proved by integration by parts. Similarly,
we have alsoNow we can state the
following
theorems(trace theorems)
which charac- terize the traces of functions u of class Z or W on the coordina,te axes.THEOREM 2.1. The
following operator
is a
densely defined
boundedoperator f rom Z(X+)
intoHI/2(R).
Letbe its
(uniquely de f ined)
linear bounded extension. Then the rangeof
Y1 is-exactly H1/2(R)
and Yl is THEOREM 2.2. Theoperator
is a
densely defined
boundedoperator from 2(X..~)
intoH1/6(R).
..Letbe its
(uniquely de f ined)
linear bounded extension. Then the rangeo f
Y2 isexactly H1/6(R)
and Y2 is anisomorphism
betweenZ(X,)/(Kernel Y2)
andHl~g(1~).
The boundedness of
(2.4)
and(2.5)
will follow from Lemma 2.3 below.The other assertion
(about
theinvertibility)
will follow from Ths. 3.2 and 3.3 of Sec. 3.LIEMMA 2.1. Let v be a
complex-valued function belonging
toC2(R+)
andsuch that xv and
v" belong
toL2(R+) .
Then thefollowing inequality
holdswhere the constant
01 (which
is the bestpossible)
isgiven by formula (2.13)
below.
PROOF. The
proof
of this lemma can be achievedby
a standardprocedure.
We
give
hereonly
few accounts of the calculations.Let v be real-valued
(for convenience).
If one is not interested infinding
the best
possible constant,
aninequality
like(2.6)
can be showeddirectly by putting g
= andrepresenting v
in the formhere v, is the solution of the
homogeneous equation v1-xv1=
0 such thatxvl E
LZ(.1~+)
and so normalized thatri(0)
=1 : weget
Hg
is the solution of thenon-homogeneous equation vanishing
at theorigin:
.g is the
integral operator
whose kernel isgiven by
-change x with y
to obtain theexpression
ofH(x, y)
for x > y.Iv(z)
andK,,(z)
are the modified Bessel functions of order v.By
Schwartzinequality
one
easily
proves thatwhere
11
IIL’(R+ :x2)
meansand
(2.8)
we haveThen,
from therepresentation (2.7)
thus we
get,
sinceFrom
(2.9)
one obtains aninequality
like(2.6)
via standardarguments
of dimensional
analysis.
In order to
get
the bestpossible
constantappearing
in(2.9)
one hasto look for the maximum of the functional
This
problem
has aunique solution; for,
let Tr be the set of functions vc~~ 00
belonging
toC2(-R,)
and such thatjs2 v2 ds
and 00 and X the seto 0
of v E V with = 1. TT is an Hilbert space
(with
obvious definition of the scalarproduct)
and X is a convex subset ofV ;
.X is also closed(thanks
to(2.9)).
Now theproblem
to find the maximum of J isequivalent
to
project
theorigin
of V onX ; thus,
thisproblem
has aunique
solution.The
maximizing
function is the solution(in V)
of thefollowing problem
(Euler equation):
let us
put
A= maxJ(v);
A is thengiven by v(O)lv’(0).
The mostgeneral
solution v of the differential
equation appearing
in(2.10)
isgiven by
Here is the first Bessel function of the third kind
(Hankel’s function);
a is an
arbitrary complex constant;
is thepath
in thecomplex z-plane
sketched in the
figure.
Figure
1.Now, imposing
the conditionv~~ ( 0 )
= 0 andcalculating 2 gives
where we
put
These numbers can be
explicitly calculated,
y and weget
The constant
01 appearing
in(2.6)
is related to Aby
LEMMA 2.2. Let v be as in lemma
2.1;
then thefollowing inequality
holds-where the constant
O2
is the bestpossible)
isgiven by formula (2.15),
below.
PROOF. It is
quite analogous
to theprevious
one. To calculateO2
onehas now to solve the
problem (in
the same class asbefore)
and calculate One finds
where the numbers
The constant
O2
is thengiven by
LEMMA 2.3. Let the
transform (with
variable
y) of u(x, y);
we havewhere
01
andO2
are the constantsdefined
in( 2 .13 )
and(2.15).
PROOF. Let us
apply inequality (2.6)
to the functionv( ~, ~);
squareboth members of the
inequality, multiply by iql I
andintegrate
over q;by applying
Holder’sinequality
onegets
Now,
from Parseval’sformula,
onegets (2.16). (2.17)
isproved
in asimilar way.
THEOREM 2.3. The
operator
~(2.18)
u -->-[restriction
of u to the half-line y =0,
x >0]
is a
densely defined
boundedoperator from W+(Y+)
intoLet
be its
(uniquely defined)
linear bounded extension. Then the rangeof
y+ isexactly
and y+ is anisomorphism
betweenW+(Y+)f(Kernel y+)
andJel(R+) .
The boundedness of
(2 .18 )
will follow from Lemma 2.4 below. The other assertion will follow from Thm. 4.3.LEMMA 2.4. Let u E
Co (Y+);
let h be the restrictionof
u to the x-axis.Then the
following inequalities
holdPROOF.
(2.19) immediately
follows from theidentity
by applying
Schwartzinequality.
To prove
(2.20)
we need a result that will beproved
later. For themoment,
let us suppose~(0,’)===0; by integration by parts,
weget
By using
now Schwartz andHardy’s inequalities,
weget
From this
inequality
weimmediately
derive(2.21)
if~(0,’)=0;
for thegeneral
case, theproof
will begiven
in thecorollary
to Th. 3.3.3. - The forward
equation.
This
section,
which has anauxiliary
character in view of the next one, contains some results for theparabolic equation (1.4),
which may be of some interestby
themselves.First,
in the subsection3.1,
we will find solutions of(1.4)
defined in all.R2 ; then,
for the associatedhomogeneous equation,
we will consider the first
(and second) boundary
valueproblem
inX+
(subsect. 3.2)
and theCauchy problem
in thehalf-plane Y+ (subsect. 3.3).
Let us notice that we are
dealing
with aparabolic equation
of evolutiontype degenerating
on a line and theproblems
we willspeak
about are theclassical ones.
Let us write eq.
(1.4)
in the standard formHere A is the linear second-order differential
operator -
(k > 0), acting
in some Hilbert space H. In[9],
where a class ofequations
similar to
(3.1 )
wereconsidered,
we choosed as H the spaceL2(.R; i. e. ,
thespace of square summable functions with the
weight lxl.
This choiceis,
insome sense, « natural », since A is
symmetric
in.L2(.R;
but it turns not to be thegood
one in view of theapplications
of eq.(3.1 )
to thenonstationary equation ( 1.1 ) .
Now we will take for .g the space
L2(R; x2) (our
solutions u must have theproperty
that suy is squaresummable). Unfortunately,
in this space, A is notsymmetric
nordissipative;
and this fact willgive
somecomplica--
tions in the discussion of the
Cauchy problem
for eq.(3.1) (see
subsect.3.3).
3.1. Solutions
defined
in the wholeplane.
THEOREM 3.1. Let
given function ;
there existsonly
onesolution u to eq.
(1.4) belonging
toZ(R2).
This solutionsatisfies
theinequality,
Moreover, if f
= 0for
y0,
this solution vanishes on the lowerhal f -plane
and its restriction to the upper
half-plane belongs
toW( Y+)
andsatisfies
theinequality
PROOF.
Uniqueness.
It follows from theinequality
which can be derived a
priori. For,
letu E Co00 (R2)
andput f =
ku -By taking
the square of the modulus off
andneglecting
the non--negative
term weget
(*)
Allalong
this paper, const means an absolute constant; sometimes we will writeconst (k)
to indicate a constantdepending
on k.Integrating
now over thehalf-plane using integration by parts, gives
from
this,
we deriveimmediately (3.4).
Thisinequality,
established forfunctions,
holds inZ(R2), owing
todensity arguments.
Existence.
Elementary
calculations showthat,
if a solutionof eq.
(1.4) exists,
its Fourier transform(with respect
toy), v
say, can berepresented
in thefollowing
waywhere
g(x, ~ )
is the Fourier transform off (x, ~ )
andV(q)
is aone-parameter family
ofintegral operators
whose Kernels areHere
vo(x,r¡)
is that solution ofequation
v = 0 whichvanishes for
x ~ -~-
oo, andW(q)
is the wronskian of the twoindependent
solutions
vo(x, q )
andvo(-x, 27).
Let us
put
Then, if q >0,
so that we havewhere a and b are
given by
In
(3.11 )
and(3.12 )
the notation means that thearguments
of the Besselfunctions H(l)(2) is
3 ( k3~2/ I~I ~ )’
If we have
Finally,
y the wronskian W isgiven by:
We need now the
following
lemma :3.1. Let
~T~‘(r~ )
be thepreviously defined family of integral operators.
For every c~ and every
real q
weWe can now
complete
theproof
of Th. 3.1. From(3.5), taking
accountof
( 3 .16 ) ,
weget
By integrating
now over q andtaking
account of Plancherel theorem for Fourier transforms we obtainFrom
(3.15)
we have alsofrom which we
get
16 - Annali della Scuola Norm. Sup. di Pisa
Then,
let us writeIf 99
is a testfunction,
we canwrite, by changing
the order ofintegration (all
the variables run from -00 to+ oo)
By integration by parts, taking
account that the function(z, q)
2013>-7fexp [iyn] 99 (x, y) dy
hascompact support
withrespect
to x and is fastdecreasing
withrespect to q,
one hasThus u is twice differentiable
(with respect
tox)
in the sense of distribu- tions and uzz isgiven by
Then u solves eq.(1.4~
and,
from(3.17)
and(3.18), inequality (3.2)
iseasily
derived. Let now bef
= 0for y
0. Thepreviously
foundsolution u,
which vanishesfor y
0(remember (3.14)),
satisfies theequation ix lx| +
ku =F,
where F ==uzz - f ;
then it may be
represented
as follows:’
Thus, by
Schwartzinequality,
y weget
Then we have
Now
inequality (3.3) follows,
since and the norm ofcan be estimated
by
the norm off.
Theorem 3.1 iscompletely proved-
LElBIlBI.A. 3.2. Let
E(x, x’ ; q)
bedefined by (3.6)
andsubsequent
The
following majorizations
hold:for
every x,x’, r~
reals.LEMMA 3.3. The
functions,
are bounded
by positive
constants.PROOF OF LEM1tiA 3.]. Let us write out
Then, by
Schwartzinequality,
andchanging
the order ofintegration,
we
get
If we estimate the
integrals
in dxand dx" by making
use of(3.19)
weget easily (3.15).
With a similar
procedure
we haveLet us make use now of estimate
(3.20);
let usput
2 - t andwe obtain
u
By estimating integrals
in dt anddt" by
the aid of lemma 3.3 weget (3.16).
PROOF oF LEMMA 3.2. Let us recall a bound for Hankel functions:
Then we see that
la(r¡)1
andIb(r¡)1
are boundedby positive
constants(independent
onk) :
Now a bound for v+ and v_, and thus for is
easily
derivedwhere z is
given by (3.8).
Then weget,
forE,
thefollowing
bounds:where we
put
z’ =z(x’ ) .
and we still
get
a bound like theprevious
oneSimilar bounds can be written when
summarizing
the resultsgives
and, finally,
we can use, in any case, the bound(3.21).
Now,
from the definition of z, we haveLet us now
provide
an estimate forputting
Now let us suppose, for the
moment,
x’> x(so that
we have
Now we can see that the bracked
expression,
as a functionof k,
attainsits minimum value at the
origin (for
every~) and,
as a functionof q,
itattains its minimum value still at the
origin (for
everyk).
Similar con-clusions can be deduced in case x’ x. Thus we
got
Now,
from(3.21), by using (3.23)
and(3.25),
bounds(3.19)
and(3.20)
can be derived without
difficulty.
PROOF OF LEMMA 3.3. Let us notice
that,
since Iq-
tt’>tt’,
we can writeThen the first
integral appearing
in the statement of the lemma ismajorized by
The other assertion of the lemma is a
special
case of lemma(3.4)
in[9].
3.2.
Boundary
valueproblems
inTHEOREM 3.2. Let 1p E
H112 (R)
be agiven function.
Z’here exists aunique f unction
u such that(in
the senseof traces)
Such a
f unction
can berepresented by
theformula
where
A(x, ~),
the Fouriertransform (with respect
toy) of A(x, y),
isgiven by
Moreover,
thefollowing
bounds on u holdwhere is the
fractional
derivative(order 2 ) of
1p.PROOF. The
uniqueness
assertion follows frominequality (3.32)
whichcan be deduced a
priori
from the factthat,
if u is a well-behaved(e.g., u
is+00
of class
Z)
solution of eq.(3.27),
the function is convex;+00 -00
if u E
Z(X+),
it is boundedtoo;
for 0x -fiu(x, y)j2dy belongs
to-oo
together
with itsderivative;
thus this function takes its maximum value at theboundary.
Representation (3.29)
can beeasily
checkedby straightforward
calcula-tions ; for, by taking
the Fourier transform withrespect
to y, anddenoting by "
the transformedfunctions,
weget
where 1
isgiven by (3.30). Now,
we can estimate veryroughly
in the
following
wayTo
get (3.35),
remember(3.24)
and usual bounds for Hankel functions recalled in theproof
of lemma 3.2.From
representation (3.29) by taking
account of(3.35),
one can show thatu is a C°°-solution of eq.
(3.27).
Let us show that u EZ(X+).
From(3.34)
and
(3.35)
we obtainThus we have
and,
from Planchereltheorem,
Now, taking
account of(3.36), directly
from theequation
7 weget (3.33).
Besides the
problem (3.26), (3.27), (3.28),
we will need to consider also thefollowing
one: to find u such that(3.26)
and(3.27)
hold andSince the treatment of this
problem
iscompletely analogous
to theprevious
one, we limit ourselves to the statement of the results in the fol-lowing
theorem.THEOREM 3.3. be a
given function.
There exists aunique
solution u to
problem (3.26), (3.27), (3.37) ;
this solution can berepresented by
theformula
where
.~(x, r~),
the Fouriertransform of B(x, y)
isgiven by
take the
complex coniugate
of the aboveexpression
for 27 0 and z, zo are the same as in Th. 3.2. Thefollowing
bounds on u holdIn
carrying
out theproof,
thefollowing
estimate is usefullWe can
complete
now theproof
ofinequality (2.20).
Moreprecisely,
wewant to prove the
following
statement:COROLLARY. Let u be a
given function belonging
toCo (~’~. ) ;
let h be its-restriction to the
x-axis ;
thusinequality (2.20)
is valid.PROOF. Let us
put 99(y)
==ux(O, y) for y >
0 and = 0 for y 0.Then 99
EH8(R)
for s.1,
inparticular
for s =g .
Let now w be a solu-tion of the
equation
for x > 0 such that:+00
sup ju,,(x, y) /2dy
oo andsatisfying
the condition : = Q.x>O
-00Such a solution exists and is
unique,
thanks to aslight
modification of Th.3.3; obviously,
this solution vanishes fory = 0. Now,
let usput
v = u - w in
Y+;
then it will be andv( ~ , 0) = h. Now,
we canapply
to vinequality (2.20),
which wealready proved
for functions whose x-derivative vanishes on they-axis ;
weget
Now we can estimate the norms of and wxx
by
the norm ofD 116 99 (thanks
toinequality (3.41)
with k =0)
andDi’6q by
means of the normsof ruv and uzz
(thanks
toinequality (2.17)).
The sameproof
can becarried out for x 0 and so
(2.20)
iscompletely proved.
LEMMA 3.4. The restrictions
of
the solutions topbs. (3.26) (3.27), (3.28)
and
(3.26), (3.27), (3.37)
to thefirst quadrant X+
r1Y, belong
toW+(X+
r1Y+).
00
PROOF. The function
0~ -~ .r~(~y
is bounded. For(let
us carryo
out the
proof
for the firstproblem)
we have from therepresentation (3.34)~
Now,
let0x~; by
Schwartzinequality,
andby using
estimate(3.35 ),
we obtain
The
integral
in dx may be estimatedby omothety,
andequals
,const
(|n|1/2 -+- 1n1| 1/2)-2.
Now the same we can do for theintegral
indn1
and we obtain
finally
An
analogous procedure
for the secondproblem gives
us3.3. Initial value
problem
inY+ .
Let us consider now the
Cauchy problem
for the evolutionequation
in the
half-plane Y+.
Because of thesymmetry properties
of eq.(3.45),
which is invariant under transformations x -> - x, we will consider sepa-
rately
the cases that the initialvalue, assigned at y
=0,
is an even or anodd function. There are some reasons for a
separate
treatment:i)
the cal-culations will be more
clearly understandable, ii)
in Sec.4,
when the results of this section will beused,
we will needonly
even solutions of eq.(3.45);
iii) finally,
the results in the two cases are different.a)
.Even solutions.THEOREM 3.4. Let
given function;
there existsonly
onefunction u belonging
to~’+(X+
r1Y+),
solution to eq.(3.45)
andverifying
theconditions
u(., 0)
= h and~cx(0, ~ )
= 0 in the senseof
traces. Such afunction satis f ies
theinequality
The
proof
of Th. 3.4 will beeasily
deduced from thefollowing
lemma :LEMMA 3. 5. Let
h EJel(R+)
be agiven function ;
there exists afunction
wdefined
inX +
f1Y,
with theproperties
Proof of TH. 3.4: let us introduce a new unknown 0 == I where w is the function considered in lemma 3.5. It is clear that w ~ exp
(- ky) belongs
to r1Y+)
and satisfies the same conditions as w at the boundaries. Then in order u to be a solution ofpb. (3.47) ... (3.49),
it iscient that 0 must be a
solution, belonging
toW+(X+
r1Y+),
of theequation
with
homogeneous boundary
conditions. Because of(3.50),
we see that theright
member of(3.52)
is anY+) function; then,
afterhaving
con-tinued this function into the whole
plane
as an even function of xvanishing
for
y 0,
we canapply
Th. 3.1. We deduce that the function 0 existsand,
from(3.2), (3.3),
it satisfies theinequalities
From
these, by taking
account of(3.50)
and(3.51),
we obtainThen we
proved (3.46), by showing
also thedependence
on k of the constantappearing
there.The
uniqueness
of the solution may be derived in the same way as we did in theproof
of Th. 3.1.For, by following
the sameprocedure
there usedto
get inequality (3.4),
we canobtain, by integrating
on the firstquadrant
and
putting f
=0,
This a
priori inequality implies uniqueness
for ourproblem.
Theorem 3.4is
proved.
To prove lemma
3.5,
we need thefollowing
lemma.the
operator represented by
is a bounded one-to-one
operator of L2(.R+)
intoitself,
whose range isthe inverse
operator (which
isunbounded)
isrepresented by
the
integrals appearing
in(3.56)
and(3.58)
converge in L2 -norm. Moreoverwe have
Here flf
is the Mellin transform off
andJ,
is the first Bessel function of order v.REMARK
i).
As it is known from thetheory
of Mellintransform,
ifits Mellin
transform,
is a square summable function onthe line
s = 2 -~- i~O,
and the Parseval formula holds:As it is easy to see, the Mellin transform e
-~. (~(,t) ( 2 -f- ie)
is the Fourier transform of the so that the conditionthat f
belongs
to B meansbelongs
toREMARK
ii).
The caseot .1
is the well-known case of Hankel trans-10rms;
the set ~3 coincides with the whole spaceL2(R+)
andSJv.l/2
=is a
unitary operator
ofL2(R+).
PROOF of lemma 3.5. Let us
represent
therequested
function w as a~superposition
ofseparated
variables solutions of eq.(3.47):
where
W(’ t)
is a boundedcontinuously
differentiable solution of the equa- tionWIT + it2r
~‘ = 0satisfying
the conditionW’(o, t)
=0 ;
we havewhere b is an
arbitrary
function.Requiring
that the condition at y = 0~to be
satisfied, gives
and
putting
we can write
formally
this condition as followsthis
notation means that I is theI~-I 1/3.1/6
transform of b.Now, since
00
by hypothesis,
thenlEL2(R+)
and(3.65)
has aunique solu-
o
tion
(thanks
to lemma3.6) given by
thus we can write the function w as follows
Then the
function $
-*~1~2w(~2~~, y)
isgiven by
thecomposition
of threeoperators applied
to 1: theoperator ~-113.1/s
which acts fromL2(.R+)
to themultiplication by
exp[- 4 ( ~ ) 2y],
which is a continuousoperation
from ~3.to ~3
(see below) and, finally,
which acts from 93 toL2(R+) .
Thusthe
composition
of these threeoperators
is defined on allL2(R+)
and it is.also
continuous; for,
lety > 0
befixed ; by taking
account of(3.60),
weget
Then, by taking
account of thechange
of variables(3.64),
weget
Let us consider now w,,; from
(3.67)
we obtainFirst,
let us show thatFor, by integration by parts, taking
account thatwe
get
(*) Remember the remark
i)
after lemma 3.6 and the fact that themultiplication by
the functioni - exp (- 9/4y e2’)
is a continuousoperation
in for it iscontinuous in
.L2(R)
and inH1(R),
because of the uniform boundedness of the above written function and of its derivative.The finite
part
vanishes(it
can be checkedby
easycalculations;
in[12],
Sect.
2.2,
calculations of thistype
areextensively
done in a moregeneral
situation);
thus weget (3.70). Now,
since11 (~) - 1$-11($)
=ZXI/4 h’(x),
themeans $ -~ Z’(~) - ~~-1 l(~) E L2(I~+);
thus thefunction t
- tb(t) belongs
to 93. With calculations and considerations ana,l- ogous to thosepreviously
done to prove(3.68)
weget
Taking
account of thechange
of variables(3.64),
we obtainBesides
(3.68)
and(3.71)
we can obtain also thefollowing
estimateTo
get (3.72 ),
let us rewrite condition(3.63)
in thefollowing
way:where we
put b(t)
=tllsb(t); putting
also1($)
= the same condi-tion writes
Then we can
proceed
asbefore,
when we deduced(3.68);
the calculationsare even
simpler,
since we aredealing
now with the case a= 2
of lemma3.6,
which is the well-known case of Hankel transforms.
Finally,
from(3.71)
and(3.72)
we deduce(remeber
the remark after the definition of the space in Sec.2):
Now,
from(3.68)
and(3.37),
we deduce(3.50).
Let us consider now wy ; from(3.67)
we haveThen, by applying inequality (3.60),
weget
By integrating
over y from 0 to 0o we obtainBy putting,
in the firstintegral,
tl = t-r andchanging
the order ofintegra- tions, we get
17 - Annali della Scuola Norm. Sup. di Pisa
Finally, by applying
the convolution theorem for Mellintransforms,
weget
(integrate
first on T, then on e,finally
ona)
Thus, by returning
to the oldvariables,
weproved
Then
(3.51)
is alsoproved
and thiscomplete
theproof
of lemma 3.5.PROOF of lemma 3.6 : the
proof
is anapplications,
with somemodifications,
of Th. C
given
in[9],
sec.2.3;
we refer to[9]
for details.Let
P(s)
be definedby
On the line Res
= 2
we haveBy taking
account of theasymptotic
behaviour of Tfunction,
weget
the bounds
where C1
and c2
arepositive
constants.Now, given
let g be definedby
and e -+ (1 + I |g|1/2 ) 1/2- a p(
is bounded. Then thecorrespondence f ->
g establishedby (3.79)
defines anoperator
from-L~(J~+)
intowhich is bounded in
_L2(_R+).
Therepresentation (3.56)
followsby taking
the inverse Mellin transform of
(3.79):
it is easy to see,by
the calculus ofresidues,
y thatInequality (3.59)
follows from(3.80)
and(3.68). Conversely,
letand
f
be definedby
where
Then,
since e-¿. (1 -~-- + ie)
isbounded, -f-
isL2(R)
andf E L2(R+).
Thecorrespondence
g -f established by (3.82)
defines an oper- ator from 93 intoL2(.R+);
thisoperator
is the inverse of reason of(3.83),
and
inequality (3.60)
follows from(3.78).
Therepresentation (3.58)
alsofollows
by taking
the inverse Mellin transform of(3.82).
b)
OddTHEOREM 3.5. Let be a
given function,
there exists at mostone
function u
EW~.(X+
r1Y+),
solution to eq.(3.45)
andverifying
the condi-tions
u(., 0)
= handu(O,.)
= 0 in the senseof
traces. Such a solutionactually
exists, provided
that hverifies
thefollowing
condition :To prove the
theorem,
we need thefollowing
lemma.to the set 9)
given by (3.57).
For the inverseoperator,
the samerepresentation (3. ~8; J
holds. Moreover we have
REMARK. Let c > 0 and
g(t)
=t,- 112
for 0 « 1 andg(t)
= 0 f or0 1;
then and
(~I(~g) (s) _ (s - 2 -~- s)-1. By applying
the convolution theorem for Mellintransforms,
weget
Thus the condition that
L2(R+):3 f belongs
to A means that the one-para-1
meter
family
of functions be bounded inmust be
uniformly
bounded in.L2(R+).
PROOF oF TH. 3.5. This
proof
followsclosely
theproof
of Th. 3.4.The
proof
ofuniqueness
is the same as in Th.3.4;
theproof
of the existencecan be
reduced,
as we did in the casea)
to the determination of a solution wto eq.
(3.47) satisfying
the sameboundary
conditions as therequested
solution u.
In order to
determinate w,
let us look for arepresentation
like(3.61)
where now
We , t), having
tosatisfy
the conditionW(0, t)
=0,
isgiven by
where a is an
arbitrary
function.Imposing
the conditionat y
=0,
andusing
the samechange
of variables(3.64), gives
eq.
(3.89)
has aunique
solution and we can write the function wgiven by (3.61)
as followsFrom
(3.90),
aninequality
like(3.68)
can be derived as we did in a.Let us consider now Wx; from the
representation
of zv weget
Thus we can
write,
with thechange
of variables(3.64)
Let us consider now the function t
--~-ta(t).
As we did ina)
inderiving (3.?’0)
we
get
nowThus,
in order t -ta(t)
tobelong
to the set93,
thefunction e
-*1/2 -e l(e)
-- 1’($)
mustbelong
to theset A
definedby (3.85).
This means(remember
the remark after lemma
3.7),
thatbe
uniformley
bounded inZ2(.I~+). Returning
to the oldvariables,
we have(by
anintegration by parts)
The second term may be estimated
by
an extension ofHardy’s inequality
(see [15],
formula(9.9.10))
and weget
Then, by hypothesis,
thisintegral
is finite and wehave,
from(3.86)
(from
the Parseval formula for Mellintransform)
Now,
from(3.91),
we see that also thebelongs
to the set
A, and,
from(3.87)
and(3.94),
weget
the estimate00
Now an estimate of follows from the estimates of
v>o 0
Finally, let
us consider tcv ;directly
fromthe