1) The document discusses using Green's functions to solve the Euler-Lagrange equation that arises from minimizing the Tikhonov functional in regularization theory. 2) It defines Green's functions and shows that solving the differential equation with a Green's function leads to a solution of the form of a weighted sum involving the Green's function. 3) This solution is then applied to the specific Euler-Lagrange equation from regularization theory to arrive at an expression for the regularized solution in terms of the Green's function and the data.

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