Distributions in Spaces with Thick Points II
Yunyun Yang
Louisiana State University
yyang18@math.lsu.edu
May 18, 2012
Yunyun Yang (Louisiana State University)
Distributions in Spaces with Thick Points II
May 18, 2012
1 / 44
Preliminaries
We shall need to consider the di¤erentiation of functions and distributions de…ned
only on a smooth hypersurface of R n : Naturally, if (v )1
n 1 is a local
Gaussian coordinate system and f is de…ned on then one may consider the
derivatives @f =@v ; 1
n 1: However, it is many times convenient and
necessary to consider derivatives with respect to the variables (xj )1 j n of the
surrounding space R n : The
derivatives are de…ned as follows.
De…nition
Suppose f is a smooth function de…ned in and let F be any smooth extension
of f to an open neighborhood of in R n ; the derivatives @F =@xj will exist, but
their restriction to will depend not only on f but also on the extension
employed. However, it can be shown that the formulas
f
=
xj
@F
@xj
nj
dF
dn
(1)
;
where n = (nj ) is the normal unit vector to and where dF =dn = nk @F =@xk
is the derivative of F in the normal direction.
Yunyun Yang (Louisiana State University)
Distributions in Spaces with Thick Points II
May 18, 2012
2 / 44
Preliminaries
Fact
Delta derivatives f = xj ; 1
extension.
Yunyun Yang (Louisiana State University)
j
n; that depend only on f and not on the
Distributions in Spaces with Thick Points II
May 18, 2012
3 / 44
It can be shown that
f
@f @v
=
;
xj
@v @xj
(2)
In this talk, we suppose now that the surface is Sn 1 ; the unit sphere in
Rn : Let f be a smooth function de…ned in Sn 1 ; that is, f (w) is de…ned if
w 2 Rn satis…es jwj = 1: While (1) can be applied for any extension F of
f ; the fact that our surface is S allows us to consider some rather natural
extensions. In particular, there is an extension to Rn n f0g that is
homogeneous of degree 0; namely,
F0 (x) = f
x
r
(3)
;
where r = jxj : Since dF0 =dn = 0 we obtain
f
@F0
=
xj
@xj
Yunyun Yang (Louisiana State University)
(4)
:
S
Distributions in Spaces with Thick Points II
May 18, 2012
4 / 44
Preliminaries
(
)
xj
=
+
xj
xj
(5)
;
Lemma
T
T
( )
=
xj
xj
+
(6)
;
xj
Proof.
*
T
( )
;
xj
+
=
=
Yunyun Yang (Louisiana State University)
;
;
=
xj
(
;
)
xj
xj
=
xj
*
Distributions in Spaces with Thick Points II
T
xj
+
xj
;
+
May 18, 2012
:
5 / 44
Preliminaries
Thus
T
xj
T
=
And
(
1)
xj
T
1
xj
=
T
=
(n
1
xj
+
xj
;
1) nj
So
|
ni
=
xj
ij
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(ni nj )
(n
1) (ni nj ) =
Distributions in Spaces with Thick Points II
ij
n
1 (ni nj )
May 18, 2012
6 / 44
A Short Review
Problem
Z
1
cos (2kx) dx:
0
Proof.
[1]
Z
0
1
Z
1
cos (2kx) dx: =
2
Z
1
=
2
=
Yunyun Yang (Louisiana State University)
+1
cos (2kx) dx
1
+1
e 2ikx dx
1
(2k) =
2
Distributions in Spaces with Thick Points II
(k)
May 18, 2012
7 / 44
A Short Review
On the other hand,
Proof.
[2]
Z
1
cos (2kx) dx =
0
sin (2kx)
2k
1
x =0
sin (2kx)
:
x !1
2k
= lim
By de…nition of a distribution, we must evaluate this limit on a test
function, f (k), with support in [0; 1) :
Z 1
Z 1
sin (2kx)
sin (2kx)
f (k) dk = lim
f (k) dk;
lim
x !1 0
x
!1
2k
2k
0
Z 1
1
sin (u)
= f (0)
du = f (0) :
2
u
4
0
R1
So 0 cos (2kx) dx = 4 (k) :
Yunyun Yang (Louisiana State University)
Distributions in Spaces with Thick Points II
May 18, 2012
8 / 44
A Short Review
Solution
[Correction of proof 1]
Z
Z 1
1
cos (2kx) dx: =
2
0
Z
1
=
2
Here
Z
+1
1
+1
cos (2kx) dx
1
+1
1
1
1; 2kg;
e 2ikx dx = Ffe
2
= e (2k) =
2
e (k) :
(7)
e 2ikx dx = Ffe
1; 2kg = 2 e (2k) = e (k) :
is the Fourier transform of the funcion 1 in the space W 0 and result in S 0 ;
the thick point space. This result holds for k positive or negative.
Yunyun Yang (Louisiana State University)
Distributions in Spaces with Thick Points II
May 18, 2012
9 / 44
A Short Review
Solution
[Correction of proof 2]
As were pointed out, in solution 2, we have ”secretly” multiplied H (k) :
Z 1
Z +1
sin (2kx)
sin (2kx)
H (k) f (k) dk = lim
f (k) dk
lim
x !1 0
x !1
2k
2k
1
In fact if we want the result for k > 0, we need to apply the projection
multiplicaiton operator MH0 : S 0 ! S 0 :
Z 1
H (k)
cos (2kx) dx = MH0 e (k) =
(k)
(8)
2
4
0
Now the consistency of the results holds.
Yunyun Yang (Louisiana State University)
Distributions in Spaces with Thick Points II
May 18, 2012
10 / 44
Space of Test Functions on Rn with a Thick Point
De…nition
A function
de…ned on Rn is in D
(a + x) =
;a
(Rn ) i¤
(a + r w)
1
X
aJ (w) r J
J =N
where N is an integer, and w 2Sn 1 ; aJ (w) 2 D Sn 1 .
Moreover, we require the the asymptotic development to be "strong".
Namely, for any di¤erentiation operator (@=@x)p = (@ p1 :::@ pn ) =@x1p1 :::@xnpn
, the asymptotic development of (@=@x)p (x) exists and is equal to the
1
P
term-by-term di¤erentiation of
aJ (w) r J :
J =N
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Distributions in Spaces with Thick Points II
May 18, 2012
11 / 44
We use D (Rn ) to denote D
Yunyun Yang (Louisiana State University)
;0
(Rn ) :
Distributions in Spaces with Thick Points II
May 18, 2012
12 / 44
Space of Test Functions on Rn with a Thick Point
De…nition
De…ne D
[k ]
(Rn ) as the subspace consists of test functions
(r w)
1
X
aJ (w) r J
J =k
Notice that D
[k ]
(Rn ) is not closed under di¤erentiation.
Note: In particular, if is a smooth function, then
1
P
[0]
(a + r !) a0 +
aJ (!) r J 2 D (Rn ). So Da (Rn )
J =1
Yunyun Yang (Louisiana State University)
Distributions in Spaces with Thick Points II
D
;a (R
n) :
May 18, 2012
13 / 44
The Topology–to have a TVS
De…nitions
De…ne a seminorm.
(@=@x)p (x)
lP1
aj ;p (w) r j
j =N jpj
rl
jj jjl ;m = sup
jpj m
where
(@=@x)p (x)
1
X
;l
N
jpj
aJ ;p (w) r J
(9)
(10)
J =N jpj
A sequence f g in D (Rn ) converges to ' i¤ there exists an integer N
[N ]
such that ' 2 D (Rn ) ;and a compact set K such that for any l; m; we
have jj
jjl ;m ! 0 as ! 1:
Yunyun Yang (Louisiana State University)
Distributions in Spaces with Thick Points II
May 18, 2012
14 / 44
Space of Test Functions on Rn with a Thick Point
Fact
D (Rn ) is a subspace of D (Rn )
i
: D (Rn ) ! D (Rn )
(11)
,!
Yunyun Yang (Louisiana State University)
Distributions in Spaces with Thick Points II
May 18, 2012
15 / 44
Space of Distributions on Rn with a Thick Point
De…nitions
The space of distributions on Rn with a thick point is the dual space of
that contains all the continous linear functionals of the test functions. We
denote it D0 (Rn ) :
Yunyun Yang (Louisiana State University)
Distributions in Spaces with Thick Points II
May 18, 2012
16 / 44
Space of Distributions on Rn with a Thick Point
De…nitions
The space of distributions on Rn with a thick point is the dual space of
that contains all the continous linear functionals of the test functions. We
denote it D0 (Rn ) :
Theorem
i
D ,! D
(12)
;a :
D0 ;a ! D0 :
, the projection operator is given explicitly as
h (f ) ; iD0
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D
= hf ; i ( )iD0 ;a
Distributions in Spaces with Thick Points II
D
;a
(13)
:
May 18, 2012
17 / 44
Space of Distributions on Rn with a Thick Point
Theorem
i
D ,! D
D
0
;a
;a :
! D0 :
, the projection operator is given explicitly as
h (f ) ; iD0
D
= hf ; i ( )iD0 ;a
D
;a
:
Theorem
Let g 2 D0 ; there exists a distribution f 2 D0 ;a ,s.t.
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Distributions in Spaces with Thick Points II
(f ) = g .
May 18, 2012
18 / 44
Space of Distributions on Rn with a Thick Point
Example
Suppose f (x) is a locally integrable funcion in Rn ; homogeneous of degree
0 Now let’s de…ne a "thick delta function" f 2 D0 (Rn ):
Let be a test function in D (Rn ), thus by de…nition could be
asymptotically expanded as
1
P
aJ (w) r J = aN (w) r N + ::: + a0 (w) + a1 (w) r + :::
J =N
Then f
hf
is given by
; iD0 (Rn )
D (Rn )
Yunyun Yang (Louisiana State University)
:=
1
Cn
1
=
Cn
1
1
hf (w) ; a0 (w)iD0 (Sn 1 ) D
Z
f (w) a0 (w) d (w)
Sn
(Sn
1)
(14)
1
Distributions in Spaces with Thick Points II
May 18, 2012
19 / 44
Space of Distributions on Rn with a Thick Point
Example
When n = 3;
; i=
hf
1
4
Z
Sn
f (w) a0 (w) d (w)
1
Example
In particular, if f (x)
h ; iD0 (Rn )
1; thenf
D (Rn )
:
=
: We may call
=
=
:
1
Cn 1
Z
1
Cn
1
h1; a0 (w)iD0 (Sn
Sn
1
1)
D (Sn
1)
a0 (w) d (w)
the "plain thick delta function".
Yunyun Yang (Louisiana State University)
Distributions in Spaces with Thick Points II
May 18, 2012
20 / 44
Projection of a Thick Delta Function onto the Usual
Distribution Space
Example
2 D0 (Rn ) can be asymptotically expanded as
1
P
(r w) a0 +
aJ (w) r J ; so
Since a usual test function
it’s Taylor expansion:
J =1
h (f
) ; i = hf
1
Cn 1 Sn
Z
a0
=
Cn 1 Sn
=
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; i ( )i
Z
1
(15)
f (w) a0 d (w)
f (w) d (w)
(16)
1
Distributions in Spaces with Thick Points II
May 18, 2012
21 / 44
Projection of a Thick Delta Function onto the Usual
Distribution Space
Example
In particular, for a plain thick delta function, it projects onto the usual
delta function:
Z
a0
d (w)
(17)
h ( ); i =
Cn 1 Sn 1
= a0 = (0)
=h ; i
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Distributions in Spaces with Thick Points II
May 18, 2012
22 / 44
Space of Distributions on Rn with a Thick Point
De…nition
(thick delta functions of degree m) A thick delta functions of degree m,
[m]
denoted f
; acting on a thick test function (x) is de…ned as the action
of f on am (w) in the corresponding asymptotic expansion divide by the
surface area of Sn 1 : Namely,
D
f
[m]
;
E
D0 (Rn ) D (Rn )
=
1
Cn
1
hf ; am (w)iD0 (Sn
1)
D (Sn
1)
Example
If f is a locally integrable funcion in Rn ; homogeneous of degree 0, a
natural example would be
D
f
[m]
;
E
D0 (Rn ) D (Rn )
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1
Cn
1
=
Cn
=
1
1
hf ; am (w)iD0 (Sn 1 ) D (Sn 1 )
Z
f (w) am (w) d (w)
Sn
1
Distributions in Spaces with Thick Points II
May 18, 2012
23 / 44
Space of Distributions on Rn with a Thick Point
De…nitions
Let f ; g 2 D0 (Rn );and (x) 2 D (Rn ) is a test function. We de…ne the
following algebraic operators:
1
2
3
4
hf + g ; i = hf ; i + hg ; i :
hf (Ax) ; (x)i = jdet1 Aj f (x) ; A 1 x : where A is a non-singular
n n matrix. In particular, hf ( x) ; (x)i = hf (x) ; ( x)i
hf (x + c) ; (x)i = hf (x) ; (x
h f ; i = hf ;
8 2 D ;a
i ; where
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c)i ; where c 2 Rn :
is a multiplier of D
Distributions in Spaces with Thick Points II
;a ;
i.e.
2D
;a ,
May 18, 2012
24 / 44
Derivatives on Thick Distributions
De…nition
The p
@
@x
th order derivative of a thick distribution f 2 D0 is given by
p
f;
= ( 1)jpj f ;
@
@x
p
= ( 1)jpj f ;
(@ p1 :::@ pn )
@x1p1 :::@xnpn
,
We can call it "thick distributional derivative" to indicate the space D0 ; in
which f sits.
Example
a …rst order parital derivative on f may be given by
@ f
;
@xj
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=
f;
@
@xj
Distributions in Spaces with Thick Points II
May 18, 2012
25 / 44
Derivatives on Thick Distributions
Lemma
Suppose f 2 D0 ; and the projection of f onto D0 is
p
(@ =@x)p f = @=@x g :
(f ) = g . Then
Proof.
if is an ordinary test function that is in D (Rn ); i denotes the inclusion
map from D to D ; the projection of f from D0 to D0 is (f ) = g , then
we have,
@
@x
p
f ;i ( )
= ( 1)jpj f ;
= ( 1)jpj
Yunyun Yang (Louisiana State University)
@
@x
(f ) ;
p
@ p
= ( 1)jpj f ; i
@x
*
+
p
@
@ p
=
g;
@x
@x
Distributions in Spaces with Thick Points II
May 18, 2012
26 / 44
More about derivatives
Because aJ (w)0 s are …nite on Sn 1 , the asymptotic expansion,
1
1
P
P
aJ (x=r ) r J :etc.
aJ (w) r J =
(r w)
J =k
J =k
So
1
X @aJ (x=r )
@
=
r J + JaJ (x=r ) nj r j
@xj
@xj
1
J =k
Yunyun Yang (Louisiana State University)
Distributions in Spaces with Thick Points II
May 18, 2012
27 / 44
Distributional Derivative of 1=r
Lemma
[m]
The partial derivative of the thick delta function wi
is
[m]
@ ni
= [ ij + ( m 1 n) ni nj ]
@xj
with respect to xj
[m+1]
where ij is the Kronecker delta function, m is the degree of wi
the dimension of Rn ; wi = xi =r ; wi = xi =r :
[m]
; n is
Proof.
*
@ ni
Yunyun Yang (Louisiana State University)
[m]
@xj
;
+
=
ni
[m]
;
Distributions in Spaces with Thick Points II
@
@xj
May 18, 2012
28 / 44
Distributional Derivative of 1=r
Proof.
1
=
Cn
1
Cn
=
1
Cn 1
1
=
Cn 1
1
1
Z
am+1 (w)
+ (m + 1) nj am+1 (w) d (w)
xj
Sn 1
am+1 (w)
ni ;
xj
D0 (Sn 1 ) D (Sn 1 )
ni
hni ; (m + 1) wj am+1 (w)iD0 (Sn
1)
|
ni
; am+1 (w)
xj
Since
D0 (Sn
1)
D (Sn
D (Sn
1)
1)
D
(m + 1) ni nj
[m+1]
;
E
D0 (Rn
|
ni
=
xj
ij
n (ni nj )
The result is obtained.
Yunyun Yang (Louisiana State University)
Distributions in Spaces with Thick Points II
May 18, 2012
29 / 44
Distributional Derivative of r
In his paper, [4,Franklin] brought up a question: As a distribution, the
well-known formula of the second derivative of 1=r
2
@
@xi @xj
1
r
=
r2
3xi xj
4
3
ij
r5
ij
(x)
cannot act on functions that are not smooth at the orign.
In other words,
2
@
@xi @xj
1
r
=
Yunyun Yang (Louisiana State University)
r2
3xi xj
r5
ij
4
3
ij
(x) 2 D0 but 2
= D0
Distributions in Spaces with Thick Points II
May 18, 2012
30 / 44
Distributional Derivative of 1=r
De…nition
Let
2 D (Rn ) ;
D
Pf r
;
E
= F :p: lim
Z
"!1 jxj "
r
(x) dx
Hence
@
@x
p
Pf r
;
jpj
= ( 1)
Pf r
= ( 1)jpj F :p: lim
;
Z
@
@x
"!1 jxj "
Yunyun Yang (Louisiana State University)
Distributions in Spaces with Thick Points II
r
p
@
@x
p
(x) dx
May 18, 2012
31 / 44
Distributional Derivative of r
Lemma
We denote Sn" 1 the n-1 sphere Rwith radius ":
De…ne r nj Sn" 1 ; (x) = Sn" 1 " nj (x) dx:for any thick test
function ;then
D
E
lim r wj Sn" 1 ; (x)
"!0
(
= 0D
if 2
=Z
E
[1 n ]
= Cn 1 nj
; (x) 0 n 1
if 2 Z
n 1
D (S
) D (S
)
Proof.
D
r nj
Sn" 1
; (x)
E
=
Z
Sn"
Z
" nj (x) dx
" nj ("w) "n
Z
Distributions in Spaces with Thick Points II
=
Sn
Yunyun Yang (Louisiana State University)
1
1
d (w)
1
May 18, 2012
32 / 44
Distributional Derivative of r
Theorem
@
pf r
@xj
where Cn
1
=
(
xj Pf r
xj Pf r
2
2
;
+ Cn
is the surface area of the n
Yunyun Yang (Louisiana State University)
1 nj
[
n+1]
2
=Z
2Z
(18)
1 unit sphere.
Distributions in Spaces with Thick Points II
May 18, 2012
33 / 44
Distributional Derivative of r
Proof.
By de…nition,
@
Pf r
@xj
;
=
Pf r
= F :p: lim
Z
@
;
@xj
"!1 jxj "
=
F :p: lim
Z
"!1 jxj "
@H (r ") r
@xj
dx =
We already know the usual distributional derivative of H (r
by [2,Kanwal]
@
H (r
@xj
Yunyun Yang (Louisiana State University)
") r
= xj r
2
H (r
r
@
dx
@xj
(19)
@H (r ") r
;
@xj
") r is given
") + r nj (S" )
Distributions in Spaces with Thick Points II
May 18, 2012
34 / 44
Distributional Derivative of r
Proof.
So equation 19 becomes
@
H (r ") r ;
"!1 @xj
D
E
= F :p: lim
xj r 2 H (r ") + r nj (S" ) ;
"!1
D
E
D
E
= xj Pf r 2 ;
+ F :p: lim r nj (S" ) ;
F :p: lim
"!1
By 29, lim r nj
"!0
Sn"
1
; (x) = Cn
1 nj
[1 n
]
So the theorem
holds.
Example
When n = 3;
Yunyun Yang
=
1;the …rst derivative of 1=r is
@
pf r 1 = xj Pf r 3 + 4 nj
@x
j
(Louisiana State University) Distributions in Spaces with Thick Points II
[ 1]
May 18, 2012
35 / 44
Second Order Distributional Derivative of r
Lemma
xj
[Q ]
= wj
[Q 1]
:
Theorem
If
is an integer,
@ 2
@xj @xk
pf r
=
jk
Pf r
+Cn
Yunyun Yang (Louisiana State University)
1
(2
2
+ (
2) nj nk
2) xj xk Pf r
[
Distributions in Spaces with Thick Points II
n+2]
+ Cn
4
1 jk
[
May 18, 2012
n+2]
36 / 44
Second Order Distributional Derivative of r
Proof.
Take the derivative of the …rst order derivative
[
n+1]
@
= xj Pf r 2 + Cn 1 nj
, we have
@xj pf r
@ 2
@xj @xk
pf r
=
@ (xj )
Pf r
@xk
=
jk
1
Yunyun Yang (Louisiana State University)
Pf r
@xk
2
2) Pf r
4
@
@xk
Pf r
+ x j Cn
+ xj
n+1]
[
nj
@
+Cn
2
1 nk
2
+ xj (
[
n+3]
@
+ Cn
Distributions in Spaces with Thick Points II
1
nj
[
n+1]
@xk
May 18, 2012
37 / 44
Second Order Distributional Derivative of 1=r
Proof.
Together with the lemma 27,
@ nj
[
n+1]
=[
@xk
jk
+(
And by lemma 32, xj
So,
[
2) nj nk ]
n+3]
= nj
[
n+2]
[
n+2]
:
:
Theorem
If
is an integer,
@ 2
@xj @xk
pf r
=
jk
Pf r
+Cn
Yunyun Yang (Louisiana State University)
1
( n
2
+ (
1) nj nk
2) xj xk Pf r
[
Distributions in Spaces with Thick Points II
n+2]
4
[
n+2]
May 18, 2012
38 / 44
+ Cn
1 jk
Second Order Distributional Derivative of 1=r
Theorem
If
is an integer,
@ 2
@xj @xk
pf r
=
jk
Pf r
+Cn
1
(2
2
+ (
2) nj nk
4
2) xj xk Pf r
n+2]
[
+ Cn
1 jk
[
n+2]
Example
If n = 3;
=
1;
@ 2
pf r
@xj @xk
1
= 3xj xk Pf r
16 nj nk
Yunyun Yang (Louisiana State University)
5
+4
Distributions in Spaces with Thick Points II
jk Pf
r
3
jk
May 18, 2012
39 / 44
Second Order Distributional Derivative of 1=r
So in the thick point spaces,
@
@xj @xk
1
r
In particular, if
Z
16 xj xk
r2
R3
=
r2
3xj xk
r5
jk
16 xj xk
r2
(x)
+4
(x) 2 D (Rn ) ; i:e:; a0 is a constant, then
Z
(x) (x) dx = 4a0
ni nj d (!)+4 jk a0 =
S2
Z
R3
4
jk
(x) (x) dx = 4
(20)
jk
a0
16
3
jk
jk a0
Hence the projection is given explicitly as:
(
@
@xi @xj
Yunyun Yang (Louisiana State University)
3xi xj r 2
1
)=
r
r5
ij
4
3
Distributions in Spaces with Thick Points II
ij
(x)
May 18, 2012
(21)
40 / 44
Second Order Distributional Derivative of 1=r
Conclusion:
@
@xi @xj
1
r
=
r2
3xi xj
r5
ij
16 xj xk
r2
(x)
+4
jk
(22)
is a thick distribution.
(
@
@xi @xj
3xi xj r 2
1
)=
r
r5
ij
4
3
ij
(x)
(23)
is a usual distribution.
Yunyun Yang (Louisiana State University)
Distributions in Spaces with Thick Points II
May 18, 2012
41 / 44
References
Vibet, C. (1999)
Transient analysis of energy equation of dynamical systems,
IEEE Trans. Edu. 42, 217-219.
Kanwal, R.P. (1998)
Generalized Functions: Theory and Technique, second edition.
Birkhauser, Boston.
Ricardo Estrada and Stephen A. Fulling. (2007)
Functions and Distributions in Spaces with Thick Points.
International Journal of Applied Mathematics & Statistics, Vol 10, No. S07, 25-37.
Jerrold Franklin. (2010)
Comment in "Some novel delta-function identities".
American Journal of Physics, November 2010 – Volume 78, Issue 11, pp. 1225
Yunyun Yang (Louisiana State University)
Distributions in Spaces with Thick Points II
May 18, 2012
42 / 44
References
Estrada, R., Essays in the Theory and Applications of Generalized Functions, Ph.
D. Dissertation, Pennsylvania State University, 1980.
[?]Estrada, R. and Kanwal, R. P., Distributional analysis of discontinuous …elds,
J.Math.Anal.Appl. 105 (1985), 478-490.
[?]Estrada, R. and Kanwal, R. P., Higher order fundamental forms of a surface and
their applications to wave propagation and distributional derivatives, Rend. Cir.
Mat. Palermo 36 (1987), 27-62.
Yunyun Yang (Louisiana State University)
Distributions in Spaces with Thick Points II
May 18, 2012
43 / 44
Thank you!
Yunyun Yang (Louisiana State University)
Distributions in Spaces with Thick Points II
May 18, 2012
44 / 44