A NNALI DELLA
S CUOLA N ORMALE S UPERIORE DI P ISA Classe di Scienze
G IORGIO T ALENTI
On functions, whose lines of steepest descent bend proportionally to level lines
Annali della Scuola Normale Superiore di Pisa, Classe di Scienze 4
esérie, tome 10, n
o4 (1983), p. 587-605
<http://www.numdam.org/item?id=ASNSP_1983_4_10_4_587_0>
© Scuola Normale Superiore, Pisa, 1983, tous droits réservés.
L’accès aux archives de la revue « Annali della Scuola Normale Superiore di Pisa, Classe di Scienze » (http://www.sns.it/it/edizioni/riviste/annaliscienze/) implique l’accord avec les conditions générales d’utilisation (http://www.numdam.org/conditions). Toute utilisa- tion commerciale ou impression systématique est constitutive d’une infraction pénale.
Toute copie ou impression de ce fichier doit contenir la présente mention de copyright.
Article numérisé dans le cadre du programme Numérisation de documents anciens mathématiques
http://www.numdam.org/
Steepest Proportionally
toLevel Lines.
GIORGIO TALENTI
1. - Introduction. , .
Suppose u
is a smooth real-valued function of two real variables x and y;consider:
Here
subscripts
stand fordifferentiation,
D stands forgradient, f (U2 +
=’V - 1.
In otherwords,
k = curvature of the level lines of u .
More
precisely,
the valueof at
anypoint ( x, y),
where thegradient
of udoes not
vanish,
is the curvature at(x, y)
ofu-11(u(x, y) ),
the level line of upassing through (x, y) ;
thesign
of k is that which makeskDu,
a normalvector field to the level lines
of u,
orientate towards the center of curvature.Analogously,
h = curvature of the lines of
steepest
descentof u ,
where lines of
steepest
descent =trajectories
of thegradient
=orthogonal trajectories
of the level lines.Pervenuto alla Redazione il 2
Maggio
1983.In section 2 we prove THEOREM 1.
at any
point
where Du is not zero.at any
point
where Du & are not zero.As
usual,
we have setOne aim of the
present
paper is toinvestigate
thegeometry
of solutions tospecial partial
differentialequations
via theorem 1. Theorem 1 tells usthat k and h are
automatically
constrainedby
asystem
of twopartial
dif-ferential
equations
of the secondorder,
whatever u is(provided u
is smoothenough
and has no criticalpoints
in theregion
ofinterest).
On the otherhand, partial
differentialequations,
of asuitably
restrictedtype,
shouldconstrain
further,
in a readable way, the curvatures of the level linesand/or
the lines of
steepest
descent of solutions. As we showlater,
this is true forfirst order
partial
differentialequations
forinstance,
or forLaplace equation.
Thus we believe that
coupling
theorem 1 with suitablepartial
differentialequations
may lead tosignificant
conclusions about solutions to those equa- tions.Although
some comments are made in section 3 about harmonicfunctions,
we
implement
the above ideas in the case of thefollowing partial
differentialequation:
which is a very convenient test for our purposes.
Here A is a constant.
Equation (1.7)
readsso that it characterizes those functions whose lines of
steepest
descent bendproportionally
to levellines;
in case £= 0, equation (1.7)
characterizesfunctions whose lines of
steepest
discent arestraight. (1.7)
is aquasi-linear equation
of the second order withquadratic nonlinearities;
note that theequation
ishyperbolic,
sinceif
a, b,
c denote the coefficients of u,,.(1.7) belongs
to thefollowing
class of
equations:
As is easy to see, any
equation
of the form(1.8) enjoys
theproperty
whatever A is. In section 5 we sketch some considerations about equa- tions
(1.8). Incidentally, (1.7)
is the Eulerequation
of thefollowing integral:
(in
the case A =0,
theintegrand
issimply arctan).
In section 4 we prove
THEOREM 2. Let u be a solution to
equation (1.7). Suppose u
has nocritical
points ;
supposefurther
u is smooth near a bounded closed set E. Then thelargest
r such that u is twicecontinuously differentiable
incannot exceed
Theorem 2 tells us that solutions to
equation (1.7),
which are notplane
waves, either have critical
points
ordevelop singularities.
Ourproof
isbased on a
priori
estimates ofintegral type,
see subsections 43 and 44.Theorem 2 is
sharp.
Infact,
we can showsolutions,
whosesingularities
are
exactly
aspredicted by
theorem 2. A convenientexample
is the distance from apoint
set.Specifically, let
E’ be a smooth arc without inflectionpoints
and letbe a
parametric representation
ofE ;
the formuladefines a
system
of smooth curvilinearcoordinates u, v
in aneighbourhood
of E.
Clearly,
the value oflul ]
at anypoint (x, y) sufficiently
close to E isthe distance of
(x, y)
from E. We have for the functional matrix of the derivativeswhere K = a’ b" - a" b’ is the curvature of E. Hence
furthermore u is a solution to
the
equation
ofgeometrical optics.
The lines ofsteepest
descent of a twicecontinuously
differentiable solution toequation (1.9)
must bestraight
(see
e.g.[1 ],
or section6).
Thus u is a solution to thefollowing equation:
a
special
case ofequation (1.7).
The above formula for thejacobian
ofu, v and the
following
formulas:tell us that u and the level lines of u cannot be smooth
beyond
the curvethe evolute of E. The same conclusion is drawn from theorem 2.
2. - Proof of theorem 1.
From
( 1.1 )
and(1.2)
weget
By
the way, we have alsoFormulas
(2.1)
and(2.2)
can berewritten
this wayif notations
(1.6)
are used. Thus(1.3) yields
Let us denote
by
co theangle
between Du and a fixeddirection; specifi- cally
From
(2.3)
and(2.4)
we obtainhence
Formula
(2.6)
has an obviousgeometrical meaning.
Infact,
from(2.6)
we infer
the derivative of m
(with respect
to thearclength) along
the level lines of u.Similarly, (2.6) implies
Since w is
real-valued, (2.6) gives
equivalently
Using (2.7)
twicegives
Obviously
Then
Equation (2.8) implies (1.4) trivially. Equation (2.8)
isquadratic
ine~~’.
Thus(2.8) gives
provided
The lastequation
and(2.5) give (1.5).
3. - Harmonic functions.
THEOREM 3.
Suppose
u is harmonic and has no criticalpoints.
Then:where
REMARKS.
Property (ii) implies
a minimumpr2ncipte
for the curvature7~ of level lines of a harmonic function u : the minimum
of k,
in anyregion
where Du has no zeros, is attained on the
boundary.
A somewhat related statement is in[3].
Statement(vi)
tells us that the curvature of level linesof a harmonic function
obeys
apartial
differentialequation
of the fourthorder.
Analogous
remarks holdfor h,
of course. Recall that the lines ofsteepest
descent of a harmonic function are level lines of the harmonicconjugate
function.PROOF. Let us introduce the
complex gradient
of u.
By hypothesis, f
has no zeros in theregion
of interest. From for- mula(2.3)
we Since u isharmonic, f
is a holo-morphic
function of thecomplex
variable z = x-!- iy.
Hencewhere ’ denotes differentiation with
respect
to z.Formula
(3.1) yields
a
holomorphic
function. Assertion(i)
follows. Assertion(ii)
follows from(i),
since
=
InIDu 1- In
InI
is harmonic and thelogarithm
of a harmonic function is
superharmonic.
From(3.1)
weget =
hence
(iii)
follows. Assertion(iv)
is an easy consequence of(iii)
and theo-rem 1
(in agreement
with theorem1,
onemight
check that any function q,having
the form(3.1),
satisfies(iv)
wheneverf
isholomorphic).
Note thatMultiplying
both sides ofq4q = g~~ + q§ by if,
thensplitting
into real andimaginary parts, give (v).
From
(v)
we haveOn the other hand
since
(i) gives
The lastquantity equals
I
is aholomorphic
function and(3.2)
holds. We conclude thatfor the zeros
of 99
are isolated.Formula
(3.3) implies (vi),
= 0whenever g
is aholomorphic
function.
4. - Proof of theorem 2.
41. PRELIMINARIES.
Equation (1.7) yields
as
long
as Du is different from zero. We assumethroughout
this section that u is a smooth solution toequation (1.7)
and u has no criticalpoint.
Equation (4.1)
and theorem 1give
at once results(i) (ii) (iii)
below.(i)
Thefollowing inequality
holds:(ii)
Eitheror
at any
point
whereDk =1=
0 .(iii)
Thefollowing equation:
holds at any
point
where 0.42. REMARKS ON EQUATION
(4.3). Equation (4.3)
andinequality (4.2) yield
at any
point
where k ~ 0.Equation (4.5)
is the Eulerequation
of the fol-lowing integral:
Here
and
is a solution to the
equation
ofgeometrical optics
whose lines of
steepest
descent aretangent
rays to the circleClearly
so that
equation (4.5)
should bequalified
asparabolic.
43. LEMMA. The
following
lemma is thekey
to aproof
of theorem 2 and will be derived fromequation (4.4).
LEMMA. Let p be any and let G be any bounded open set with a
rectifiable boundary.
Then44. CONCLUSIVE
steps.
Setand
for every
sufficiently
small t > 0. As is well-known and easy to see,at almost every
point (x, y)
out of E.Consequently, Federer’s
coarea formula(see [2),
section3.2)
ensures thatUp
isabsolutely
continuous andfor almost every t.
Analogously,
if we setwe have
Schwarz
inequality yields
where q = p/(p -1 ).
Theright-hand
side of the lastinequality
can beestimated from below with the
help
of(4.7).
Thus we obtaina differential
inequality
forU,.
From
(4.9)
weget
if Since and
we see that
Letting
p -+
oo and 8 -~0,
we infer from(4.10)
that if k is bounded inthen
Theorem 2 is
proved.
Noteincidentally
that the solutions toequation (1.7) obey
thefollowing inequality:
45. PROOF OF THE LEMMA.
Equation (4.4)
readswhere
A(p, q)
andB(p, q)
are solutions of thefollowing system:
Inequality (4.2)
ensures that(4.11)
holds at everypoint
where k =1= 0.Consider
a level set of k. Here t > 0.
Inequality (4.2)
ensures that(4.13)
is free fromcritical
points
of k. Hence(4.13)
must reach theboundary
ofG ;
on theother
hand,
thatpart
of theboundary
of(4.13),
which is in the interior ofG,
isexactly y)
E G:= tl,
a collection of smooth arcs.Integrating
both sides ofequation (4.11)
over level set(4.13) gives
where
v2 )
is the exterior normal to a G.Thanks to
(4.12 ),
theintegrand
at theright-hand
side of(4.14) Ik 1,
and the
integrand
at the left-hand sideequals
Hence(4.14) gives
where
is the distribution function of k. In
fact,
Federer’s coarea formula( [2 ],
sec-tion
3.2) yields
From
(4.15)
we deduce:The lemma is
proved.
5. - Remarks on
equations (1.8).
51. Consider first
the A = 0 case of
equation (1.7).
Let u be a solution without criticalpoints.
By (1.2)
and(2.2)
we haveso that
is an exact differential. Let v be defined
by
Clearly
and
Thus the
following proposition
isproved:
if u satisfiesequation (5.1)
and has no critical
point,
then a solution v to theequation
ofgeometrical optics (5.2)
exists such that u and v are functions of each other(i.e.
inter-dependent).
Recall that
equation (5.1)
characterizes functions whose lines ofsteepest
descent are
straight.
52. Consider
the
(1/2)
= 0 case ofequation (1.7),
anda first-order
partial
differentialequation having
the form of a conservation law.According
to formula(1.1 j, equation (5.3)
characterizes functions ’U whose level lines arestraight.
Note that(5.4)
becomesif v = tanw.
The
following proposition
holds: if u satisfies(5.3)
and has no criticalpoint,
then a solution m toequation (5.4)
exists such that u and ware functions of each other.PROOF. Let w be defined
by (2.4),
i.e. theangle
between Du and the z-axis. Since u satisfies(5.3),
formulas(1.1)
and(2.6i)
tell us thatOn the other
hand,
we learn from(1.3)
and(2.7)
thatmv = - h sin (o; thus
53. Consider now any
equation of
theform (1.8) .
Since the case is covered in subsection
~2,
we assume hereLet
f(p, q )
be definedby
Note that the
right-hand
side of(5.5)
is an exact differential(provided
theunderlying region
does not wind round theorigin).
In faotfor a and b are
homogeneous
of the samedegree.
In the case of
equation (1.7),
one haswhere e,
cv arepolar
coordinates in thepq-plane (p
= ~ cos w, q = e sinco).
The
following propositions
hold:(i) Any (sufficiently smooth)
solution u to the first-orderpartial
dif-ferential
equation ~~)
= 1 is a solution toequation (1.8).
(ii)
If u is a solution toequation (1.8),
such that Du ranges over the domain off
and avoids the zerosof f,
thenv~)
= 1 has a solution v such that u and v are functions of each other.PROOF. From
(5.5)
weget
and
By
virtue of(5.5i), equation (1.8)
can be rewritten this way:The left-hand side of the last
equation
isexactly
thenegative
of thejacobian
of u and Thus
equation (1.8)
readsProposition (i)
follows.Let u be as in
(ii).
Asand
(5.6) holds,
is an exact differential. Let v be defined
by
Obviously,
y thejacobian
of u and v vanishes. On the otherhand,
equa- tion(5.5ii)
tells usthat f
ishomogeneous
ofdegree
1. Thuswy)
= 1.Proposition (ii)
isproved.
6. - Further comments on theorem 3.
If u is a twice
continuously
differentiable solution to a first-orderpartial
differential
equation
of the formthe lines of
steepest
descent and the level lines of u are constrainedby
In
fact,
the left-hand side of(6.2)
is thenegative
ofthe derivative of
f(u, u,, U1/) along
the level 1 ines of u.More
precisely,
differentiations of the left-hand side of(6.1)y
and for-mulas
(1.1)
and(1.2), give
- 1 11a linear
system
in the second-order derivatives of u.Here
standfor
u.,,, u,),
etc.Solving
withrespeot
to and thenforming yields
Consider for instance the
equation
ofgeometrical optics
In this case, formulas
(6.2)
and(6.3) give
Furthermore
(11u)2
=uxx + + f or
the hessian vanishes(recall
that thegraph
of a twicecontinuously
differentiable solution to(6.1)
is a
developable surface,
ifayau - 0).
Thus thefollowing proposition
iscomprised
in theorem 2.Suppose
u is a solution to(6.4)
and 2~ is smoothnear a bounded closed set E. Then u cannot be twice
continuously differentiable
in the whok
of
E+ I(X, y) :
x2+ y2 r2 }
unlessSimilar remarks can be made for the conservation law
For
(6.2)
and(6.3)
tells us that the twicecontinuously
differentiable solutions to(6.5) obey
On the other
hand,
theorem 2 has thefollowing corollary.
Let usatisfy
and let u be smooth near a bounded closed set E.
If
u has no criticalpoint,
then u cannot be smooth in E
+ ~(x, y) :
X2+
unlessThe above statements about
equations (6.4), (6.5)
and(6.6)
can be alter-natively
derived via astraightforward geometric argument (we
are indebtedto
professor
J. C. C. Nitsche for remarks on thismatter). They
arequoted
here both for
testing
thesharpness
of theorem 2 and forshowing
how ourmethod
might
work for first-orderpartial
differentialequations.
For
instance,
let u be a solution toequation (6.4)
and suppose that u is twicecontinuously
differentiable in the disk X2+ y 2
r2. Anelementary argument
shows that theperpendiculars
to a smooth arc, whose foots havean infinitesimal distance from each
other,
meet atpoints
of the evolute.On the other
hand,
aperpendicular
to a level line of our solution u is a line ofsteepest
descent of Thus the evolute of~c-1 (u(o, 0)),
the level line of upassing through
theorigin,
must lie out of the disk x2+ y2
r2. Inparticu-
lar we must
have lk(O, 0 )~1 ~ c 1,
that isThis is
essentially
what we stated above.However,
theproof
of theorem 2 leads to thefollowing stronger
result. Let u be a solutions to theequation of geometrical optics (6.4)
and suppose that the restrictionof
u to a disk x2-~- y2
e2belongs
to Sobolev spacew2,f) for
some p > 2. Then u cannotbelong
to~Y2~~
in a
larger
disk x 2+ y2
r 2 unlessREFERENCES
A)
Quoted
papers[1] R. COURANT - D. HILBERT, Methods
of
mathematicalphysics,
Interscience, 1962.[2] H. FEDERER, Geometric measure
theory, Springer-Verlag,
1969.[3] M. LONGINETTI, Sulla convessità delle linee di livello di
funzioni
armoniche, Boll.Un. Mat. It., ser. 6, vol. 2-A (1983),
pp. 71-75.
B)
Related papers[4] F. JOHN, Partial
differential equations, Springer-Verlag,
1971.[5] F. JOHN, Formation
of singularities
in one-dimensional wavepropagation,
Comm.Pure
Appl.
Math., 27(1974),
pp. 377-405.[6] S. KLAINERMAN - A. MAJDA, Formation
of singularities for
waveequation
in-cluding
the nonlinearvibrating string,
Comm. PureAppl.
Math., 33(1980),
pp. 241-263.
[7] P. D. LAX,
Development of singularities of
solutionsof
nonlinearhyperbolic partial differential equations,
J. Math.Phys.,
5 (1964), pp. 611-613.[8]
S. A. LEVIN, Nonlinearboundary problems for
aquasilinear parabolic equation,
J. Diff.
Eq.,
5 (1969), pp. 32-37.[9]
G. TALENTI, A note on the Gauss curvatureof
harmonic and minimalsurfaces,
Pacific J. Math., 101 (1982), pp. 477-492.
Istituto Matematico dell’Universita viale
Morgagni, 67/A
50134 Firenze