Distributional radius of curvature

MATHEMATICAL METHODS IN THE APPLIED SCIENCESMath. Meth. Appl. Sci. 2006; 29:427–444Published online 8 December 2005 in Wiley InterScience (www.interscience.wiley.com). DOI: 10.1002/mma.692MOS subject classi�cation: primary 46 F 12

Distributional radius of curvature

Ricardo Estrada∗;†

Department of Mathematics; Louisiana State University; Baton Rouge; LA 70803; U.S.A.

Communicated by M. Mitrea

SUMMARY

We show that any continuous plane path that turns to the left has a well-de�ned distribution thatcorresponds to the radius of curvature of smooth paths. We show that the distributional radius ofcurvature determines the path uniquely except for a translation. We show that Dirac delta contributionsin the radius of curvature correspond to facets, that is, �at sections of the path, and show how a pathcan be deformed into a facet by letting the radius of curvature approach a delta function. Copyright? 2005 John Wiley & Sons, Ltd.

KEY WORDS: radius of curvature; distributions

1. DIVISION BY ZERO

Even in the very �rst stages of the teaching of mathematics, starting in elementary school,students are taught that one cannot divide by zero. This is important, of course, since divisionby zero is not well de�ned, and can yield non-sensical results.There are, however, several mathematical frameworks where one can consider the division

by zero in a mathematically sound fashion. For instance, when studying indeterminate formsof limits in calculus, one uses the symbolic expression K=0= ± ∞, which is a shorthand tothe fact that if limx→a f(x)=K �=0, and limx→a g(x)=0, with g(x) �=0 for x close to a butdi�erent from a, then limx→a f(x)=g(x)= ± ∞, the sign being determined by the sign of Kand the sign of g(x) near x= a. Similarly, in the theory of analytic functions it is convenientto consider the range of a meromorphic function to be �C=C∪{∞}, the Riemann sphere, andde�ne the value of a meromorphic function at a pole to be ∞; thus, if f and g are analyticin a region �, a∈�, and f(a)=K �=0, while g(a)=0, then the value of h(z)=f(z)=g(z) atz= a is ∞, which in a colourful way can be written as K=0=∞.Our aim in the present article is, however, to point out that in several cases the division

by 0 could be interpreted not as ∞ but as a suitable Dirac delta function. That this is the

∗Correspondence to: Ricardo Estrada, Department of Mathematics, Louisiana State University, Baton Rouge,LA 70803, U.S.A.

†E-mail: [email protected]

Received 5 December 2004Revised 10 February 2005

Copyright ? 2005 John Wiley & Sons, Ltd. Accepted 23 July 2005

428 R. ESTRADA

case can already be seen in the simplest division problems in the theory of distributions. Forinstance, the distributional solutions of the division problem

xf(x)=0 (1)

are precisely [1]

f(x)= c�(x) (2)

where c is an arbitrary constant, while if is a smooth function in Rn that vanishes exactlyon a smooth surface �, and if |∇ (x)| �=0 for all x∈�, then the solution of the divisionproblem

(x)f(x)=0 (3)

is given by the single layers concentrated on � [2]

f(x)=�(x)�(�) (4)

A much more striking example of a delta function arising from a division by zero isprovided by the work of Xin and Wong [3–5] on crystal solids. Indeed, in their work, theyconsider a two-dimensional crystal whose boundary surface contains one or more facets, thatis, �at sections. Along a facet, the curvature � vanishes and, consequently, the radius ofcurvature, R=1=�, is not de�ned. Nevertheless, they show that by taking

R(�)= s�(� − �0) (5)

for a suitable constant s, where � is the angle of the normal direction to the boundary andthe positive x-axis, and where �0 is the angle corresponding to the facet, then a most usefuldescription of the crystal is obtained. Indeed, the reduced energy of the surface of the crystalsatis�es a di�erential equation whose source term is proportional to the radius of curvature,

d2�d�2

+ �=A(�)R(�) (6)

and the use of (5) yields the proper solution. Their approach allows them to avoid the ill-possedness encountered by other authors [6,7]. Notice, in particular, that writing (6) in theform

�(d2�d�2

+ �)=A(�) (7)

is not very useful along a facet!

2. INVERSE FUNCTIONS AND DIRAC DELTA FUNCTIONS

In order to understand the appearance of distributional terms in the radius of curvature of apath, it is convenient to start by considering a simpler problem, that of inverse functions.Let � : [a; b] −→ [c; d] be an increasing but not necessarily strictly increasing function.

Namely, we assume that if x16 x2 then �(x1)6�(x2), but the possibility of �(x1)=�(x2)when x1¡x2 is not ruled out. We do assume that �(a)= c and �(b)=d.

Copyright ? 2005 John Wiley & Sons, Ltd. Math. Meth. Appl. Sci. 2006; 29:427–444

RADIUS OF CURVATURE 429

Let us �rst suppose that � is surjective, that is, if y∈ [c; d] then�(x)=y (8)

has at least one solution in [a; b]. If � is strictly increasing then we can de�ne an inversefunction, � : [c; d] −→ [a; b] by setting �(y)= x, where x is the unique solution of (8).However, if � is not strictly increasing, then � is not uniquely de�ned since when � isconstant, equal to y, over a subinterval [a′; b′] of [a; b] then �(y) can be chosen arbitrarily asany value on [a′; b′]. There is a smallest and a largest inverse functions, �− and �+, obtainedby taking �(y) as the smallest or the largest solution of (8), respectively, and all othersolutions satisfy �−6�6�+. It is to be observed, nevertheless, that all inverses are equalalmost everywhere, and thus they all correspond to one and only distribution, which unlessthere is danger of confusion we denote by �. The distribution � corresponds to an increasingbut possibly discontinuous function, the discontinuities arising from the �at portions of thegraph of �.It is interesting to notice that one can de�ne the notion of an increasing distribution, without

reference to point values, since distributions in general do not have point values. Indeed, iff∈D′(R), we say that f is increasing if for every �∈D(R) with �(x)¿ 0 ∀x∈R theparametric function

�(t)= 〈f(x + t); �(x)〉 (9)

is increasing. Then it is not hard to show that a distribution is increasing if and only if it isa regular distribution corresponding to an increasing function.Let us now study the derivative, d�=dx. We consider this derivative in the distributional

sense, since then � can be recovered from d�=dx by integration:

�(�)=�(a) +∫ �

a

d�dxdx (10)

for almost all �∈ [a; b], where we use the de�nition of the de�nite integral of a distribution[8]. Observe that the ordinary derivative of � exists almost everywhere since � is increasing,but in general (10) is not satis�ed when the ordinary derivative is used.The derivative of the inverse function, d�=dy, on the other hand, can be computed from

d�=dx by the elementary formula

d�dy=

1(d�=dx)

(11)

or more clearly

dxdy=

1(dy=dx)

(12)

as long as d�=dx �=0.Suppose now that � is constant in the subinterval [a′; b′] of [a; b], but not in any larger

subinterval; let y0 =�(x) for a′6 x6 b′. Then the right-hand side of (11) is certainly notde�ned for a′6 x6 b′: the right-hand side of the equation reads 1=0. However, the left-handside of (11) is well de�ned, and for a′6 x6 b′ it gives a contribution of

(b′ − a′)�(y − y0) (13)


430 R. ESTRADA

Loosely speaking, a Dirac delta function is obtained from (11) when one needs to divide byzero.

ExampleLet

�(x)=

⎧⎪⎪⎨⎪⎪⎩x + 1; x¡ − 10; −16 x6 1

x − 1; x¿1

Then

d�dx=

{1; |x|¿10; |x|¡1

The inverse is

�(y)=

{y − 1; y¡0

y + 1; y¿0

We also observe that the derivative is d�=dy=1 + 2�(y), that has a delta function at y=0of strength 2, arising from the fact that d�=dx=0 on the interval |x|¡1, whose length is 2,that is, from the horizontal part of the graph of �, that has length 2 (Figures 1 and 2).In our previous analysis � was surjective or, equivalently, continuous. A more symmetric

situation is obtained when both � and � are increasing, but not necessarily strictly increasing,and both can have several jump discontinuities. Then both derivatives d�=dx and d�=dy willhave Dirac delta function components arising from the horizontal portions of the graphs of �and �, respectively, or by ‘dividing by zero’ in (11) or in d�=dx=1=(d�=dy).

0-2.5-5

4

2

0

-2

-4

y

Figure 1.



0-2.5-5

5

2.5

0

-2.5

-5

y

Figure 2.

3. RADIUS OF CURVATURE

Let C be an oriented path in the Cartesian two-dimensional plane. This means that C is givenby a continuous parametrization x= x(t), y=y(t), a6 t6 b.Suppose initially that the path is smooth, in the sense that both functions x(t) and y(t)

have �rst-order derivatives that do not vanish simultaneously. Then at every point P of thepath there is a well-de�ned unit tangent vector T=T(P) that points in the positive directionalong C. The path locally divides the plane into two parts, one to the left and one to theright of C. At every point P of C there is a well-de�ned normal direction and, consequently,there are two unit normal vectors, one pointing to the right and one to the left of the path;on a smooth path one can choose a continuous unit vector, N=N(P), that always points tothe left or always points to the right, but, as we shall see, this not necessarily the case fornon-smooth paths.We now assume that the path turns to the left as one moves along C in the positive

direction. This can be described in a variety of ways. We can say that locally the path liesto the left of the tangent line at every point, or we can say that locally the chords joiningtwo points of the path lie to the left of the curve.Alternatively, let �(P) be the angle that the continuously chosen unit normal vector N(P)

makes with the positive real axis; then the curve turns to the left if and only if and onlyif the function �(P) is increasing. Observe that if the path contains a �at section, a facet,then the angle function remains constant along the facet. The function �(P) will be strictlyincreasing if and only if the path does not contain facets, if and only if the path turns strictlyto the left. Observe that the function �(P) is not unique, since �(P) + 2k is also an anglefunction for any k ∈Z, but it becomes unique once a particular value �(P0) is chosen fora �xed point P0 of C. Notice also that �(P) + is the angle function corresponding to theelection of the other orientation for the unit normal vector.Next we consider non-smooth paths that turn to the left. They are given in terms on

an increasing angle function �(P) that is de�ned except at a denumerable set of points.If P0 is a corner of the path then �(P) is not de�ned at P=P0, but the lateral limits


432 R. ESTRADA

�(P+0 )= limP→P+0�(P) and �(P−

0 )= limP→P−0�(P) both exist and #=�(P+0 )−�(P−

0 ) isthe interior angle at the corner. The function �(P) is called a regulated function [9,10].In terms of a parametrization, P=P(t), a6 t6 b, we obtain an increasing function

�(t)=�(P(t)) with the property that �(t0) is the angle the normal vector makes with thepath when t= t0.When a smooth path turns strictly to the left, we can use the angle �=�(P) as a new

parameter: P=P(�). The function P(�) is then an increasing continuous function of thenormal angle �, the inverse of the angle function �(P). When the path contains facets, oris non-smooth, then P(�) will no longer be a function, but rather a distribution. Then �becomes a distributional parameter, and the generalized function P(�) is the normal angledistributional parametrization of the path; in this case the values of P(�) are de�ned onlyalmost everywhere, and �=�(P(�)) almost everywhere, while at the points where P(�0) isdiscontinuous, the lateral limits P(�+0 ) and P(�−

0 ) both exist, and the path has a facet of angle�0 going from P(�−

0 ) to P(�+0 ). The distributional parametrization P(�) and the angle function�(P) are inverse distributions.It is convenient to stop to have this clearly stated.

De�nition 1A path that turns to the left is a path that admits the normal angle distributional parametriza-tion.

Let r=−→OP be the position vector along the path, and N and T the unit normal and unit

tangent vectors, as before. Then, in terms of the distributional parameter �, we have

N(�)= cos � i+ sin � j (14)

whileT(�)= j(− sin � i+ cos � j); j= ± 1 (15)

If the normal vector points to the right then j=1 and if points to the left then j=−1.Observe also that

dNd�= jT; dT

d�=−jN (16)

We now consider the curvature and the radius of curvature. We start with the case ofrecti�able paths, that is, paths of �nite length. Let s be arc length measured along the path.When the curve is smooth, the curvature � is given by the equation [11]

dTds=−�N (17)

and if � �=0, the radius of curvature is R=1=�. If N points to the right then �¿0 and R¿0,while if N points to the left then �¡0 and R¡0. In any case, however, the centre of curvatureis located at Q, where

−→OQ= r − RN; the circle with centre at Q and with radius |R| is the

one that gives the best �t to the path.Alternatively, we could de�ne the curvature as

�= jd�ds

(18)



and de�ne the radius of curvature as

R= j dsd�

(19)

These two last equations make sense even for non-smooth recti�able curves, even if � isa distributional parameter. Equation (19) gives the radius of curvature as a distributionalderivative, and therefore, the radius of curvature is a distribution.Notice that for a recti�able path that turns to the left d�=ds and ds=d� are both positive

distributions, that is, positive Radon measures.Observe now that since

drds=T (20)

it follows that

drd�= jRT (21)

or

dxd�=−R sin �;

dyd�=R cos � (22)

Use of Equations (21) or (22) allows us to see that there is no need to assume that the curveis recti�able in order to de�ne the radius of curvature. Indeed, we can solve for R in any ofthem, without using the arc length parameter s, that perhaps is not de�ned.

De�nition 2Let C be a path that turns to the left. Then the radius of curvature is the distribution

R(�)=drd�

· dNd�

(23)

Notice that the radius of curvature is a generalized function of the distributional parameter �.Observe also that (23) gives R(�) as the product of a C∞ function, dN=d�, and the derivativeof the function r(�), which is de�ned except on a denumerable set of points, where it hasjump discontinuities, a so-called regulated function. Therefore,

R(�)=R1(�) + R2(�) (24)

where R1(�) is a continuously integrable distribution, that is, a distribution whose integral iscontinuous [12], while R2(�) is a sum of Dirac delta functions concentrated at the points ofdiscontinuity. Actually, this is a characterization of all the radius of curvature distributions.

Theorem 1Let R(�) be a distribution with support in the closed interval [�0; �1]. Suppose it admits adecomposition of form (24). Let P0 be any point in the plane. Then there exists a uniquepath that turns to the left whose radius of curvature is R(�) and such that P(�0)=P0.


434 R. ESTRADA

ProofIndeed, using (22) we see that the path has to be de�ned as having the normal angleparametrization

x(�) = x0 −∫ �

−∞R( ) sin d (25)

y(�) = y0 −∫ �

−∞R( ) cos d (26)

where P0 has coordinates (x0; y0). The fact that R has the decomposition (24) implies thatthe integrals are de�ned except on a denumerable set of points, where the functions x(�) andy(�) have jump discontinuities, as required.

We can then say that the radius of curvature distribution determines a path that turns tothe left uniquely, except for a translation, and conversely.

4. ILLUSTRATIONS

We now present several illustrations that show how the properties of the radius of curvaturedistribution R(�) and the properties of the path are related. Recall that given R(�) the pathis determined save for a translation.

ExampleIf R(�)=R0, a positive constant, then the path is an arc of a circle of radius R. Figure 3shows the arc corresponding to R(�)=R0, 06 �6 3=2.

ExampleIf R(�)=

∑nj=1 cj�(� − �j), then the path is a polygonal line whose sides are of lengths

c1; : : : ; cn and whose normal angles are �1; : : : ; �n. Figure 4 shows the square corresponding tothe radius of curvature distribution

R(�)= c(�(�) + �(� − =2) + �(� − ) + �(� − 3=2)) (27)

for 06 �6 2.

ExampleWhen the two previous examples are combined we obtain

R(�)=R0 +n∑

j=1cj�(� − �j) (28)

the radius of curvature of a polygonal line whose corners have been rounded. Figure 5 showsa square with rounded corners with radius

R(�)=R0 + c(�(�) + �(� − =2) + �(� − ) + �(� − 3=2)) (29)

for 06 �6 2.



Figure 3.

Figure 4.

Figure 5.


436 R. ESTRADA

Figure 6.

Observe that in the intervals where R(�)¿0 the path is smooth, while if R(�)=0 onan interval (; �), where both and � are in the support of R, then the path will havea corner of angle − �. Observe also that the path contains a facet of length c and nor-mal angle �0 corresponding to each delta function contribution of the form c�(� − �0) inthe radius of curvature distribution R(�). Paths where R(�)¿0 are locally simple, i.e. willnot intersect themselves over intervals of length less than 2. A path corresponding to theradius of curvature R(�) de�ned over an interval [; + 2] of length 2 is closed if andonly if

∫ +2

R(�) sin � d�=0;

∫ +2

R(�) cos � d�=0 (30)

Next we consider paths whose radius of curvature is negative. If R(�)¡0 for � in theinterval [; �], then the normal vector will point to the left of the path there, and therefore,we obtain the same path as that of −R(� − ) over [ + ; � + ]. Notice, however, thatattention needs to be paid to the initial point of the path.

ExampleFigure 6 shows the path with radius of curvature distribution R(�)=−R0 for 06 �6 3=2,with the same initial point as that of Figure 3.In general, the radius of curvature will be positive in some intervals and negative in others.

There will be corners whenever the curve changes sign.

ExampleLet us consider the radius of curvature given as

R(�)=

{R0; 06 �6

−R0; 6 �6 2(31)

and with initial point (R0; 0). Then the path consists of two arcs of circle of radius R0 thatmake a corner at (−R0; 0), as Figure 7 shows.



Figure 7.

Figure 8.

The radius of curvature distribution

R(�)=

⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

R0; 06 �6=2

−R0; =26 �6

R0; 6 �6 3=2

−R0; 3=26 �6 2

(32)

yields the simple closed path shown in Figure 8, where we chose the initial point at (−R0;−R0)in order to have the origin at the centre of the path.Starting with an initial radius of curvature like (32) we can construct non-smooth paths by

repeating the same construction on each of the smooth segments. Analytically, we constructa new radius of curvature distribution

R1(�)=∞∑n=1

R(�n�) (33)


438 R. ESTRADA

Figure 9.

if {�n}∞n=1 is a sequence of strictly positive numbers that satisfy

∞∑n=1

1�n

¡∞ (34)

We notice that the series of integrals in (33) is a uniformly convergent series of continuousfunctions, by the Weierstrass M-test. In particular, if �n=2n, we basically obtain van derWaerden example [13] of a continuous function whose derivative does not exist at any point:then R1(�), the derivative, is a distribution without point values anywhere, that is not equalto a Radon measure in any interval while the corresponding path is not smooth at any pointnor recti�able in any interval.When the radius of curvature contains a contribution of −c�(� − �0), where c¿0, then

the path will contain a facet of length c and of normal angle − �0. Figure 9 shows thepath corresponding to the radius of curvature distribution c0�(� − �0) − c1�(� − �1), wherec1; c2¿0.Interestingly, we can consider the path corresponding to the radius of curvature distribution

”−1(�(�−�0)−�(�+”−�0)), and then let ”→ 0. On the one hand, we obtain the distribution�′(� − �0), the derivative of the Dirac delta function, while on the other we obtain a pathconsisting of a ray from a point to in�nite, and back.

5. DEFORMATIONS AND APPROXIMATIONS

We shall now show how Dirac delta function contributions in the radius of curvature arisewhen a path is deformed into a facet.Let us start with the simplest situation. Let P0 and P1 be two given points, with coordinates

(x0; y0) and (x1; y1). Let s be the length of the segment from P0 to P1 and let �0 be the angleof the normal to the segment that points to the right.We will approximate the directed segment from P0 to P1 by the smaller arc of a cir-

cle of large radius, with centre to the left of the segment, that goes through P0 and P1.See Figure 10. Let � be the radius of the circle. Then the radius of curvature function of



Figure 10.

the arc is

R�(�)=

{�; |� − �0|¡�

0; |� − �0|¿�(35)

where � is chosen in such a way that the circle goes through the points P0 and P1. A simplecomputation shows that

sin �=s2�

(36)

Let f(�)=H (1 − |�|) be the characteristic function of the interval (−1; 1). Here H (�) isthe Heaviside function, given by H (�)=0 if �¡0 and H (�)=1 if �¿0. Then

R�(�)=s

2 sin �f

(� − �0

�

)(37)

Observe now that since the function f has compact support, it satis�es the moment asymp-totic expansion [14,15],

f(tx)∼ 0�(x)t

− 1�′(x)t2

+ 2�′′(x)

t3− · · · ; as t → ∞ (38)

where

k =∫ ∞

−∞xkf(x) dx (39)

are the moments of the function f. In this case 0 = 2, hence

f(tx)=2�(x)

t+ o

(1t

); t → ∞ (40)


440 R. ESTRADA

This yields, with t=1=�, after observing that �→ 0 if and only if �→ ∞, that

R�(�) =s

2 sin �f

(� − �0

�

)

=s�sin �

�(� − �0) + o(

�sin �

)

= s�(� − �0) + o(1); �→ ∞ (41)

Thus

lim�→∞

R�(�)= s�(� − �0) (42)

which is the distributional radius of curvature of the facet, the segment from P0 to P1.More generally, we may start with any initial path C1, with initial point P0 and with radius

of curvature R1(�), a6 �6 b. We extend R1(�) to be 0 outside of the interval [a; b]. Let C�

be the path with the same initial point P0 and with radius of curvature

R�(�)=K�R1(�0 + �(� − �0)) (43)

where �0 is a �xed angle and where the constant K� is chosen in such a way that

K�=(

ss1

)�+ o(�); �→ ∞ (44)

here s is a constant and

s1 =∫ ∞

−∞R1(�) d� (45)

Then, use of the moment expansion yields

lim�→∞

R�(�) = lim�→∞

K�R1(�0 + �(� − �0))

= lim�→∞

K�

{s1��(� − �0) + o

(1�

)}

= s�(� − �0) (46)

Therefore, the path C� approaches in the limit a facet of length s, normal angle �0, and initialpoint P0. See Figure 11, where an initial path approximates a horizontal facet in the limit.Conversely, if we start with any path with normal-angle parametrization P(�), a6 �6 b,

we can approximate it by a polygonal line with vertices P0; n, P1; n; : : : ; Pn; n, where Pk; n=P(a+ k(b− a)=n), and consequently the radius of curvature R(�) of the path is approximatedby

Rn(�)=n−1∑k=0

sk; n�(� − �k; n) (47)



Figure 11.

Figure 12.

where sk; n is the length of the segment from Pk; n to Pk+1; n and where �k; n is the correspondingnormal angle. Observe that Rn(�) tends to R(�) in the distributional sense even if the path isnot recti�able.Let C0 be any path that turns to the left. We shall now consider the paths obtained from

C0 by parallel propagation. For any �∈R the path C� is obtained from C0 by drawing at eachpoint the normal direction and moving along this direction c� units, where c is a constant,the speed of the parallel propagation. This geometric construction breaks down, of course, atcorners and non-smooth parts of the path. Figure 12 shows how a three petal rose evolveswhen subject to parallel propagation.


442 R. ESTRADA

When we describe the initial path by its distributional radius of curvature,

R0(�); a6 �6 b (48)

then we can de�ne the path C� without any restrictions by setting

R�(�)=R0(�) + c�; a6 �6 b (49)

In this case a path with a corner will propagate into a path with similar shape but whosecorner has been rounded. An example is provided by Figure 5 that shows the path obtainedwhen parallel propagation is applied to the square in Figure 4.It is interesting to observe that parallel propagation arises when considering the solutions

of the wave equation. Indeed, if u(x; y; t) is a solution of the wave equation

2u=O2u − 1c2

@2u@t2

= 0 (50)

that when t=0 vanishes outside a convex region �0 bounded by a path C0, then for anyt¿0 the solution will vanish outside a region �t bounded by a path Ct , obtained from C0 byparallel propagation. If A is the jump of u across the front Ct , then [2,16,17]

dAd�=�A (51)

In particular, along the parallel propagation of a facet, the strength of the jump remainsconstant.We would also like to point out that the approximation of delta functions by spike functions

in the radius of curvature distributions has been studied by Xin and Wong [5].

6. DIFFERENTIAL EQUATIONS

We now consider the solution of some di�erential equations along a path with a distributionalparametrization. Namely, let us study

d2�d�2

+!2�= cR(�) (52)

where c is a constant and where !¿0, along a path C with normal angle distributionalparameter �. We observe that the equation for the reduced surface energy on the boundary ofa solid crystal satis�es (52) with !=1 and c= e=�, where e is the equilibrium chemicalpotential and where � is the atomic volume [3].The solution of (52) follows by applying the usual variation of parameters method. The

result is

� = a1 cos!�+ a2 sin!�+1!

∫ �

−∞cR( ){sin!� cos! − cos!� sin! } d

where a1 and a2 are arbitrary constants, or

�=(a1 +

c!x!

)cos!�+

(a2 +

c!y!

)sin!� (53)



where

x!=−∫ �

−∞R( ) sin! d ; y!=

∫ �

−∞R( ) cos! d (54)

are the coordinates along a new path C(!) obtained from the original path C by a contractionor stretching in the positive � direction by multiplying by the constant !. When !=1 thenx!= x and y!=y, and (53) reduces to the solution found in Reference [3].Equation (53) is very interesting along a facet or at a corner. Indeed, along a facet of C,

the corresponding path C(!) will also have a facet; if �0 is the normal angle on the facetthen x! and y! are not de�ned for �= �0 and so �(�) is not de�ned when �= �0. However,� is actually de�ned as a function of (x; y), a point moving along the facet: �= �(x; y). At acorner of the path, on the other hand, (53) gives � as a function of �, since for the anglescorresponding to the corner both x! and y! remain constant, since the path C(!) has alsoa corresponding corner; but � will not be a function of (x; y) since several values of � areobtained at the corner.

7. INTEGRALS

In this section we explain how one needs to handle integrals given in terms of the normalangle distributional parametrization.Let C be a path that turns to the left, with normal angle parameter �, a6 �6 b, so that

x(�) and y(�) are distributions. Consider the path integral

I =∫Cp(x; y) dx + q(x; y) dy (55)

where p and q are continuous functions of x and y in a region that contains the path. Then,using (22), we formally obtain

I =∫ ∞

−∞{−p(x(�); y(�)) sin �+ q(x(�); y(�)) cos �}R(�) d� (56)

Observe however that if the radius of curvature R(�) contains a delta function term s�(�−�1),corresponding to a facet of strength s with normal angle �1, then (56) cannot be evaluatedsince the expression inside the brackets is not de�ned at �= �1, and actually it has a jumpdiscontinuity at �= �1. It turns out that, because I is also given by (55), that the evaluationof the delta function at this function of x and y is given by

{−p̂ sin �1 + q̂ cos �1}s (57)

where p̂ and q̂ are the averages of p(x; y) and q(x; y), respectively, over the facet.

ExampleConsider the line integral

I =∫Cx dy (58)


444 R. ESTRADA

Suppose the path C is a polygonal line with radius of curvature

R(�)=N∑

n=1sn�(� − �n) (59)

Then

I =∫ ∞

−∞x cos �

N∑n=1

sn�(� − �n) d�

=N∑

n=1x̂nsn cos �n

=N∑

n=1x̂n[y]n (60)

where x̂n is the average of x along the facet corresponding to �n and where [y]n= sn cos �n

is the increment of y along this facet.In particular, if the polygonal path C is simple and closed, enclosing a region �, then I

would be the area of �, and we obtain

area(�)=N∑

n=1x̂n[y]n (61)

the expected formula.

REFERENCES

1. Kanwal RP. Generalized Functions: Theory and Technique (2nd edn). Birkhauser: Boston, 1998.2. Estrada R, Kanwal RP. Applications of distributional derivatives to wave propagation. Journal of Institute ofMathematics and its Applications 1980; 26:39–63.

3. Xin T, Wong H. A �-function model of facets. Surface Science Letters 2002; 487:L529–L533.4. Xin T, Wong H. Grain-boundary grooving by surface di�usion with strong surface energy anisotropy. ActaMaterialia 2003; 51:2305–2317.

5. Xin T, Wong H. A spike-function model of facets. Materials Science and Engineering A 2004; 364:287–295.6. Golovin AA, Davis SH, Nepomnyashchy AA. A convective Cahn–Hilliard model for the formation of facetsand corners in crystal growth. Physica D 1998; 122:202–230.

7. Taylor JE, Cahn JW. Di�use interfaces with sharp corners and facets: phase �eld models with strong anisotropicsurfaces. Physica D 1998; 112:381–411.

8. Campos Ferreira J. Introduction to the Theory of Distributions. Longman: London, 1997.9. Dieudonn J. Foundations of Modern Analysis. Academic Press: New York, 1969.10. Pommerenke Ch. Boundary Behaviour of Conformal Maps. Springer: Berlin, 1992.11. Struik DJ. Lectures on Classical Di�erential Geometry (2nd edn). Dover: New York, 1988.12. Baeumer B, Lumer G, Neubrander F. Convolution kernels and generalized functions. Generalized Functions,

Operator Theory, and Dynamical Systems, C.R.C. Research Notes in Mathematics. Chapman & Hall: London,1998.

13. van der Waerden BL. Ein einfaches Beispiel einer nicht-di�ernzierbaren stetigen Funktion. MathematischeZeitschrift 1930; 32:474–475.

14. Estrada R, Kanwal RP. A distributional theory for asymptotic expansions. Proceedings of the Royal Society ofLondon, Series A 1990; 428:399–430.

15. Estrada R, Kanwal RP. A Distributional Approach to Asymptotics. Birkhauser: Boston, 2002.16. Kanwal RP. Expansion coecients of wave fronts in continuum mechanics. Applied Scienti�c Research 1964;

10:435–440.17. Kline M. A note on the expansion coecient of geometrical optics. Communications on Pure and Applied

Mathematics 1960; 14:473–479.


Distributional radius of curvature

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