Boundary control of a singular reaction-diffusion equation on a disk

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by  Rafael Vazquez

Boundary control of a singular
reaction-diffusion equation on a disk
Rafael Vazquez∗ Miroslav Krstic∗∗
∗Department of Aerospace Engineering, Universidad de Sevilla,
6 Camino de los Descubrimiento s.n., 41092 Sevilla, Spain (e-mail:
1 [email protected]).
0 ∗∗Department of Mechanical and Aerospace Engineering, University of
2 California San Diego, La Jolla, CA 92093-0411, USA (e-mail:
[email protected]).
n
a
J
Abstract: Recently, the problem of boundary stabilization for unstable linear constant-
8
coefficient reaction-diffusion equation on n-balls (in particular, disks and spheres) has been

solvedby means ofthe backsteppingmethod. However,the extensionofthis resultto spatially-
]
C varying coefficients is far from trivial. As a first step, this work deals with radially-varying
reaction coefficients under revolution symmetry conditions on a disk (the 2-D case). Under
O
these conditions,the equationsbecome singularinthe radius.When applying the backstepping
.
h method,thesametypeofsingularityappearsinthebacksteppingkernelequations.Traditionally,
t well-posedness of the kernel equations is proved by transforming them into integral equations
a
and then applying the method of successive approximations.In this case, the resulting integral
m
equation is singular. A successive approximation series can still be formulated, however its
[ convergence is challenging to show due to the singularities. The problem is solved by a rather

non-standard proof that uses the properties of the Catalan numbers, a well-known sequence
1
v frequently appearing in combinatorialmathematics.
0
1 1. INTRODUCTION steppingmethod[8].Thebacksteppingmethodhasproved
0
itself to be an ubiquitous method for PDE control, with
2
many other applications including, among others, flow
0 In a series of previous results, the problem boundary sta-
control [19, 25], nonlinear PDEs [20], hyperbolic 1-D sys-
1. bilization for unstable linear constant-coefficientreaction- tems [5, 6, 10], adaptive control [18], wave equations [17],
0 diffusion equation on n-balls has been solved. In partic-
and delays [9]. The main idea of backstepping is finding
6 ular [22] and [23] describe, respectively, the (full-state)
an invertible transformation that maps the system into a
1 control design for the particular case of a 2-D disk and
stable target system which needs to be chosenjudiciously.
: a general n-ball; that same design, augmented with an
v Tofindthetransformation,ahyperbolicpartialdifferential
observer,isappliedin[15]tomulti-agentdeploymentin3-
i equation (called the kernel equation) needs to be solved.
X Dspace,withtheagentsdistributedonadisk-shapedgrid
Typically, the well-posedness of the kernel equation is
r andcommandedbyleaderagentslocatedattheboundary.
studied by transforming it into an integral equation and
a Theoutput-feedbackgeneralizationton-ballsispresented
then applying successive approximations to construct a
in[24].Older,relatedresultsthatusebacksteppinginclude
solution.Theconvergenceofthesuccessiveapproximation
thedesignanoutputfeedbacklawforaconvectionproblem
seriesguaranteesthatasolutionalwaysexists,itisunique,
on an annular domain (see [21], also [11]), or observer
and it is bounded. However, in the problem posed in this
design on cuboid domains [7]. However, going from an
paper, one obtains a singular kernel equation. Following
annulardomainto adisk (whichincludesthe origin)com-
the previously-outlined procedure, one can transform it
plicatesthedesign,assingularitiesappearontheequations
into a (singular) integral equation and then apply the
and have to be dealt with.
method of successive approximations. However, proving
Thisworkcanbeseenasafirststeptowardsextendingthis the convergence of the resulting series is challenging. The
family of previous results to the non-constant coefficient main technical contribution of this paper is tackling this
case, by assuming a certain symmetry for the initial issue. We use a rather non-standard proof based on a
conditions,whichsimplifies the problem.Therehavebeen combinatorialsequenceofintegers(theCatalannumbers).
specific results ondisk- orspherical-shapeddomains,such
The structure of the paper is as follows. In Section 2
as [3] and [14], which have assumed these same symmetry
we introduce the problem and state our main result. We
conditions.
explain our design method and find the control kernel
Based on the domain shape we use polar coordinates, equation in Section 3. Next, we prove its well-posedness
and using the symmetry of the initial conditions and inSection4.Weconcludethe paperwithsomeremarksin
imposing an equally symmetric controller, the system is Section 5.
transformed into a single 1-D system with singular terms.
We design a feedback law for this system using the back-
2. 2-D REACTION-DIFFUSION SYSTEM ON A DISK K K K λ(ρ)
r ρ
K + K + = K (8)
rr r − ρρ ρ − ρ2 ǫ
Considerthefollowingreaction-diffusionsystemonadisk,
with boundary conditions
written in polar coordinates (r,θ):
ǫ ǫ K(r,0)=0, (9)
u = (ru ) + u +λ(r)u, (1)
t r r r r2 θθ r λ(ρ)
K(r,r)= dρ, (10)
evolving in the disk = (r,θ) : r [0,R],θ [0,2π) , − 2ǫ
DR { ∈ ∈ } Z0
for t>0, with boundary conditions
in the domain = (r,ρ):0 ρ r R . Then system
(1)–(2)hasaunTique{L2(0,R)s≤olut≤ion,≤and}theequilibrium
u(t,R,θ)=U(t,θ), (2) profile u 0 is exponentially stable in the L2 norm, i.e.,
≡
where U(t) is the actuation (we assume we cancontrolall there exists c1,c2 >0 such that
the boundary). Note that the system will be unstable for u(t, ) c e−c2t u . (11)
L2(0,R) 1 0 L2(0,R)
k · k ≤ k k
large values of λ.
In the next sections we prove Theorem 2. First, Section 3
Denote by L2( ) the space of L2 functions on the disk
R applies the backstepping method, finding a feedback law
D
defined as usual. For the case when λ does not depend on
whose kernel is the solution of the singular hyperbolic
r, the following result was shown in [22]:
PDE (8)–(10), and shows the Theorem, assuming that
Theorem 1. Consider (1)–(2) with constant λ > 0, with kernel equation is well-posed and has a bounded solution.
initialconditionsu L2( )andthefollowing(explicit) Section4dealswiththewell-posednessofthekernelPDE.
0 R
∈ D
full-state feedback law for U:
3. CONTROL LAW DESIGN AND STABILITY
I λ(R2 ρ2) RESULT
1 λ R 1 ǫ −
U(t,θ)= ρ (cid:20)q (cid:21)
−2π ǫ Z0 λ(R2 ρ2) Duetolackofspace,wesketchthedetailsofhowtoobtain
ǫ − (8)–(10) by borrowing the results from a previous, more
π (R2 qρ2)u(t,ρ,ψ) general publication.
− dψdρ, (3)
× R2+ρ2 2Rρcos(θ ψ)
Z−π − − 3.1 Target system and backstepping transformation
where I is the first-order modified Bessel function of
1
the first kind. Then system (1)–(2) has a unique L2( )
DR In [22] the problem was solved by posing a backstepping
solution,andtheequilibriumprofileu 0isexponentially
stableintheL2( )norm,i.e.,theree≡xistsc ,c >0such transformation from (1)–(2) to the target system
R 1 2
D
that ǫ ǫ
ku(t,·)kL2(DR) ≤c1e−c2tku0kL2(DR). (4) wt= r (rwr)r+ r2wθθ, (12)
w(t,R,θ)=0, (13)
Now we try to extend this result to the case when λ is a
function ofr. As a firstapproach,considerthat the initial a well-posed and stable target system (one just needs to
conditions do not depend on the angle (have revolution take the mean value of the target system in [22]). The
symmetry) and fix U as a constant (also not depending transformation had the form
on the angle θ). Then, by symmetry, there is no angular r π
w(t,r,θ)=u(t,r,θ) K(r,ρ,θ,ψ)u(t,ρ,ψ)dψdρ.
dependence and one can drop the θ derivative in (1)–(2), −
finding the following 1-D problem: Z0 Z−π (14)
ǫ Proceedinganalogously,weposeatransformationwithout
u = (ru ) +λ(r)u, (5)
t r r r angular dependences
r
for r [0,R), t>0, with boundary conditions
∈ w(t,r)=u(t,r) K(r,ρ)u(t,ρ)dρ, (15)
−
Z0
u(t,R)=U(t), (6) to reach the target system
Stabilizationof(5)–(6)issimplerthan(1)–(2)becausethe ǫ
w = (rw ) , (16)
systemisnow1-D,butstillchallengingduetothesingular t r r r
terms in (5). In addition, an explicit expression for the w(t,R)=0. (17)
controller is not possible as in (3) due to the spatially-
varying λ(r). Our main result is as follows. The system(16)–(17)inherits its stability propertiesfrom
Theorem 2. Consider (5)–(6) with constant λ > 0, with (12)–(13), because it is a particular case of it (with ini-
initialconditionsu L2(0,R)notdependingontheangle tial conditions having revolution symmetry). Similarly,
0
and the following ful∈l-state feedback law for U: the transformation (15) is a particular case of (14) and
the resulting kernel equations can be directly extracted
R from [22]. To do this, it must be noted that [22] de-
U(t)= K(R,ρ)u(t,ρ)dρ, (7) composed the system equation (1)–(2) and the transfor-
Z0 mation (14) in its Fourier components, and that having
where the kernel K(r,ρ) is obtained as the solution of revolution symmetry is equivalent to only considering the
meanFouriercomponent(n=0).Thus,theequationthat G(r,0)=0, (23)
K(r,ρ) in (15) has to verify to map (5)–(6)into (16)–(17) r λ(ρ)
is directly obtained from [22] as (8)–(10). G(r,r)= dρ. (24)
− 2ǫ
Z0
Finally, applying the transformation (15) at r = R and
using the boundary conditions of both targetand original Following [8], define α = r+ρ, β = r ρ. Then, the G
−
systems,(17),(6)respectively,weobtainthe feedbacklaw equation in α,β variables becomes
(7).
λ α−β
G G 2
3.2 Invertibility of the transformation 4Gαβ + (α+β)2 − (α β)2 = (cid:16) ǫ (cid:17)G (25)
−
It can be shown that if we pose an inverse transformation inthedomain ′ = (α,β):0 β α 2R,β 2R α
T { ≤ ≤ ≤ ≤ − }
of the form with boundary conditions
r
u(t,r)=w(t,r)+ L(r,ρ)u(t,ρ,ψ)dρ, (18)
G(β,β)=0, (26)
Z0
wecanfind(usingthesameprocedureofSection3.1)that α/2 λ(ρ)
G(α,0)= dρ. (27)
the inverse kernel L verifies the following hyperbolic PDE − 2ǫ
Z0
L L L λ(r)
r ρ
Lrr+ r −Lρρ+ ρ − ρ2 =− ǫ L. (19) This can be transformed into an integral equation as
typical in backstepping [8]. First, we find
with boundary conditions
λ α−β
L(r,0)=0, (20) 2 αβ
G = G+ G (28)
r λ(ρ) αβ (cid:16)4ǫ (cid:17) (α2 β2)2
L(r,r)= dρ. (21) −
− 2ǫ now, integrating in β from 0 to β:
Z0
These equations are very similar to (8)–(10) with slight
β λ α−σ
differences. The proof that we will show in Section 4 can G (α,β) G (α,0)= 2 G(α,σ)dσ
α α
beappliedtoshowthereexistsaboundedsolutionto(19)– − 4ǫ
Z0 (cid:0) (cid:1)
(21). β ασ
+ G(α,σ)dσ. (29)
Onceithasbeenestablishedthatthereareboundeddirect (α2 σ2)2
Z0 −
and inverse backstepping transformations, then it is easy
and integrating again in α from β to α:
to show that both transformations map L2 functions into
L2 functions (see e.g. [24]). Then, the well-posedness and
stability properties of the target system (16)–(17) are G(α,β) G(β,β) G(α,0)+G(β,0)
− −
mapped totheoriginalsystem(5)–(6),provingTheorem2. α β λ η−σ
It only remains to show that the kernel equations have a = 2 G(η,σ)dσdη
4ǫ
bounded solution, which is done next in Section 4. Zβ Z0 (cid:0) (cid:1)
α β ησ
+ G(η,σ)dσdη. (30)
4. WELL-POSEDNESS OF THE KERNEL PDE Zβ Z0 (η2−σ2)2
Using the boundary conditions:
Next we show the well-posedness of (8)–(10), showing in
particular that there exists a bounded solution by using α/2 λ(ρ) α β λ η−σ
a constructive method. This result is the main technical G(α,β)= dρ+ 2 G(η,σ)dσdη
− 2ǫ 4ǫ
development of this paper which is necessary to complete Zβ/2 Zβ Z0 (cid:0) (cid:1)
the proof of Theorem 2. α β ησ
+ G(η,σ)dσdη. (31)
(η2 σ2)2
Zβ Z0 −
4.1 TransformingthekernelPDEintoanintegralequation
Thisisasingularintegralequationduetothetermsinthe
last integral.
To better analyze the kernel equation, define G = rK.
ρ
ThisisanallowedtransformationgiventhatK isasqsumed 4.2 Successive approximations series
tobe differentiableandzeroatρ=0.ThusK,whenclose
to ρ=0,behaveslike ρ andthereforeit canbe dividedby The method of successive approximations applied in [8]
√ρ. and posterior works to show that (31) has a solution can
be applied. Thus, define
The equation verified by G is
G G λ(ρ) α/2 λ(ρ)
Grr−Gρρ+ 4r2 − 4ρ2 = ǫ G (22) G0(α,β)=−Zβ/2 2ǫ dρ (32)
with boundary conditions and for k >0,
α β λ η−σ α β ∂2
Gk(α,β)= 2 Gk−1(η,σ)dσdη Fnk(η,σ)dσdη
4ǫ ∂α∂β
Zβ Z0 (cid:0) (cid:1) Zβ Z0
α β ησ =Fnk(α,β) Fnk(β,β) Fnk(α,0)+Fnk(β,0)
+Zβ Z0 (η2−σ2)2Gk−1(η,σ)dσdη.(33) =4 α β η−σ Fn(k−1)(−η,σ)−Fn(k−2)(η,σ) dσdη
(η2 σ2)2
Zβ Z0 (cid:0) − (cid:1)
Then, the solution to the integral equation is α β
+ λ¯F(n−1)k(η,σ)dσdη, (38)
∞ Zβ Z0
G= G (α,β), (34) and solving for F (α,β) and substituting the values at
k nk
the boundaries, we reach the formula. ✷
k=0
X
From the previous lemma, the following result is straight-
assuming the series converges.
forward.
However, proving convergence of the series is harder than
Lemma 4.2. For n 1 or k 1
usual. The typical procedure (see [8] and posterior works) ≥ ≥
is to assume a functional bound for G and show by
k i=k α β i=k
recursionit is verified for every k. In this case we follow a Fni= λ¯ F(n−1)i(η,σ)dσdη
different method.
i=1 Zβ Z0 i=1
X X
λ α−β α β ησ
Callλ¯ =max(α,β)∈T′(cid:12)(cid:12) (cid:16)4ǫ2 (cid:17)(cid:12)(cid:12).Thenoneclearlyobtains +4Zβ Z0 (η2−σ2)2Fn(k−1)(η,σ)dσdη. (39)
(cid:12) (cid:12)
G (α,β) λ¯(α β)(cid:12). However(cid:12)when trying to substitute WecanusetheformulasfromLemma4.2tofindthevalues
| 0 |≤ − (cid:12) (cid:12)
thisboundintheexpr(cid:12)essionofG(cid:12)1 wefindanintegralthat ofsomeboundsforGk.Forillustration,letusfindthefirst
isnotsoeasytocompute.Instead,weformulateaseriesof values. Obviously
technicalresultsthatwillhelpderivingafunctionalbound G F . (40)
0 00
| |≤
for G .
k Then
Lemma 4.1. Define, for n 0,k 0,
≥ ≥ α β
G λ¯F (η,σ)dσdη
λ¯n+1αnβn logk αα−+ββ | 1|≤Zβ Z0 00
F (α,β)= (α β) . (35)
nk n!(n+1)! − (cid:16)k! (cid:17) α β ησ
+ F (η,σ)dσdη
(η2 σ2)2 00
and F =0 if n<0 or k <0. Then: Zβ Z0 −
nk F
01
(a) Fnk is well-defined and nonnegative in ′ for all n,k =F10+ 4 (41)
T
(b) F (β,β)=0 for all n and k
nk where we have used the formulas of Lemma 4.1. The next
(c) F (α,0)=0 if n 1 or k 1
nk term is
≥ ≥
(d) F (α,0)=α
00
(e) The following identity is valid for n 1 or k 1. α β F
≥ ≥ G λ¯ F + 01 dσdη
2 10
| |≤ 4
Zβ Z0 (cid:18) (cid:19)
α β ησ F
Fnk=4ZβαZ0β ησ(cid:0)Fn(k−1)((ηη,2σ−)−σ2F)2n(k−2)(η,σ)(cid:1)dσdη +Zβ ZF0 (η2F−σ+2)F2 (cid:18)F10+ 401(cid:19)dσdη
α β =F + 11 + 01 02. (42)
+ λ¯F(n−1)k(η,σ)dσdη. (36) 20 4 16
Zβ Z0 Similarly we find
Proof First, it is easy to see that since α+β > 1, F21 F11+F12 2F01+2F02+F03
α−β G3 <F30+ + + (43)
the logarithm is always nonnegative. Also, for any k, | | 4 16 64
(α β)logk α+β is bounded since a linear term always Some particular number appear in these expressions. To
− α−β try to identify the pattern, call:
dominates a(cid:16)logar(cid:17)ithm (no matter the exponent of the
logarithm). Thus Fnk is well-defined in ′. The values at α β
the boundaries of ′ are found by simTple substitution. H [F]= λ¯F(η,σ)dσdη
1
T
Finally, for the integral, we have the following identity Zβ Z0
which is found by differentiation: α β ησ
H [F]= F(η,σ)dσdη, (44)
2 (η2 σ2)2
∂2F∂nαk∂(αβ,β) =F(n−1)k+4αβ Fn((αk−21)−β2F)2n(k−2) . (37) which are the twoZoβpeZr0ations−to find a successive approxi-
(cid:0) − (cid:1) mationtermfromthepreviousone.TofindaboundonG4
we have to apply H and H to (43) and use Lemmas 4.1
Thus 1 2
and 4.2, finding:
Table 1. Catalan’s Triangle
Cij j=1 j=2 j=3 j=4 j=5 j=6 j=7 j=8 j=9 j=10
i=1 1
i=2 1 1
i=3 2 2 1
i=4 5 5 3 1
i=5 14 14 9 4 1
i=6 42 42 28 14 5 1
i=7 132 132 90 48 20 6 1
i=8 429 429 297 165 75 27 7 1
i=9 1430 1430 1001 572 275 110 35 8 1
i=10 4862 4862 3432 2002 1001 429 154 44 9 1
H [F ]=F Let us establish some properties about these numbers
1 30 40
before proving Lemma 4.3.
H [F ] F
1 21 31
H [F ]+ =
2 30 Lemma 4.4. For 0<j i it holds that
4 4
≤
H2[F21] + H1[F11+F12] = F21+F22 (a) Cii =1.
4 16 16 (b) Cij = ik−=1j−1C(i−1)k.
H [F +F ] H [2F +2F +F ]
2 11 12 1 01 02 03
16 + 64 Proof WePshow (a) by induction on i. For i=1, C11 =1
by definition. Assuming it true for i, then C =
2F +2F +F (i+1)(i+1)
11 12 13
= C +C =C =1.
64 ii (i+1)(i+2) ii
H [2F +2F +F ] 5F +5F +3F +F We show (b) by descending induction on 1 j i.
2 01 02 03 01 02 03 04
64 = 256 .(45) For j = i, we obviously have Cii = ik−=1i−1≤C(i−1)≤k =
Thus, we obtain C(i−1)(i−1) =1,as wejust showed.Assuming ittrue forj,
weshowitforj 1byusingthedefinitioPnofthenumbers:
−
F F +F 2F +2F +F
31 21 22 11 12 13
G F + + + +
4 40
| |≤ 4 16 64
5F +5F +3F +F Ci(j−1)=C(i−1)(j−2)+Cij
01 02 03 04
256 . (46) i−1
=C(i−1)(j−2)+ C(i−1)k
By extending this structure to the general case, we find k=j−1
X
the following recursive formula for n > 0, expressed as a i−1
lemma. = C(i−1)k, (48)
Lemma 4.3. For n>0, it holds that k=j−2
X
n−1j=n−iC(n−i)j thus proving the Lemma. ✷
G F + F , (47)
| n|≤ n0 4n−i ij Proof of Lemma 4.3 Weestablishthelemmabyinduc-
i=0 j=1
X X tiononn 2.ForG itisalreadyestablished.Nowassume
2
where the numbers Cij verify it is true f≥or Gn and prove it for Gn+1. We have:
(a) C =1
11
(b) C =0
i0
(c) C =0 if j >i, for all i. G H [G ]+H [G ] (49)
ij | n+1|≤ 1 | n| 2 | n|
(d) Cij =C(i−1)(j−1)+Ci(j+1) forallothervaluesofiand
j.
Thus
The set of numbers in Lemma 4.3, known as the “Cata-
lan’s Triangle”, or the ballot numbers (see. e.g. [16] and
references therein, even though the numbers are written n−1j=n−iC(n−i)j
in a slightly different ordering).The firstfew numbers are |Gn+1|≤H1Fn0+ 4n−i Fij
showninTable1.Inparticular,thefirstcolumnofTable1, i=0 j=1
X X
tahsisthise,ywahraetuwsueahlalyvedceafilnleeddC[4i1,,pa.2re65t]h.eBCoathtaltahnenCuamtablearns  n−1j=n−iC(n−i)j 
+H F + F , (50)
numbers and Catalan’s Triangle verify many interesting 2 n0 4n−i ij
properties, and are connected to a wide set of combina- Xi=0 Xj=1
torial problems as well as other number sets, such as the  
coefficients of certain Chebyshev polynomials (see e.g. [1,
and since the integral H , H operatorsare linear,we can
1 2
p. 797], where the nonnegative rows of table 22.8 are the
express this inequality in a convenient way as follows
columns of Table 1).
|Gn+1|≤H1[Fn0]+(cid:18)H2[Fn0]+H1(cid:20)C11F4(n−1)1(cid:21)(cid:19) |Gn+1|≤F(n+1)0+n−1j=n−i+1C4(nn−+i1+−1i)jFij
+n−1 H j=n−iC(n−i)jF Xi=1 jXj==n1+1 
 2 4n−i ij C11Fn1 1
+ + C F
Xi=1 Xj=1 4 4n+1 (n+1)l 0l
j=n−i+1  Xl=1
+H1 j=1 C4(nn−−ii++11)jFi−1j =F(n+1)0+ n j=n+1−iC4(nn++11−−ii)jFij, (54)
X i=0 j=1
  X X
j=nC proving the Lemma. ✷
nj
+H F . (51)
2 4n 0j Therefore, since the solution of the successive approxima-
j=1
X tions series verifies
  ∞
We next manipulate some of the lines of (51) to reachthe
G G (α,β) (55)
n
finalresult.First,for the secondline of(51),we havethat | |≤ | |
n=0
X
if we can prove the convergence of the series with the
estimatesofLemma4.3thatwejustderived,wecanprove
j=n−i
C(n−i)j the existence of a solution to the integral equation and
H F
2 4n−i ij therefore to the kernel equation.
j=1
X
j=n−i  4.3 Convergence of the successive approximation series
1
= 4n−i C(n−i)jH2[Fij]
j=1 By Lemma 4.3, we find that the series whose convergence
X
= 4n1−i j=Xj=n−1iC(n−i)j Pjl=+11(cid:0)Fil−4H1[F(i−1)l](cid:1)! we nee|dG|to≤st∞udyFnc0an+b∞e wnr−it1tej=nna−siC4(nn−−ii)jFij. (56)
j=n−ij+1 nX=0 nX=1Xi=0 Xj=1
1
= 4n+1−i C(n−i)j Fil−H1[F(i−1)l] The first term is easy to compute (see e.g. [1, p.375]):
j=1 l=1
X X (cid:0) (cid:1)
1 l=n−i+1 n−i ∞ ∞ λ¯n+1αnβn
= 4n+1−i Fil−H1[F(i−1)l] C(n−i)j Fn0= n!(n+1)!(α−β)
Xl=1 (cid:0) (cid:1)j=Xl−1 nX=0 nX=0
1 l=n−i+1 I1 2 λ¯αβ
= 4n+1−i Fil−H1[F(i−1)l] C(n−i+1)l. (52) = λ¯(α−β) h√pαβ i, (57)
Xl=1 (cid:0) (cid:1) where I is the first-orpder modified Bessel function of the
1
and we can see that the sum of the second term in the first kind. For the next term, we use the fact that
parenthesis of (52) cancels the third line of (51). ∞ n−1 ∞ ∞
H(n,i)= H(l+i,i), (58)
For the last line of (51), we have that
n=1i=0 i=0 l=1
XX XX
therefore
H2jj==n1 C4nnjF0j=jj==n1 C4nnjH2[F0j] ∞ n−1j=n−iC4(nn−−ii)jFij
X X n=1i=0 j=1
XX X
  j=nC j+1 F ∞ ∞ j=l
nj 0l C
= lj
= F
4n 4 4l ij
Xj=1 Xl=1 Xi=0Xl=1Xj=1
j=nj+1
1 ∞ j=∞ ∞
= 4n+1 j=1 l=1CnjF0l =  C4lljFij. (59)
XX i=0 j=1 l=j
X X X
l=n+1 j=n
1 It turns out that the parenthesis can be calculated and
= F C
4n+1 0l nj gives an exact sum for each j.
l=1 j=l−1
X X
For that, we need only the fact (see any combinatorics
l=n+1
1
book, e.g. [26, p.44]) that the generating function of the
= C F . (53)
4n+1 (n+1)l 0l Catalan numbers C is given by1
l=1 l1
X
1 This generating function, touted as one of the most celebrated
generating functions in combinatorics, is typically expressed as
Thus we find that
1−√21x−4x,whichiseasilyconverted to(60).
2 Proof We prove the Lemma by induction. It is already
f (x)= (60)
1 1+√1 4x provedforj =1,2.Assumeitforj 1andj 2andprove
− it for j. As we have shown − −
Remember that a generating function of a sequence of
numberisafunctionsuchthatthecoefficientsofitspower ∞ Clj = fj(41) (71)
series is exactly those of the sequence of numbers. Thus, 4l 4
∞ Xl=j
f1(x)=C11+C21x+C31x2+...= Cl1xl−1 (61) But on the other hand fj(1/4) = fj−1(1/4)− fj−24(1/4),
and substituting
l=1
X
Therefore if we evaluate the function at x = 1/4 we find
that ∞ ∞ Clj = fj−1(1/4)− fj−24(1/4)
1 1 4l 4
f ( )= C , (62) l=j
1 4 l14l−1 X ∞ ∞
l=1 C 1 C
X lj lj
thus we obtain =
4l − 4 4l
∞ Cl1 = 1 ∞ Clj = f1(14) = 1. (63) l=X1j−1 1 1 l=Xj−2
4l 4 4l−1 4 2 =
Xl=1 Xl=1 2j−1 − 42j−2
Following this argument, it is clear that 1
= , (72)
∞ C 1 ∞ C f (1) 2j
lj = lj = j 4 , (64) thus proving the Lemma. ✷
4l 4 4l−1 4
l=j l=j
X X Thuswefinallyobtainsomepartialsumsin(56)asfollows
wherewedefinethesequenceofgeneratingfunctionsf as
j
∞ I 2 λ¯αβ ∞ j=∞
fj(x)= Cljxl−1. (65) G λ¯(α β) 1 + Fij
| |≤ − h√pαβ i 2j
l=j
X p Xi=0 Xj=1
NowsinceCl2 =Cl1 butobviouslyC12 =0,itisclearthat I1 2 λ¯αβ
f =f C =f 1. Thus f (1/4)=1 and we find = λ¯(α β)
2 1− 11 1− 2 − h√pαβ i
∞ C4ll2f24(14) = 14. (66) p∞ j=∞λ¯i+1αiβi logj αα+−ββ
l=2 + (α β) , (73)
X i!(i+1)! − 2(cid:16)jj! (cid:17)
i=0 j=1
To find successive generating functions we use the prop- X X
which is a summable series both in i and in j. Summing
erties of the Catalan’s Triangle and make the following
first in i we find the same term as in (57), thus
claim:
Lemma 4.5. For n>1
I 2 λ¯αβ j=∞logj α+β
G λ¯(α β) 1 α−β ,(74)
fn(x)=fn−1(x)−xfn−2(x) (67) | |≤p − h√pαβ iXj=0 2(cid:16)jj! (cid:17)
Proof To proveit, we write the definition of the generat- and the second term is the series of an exponential,
ing function fn−1 therefore we finally reach
∞
fn−1(x)=l=Xn−1Cl(n−1)xl−1, (68) |G|≤ λ¯(α−β)I1h2√pαλ¯βαβielog(cid:16)pαα−+ββ(cid:17)
and using the properties of the numbers p
I 2 λ¯αβ
1
∞ ∞ = λ¯(α2 β2) . (75)
fn−1(x)= Clnxl−1+ C(l−1)(n−2)xl−1 q − h√pαβ i
Substituting α and β by the physical variables r, ρ, it is
l=n−1 l=n−1
∞X X∞ found that
= Clnxl−1+x Cl(n−2)xl−1 I 2 λ¯(r2 ρ2)
Xl=n l=Xn−2 G λ¯rρ 1 − , (76)
=fn(x)+xfn−1x, (69) | |≤ h pr2 ρ2 i
q −
thus proving the claim. ✷ and going back to the original kernel K, we finally found
p
that the successive approximation series converges and
Based on this fact, we can now prove another result.
defines a kernel satisfying the following bound
Lemma 4.6. For j 1, there holds
≥
I 2 λ¯(r2 ρ2)
∞ C 1 K(r,ρ) ρ λ¯ 1 − , (77)
lj = (70) | |≤ h pr2 ρ2 i
4l 2j p −
l=j thus completing the proof of Theorem 2.
X p
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deployment: A nonlinear PDE motion planning ap-
Thispaperisafirststeptowardsextendingboundarysta- proach,” Automatica, vol. 47, pp. 2534–2542,2011.
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paper, a method of proof based on the properties of the of the 2012 American Control Conference, 2012.
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