Radial Basis Function

1. Green’s Function in Regularization of RBF
Chaand Chopra
17MCMC34
April 4, 2019
Chaand Chopra (17MCMC34) Green’s Function in Regularization of RBF April 4, 2019 1 / 13

2. Frechet Differential of Tikhonov functional
Standard Error Term:
Es(F) =
1
2
N
i=1
(di − yi )2
(1)
Regularizing Term:
Ec(F) =
1
2
||DF||2
(2)
where, Dis a linear differential operator.
Chaand Chopra (17MCMC34) Green’s Function in Regularization of RBF April 4, 2019 2 / 13

3. Definition
The principle of regularization may now be stated as: Find the function
F1.. (x) that minimizes the Tikhonov functional E(F), defined by
E(F, h) = Es(F, h) + λEc(F, h) (3)
where Es(F) is the standard error term, Ec(F) is the regularizing term, and
λ is the regularization parameter.
So, quantity to be minimized in regularization theory is
E(F) = Es(F) + Ec(F) (4)
E(F) =
1
2
N
i=1
(di − yi )2
+ λ
1
2
||DF||2
(5)
The minimization of the cost functional E(F), using Frechet
Differential.
Chaand Chopra (17MCMC34) Green’s Function in Regularization of RBF April 4, 2019 3 / 13

4. The Frechet differential of a functional may be interpreted as the best
local linear approximation. Thus the Frechet differential of the
functional E(F) is formally defined by
dE(F) = [
d
dβ
E(F + βh)] (6)
dE(F, h) = dEs(F, h) + λdEc(F, h) = 0 (7)
Evaluating the Frechet differential of the standard error term Es(F, h)
of Eq. (1),we have
dEs(F) = [
d
dβ
E(F + βh)]β=0
= [
1
2
d
dβ
N
i=1
[di − F(xi ) − βh(xi )]2
]β=0
(8)
Chaand Chopra (17MCMC34) Green’s Function in Regularization of RBF April 4, 2019 4 / 13

5. dEs(F) = −
N
i=1
[[di − F(xi ) − βh(xi )]2
]h(xi )|β=0
= −
N
i=1
[[di − F(xi )]h(xi )
(9)
By using the Riesz Representation Theorem on (9), we get
dE(F) = h,
N
i=1
(di − F)δxi H (10)
where δxi is Dirac delta distribution of x, centered at xi .
., . represents the inner (scalar) product of two func. in (H).
Chaand Chopra (17MCMC34) Green’s Function in Regularization of RBF April 4, 2019 5 / 13

6. Now evaluation of the Frechet differential of dEc(F)
dEc(F) = [
d
dβ
Ec(F + βh)]β=0
=
1
2
d
dβ Rm
(D(F + βh))2
dx|β=0
=
Rm
(D(F + βh))Dhdx|β=0 =
R
DFDhdx
(11)
By using the Riesz Representation Theorem on (11), we get
dEc(F) = DF, Dh H (12)
where DF, Dh is the inner product of the two functions DF(x) and
Dh(x) that result from the action of the differential operator D on
h(x) and F(x), respectively.
Chaand Chopra (17MCMC34) Green’s Function in Regularization of RBF April 4, 2019 6 / 13

7. Eular Lagrange Function
Definition
Given a linear differential operator D, we can find a uniquely determined
adjoint operator, denoted by D , such that for any pair of functions u(x)
and v(x) which are sufficiently differentiable
Rm
u(x)Dv(x)dx =
Rm
v(x)D u(x)dx (13)
This equation is called Green’s Identity.
Comparing the left-hand side of Eq. (13) with the fourth line of Eq.
(11), we may make the following identifications:
u(x) = DF(x) (14)
Dv(x) = Dh(x) (15)
Chaand Chopra (17MCMC34) Green’s Function in Regularization of RBF April 4, 2019 7 / 13

8. Using Green’s identity, we may rewrite Eq. (10)
dEc(F) =
Rm
h(x)DD F(x)dx = h, DD F (16)
where D is adjoint of D.
We may now express the Frechet differential dE(F, h) using Eq. (16)
and Eq. (10) as
dE(F, h) = h, DD F(x) −
1
λ
N
i=1
(di − F)δxi (17)
Chaand Chopra (17MCMC34) Green’s Function in Regularization of RBF April 4, 2019 8 / 13

9. Now dE(F, h) = 0
DD F(x) −
1
λ
N
i=1
(di − F)δxi = 0 (18)
Equivalently,
DD F(x) =
1
λ
N
i=1
(di − F)δ(x − xi ) (19)
Eq. (19) is the Euler-Lagrange equation for the Tikhonov functional
E(F).
Chaand Chopra (17MCMC34) Green’s Function in Regularization of RBF April 4, 2019 9 / 13

10. Greens Function
Let G(x, ξ) denote a function in which x is a parameter and ξ as an
argument. For a given linear differential operator L, we stipulate that the
function G(x, ξ) satisfies the following conditions:
For a fixed ξ, G(x, ξ) is a function of x and ξ satisfies the prescribed
boundary conditions.
Except at the point x = ξ, the derivatives of G(x, ξ) with respect to x
are all continuous; the number of derivatives is determined by the
order of the operator L.
LG(x, ξ) =
0, if x = ξ
δ(x − ξ), if x = ξ
(20)
Chaand Chopra (17MCMC34) Green’s Function in Regularization of RBF April 4, 2019 10 / 13

11. Let ϕ(x) denote a continuous or piecewise continuous function of
x ∈ Rm. Then the function
F(x) =
Rm
G(x, ξ)ϕ(ξ)dξ (21)
is a solution of the differential equation
LF(x) = ϕ(x) (22)
Chaand Chopra (17MCMC34) Green’s Function in Regularization of RBF April 4, 2019 11 / 13

12. Solution to the Regularization Problem
Returning to the issue at hand, namely, that of solving the
Euler-Lagrange equation(19), set
L = DD (23)
ϕ(ξ) =
1
λ
N
i=1
(di − F)δ(x − xi ) (24)
is a solution of the differential equation
Fλ(x) =
Rm
G(x, ξ)
1
λ
N
i=1
(di − F)δ(x − xi )dξ (25)
Chaand Chopra (17MCMC34) Green’s Function in Regularization of RBF April 4, 2019 12 / 13

13. Fλ(x) =
1
λ
N
i=1
(di − F)
Rm
G(x, ξ)δ(x − xi )ξ (26)
Definition
Sifting Property of Dirac Delta Function:
Rm
δ(x − ξ)ϕ(ξ)dξ = ϕ(x) (27)
Finally, using the sifting property of the Dirac delta function, we get the
desired solu tion to the Euler-Lagrange equation (19) as follows:
Fλ(x) =
1
λ
N
i=1
(di − F(xi ))G(x, xi ) (28)
Chaand Chopra (17MCMC34) Green’s Function in Regularization of RBF April 4, 2019 13 / 13