An Algorithm for Solving Nonlinear Equations Using Distributions

In this paper we will present a new algorithm to solve the nonlinear equation f (x) = 0 where x is a scalar, indeed using the theory of distributions, we will be able to construct a sequence with an explicit formula that converges to the solution of f (x) = 0.


Introduction
Many methods have been developed in order to solve a nonlinear equation ( f (x) = 0) and find its solution, we can cite for example the Newton's method. Many of these methods lead to an iterative sequence that converges to the solution, the disadvantage of these methods is the dependency of the first value of the sequence, in other words the sequence converges to the solution if the first value is close enough to the solution.
The goal of this paper is to create a sequence with an explicit formula, which converges to the solution of a given nonlinear equation where the solution is a scalar. Therefore in the following we will present a brief recall of the theory of distributions, and by manipulating the distributions we will be able to create this sequence which converges to the solution, and we will also present an example to illustrate the algorithm.
In addition to the explicit formula of the sequence, the advantage of this method is that it requires only a few conditions for the function f in a neighborhood of the solution. (Dreyfuss, 2012), (Schwartz, 1963), (Schwartz, 1966)

Test Functions
Let Ω ⊆ R d be an open set where d ∈ N * , and let ϕ an application ϕ : Ω → R, then ϕ is a test function if and only if: 1. The support of ϕ: supp(ϕ) is compact in Ω 2. The function ϕ is infinitely differentiable: ϕ ∈ C ∞ (Ω) The space of the test functions on Ω is called D(Ω).
Let ϕ n be a sequence in D(Ω) and let ϕ ∈ D(Ω) then we say that ϕ n converges to ϕ in D(Ω): ϕ n D(Ω) −→ ϕ if and only if:

Example
The function ϕ defined by:

Distributions
Let Ω ⊆ R d be an open set, a distribution is a linear application T : D(Ω) → R such that: If ϕ n ∈ D(Ω) and ϕ n < T, ϕ > denotes T (ϕ), and D ′ (Ω) denotes the space of the distributions on Ω.

Example
Let Ω ⊆ R d be an open set, and let f a function f : Ω → R such that f is locally integrable on Ω: f ∈ L 1 loc (Ω) . If T is an application such that: Where λ is the Lebesgue measure, then T is a distribution: T ∈ D ′ (Ω). Otherwise T is also noted by f , in other words < f, ϕ >=< T, ϕ >.

Creating a Sequence of Distributions that Converges to a Dirac
We consider in the following a function g :Ī → R where I is a non-empty, bounded and open interval I ⊂ R. We suppose that g is a positive function g(x) ≥ 0 ∀x ∈Ī. Using some additional properties on g we will be able to create a sequence of distributions which converges to the Dirac distribution, this creation is done by the theorem 1.
The following lemma is used in the proof of the theorem 1.
Lemma 1 Let I 0 ⊂ I be an open interval. We suppose that g ∈ C 2 (I 0 ), g(x) ≥ 0 ∀x ∈ I 0 and there exists a zero x * ∈ I 0 for the function g: g(x * ) = 0, let g ′ denotes the derivative of the function g, we suppose in addition that there exists a finite number of zeros for the function g ′ in a neighborhood of x * . Then there exists a neighborhood of Proof. First we will prove that ∃ r > 0 such that ∀x ∈ [x * − r, x * [⇒ g ′ (x) ≤ 0. In order to prove this we will proceed by contradiction, in other words we suppose that: Let the interval J ⊂ I 0 be the neighborhood of x * which contains a finite number of zeros of the function g ′ , and let N be the number of these zeros. we take r * > 0 such that [x * − r * , x * [⊂ J and such that [x * − r * , x * [ contains no zeros of g ′ .
Therefore there exist a N + 1 zeros of g ′ in the neighborhood J of x * , which contradicts our hypothesis.
We conclude that there exist a neighborhood of Theorem 1 Let the function g have only one zero x * inĪ and x * ∈ I such that g is C 2 in a neighborhood of x * and g ′′ (x * ) > 0, we suppose in addition that there exists a finite number of zeros for the function g ′ in a neighborhood of x * , if we consider the sequence of functions g n defined by: g n (x) = 1 I (x) n √ π e −n 2 g(x) then: , δ x * the Dirac distribution at x * . The convergence in the sense of distributions means that: .
Proof. First the function g reaches its minimum at x * because g ≥ 0 thus g ′ (x * ) = 0. Using the Taylor's theorem we can express g by: 0.
Let f n denotes the sequence of functions defined by: f n (x) = n √ π e −cn 2 (x−x * ) 2 . In the following of this proof we will prove that: In order to prove this convergence we will study it on a different intervals, thus we begin by the interval We make a substitution: Using another substitution: y = nz we obtain: ∫ Otherwise we know that ϵ(x) −→ x→0 0 therefore: For n ≥ N 1 we have two cases to treat: We conclude that: The term in integral in the equation (1) is majored by: n] (y)e −cy 2 |1 − e −y 2 ϵ( y n ) | ≤ 1 √ π (e −y 2 (c−η) − e −y 2 (c+η) ) ∀y ∈ R, ∀n ≥ N 1 n] (y)e −cy 2 |1 − e −y 2 ϵ( y n ) | ∀y ∈ R, by the choice that we made: c − η > 0 and c + η > 0 then: In the other hand if we fix y then ϵ( y n ) −→ n→+∞ 0 ⇒ h n (y) −→ n→+∞ 0. Therefore using the dominated convergence theorem we obtain: From lemma 1 we conclude that there exist a neighborhood of then g n is an increasing function on J n and a decreasing function on K n , same thing for f n . Now we study the quantity: Because g n and f n are an increasing functions on J n then: The same procedure is used for the quantity: Indeed g n and f n are an decreasing functions on K n then: As c − η 2 > 0 we conclude that: In this final part of the proof we know from the hypothesis of this theorem that g > 0 onĪ\{x * }, in order to finish the proof it remains to compute the 2 quantities: Let L 1 be the minimum of g on [x 1 , D] then L 1 > 0 and: Let L 2 be the minimum of g on [E, x 2 ] then L 2 > 0 and: As before E − x * > 0 then: Finally we conclude that: We know that f n converges in the sense of distributions to 1 √ c δ x * then: Otherwise if ϕ ∈ D(R) then ϕ is bounded: Research Vol. 10, No. 5;2018 In conclusion: lim In general case if g has several zeros in I we have the following theorem: Theorem 2 If the function g has only k zeros x * 1 , . . . , x * k inĪ and x * 1 , . . . , x * k ∈ I such that g satisfies the hypotheses of the theorem 1 at x i ∀i = 1, . . . , k , and if we consider the sequence of functions g n defined by: g n (x) = 1 I (x) n √ π e −n 2 g(x) then: In the sense of distributions Proof. It's enough to write I =

Application
Let a function f :Ī → R where I ⊂ R is an open and bounded interval. We suppose that the function f has only one zero x * inĪ and x * ∈ I such that f is C 2 in a neighborhood of x * and f ′ (x * ) 0, let g(x) = f 2 (x) we suppose in addition that there exists a finite number of zeros for the function g ′ in a neighborhood of x * , then we can apply the theorem 1.

Algorithm to Find the Solution
In this section we will present an algorithm to find the zero x * of the function g which satisfies all conditions presented in the theorem 1.
We consider then the sequence of functions g n (x) = 1 I (x) n √ π e −n 2 g(x) where I =]X 1 , X 2 [ with X 1 , X 2 ∈ R . From theorem 1 we know that: lim Let X 0 be the midpoint of I: X 0 = 1 2 (X 1 + X 2 ), and R the radius of I:R = 1 2 (X 2 − X 1 ), let y 1 = X 2 − X.R and y 2 = X 2 + X.R where X is a positive real value such that X ≥ 2, we can choose for example X = 3.
From the space D(R) we will consider the function: Because ϕ and ϕ 2 belong to D(R) and because x * ∈ I ⊂]y 1 , y 2 [ therefore: Considering the 2 sequences y n = ∫ R g n ϕ 2 dλ ∫ R g n ϕ dλ and ρ n = √ ( y 1 −y 2 2 ) 2 + 1 Ln(y n ) , the sequence x * n will be defined by: Otherwise y 1 +y 2 2 = X 2 therefore x * − y 1 +y 2 2 < 0 we obtain then: We conclude that: lim In other words we only have to compute y n for a big value of n in order to have an approximation of the solution x * .
This approximation can be used for another algorithm for solving nonlinear equation.
Remark 1 If we don't know the properties of the function g at the solution x * , in other words if we don't know if g is C 2 in the neighborhood of x * and g ′′ (x * ) > 0, then we apply the algorithm for a big value of n and we verify these properties at the neighborhood of the approximated solution x * n .

Example
Let us take a simple equation: The exact solution of this equation is: x = 2.7182.
In order to apply our algorithm we take the function g(x) = (ln(x) − 1) 2 and its corresponding sequence of functions g n (x) = 1 I (x) n √ π e −n 2 g(x) where I =]2, 4[, and for the test function ϕ we take y 1 = 4 − 3 = 1 and y 2 = 4 + 3 = 7. The function g satisfies all conditions presented in the theorem 1 therefore by taking n = 10 we obtain: x * 10 = 2.7138

Algorithm to Localize the Solutions
In the case of the existence of several zeros x * 1 , . . . , x * k of a function g which satisfies all conditions of the theorem 2, then the following algorithm helps to localize the zeros of this function.