A Generalized Algorithm for Solving Nonlinear Equations Using Distributions

In this paper we introduce a method for solving the nonlinear equation f (x) = 0 where x, f (x) are two vectors. This method uses the theory of distributions, and it’s a non-iterative method, indeed it creates a sequence of vectors with an explicit formula and this sequence will converge to the solution of the nonlinear equation.


Introduction
In general, iterative methods are used for solving nonlinear equations ( f (x) = 0), thus a starting point of the iterative sequence is required for these methods. The starting point must be in a neighborhood and close enough to the solution of the nonlinear equation so that the sequence converges to the solution, which is a disadvantage for these methods. This paper is a generalization of (Houssein, 2018) in order to treat the case where x and f (x) are two vectors. therefore the method in this paper will produce a sequence of vectors with an explicit formula, where the solution of the nonlinear equation is the limit of this sequence. In the following we will study a sequence of distributions in order to present the algorithm responsible of the creation of this sequence which converges to the solution, in addition two examples are presented to illustrate the method.
The explicit formula of the sequence and the requirement of only few conditions for the function f in a neighborhood of the solution are the advantages of this method. (Houssein, 2018), (Schwartz, 1963), (Schwartz, 1966) Let Ω ⊆ R d be an open set with d ∈ N * , and let f a locally integrable function on Ω, f : Ω → R, then the considered distribution T is the following:

Distributions and Test Functions
Where λ is the Lebesgue measure, D(Ω) the space of the test functions on Ω. T is also noted by f , in other words < f, ϕ >=< T, ϕ >.
An example of a test function belonging to D(R d ) is the following:

Building a Sequence of Distributions That Converges to a Dirac
In the following we consider a function g :Ω → R where Ω ⊂ R d is a non-empty, bounded and open set with d ∈ N * . The function g is supposed to be a positive function g(x) ≥ 0 ∀x ∈Ω. Using the theorem 1 we will be able to create a sequence of distributions which converges to the Dirac distribution.
Theorem 1 We suppose that the function g has only one zero x * inΩ and x * ∈ Ω such that g is C 2 in a neighborhood of x * and the Hessian matrix at x * : Hess x * (g) is positive definite, in addition we suppose that exists a neighborhood of x * which contains a finite number of critical points of g , if we consider the sequence of functions g n defined by: ) d e −n 2 g(x) then: In the sense of distributions Where c = det(Hess x * (g)) , δ x * the Dirac distribution at x * . The convergence in the sense of distributions means that: Proof. At first because g ≥ 0, the function g reaches its minimum at x * , thus: Using the Taylor's theorem g can be expressed by: 0 and where: The Hessian matrix at x * : Hess x * (g) is symmetric and positive definite therefore: We obtain that: For a ϕ ∈ D(R d ) we consider the distribution sequence: If we make a substitution y = x − x * we obtain: Where Another substitution is made x = Oy thus we obtain: Where OΩ * = {O.y | y ∈ Ω * }, OΩ * is also a neighborhood of 0, indeed because Ω * is a a neighborhood of 0 then there exist a ball In addition we have Then ∥Ox∥ = ∥x∥ and In the other sens let y ∈ R d such that ∥y∥ < r then y Back to the substitution x = Oy , the Jacobian matrix of this transforamtion is equal to Jac = O −1 , therefore |det(Jac)| = 1. We know also that Let the sequence E n be defined by: we will prove next that: ) d e − 1 2 n 2 ( t x.D.x) dλ(x) then: Let the substitution be y = nx ⇐⇒ (y 1 , . . . , y d ) = n(x 1 , . . . , x d ) then the Jacobian matrix of this transformation is equal to Jac = 1 n I d and |det(Jac)| = ( 1 n ) d . We will obtain: Using Fubini's theorem: Using the substitution y = nx for E n : Otherwise we have: The application e − 1 2 (c 1 y 2 1 +...+c d y 2 d ) .ϕ(O −1 . y n + x * ) is measurable and the application e − 1 2 (c 1 y 2 1 +...+c d y 2 d ) is integrable, therefore using the Dominated Convergence theorem we deduce that: Let G n and F n be two sequence of functions defined on R d by: We will prove next that: By making the substitution y = nx: Besides we have ϵ(y) −→ y→0 0 and: Let 0 < η < 1 2 Min(c 1 , . . . , c d ) then: Vol. 10, No. 6;2018 We have ∥y∥ For n ≥ N 3 we have two cases to treat: We conclude that: We deduce that ∀n ≥ N 3 : By supposition η < 1 2 Min(c 1 , . . . , c d ) then: In the other hand for y fixed: Using the Dominated Convergence theorem, we deduce that: ∫ The second part of the proof will use the lemma 1 presented after the proof. The lemma 1 says that there exists a R > 0 where Otherwise let 0 < η < 1 2 k then: For n ≥ N 5 we have two cases to treat: Finally onΩ\B(x * , R) the application g is strictly positive g > 0 then g reaches its minimum at σ ∈Ω where g(σ) > 0.
Using the substitution y = nx: We conclude that: As ϕ is bounded then : The following lemma is used in the proof of the theorem 1.
Lemma 1 Let g be a positive application g : R d → R + , we suppose that there exists x * ∈ R d such that g(x * ) = 0, and that exists a neighborhood Ω of x * such that g ∈ C 1 (Ω) and such that the number of critical points of g in Ω is finite. Then Proof. We will proceed by contradiction, we will suppose that it's not true thus we have: we will construct a sequence, indeed let R n > 0 for n ∈ N thus: The application g is continuous at x * : Let ζ n such that g(ζ n ) = in f x∈C(x * ,r n ) (g(x)) therefore ∃ x n such that r n < ∥x n − x * ∥ ≤ R n and g(x n ) < g(ζ n ) ≤ g(ζ) ∀ζ ∈ C(x * , r n ) Therefore we conclude that there exists x n ∈ B(x * , R * n )\B(x * , r n ) such that: x * x n B(x * ,r n ) B(x * ,R n ) B(x * ,R n * )

Figure 1. The neighborhood Ω
Otherwise the set O n = {x ∈ R d | r n ≤ ∥x − x * ∥ ≤ R * n } is closed and bounded thus O n is a compact set. The application g is continuous on O n thus the minimum of g is reached on O n , let x * n denotes the vector where g reaches its minimum on O n . x * n does not belong to ∂O n because as we saw above, ∃ x n belongs to the interior of O n and g(x n ) < g(y) ∀y ∈ ∂O n . Therefore we deduce that x * n is a critical point.
Second we will repeat the process for the step n + 1 by taking R * n+1 such that R * n+1 < ∥x * n − x * ∥, therefore we will obtain an infinity of critical points in Ω, which leads to a contradiction.
In general we have the following theorem if g has several zeros in Ω: Theorem 2 If we suppose that the function g has only k zeros x * 1 , . . . , x * k inΩ and x * 1 , . . . , x * k ∈ Ω such that g satisfies the hypotheses of the theorem 1 at each x * i , then by considering the sequence of functions g n defined by: ) d e −n 2 g(x) we have: where Ω i are disjoints and x * i ∈ • Ω i .

Application
Let (S ) be a system of equations defined by: Where ∀i = 1, . . . , d the application f i :Ω → R is a function defined onΩ where Ω ⊂ R d an open and bounded set with d ∈ N * . In addition we suppose that there exists only one zero x * of the system of equations inΩ and x * ∈ Ω, we suppose also that ∀i = 1, . . . , d the application f i is C 2 in a neighborhood of x * . Let g the application g :Ω → R defined by g(x) = f 2 1 (x) + . . . + f 2 d (x) ∀x ∈Ω, we suppose that exists a neighborhood of x * which contains a finite number of critical points of g, let g n be a sequence of functions defined by Where Jac x * ( f i ) denotes the Jacobian matrix of the application f i .
Let Jac x * (S ) be the Jacobian of the system (S ) at x * in other words: If Jac x * (S ) is invertible then: Proof. It results from the theorem 1. In fact And for two integers i and j less or equal than d As Jac x * (S ) is invertible then is a basis of R d , therefore t x.Hess x * (g).x > 0 because if t x.Hess x * (g).x = 0 ⇒ (Jac x * ( f i )) .x = 0 ∀i = 1, . . . , d then x is orthogonal to all vectors of the basis thus x is the zero vector. We conclude that Hess x * (g) is positive definite.

Climbing the Dimensions Algorithm for Finding the Solution
This section will present an algorithm to find the zero x * of the function g which satisfies all conditions presented in the theorem 1.
We take into consideration the sequence of functions g n (x) = 1 Ω (x) . Using theorem 1 we have: Where c = det(Hess x * (g)).
We consider a several test functions for this algorithm, the idea is to trap the solution each time in a sphere to finish in the intersection of these spheres, where we will have in the end 2 points where one is the solution and the other is outside the domain.
We will consider the following test function from the space D(R d ): For each test function ϕ r,a and for each n ∈ N * we call S ϕ r,a (a, ρ n ) the sphere of center a and of radius ρ n where ρ n = √ r 2 − 1 Ln(y n ) and y n = ∫ R d g n ϕ r,a dλ ∫ R d g n ϕ 2 r,a dλ . Besides if we have 2 spheres S (ξ, r) ⊂ R d and S (η, R) ⊂ R d of centers respectively ξ and η, of radius respectively r and R such that ξ η and S (ξ, r) ∩ S (η, R) ∅, then S (ξ, r) ∩ S (η, R) is a sphere in R d−1 with a radius equal to √ r 2 − ( L 2 +r 2 −R 2 2L ) 2 and with a center χ belongs to the line (ξ, η) such that χ = x 1 η−ξ L +ξ, where L = ∥η−ξ∥ and x 1 = L 2 +r 2 −R 2 2L . Now we begin the algorithm, let n ∈ N * a bit big, let a ∈ Ω and R * > 0 such that Ω ⊂ B(a, R * ). Let (e 1 , . . . , e d ) the orthonormal basis of R d , we choose η 0 such that η 0 − a = Ke 1 where K > R * .
Next we consider the intersection of the 2 spheres: Where η 1 and R 1 are easily obtained from before and η 1 ∈ (η 0 , χ 0 ).
Next we choose χ 1 such that The symbol < .|. > denotes the scalar product in R d .
, and η 0 − a is collinear to e 1 , η 0 − a = Ke 1 thus: In conclusion we consider the sequence Let r > 0 and c ∈ R d , let ϕ(x) = e −1 r 2 −∥x−c∥ 2 1 B(c,r) (x) , we take two sequences r n > 0 and c n ∈ R d such that r n −→ n→+∞ r and c n −→ n→+∞ c therefore: If we take the same notations as in the proof of the theorem 1. then: As ϕ n and ϕ are bounded and by using the Dominated Convergence theorem, we obtain If x * ∈ B(c, r) then < g n , ϕ n > < g n , ϕ 2 n > −→ n→+∞ e 1 r 2 −∥x * −c∥ 2 Let ξ n = <g n ,ϕ n > <g n ,ϕ 2 n > and H). Back to the algorithm we have: Therefore x * ∈ S (η 0 , R 0 ) and x * ∈ S (χ 0 , ρ 1 ) ⇒ x * ∈ S (η 0 , R 0 ) ∩ S (χ 0 , ρ 1 ) = S (η 1 , R 1 ). We repeat the procedure to obtain: In other words we can obtain an approximation of the solution x * by only computing x n for a big value of n.
Otherwise we can use this approximation for another algorithm for solving nonlinear systems of equations.

Example in 2D
Let us take a simple system of equations: The exact solution of this system is: (x, y) = (0, 0).

Conclusion
The method presented above is a generalization of (Houssein, 2018) and it demands only to make an integration in order to find the solution of the nonlinear equation, it presents an advantages compared to some other methods. Otherwise the approximate solution can be used as a starting point for iterative methods.